-{% capture content_after_katex %}
-{% katexmm %}
-{{ page.content }}
-{% endkatexmm %}
-{% endcapture %}
-
-{{ content_after_katex | markdownify }}
+{{ content }}
diff --git a/source/know/concept/alfven-waves/index.md b/source/know/concept/alfven-waves/index.md
index c560c67..31576f3 100644
--- a/source/know/concept/alfven-waves/index.md
+++ b/source/know/concept/alfven-waves/index.md
@@ -10,11 +10,11 @@ layout: "concept"
---
In the [magnetohydrodynamic](/know/concept/magnetohydrodynamics/) description of a plasma,
-we split the velocity $\vb{u}$, electric current $\vb{J}$,
-[magnetic field](/know/concept/magnetic-field/) $\vb{B}$
-and [electric field](/know/concept/electric-field/) $\vb{E}$ like so,
-into a constant uniform equilibrium (subscript $0$)
-and a small unknown perturbation (subscript $1$):
+we split the velocity $$\vb{u}$$, electric current $$\vb{J}$$,
+[magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$
+and [electric field](/know/concept/electric-field/) $$\vb{E}$$ like so,
+into a constant uniform equilibrium (subscript $$0$$)
+and a small unknown perturbation (subscript $$1$$):
$$\begin{aligned}
\vb{u}
@@ -41,7 +41,7 @@ $$\begin{aligned}
\end{aligned}$$
We do this for the momentum equation too,
-assuming that $\vb{J}_0 \!=\! 0$ (to be justified later).
+assuming that $$\vb{J}_0 \!=\! 0$$ (to be justified later).
Note that the temperature is set to zero, such that the pressure vanishes:
$$\begin{aligned}
@@ -49,8 +49,8 @@ $$\begin{aligned}
= \vb{J}_1 \cross \vb{B}_0
\end{aligned}$$
-Where $\rho$ is the uniform equilibrium density.
-We would like an equation for $\vb{J}_1$,
+Where $$\rho$$ is the uniform equilibrium density.
+We would like an equation for $$\vb{J}_1$$,
which is provided by the magnetohydrodynamic form of Ampère's law:
$$\begin{aligned}
@@ -62,14 +62,14 @@ $$\begin{aligned}
\end{aligned}$$
Substituting this into the momentum equation,
-and differentiating with respect to $t$:
+and differentiating with respect to $$t$$:
$$\begin{aligned}
\rho \pdvn{2}{\vb{u}_1}{t}
= \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \pdv{}{\vb{B}1}{t} \Big) \cross \vb{B}_0 \bigg)
\end{aligned}$$
-For which we can use Faraday's law to rewrite $\ipdv{\vb{B}_1}{t}$,
+For which we can use Faraday's law to rewrite $$\ipdv{\vb{B}_1}{t}$$,
incorporating Ohm's law too:
$$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
= \nabla \cross (\vb{u}_1 \cross \vb{B}_0)
\end{aligned}$$
-Inserting this into the momentum equation for $\vb{u}_1$
+Inserting this into the momentum equation for $$\vb{u}_1$$
thus yields its final form:
$$\begin{aligned}
@@ -86,9 +86,9 @@ $$\begin{aligned}
= \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg)
\end{aligned}$$
-Suppose the magnetic field is pointing in $z$-direction,
-i.e. $\vb{B}_0 = B_0 \vu{e}_z$.
-Then Faraday's law justifies our earlier assumption that $\vb{J}_0 = 0$,
+Suppose the magnetic field is pointing in $$z$$-direction,
+i.e. $$\vb{B}_0 = B_0 \vu{e}_z$$.
+Then Faraday's law justifies our earlier assumption that $$\vb{J}_0 = 0$$,
and the equation can be written as:
$$\begin{aligned}
@@ -96,7 +96,7 @@ $$\begin{aligned}
= v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
\end{aligned}$$
-Where we have defined the so-called **Alfvén velocity** $v_A$ to be given by:
+Where we have defined the so-called **Alfvén velocity** $$v_A$$ to be given by:
$$\begin{aligned}
\boxed{
@@ -105,24 +105,24 @@ $$\begin{aligned}
}
\end{aligned}$$
-Now, consider the following plane-wave ansatz for $\vb{u}_1$,
-with wavevector $\vb{k}$ and frequency $\omega$:
+Now, consider the following plane-wave ansatz for $$\vb{u}_1$$,
+with wavevector $$\vb{k}$$ and frequency $$\omega$$:
$$\begin{aligned}
\vb{u}_1(\vb{r}, t)
&= \vb{u}_1 \exp(i \vb{k} \cdot \vb{r} - i \omega t)
\end{aligned}$$
-Inserting this into the above differential equation for $\vb{u}_1$ leads to:
+Inserting this into the above differential equation for $$\vb{u}_1$$ leads to:
$$\begin{aligned}
\omega^2 \vb{u}_1
= v_A^2 \bigg( \Big( \vb{k} \cross \big( \vb{k} \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
\end{aligned}$$
-To evaluate this, we rotate our coordinate system around the $z$-axis
-such that $\vb{k} = (0, k_\perp, k_\parallel)$,
-i.e. the wavevector's $x$-component is zero.
+To evaluate this, we rotate our coordinate system around the $$z$$-axis
+such that $$\vb{k} = (0, k_\perp, k_\parallel)$$,
+i.e. the wavevector's $$x$$-component is zero.
Calculating the cross products:
$$\begin{aligned}
@@ -149,7 +149,7 @@ $$\begin{aligned}
\end{aligned}$$
We rewrite this equation in matrix form,
-using that $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$:
+using that $$k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$$:
$$\begin{aligned}
\begin{bmatrix}
@@ -161,9 +161,9 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-This has the form of an eigenvalue problem for $\omega^2$,
+This has the form of an eigenvalue problem for $$\omega^2$$,
meaning we must find non-trivial solutions,
-where we cannot simply choose the components of $\vb{u}_1$ to satisfy the equation.
+where we cannot simply choose the components of $$\vb{u}_1$$ to satisfy the equation.
To achieve this, we demand that the matrix' determinant is zero:
$$\begin{aligned}
@@ -171,12 +171,12 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-This equation has three solutions for $\omega^2$,
+This equation has three solutions for $$\omega^2$$,
one for each of its three factors being zero.
-The simplest case $\omega^2 = 0$ is of no interest to us,
+The simplest case $$\omega^2 = 0$$ is of no interest to us,
because we are looking for waves.
-The first interesting case is $\omega^2 = v_A^2 k_\parallel^2$,
+The first interesting case is $$\omega^2 = v_A^2 k_\parallel^2$$,
yielding the following dispersion relation:
$$\begin{aligned}
@@ -188,10 +188,10 @@ $$\begin{aligned}
The resulting waves are called **shear Alfvén waves**.
From the eigenvalue problem, we see that in this case
-$\vb{u}_1 = (u_{1x}, 0, 0)$, meaning $\vb{u}_1 \cdot \vb{k} = 0$:
+$$\vb{u}_1 = (u_{1x}, 0, 0)$$, meaning $$\vb{u}_1 \cdot \vb{k} = 0$$:
these waves are **transverse**.
-The phase velocity $v_p$ and group velocity $v_g$ are as follows,
-where $\theta$ is the angle between $\vb{k}$ and $\vb{B}_0$:
+The phase velocity $$v_p$$ and group velocity $$v_g$$ are as follows,
+where $$\theta$$ is the angle between $$\vb{k}$$ and $$\vb{B}_0$$:
$$\begin{aligned}
v_p
@@ -204,7 +204,7 @@ $$\begin{aligned}
= v_A
\end{aligned}$$
-The other interesting case is $\omega^2 = v_A^2 k^2$,
+The other interesting case is $$\omega^2 = v_A^2 k^2$$,
which leads to so-called **compressional Alfvén waves**,
with the simple dispersion relation:
@@ -215,10 +215,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Looking at the eigenvalue problem reveals that $\vb{u}_1 = (0, u_{1y}, 0)$,
-meaning $\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$,
-so these waves are not necessarily transverse, nor longitudinal (since $k_\parallel$ is free).
-The phase velocity $v_p$ and group velocity $v_g$ are given by:
+Looking at the eigenvalue problem reveals that $$\vb{u}_1 = (0, u_{1y}, 0)$$,
+meaning $$\vb{u}_1 \cdot \vb{k} = u_{1y} k_\perp$$,
+so these waves are not necessarily transverse, nor longitudinal (since $$k_\parallel$$ is free).
+The phase velocity $$v_p$$ and group velocity $$v_g$$ are given by:
$$\begin{aligned}
v_p
diff --git a/source/know/concept/archimedes-principle/index.md b/source/know/concept/archimedes-principle/index.md
index dc02d12..364461f 100644
--- a/source/know/concept/archimedes-principle/index.md
+++ b/source/know/concept/archimedes-principle/index.md
@@ -23,7 +23,7 @@ which has a pressure and thus affects it.
The right thing to do is treat the entire body as being
submerged in a fluid with varying properties.
-Let us consider a volume $V$ completely submerged in such a fluid.
+Let us consider a volume $$V$$ completely submerged in such a fluid.
This volume will experience a downward force due to gravity, given by:
$$\begin{aligned}
@@ -31,10 +31,10 @@ $$\begin{aligned}
= \int_V \va{g} \rho_\mathrm{b} \dd{V}
\end{aligned}$$
-Where $\va{g}$ is the gravitational field,
-and $\rho_\mathrm{b}$ is the density of the body.
-Meanwhile, the pressure $p$ of the surrounding fluid exerts a force
-on the entire surface $S$ of $V$:
+Where $$\va{g}$$ is the gravitational field,
+and $$\rho_\mathrm{b}$$ is the density of the body.
+Meanwhile, the pressure $$p$$ of the surrounding fluid exerts a force
+on the entire surface $$S$$ of $$V$$:
$$\begin{aligned}
\va{F}_p
@@ -44,7 +44,7 @@ $$\begin{aligned}
Where we have used the divergence theorem.
Assuming [hydrostatic equilibrium](/know/concept/hydrostatic-pressure/),
-we replace $\nabla p$,
+we replace $$\nabla p$$,
leading to the definition of the **buoyant force**:
$$\begin{aligned}
@@ -54,7 +54,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$.
+For the body to be at rest, we require $$\va{F}_g + \va{F}_p = 0$$.
Concretely, the equilibrium condition is:
$$\begin{aligned}
@@ -64,8 +64,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-It is commonly assumed that $\va{g}$ is constant everywhere, with magnitude $\mathrm{g}$.
-If we also assume that $\rho_\mathrm{f}$ is constant on the "submerged" side,
+It is commonly assumed that $$\va{g}$$ is constant everywhere, with magnitude $$\mathrm{g}$$.
+If we also assume that $$\rho_\mathrm{f}$$ is constant on the "submerged" side,
and zero on the "non-submerged" side, we find:
$$\begin{aligned}
@@ -73,12 +73,12 @@ $$\begin{aligned}
= \mathrm{g} (m_\mathrm{b} - m_\mathrm{f})
\end{aligned}$$
-In other words, the mass $m_\mathrm{b}$ of the entire body
-is equal to the mass $m_\mathrm{f}$ of the fluid it displaces.
+In other words, the mass $$m_\mathrm{b}$$ of the entire body
+is equal to the mass $$m_\mathrm{f}$$ of the fluid it displaces.
This is the best-known version of Archimedes' principle.
-Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$,
-then the displaced mass $m_\mathrm{f} < m_\mathrm{b}$
+Note that if $$\rho_\mathrm{b} > \rho_\mathrm{f}$$,
+then the displaced mass $$m_\mathrm{f} < m_\mathrm{b}$$
even if the entire body is submerged,
and the object will therefore continue to sink.
diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md
index 1091773..0f75930 100644
--- a/source/know/concept/bb84-protocol/index.md
+++ b/source/know/concept/bb84-protocol/index.md
@@ -26,8 +26,8 @@ because the later stages of the protocol involve revealing parts of the data
over the (insecure) classical channel.
For each bit, Alice randomly chooses a qubit basis,
-either $\{ \Ket{0}, \Ket{1} \}$ (eigenstates of the $z$-spin $\hat{\sigma}_z$)
-or $\{ \Ket{-}, \Ket{+} \}$ (eigenstates of the $x$-spin $\hat{\sigma}_x$).
+either $$\{ \Ket{0}, \Ket{1} \}$$ (eigenstates of the $$z$$-spin $$\hat{\sigma}_z$$)
+or $$\{ \Ket{-}, \Ket{+} \}$$ (eigenstates of the $$x$$-spin $$\hat{\sigma}_x$$).
Using the basis she chose, she then transmits the bits to Bob over the quantum channel,
encoding them as follows:
@@ -38,7 +38,7 @@ $$\begin{aligned}
\end{aligned}$$
Crucially, Bob has no idea which basis Alice used for any of the bits.
-For every bit, he chooses $\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random,
+For every bit, he chooses $$\hat{\sigma}_z$$ or $$\hat{\sigma}_x$$ at random,
and makes a measurement of the qubit, yielding 0 or 1.
If he guessed the basis correctly, he gets the bit value intended by Alice,
but if he guessed incorrectly, he randomly gets 0 or 1 with a 50-50 probability:
@@ -67,7 +67,7 @@ Suppose that Eve is performing an *intercept-resend attack*
(not very effective, but simple),
where she listens on the quantum channel.
For each qubit received from Alice, Eve chooses
-$\hat{\sigma}_z$ or $\hat{\sigma}_x$ at random and measures it.
+$$\hat{\sigma}_z$$ or $$\hat{\sigma}_x$$ at random and measures it.
She records her results and resends the qubits to Bob
using the basis she chose, which may or may not be what Alice intended.
@@ -105,17 +105,17 @@ In practice, even without Eve, quantum channels are imperfect,
and will introduce some errors in the qubits received by Bob.
Suppose that after basis reconciliation,
Alice and Bob have the strings
-$\{a_1, ..., a_N\}$ and $\{b_1, ..., b_N\}$, respectively.
-We define $p$ as the probability that Alice and Bob agree on the $n$th bit,
+$$\{a_1, ..., a_N\}$$ and $$\{b_1, ..., b_N\}$$, respectively.
+We define $$p$$ as the probability that Alice and Bob agree on the $$n$$th bit,
which we assume to be greater than 50%:
$$\begin{aligned}
p = P(a_n = b_n) > \frac{1}{2}
\end{aligned}$$
-Ideally, $p = 1$. To improve $p$, the following simple scheme can be used:
-starting at $n = 1$, Alice and Bob reveal $A$ and $B$ over the classical channel,
-where $\oplus$ is an XOR:
+Ideally, $$p = 1$$. To improve $$p$$, the following simple scheme can be used:
+starting at $$n = 1$$, Alice and Bob reveal $$A$$ and $$B$$ over the classical channel,
+where $$\oplus$$ is an XOR:
$$\begin{aligned}
A = a_n \oplus a_{n+1}
@@ -123,12 +123,12 @@ $$\begin{aligned}
B = b_n \oplus b_{n+1}
\end{aligned}$$
-If $A = B$, then $a_{n+1}$ and $b_{n+1}$ are discarded to prevent
+If $$A = B$$, then $$a_{n+1}$$ and $$b_{n+1}$$ are discarded to prevent
a listener on the classical channel from learning anything about the string.
-If $A \neq B$, all of $a_n$, $b_n$, $a_{n+1}$ and $b_{n+1}$ are discarded,
-and then Alice and Bob move on to $n = 3$, etc.
+If $$A \neq B$$, all of $$a_n$$, $$b_n$$, $$a_{n+1}$$ and $$b_{n+1}$$ are discarded,
+and then Alice and Bob move on to $$n = 3$$, etc.
-Given that $A = B$, the probability that $a_n = b_n$,
+Given that $$A = B$$, the probability that $$a_n = b_n$$,
which is what we want, is given by:
$$\begin{aligned}
@@ -140,7 +140,7 @@ $$\begin{aligned}
&= \frac{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1})}{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n}) \: P(a_{n+1} \neq b_{n+1})}
\end{aligned}$$
-We use the definition of $p$ to get the following inequality,
+We use the definition of $$p$$ to get the following inequality,
which can be verified by plotting:
$$\begin{aligned}
@@ -150,9 +150,9 @@ $$\begin{aligned}
\end{aligned}$$
Alice and Bob can repeat this error correction scheme multiple times,
-until their estimate of $p$ is satisfactory.
+until their estimate of $$p$$ is satisfactory.
This involves discarding many bits,
-so the length $N_\mathrm{new}$ of the string they end up with
+so the length $$N_\mathrm{new}$$ of the string they end up with
after one iteration is given by:
$$\begin{aligned}
@@ -166,11 +166,11 @@ More efficient schemes exist, which do not consume so many bits.
## Privacy amplification
-Suppose that after the error correction step, $p = 1$,
+Suppose that after the error correction step, $$p = 1$$,
so Alice and Bob fully agree on the random string.
However, in the meantime, Eve has been listening,
and has been doing a good job
-building up her own string $\{e_1, ..., e_N\}$,
+building up her own string $$\{e_1, ..., e_N\}$$,
such that she knows more that 50% of the bits:
$$\begin{aligned}
@@ -178,9 +178,9 @@ $$\begin{aligned}
\end{aligned}$$
**Privacy amplification** is an optional final step of the BB84 protocol
-which aims to reduce Eve's $q$.
+which aims to reduce Eve's $$q$$.
Alice and Bob use their existing strings to generate a new one
-$\{a_1', ..., a_M'\}$:
+$$\{a_1', ..., a_M'\}$$:
$$\begin{aligned}
a_1'
@@ -197,8 +197,8 @@ more efficient schemes exist, which consume less.
To see why this improves Alice and Bob's privacy,
suppose that Eve is following along,
-and creates a new string $\{e_1', ..., e_M'\}$
-where $e_m' = e_{2m - 1} \oplus e_{2m}$.
+and creates a new string $$\{e_1', ..., e_M'\}$$
+where $$e_m' = e_{2m - 1} \oplus e_{2m}$$.
The probability that Eve's result agrees with
Alice and Bob's string is given by:
@@ -209,7 +209,7 @@ $$\begin{aligned}
&= P(e_1 = a_1) \: P(e_2 = a_2) + P(e_1 \neq a_1) \: P(e_2 \neq a_2)
\end{aligned}$$
-Recognizing $q$ 's definition,
+Recognizing $$q$$ 's definition,
we find the following inequality,
which can be verified by plotting:
@@ -219,8 +219,8 @@ $$\begin{aligned}
< q
\end{aligned}$$
-After repeating this step several times, $q$ will be close to 1/2,
-which is the ideal value: for $q =$ 0.5,
+After repeating this step several times, $$q$$ will be close to 1/2,
+which is the ideal value: for $$q =$$ 0.5,
Eve would only know 50% of the bits,
which is equivalent to her guessing at random.
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
index 5f333a2..f454264 100644
--- a/source/know/concept/bell-state/index.md
+++ b/source/know/concept/bell-state/index.md
@@ -24,14 +24,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where e.g. $\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$
-is the tensor product of qubit $A$ in state $\Ket{0}$ and $B$ in $\Ket{1}$.
+Where e.g. $$\Ket{0}_A \Ket{1}_B = \Ket{0}_A \otimes \Ket{1}_B$$
+is the tensor product of qubit $$A$$ in state $$\Ket{0}$$ and $$B$$ in $$\Ket{1}$$.
These states form an orthonormal basis for the two-qubit
[Hilbert space](/know/concept/hilbert-space/).
More importantly, however,
is that the Bell states are maximally entangled,
-which we prove here for $\ket{\Phi^{+}}$.
+which we prove here for $$\ket{\Phi^{+}}$$.
Consider the following pure [density operator](/know/concept/density-operator/):
$$\begin{aligned}
@@ -40,7 +40,7 @@ $$\begin{aligned}
&= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big)
\end{aligned}$$
-The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows:
+The reduced density operator $$\hat{\rho}_A$$ of qubit $$A$$ is then calculated as follows:
$$\begin{aligned}
\hat{\rho}_A
@@ -54,14 +54,14 @@ $$\begin{aligned}
= \frac{1}{2} \hat{I}
\end{aligned}$$
-This result is maximally mixed, therefore $\ket{\Phi^{+}}$ is maximally entangled.
+This result is maximally mixed, therefore $$\ket{\Phi^{+}}$$ is maximally entangled.
The same holds for the other three Bell states,
-and is equally true for qubit $B$.
+and is equally true for qubit $$B$$.
-This means that a measurement of qubit $A$
-has a 50-50 chance to yield $\Ket{0}$ or $\Ket{1}$.
+This means that a measurement of qubit $$A$$
+has a 50-50 chance to yield $$\Ket{0}$$ or $$\Ket{1}$$.
However, due to the entanglement,
-measuring $A$ also has consequences for qubit $B$:
+measuring $$A$$ also has consequences for qubit $$B$$:
$$\begin{aligned}
\big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
@@ -81,10 +81,10 @@ $$\begin{aligned}
= \frac{1}{2}
\end{aligned}$$
-As an example, if $A$ collapses into $\Ket{0}$ due to a measurement,
-then $B$ instantly also collapses into $\Ket{0}$, never $\Ket{1}$,
+As an example, if $$A$$ collapses into $$\Ket{0}$$ due to a measurement,
+then $$B$$ instantly also collapses into $$\Ket{0}$$, never $$\Ket{1}$$,
even if it was not measured.
-This was a specific example for $\ket{\Phi^{+}}$,
+This was a specific example for $$\ket{\Phi^{+}}$$,
but analogous results can be found for the other Bell states.
diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md
index 3b71dbf..a01bf9e 100644
--- a/source/know/concept/bells-theorem/index.md
+++ b/source/know/concept/bells-theorem/index.md
@@ -13,7 +13,7 @@ layout: "concept"
cannot be explained by theories built on
so-called **local hidden variables** (LHVs).
-Suppose that we have two spin-1/2 particles, called $A$ and $B$,
+Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$,
in an entangled [Bell state](/know/concept/bell-state/):
$$\begin{aligned}
@@ -22,23 +22,23 @@ $$\begin{aligned}
\end{aligned}$$
Since they are entangled,
-if we measure the $z$-spin of particle $A$, and find e.g. $\Ket{\uparrow}$,
-then particle $B$ immediately takes the opposite state $\Ket{\downarrow}$.
+if we measure the $$z$$-spin of particle $$A$$, and find e.g. $$\Ket{\uparrow}$$,
+then particle $$B$$ immediately takes the opposite state $$\Ket{\downarrow}$$.
The point is that this collapse is instant,
-regardless of the distance between $A$ and $B$.
+regardless of the distance between $$A$$ and $$B$$.
Einstein called this effect "action-at-a-distance",
and used it as evidence that quantum mechanics is an incomplete theory.
-He said that there must be some **hidden variable** $\lambda$
-that determines the outcome of measurements of $A$ and $B$
+He said that there must be some **hidden variable** $$\lambda$$
+that determines the outcome of measurements of $$A$$ and $$B$$
from the moment the entangled pair is created.
However, according to Bell's theorem, he was wrong.
-To prove this, let us assume that Einstein was right, and some $\lambda$,
+To prove this, let us assume that Einstein was right, and some $$\lambda$$,
which we cannot understand, let alone calculate or measure, controls the results.
We want to know the spins of the entangled pair
-along arbitrary directions $\vec{a}$ and $\vec{b}$,
-so the outcomes for particles $A$ and $B$ are:
+along arbitrary directions $$\vec{a}$$ and $$\vec{b}$$,
+so the outcomes for particles $$A$$ and $$B$$ are:
$$\begin{aligned}
A(\vec{a}, \lambda) = \pm 1
@@ -46,8 +46,8 @@ $$\begin{aligned}
B(\vec{b}, \lambda) = \pm 1
\end{aligned}$$
-Where $\pm 1$ are the eigenvalues of the Pauli matrices
-in the chosen directions $\vec{a}$ and $\vec{b}$:
+Where $$\pm 1$$ are the eigenvalues of the Pauli matrices
+in the chosen directions $$\vec{a}$$ and $$\vec{b}$$:
$$\begin{aligned}
\hat{\sigma}_a
@@ -59,8 +59,8 @@ $$\begin{aligned}
= b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z
\end{aligned}$$
-Whether $\lambda$ is a scalar or a vector does not matter;
-we simply demand that it follows an unknown probability distribution $\rho(\lambda)$:
+Whether $$\lambda$$ is a scalar or a vector does not matter;
+we simply demand that it follows an unknown probability distribution $$\rho(\lambda)$$:
$$\begin{aligned}
\int \rho(\lambda) \dd{\lambda} = 1
@@ -68,8 +68,8 @@ $$\begin{aligned}
\rho(\lambda) \ge 0
\end{aligned}$$
-The product of the outcomes of $A$ and $B$ then has the following expectation value.
-Note that we only multiply $A$ and $B$ for shared $\lambda$-values:
+The product of the outcomes of $$A$$ and $$B$$ then has the following expectation value.
+Note that we only multiply $$A$$ and $$B$$ for shared $$\lambda$$-values:
this is what makes it a **local** hidden variable:
$$\begin{aligned}
@@ -83,7 +83,7 @@ which both prove Bell's theorem.
## Bell inequality
-If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins:
+If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins:
$$\begin{aligned}
A(\vec{a}, \lambda)
@@ -98,8 +98,8 @@ $$\begin{aligned}
= - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
\end{aligned}$$
-Next, we introduce an arbitrary third direction $\vec{c}$,
-and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$:
+Next, we introduce an arbitrary third direction $$\vec{c}$$,
+and use the fact that $$( A(\vec{b}, \lambda) )^2 = 1$$:
$$\begin{aligned}
\Expval{A_a B_b} - \Expval{A_a B_c}
@@ -109,7 +109,7 @@ $$\begin{aligned}
\end{aligned}$$
Inside the integral, the only factors that can be negative
-are the last two, and their product is $\pm 1$.
+are the last two, and their product is $$\pm 1$$.
Taking the absolute value of the whole left,
and of the integrand on the right, we thus get:
@@ -121,7 +121,7 @@ $$\begin{aligned}
&\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda}
\end{aligned}$$
-Since $\rho(\lambda)$ is a normalized probability density function,
+Since $$\rho(\lambda)$$ is a normalized probability density function,
we arrive at the **Bell inequality**:
$$\begin{aligned}
@@ -131,18 +131,18 @@ $$\begin{aligned}
}
\end{aligned}$$
-Any theory involving an LHV $\lambda$ must obey this inequality.
+Any theory involving an LHV $$\lambda$$ must obey this inequality.
The problem, however, is that quantum mechanics dictates the expectation values
-for the state $\Ket{\Psi^{-}}$:
+for the state $$\Ket{\Psi^{-}}$$:
$$\begin{aligned}
\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}
\end{aligned}$$
Finding directions which violate the Bell inequality is easy:
-for example, if $\vec{a}$ and $\vec{b}$ are orthogonal,
-and $\vec{c}$ is at a $\pi/4$ angle to both of them,
-then the left becomes $0.707$ and the right $0.293$,
+for example, if $$\vec{a}$$ and $$\vec{b}$$ are orthogonal,
+and $$\vec{c}$$ is at a $$\pi/4$$ angle to both of them,
+then the left becomes $$0.707$$ and the right $$0.293$$,
which clearly disagrees with the inequality,
meaning that LHVs are impossible.
@@ -152,8 +152,8 @@ meaning that LHVs are impossible.
The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality**
takes a slightly different approach, and is more useful in practice.
-Consider four spin directions, two for $A$ called $\vec{a}_1$ and $\vec{a}_2$,
-and two for $B$ called $\vec{b}_1$ and $\vec{b}_2$.
+Consider four spin directions, two for $$A$$ called $$\vec{a}_1$$ and $$\vec{a}_2$$,
+and two for $$B$$ called $$\vec{b}_1$$ and $$\vec{b}_2$$.
Let us introduce the following abbreviations:
$$\begin{aligned}
@@ -196,7 +196,7 @@ $$\begin{aligned}
+ \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
\end{aligned}$$
-Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$,
+Using the fact that the product of $$A$$ and $$B$$ is always either $$-1$$ or $$+1$$,
we can reduce this to:
$$\begin{aligned}
@@ -209,7 +209,7 @@ $$\begin{aligned}
\end{aligned}$$
Evaluating these integrals gives us the following inequality,
-which holds for both choices of $\pm$:
+which holds for both choices of $$\pm$$:
$$\begin{aligned}
\Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
@@ -235,8 +235,8 @@ $$\begin{aligned}
&\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} + \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
\end{aligned}$$
-The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$,
-and measures the correlation between the spins of $A$ and $B$:
+The quantity on the right-hand side is sometimes called the **CHSH quantity** $$S$$,
+and measures the correlation between the spins of $$A$$ and $$B$$:
$$\begin{aligned}
\boxed{
@@ -244,7 +244,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The CHSH inequality places an upper bound on the magnitude of $S$
+The CHSH inequality places an upper bound on the magnitude of $$S$$
for LHV-based theories:
$$\begin{aligned}
@@ -258,15 +258,15 @@ $$\begin{aligned}
Quantum physics can violate the CHSH inequality, but by how much?
Consider the following two-particle operator,
-whose expectation value is the CHSH quantity, i.e. $S = \expval{\hat{S}}$:
+whose expectation value is the CHSH quantity, i.e. $$S = \expval{\hat{S}}$$:
$$\begin{aligned}
\hat{S}
= \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2
\end{aligned}$$
-Where $\otimes$ is the tensor product,
-and e.g. $\hat{A}_1$ is the Pauli matrix for the $\vec{a}_1$-direction.
+Where $$\otimes$$ is the tensor product,
+and e.g. $$\hat{A}_1$$ is the Pauli matrix for the $$\vec{a}_1$$-direction.
The square of this operator is then given by:
$$\begin{aligned}
@@ -292,15 +292,15 @@ $$\begin{aligned}
\end{aligned}$$
Spin operators are unitary, so their square is the identity,
-e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to:
+e.g. $$\hat{A}_1^2 = \hat{I}$$. Therefore $$\hat{S}^2$$ reduces to:
$$\begin{aligned}
\hat{S}^2
&= 4 \: (\hat{I} \otimes \hat{I}) + \comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}
\end{aligned}$$
-The *norm* $\norm{\hat{S}^2}$ of this operator
-is the largest possible expectation value $\expval{\hat{S}^2}$,
+The *norm* $$\norm{\hat{S}^2}$$ of this operator
+is the largest possible expectation value $$\expval{\hat{S}^2}$$,
which is the same as its largest eigenvalue.
It is given by:
@@ -321,7 +321,7 @@ $$\begin{aligned}
\le 2
\end{aligned}$$
-And $\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason.
+And $$\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$$ for the same reason.
The norm is the largest eigenvalue, therefore:
$$\begin{aligned}
@@ -336,7 +336,7 @@ $$\begin{aligned}
We thus arrive at **Tsirelson's bound**,
which states that quantum mechanics can violate
-the CHSH inequality by a factor of $\sqrt{2}$:
+the CHSH inequality by a factor of $$\sqrt{2}$$:
$$\begin{aligned}
\boxed{
@@ -359,8 +359,8 @@ $$\begin{aligned}
\hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}}
\end{aligned}$$
-Using the fact that $\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$,
-it can then be shown that $S = 2 \sqrt{2}$ in this case.
+Using the fact that $$\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$$,
+it can then be shown that $$S = 2 \sqrt{2}$$ in this case.
diff --git a/source/know/concept/beltrami-identity/index.md b/source/know/concept/beltrami-identity/index.md
index 3fa566c..be9a344 100644
--- a/source/know/concept/beltrami-identity/index.md
+++ b/source/know/concept/beltrami-identity/index.md
@@ -8,26 +8,26 @@ categories:
layout: "concept"
---
-Consider a general functional $J[f]$ of the following form,
-with $f(x)$ an unknown function:
+Consider a general functional $$J[f]$$ of the following form,
+with $$f(x)$$ an unknown function:
$$\begin{aligned}
J[f]
= \int_{x_0}^{x_1} L(f, f', x) \dd{x}
\end{aligned}$$
-Where $L$ is the Lagrangian.
-To find the $f$ that maximizes or minimizes $J[f]$,
+Where $$L$$ is the Lagrangian.
+To find the $$f$$ that maximizes or minimizes $$J[f]$$,
the [calculus of variations](/know/concept/calculus-of-variations/)
-states that the Euler-Lagrange equation must be solved for $f$:
+states that the Euler-Lagrange equation must be solved for $$f$$:
$$\begin{aligned}
0
= \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f'} \Big)
\end{aligned}$$
-We now want to know exactly how $L$ depends on the free variable $x$,
-since it is a function of $x$, $f(x)$ and $f'(x)$.
+We now want to know exactly how $$L$$ depends on the free variable $$x$$,
+since it is a function of $$x$$, $$f(x)$$ and $$f'(x)$$.
Using the chain rule:
$$\begin{aligned}
@@ -44,16 +44,16 @@ $$\begin{aligned}
&= \dv{}{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x}
\end{aligned}$$
-Although we started from the "hard" derivative $\idv{L}{x}$,
-we arrive at an expression for the "soft" derivative $\ipdv{L}{x}$,
-describing the *explicit* dependence of $L$ on $x$:
+Although we started from the "hard" derivative $$\idv{L}{x}$$,
+we arrive at an expression for the "soft" derivative $$\ipdv{L}{x}$$,
+describing the *explicit* dependence of $$L$$ on $$x$$:
$$\begin{aligned}
- \pdv{L}{x}
= \dv{}{x} \bigg( f' \pdv{L}{f'} - L \bigg)
\end{aligned}$$
-What if $L$ does not explicitly depend on $x$, i.e. $\ipdv{L}{x} = 0$?
+What if $$L$$ does not explicitly depend on $$x$$, i.e. $$\ipdv{L}{x} = 0$$?
In that case, the equation can be integrated to give the **Beltrami identity**:
$$\begin{aligned}
@@ -63,19 +63,19 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $C$ is a constant.
-This says that the left-hand side is a conserved quantity in $x$,
+Where $$C$$ is a constant.
+This says that the left-hand side is a conserved quantity in $$x$$,
which could be useful to know.
-If we insert a concrete expression for $L$,
-the Beltrami identity might be easier to solve for $f$ than the full Euler-Lagrange equation.
-The assumption $\ipdv{L}{x} = 0$ is justified;
-for example, if $x$ is time, it means that the potential is time-independent.
+If we insert a concrete expression for $$L$$,
+the Beltrami identity might be easier to solve for $$f$$ than the full Euler-Lagrange equation.
+The assumption $$\ipdv{L}{x} = 0$$ is justified;
+for example, if $$x$$ is time, it means that the potential is time-independent.
## Higher dimensions
-Above, a 1D problem was considered, i.e. $f$ depended only on a single variable $x$.
-Consider now a 2D problem, such that $J[f]$ is given by:
+Above, a 1D problem was considered, i.e. $$f$$ depended only on a single variable $$x$$.
+Consider now a 2D problem, such that $$J[f]$$ is given by:
$$\begin{aligned}
J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
@@ -87,7 +87,7 @@ $$\begin{aligned}
0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{y} \Big( \pdv{L}{f_y} \Big)
\end{aligned}$$
-Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous):
+Once again, we calculate the hard $$x$$-derivative of $$L$$ (the $$y$$-derivative is analogous):
$$\begin{aligned}
\dv{L}{x}
@@ -99,7 +99,7 @@ $$\begin{aligned}
&= \dv{}{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
\end{aligned}$$
-This time, we arrive at the following expression for the soft derivative $\ipdv{L}{x}$:
+This time, we arrive at the following expression for the soft derivative $$\ipdv{L}{x}$$:
$$\begin{aligned}
- \pdv{L}{x}
@@ -109,9 +109,9 @@ $$\begin{aligned}
Due to the derivatives, this cannot be cleanly turned into an analogue of the 1D Beltrami identity,
and therefore we use that name only in the 1D case.
-However, if $\ipdv{L}{x} = 0$, this equation is still useful.
+However, if $$\ipdv{L}{x} = 0$$, this equation is still useful.
For an off-topic demonstration of this fact,
-let us choose $x$ as the transverse coordinate, and integrate over it to get:
+let us choose $$x$$ as the transverse coordinate, and integrate over it to get:
$$\begin{aligned}
0
@@ -123,7 +123,7 @@ $$\begin{aligned}
\end{aligned}$$
If our boundary conditions cause the boundary term to vanish (as is often the case),
-then the integral on the right is a conserved quantity with respect to $y$.
+then the integral on the right is a conserved quantity with respect to $$y$$.
While not as elegant as the 1D Beltrami identity,
the above 2D counterpart still fulfills the same role.
diff --git a/source/know/concept/bernoullis-theorem/index.md b/source/know/concept/bernoullis-theorem/index.md
index 12bd0ca..6b933d2 100644
--- a/source/know/concept/bernoullis-theorem/index.md
+++ b/source/know/concept/bernoullis-theorem/index.md
@@ -10,8 +10,8 @@ layout: "concept"
---
For inviscid fluids, **Bernuilli's theorem** states
-that an increase in flow velocity $\va{v}$ is paired
-with a decrease in pressure $p$ and/or potential energy.
+that an increase in flow velocity $$\va{v}$$ is paired
+with a decrease in pressure $$p$$ and/or potential energy.
For a qualitative argument, look no further than
one of the [Euler equations](/know/concept/euler-equations/),
with a [material derivative](/know/concept/material-derivative/):
@@ -22,16 +22,16 @@ $$\begin{aligned}
= \va{g} - \frac{\nabla p}{\rho}
\end{aligned}$$
-Assuming that $\va{v}$ is constant in $t$,
-it becomes clear that a higher $\va{v}$ requires a lower $p$.
+Assuming that $$\va{v}$$ is constant in $$t$$,
+it becomes clear that a higher $$\va{v}$$ requires a lower $$p$$.
## Simple form
For an incompressible fluid
-with a time-independent velocity field $\va{v}$ (i.e. **steady flow**),
+with a time-independent velocity field $$\va{v}$$ (i.e. **steady flow**),
Bernoulli's theorem formally states that the
-**Bernoulli head** $H$ is constant along a streamline:
+**Bernoulli head** $$H$$ is constant along a streamline:
$$\begin{aligned}
\boxed{
@@ -40,8 +40,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Phi$ is the gravitational potential, such that $\va{g} = - \nabla \Phi$.
-To prove this theorem, we take the material derivative of $H$:
+Where $$\Phi$$ is the gravitational potential, such that $$\va{g} = - \nabla \Phi$$.
+To prove this theorem, we take the material derivative of $$H$$:
$$\begin{aligned}
\frac{\mathrm{D} H}{\mathrm{D} t}
@@ -63,7 +63,7 @@ $$\begin{aligned}
+ \va{v} \cdot \big( \va{g} + \nabla \Phi \big) + \va{v} \cdot \Big( \frac{\nabla p}{\rho} - \frac{\nabla p}{\rho} \Big)
\end{aligned}$$
-Using the fact that $\va{g} = - \nabla \Phi$,
+Using the fact that $$\va{g} = - \nabla \Phi$$,
we are left with the following equation:
$$\begin{aligned}
@@ -72,12 +72,12 @@ $$\begin{aligned}
\end{aligned}$$
Assuming that the flow is steady, both derivatives vanish,
-leading us to the conclusion that $H$ is conserved along the streamline.
+leading us to the conclusion that $$H$$ is conserved along the streamline.
In fact, there exists **Bernoulli's stronger theorem**,
-which states that $H$ is constant *everywhere* in regions with
-zero [vorticity](/know/concept/vorticity/) $\va{\omega} = 0$.
-For a proof, see the derivation of $\va{\omega}$'s equation of motion.
+which states that $$H$$ is constant *everywhere* in regions with
+zero [vorticity](/know/concept/vorticity/) $$\va{\omega} = 0$$.
+For a proof, see the derivation of $$\va{\omega}$$'s equation of motion.
## References
diff --git a/source/know/concept/bernstein-vazirani-algorithm/index.md b/source/know/concept/bernstein-vazirani-algorithm/index.md
index af49841..f91c0ba 100644
--- a/source/know/concept/bernstein-vazirani-algorithm/index.md
+++ b/source/know/concept/bernstein-vazirani-algorithm/index.md
@@ -17,10 +17,10 @@ It is extremely similar to the
and even uses the same circuit.
It solves a very artificial problem:
-we are given a "black box" function $f(x)$
-that takes an $N$-bit $x$ and returns a single bit,
+we are given a "black box" function $$f(x)$$
+that takes an $$N$$-bit $$x$$ and returns a single bit,
which we are promised is the lowest bit of the bitwise dot product
-of $x$ with an unknown $N$-bit string $s$:
+of $$x$$ with an unknown $$N$$-bit string $$s$$:
$$\begin{aligned}
f(x)
@@ -28,10 +28,10 @@ $$\begin{aligned}
= (s_1 x_1 + s_2 x_2 + \:...\: + s_N x_N) \:\:(\bmod \: 2)
\end{aligned}$$
-The goal is to find $s$.
+The goal is to find $$s$$.
To solve this problem,
-a classical computer would need to call $f(x)$ exactly $N$ times
-with $x = 2^n$ for $n \in \{ 0, ..., N \!-\! 1\}$.
+a classical computer would need to call $$f(x)$$ exactly $$N$$ times
+with $$x = 2^n$$ for $$n \in \{ 0, ..., N \!-\! 1\}$$.
However, the Bernstein-Vazirani algorithm
allows a quantum computer to do it with only a single query.
It uses the following circuit:
@@ -40,9 +40,9 @@ It uses the following circuit:
-Where $U_f$ is a phase oracle,
+Where $$U_f$$ is a phase oracle,
whose action is defined as follows,
-where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$:
+where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$:
$$\begin{aligned}
\Ket{x}
@@ -51,14 +51,14 @@ $$\begin{aligned}
= (-1)^{s \cdot x} \Ket{x}
\end{aligned}$$
-That is, it introduces a phase flip based on the value of $f(x)$.
+That is, it introduces a phase flip based on the value of $$f(x)$$.
For an example implementation of such an oracle,
see the Deutsch-Jozsa algorithm:
its circuit is identical to this one,
-but describes $U_f$ in a different (but equivalent) way.
+but describes $$U_f$$ in a different (but equivalent) way.
-Starting from the state $\Ket{0}^{\otimes N}$,
-applying the [Hadamard gate](/know/concept/quantum-gate/) $H$
+Starting from the state $$\Ket{0}^{\otimes N}$$,
+applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
to all qubits yields:
$$\begin{aligned}
@@ -68,7 +68,7 @@ $$\begin{aligned}
= \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x}
\end{aligned}$$
-This is an equal superposition of all candidates $\Ket{x}$,
+This is an equal superposition of all candidates $$\Ket{x}$$,
which we feed to the oracle:
$$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
\end{aligned}$$
Then, thanks to the definition of the Hadamard transform,
-a final set of $H$-gates leads us to:
+a final set of $$H$$-gates leads us to:
$$\begin{aligned}
\frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \Ket{x}
@@ -87,9 +87,9 @@ $$\begin{aligned}
= \Ket{s_1} \cdots \Ket{s_N}
\end{aligned}$$
-Which, upon measurement, gives us the desired binary representation of $s$.
-For comparison, the Deutsch-Jozsa algorithm only cares whether $s = 0$ or $s \neq 0$,
-whereas this algorithm is interested in the exact value of $s$.
+Which, upon measurement, gives us the desired binary representation of $$s$$.
+For comparison, the Deutsch-Jozsa algorithm only cares whether $$s = 0$$ or $$s \neq 0$$,
+whereas this algorithm is interested in the exact value of $$s$$.
diff --git a/source/know/concept/berry-phase/index.md b/source/know/concept/berry-phase/index.md
index eedc548..d237ea5 100644
--- a/source/know/concept/berry-phase/index.md
+++ b/source/know/concept/berry-phase/index.md
@@ -8,8 +8,8 @@ categories:
layout: "concept"
---
-Consider a Hamiltonian $\hat{H}$ that does not explicitly depend on time,
-but does depend on a given parameter $\vb{R}$.
+Consider a Hamiltonian $$\hat{H}$$ that does not explicitly depend on time,
+but does depend on a given parameter $$\vb{R}$$.
The Schrödinger equations then read:
$$\begin{aligned}
@@ -20,9 +20,9 @@ $$\begin{aligned}
&= E_n(\vb{R}) \Ket{\psi_n(\vb{R})}
\end{aligned}$$
-The general full solution $\Ket{\Psi_n}$ has the following form,
-where we allow $\vb{R}$ to evolve in time,
-and we have abbreviated the traditional phase of the "wiggle factor" as $L_n$:
+The general full solution $$\Ket{\Psi_n}$$ has the following form,
+where we allow $$\vb{R}$$ to evolve in time,
+and we have abbreviated the traditional phase of the "wiggle factor" as $$L_n$$:
$$\begin{aligned}
\Ket{\Psi_n(t)}
@@ -31,11 +31,11 @@ $$\begin{aligned}
L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'}
\end{aligned}$$
-The **geometric phase** $\gamma_n(t)$ is more interesting.
-It is not included in $\Ket{\psi_n}$,
-because it depends on the path $\vb{R}(t)$
-rather than only the present $\vb{R}$ and $t$.
-Its dynamics can be found by inserting the above $\Ket{\Psi_n}$
+The **geometric phase** $$\gamma_n(t)$$ is more interesting.
+It is not included in $$\Ket{\psi_n}$$,
+because it depends on the path $$\vb{R}(t)$$
+rather than only the present $$\vb{R}$$ and $$t$$.
+Its dynamics can be found by inserting the above $$\Ket{\Psi_n}$$
into the time-dependent Schrödinger equation:
$$\begin{aligned}
@@ -58,8 +58,8 @@ $$\begin{aligned}
&= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
\end{aligned}$$
-Front-multiplying by $i \Bra{\Psi_n}$ gives us
-the equation of motion of the geometric phase $\gamma_n$:
+Front-multiplying by $$i \Bra{\Psi_n}$$ gives us
+the equation of motion of the geometric phase $$\gamma_n$$:
$$\begin{aligned}
\boxed{
@@ -68,7 +68,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows:
+Where we have defined the so-called **Berry connection** $$\vb{A}_n$$ as follows:
$$\begin{aligned}
\boxed{
@@ -77,9 +77,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Importantly, note that $\vb{A}_n$ is real,
-provided that $\Ket{\psi_n}$ is always normalized for all $\vb{R}$.
-To prove this, we start from the fact that $\nabla_\vb{R} 1 = 0$:
+Importantly, note that $$\vb{A}_n$$ is real,
+provided that $$\Ket{\psi_n}$$ is always normalized for all $$\vb{R}$$.
+To prove this, we start from the fact that $$\nabla_\vb{R} 1 = 0$$:
$$\begin{aligned}
0
@@ -91,14 +91,14 @@ $$\begin{aligned}
= 2 \Imag\{ \vb{A}_n \}
\end{aligned}$$
-Consequently, $\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real,
-because $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary.
+Consequently, $$\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real,
+because $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary.
-Suppose now that the parameter $\vb{R}(t)$ is changed adiabatically
+Suppose now that the parameter $$\vb{R}(t)$$ is changed adiabatically
(i.e. so slow that the system stays in the same eigenstate)
-for $t \in [0, T]$, along a circuit $C$ with $\vb{R}(0) \!=\! \vb{R}(T)$.
-Integrating the phase $\gamma_n(t)$ over this contour $C$ then yields
-the **Berry phase** $\gamma_n(C)$:
+for $$t \in [0, T]$$, along a circuit $$C$$ with $$\vb{R}(0) \!=\! \vb{R}(T)$$.
+Integrating the phase $$\gamma_n(t)$$ over this contour $$C$$ then yields
+the **Berry phase** $$\gamma_n(C)$$:
$$\begin{aligned}
\boxed{
@@ -107,9 +107,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-But we have a problem: $\vb{A}_n$ is not unique!
+But we have a problem: $$\vb{A}_n$$ is not unique!
Due to the Schrödinger equation's gauge invariance,
-any function $f(\vb{R}(t))$ can be added to $\gamma_n(t)$
+any function $$f(\vb{R}(t))$$ can be added to $$\gamma_n(t)$$
without making an immediate physical difference to the state.
Consider the following general gauge transformation:
@@ -118,9 +118,9 @@ $$\begin{aligned}
\equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})}
\end{aligned}$$
-To find $\vb{A}_n$ for a particular choice of $f$,
+To find $$\vb{A}_n$$ for a particular choice of $$f$$,
we need to evaluate the inner product
-$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:
+$$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$:
$$\begin{aligned}
\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}
@@ -131,16 +131,16 @@ $$\begin{aligned}
&= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
\end{aligned}$$
-Unfortunately, $f$ does not vanish as we would have liked,
-so $\vb{A}_n$ depends on our choice of $f$.
+Unfortunately, $$f$$ does not vanish as we would have liked,
+so $$\vb{A}_n$$ depends on our choice of $$f$$.
However, the curl of a gradient is always zero,
-so although $\vb{A}_n$ is not unique,
-its curl $\nabla_\vb{R} \cross \vb{A}_n$ is guaranteed to be.
-Conveniently, we can introduce a curl in the definition of $\gamma_n(C)$
+so although $$\vb{A}_n$$ is not unique,
+its curl $$\nabla_\vb{R} \cross \vb{A}_n$$ is guaranteed to be.
+Conveniently, we can introduce a curl in the definition of $$\gamma_n(C)$$
by applying Stokes' theorem, under the assumption
-that $\vb{A}_n$ has no singularities in the area enclosed by $C$
-(fortunately, $\vb{A}_n$ can always be chosen to satisfy this):
+that $$\vb{A}_n$$ has no singularities in the area enclosed by $$C$$
+(fortunately, $$\vb{A}_n$$ can always be chosen to satisfy this):
$$\begin{aligned}
\boxed{
@@ -149,10 +149,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we defined $\vb{B}_n$ as the curl of $\vb{A}_n$.
-Now $\gamma_n(C)$ is guaranteed to be unique.
-Note that $\vb{B}_n$ is analogous to a magnetic field,
-and $\vb{A}_n$ to a magnetic vector potential:
+Where we defined $$\vb{B}_n$$ as the curl of $$\vb{A}_n$$.
+Now $$\gamma_n(C)$$ is guaranteed to be unique.
+Note that $$\vb{B}_n$$ is analogous to a magnetic field,
+and $$\vb{A}_n$$ to a magnetic vector potential:
$$\begin{aligned}
\vb{B}_n(\vb{R})
@@ -160,9 +160,9 @@ $$\begin{aligned}
= \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\}
\end{aligned}$$
-Unfortunately, $\nabla_\vb{R} \psi_n$ is difficult to evaluate explicitly,
-so we would like to rewrite $\vb{B}_n$ such that it does not enter.
-We do this as follows, inserting $1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$ along the way:
+Unfortunately, $$\nabla_\vb{R} \psi_n$$ is difficult to evaluate explicitly,
+so we would like to rewrite $$\vb{B}_n$$ such that it does not enter.
+We do this as follows, inserting $$1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$$ along the way:
$$\begin{aligned}
i \vb{B}_n
@@ -172,9 +172,9 @@ $$\begin{aligned}
&= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
\end{aligned}$$
-The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary
+The fact that $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary
means it is parallel to its complex conjugate,
-and thus the cross product vanishes, so we exclude $n$ from the sum:
+and thus the cross product vanishes, so we exclude $$n$$ from the sum:
$$\begin{aligned}
\vb{B}_n
@@ -200,8 +200,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Which only involves $\nabla_\vb{R} \hat{H}$,
-and is therefore easier to evaluate than any $\Ket{\nabla_\vb{R} \psi_n}$.
+Which only involves $$\nabla_\vb{R} \hat{H}$$,
+and is therefore easier to evaluate than any $$\Ket{\nabla_\vb{R} \psi_n}$$.
diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md
index 14ba4cb..1193a93 100644
--- a/source/know/concept/binomial-distribution/index.md
+++ b/source/know/concept/binomial-distribution/index.md
@@ -9,11 +9,11 @@ layout: "concept"
---
The **binomial distribution** is a discrete probability distribution
-describing a **Bernoulli process**: a set of independent $N$ trials where
+describing a **Bernoulli process**: a set of independent $$N$$ trials where
each has only two possible outcomes, "success" and "failure",
-the former with probability $p$ and the latter with $q = 1 - p$.
+the former with probability $$p$$ and the latter with $$q = 1 - p$$.
The binomial distribution then gives the probability
-that $n$ out of the $N$ trials succeed:
+that $$n$$ out of the $$N$$ trials succeed:
$$\begin{aligned}
\boxed{
@@ -22,8 +22,8 @@ $$\begin{aligned}
\end{aligned}$$
The first factor is known as the **binomial coefficient**, which describes the
-number of microstates (i.e. permutations) that have $n$ successes out of $N$ trials.
-These happen to be the coefficients in the polynomial $(a + b)^N$,
+number of microstates (i.e. permutations) that have $$n$$ successes out of $$N$$ trials.
+These happen to be the coefficients in the polynomial $$(a + b)^N$$,
and can be read off of Pascal's triangle.
It is defined as follows:
@@ -33,10 +33,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-The remaining factor $p^n (1 - p)^{N - n}$ is then just the
+The remaining factor $$p^n (1 - p)^{N - n}$$ is then just the
probability of attaining each microstate.
-The expected or mean number of successes $\mu$ after $N$ trials is as follows:
+The expected or mean number of successes $$\mu$$ after $$N$$ trials is as follows:
$$\begin{aligned}
\boxed{
@@ -49,7 +49,7 @@ $$\begin{aligned}
-The trick is to treat $p$ and $q$ as independent until the last moment:
+The trick is to treat $$p$$ and $$q$$ as independent until the last moment:
$$\begin{aligned}
\mu
@@ -61,12 +61,12 @@ $$\begin{aligned}
= N p (p + q)^{N - 1}
\end{aligned}$$
-Inserting $q = 1 - p$ then gives the desired result.
+Inserting $$q = 1 - p$$ then gives the desired result.
-Meanwhile, we find the following variance $\sigma^2$,
-with $\sigma$ being the standard deviation:
+Meanwhile, we find the following variance $$\sigma^2$$,
+with $$\sigma$$ being the standard deviation:
$$\begin{aligned}
\boxed{
@@ -79,7 +79,7 @@ $$\begin{aligned}
-We use the same trick to calculate $\overline{n^2}$
+We use the same trick to calculate $$\overline{n^2}$$
(the mean squared number of successes):
$$\begin{aligned}
@@ -96,7 +96,7 @@ $$\begin{aligned}
&= N p + N^2 p^2 - N p^2
\end{aligned}$$
-Using this and the earlier expression $\mu = N p$, we find the variance $\sigma^2$:
+Using this and the earlier expression $$\mu = N p$$, we find the variance $$\sigma^2$$:
$$\begin{aligned}
\sigma^2
@@ -105,11 +105,11 @@ $$\begin{aligned}
= N p (1 - p)
\end{aligned}$$
-By inserting $q = 1 - p$, we arrive at the desired expression.
+By inserting $$q = 1 - p$$, we arrive at the desired expression.
-As $N \to \infty$, the binomial distribution
+As $$N \to \infty$$, the binomial distribution
turns into the continuous normal distribution,
a fact that is sometimes called the **de Moivre-Laplace theorem**:
@@ -124,8 +124,8 @@ $$\begin{aligned}
-We take the Taylor expansion of $\ln\!\big(P_N(n)\big)$
-around the mean $\mu = Np$:
+We take the Taylor expansion of $$\ln\!\big(P_N(n)\big)$$
+around the mean $$\mu = Np$$:
$$\begin{aligned}
\ln\!\big(P_N(n)\big)
@@ -134,7 +134,7 @@ $$\begin{aligned}
D_m(n) = \dvn{m}{\ln\!\big(P_N(n)\big)}{n}
\end{aligned}$$
-We use Stirling's approximation to calculate the factorials in $D_m$:
+We use Stirling's approximation to calculate the factorials in $$D_m$$:
$$\begin{aligned}
\ln\!\big(P_N(n)\big)
@@ -143,8 +143,8 @@ $$\begin{aligned}
&\approx \ln(N!) - n \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) - (N\!-\!n) \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big)
\end{aligned}$$
-For $D_0(\mu)$, we need to use a stronger version of Stirling's approximation
-to get a non-zero result. We take advantage of $N - N p = N q$:
+For $$D_0(\mu)$$, we need to use a stronger version of Stirling's approximation
+to get a non-zero result. We take advantage of $$N - N p = N q$$:
$$\begin{aligned}
D_0(\mu)
@@ -161,7 +161,7 @@ $$\begin{aligned}
= \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big)
\end{aligned}$$
-Next, we expect that $D_1(\mu) = 0$, because $\mu$ is the maximum.
+Next, we expect that $$D_1(\mu) = 0$$, because $$\mu$$ is the maximum.
This is indeed the case:
$$\begin{aligned}
@@ -176,7 +176,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-For the same reason, we expect that $D_2(\mu)$ is negative.
+For the same reason, we expect that $$D_2(\mu)$$ is negative.
We find the following expression:
$$\begin{aligned}
@@ -189,7 +189,7 @@ $$\begin{aligned}
= - \frac{1}{\sigma^2}
\end{aligned}$$
-The higher-order derivatives tend to zero for $N \to \infty$, so we discard them:
+The higher-order derivatives tend to zero for $$N \to \infty$$, so we discard them:
$$\begin{aligned}
D_3(n)
@@ -201,7 +201,7 @@ $$\begin{aligned}
\cdots
\end{aligned}$$
-Putting everything together, for large $N$,
+Putting everything together, for large $$N$$,
the Taylor series approximately becomes:
$$\begin{aligned}
@@ -210,7 +210,7 @@ $$\begin{aligned}
= \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) - \frac{(n - \mu)^2}{2 \sigma^2}
\end{aligned}$$
-Taking $\exp$ of this expression then yields a normalized Gaussian distribution.
+Taking $$\exp$$ of this expression then yields a normalized Gaussian distribution.
diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md
index fd4af57..de80f96 100644
--- a/source/know/concept/blasius-boundary-layer/index.md
+++ b/source/know/concept/blasius-boundary-layer/index.md
@@ -12,15 +12,15 @@ layout: "concept"
In fluid dynamics, the **Blasius boundary layer** is an application of
the [Prandtl equations](/know/concept/prandtl-equations/),
which govern the flow of a fluid
-at large Reynolds number $\mathrm{Re} \gg 1$
+at large Reynolds number $$\mathrm{Re} \gg 1$$
close to a surface.
Specifically, the Blasius layer is the solution
for a half-plane approached from the edge by a fluid.
-A fluid with velocity field $\va{v} = U \vu{e}_x$ flows to the plane,
-which starts at $y = 0$ and exists for $x \ge 0$.
+A fluid with velocity field $$\va{v} = U \vu{e}_x$$ flows to the plane,
+which starts at $$y = 0$$ and exists for $$x \ge 0$$.
To describe this, we make an ansatz
-for the *slip-flow* region's $x$-velocity $v_x(x, y)$:
+for the *slip-flow* region's $$x$$-velocity $$v_x(x, y)$$:
$$\begin{aligned}
v_x
@@ -30,13 +30,13 @@ $$\begin{aligned}
\equiv \frac{y}{\delta(x)}
\end{aligned}$$
-Note that $f'(s)$ is the derivative of an unknown $f(s)$,
-and that it obeys the boundary conditions $f'(0) = 0$ and $f'(\infty) = 1$.
-Furthermore, $\delta(x)$ is the thickness of the stationary boundary layer at the surface.
+Note that $$f'(s)$$ is the derivative of an unknown $$f(s)$$,
+and that it obeys the boundary conditions $$f'(0) = 0$$ and $$f'(\infty) = 1$$.
+Furthermore, $$\delta(x)$$ is the thickness of the stationary boundary layer at the surface.
To derive the Prandtl equations,
-the estimate $\delta(x) = \sqrt{\nu x / U}$ was used,
+the estimate $$\delta(x) = \sqrt{\nu x / U}$$ was used,
which we will stick with.
-For later use, it is worth writing the derivatives of $s$:
+For later use, it is worth writing the derivatives of $$s$$:
$$\begin{aligned}
\pdv{s}{x}
@@ -47,7 +47,7 @@ $$\begin{aligned}
= \frac{1}{\delta}
\end{aligned}$$
-Inserting the ansatz for $v_x$ into the incompressibility condition then yields:
+Inserting the ansatz for $$v_x$$ into the incompressibility condition then yields:
$$\begin{aligned}
\pdv{v_y}{y}
@@ -55,7 +55,7 @@ $$\begin{aligned}
= U s f'' \frac{\delta'}{\delta}
\end{aligned}$$
-Which we integrate to get an expression for the $y$-velocity $v_y$, namely:
+Which we integrate to get an expression for the $$y$$-velocity $$v_y$$, namely:
$$\begin{aligned}
v_y
@@ -64,21 +64,21 @@ $$\begin{aligned}
\end{aligned}$$
Now, consider the main Prandtl equation,
-assuming that the attack velocity $U$ is constant:
+assuming that the attack velocity $$U$$ is constant:
$$\begin{aligned}
v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
= \nu \pdvn{2}{v_x}{y}
\end{aligned}$$
-Inserting our expressions for $v_x$ and $v_y$ into this leads us to:
+Inserting our expressions for $$v_x$$ and $$v_y$$ into this leads us to:
$$\begin{aligned}
- U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
= \nu U \frac{1}{\delta^2} f'''
\end{aligned}$$
-After multiplying it by $\delta^2 / U$ and cancelling out some terms,
+After multiplying it by $$\delta^2 / U$$ and cancelling out some terms,
it reduces to:
$$\begin{aligned}
@@ -86,7 +86,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Then, substituting $\delta(x) = \sqrt{\nu x / U}$ and $\delta'(x) = (1/2) \sqrt{\nu / (U x)}$ yields:
+Then, substituting $$\delta(x) = \sqrt{\nu x / U}$$ and $$\delta'(x) = (1/2) \sqrt{\nu / (U x)}$$ yields:
$$\begin{aligned}
\nu f''' + U \frac{\nu}{2 U} f'' f
@@ -94,7 +94,7 @@ $$\begin{aligned}
\end{aligned}$$
Simplifying this leads us to the **Blasius equation**,
-which is a nonlinear ODE for $f(s)$:
+which is a nonlinear ODE for $$f(s)$$:
$$\begin{aligned}
\boxed{
@@ -103,7 +103,7 @@ $$\begin{aligned}
\end{aligned}$$
Unfortunately, this cannot be solved analytically, only numerically.
-Nevertheless, the result shows a boundary layer $\delta(x)$
+Nevertheless, the result shows a boundary layer $$\delta(x)$$
exhibiting the expected downstream thickening.
diff --git a/source/know/concept/bloch-sphere/index.md b/source/know/concept/bloch-sphere/index.md
index 5d747f7..2cb7742 100644
--- a/source/know/concept/bloch-sphere/index.md
+++ b/source/know/concept/bloch-sphere/index.md
@@ -17,7 +17,7 @@ All pure qubit states are represented by a point on the sphere's surface:
-The $x$, $y$ and $z$-axes represent the components of a spin-1/2-alike system,
+The $$x$$, $$y$$ and $$z$$-axes represent the components of a spin-1/2-alike system,
and their extremes are the eigenstates of the Pauli matrices:
$$\begin{aligned}
@@ -31,7 +31,7 @@ $$\begin{aligned}
\to \{\Ket{+i}, \Ket{-i}\}
\end{aligned}$$
-Where the latter two states are expressed as follows in the conventional $z$-basis:
+Where the latter two states are expressed as follows in the conventional $$z$$-basis:
$$\begin{aligned}
\Ket{\pm}
@@ -42,15 +42,15 @@ $$\begin{aligned}
\end{aligned}$$
More generally, every point on the surface of the sphere
-describes a pure qubit state in terms of the angles $\theta$ and $\varphi$,
+describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
respectively the elevation and azimuth:
$$\begin{aligned}
\Ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \Ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \Ket{1}
\end{aligned}$$
-We can generalize this further by describing points using the **Bloch vector** $\vec{r}$,
-with radius $r \le 1$:
+We can generalize this further by describing points using the **Bloch vector** $$\vec{r}$$,
+with radius $$r \le 1$$:
$$\begin{aligned}
\boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that $\vec{r}$ is not actually a qubit state,
+Note that $$\vec{r}$$ is not actually a qubit state,
but rather an implicit description of one,
meaning that it does not need to be normalized.
The main point of the Bloch vector is that it allows us
@@ -73,9 +73,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$ is the Pauli "vector".
-Now, we know that $\hat{\rho}$ represents a pure ensemble
-if and only if it is idempotent, i.e. $\hat{\rho}^2 = \hat{\rho}$:
+Where $$\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$$ is the Pauli "vector".
+Now, we know that $$\hat{\rho}$$ represents a pure ensemble
+if and only if it is idempotent, i.e. $$\hat{\rho}^2 = \hat{\rho}$$:
$$\begin{aligned}
\hat{\rho}^2
@@ -83,8 +83,8 @@ $$\begin{aligned}
= \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
\end{aligned}$$
-You can easily convince yourself that if $(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$,
-then we get $\hat{\rho}$ again, and the state is pure:
+You can easily convince yourself that if $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
+then we get $$\hat{\rho}$$ again, and the state is pure:
$$\begin{aligned}
(\vec{r} \cdot \vec{\sigma})^2
@@ -105,13 +105,13 @@ $$\begin{aligned}
= r^2 \hat{I}
\end{aligned}$$
-Therefore, if the radius $r = 1$, the ensemble is pure,
-else if $r < 1$ it is mixed.
+Therefore, if the radius $$r = 1$$, the ensemble is pure,
+else if $$r < 1$$ it is mixed.
Another useful property of the Bloch vector
is that the expectation value of the Pauli matrices
-are given by the corresponding component of $\vec{r}$,
-for example for $\hat{\sigma}_z$:
+are given by the corresponding component of $$\vec{r}$$,
+for example for $$\hat{\sigma}_z$$:
$$\begin{aligned}
\Expval{\hat{\sigma}_z}
diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
index e049a71..6f445f1 100644
--- a/source/know/concept/blochs-theorem/index.md
+++ b/source/know/concept/blochs-theorem/index.md
@@ -8,14 +8,14 @@ layout: "concept"
---
In quantum mechanics, **Bloch's theorem** states that,
-given a potential $V(\vb{r})$ which is periodic on a lattice,
-i.e. $V(\vb{r}) = V(\vb{r} + \vb{a})$
-for a primitive lattice vector $\vb{a}$,
-then it follows that the solutions $\psi(\vb{r})$
+given a potential $$V(\vb{r})$$ which is periodic on a lattice,
+i.e. $$V(\vb{r}) = V(\vb{r} + \vb{a})$$
+for a primitive lattice vector $$\vb{a}$$,
+then it follows that the solutions $$\psi(\vb{r})$$
to the time-independent Schrödinger equation
take the following form,
-where the function $u(\vb{r})$ is periodic on the same lattice,
-i.e. $u(\vb{r}) = u(\vb{r} + \vb{a})$:
+where the function $$u(\vb{r})$$ is periodic on the same lattice,
+i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$:
$$
\begin{aligned}
@@ -30,8 +30,8 @@ the solutions are simply plane waves with a periodic modulation,
known as **Bloch functions** or **Bloch states**.
This is suprisingly easy to prove:
-if the Hamiltonian $\hat{H}$ is lattice-periodic,
-then both $\psi(\vb{r})$ and $\psi(\vb{r} + \vb{a})$
+if the Hamiltonian $$\hat{H}$$ is lattice-periodic,
+then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$
are eigenstates with the same energy:
$$
@@ -42,8 +42,8 @@ $$
\end{aligned}
$$
-Now define the unitary translation operator $\hat{T}(\vb{a})$ such that
-$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$.
+Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that
+$$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$$.
From the previous equation, we then know that:
$$
@@ -55,10 +55,10 @@ $$
\end{aligned}
$$
-In other words, if $\hat{H}$ is lattice-periodic,
-then it will commute with $\hat{T}(\vb{a})$,
-i.e. $[\hat{H}, \hat{T}(\vb{a})] = 0$.
-Consequently, $\hat{H}$ and $\hat{T}(\vb{a})$ must share eigenstates $\psi(\vb{r})$:
+In other words, if $$\hat{H}$$ is lattice-periodic,
+then it will commute with $$\hat{T}(\vb{a})$$,
+i.e. $$[\hat{H}, \hat{T}(\vb{a})] = 0$$.
+Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ must share eigenstates $$\psi(\vb{r})$$:
$$
\begin{aligned}
@@ -68,10 +68,10 @@ $$
\end{aligned}
$$
-Since $\hat{T}$ is unitary,
-its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real.
-Therefore a translation by $\vb{a}$ causes a phase shift,
-for some vector $\vb{k}$:
+Since $$\hat{T}$$ is unitary,
+its eigenvalues $$\tau$$ must have the form $$e^{i \theta}$$, with $$\theta$$ real.
+Therefore a translation by $$\vb{a}$$ causes a phase shift,
+for some vector $$\vb{k}$$:
$$
\begin{aligned}
@@ -83,7 +83,7 @@ $$
$$
Let us now define the following function,
-keeping our arbitrary choice of $\vb{k}$:
+keeping our arbitrary choice of $$\vb{k}$$:
$$
\begin{aligned}
@@ -92,7 +92,7 @@ $$
\end{aligned}
$$
-As it turns out, this function is guaranteed to be lattice-periodic for any $\vb{k}$:
+As it turns out, this function is guaranteed to be lattice-periodic for any $$\vb{k}$$:
$$
\begin{aligned}
@@ -108,4 +108,4 @@ $$
$$
Then Bloch's theorem follows from
-isolating the definition of $u(\vb{r})$ for $\psi(\vb{r})$.
+isolating the definition of $$u(\vb{r})$$ for $$\psi(\vb{r})$$.
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 20399df..9ed2fd2 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -10,17 +10,17 @@ layout: "concept"
---
Consider a collection of particles,
-each with its own position $\vb{r}$ and velocity $\vb{v}$.
-We can thus define a probability density function $f(\vb{r}, \vb{v}, t)$
-describing the expected number of particles at $(\vb{r}, \vb{v})$ at time $t$.
-Let the total number of particles $N$ be conserved, then clearly:
+each with its own position $$\vb{r}$$ and velocity $$\vb{v}$$.
+We can thus define a probability density function $$f(\vb{r}, \vb{v}, t)$$
+describing the expected number of particles at $$(\vb{r}, \vb{v})$$ at time $$t$$.
+Let the total number of particles $$N$$ be conserved, then clearly:
$$\begin{aligned}
N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}}
\end{aligned}$$
At equilibrium, all processes affecting the particles
-no longer have a net effect, so $f$ is fixed:
+no longer have a net effect, so $$f$$ is fixed:
$$\begin{aligned}
\dv{f}{t}
@@ -35,7 +35,7 @@ $$\begin{aligned}
= \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
\end{aligned}$$
-Where the right-hand side simply means "all changes in $f$ due to collisions".
+Where the right-hand side simply means "all changes in $$f$$ due to collisions".
Applying the chain rule to the left-hand side then yields:
$$\begin{aligned}
@@ -49,8 +49,8 @@ $$\begin{aligned}
&= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}}
\end{aligned}$$
-Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$.
-Inserting Newton's second law $\vb{F} = m \vb{a}$
+Where we have introduced the shorthand $$\ipdv{f}{\vb{v}}$$.
+Inserting Newton's second law $$\vb{F} = m \vb{a}$$
leads us to the **Boltzmann equation** or
**Boltzmann transport equation** (BTE):
@@ -64,41 +64,41 @@ $$\begin{aligned}
But what about the collision term?
Expressions for it exist, which are almost exact in many cases,
but unfortunately also quite difficult to work with.
-In addition, $f$ is a 7-dimensional function,
+In addition, $$f$$ is a 7-dimensional function,
so the BTE is already hard to solve without collisions.
We only present the simplest case,
known as the **Bhatnagar-Gross-Krook approximation**:
-if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known,
-then each collision brings the system closer to $f_0$:
+if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known,
+then each collision brings the system closer to $$f_0$$:
$$\begin{aligned}
\pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}}
= \frac{f_0 - f}{\tau}
\end{aligned}$$
-Where $\tau$ is the average collision period.
+Where $$\tau$$ is the average collision period.
The right-hand side is called the **Krook term**.
## Moment equations
-From the definition of $f$,
-we see that integrating over all $\vb{v}$ yields the particle density $n$:
+From the definition of $$f$$,
+we see that integrating over all $$\vb{v}$$ yields the particle density $$n$$:
$$\begin{aligned}
n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
\end{aligned}$$
-Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so:
+Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so:
$$\begin{aligned}
\Expval{Q}
= \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
\end{aligned}$$
-With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate,
-assuming that $\vb{F}$ does not depend on $\vb{v}$:
+With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate,
+assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$:
$$\begin{aligned}
0
@@ -110,10 +110,10 @@ $$\begin{aligned}
+ \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{}{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}}
\end{aligned}$$
-The first integral is simply $n \Expval{Q}$.
-In the second integral, note that $\vb{v}$ is a coordinate
-and hence not dependent on $\vb{r}$, so $\nabla \cdot \vb{v} = 0$.
-Since $f$ is a probability density, $f \to 0$ for $\vb{v} \to \pm\infty$,
+The first integral is simply $$n \Expval{Q}$$.
+In the second integral, note that $$\vb{v}$$ is a coordinate
+and hence not dependent on $$\vb{r}$$, so $$\nabla \cdot \vb{v} = 0$$.
+Since $$f$$ is a probability density, $$f \to 0$$ for $$\vb{v} \to \pm\infty$$,
so the first term in the third integral vanishes after it is integrated:
$$\begin{aligned}
@@ -134,9 +134,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-If we set $Q = m$, then the mass density $\rho = n \Expval{Q}$,
+If we set $$Q = m$$, then the mass density $$\rho = n \Expval{Q}$$,
and we find that the **zeroth moment** of the BTE describes conservation of mass,
-where $\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$ is the fluid velocity:
+where $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity:
$$\begin{aligned}
\boxed{
@@ -150,8 +150,8 @@ $$\begin{aligned}
-We insert $Q = m$ into our prototype,
-and since $m$ is constant, the rest is trivial:
+We insert $$Q = m$$ into our prototype,
+and since $$m$$ is constant, the rest is trivial:
$$\begin{aligned}
0
@@ -159,12 +159,13 @@ $$\begin{aligned}
\\
&= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
\end{aligned}$$
+
-If we instead choose the momentum $Q = m \vb{v}$,
+If we instead choose the momentum $$Q = m \vb{v}$$,
we find that the **first moment** of the BTE describes conservation of momentum,
-where $\hat{P}$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/):
+where $$\hat{P}$$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/):
$$\begin{aligned}
\boxed{
@@ -178,7 +179,7 @@ $$\begin{aligned}
-We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible:
+We insert $$Q = m \vb{v}$$ into our prototype and recognize $$\rho$$ wherever possible:
$$\begin{aligned}
0
@@ -189,11 +190,11 @@ $$\begin{aligned}
- \vb{F} \cdot \bigg( n \Expval{\pdv{\vb{v}}{\vb{v}}} \bigg)
\end{aligned}$$
-With $\vb{v} \vb{v}$ being a dyadic product.
+With $$\vb{v} \vb{v}$$ being a dyadic product.
To give it a physical interpretation,
-we split $\vb{v} = \vb{V} \!+\! \vb{w}$,
-where $\vb{V}$ is the average velocity vector,
-and $\vb{w}$ is the local deviation from $\vb{V}$:
+we split $$\vb{v} = \vb{V} \!+\! \vb{w}$$,
+where $$\vb{V}$$ is the average velocity vector,
+and $$\vb{w}$$ is the local deviation from $$\vb{V}$$:
$$\begin{aligned}
\Expval{\vb{v} \vb{v}}
@@ -202,7 +203,7 @@ $$\begin{aligned}
= \vb{V} \vb{V} + 2 \vb{V} \Expval{\vb{w}} + \Expval{\vb{w} \vb{w}}
\end{aligned}$$
-Since $\vb{w}$ represents a deviation from the mean, $\Expval{\vb{w}} = 0$.
+Since $$\vb{w}$$ represents a deviation from the mean, $$\Expval{\vb{w}} = 0$$.
We define the pressure tensor:
$$\begin{aligned}
@@ -212,19 +213,20 @@ $$\begin{aligned}
\end{aligned}$$
This leads to the expected result,
-where $\nabla \cdot (\rho \vb{V}\vb{V})$ represents the fluid momentum,
-and $\nabla \cdot \hat{P}$ the viscous/pressure momentum:
+where $$\nabla \cdot (\rho \vb{V}\vb{V})$$ represents the fluid momentum,
+and $$\nabla \cdot \hat{P}$$ the viscous/pressure momentum:
$$\begin{aligned}
0
&= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
\end{aligned}$$
+
-Finally, if we choose the kinetic energy $Q = m |\vb{v}|^2 / 2$,
+Finally, if we choose the kinetic energy $$Q = m |\vb{v}|^2 / 2$$,
we find that the **second moment** gives conservation of energy,
-where $U$ is the thermal energy density and $\vb{J}$ is the heat flux:
+where $$U$$ is the thermal energy density and $$\vb{J}$$ is the heat flux:
$$\begin{aligned}
\boxed{
@@ -240,7 +242,7 @@ $$\begin{aligned}
-We insert $Q = m |\vb{v}|^2 / 2$ into our prototype and recognize $\rho$ wherever possible:
+We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:
$$\begin{aligned}
0
@@ -253,7 +255,7 @@ $$\begin{aligned}
- \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg)
\end{aligned}$$
-We handle these terms one by one. Substituting $\vb{v} = \vb{V} + \vb{w}$ in the first gives:
+We handle these terms one by one. Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:
$$\begin{aligned}
\Expval{|\vb{v}|^2}
@@ -265,7 +267,7 @@ $$\begin{aligned}
\end{aligned}$$
And likewise for the second term,
-where we recognize the stress tensor $\Expval{\vb{w} \vb{w}}$:
+where we recognize the stress tensor $$\Expval{\vb{w} \vb{w}}$$:
$$\begin{aligned}
\Expval{|\vb{v}|^2 \vb{v}}
@@ -294,7 +296,7 @@ $$\begin{aligned}
\end{aligned}$$
To clarify the physical interpretation,
-we define $U$, $\vb{J}$ and $\hat{P}$ as follows:
+we define $$U$$, $$\vb{J}$$ and $$\hat{P}$$ as follows:
$$\begin{aligned}
U
@@ -347,6 +349,7 @@ $$\begin{aligned}
\end{bmatrix}
= \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
\end{aligned}$$
+
diff --git a/source/know/concept/boltzmann-relation/index.md b/source/know/concept/boltzmann-relation/index.md
index b7f82b7..d5409d2 100644
--- a/source/know/concept/boltzmann-relation/index.md
+++ b/source/know/concept/boltzmann-relation/index.md
@@ -10,12 +10,12 @@ layout: "concept"
In a plasma where the ions and electrons are both in thermal equilibrium,
and in the absence of short-lived induced electromagnetic fields,
-their densities $n_i$ and $n_e$ can be predicted.
+their densities $$n_i$$ and $$n_e$$ can be predicted.
-By definition, a particle in an [electric field](/know/concept/electric-field/) $\vb{E}$
-experiences a [Lorentz force](/know/concept/lorentz-force/) $\vb{F}_e$.
-This corresponds to a force density $\vb{f}_e$,
-such that $\vb{F}_e = \vb{f}_e \dd{V}$.
+By definition, a particle in an [electric field](/know/concept/electric-field/) $$\vb{E}$$
+experiences a [Lorentz force](/know/concept/lorentz-force/) $$\vb{F}_e$$.
+This corresponds to a force density $$\vb{f}_e$$,
+such that $$\vb{F}_e = \vb{f}_e \dd{V}$$.
For the electrons, we thus have:
$$\begin{aligned}
@@ -25,8 +25,8 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, if we treat the electrons as a gas
-obeying the ideal gas law $p_e = k_B T_e n_e$,
-then the pressure $p_e$ leads to another force density $\vb{f}_p$:
+obeying the ideal gas law $$p_e = k_B T_e n_e$$,
+then the pressure $$p_e$$ leads to another force density $$\vb{f}_p$$:
$$\begin{aligned}
\vb{f}_p
@@ -34,8 +34,8 @@ $$\begin{aligned}
= - k_B T_e \nabla n_e
\end{aligned}$$
-At equilibrium, we demand that $\vb{f}_e = - \vb{f}_p$,
-and isolate this equation for $\nabla n_e$, yielding:
+At equilibrium, we demand that $$\vb{f}_e = - \vb{f}_p$$,
+and isolate this equation for $$\nabla n_e$$, yielding:
$$\begin{aligned}
k_B T_e \nabla n_e
@@ -47,7 +47,7 @@ $$\begin{aligned}
\end{aligned}$$
This equation is straightforward to integrate,
-leading to the following expression for $n_e$,
+leading to the following expression for $$n_e$$,
known as the **Boltzmann relation**,
due to its resemblance to the statistical Boltzmann distribution
(see [canonical ensemble](/know/concept/canonical-ensemble/)):
@@ -59,10 +59,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the linearity factor $n_{e0}$ represents
-the electron density for $\phi = 0$.
+Where the linearity factor $$n_{e0}$$ represents
+the electron density for $$\phi = 0$$.
We can do the same for ions instead of electrons,
-leading to the following ion density $n_i$:
+leading to the following ion density $$n_i$$:
$$\begin{aligned}
\boxed{
@@ -75,7 +75,7 @@ However, due to their larger mass,
ions are much slower to respond to fluctuations in the above equilibrium.
Consequently, after a perturbation,
the ions spend much more time in a transient non-equilibrium state
-than the electrons, so this formula for $n_i$ is only valid
+than the electrons, so this formula for $$n_i$$ is only valid
if the perturbation is sufficiently slow,
allowing the ions to keep up.
Usually, electrons do not suffer the same issue,
diff --git a/source/know/concept/bose-einstein-distribution/index.md b/source/know/concept/bose-einstein-distribution/index.md
index 594d6e0..e420d7c 100644
--- a/source/know/concept/bose-einstein-distribution/index.md
+++ b/source/know/concept/bose-einstein-distribution/index.md
@@ -14,13 +14,13 @@ which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-
will distribute themselves across the available states
in a system at equilibrium.
-Consider a single-particle state $s$,
+Consider a single-particle state $$s$$,
which can contain any number of bosons.
-Since the occupation number $N$ is variable,
+Since the occupation number $$N$$ is variable,
we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
-whose grand partition function $\mathcal{Z}$ is as follows,
-where $\varepsilon$ is the energy per particle,
-and $\mu$ is the chemical potential:
+whose grand partition function $$\mathcal{Z}$$ is as follows,
+where $$\varepsilon$$ is the energy per particle,
+and $$\mu$$ is the chemical potential:
$$\begin{aligned}
\mathcal{Z}
@@ -29,7 +29,7 @@ $$\begin{aligned}
\end{aligned}$$
The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
-is the Landau potential $\Omega$, given by:
+is the Landau potential $$\Omega$$, given by:
$$\begin{aligned}
\Omega
@@ -37,8 +37,8 @@ $$\begin{aligned}
= k T \ln\!\Big( 1 - \exp(- \beta (\varepsilon - \mu)) \Big)
\end{aligned}$$
-The average number of particles $\Expval{N}$
-is found by taking a derivative of $\Omega$:
+The average number of particles $$\Expval{N}$$
+is found by taking a derivative of $$\Omega$$:
$$\begin{aligned}
\Expval{N}
@@ -47,8 +47,8 @@ $$\begin{aligned}
= \frac{\exp(- \beta (\varepsilon - \mu))}{1 - \exp(- \beta (\varepsilon - \mu))}
\end{aligned}$$
-By multitplying both the numerator and the denominator by $\exp(\beta(\varepsilon \!-\! \mu))$,
-we arrive at the standard form of the **Bose-Einstein distribution** $f_B$:
+By multitplying both the numerator and the denominator by $$\exp(\beta(\varepsilon \!-\! \mu))$$,
+we arrive at the standard form of the **Bose-Einstein distribution** $$f_B$$:
$$\begin{aligned}
\boxed{
@@ -58,9 +58,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-This tells the expected occupation number $\Expval{N}$ of state $s$,
-given a temperature $T$ and chemical potential $\mu$.
-The corresponding variance $\sigma^2$ of $N$ is found to be:
+This tells the expected occupation number $$\Expval{N}$$ of state $$s$$,
+given a temperature $$T$$ and chemical potential $$\mu$$.
+The corresponding variance $$\sigma^2$$ of $$N$$ is found to be:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/calculus-of-variations/index.md b/source/know/concept/calculus-of-variations/index.md
index 0b1d070..7701358 100644
--- a/source/know/concept/calculus-of-variations/index.md
+++ b/source/know/concept/calculus-of-variations/index.md
@@ -11,24 +11,24 @@ layout: "concept"
The **calculus of variations** lays the mathematical groundwork
for [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
-Consider a **functional** $J$, mapping a function $f(x)$ to a scalar value
-by integrating over the so-called **Lagrangian** $L$,
-which represents an expression involving $x$, $f$ and the derivative $f'$:
+Consider a **functional** $$J$$, mapping a function $$f(x)$$ to a scalar value
+by integrating over the so-called **Lagrangian** $$L$$,
+which represents an expression involving $$x$$, $$f$$ and the derivative $$f'$$:
$$\begin{aligned}
J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
\end{aligned}$$
-If $J$ in some way measures the physical "cost" (e.g. energy) of
-the path $f(x)$ taken by a physical system,
-the **principle of least action** states that $f$ will be a minimum of $J[f]$,
+If $$J$$ in some way measures the physical "cost" (e.g. energy) of
+the path $$f(x)$$ taken by a physical system,
+the **principle of least action** states that $$f$$ will be a minimum of $$J[f]$$,
so for example the expended energy will be minimized.
In practice, various cost metrics may be used,
-so maxima of $J[f]$ are also interesting to us.
+so maxima of $$J[f]$$ are also interesting to us.
-If $f(x, \varepsilon\!=\!0)$ is the optimal route, then a slightly
+If $$f(x, \varepsilon\!=\!0)$$ is the optimal route, then a slightly
different (and therefore worse) path between the same two points can be expressed
-using the parameter $\varepsilon$:
+using the parameter $$\varepsilon$$:
$$\begin{aligned}
f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x)
@@ -36,19 +36,19 @@ $$\begin{aligned}
\delta f = \varepsilon \eta(x)
\end{aligned}$$
-Where $\eta(x)$ is an arbitrary differentiable deviation.
-Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$,
+Where $$\eta(x)$$ is an arbitrary differentiable deviation.
+Since $$f(x, \varepsilon)$$ must start and end in the same points as $$f(x,0)$$,
we have the boundary conditions:
$$\begin{aligned}
\eta(x_0) = \eta(x_1) = 0
\end{aligned}$$
-Given $L$, the goal is to find an equation for the optimal path $f(x,0)$.
+Given $$L$$, the goal is to find an equation for the optimal path $$f(x,0)$$.
Just like when finding the minimum of a real function,
-the minimum $f$ of a functional $J[f]$ is a stationary point
-with respect to the deviation weight $\varepsilon$,
-a condition often written as $\delta J = 0$.
+the minimum $$f$$ of a functional $$J[f]$$ is a stationary point
+with respect to the deviation weight $$\varepsilon$$,
+a condition often written as $$\delta J = 0$$.
In the following, the integration limits have been omitted:
$$\begin{aligned}
@@ -63,14 +63,14 @@ $$\begin{aligned}
\end{aligned}$$
The boundary term from partial integration vanishes due to the boundary
-conditions for $\eta(x)$. We are thus left with:
+conditions for $$\eta(x)$$. We are thus left with:
$$\begin{aligned}
0
= \int \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f'} \Big) \bigg) \dd{x}
\end{aligned}$$
-This holds for all $\eta$, but $\eta$ is arbitrary, so in fact
+This holds for all $$\eta$$, but $$\eta$$ is arbitrary, so in fact
only the parenthesized expression matters:
$$\begin{aligned}
@@ -79,27 +79,27 @@ $$\begin{aligned}
}
\end{aligned}$$
-This is known as the **Euler-Lagrange equation** of the Lagrangian $L$,
-and its solutions represent the optimal paths $f(x, 0)$.
+This is known as the **Euler-Lagrange equation** of the Lagrangian $$L$$,
+and its solutions represent the optimal paths $$f(x, 0)$$.
## Multiple functions
-Suppose that the Lagrangian $L$ depends on multiple independent functions
-$f_1, f_2, ..., f_N$:
+Suppose that the Lagrangian $$L$$ depends on multiple independent functions
+$$f_1, f_2, ..., f_N$$:
$$\begin{aligned}
J[f_1, ..., f_N] = \int_{x_0}^{x_1} L(f_1, ..., f_N, f_1', ..., f_N', x) \dd{x}
\end{aligned}$$
-In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$,
-satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$:
+In this case, every $$f_n(x)$$ has its own deviation $$\eta_n(x)$$,
+satisfying $$\eta_n(x_0) = \eta_n(x_1) = 0$$:
$$\begin{aligned}
f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x)
\end{aligned}$$
-The derivation procedure is identical to the case $N = 1$ from earlier:
+The derivation procedure is identical to the case $$N = 1$$ from earlier:
$$\begin{aligned}
0
@@ -113,8 +113,8 @@ $$\begin{aligned}
+ \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \dv{}{x}\Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x}
\end{aligned}$$
-Once again, $\eta_n(x)$ is arbitrary and disappears at the boundaries,
-so we end up with $N$ equations of the same form as for a single function:
+Once again, $$\eta_n(x)$$ is arbitrary and disappears at the boundaries,
+so we end up with $$N$$ equations of the same form as for a single function:
$$\begin{aligned}
\boxed{
@@ -127,7 +127,7 @@ $$\begin{aligned}
## Higher-order derivatives
-Suppose that the Lagrangian $L$ depends on multiple derivatives of $f(x)$:
+Suppose that the Lagrangian $$L$$ depends on multiple derivatives of $$f(x)$$:
$$\begin{aligned}
J[f] = \int_{x_0}^{x_1} L(f, f', f'', ..., f^{(N)}, x) \dd{x}
@@ -144,9 +144,9 @@ $$\begin{aligned}
&= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x}
\end{aligned}$$
-The goal is to turn each $\eta^{(n)}(x)$ into $\eta(x)$, so we need to
-partially integrate the $n$th term of the sum $n$ times. In this case,
-we will need some additional boundary conditions for $\eta(x)$:
+The goal is to turn each $$\eta^{(n)}(x)$$ into $$\eta(x)$$, so we need to
+partially integrate the $$n$$th term of the sum $$n$$ times. In this case,
+we will need some additional boundary conditions for $$\eta(x)$$:
$$\begin{aligned}
\eta'(x_0) = \eta'(x_1) = 0
@@ -161,7 +161,7 @@ $$\begin{aligned}
&= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dvn{n}{}{x}\Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x}
\end{aligned}$$
-Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes:
+Once again, because $$\eta(x)$$ is arbitrary, the Euler-Lagrange equation becomes:
$$\begin{aligned}
\boxed{
@@ -172,16 +172,16 @@ $$\begin{aligned}
## Multiple coordinates
-Suppose now that $f$ is a function of multiple variables.
-For brevity, we only consider two variables $x$ and $y$,
+Suppose now that $$f$$ is a function of multiple variables.
+For brevity, we only consider two variables $$x$$ and $$y$$,
but the results generalize effortlessly to larger amounts.
-The Lagrangian now depends on all the partial derivatives of $f(x, y)$:
+The Lagrangian now depends on all the partial derivatives of $$f(x, y)$$:
$$\begin{aligned}
J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
\end{aligned}$$
-The arbitrary deviation $\eta$ is then also a function of multiple variables:
+The arbitrary deviation $$\eta$$ is then also a function of multiple variables:
$$\begin{aligned}
f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y)
@@ -199,7 +199,7 @@ $$\begin{aligned}
&= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y}
\end{aligned}$$
-We partially integrate for both $\eta_x$ and $\eta_y$, yielding:
+We partially integrate for both $$\eta_x$$ and $$\eta_y$$, yielding:
$$\begin{aligned}
0
@@ -208,7 +208,7 @@ $$\begin{aligned}
&\quad + \iint \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big) \bigg) \dd{x} \dd{y}
\end{aligned}$$
-But now, to eliminate these boundary terms, we need extra conditions for $\eta$:
+But now, to eliminate these boundary terms, we need extra conditions for $$\eta$$:
$$\begin{aligned}
\forall y: \eta(x_0, y) = \eta(x_1, y) = 0
@@ -216,15 +216,15 @@ $$\begin{aligned}
\forall x: \eta(x, y_0) = \eta(x, y_1) = 0
\end{aligned}$$
-In other words, the deviation $\eta$ must be zero on the whole "box".
-Again relying on the fact that $\eta$ is arbitrary, the Euler-Lagrange
+In other words, the deviation $$\eta$$ must be zero on the whole "box".
+Again relying on the fact that $$\eta$$ is arbitrary, the Euler-Lagrange
equation is:
$$\begin{aligned}
0 = \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big)
\end{aligned}$$
-This generalizes nicely to functions of even more variables $x_1, x_2, ..., x_N$:
+This generalizes nicely to functions of even more variables $$x_1, x_2, ..., x_N$$:
$$\begin{aligned}
\boxed{
@@ -235,14 +235,14 @@ $$\begin{aligned}
## Constraints
-So far, for multiple functions $f_1, ..., f_N$,
-we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$.
-Suppose that we now have $M < N$ constraints $\phi_m$
-that all $f_n$ need to obey, introducing implicit dependencies between them.
+So far, for multiple functions $$f_1, ..., f_N$$,
+we have been assuming that all $$f_n$$ are independent, and by extension all $$\eta_n$$.
+Suppose that we now have $$M < N$$ constraints $$\phi_m$$
+that all $$f_n$$ need to obey, introducing implicit dependencies between them.
-Let us consider constraints $\phi_m$ of the two forms below.
+Let us consider constraints $$\phi_m$$ of the two forms below.
It is important that they are **holonomic**,
-meaning they do not depend on any derivatives of any $f_n(x)$:
+meaning they do not depend on any derivatives of any $$f_n(x)$$:
$$\begin{aligned}
\phi_m(f_1, ..., f_N, x) = 0
@@ -250,14 +250,14 @@ $$\begin{aligned}
\int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m
\end{aligned}$$
-Where $C_m$ is a constant.
-Note that the first form can also be used for $\phi_m = C_m \neq 0$,
-by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.
+Where $$C_m$$ is a constant.
+Note that the first form can also be used for $$\phi_m = C_m \neq 0$$,
+by simply redefining the constraint as $$\phi_m^0 = \phi_m - C_m = 0$$.
-To solve this constrained optimization problem for $f_n(x)$,
-we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$.
-In the former case $\lambda_m(x)$ is a function of $x$, while in the
-latter case $\lambda_m$ is constant:
+To solve this constrained optimization problem for $$f_n(x)$$,
+we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $$\lambda_m$$.
+In the former case $$\lambda_m(x)$$ is a function of $$x$$, while in the
+latter case $$\lambda_m$$ is constant:
$$\begin{aligned}
\int \lambda_m(x) \: \phi_m(\{f_n\}, x) \dd{x} = 0
@@ -265,15 +265,15 @@ $$\begin{aligned}
\lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m
\end{aligned}$$
-The reason for this distinction in $\lambda_m$
-is that we need to find the stationary points with respect to $\varepsilon$
+The reason for this distinction in $$\lambda_m$$
+is that we need to find the stationary points with respect to $$\varepsilon$$
of both constraint types. Written in the variational form, this is:
$$\begin{aligned}
\delta \int \lambda_m \: \phi_m \dd{x} = 0
\end{aligned}$$
-From this, we define a new Lagrangian $\Lambda$ for the functional $J$,
+From this, we define a new Lagrangian $$\Lambda$$ for the functional $$J$$,
with the contraints built in:
$$\begin{aligned}
@@ -283,7 +283,7 @@ $$\begin{aligned}
&= \int L + \sum_{m} \lambda_m \phi_m \dd{x}
\end{aligned}$$
-Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$:
+Then we derive the Euler-Lagrange equation as usual for $$\Lambda$$ instead of $$L$$:
$$\begin{aligned}
0
@@ -297,15 +297,15 @@ $$\begin{aligned}
+ \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x}
\end{aligned}$$
-Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$:
+Using the same logic as before, we end up with a set of Euler-Lagrange equations with $$\Lambda$$:
$$\begin{aligned}
0
= \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big)
\end{aligned}$$
-By inserting the definition of $\Lambda$, we then get the following.
-Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$:
+By inserting the definition of $$\Lambda$$, we then get the following.
+Recall that $$\phi_m$$ is holonomic, and thus independent of all derivatives $$f_n'$$:
$$\begin{aligned}
\boxed{
@@ -316,18 +316,18 @@ $$\begin{aligned}
These are **Lagrange's equations of the first kind**,
with their second-kind counterparts being the earlier Euler-Lagrange equations.
-Note that there are $N$ separate equations, one for each $f_n$.
+Note that there are $$N$$ separate equations, one for each $$f_n$$.
-Due to the constraints $\phi_m$, the functions $f_n$ are not independent.
-This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold,
-i.e. solving a system of $M$ equations for $\lambda_m$:
+Due to the constraints $$\phi_m$$, the functions $$f_n$$ are not independent.
+This is solved by choosing $$\lambda_m$$ such that $$M$$ of the $$N$$ equations hold,
+i.e. solving a system of $$M$$ equations for $$\lambda_m$$:
$$\begin{aligned}
\dv{}{x}\Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n}
= \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
\end{aligned}$$
-And then the remaining $N - M$ equations can be solved in the normal unconstrained way.
+And then the remaining $$N - M$$ equations can be solved in the normal unconstrained way.
diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md
index dd7fe90..8a96e91 100644
--- a/source/know/concept/canonical-ensemble/index.md
+++ b/source/know/concept/canonical-ensemble/index.md
@@ -12,25 +12,25 @@ layout: "concept"
The **canonical ensemble** or **NVT ensemble** builds on
the [microcanonical ensemble](/know/concept/microcanonical-ensemble/),
by allowing the system to exchange energy with a very large heat bath,
-such that its temperature $T$ remains constant,
-but internal energy $U$ does not.
+such that its temperature $$T$$ remains constant,
+but internal energy $$U$$ does not.
The conserved state functions are
-the temperature $T$, the volume $V$, and the particle count $N$.
+the temperature $$T$$, the volume $$V$$, and the particle count $$N$$.
-We refer to the system of interest as $A$, and the heat bath as $B$.
-The combination $A\!+\!B$ forms a microcanonical ensemble,
-i.e. it has a fixed total energy $U$,
+We refer to the system of interest as $$A$$, and the heat bath as $$B$$.
+The combination $$A\!+\!B$$ forms a microcanonical ensemble,
+i.e. it has a fixed total energy $$U$$,
and eventually reaches an equilibrium
-with a uniform temperature $T$ in both $A$ and $B$.
+with a uniform temperature $$T$$ in both $$A$$ and $$B$$.
Assuming that this equilibrium has been reached,
-we want to know which microstates $A$ prefers in that case.
-Specifically, if $A$ has energy $U_A$, and $B$ has $U_B$,
-which $U_A$ does $A$ prefer?
+we want to know which microstates $$A$$ prefers in that case.
+Specifically, if $$A$$ has energy $$U_A$$, and $$B$$ has $$U_B$$,
+which $$U_A$$ does $$A$$ prefer?
-Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
-Then the probability that $A$ is in a specific microstate $s_A$ is as follows,
-where $U_A(s_A)$ is the resulting energy:
+Let $$c_B(U_B)$$ be the number of $$B$$-microstates with energy $$U_B$$.
+Then the probability that $$A$$ is in a specific microstate $$s_A$$ is as follows,
+where $$U_A(s_A)$$ is the resulting energy:
$$\begin{aligned}
p(s_A)
@@ -39,12 +39,12 @@ $$\begin{aligned}
D \equiv \sum_{s_A} c_B(U - U_A(s_A))
\end{aligned}$$
-In other words, we choose an $s_A$,
-and count the number $c_B$ of compatible $B$-microstates.
+In other words, we choose an $$s_A$$,
+and count the number $$c_B$$ of compatible $$B$$-microstates.
-Since the heat bath is large, let us assume that $U_B \gg U_A$.
-We thus approximate $\ln{p(s_A)}$ by
-Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$:
+Since the heat bath is large, let us assume that $$U_B \gg U_A$$.
+We thus approximate $$\ln{p(s_A)}$$ by
+Taylor-expanding $$\ln{c_B(U_B)}$$ around $$U_B = U$$:
$$\begin{aligned}
\ln{p(s_A)}
@@ -53,8 +53,8 @@ $$\begin{aligned}
&\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A)
\end{aligned}$$
-Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$,
-and that its $U_B$-derivative is $1/T$:
+Here, we use the definition of entropy $$S_B \equiv k \ln{c_B}$$,
+and that its $$U_B$$-derivative is $$1/T$$:
$$\begin{aligned}
\ln{p(s_A)}
@@ -63,7 +63,7 @@ $$\begin{aligned}
&\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T}
\end{aligned}$$
-We now define the **partition function** or **Zustandssumme** $Z$ as follows,
+We now define the **partition function** or **Zustandssumme** $$Z$$ as follows,
which will act as a normalization factor for the probability:
$$\begin{aligned}
@@ -74,8 +74,8 @@ $$\begin{aligned}
= \frac{D}{c_B(U)}
\end{aligned}$$
-Where $\beta \equiv 1/ (k T)$.
-The probability of finding $A$ in a microstate $s_A$ is thus given by:
+Where $$\beta \equiv 1/ (k T)$$.
+The probability of finding $$A$$ in a microstate $$s_A$$ is thus given by:
$$\begin{aligned}
\boxed{
@@ -84,33 +84,33 @@ $$\begin{aligned}
\end{aligned}$$
This is the **Boltzmann distribution**,
-which, as it turns out, maximizes the entropy $S_A$
-for a fixed value of the average energy $\Expval{U_A}$,
-i.e. a fixed $T$ and set of microstates $s_A$.
+which, as it turns out, maximizes the entropy $$S_A$$
+for a fixed value of the average energy $$\Expval{U_A}$$,
+i.e. a fixed $$T$$ and set of microstates $$s_A$$.
-Because $A\!+\!B$ is a microcanonical ensemble,
+Because $$A\!+\!B$$ is a microcanonical ensemble,
we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/)
-is the entropy $S$.
-But what about the canonical ensemble, just $A$?
+is the entropy $$S$$.
+But what about the canonical ensemble, just $$A$$?
The solution is a bit backwards.
-Note that the partition function $Z$ is not a constant;
-it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$).
+Note that the partition function $$Z$$ is not a constant;
+it depends on $$T$$ (via $$\beta$$), $$V$$ and $$N$$ (via $$s_A$$).
Using the same logic as for the microcanonical ensemble,
-we define "equilibrium" as the set of microstates $s_A$
-that $A$ is most likely to occupy,
-which must be the set (as a function of $T,V,N$) that maximizes $Z$.
+we define "equilibrium" as the set of microstates $$s_A$$
+that $$A$$ is most likely to occupy,
+which must be the set (as a function of $$T,V,N$$) that maximizes $$Z$$.
-However, $T$, $V$ and $N$ are fixed,
-so how can we maximize $Z$?
+However, $$T$$, $$V$$ and $$N$$ are fixed,
+so how can we maximize $$Z$$?
Well, as it turns out,
the Boltzmann distribution has already done it for us!
We will return to this point later.
-Still, $Z$ does not have a clear physical interpretation.
+Still, $$Z$$ does not have a clear physical interpretation.
To find one, we start by showing that the ensemble averages
-of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$
-can be calculated by differentiating $Z$.
+of the energy $$U_A$$, pressure $$P_A$$ and chemical potential $$\mu_A$$
+can be calculated by differentiating $$Z$$.
As preparation, note that:
$$\begin{aligned}
@@ -118,7 +118,7 @@ $$\begin{aligned}
\end{aligned}$$
With this, we can find the ensemble averages
-$\Expval{U_A}$, $\Expval{P_A}$ and $\Expval{\mu_A}$ of the system:
+$$\Expval{U_A}$$, $$\Expval{P_A}$$ and $$\Expval{\mu_A}$$ of the system:
$$\begin{aligned}
\Expval{U_A}
@@ -141,7 +141,7 @@ $$\begin{aligned}
= - \frac{1}{Z \beta} \pdv{Z}{N}
\end{aligned}$$
-It will turn out more convenient to use derivatives of $\ln{Z}$ instead,
+It will turn out more convenient to use derivatives of $$\ln{Z}$$ instead,
in which case:
$$\begin{aligned}
@@ -155,15 +155,15 @@ $$\begin{aligned}
= - \frac{1}{\beta} \pdv{\ln{Z}}{N}
\end{aligned}$$
-Now, to find a physical interpretation for $Z$.
-Consider the quantity $F$, in units of energy,
-whose minimum corresponds to a maximum of $Z$:
+Now, to find a physical interpretation for $$Z$$.
+Consider the quantity $$F$$, in units of energy,
+whose minimum corresponds to a maximum of $$Z$$:
$$\begin{aligned}
F \equiv - k T \ln{Z}
\end{aligned}$$
-We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element:
+We rearrange the equation to $$\beta F = - \ln{Z}$$ and take its differential element:
$$\begin{aligned}
\dd{(\beta F)}
@@ -198,7 +198,7 @@ $$\begin{aligned}
\end{aligned}$$
As was already suggested by our notation,
-$F$ turns out to be the **Helmholtz free energy**:
+$$F$$ turns out to be the **Helmholtz free energy**:
$$\begin{aligned}
\boxed{
@@ -209,7 +209,7 @@ $$\begin{aligned}
\end{aligned}$$
We can therefore reinterpret
-the partition function $Z$ and the Boltzmann distribution $p(s_A)$
+the partition function $$Z$$ and the Boltzmann distribution $$p(s_A)$$
in the following "more physical" way:
$$\begin{aligned}
@@ -220,17 +220,17 @@ $$\begin{aligned}
= \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big)
\end{aligned}$$
-Finally, by rearranging the expressions for $F$,
-we find the entropy $S_A$ to be:
+Finally, by rearranging the expressions for $$F$$,
+we find the entropy $$S_A$$ to be:
$$\begin{aligned}
S_A
= k \ln{Z} + \frac{\Expval{U_A}}{T}
\end{aligned}$$
-This is why $Z$ is already maximized:
-the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\Expval{U_A}$,
-leaving $Z$ as the only "variable".
+This is why $$Z$$ is already maximized:
+the Boltzmann distribution maximizes $$S_A$$ for fixed values of $$T$$ and $$\Expval{U_A}$$,
+leaving $$Z$$ as the only "variable".
diff --git a/source/know/concept/capillary-action/index.md b/source/know/concept/capillary-action/index.md
index 1ee3cd4..4b9e76c 100644
--- a/source/know/concept/capillary-action/index.md
+++ b/source/know/concept/capillary-action/index.md
@@ -16,7 +16,7 @@ It occurs when the [Laplace pressure](/know/concept/young-laplace-law/)
from surface tension is much larger in magnitude than the
[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity.
-Consider a spherical droplet of liquid with radius $R$.
+Consider a spherical droplet of liquid with radius $$R$$.
The hydrostatic pressure difference
between the top and bottom of the drop
is much smaller than the Laplace pressure:
@@ -25,17 +25,17 @@ $$\begin{aligned}
2 R \rho g \ll 2 \frac{\alpha}{R}
\end{aligned}$$
-Where $\rho$ is the density of the liquid,
-$g$ is the acceleration due to gravity,
-and $\alpha$ is the energy cost per unit surface area.
+Where $$\rho$$ is the density of the liquid,
+$$g$$ is the acceleration due to gravity,
+and $$\alpha$$ is the energy cost per unit surface area.
Rearranging the inequality yields:
$$\begin{aligned}
R^2 \ll \frac{\alpha}{\rho g}
\end{aligned}$$
-From the right-hand side we define the **capillary length** $L_c$,
-so gravity is negligible if $R \ll L_c$:
+From the right-hand side we define the **capillary length** $$L_c$$,
+so gravity is negligible if $$R \ll L_c$$:
$$\begin{aligned}
\boxed{
@@ -44,10 +44,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-In general, for a system with characteristic length $L$,
+In general, for a system with characteristic length $$L$$,
the relative strength of gravity compared to surface tension
-is described by the **Bond number** $\mathrm{Bo}$
-or **Eötvös number** $\mathrm{Eo}$:
+is described by the **Bond number** $$\mathrm{Bo}$$
+or **Eötvös number** $$\mathrm{Eo}$$:
$$\begin{aligned}
\boxed{
@@ -58,16 +58,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-The right-most side gives an alternative way of understanding $\mathrm{Bo}$:
-$m$ is the mass of a cube with side $L$, such that the numerator is the weight force,
+The right-most side gives an alternative way of understanding $$\mathrm{Bo}$$:
+$$m$$ is the mass of a cube with side $$L$$, such that the numerator is the weight force,
and the denominator is the tension force of the surface.
-In any case, capillary action can be observed when $\mathrm{Bo \ll 1}$.
+In any case, capillary action can be observed when $$\mathrm{Bo \ll 1}$$.
The most famous example of capillary action is **capillary rise**,
-where a liquid "climbs" upwards in a narrow vertical tube with radius $R$,
+where a liquid "climbs" upwards in a narrow vertical tube with radius $$R$$,
apparently defying gravity.
Assuming the liquid-air interface is a spherical cap
-with constant [curvature](/know/concept/curvature/) radius $R_c$,
+with constant [curvature](/know/concept/curvature/) radius $$R_c$$,
then we know that the liquid is at rest
when the hydrostatic pressure equals the Laplace pressure:
@@ -77,12 +77,12 @@ $$\begin{aligned}
= 2 \alpha \frac{\cos\theta}{R}
\end{aligned}$$
-Where $\theta$ is the liquid-tube contact angle,
-and we are neglecting variations of the height $h$ due to the curvature
+Where $$\theta$$ is the liquid-tube contact angle,
+and we are neglecting variations of the height $$h$$ due to the curvature
(i.e. the [meniscus](/know/concept/meniscus/)).
-By isolating the above equation for $h$,
+By isolating the above equation for $$h$$,
we arrive at **Jurin's law**,
-which predicts the height climbed by a liquid in a tube with radius $R$:
+which predicts the height climbed by a liquid in a tube with radius $$R$$:
$$\begin{aligned}
\boxed{
@@ -91,7 +91,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Depending on $\theta$, $h$ can be negative,
+Depending on $$\theta$$, $$h$$ can be negative,
i.e. the liquid might descend below the ambient level.
@@ -106,7 +106,7 @@ $$\begin{aligned}
\approx 2 \pi R (\alpha_{sg} - \alpha_{sl})
\end{aligned}$$
-Where $\alpha_{sg}$ and $\alpha_{sl}$ are the energy costs
+Where $$\alpha_{sg}$$ and $$\alpha_{sl}$$ are the energy costs
of the solid-gas and solid-liquid interfaces.
Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/),
we can rewrite this as follows:
@@ -116,7 +116,7 @@ $$\begin{aligned}
= 2 \alpha \cos\theta
\end{aligned}$$
-Isolating this for $h$ simply yields Jurin's law again, as expected.
+Isolating this for $$h$$ simply yields Jurin's law again, as expected.
diff --git a/source/know/concept/cauchy-principal-value/index.md b/source/know/concept/cauchy-principal-value/index.md
index a2582f2..f09611b 100644
--- a/source/know/concept/cauchy-principal-value/index.md
+++ b/source/know/concept/cauchy-principal-value/index.md
@@ -7,15 +7,15 @@ categories:
layout: "concept"
---
-The **Cauchy principal value** $\mathcal{P}$,
+The **Cauchy principal value** $$\mathcal{P}$$,
or just **principal value**,
is a method for integrating problematic functions,
i.e. functions with singularities,
whose integrals would otherwise diverge.
-Consider a function $f(x)$ with a singularity at some finite $x = b$,
+Consider a function $$f(x)$$ with a singularity at some finite $$x = b$$,
which is hampering attempts at integrating it.
-To resolve this, we define the Cauchy principal value $\mathcal{P}$ as follows:
+To resolve this, we define the Cauchy principal value $$\mathcal{P}$$ as follows:
$$\begin{aligned}
\boxed{
@@ -24,8 +24,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $f(x)$ instead has a singularity at postive infinity $+\infty$,
-then we define $\mathcal{P}$ as follows:
+If $$f(x)$$ instead has a singularity at postive infinity $$+\infty$$,
+then we define $$\mathcal{P}$$ as follows:
$$\begin{aligned}
\boxed{
@@ -34,10 +34,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-And analogously for $-\infty$.
-If $f(x)$ has singularities both at $+\infty$ and at $b$,
+And analogously for $$-\infty$$.
+If $$f(x)$$ has singularities both at $$+\infty$$ and at $$b$$,
then we simply combine the two previous cases,
-such that $\mathcal{P}$ is given by:
+such that $$\mathcal{P}$$ is given by:
$$\begin{aligned}
\mathcal{P} \int_{a}^\infty f(x) \:dx
@@ -49,5 +49,5 @@ And so on, until all problematic singularities have been dealt with.
In some situations, for example involving
the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/),
-the symbol $\mathcal{P}$ is written without an integral,
+the symbol $$\mathcal{P}$$ is written without an integral,
in which case the calculations are implicitly integrated.
diff --git a/source/know/concept/cauchy-strain-tensor/index.md b/source/know/concept/cauchy-strain-tensor/index.md
index a628820..bae7bb8 100644
--- a/source/know/concept/cauchy-strain-tensor/index.md
+++ b/source/know/concept/cauchy-strain-tensor/index.md
@@ -11,9 +11,9 @@ layout: "concept"
**Strain** quantifies the deformation of a solid object.
If the body has been deformed, e.g. by pulling or bending,
its constituent particles have moved a bit.
-Let $\va{X}$ be the original location of a particle,
-and $\va{x}$ its new location after the deformation.
-We can thus define the **displacement field** $\va{u}$:
+Let $$\va{X}$$ be the original location of a particle,
+and $$\va{x}$$ its new location after the deformation.
+We can thus define the **displacement field** $$\va{u}$$:
$$\begin{aligned}
\va{u}
@@ -21,7 +21,7 @@ $$\begin{aligned}
\end{aligned}$$
We restrict ourselves to **infinitesimal strain**,
-where $\va{u}$ is so tiny that the material's properties are unchanged,
+where $$\va{u}$$ is so tiny that the material's properties are unchanged,
and a **slowly-varying strain**,
where the particle's neighbourhood has been distorted,
but not completely changed.
@@ -31,15 +31,15 @@ is that we need to somehow exclude movements of the *entire* body:
for example, you can bend a twig in your hands while walking or dancing,
but we are only interested in the twig's shape change,
not in your movements.
-The above definition of $\vu{u}$ includes both,
+The above definition of $$\vu{u}$$ includes both,
so we should be careful how we extract the strain from it.
## Definition
We use the **Eulerian description** of deformation,
-where the new position $\va{x}$ is the reference,
-and the old position $\va{X}$ is expressed as a function of $\va{x}$:
+where the new position $$\va{x}$$ is the reference,
+and the old position $$\va{X}$$ is expressed as a function of $$\va{x}$$:
$$\begin{aligned}
\va{u}(\va{x})
@@ -47,10 +47,10 @@ $$\begin{aligned}
\end{aligned}$$
Let us choose two nearby points in the deformed solid,
-and call them $\va{x}$ and $\va{x} + \va{a}$,
-where $\va{a}$ is a tiny vector pointing from one to the other.
+and call them $$\va{x}$$ and $$\va{x} + \va{a}$$,
+where $$\va{a}$$ is a tiny vector pointing from one to the other.
Before the displacement, those points respectively had these positions,
-where we define $\va{A}$ as the "old" version of $\va{a}$:
+where we define $$\va{A}$$ as the "old" version of $$\va{a}$$:
$$\begin{aligned}
\va{X} = \va{X}(\va{x})
@@ -58,9 +58,9 @@ $$\begin{aligned}
\va{X} + \va{A} = \va{X}(\va{x} + \va{a})
\end{aligned}$$
-Because the new positions $\va{x}$ are our reference,
-we would like to write $\va{A}$ without $\va{X}$.
-To do so, we use the definition of $\va{u}(\va{x})$, yielding:
+Because the new positions $$\va{x}$$ are our reference,
+we would like to write $$\va{A}$$ without $$\va{X}$$.
+To do so, we use the definition of $$\va{u}(\va{x})$$, yielding:
$$\begin{aligned}
\va{A}
@@ -71,8 +71,8 @@ $$\begin{aligned}
&= \va{a} - \va{u}(\va{x} + \va{a}) - \va{u}(\va{x})
\end{aligned}$$
-Using the fact that $\va{a}$ is tiny by definition,
-we expand the middle term to first order in $\va{a}$:
+Using the fact that $$\va{a}$$ is tiny by definition,
+we expand the middle term to first order in $$\va{a}$$:
$$\begin{aligned}
\va{u}(\va{x} + \va{a})
@@ -80,8 +80,8 @@ $$\begin{aligned}
= \va{u}(\va{x}) + (\va{a} \cdot \nabla) \va{u}(\va{x})
\end{aligned}$$
-With this, we can now define the "shift" $\delta\va{a}$
-as the difference between $\va{a}$ and $\va{A}$ like so:
+With this, we can now define the "shift" $$\delta\va{a}$$
+as the difference between $$\va{a}$$ and $$\va{A}$$ like so:
$$\begin{aligned}
\delta{\va{a}}
@@ -90,24 +90,24 @@ $$\begin{aligned}
\end{aligned}$$
In index notation, we write this expression as follows,
-with $\nabla_j \equiv \ipdv{}{x_j}$ simply being the partial derivative
-with respect to the $j$th coordinate:
+with $$\nabla_j \equiv \ipdv{}{x_j}$$ simply being the partial derivative
+with respect to the $$j$$th coordinate:
$$\begin{aligned}
\delta a_i
= \sum_{j} a_j \nabla_j u_i
\end{aligned}$$
-Where $\nabla_j u_i$ are called the **displacement gradients**,
+Where $$\nabla_j u_i$$ are called the **displacement gradients**,
and are just one step away from the desired definition of strain.
Note that these gradients are dimensionless,
-so we can more formally define a *slowly-varying* displacement $\va{u}(\va{x})$
-as one where $|\nabla_j u_i| \ll 1$.
+so we can more formally define a *slowly-varying* displacement $$\va{u}(\va{x})$$
+as one where $$|\nabla_j u_i| \ll 1$$.
Now, to solve the problem of macroscopic movements,
-we take another tiny vector $\va{b}$ starting in the same point $\va{x}$ as $\va{a}$.
+we take another tiny vector $$\va{b}$$ starting in the same point $$\va{x}$$ as $$\va{a}$$.
Here is the trick: if the whole body is uniformly translated or rotated,
-the scalar product $\va{a} \cdot \va{b}$ is unchanged,
+the scalar product $$\va{a} \cdot \va{b}$$ is unchanged,
but if there is a non-uniform distortion, it changes.
We thus define the scalar product's difference like so:
@@ -116,7 +116,7 @@ $$\begin{aligned}
\equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B}
\end{aligned}$$
-Where $\va{B}$ is the old version of $\va{b}$.
+Where $$\va{B}$$ is the old version of $$\va{b}$$.
Since these vectors are all tiny, we apply the product rule:
$$\begin{aligned}
@@ -125,7 +125,7 @@ $$\begin{aligned}
\end{aligned}$$
It is more informative to switch to index notation here.
-Inserting $\delta\va{a}$ and $\delta\va{b}$ yields:
+Inserting $$\delta\va{a}$$ and $$\delta\va{b}$$ yields:
$$\begin{aligned}
\delta(\va{a} \cdot \va{b})
@@ -136,8 +136,8 @@ $$\begin{aligned}
&= \sum_{ij} \big( \nabla_i u_j + \nabla_j u_i \big) \: a_i b_j
\end{aligned}$$
-At last, we define the **Cauchy infinitesimal strain tensor** $\hat{u}$
-such that it has $u_{ij}$ as components:
+At last, we define the **Cauchy infinitesimal strain tensor** $$\hat{u}$$
+such that it has $$u_{ij}$$ as components:
$$\begin{aligned}
\boxed{
@@ -154,7 +154,7 @@ $$\begin{aligned}
= 2 \va{a} \cdot \hat{u} \cdot \va{b}
\end{aligned}$$
-The Cauchy strain tensor $\hat{u}$ is a second-rank tensor,
+The Cauchy strain tensor $$\hat{u}$$ is a second-rank tensor,
and can alternatively be expressed like so:
$$\begin{aligned}
@@ -164,17 +164,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\top$ is the transpose. Being defined from the scalar product,
+Where $$\top$$ is the transpose. Being defined from the scalar product,
all macroscopic movements of the body are removed from the tensor,
-which turns out to make it symmetric, i.e. $u_{ij} = u_{ji}$.
+which turns out to make it symmetric, i.e. $$u_{ij} = u_{ji}$$.
## Geometry
So far we have used Cartesian coordinates,
-but we can choose any three vectors $\va{a}$, $\va{b}$ and $\va{c}$,
-and **project** $\hat{u}$ onto this basis.
-For example, the component $u_{ab}$ then becomes:
+but we can choose any three vectors $$\va{a}$$, $$\va{b}$$ and $$\va{c}$$,
+and **project** $$\hat{u}$$ onto this basis.
+For example, the component $$u_{ab}$$ then becomes:
$$\begin{aligned}
\boxed{
@@ -184,12 +184,12 @@ $$\begin{aligned}
\end{aligned}$$
And so forth, for the other eight components.
-The basis in which $\hat{u}$ is diagonal is the one formed by its eigenvectors,
+The basis in which $$\hat{u}$$ is diagonal is the one formed by its eigenvectors,
and their directions are the **principal axes of strain**
at that point in the solid.
-Because $\hat{u}$ is symmetric, such a basis always exists.
+Because $$\hat{u}$$ is symmetric, such a basis always exists.
-Given a vector $\va{a}$, its relative length change
+Given a vector $$\va{a}$$, its relative length change
due to the deformation is simply given by:
$$\begin{aligned}
@@ -199,8 +199,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-To find the angle change $\delta\theta$
-between two vectors $\va{a}$ and $\va{b}$,
+To find the angle change $$\delta\theta$$
+between two vectors $$\va{a}$$ and $$\va{b}$$,
we start with the product rule:
$$\begin{aligned}
@@ -211,9 +211,9 @@ $$\begin{aligned}
- \big|\va{a}\big| \big|\va{b}\big| \sin\theta \: \delta\theta
\end{aligned}$$
-We isolate this for $\delta\theta$, using the fact that
-$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$
-thanks to the projection $u_{ab}$:
+We isolate this for $$\delta\theta$$, using the fact that
+$$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$$
+thanks to the projection $$u_{ab}$$:
$$\begin{aligned}
\delta\theta
@@ -223,7 +223,7 @@ $$\begin{aligned}
{\big|\va{a}\big| \big|\va{b}\big| \sin\theta}
\end{aligned}$$
-By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$,
+By recognizing the length change $$\delta|\va{a}|/|\va{a}| = u_{aa}$$,
we arrive at the following expression:
$$\begin{aligned}
@@ -234,8 +234,8 @@ $$\begin{aligned}
\end{aligned}$$
Now, everything so far has been about tiny vectors,
-so the change of the line element $\dd{\va{l}}$
-is easy to express using the displacement field $\va{u}$:
+so the change of the line element $$\dd{\va{l}}$$
+is easy to express using the displacement field $$\va{u}$$:
$$\begin{aligned}
\boxed{
@@ -244,9 +244,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Next, we calculate the change of the differential volume element $\dd{V}$
+Next, we calculate the change of the differential volume element $$\dd{V}$$
by treating it as the volume of a tiny parallelepiped
-spanned by $\va{a}$, $\va{b}$ and $\va{c}$:
+spanned by $$\va{a}$$, $$\va{b}$$ and $$\va{c}$$:
$$\begin{aligned}
\delta(\dd{V})
@@ -280,7 +280,7 @@ $$\begin{aligned}
&= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u})
\end{aligned}$$
-Here, we recognize the definition of $\dd{V}$,
+Here, we recognize the definition of $$\dd{V}$$,
leading to the following infinitesimal volume change:
$$\begin{aligned}
@@ -290,8 +290,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$,
-we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$:
+Finally, for the surface element $$\dd{\va{S}} = \va{a} \cross \va{b}$$,
+we use that the volume element $$\dd{V} = \va{c} \cdot \dd{\va{S}}$$:
$$\begin{aligned}
\delta(\dd{V})
@@ -300,7 +300,7 @@ $$\begin{aligned}
= (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
\end{aligned}$$
-By comparing this to the previous result for $\delta(\dd{V})$,
+By comparing this to the previous result for $$\delta(\dd{V})$$,
we arrive at the following equation:
$$\begin{aligned}
@@ -308,8 +308,8 @@ $$\begin{aligned}
= (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
\end{aligned}$$
-Since $\va{c}$ is dot-multiplied at the front of each term,
-we remove it, and isolate the rest for $\delta(\dd{\va{S}})$:
+Since $$\va{c}$$ is dot-multiplied at the front of each term,
+we remove it, and isolate the rest for $$\delta(\dd{\va{S}})$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/cauchy-stress-tensor/index.md b/source/know/concept/cauchy-stress-tensor/index.md
index 1d9ad0d..d83f430 100644
--- a/source/know/concept/cauchy-stress-tensor/index.md
+++ b/source/know/concept/cauchy-stress-tensor/index.md
@@ -22,10 +22,10 @@ but it is arguably most intuitive for solids.
## Definition
In the solid, imagine an infinitesimal cube
-whose sides, $\dd{S}_x$, $\dd{S}_y$ and $\dd{S}_z$,
-are orthogonal to the $x$, $y$ and $z$ axes, respectively.
-There is a force $\dd{\va{F}}_1$ acting on $\dd{S}_x$,
-$\dd{\va{F}}_2$ on $\dd{S}_y$, and $\dd{\va{F}}_3$ on $\dd{S}_z$.
+whose sides, $$\dd{S}_x$$, $$\dd{S}_y$$ and $$\dd{S}_z$$,
+are orthogonal to the $$x$$, $$y$$ and $$z$$ axes, respectively.
+There is a force $$\dd{\va{F}}_1$$ acting on $$\dd{S}_x$$,
+$$\dd{\va{F}}_2$$ on $$\dd{S}_y$$, and $$\dd{\va{F}}_3$$ on $$\dd{S}_z$$.
Then we can decompose each of these forces, for example:
$$\begin{aligned}
@@ -33,24 +33,24 @@ $$\begin{aligned}
= \va{e}_x F_{x1} + \va{e}_y F_{y1} + \va{e}_z F_{z1}
\end{aligned}$$
-Where $\va{e}_x$, $\va{e}_y$ and $\va{e}_z$ are the basis unit vectors.
-If we divide each of the force components by the area $\dd{S}_x$
+Where $$\va{e}_x$$, $$\va{e}_y$$ and $$\va{e}_z$$ are the basis unit vectors.
+If we divide each of the force components by the area $$\dd{S}_x$$
(like in a fluid, in order to get the pressure),
-we find the stresses $\sigma_{xx}$, $\sigma_{yx}$ and $\sigma_{zx}$
-that are being "felt" by the $x$ surface element $\dd{S}_x$:
+we find the stresses $$\sigma_{xx}$$, $$\sigma_{yx}$$ and $$\sigma_{zx}$$
+that are being "felt" by the $$x$$ surface element $$\dd{S}_x$$:
$$\begin{aligned}
\dd{\va{F}}_1
= \big( \va{e}_x \sigma_{xx} + \va{e}_y \sigma_{yx} + \va{e}_z \sigma_{zx} \big) \dd{S}_x
\end{aligned}$$
-The perpendicular component $\sigma_{xx}$ is called a **tensile stress**,
+The perpendicular component $$\sigma_{xx}$$ is called a **tensile stress**,
and its sign is always chosen so that a positive value corresponds to a tension,
-i.e. the $x$-side is pulled away from the rest of the cube.
-The tangential components $\sigma_{yx}$ and $\sigma_{zx}$
+i.e. the $$x$$-side is pulled away from the rest of the cube.
+The tangential components $$\sigma_{yx}$$ and $$\sigma_{zx}$$
are called **shear stresses**.
-Evidently, the other two forces $\dd{\va{F}}_2$ and $\dd{\va{F}}_3$
+Evidently, the other two forces $$\dd{\va{F}}_2$$ and $$\dd{\va{F}}_3$$
can be decomposed in the exact same way,
yielding nine stress components in total:
@@ -64,7 +64,7 @@ $$\begin{aligned}
= \big( \va{e}_x \sigma_{xz} + \va{e}_y \sigma_{yz} + \va{e}_z \sigma_{zz} \big) \dd{S}_z
\end{aligned}$$
-The total force $\dd{\va{F}}$ on the entire infinitesimal cube
+The total force $$\dd{\va{F}}$$ on the entire infinitesimal cube
is simply the sum of the previous three:
$$\begin{aligned}
@@ -72,8 +72,8 @@ $$\begin{aligned}
= \dd{\va{F}}_1 + \dd{\va{F}}_2 + \dd{\va{F}}_3
\end{aligned}$$
-We can then decompose $\dd{\va{F}}$ into its net components
-along the $x$, $y$ and $z$ axes:
+We can then decompose $$\dd{\va{F}}$$ into its net components
+along the $$x$$, $$y$$ and $$z$$ axes:
$$\begin{aligned}
\dd{\va{F}}
@@ -94,7 +94,7 @@ $$\begin{aligned}
\end{aligned}$$
We can write this much more compactly using index notation,
-where $i, j \in \{x, y, z\}$:
+where $$i, j \in \{x, y, z\}$$:
$$\begin{aligned}
\boxed{
@@ -103,9 +103,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-The stress components $\sigma_{ij}$ can be written as a second-rank tensor
+The stress components $$\sigma_{ij}$$ can be written as a second-rank tensor
(i.e. a matrix that transforms in a certain way),
-called the **Cauchy stress tensor** $\hat{\sigma}$:
+called the **Cauchy stress tensor** $$\hat{\sigma}$$:
$$\begin{aligned}
\boxed{
@@ -119,8 +119,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Then $\dd{\va{F}}$ is written even more compactly
-using the dot product, with $\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z)$:
+Then $$\dd{\va{F}}$$ is written even more compactly
+using the dot product, with $$\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z)$$:
$$\begin{aligned}
\boxed{
@@ -132,13 +132,13 @@ $$\begin{aligned}
All forces on the cube's sides can be written in this form.
**Cauchy's stress theorem** states that the force on *any*
surface element inside the solid can be written like this,
-simply by projecting it onto the $x$, $y$ and $z$ zero-planes
-to get the areas $\dd{S}_x$, $\dd{S}_y$ and $\dd{S}_z$.
+simply by projecting it onto the $$x$$, $$y$$ and $$z$$ zero-planes
+to get the areas $$\dd{S}_x$$, $$\dd{S}_y$$ and $$\dd{S}_z$$.
-Note that for fluids, the pressure $p$ was defined
-such that $\dd{\va{F}} = - p \dd{\va{S}}$.
-If we wanted to define $p$ for solids in the same way,
-we would need $\hat{\sigma}$ to be diagonal *and*
+Note that for fluids, the pressure $$p$$ was defined
+such that $$\dd{\va{F}} = - p \dd{\va{S}}$$.
+If we wanted to define $$p$$ for solids in the same way,
+we would need $$\hat{\sigma}$$ to be diagonal *and*
all of its diagonal elements to be identical.
Since this is almost never the case,
the scalar pressure is ill-defined in solids.
@@ -146,9 +146,9 @@ the scalar pressure is ill-defined in solids.
## Equilibrium
-The total force $\va{F}$ acting on a (non-infinitesimal) volume $V$ of the solid
-is given by the sum of the total body force $\va{F}_b$ and total surface force $\va{F}_s$,
-where $\vec{f}$ is the body force density:
+The total force $$\va{F}$$ acting on a (non-infinitesimal) volume $$V$$ of the solid
+is given by the sum of the total body force $$\va{F}_b$$ and total surface force $$\va{F}_s$$,
+where $$\vec{f}$$ is the body force density:
$$\begin{aligned}
\va{F}
@@ -157,7 +157,7 @@ $$\begin{aligned}
\end{aligned}$$
We can rewrite the surface term using the divergence theorem,
-where $\top$ is the transpose:
+where $$\top$$ is the transpose:
$$\begin{aligned}
\va{F}_s
@@ -166,7 +166,7 @@ $$\begin{aligned}
\end{aligned}$$
For some people, this equation may be more enlightening in index notation,
-where $\nabla_j \equiv \ipdv{}{x_j}$ is the partial derivative with respect to the $j$th coordinate:
+where $$\nabla_j \equiv \ipdv{}{x_j}$$ is the partial derivative with respect to the $$j$$th coordinate:
$$\begin{aligned}
F_{s, i}
@@ -174,8 +174,8 @@ $$\begin{aligned}
= \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V}
\end{aligned}$$
-In any case, the total force $\va{F}$ can then be expressed
-as a single volume integral over $V$:
+In any case, the total force $$\va{F}$$ can then be expressed
+as a single volume integral over $$V$$:
$$\begin{aligned}
\va{F}
@@ -183,7 +183,7 @@ $$\begin{aligned}
= \int_V \va{f^*} \dd{V}
\end{aligned}$$
-Where we have defined the **effective force density** $\va{f^*}$ as follows:
+Where we have defined the **effective force density** $$\va{f^*}$$ as follows:
$$\begin{aligned}
\boxed{
@@ -192,14 +192,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-The volume $V$ is in **mechanical equilibrium** if the net force acting on it amounts to zero:
+The volume $$V$$ is in **mechanical equilibrium** if the net force acting on it amounts to zero:
$$\begin{aligned}
\va{F}
= 0
\end{aligned}$$
-However, because $V$ is abritrary, the equilibrium condition for the whole solid is in fact:
+However, because $$V$$ is abritrary, the equilibrium condition for the whole solid is in fact:
$$\begin{aligned}
\boxed{
@@ -215,7 +215,7 @@ which needs boundary conditions at the object's surface.
Newton's third law states that the two sides of the boundary
exert opposite forces on each other,
so the boundary condition is continuity of the **stress vector**
-$\hat{\sigma} \cdot \va{n}$:
+$$\hat{\sigma} \cdot \va{n}$$:
$$\begin{aligned}
\boxed{
@@ -224,10 +224,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the normal of the outer surface is $\va{n}$,
-and the normal of the inner surface is $-\va{n}$.
+Where the normal of the outer surface is $$\va{n}$$,
+and the normal of the inner surface is $$-\va{n}$$.
Note that the above equation does *not* mean
-that $-\hat{\sigma}_{\mathrm{inner}}$ equals $\hat{\sigma}_{\mathrm{outer}}$:
+that $$-\hat{\sigma}_{\mathrm{inner}}$$ equals $$\hat{\sigma}_{\mathrm{outer}}$$:
the tensors are allowed to be very different,
as long as the stress vector's three components are equal.
diff --git a/source/know/concept/cavitation/index.md b/source/know/concept/cavitation/index.md
index 346edc4..679e66f 100644
--- a/source/know/concept/cavitation/index.md
+++ b/source/know/concept/cavitation/index.md
@@ -20,11 +20,11 @@ the [Rayleigh-Plesset equation](/know/concept/rayleigh-plesset-equation/)
for an inviscid liquid without surface tension.
Note that the RP equation assumes incompressibility.
-We assume that the whole liquid is at a constant pressure $p_\infty$,
-and the bubble is empty, such that the interface pressure $P = 0$,
-meaning $\Delta p = - p_\infty$.
-At first, the radius is stationary $R'(0) = 0$,
-and given by a constant $R(0) = a$.
+We assume that the whole liquid is at a constant pressure $$p_\infty$$,
+and the bubble is empty, such that the interface pressure $$P = 0$$,
+meaning $$\Delta p = - p_\infty$$.
+At first, the radius is stationary $$R'(0) = 0$$,
+and given by a constant $$R(0) = a$$.
The simple Rayleigh-Plesset equation is then:
$$\begin{aligned}
@@ -32,7 +32,7 @@ $$\begin{aligned}
= - \frac{p_\infty}{\rho}
\end{aligned}$$
-To solve it, we multiply both sides by $R^2 R'$
+To solve it, we multiply both sides by $$R^2 R'$$
and rewrite it in the following way:
$$\begin{aligned}
@@ -44,7 +44,7 @@ $$\begin{aligned}
\end{aligned}$$
It is then straightforward to integrate both sides
-with respect to time $\tau$, from $0$ to $t$:
+with respect to time $$\tau$$, from $$0$$ to $$t$$:
$$\begin{aligned}
- \frac{2 p_\infty}{3 \rho} \int_0^t \dv{}{\tau}\Big( R^3 \Big) \dd{\tau}
@@ -58,7 +58,7 @@ $$\begin{aligned}
\end{aligned}$$
Rearranging this equation yields the following expression
-for the derivative $R'$:
+for the derivative $$R'$$:
$$\begin{aligned}
(R')^2
@@ -66,10 +66,10 @@ $$\begin{aligned}
\end{aligned}$$
This equation is nasty to integrate.
-The trick is to invert $R(t)$ into $t(R)$,
+The trick is to invert $$R(t)$$ into $$t(R)$$,
and, because we are only interested in collapse,
-we just need to consider the case $R' < 0$.
-The time of a given radius $R$ is then as follows,
+we just need to consider the case $$R' < 0$$.
+The time of a given radius $$R$$ is then as follows,
where we are using slightly sloppy notation:
$$\begin{aligned}
@@ -79,8 +79,8 @@ $$\begin{aligned}
= \int_{R}^{a} \frac{\dd{R}}{R'}
\end{aligned}$$
-The minus comes from the constraint that $R' < 0$, but $t \ge 0$.
-We insert the expression for $R'$:
+The minus comes from the constraint that $$R' < 0$$, but $$t \ge 0$$.
+We insert the expression for $$R'$$:
$$\begin{aligned}
t
@@ -89,9 +89,9 @@ $$\begin{aligned}
\end{aligned}$$
This integral needs to be looked up,
-and involves the hypergeometric function ${}_2 F_1$.
-However, we only care about *collapse*, which is when $R = 0$.
-The time $t_0$ at which this occurs is:
+and involves the hypergeometric function $${}_2 F_1$$.
+However, we only care about *collapse*, which is when $$R = 0$$.
+The time $$t_0$$ at which this occurs is:
$$\begin{aligned}
t_0
@@ -101,9 +101,9 @@ $$\begin{aligned}
With our assumptions, a bubble will always collapse.
However, unsurprisingly, reality turns out to be more complicated:
-as $R \to 0$, the interface velocity $R' \to \infty$.
+as $$R \to 0$$, the interface velocity $$R' \to \infty$$.
By looking at the derivation of the Rayleigh-Plesset equation,
-it can be shown that the pressure just outside the bubble diverges due to $R'$.
+it can be shown that the pressure just outside the bubble diverges due to $$R'$$.
This drastically changes the liquid's properties, and breaks our assumptions.
diff --git a/source/know/concept/central-limit-theorem/index.md b/source/know/concept/central-limit-theorem/index.md
index 0c08a6f..595cee7 100644
--- a/source/know/concept/central-limit-theorem/index.md
+++ b/source/know/concept/central-limit-theorem/index.md
@@ -10,16 +10,16 @@ layout: "concept"
In statistics, the **central limit theorem** states that
the sum of many independent variables tends towards a normal distribution,
-even if the individual variables $x_n$ follow different distributions.
+even if the individual variables $$x_n$$ follow different distributions.
-For example, by taking $M$ samples of size $N$ from a population,
-and calculating $M$ averages $\mu_m$ (which involves summing over $N$),
-the resulting means $\mu_m$ are normally distributed
-across the $M$ samples if $N$ is sufficiently large.
+For example, by taking $$M$$ samples of size $$N$$ from a population,
+and calculating $$M$$ averages $$\mu_m$$ (which involves summing over $$N$$),
+the resulting means $$\mu_m$$ are normally distributed
+across the $$M$$ samples if $$N$$ is sufficiently large.
-More formally, for $N$ independent variables $x_n$ with probability distributions $p(x_n)$,
+More formally, for $$N$$ independent variables $$x_n$$ with probability distributions $$p(x_n)$$,
the central limit theorem states the following,
-where we define the sum $S$:
+where we define the sum $$S$$:
$$\begin{aligned}
S = \sum_{n = 1}^N x_n
@@ -29,8 +29,8 @@ $$\begin{aligned}
\sigma_S^2 = \sum_{n = 1}^N \sigma_n^2
\end{aligned}$$
-And crucially, it states that the probability distribution $p_N(S)$ of $S$ for $N$ variables
-will become a normal distribution when $N$ goes to infinity:
+And crucially, it states that the probability distribution $$p_N(S)$$ of $$S$$ for $$N$$ variables
+will become a normal distribution when $$N$$ goes to infinity:
$$\begin{aligned}
\boxed{
@@ -41,14 +41,14 @@ $$\begin{aligned}
We prove this below,
but first we need to introduce some tools.
-Given a probability density $p(x)$, its [Fourier transform](/know/concept/fourier-transform/)
-is called the **characteristic function** $\phi(k)$:
+Given a probability density $$p(x)$$, its [Fourier transform](/know/concept/fourier-transform/)
+is called the **characteristic function** $$\phi(k)$$:
$$\begin{aligned}
\phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x}
\end{aligned}$$
-Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$.
+Note that $$\phi(k)$$ can be interpreted as the average of $$\exp(i k x)$$.
We take its Taylor expansion in two separate ways,
where an overline denotes the mean:
@@ -67,7 +67,7 @@ $$\begin{aligned}
\phi^{(n)}(0) = i^n \: \overline{x^n}
\end{aligned}$$
-Next, the **cumulants** $C^{(n)}$ are defined from the Taylor expansion of $\ln\!\big(\phi(k)\big)$:
+Next, the **cumulants** $$C^{(n)}$$ are defined from the Taylor expansion of $$\ln\!\big(\phi(k)\big)$$:
$$\begin{aligned}
\ln\!\big( \phi(k) \big)
@@ -76,7 +76,7 @@ $$\begin{aligned}
C^{(n)} = \frac{1}{i^n} \: \dvn{n}{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0}
\end{aligned}$$
-The first two cumulants $C^{(1)}$ and $C^{(2)}$ are of particular interest,
+The first two cumulants $$C^{(1)}$$ and $$C^{(2)}$$ are of particular interest,
since they turn out to be the mean and the variance respectively,
using our earlier relation:
@@ -92,14 +92,14 @@ $$\begin{aligned}
= - \overline{x}^2 + \overline{x^2} = \sigma^2
\end{aligned}$$
-Let us now define $S$ as the sum of $N$ independent variables $x_n$, in other words:
+Let us now define $$S$$ as the sum of $$N$$ independent variables $$x_n$$, in other words:
$$\begin{aligned}
S = \sum_{n = 1}^N x_n = x_1 + x_2 + ... + x_N
\end{aligned}$$
-The probability density of $S$ is then as follows, where $p(x_n)$ are
-the densities of all the individual variables and $\delta$ is
+The probability density of $$S$$ is then as follows, where $$p(x_n)$$ are
+the densities of all the individual variables and $$\delta$$ is
the [Dirac delta function](/know/concept/dirac-delta-function/):
$$\begin{aligned}
@@ -109,9 +109,9 @@ $$\begin{aligned}
&= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(S)
\end{aligned}$$
-In other words, the integrals pick out all combinations of $x_n$ which
-add up to the desired $S$-value, and multiply the probabilities
-$p(x_1) p(x_2) \cdots p(x_N)$ of each such case. This is a convolution,
+In other words, the integrals pick out all combinations of $$x_n$$ which
+add up to the desired $$S$$-value, and multiply the probabilities
+$$p(x_1) p(x_2) \cdots p(x_N)$$ of each such case. This is a convolution,
so the [convolution theorem](/know/concept/convolution-theorem/)
states that it is a product in the Fourier domain:
@@ -128,22 +128,22 @@ $$\begin{aligned}
= \sum_{n = 1}^N \sum_{m = 1}^{\infty} \frac{(ik)^m}{m!} C_n^{(m)}
\end{aligned}$$
-Consequently, the cumulants $C^{(m)}$ stack additively for the sum $S$
-of independent variables $x_m$, and therefore
-the means $C^{(1)}$ and variances $C^{(2)}$ do too:
+Consequently, the cumulants $$C^{(m)}$$ stack additively for the sum $$S$$
+of independent variables $$x_m$$, and therefore
+the means $$C^{(1)}$$ and variances $$C^{(2)}$$ do too:
$$\begin{aligned}
C_S^{(m)} = \sum_{n = 1}^N C_n^{(m)} = C_1^{(m)} + C_2^{(m)} + ... + C_N^{(m)}
\end{aligned}$$
-We now introduce the scaled sum $z$ as the new combined variable:
+We now introduce the scaled sum $$z$$ as the new combined variable:
$$\begin{aligned}
z = \frac{S}{\sqrt{N}} = \frac{1}{\sqrt{N}} (x_1 + x_2 + ... + x_N)
\end{aligned}$$
-Its characteristic function $\phi_z(k)$ is then as follows,
-with $\sqrt{N}$ appearing in the arguments of $\phi_n$:
+Its characteristic function $$\phi_z(k)$$ is then as follows,
+with $$\sqrt{N}$$ appearing in the arguments of $$\phi_n$$:
$$\begin{aligned}
\phi_z(k)
@@ -158,9 +158,9 @@ $$\begin{aligned}
&= \prod_{n = 1}^N \phi_n\Big(\frac{k}{\sqrt{N}}\Big)
\end{aligned}$$
-By expanding $\ln\!\big(\phi_z(k)\big)$ in terms of its cumulants $C^{(m)}$
-and introducing $\kappa = k / \sqrt{N}$, we see that the higher-order terms
-become smaller for larger $N$:
+By expanding $$\ln\!\big(\phi_z(k)\big)$$ in terms of its cumulants $$C^{(m)}$$
+and introducing $$\kappa = k / \sqrt{N}$$, we see that the higher-order terms
+become smaller for larger $$N$$:
$$\begin{gathered}
\ln\!\big( \phi_z(k) \big)
@@ -171,7 +171,7 @@ $$\begin{gathered}
= \frac{1}{i^m N^{m/2}} \dvn{m}{}{\kappa} \sum_{n = 1}^N \ln\!\big( \phi_n(\kappa) \big)
\end{gathered}$$
-For sufficiently large $N$, we can therefore approximate it using just the first two terms:
+For sufficiently large $$N$$, we can therefore approximate it using just the first two terms:
$$\begin{aligned}
\ln\!\big( \phi_z(k) \big)
@@ -182,7 +182,7 @@ $$\begin{aligned}
&\approx \exp(i k \overline{z}) \exp(- k^2 \sigma_z^2 / 2)
\end{aligned}$$
-We take its inverse Fourier transform to get the density $p(z)$,
+We take its inverse Fourier transform to get the density $$p(z)$$,
which turns out to be a Gaussian normal distribution,
which is even already normalized:
diff --git a/source/know/concept/conditional-expectation/index.md b/source/know/concept/conditional-expectation/index.md
index 7b13a4a..f64fa72 100644
--- a/source/know/concept/conditional-expectation/index.md
+++ b/source/know/concept/conditional-expectation/index.md
@@ -10,24 +10,24 @@ categories:
layout: "concept"
---
-Recall that the expectation value $\mathbf{E}[X]$
-of a [random variable](/know/concept/random-variable/) $X$
-is a function of the probability space $(\Omega, \mathcal{F}, P)$
-on which $X$ is defined, and the definition of $X$ itself.
-
-The **conditional expectation** $\mathbf{E}[X|A]$
-is the expectation value of $X$ given that an event $A$ has occurred,
-i.e. only the outcomes $\omega \in \Omega$
-satisfying $\omega \in A$ should be considered.
-If $A$ is obtained by observing a variable,
-then $\mathbf{E}[X|A]$ is a random variable in its own right.
-
-Consider two random variables $X$ and $Y$
-on the same probability space $(\Omega, \mathcal{F}, P)$,
-and suppose that $\Omega$ is discrete.
-If $Y = y$ has been observed,
-then the conditional expectation of $X$
-given the event $Y = y$ is as follows:
+Recall that the expectation value $$\mathbf{E}[X]$$
+of a [random variable](/know/concept/random-variable/) $$X$$
+is a function of the probability space $$(\Omega, \mathcal{F}, P)$$
+on which $$X$$ is defined, and the definition of $$X$$ itself.
+
+The **conditional expectation** $$\mathbf{E}[X|A]$$
+is the expectation value of $$X$$ given that an event $$A$$ has occurred,
+i.e. only the outcomes $$\omega \in \Omega$$
+satisfying $$\omega \in A$$ should be considered.
+If $$A$$ is obtained by observing a variable,
+then $$\mathbf{E}[X|A]$$ is a random variable in its own right.
+
+Consider two random variables $$X$$ and $$Y$$
+on the same probability space $$(\Omega, \mathcal{F}, P)$$,
+and suppose that $$\Omega$$ is discrete.
+If $$Y = y$$ has been observed,
+then the conditional expectation of $$X$$
+given the event $$Y = y$$ is as follows:
$$\begin{aligned}
\mathbf{E}[X | Y \!=\! y]
@@ -37,12 +37,12 @@ $$\begin{aligned}
= \frac{P(X \!=\! x \cap Y \!=\! y)}{P(Y \!=\! y)}
\end{aligned}$$
-Where $Q$ is a renormalized probability function,
-which assigns zero to all events incompatible with $Y = y$.
-If we allow $\Omega$ to be continuous,
-then from the definition $\mathbf{E}[X]$,
+Where $$Q$$ is a renormalized probability function,
+which assigns zero to all events incompatible with $$Y = y$$.
+If we allow $$\Omega$$ to be continuous,
+then from the definition $$\mathbf{E}[X]$$,
we know that the following Lebesgue integral can be used,
-which we call $f(y)$:
+which we call $$f(y)$$:
$$\begin{aligned}
\mathbf{E}[X | Y \!=\! y]
@@ -50,9 +50,9 @@ $$\begin{aligned}
= \int_\Omega X(\omega) \dd{Q(\omega)}
\end{aligned}$$
-However, this is only valid if $P(Y \!=\! y) > 0$,
-which is a problem for continuous sample spaces $\Omega$.
-Sticking with the assumption $P(Y \!=\! y) > 0$, notice that:
+However, this is only valid if $$P(Y \!=\! y) > 0$$,
+which is a problem for continuous sample spaces $$\Omega$$.
+Sticking with the assumption $$P(Y \!=\! y) > 0$$, notice that:
$$\begin{aligned}
f(y)
@@ -60,9 +60,9 @@ $$\begin{aligned}
= \frac{\mathbf{E}[X \cdot I(Y \!=\! y)]}{P(Y \!=\! y)}
\end{aligned}$$
-Where $I$ is the indicator function,
-equal to $1$ if its argument is true, and $0$ if not.
-Multiplying the definition of $f(y)$ by $P(Y \!=\! y)$ then leads us to:
+Where $$I$$ is the indicator function,
+equal to $$1$$ if its argument is true, and $$0$$ if not.
+Multiplying the definition of $$f(y)$$ by $$P(Y \!=\! y)$$ then leads us to:
$$\begin{aligned}
\mathbf{E}[X \cdot I(Y \!=\! y)]
@@ -71,22 +71,22 @@ $$\begin{aligned}
&= \mathbf{E}[f(Y) \cdot I(Y \!=\! y)]
\end{aligned}$$
-Recall that because $Y$ is a random variable,
-$\mathbf{E}[X|Y] = f(Y)$ is too.
-In other words, $f$ maps $Y$ to another random variable,
+Recall that because $$Y$$ is a random variable,
+$$\mathbf{E}[X|Y] = f(Y)$$ is too.
+In other words, $$f$$ maps $$Y$$ to another random variable,
which, thanks to the *Doob-Dynkin lemma*
(see [random variable](/know/concept/random-variable/)),
-means that $\mathbf{E}[X|Y]$ is measurable with respect to $\sigma(Y)$.
+means that $$\mathbf{E}[X|Y]$$ is measurable with respect to $$\sigma(Y)$$.
Intuitively, this makes sense:
-$\mathbf{E}[X|Y]$ cannot contain more information about events
-than the $Y$ it was calculated from.
+$$\mathbf{E}[X|Y]$$ cannot contain more information about events
+than the $$Y$$ it was calculated from.
This suggests a straightforward generalization of the above:
-instead of a specific value $Y = y$,
-we can condition on *any* information from $Y$.
-If $\mathcal{H} = \sigma(Y)$ is the information generated by $Y$,
-then the conditional expectation $\mathbf{E}[X|\mathcal{H}] = Z$
-is $\mathcal{H}$-measurable, and given by a $Z$ satisfying:
+instead of a specific value $$Y = y$$,
+we can condition on *any* information from $$Y$$.
+If $$\mathcal{H} = \sigma(Y)$$ is the information generated by $$Y$$,
+then the conditional expectation $$\mathbf{E}[X|\mathcal{H}] = Z$$
+is $$\mathcal{H}$$-measurable, and given by a $$Z$$ satisfying:
$$\begin{aligned}
\boxed{
@@ -95,51 +95,51 @@ $$\begin{aligned}
}
\end{aligned}$$
-For any $H \in \mathcal{H}$. Note that $Z$ is almost surely unique:
+For any $$H \in \mathcal{H}$$. Note that $$Z$$ is almost surely unique:
*almost* because it could take any value
-for an event $A$ with zero probability $P(A) = 0$.
-Fortunately, if there exists a continuous $f$
-such that $\mathbf{E}[X | \sigma(Y)] = f(Y)$,
-then $Z = \mathbf{E}[X | \sigma(Y)]$ is unique.
+for an event $$A$$ with zero probability $$P(A) = 0$$.
+Fortunately, if there exists a continuous $$f$$
+such that $$\mathbf{E}[X | \sigma(Y)] = f(Y)$$,
+then $$Z = \mathbf{E}[X | \sigma(Y)]$$ is unique.
## Properties
A conditional expectation defined in this way has many useful properties,
most notably linearity:
-$\mathbf{E}[aX \!+\! bY | \mathcal{H}] = a \mathbf{E}[X|\mathcal{H}] + b \mathbf{E}[Y|\mathcal{H}]$
-for any $a, b \in \mathbb{R}$.
+$$\mathbf{E}[aX \!+\! bY | \mathcal{H}] = a \mathbf{E}[X|\mathcal{H}] + b \mathbf{E}[Y|\mathcal{H}]$$
+for any $$a, b \in \mathbb{R}$$.
-The **tower property** states that if $\mathcal{F} \supset \mathcal{G} \supset \mathcal{H}$,
-then $\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}] = \mathbf{E}[X|\mathcal{H}]$.
+The **tower property** states that if $$\mathcal{F} \supset \mathcal{G} \supset \mathcal{H}$$,
+then $$\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}] = \mathbf{E}[X|\mathcal{H}]$$.
Intuitively, this works as follows:
-suppose person $G$ knows more about $X$ than person $H$,
-then $\mathbf{E}[X | \mathcal{H}]$ is $H$'s expectation,
-$\mathbf{E}[X | \mathcal{G}]$ is $G$'s "better" expectation,
-and then $\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}]$
-is $H$'s prediction about what $G$'s expectation will be.
-However, $H$ does not have access to $G$'s extra information,
-so $H$'s best prediction is simply $\mathbf{E}[X | \mathcal{H}]$.
+suppose person $$G$$ knows more about $$X$$ than person $$H$$,
+then $$\mathbf{E}[X | \mathcal{H}]$$ is $$H$$'s expectation,
+$$\mathbf{E}[X | \mathcal{G}]$$ is $$G$$'s "better" expectation,
+and then $$\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}]$$
+is $$H$$'s prediction about what $$G$$'s expectation will be.
+However, $$H$$ does not have access to $$G$$'s extra information,
+so $$H$$'s best prediction is simply $$\mathbf{E}[X | \mathcal{H}]$$.
The **law of total expectation** says that
-$\mathbf{E}[\mathbf{E}[X | \mathcal{G}]] = \mathbf{E}[X]$,
+$$\mathbf{E}[\mathbf{E}[X | \mathcal{G}]] = \mathbf{E}[X]$$,
and follows from the above tower property
-by choosing $\mathcal{H}$ to contain no information:
-$\mathcal{H} = \{ \varnothing, \Omega \}$.
-
-Another useful property is that $\mathbf{E}[X | \mathcal{H}] = X$
-if $X$ is $\mathcal{H}$-measurable.
-In other words, if $\mathcal{H}$ already contains
-all the information extractable from $X$,
-then we know $X$'s exact value.
+by choosing $$\mathcal{H}$$ to contain no information:
+$$\mathcal{H} = \{ \varnothing, \Omega \}$$.
+
+Another useful property is that $$\mathbf{E}[X | \mathcal{H}] = X$$
+if $$X$$ is $$\mathcal{H}$$-measurable.
+In other words, if $$\mathcal{H}$$ already contains
+all the information extractable from $$X$$,
+then we know $$X$$'s exact value.
Conveniently, this can easily be generalized to products:
-$\mathbf{E}[XY | \mathcal{H}] = X \mathbf{E}[Y | \mathcal{H}]$
-if $X$ is $\mathcal{H}$-measurable:
-since $X$'s value is known, it can simply be factored out.
+$$\mathbf{E}[XY | \mathcal{H}] = X \mathbf{E}[Y | \mathcal{H}]$$
+if $$X$$ is $$\mathcal{H}$$-measurable:
+since $$X$$'s value is known, it can simply be factored out.
Armed with this definition of conditional expectation,
we can define other conditional quantities,
-such as the **conditional variance** $\mathbf{V}[X | \mathcal{H}]$:
+such as the **conditional variance** $$\mathbf{V}[X | \mathcal{H}]$$:
$$\begin{aligned}
\mathbf{V}[X | \mathcal{H}]
@@ -147,11 +147,11 @@ $$\begin{aligned}
\end{aligned}$$
The **law of total variance** then states that
-$\mathbf{V}[X] = \mathbf{E}[\mathbf{V}[X | \mathcal{H}]] + \mathbf{V}[\mathbf{E}[X | \mathcal{H}]]$.
+$$\mathbf{V}[X] = \mathbf{E}[\mathbf{V}[X | \mathcal{H}]] + \mathbf{V}[\mathbf{E}[X | \mathcal{H}]]$$.
-Likewise, we can define the **conditional probability** $P$,
-**conditional distribution function** $F_{X|\mathcal{H}}$,
-and **conditional density function** $f_{X|\mathcal{H}}$
+Likewise, we can define the **conditional probability** $$P$$,
+**conditional distribution function** $$F_{X|\mathcal{H}}$$,
+and **conditional density function** $$f_{X|\mathcal{H}}$$
like their non-conditional counterparts:
$$\begin{aligned}
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index d4655cf..742c8ff 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -9,14 +9,14 @@ layout: "concept"
The **convolution theorem** states that a convolution in the direct domain
is equal to a product in the frequency domain. This is especially useful
-for computation, replacing an $\mathcal{O}(n^2)$ convolution with an
-$\mathcal{O}(n \log(n))$ transform and product.
+for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an
+$$\mathcal{O}(n \log(n))$$ transform and product.
## Fourier transform
The convolution theorem is usually expressed as follows, where
-$\hat{\mathcal{F}}$ is the [Fourier transform](/know/concept/fourier-transform/),
-and $A$ and $B$ are constants from its definition:
+$$\hat{\mathcal{F}}$$ is the [Fourier transform](/know/concept/fourier-transform/),
+and $$A$$ and $$B$$ are constants from its definition:
$$\begin{aligned}
\boxed{
@@ -46,7 +46,7 @@ $$\begin{aligned}
\end{aligned}$$
Then we do the same again,
-this time starting from a product in the $x$-domain:
+this time starting from a product in the $$x$$-domain:
$$\begin{aligned}
\hat{\mathcal{F}}\{f(x) \: g(x)\}
@@ -63,7 +63,7 @@ $$\begin{aligned}
## Laplace transform
-For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$,
+For functions $$f(t)$$ and $$g(t)$$ which are only defined for $$t \ge 0$$,
the convolution theorem can also be stated using
the [Laplace transform](/know/concept/laplace-transform/):
@@ -71,7 +71,7 @@ $$\begin{aligned}
\boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
\end{aligned}$$
-Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is
+Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is
unpleasant, the theorem is often stated using the forward transform
instead:
@@ -85,8 +85,8 @@ $$\begin{aligned}
We expand the left-hand side.
-Note that the lower integration limit is 0 instead of $-\infty$,
-because we set both $f(t)$ and $g(t)$ to zero for $t < 0$:
+Note that the lower integration limit is 0 instead of $$-\infty$$,
+because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
$$\begin{aligned}
\hat{\mathcal{L}}\{(f * g)(t)\}
@@ -95,7 +95,7 @@ $$\begin{aligned}
&= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
\end{aligned}$$
-Then we define a new integration variable $\tau = t - t'$, yielding:
+Then we define a new integration variable $$\tau = t - t'$$, yielding:
$$\begin{aligned}
\hat{\mathcal{L}}\{(f * g)(t)\}
diff --git a/source/know/concept/coulomb-logarithm/index.md b/source/know/concept/coulomb-logarithm/index.md
index d036aa6..b843eb3 100644
--- a/source/know/concept/coulomb-logarithm/index.md
+++ b/source/know/concept/coulomb-logarithm/index.md
@@ -15,7 +15,7 @@ In any case, the particles' paths are deflected,
and it would be nice to know
whether those deflections are usually large or small.
-Let us choose $\pi/2$ as an example of a large deflection angle.
+Let us choose $$\pi/2$$ as an example of a large deflection angle.
Then Rutherford predicts:
$$\begin{aligned}
@@ -24,7 +24,7 @@ $$\begin{aligned}
= 1
\end{aligned}$$
-Isolating this for the impact parameter $b_\mathrm{large}$
+Isolating this for the impact parameter $$b_\mathrm{large}$$
then yields an effective radius of a particle:
$$\begin{aligned}
@@ -32,9 +32,9 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu}
\end{aligned}$$
-Therefore, the collision cross-section $\sigma_\mathrm{large}$
+Therefore, the collision cross-section $$\sigma_\mathrm{large}$$
for large deflections can be roughly estimated as
-the area of a disc with radius $b_\mathrm{large}$:
+the area of a disc with radius $$b_\mathrm{large}$$:
$$\begin{aligned}
\sigma_\mathrm{large}
@@ -43,7 +43,7 @@ $$\begin{aligned}
\end{aligned}$$
Next, we want to find the cross-section for small deflections.
-For sufficiently small angles $\theta$,
+For sufficiently small angles $$\theta$$,
we can Taylor-expand the Rutherford scattering formula to first order:
$$\begin{aligned}
@@ -55,33 +55,33 @@ $$\begin{aligned}
\approx \frac{q_1 q_2}{2 \pi \varepsilon_0 |\vb{v}|^2 \mu b}
\end{aligned}$$
-Clearly, $\theta$ is inversely proportional to $b$.
+Clearly, $$\theta$$ is inversely proportional to $$b$$.
Intuitively, we know that a given particle in a uniform plasma
always has more "distant" neighbours than "close" neighbours,
-so we expect that small deflections (large $b$)
+so we expect that small deflections (large $$b$$)
are more common than large deflections.
That said, many small deflections can add up to a large total.
They can also add up to zero,
so we should use random walk statistics.
-We now ask: how many $N$ small deflections $\theta_n$
-are needed to get a large total of, say, $1$ radian?
+We now ask: how many $$N$$ small deflections $$\theta_n$$
+are needed to get a large total of, say, $$1$$ radian?
$$\begin{aligned}
\sum_{n = 1}^N \theta_n^2 \approx 1
\end{aligned}$$
-Traditionally, $1$ is chosen instead of $\pi/2$ for convenience.
+Traditionally, $$1$$ is chosen instead of $$\pi/2$$ for convenience.
We are only making rough estimates,
so those two angles are close enough for our purposes.
Furthermore, the end result will turn out to be logarithmic,
and is thus barely affected by this inconsistency.
You can easily convince yourself
-that the average time $\tau$ between "collisions"
-is related like so to the cross-section $\sigma$,
-the total density $n$ of charged particles,
-and the relative velocity $|\vb{v}|$:
+that the average time $$\tau$$ between "collisions"
+is related like so to the cross-section $$\sigma$$,
+the total density $$n$$ of charged particles,
+and the relative velocity $$|\vb{v}|$$:
$$\begin{aligned}
\frac{1}{\tau}
@@ -91,9 +91,9 @@ $$\begin{aligned}
= n |\vb{v}| \tau \sigma
\end{aligned}$$
-Therefore, in a given time interval $t$,
-the expected number of collision $N_b$
-for impact parameters between $b$ and $b\!+\!\dd{b}$
+Therefore, in a given time interval $$t$$,
+the expected number of collision $$N_b$$
+for impact parameters between $$b$$ and $$b\!+\!\dd{b}$$
(imagine a ring with these inner and outer radii)
is given by:
@@ -103,9 +103,9 @@ $$\begin{aligned}
= n |\vb{v}| t \:(2 \pi b \dd{b})
\end{aligned}$$
-In this time interval $t$,
+In this time interval $$t$$,
we can thus turn our earlier sum
-into an integral of $N_b$ over $b$:
+into an integral of $$N_b$$ over $$b$$:
$$\begin{aligned}
1
@@ -114,10 +114,10 @@ $$\begin{aligned}
= n |\vb{v}| t \int 2 \pi \theta^2 b \dd{b}
\end{aligned}$$
-Using the formula $n |\vb{v}| \tau \sigma = 1$,
-we thus define $\sigma_{small}$ as the effective cross-section
-needed to get a large deflection (of $1$ radian),
-with an average period $t$:
+Using the formula $$n |\vb{v}| \tau \sigma = 1$$,
+we thus define $$\sigma_{small}$$ as the effective cross-section
+needed to get a large deflection (of $$1$$ radian),
+with an average period $$t$$:
$$\begin{aligned}
\sigma_\mathrm{small}
@@ -125,8 +125,8 @@ $$\begin{aligned}
= \int \frac{2 \pi q_1^2 q_2^2}{4 \pi^2 \varepsilon_0^2 |\vb{v}|^4 \mu^2 b^2} b \dd{b}
\end{aligned}$$
-Where we have replaced $\theta$ with our earlier Taylor expansion.
-Here, we recognize $\sigma_\mathrm{large}$:
+Where we have replaced $$\theta$$ with our earlier Taylor expansion.
+Here, we recognize $$\sigma_\mathrm{large}$$:
$$\begin{aligned}
\sigma_\mathrm{small}
@@ -135,12 +135,12 @@ $$\begin{aligned}
\end{aligned}$$
But what are the integration limits?
-We know that the deflection grows for smaller $b$,
-so it would be reasonable to choose $b_\mathrm{large}$ as the lower limit.
-For very large $b$, the plasma shields the particles from each other,
+We know that the deflection grows for smaller $$b$$,
+so it would be reasonable to choose $$b_\mathrm{large}$$ as the lower limit.
+For very large $$b$$, the plasma shields the particles from each other,
thereby nullifying the deflection,
so as upper limit we choose
-the [Debye length](/know/concept/debye-length/) $\lambda_D$,
+the [Debye length](/know/concept/debye-length/) $$\lambda_D$$,
i.e. the plasma's self-shielding length.
We thus find:
@@ -152,10 +152,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Here, $\ln\!(\Lambda)$ is known as the **Coulomb logarithm**,
-with the **plasma parameter** $\Lambda$ defined below,
-equal to $9/2$ times the number of particles
-in a sphere with radius $\lambda_D$:
+Here, $$\ln\!(\Lambda)$$ is known as the **Coulomb logarithm**,
+with the **plasma parameter** $$\Lambda$$ defined below,
+equal to $$9/2$$ times the number of particles
+in a sphere with radius $$\lambda_D$$:
$$\begin{aligned}
\boxed{
@@ -165,14 +165,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-The above relation between $\sigma_\mathrm{small}$ and $\sigma_\mathrm{large}$
+The above relation between $$\sigma_\mathrm{small}$$ and $$\sigma_\mathrm{large}$$
gives us an estimate of how much more often
small deflections occur, compared to large ones.
-In a typical plasma, $\ln\!(\Lambda)$ is between 6 and 25,
-such that $\sigma_\mathrm{small}$ is 2-3 orders of magnitude larger than $\sigma_\mathrm{large}$.
+In a typical plasma, $$\ln\!(\Lambda)$$ is between 6 and 25,
+such that $$\sigma_\mathrm{small}$$ is 2-3 orders of magnitude larger than $$\sigma_\mathrm{large}$$.
-Note that $t$ is now fixed as the period
-for small deflections to add up to $1$ radian.
+Note that $$t$$ is now fixed as the period
+for small deflections to add up to $$1$$ radian.
In more useful words, it is the time scale
for significant energy transfer between partices:
@@ -183,7 +183,7 @@ $$\begin{aligned}
\sim \frac{n}{T^{3/2}}
\end{aligned}$$
-Where we have used that $|\vb{v}| \propto \sqrt{T}$, for some temperature $T$.
+Where we have used that $$|\vb{v}| \propto \sqrt{T}$$, for some temperature $$T$$.
Consequently, in hotter plasmas, there is less energy transfer,
meaning that a hot plasma is hard to heat up further.
diff --git a/source/know/concept/coupled-mode-theory/index.md b/source/know/concept/coupled-mode-theory/index.md
index ecae1bb..6a5ec1b 100644
--- a/source/know/concept/coupled-mode-theory/index.md
+++ b/source/know/concept/coupled-mode-theory/index.md
@@ -10,21 +10,21 @@ layout: "concept"
Given an optical resonator (e.g. a photonic crystal cavity),
consider one of its quasinormal modes
-with frequency $\omega_0$ and decay rate $1 / \tau_0$.
-Its complex amplitude $A$ is governed by:
+with frequency $$\omega_0$$ and decay rate $$1 / \tau_0$$.
+Its complex amplitude $$A$$ is governed by:
$$\begin{aligned}
\dv{A}{t}
&= \bigg( \!-\! i \omega_0 - \frac{1}{\tau_0} \bigg) A
\end{aligned}$$
-We choose to normalize $A$ so that $|A(t)|^2$
-is the total energy inside the resonator at time $t$.
+We choose to normalize $$A$$ so that $$|A(t)|^2$$
+is the total energy inside the resonator at time $$t$$.
-Suppose that $N$ waveguides are now "connected" to this resonator,
-meaning that the resonator mode $A$ and the outgoing waveguide mode $S_\ell^\mathrm{out}$
-overlap sufficiently for $A$ to leak into $S_\ell^\mathrm{out}$ at a rate $1 / \tau_\ell$.
-Conversely, the incoming mode $S_\ell^\mathrm{in}$ brings energy to $A$.
+Suppose that $$N$$ waveguides are now "connected" to this resonator,
+meaning that the resonator mode $$A$$ and the outgoing waveguide mode $$S_\ell^\mathrm{out}$$
+overlap sufficiently for $$A$$ to leak into $$S_\ell^\mathrm{out}$$ at a rate $$1 / \tau_\ell$$.
+Conversely, the incoming mode $$S_\ell^\mathrm{in}$$ brings energy to $$A$$.
Therefore, we can write up the following general set of equations:
$$\begin{aligned}
@@ -36,35 +36,35 @@ $$\begin{aligned}
&= \beta_\ell S_\ell^\mathrm{in} + \gamma_\ell A
\end{aligned}$$
-Where $\alpha_\ell$ and $\gamma_\ell$ are unknown coupling constants,
-and $\beta_\ell$ represents reflection.
-We normalize $S_\ell^\mathrm{in}$
-so that $|S_\ell^\mathrm{in}(t)|^2$ is the power flowing towards $A$ at time $t$,
-and likewise for $S_\ell^\mathrm{out}$.
+Where $$\alpha_\ell$$ and $$\gamma_\ell$$ are unknown coupling constants,
+and $$\beta_\ell$$ represents reflection.
+We normalize $$S_\ell^\mathrm{in}$$
+so that $$|S_\ell^\mathrm{in}(t)|^2$$ is the power flowing towards $$A$$ at time $$t$$,
+and likewise for $$S_\ell^\mathrm{out}$$.
Note that we have made a subtle approximation here:
by adding new damping mechanisms,
-we are in fact modifying $\omega_0$;
+we are in fact modifying $$\omega_0$$;
see the [harmonic oscillator](/know/concept/harmonic-oscillator/) for a demonstration.
However, the frequency shift is second-order in the decay rate,
-so by assuming that all $\tau_\ell$ are large,
+so by assuming that all $$\tau_\ell$$ are large,
we only need to keep the first-order terms, as we did.
This is called **weak coupling**.
-If we also assume that $\tau_0$ is large
-(its effect is already included in $\omega_0$),
+If we also assume that $$\tau_0$$ is large
+(its effect is already included in $$\omega_0$$),
then we can treat the decay mechanisms separately:
-to analyze the decay into a certain waveguide $\ell$,
-it is first-order accurate to neglect all other waveguides and $\tau_0$:
+to analyze the decay into a certain waveguide $$\ell$$,
+it is first-order accurate to neglect all other waveguides and $$\tau_0$$:
$$\begin{aligned}
\dv{A}{t}
\approx \bigg( \!-\! i \omega_0 - \frac{1}{\tau_\ell} \bigg) A + \sum_{\ell' = 1}^N \alpha_\ell S_{\ell'}^\mathrm{in}
\end{aligned}$$
-To determine $\gamma_\ell$, we use energy conservation.
-If all $S_{\ell'}^\mathrm{in} = 0$,
-then the energy in $A$ decays as:
+To determine $$\gamma_\ell$$, we use energy conservation.
+If all $$S_{\ell'}^\mathrm{in} = 0$$,
+then the energy in $$A$$ decays as:
$$\begin{aligned}
\dv{|A|^2}{t}
@@ -77,7 +77,7 @@ $$\begin{aligned}
\end{aligned}$$
Since all other mechanisms are neglected,
-all this energy must go into $S_\ell^\mathrm{out}$, meaning:
+all this energy must go into $$S_\ell^\mathrm{out}$$, meaning:
$$\begin{aligned}
|S_\ell^\mathrm{out}|^2
@@ -85,32 +85,32 @@ $$\begin{aligned}
= \frac{2}{\tau_\ell} |A|^2
\end{aligned}$$
-Taking the square root, we clearly see that $|\gamma_\ell| = \sqrt{2 / \tau_\ell}$.
-Because the phase of $S_\ell^\mathrm{out}$ is arbitrarily defined,
-$\gamma_\ell$ need not be complex, so we choose $\gamma_\ell = \sqrt{2 / \tau_\ell}$.
+Taking the square root, we clearly see that $$|\gamma_\ell| = \sqrt{2 / \tau_\ell}$$.
+Because the phase of $$S_\ell^\mathrm{out}$$ is arbitrarily defined,
+$$\gamma_\ell$$ need not be complex, so we choose $$\gamma_\ell = \sqrt{2 / \tau_\ell}$$.
-Next, to find $\alpha_\ell$, we exploit the time-reversal symmetry
+Next, to find $$\alpha_\ell$$, we exploit the time-reversal symmetry
of [Maxwell's equations](/know/concept/maxwells-equations/),
which govern the light in the resonator and the waveguides.
-In the above calculation of $\gamma_\ell$, $A$ evolved as follows,
-with the lost energy ending up in $S_\ell^\mathrm{out}$:
+In the above calculation of $$\gamma_\ell$$, $$A$$ evolved as follows,
+with the lost energy ending up in $$S_\ell^\mathrm{out}$$:
$$\begin{aligned}
A(t)
= A e^{-i \omega_0 t - t / \tau_\ell}
\end{aligned}$$
-After reversing time, $A$ evolves like so,
+After reversing time, $$A$$ evolves like so,
where we have taken the complex conjugate
to preserve the meanings of the symbols
-$A$, $S_\ell^\mathrm{out}$, and $S_\ell^\mathrm{in}$:
+$$A$$, $$S_\ell^\mathrm{out}$$, and $$S_\ell^\mathrm{in}$$:
$$\begin{aligned}
A(t)
= A e^{-i \omega_0 t + t / \tau_\ell}
\end{aligned}$$
-We insert this expression for $A(t)$ into its original differential equation, yielding:
+We insert this expression for $$A(t)$$ into its original differential equation, yielding:
$$\begin{aligned}
\dv{A}{t}
@@ -118,7 +118,7 @@ $$\begin{aligned}
= \bigg( \!-\! i \omega_0 - \frac{1}{\tau_\ell} \bigg) A + \alpha_\ell S_\ell^\mathrm{in}
\end{aligned}$$
-Isolating this for $A$ leads us to the following power balance equation:
+Isolating this for $$A$$ leads us to the following power balance equation:
$$\begin{aligned}
A
@@ -129,7 +129,7 @@ $$\begin{aligned}
\end{aligned}$$
But thanks to energy conservation,
-all power delivered by $S_\ell^\mathrm{in}$ ends up in $A$, so we know:
+all power delivered by $$S_\ell^\mathrm{in}$$ ends up in $$A$$, so we know:
$$\begin{aligned}
|S_\ell^\mathrm{in}|^2
@@ -138,20 +138,20 @@ $$\begin{aligned}
\end{aligned}$$
To reconcile the two equations above,
-we need $|\alpha_\ell| = \sqrt{2 / \tau_\ell}$.
-Discarding the phase thanks to our choice of $\gamma_\ell$,
-we conclude that $\alpha_\ell = \sqrt{2 / \tau_\ell} = \gamma_\ell$.
+we need $$|\alpha_\ell| = \sqrt{2 / \tau_\ell}$$.
+Discarding the phase thanks to our choice of $$\gamma_\ell$$,
+we conclude that $$\alpha_\ell = \sqrt{2 / \tau_\ell} = \gamma_\ell$$.
-Finally, $\beta_\ell$ can also be determined using energy conservation.
+Finally, $$\beta_\ell$$ can also be determined using energy conservation.
Again using our weak coupling assumption,
-if energy is only entering and leaving $A$ through waveguide $\ell$, we have:
+if energy is only entering and leaving $$A$$ through waveguide $$\ell$$, we have:
$$\begin{aligned}
|S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
= \dv{|A|^2}{t}
\end{aligned}$$
-Meanwhile, using the differential equation for $A$,
+Meanwhile, using the differential equation for $$A$$,
we find the following relation:
$$\begin{aligned}
@@ -161,7 +161,7 @@ $$\begin{aligned}
&= - \frac{2}{\tau_\ell} |A|^2 + \alpha_\ell \Big( S_\ell^\mathrm{in} A^* + (S_\ell^\mathrm{in})^* A \Big)
\end{aligned}$$
-By isolating both of the above relations for $\idv{|A|^2}{t}$
+By isolating both of the above relations for $$\idv{|A|^2}{t}$$
and equating them, we arrive at:
$$\begin{aligned}
@@ -169,9 +169,9 @@ $$\begin{aligned}
&= - \frac{2}{\tau_\ell} |A|^2 + \alpha_\ell \Big( S_\ell^\mathrm{in} A^* + (S_\ell^\mathrm{in})^* A \Big)
\end{aligned}$$
-We insert the definition of $\gamma_\ell$ and $\beta_\ell$,
-namely $\gamma_\ell A = S_\ell^\mathrm{out} - \beta_\ell S_\ell^\mathrm{in}$,
-and use $\alpha_\ell = \gamma_\ell$:
+We insert the definition of $$\gamma_\ell$$ and $$\beta_\ell$$,
+namely $$\gamma_\ell A = S_\ell^\mathrm{out} - \beta_\ell S_\ell^\mathrm{in}$$,
+and use $$\alpha_\ell = \gamma_\ell$$:
$$\begin{aligned}
|S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
@@ -191,8 +191,8 @@ $$\begin{aligned}
&\quad\; + (1 - \beta_\ell) S_\ell^\mathrm{in} (S_\ell^\mathrm{out})^* + (1 - \beta_\ell^*) (S_\ell^\mathrm{in})^* S_\ell^\mathrm{out}
\end{aligned}$$
-This equation is only satisfied if $\beta_\ell = -1$.
-Combined with $\alpha_\ell = \gamma_\ell = \sqrt{2 / \tau_\ell}$,
+This equation is only satisfied if $$\beta_\ell = -1$$.
+Combined with $$\alpha_\ell = \gamma_\ell = \sqrt{2 / \tau_\ell}$$,
the **coupled-mode equations** take the following form:
$$\begin{aligned}
diff --git a/source/know/concept/curvature/index.md b/source/know/concept/curvature/index.md
index bf8ae25..a3aec08 100644
--- a/source/know/concept/curvature/index.md
+++ b/source/know/concept/curvature/index.md
@@ -7,9 +7,9 @@ categories:
layout: "concept"
---
-Given a curve or surface, its **curvature** $\kappa$
+Given a curve or surface, its **curvature** $$\kappa$$
describes how sharply it is bending at a given point.
-It is defined as the inverse of the **radius of curvature** $R$,
+It is defined as the inverse of the **radius of curvature** $$R$$,
which is the radius of the tangent circle
that **osculates** (i.e. best approximates)
the curve/surface at that point:
@@ -18,7 +18,7 @@ $$\begin{aligned}
\kappa = \frac{1}{R}
\end{aligned}$$
-Typically, $\kappa$ is positive for convex curves/surfaces,
+Typically, $$\kappa$$ is positive for convex curves/surfaces,
and negative for concave ones, although this distinction is somewhat arbitrary.
Below, we calculate the curvature in several general cases.
@@ -34,13 +34,13 @@ to find the curvature.
This approach relies on the fact that a circle
has the highest area-perimeter ratio of any 2D shape,
and a sphere has the highest volume-surface ratio of any 3D body.
-By the definition of curvature, these shapes have constant $\kappa$.
+By the definition of curvature, these shapes have constant $$\kappa$$.
We will thus minimize the perimeter/surface while keeping the area/volume fixed,
which will give us a shape with constant curvature,
-and from that we can extrapolate an expression for $\kappa$.
+and from that we can extrapolate an expression for $$\kappa$$.
-In 2D, for a single-variable height function $h(x)$,
+In 2D, for a single-variable height function $$h(x)$$,
the length of a small segment of the curve is:
$$\begin{aligned}
@@ -49,7 +49,7 @@ $$\begin{aligned}
= \dd{x} \sqrt{1 + h_x^2}
\end{aligned}$$
-Which leads us to define the following Lagrangian $\mathcal{L}$
+Which leads us to define the following Lagrangian $$\mathcal{L}$$
describing the "energy cost" of the curve:
$$\begin{aligned}
@@ -66,8 +66,8 @@ $$\begin{aligned}
\end{aligned}$$
By putting these things together,
-we arrive at the following energy functional $E[h]$,
-where $\kappa$ is an ominously-named [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+we arrive at the following energy functional $$E[h]$$,
+where $$\kappa$$ is an ominously-named [Lagrange multiplier](/know/concept/lagrange-multiplier/):
$$\begin{aligned}
E[h]
@@ -83,7 +83,7 @@ $$\begin{aligned}
\end{aligned}$$
We evaluate the terms of this equation
-to arrive at an expression for the curvature $\kappa$:
+to arrive at an expression for the curvature $$\kappa$$:
$$\begin{aligned}
\boxed{
@@ -92,15 +92,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-In this optimization problem, $\kappa$ is a constant,
+In this optimization problem, $$\kappa$$ is a constant,
but in fact the statement above is valid for variable curvatures too,
-in which case $\kappa$ is a function of $x$.
+in which case $$\kappa$$ is a function of $$x$$.
## 2D in general
We can parametrically describe an arbitrary plane curve
-as a function of the arc length $s$:
+as a function of the arc length $$s$$:
$$\begin{aligned}
\big( x(s), y(s) \big)
@@ -108,9 +108,9 @@ $$\begin{aligned}
\dd{s}^2 = \dd{x}^2 + \dd{y}^2
\end{aligned}$$
-If we choose the horizontal $x$-axis as a reference,
-we can furthermore define the **elevation angle** $\theta(s)$
-as the angle between the reference and the curve's tangent vector $\vu{t}$:
+If we choose the horizontal $$x$$-axis as a reference,
+we can furthermore define the **elevation angle** $$\theta(s)$$
+as the angle between the reference and the curve's tangent vector $$\vu{t}$$:
$$\begin{aligned}
\vu{t}
@@ -118,9 +118,9 @@ $$\begin{aligned}
= \big( \cos\theta(s), \sin\theta(s) \big)
\end{aligned}$$
-Where $x_s(s) = \idv{x}{s}$.
-The curvature $\kappa$ is defined as
-the $s$-derivative of this elevation angle:
+Where $$x_s(s) = \idv{x}{s}$$.
+The curvature $$\kappa$$ is defined as
+the $$s$$-derivative of this elevation angle:
$$\begin{aligned}
\kappa
@@ -128,10 +128,10 @@ $$\begin{aligned}
= \theta_s(s)
\end{aligned}$$
-We have two ways of writing $\vu{t}$:
-using the derivatives $x_s$ and $y_s$,
-or the elevation angle $\theta$.
-Now, let us take the $s$-derivative of both expressions,
+We have two ways of writing $$\vu{t}$$:
+using the derivatives $$x_s$$ and $$y_s$$,
+or the elevation angle $$\theta$$.
+Now, let us take the $$s$$-derivative of both expressions,
and equate them:
$$\begin{aligned}
@@ -147,15 +147,15 @@ $$\begin{aligned}
y_{ss} = \kappa x_s
\end{aligned}$$
-We multiply these equation by $y_s$ and $x_s$, respectively,
+We multiply these equation by $$y_s$$ and $$x_s$$, respectively,
and subtract the first from the last:
$$\begin{aligned}
y_{ss} x_s - x_{ss} y_s = \kappa x_s^2 + \kappa y_s^2
\end{aligned}$$
-Isolating this for $\kappa$ and using the fact that $x_s^2 + y_s^2 = 1$
-thanks to $s$ being the arc length:
+Isolating this for $$\kappa$$ and using the fact that $$x_s^2 + y_s^2 = 1$$
+thanks to $$s$$ being the arc length:
$$\begin{aligned}
\kappa
@@ -165,8 +165,8 @@ $$\begin{aligned}
While this result is correct,
we would like to generalize it to cases where the curve
-is parametrized by some other $t$, not necessarily the arc length.
-Let prime denote the $t$-derivative:
+is parametrized by some other $$t$$, not necessarily the arc length.
+Let prime denote the $$t$$-derivative:
$$\begin{aligned}
x_s
@@ -182,7 +182,7 @@ $$\begin{aligned}
= y'' t_s^2 + x' t_{ss}
\end{aligned}$$
-By inserting these expression into the earlier formula for $\kappa$, we find:
+By inserting these expression into the earlier formula for $$\kappa$$, we find:
$$\begin{aligned}
\kappa
@@ -194,9 +194,9 @@ $$\begin{aligned}
&= t_s^3 (x' y'' - y' x'')
\end{aligned}$$
-Since $x_s^2 + y_s^2 = 1$, we know that $(x')^2 + (y')^2 = 1 / t_s^2$,
+Since $$x_s^2 + y_s^2 = 1$$, we know that $$(x')^2 + (y')^2 = 1 / t_s^2$$,
which leads us to the following general expression for
-the curvature $\kappa$ of a plane curve:
+the curvature $$\kappa$$ of a plane curve:
$$\begin{aligned}
\boxed{
@@ -205,13 +205,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-If the curve happens to be a height function, i.e. $y(x)$,
-then $x' = 1$ and $x'' = 0$, and we arrive at our previous result again.
+If the curve happens to be a height function, i.e. $$y(x)$$,
+then $$x' = 1$$ and $$x'' = 0$$, and we arrive at our previous result again.
## 3D height functions
-The generalization to a 3D height function $h(x, y)$ is straightforward:
+The generalization to a 3D height function $$h(x, y)$$ is straightforward:
the cost of an infinitesimal portion of the surface is as follows,
using the same reasoning as before:
@@ -220,8 +220,8 @@ $$\begin{aligned}
= \sqrt{1 + h_x^2 + h_y^2}
\end{aligned}$$
-Keeping the volume $V$ constant,
-we get the following energy functional $E$ to minimize:
+Keeping the volume $$V$$ constant,
+we get the following energy functional $$E$$ to minimize:
$$\begin{aligned}
E[h]
@@ -229,14 +229,14 @@ $$\begin{aligned}
\end{aligned}$$
Which gives us an Euler-Lagrange equation
-involving the Lagrange multiplier $\lambda$:
+involving the Lagrange multiplier $$\lambda$$:
$$\begin{aligned}
0
= \pdv{\mathcal{L}}{h} - \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) - \dv{}{y}\Big( \pdv{\mathcal{L}}{h_y} \Big) + \lambda
\end{aligned}$$
-Inserting $\mathcal{L}$ into this and evaluating all the derivatives
+Inserting $$\mathcal{L}$$ into this and evaluating all the derivatives
yields a result for the (variable) curvature:
$$\begin{aligned}
@@ -247,10 +247,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-What are $\kappa_1$ and $\kappa_2$?
+What are $$\kappa_1$$ and $$\kappa_2$$?
Well, the problem in 3D is that the curvature of an osculating circle
depends on the orientation of that circle.
-The **principal curvatures** $\kappa_1$ and $\kappa_2$
+The **principal curvatures** $$\kappa_1$$ and $$\kappa_2$$
are the largest and smallest curvatures at a given point,
but finding their values and the corresponding **principal directions** is not so easy.
Fortunately, in practice, we are often only interested in their sum:
@@ -261,7 +261,7 @@ $$\begin{aligned}
= \frac{1}{R_1} + \frac{1}{R_2}
\end{aligned}$$
-These **principal radii** $R_1$ and $R_2$ are important
+These **principal radii** $$R_1$$ and $$R_2$$ are important
for e.g. the [Young-Laplace law](/know/concept/young-laplace-law/).
@@ -269,10 +269,10 @@ for e.g. the [Young-Laplace law](/know/concept/young-laplace-law/).
To find a general expression for the mean curvature of an arbitrary surface,
we "cut off" a small part of the surface that we can regard as a height function.
-We call the "cutting" reference plane $(x, y)$,
-and the surface it describes $h(x, y)$.
-We then define the unit tangent vectors $\vu{t}_x$ and $\vu{t}_y$
-to be parallel to the $x$-axis and $y$-axis, respectively:
+We call the "cutting" reference plane $$(x, y)$$,
+and the surface it describes $$h(x, y)$$.
+We then define the unit tangent vectors $$\vu{t}_x$$ and $$\vu{t}_y$$
+to be parallel to the $$x$$-axis and $$y$$-axis, respectively:
$$\begin{aligned}
\vu{t}_x
@@ -290,7 +290,7 @@ $$\begin{aligned}
Since they were chosen to lie along the axes,
these vectors are not necessarily orthogonal,
-so we need to normalize the resulting normal vector $\vu{n}$:
+so we need to normalize the resulting normal vector $$\vu{n}$$:
$$\begin{aligned}
\vu{n}
@@ -301,7 +301,7 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-Let us take a look at the divergence of $\vu{n}$,
+Let us take a look at the divergence of $$\vu{n}$$,
or to be precise, its *projection* onto the reference plane
(although this distinction is not really important for our purposes):
@@ -310,7 +310,7 @@ $$\begin{aligned}
= - \dv{}{x}\bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) - \dv{}{y}\bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg)
\end{aligned}$$
-Compare this with the expression for $\lambda$ we found earlier,
+Compare this with the expression for $$\lambda$$ we found earlier,
with the help of variational calculus:
$$\begin{aligned}
@@ -342,8 +342,8 @@ $$\begin{aligned}
&= \frac{1}{2} a r^2 \cos^2\varphi + \frac{1}{2} b r^2 \sin^2\varphi + c r^2 \cos\varphi \sin\varphi
\end{aligned}$$
-Sufficiently close to the extremum, where $h_x$ and $h_y$ are negligible,
-the curvature along a certain direction $\varphi$ is given by
+Sufficiently close to the extremum, where $$h_x$$ and $$h_y$$ are negligible,
+the curvature along a certain direction $$\varphi$$ is given by
our earlier formula for a 2D height function:
$$\begin{aligned}
@@ -352,8 +352,8 @@ $$\begin{aligned}
= a \cos^2\varphi + b \sin^2\varphi + c \sin(2 \varphi)
\end{aligned}$$
-To find the extremes of $\kappa$,
-we differentiate with respect to $\varphi$ and demand that it is zero:
+To find the extremes of $$\kappa$$,
+we differentiate with respect to $$\varphi$$ and demand that it is zero:
$$\begin{aligned}
0
@@ -370,10 +370,10 @@ $$\begin{aligned}
= \tan(2 \varphi)
\end{aligned}$$
-Since the $\tan$ function is $\pi$-periodic,
-this has two solutions, $\varphi_0$ and $\varphi_0 + \pi/2$,
+Since the $$\tan$$ function is $$\pi$$-periodic,
+this has two solutions, $$\varphi_0$$ and $$\varphi_0 + \pi/2$$,
which are clearly orthogonal,
-hence the principal directions are at an angle of $\pi/2$.
+hence the principal directions are at an angle of $$\pi/2$$.
Finally, it is also worth mentioning that
the principal directions always lie in planes
diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md
index 3012ca6..cb22e43 100644
--- a/source/know/concept/curvilinear-coordinates/index.md
+++ b/source/know/concept/curvilinear-coordinates/index.md
@@ -14,15 +14,15 @@ is known as a **coordinate surface**, and the intersections of
the surfaces of different coordinates are called **coordinate lines**.
A **curvilinear** coordinate system is one where at least one of the coordinate surfaces is curved,
-e.g. in cylindrical coordinates the line between $r$ and $z$ is a circle.
+e.g. in cylindrical coordinates the line between $$r$$ and $$z$$ is a circle.
If the coordinate surfaces are mutually perpendicular,
it is an **orthogonal** system, which is generally desirable.
-A useful attribute of a coordinate system is its **line element** $\dd{\ell}$,
+A useful attribute of a coordinate system is its **line element** $$\dd{\ell}$$,
which represents the differential element of a line in any direction.
-For an orthogonal system, its square $\dd{\ell}^2$ is calculated
-by taking the differential elements of the old Cartesian $(x, y, z)$ system
-and writing them out in the new $(x_1, x_2, x_3)$ system.
+For an orthogonal system, its square $$\dd{\ell}^2$$ is calculated
+by taking the differential elements of the old Cartesian $$(x, y, z)$$ system
+and writing them out in the new $$(x_1, x_2, x_3)$$ system.
The resulting expression will be of the form:
$$\begin{aligned}
@@ -33,7 +33,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $h_1$, $h_2$, and $h_3$ are called **scale factors**,
+Where $$h_1$$, $$h_2$$, and $$h_3$$ are called **scale factors**,
and need not be constants.
The equation above only contains quadratic terms
because the coordinate system is orthogonal by assumption.
@@ -45,20 +45,20 @@ and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coor
In the following subsections,
we derive general formulae to convert expressions
-from Cartesian coordinates to the new orthogonal system $(x_1, x_2, x_3)$.
+from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$.
## Basis vectors
-Consider the the vector form of the line element $\dd{\ell}$,
-denoted by $\dd{\vu{\ell}}$ and expressed as:
+Consider the the vector form of the line element $$\dd{\ell}$$,
+denoted by $$\dd{\vu{\ell}}$$ and expressed as:
$$\begin{aligned}
\dd{\vu{\ell}}
= \vu{e}_x \dd{x} + \vu{e}_y \dd{y} + \vu{e}_z \dd{z}
\end{aligned}$$
-We can expand the Cartesian differential elements, e.g. $\dd{y}$,
+We can expand the Cartesian differential elements, e.g. $$\dd{y}$$,
in the new basis as follows:
$$\begin{aligned}
@@ -66,17 +66,17 @@ $$\begin{aligned}
= \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3}
\end{aligned}$$
-If we write this out for $\dd{x}$, $\dd{y}$ and $\dd{z}$,
-and group the terms according to $\dd{x}_1$, $\dd{x}_2$ and $\dd{x}_3$,
-we can compare it the alternative form of $\dd{\vu{\ell}}$:
+If we write this out for $$\dd{x}$$, $$\dd{y}$$ and $$\dd{z}$$,
+and group the terms according to $$\dd{x}_1$$, $$\dd{x}_2$$ and $$\dd{x}_3$$,
+we can compare it the alternative form of $$\dd{\vu{\ell}}$$:
$$\begin{aligned}
\dd{\vu{\ell}}
= \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4}
\end{aligned}$$
-From this, we can read off $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$.
-Here we only give $\vu{e}_1$, since $\vu{e}_2$ and $\vu{e}_3$ are analogous:
+From this, we can read off $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$.
+Here we only give $$\vu{e}_1$$, since $$\vu{e}_2$$ and $$\vu{e}_3$$ are analogous:
$$\begin{aligned}
\boxed{
@@ -89,9 +89,9 @@ $$\begin{aligned}
## Gradient
In an orthogonal coordinate system,
-the gradient $\nabla f$ of a scalar $f$ is as follows,
-where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
-are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$:
+the gradient $$\nabla f$$ of a scalar $$f$$ is as follows,
+where $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$
+are the basis unit vectors respectively corresponding to $$x_1$$, $$x_2$$ and $$x_3$$:
$$\begin{gathered}
\boxed{
@@ -107,8 +107,8 @@ $$\begin{gathered}
-For a direction $\dd{\ell}$, we know that
-$\idv{f}{\ell}$ is the component of $\nabla f$ in that direction:
+For a direction $$\dd{\ell}$$, we know that
+$$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction:
$$\begin{aligned}
\dv{f}{\ell}
@@ -117,9 +117,9 @@ $$\begin{aligned}
= \nabla f \cdot \vu{u}
\end{aligned}$$
-Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$.
-We thus find the expression for the gradient $\nabla f$
-by choosing $\dd{\ell}$ to be $h_1 \dd{x_1}$, $h_2 \dd{x_2}$ and $h_3 \dd{x_3}$ in turn:
+Where $$\vu{u}$$ is simply a unit vector in the direction of $$\dd{\ell}$$.
+We thus find the expression for the gradient $$\nabla f$$
+by choosing $$\dd{\ell}$$ to be $$h_1 \dd{x_1}$$, $$h_2 \dd{x_2}$$ and $$h_3 \dd{x_3}$$ in turn:
$$\begin{gathered}
\nabla f
@@ -127,13 +127,14 @@ $$\begin{gathered}
+ \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
+ \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
\end{gathered}$$
+
## Divergence
-The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$
+The divergence of a vector $$\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$$
in an orthogonal system is given by:
$$\begin{aligned}
@@ -149,7 +150,7 @@ $$\begin{aligned}
-As preparation, we rewrite $\vb{V}$ as follows
+As preparation, we rewrite $$\vb{V}$$ as follows
to introduce the scale factors:
$$\begin{aligned}
@@ -159,7 +160,7 @@ $$\begin{aligned}
+ \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
\end{aligned}$$
-We start by taking only the $\vu{e}_1$-component of this vector,
+We start by taking only the $$\vu{e}_1$$-component of this vector,
and expand its divergence using the following vector identity:
$$\begin{gathered}
@@ -167,8 +168,8 @@ $$\begin{gathered}
= \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) f
\end{gathered}$$
-Inserting the scalar $f = h_2 h_3 V_1$
-the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$,
+Inserting the scalar $$f = h_2 h_3 V_1$$
+the vector $$\vb{U} = \vu{e}_1 / (h_2 h_3)$$,
we arrive at:
$$\begin{gathered}
@@ -178,8 +179,8 @@ $$\begin{gathered}
\end{gathered}$$
The first right-hand term is easy to calculate
-thanks to our expression for the gradient $\nabla f$.
-Only the $\vu{e}_1$-component survives due to the dot product:
+thanks to our expression for the gradient $$\nabla f$$.
+Only the $$\vu{e}_1$$-component survives due to the dot product:
$$\begin{aligned}
\frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
@@ -200,7 +201,7 @@ $$\begin{aligned}
= \frac{\vu{e}_3}{h_3}
\end{aligned}$$
-Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis,
+Because $$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$ in an orthogonal basis,
these gradients can be used to express the vector whose divergence we want:
$$\begin{aligned}
@@ -210,7 +211,7 @@ $$\begin{aligned}
\end{aligned}$$
We then apply the divergence and expand the expression using a vector identity.
-In all cases, the curl of a gradient $\nabla \cross \nabla f$ is zero, so:
+In all cases, the curl of a gradient $$\nabla \cross \nabla f$$ is zero, so:
$$\begin{aligned}
\nabla \cdot \frac{\vu{e}_1}{h_2 h_3}
@@ -219,7 +220,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-After repeating this procedure for the other components of $\vb{V}$,
+After repeating this procedure for the other components of $$\vb{V}$$,
we get the desired general expression for the divergence.
@@ -227,7 +228,7 @@ we get the desired general expression for the divergence.
## Laplacian
-The Laplacian $\nabla^2 f$ is simply $\nabla \cdot \nabla f$,
+The Laplacian $$\nabla^2 f$$ is simply $$\nabla \cdot \nabla f$$,
so we can find the general formula
by combining the two preceding results
for the gradient and the divergence:
@@ -247,7 +248,7 @@ $$\begin{aligned}
## Curl
-The curl of a vector $\vb{V}$ is as follows
+The curl of a vector $$\vb{V}$$ is as follows
in a general orthogonal curvilinear system:
$$\begin{aligned}
@@ -269,22 +270,22 @@ $$\begin{aligned}
The curl is found in a similar way as the divergence.
-We rewrite $\vb{V}$ like so:
+We rewrite $$\vb{V}$$ like so:
$$\begin{aligned}
\vb{V}
= \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3)
\end{aligned}$$
-We expand the curl of its $\vu{e}_1$-component using the following vector identity:
+We expand the curl of its $$\vu{e}_1$$-component using the following vector identity:
$$\begin{gathered}
\nabla \cross (\vb{U} \: f)
= (\nabla \cross \vb{U}) f - \vb{U} \cross (\nabla f)
\end{gathered}$$
-Inserting the scalar $f = h_1 V_1$
-and the vector $\vb{U} = \vu{e}_1 / h_1$, we arrive at:
+Inserting the scalar $$f = h_1 V_1$$
+and the vector $$\vb{U} = \vu{e}_1 / h_1$$, we arrive at:
$$\begin{gathered}
\nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
@@ -292,7 +293,7 @@ $$\begin{gathered}
\end{gathered}$$
Previously, when proving the divergence,
-we already showed that $\vu{e}_1 / h_1 = \nabla x_1$.
+we already showed that $$\vu{e}_1 / h_1 = \nabla x_1$$.
Because the curl of a gradient is zero,
the first term disappears, leaving only the second,
which contains a gradient that turns out to be:
@@ -304,8 +305,8 @@ $$\begin{aligned}
+ \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3}
\end{aligned}$$
-Consequently, the curl of the first component of $\vb{V}$ is as follows,
-using the fact that $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+Consequently, the curl of the first component of $$\vb{V}$$ is as follows,
+using the fact that $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$
are related to each other by cross products:
$$\begin{aligned}
@@ -314,7 +315,7 @@ $$\begin{aligned}
= - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3}
\end{aligned}$$
-If we go through the same process for the other components of $\vb{V}$
+If we go through the same process for the other components of $$\vb{V}$$
and add up the results, we get the desired expression for the curl.
@@ -322,14 +323,14 @@ and add up the results, we get the desired expression for the curl.
## Differential elements
-The point of the scale factors $h_1$, $h_2$ and $h_3$, as can seen from their derivation,
+The point of the scale factors $$h_1$$, $$h_2$$ and $$h_3$$, as can seen from their derivation,
is to correct for "distortions" of the coordinates compared to the Cartesian system,
-such that the line element $\dd{\ell}$ retains its length.
-This property extends to the surface $\dd{S}$ and volume $\dd{V}$.
+such that the line element $$\dd{\ell}$$ retains its length.
+This property extends to the surface $$\dd{S}$$ and volume $$\dd{V}$$.
When handling a differential volume in curvilinear coordinates,
e.g. for a volume integral,
-the size of the box $\dd{V}$ must be corrected by the scale factors:
+the size of the box $$\dd{V}$$ must be corrected by the scale factors:
$$\begin{aligned}
\boxed{
@@ -339,8 +340,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-The same is true for the isosurfaces $\dd{S_1}$, $\dd{S_2}$ and $\dd{S_3}$
-where the coordinates $x_1$, $x_2$ and $x_3$ are respectively kept constant:
+The same is true for the isosurfaces $$\dd{S_1}$$, $$\dd{S_2}$$ and $$\dd{S_3}$$
+where the coordinates $$x_1$$, $$x_2$$ and $$x_3$$ are respectively kept constant:
$$\begin{aligned}
\boxed{
@@ -354,7 +355,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Using the same logic, the normal vector element $\dd{\vu{S}}$
+Using the same logic, the normal vector element $$\dd{\vu{S}}$$
of an arbitrary surface is given by:
$$\begin{aligned}
@@ -364,7 +365,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, the tangent vector element $\dd{\vu{\ell}}$ takes the following form:
+Finally, the tangent vector element $$\dd{\vu{\ell}}$$ takes the following form:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/cylindrical-parabolic-coordinates/index.md
index 76ff756..c8e16da 100644
--- a/source/know/concept/cylindrical-parabolic-coordinates/index.md
+++ b/source/know/concept/cylindrical-parabolic-coordinates/index.md
@@ -9,11 +9,11 @@ layout: "concept"
---
**Cylindrical parabolic coordinates** are a coordinate system
-that describes a point in space using three coordinates $(\sigma, \tau, z)$.
-The $z$-axis is unchanged from the Cartesian system,
+that describes a point in space using three coordinates $$(\sigma, \tau, z)$$.
+The $$z$$-axis is unchanged from the Cartesian system,
hence it is called a *cylindrical* system.
-In the $z$-isoplane, however, confocal parabolas are used.
-These coordinates can be converted to the Cartesian $(x, y, z)$ as follows:
+In the $$z$$-isoplane, however, confocal parabolas are used.
+These coordinates can be converted to the Cartesian $$(x, y, z)$$ as follows:
$$\begin{aligned}
\boxed{
@@ -39,7 +39,7 @@ $$\begin{aligned}
Cylindrical parabolic coordinates form an orthogonal
[curvilinear system](/know/concept/curvilinear-coordinates/),
-so we would like to find its scale factors $h_\sigma$, $h_\tau$ and $h_z$.
+so we would like to find its scale factors $$h_\sigma$$, $$h_\tau$$ and $$h_z$$.
The differentials of the Cartesian coordinates are as follows:
$$\begin{aligned}
@@ -50,7 +50,7 @@ $$\begin{aligned}
\dd{z} = \dd{z}
\end{aligned}$$
-We calculate the line segment $\dd{\ell}^2$,
+We calculate the line segment $$\dd{\ell}^2$$,
skipping many terms thanks to orthogonality:
$$\begin{aligned}
@@ -58,7 +58,7 @@ $$\begin{aligned}
&= (\sigma^2 + \tau^2) \:\dd{\sigma}^2 + (\tau^2 + \sigma^2) \:\dd{\tau}^2 + \dd{z}^2
\end{aligned}$$
-From this, we can directly read off the scale factors $h_\sigma^2$, $h_\tau^2$ and $h_z^2$,
+From this, we can directly read off the scale factors $$h_\sigma^2$$, $$h_\tau^2$$ and $$h_z^2$$,
which turn out to be:
$$\begin{aligned}
@@ -131,7 +131,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The differential element of volume $\dd{V}$
+The differential element of volume $$\dd{V}$$
in cylindrical parabolic coordinates is given by:
$$\begin{aligned}
@@ -141,7 +141,7 @@ $$\begin{aligned}
\end{aligned}$$
The differential elements of the isosurfaces are as follows,
-where $\dd{S_\sigma}$ is the $\sigma$-isosurface, etc.:
+where $$\dd{S_\sigma}$$ is the $$\sigma$$-isosurface, etc.:
$$\begin{aligned}
\boxed{
@@ -155,8 +155,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-The normal element $\dd{\vu{S}}$ of a surface and
-the tangent element $\dd{\vu{\ell}}$ of a curve are respectively:
+The normal element $$\dd{\vu{S}}$$ of a surface and
+the tangent element $$\dd{\vu{\ell}}$$ of a curve are respectively:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/cylindrical-polar-coordinates/index.md
index 8673c0b..686a4ed 100644
--- a/source/know/concept/cylindrical-polar-coordinates/index.md
+++ b/source/know/concept/cylindrical-polar-coordinates/index.md
@@ -10,12 +10,12 @@ layout: "concept"
**Cylindrical polar coordinates** are an extension of polar coordinates to 3D,
which describes the location of a point in space
-using the coordinates $(r, \varphi, z)$.
-The $z$-axis is unchanged from Cartesian coordinates,
+using the coordinates $$(r, \varphi, z)$$.
+The $$z$$-axis is unchanged from Cartesian coordinates,
hence it is called a *cylindrical* system.
-Cartesian coordinates $(x, y, z)$
-and the cylindrical system $(r, \varphi, z)$ are related by:
+Cartesian coordinates $$(x, y, z)$$
+and the cylindrical system $$(r, \varphi, z)$$ are related by:
$$\begin{aligned}
\boxed{
@@ -27,8 +27,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Conversely, a point given in $(x, y, z)$
-can be converted to $(r, \varphi, z)$
+Conversely, a point given in $$(x, y, z)$$
+can be converted to $$(r, \varphi, z)$$
using these formulae:
$$\begin{aligned}
@@ -43,7 +43,7 @@ $$\begin{aligned}
The cylindrical polar coordinates form an orthogonal
[curvilinear system](/know/concept/curvilinear-coordinates/),
-whose scale factors $h_r$, $h_\varphi$ and $h_z$ we want to find.
+whose scale factors $$h_r$$, $$h_\varphi$$ and $$h_z$$ we want to find.
To do so, we calculate the differentials of the Cartesian coordinates:
$$\begin{aligned}
@@ -54,7 +54,7 @@ $$\begin{aligned}
\dd{z} = \dd{z}
\end{aligned}$$
-And then we calculate the line element $\dd{\ell}^2$,
+And then we calculate the line element $$\dd{\ell}^2$$,
skipping many terms thanks to orthogonality,
$$\begin{aligned}
@@ -68,7 +68,7 @@ $$\begin{aligned}
Finally, we can simply read off
the squares of the desired scale factors
-$h_r^2$, $h_\varphi^2$ and $h_z^2$:
+$$h_r^2$$, $$h_\varphi^2$$ and $$h_z^2$$:
$$\begin{aligned}
\boxed{
@@ -141,7 +141,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The differential element of volume $\dd{V}$
+The differential element of volume $$\dd{V}$$
takes the following form:
$$\begin{aligned}
@@ -158,7 +158,7 @@ $$\begin{aligned}
= \int_{-\infty}^{\infty} \int_0^{2\pi} \int_0^\infty f(r, \varphi, z) \: r \dd{r} \dd{\varphi} \dd{z}
\end{aligned}$$
-The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+The isosurface elements are as follows, where $$S_r$$ is a surface at constant $$r$$, etc.:
$$\begin{aligned}
\boxed{
@@ -172,7 +172,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+Similarly, the normal vector element $$\dd{\vu{S}}$$ for an arbitrary surface is given by:
$$\begin{aligned}
\boxed{
@@ -183,7 +183,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+And finally, the tangent vector element $$\dd{\vu{\ell}}$$ of a given curve is as follows:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/debye-length/index.md b/source/know/concept/debye-length/index.md
index a42d137..e226ad9 100644
--- a/source/know/concept/debye-length/index.md
+++ b/source/know/concept/debye-length/index.md
@@ -16,10 +16,10 @@ This has the effect of **shielding** the object's presence
from the rest of the plasma.
We start from [Gauss' law](/know/concept/maxwells-equations/)
-for the [electric field](/know/concept/electric-field/) $\vb{E}$,
-expressing $\vb{E}$ as the gradient of a potential $\phi$,
-i.e. $\vb{E} = -\nabla \phi$,
-and splitting the charge density into ions $n_i$ and electrons $n_e$:
+for the [electric field](/know/concept/electric-field/) $$\vb{E}$$,
+expressing $$\vb{E}$$ as the gradient of a potential $$\phi$$,
+i.e. $$\vb{E} = -\nabla \phi$$,
+and splitting the charge density into ions $$n_i$$ and electrons $$n_e$$:
$$\begin{aligned}
\nabla^2 \phi(\vb{r})
@@ -28,9 +28,9 @@ $$\begin{aligned}
The last term represents a *test particle*,
which will be shielded.
-This particle is a point charge $q_t$,
-whose density is simply a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(\vb{r})$,
-and is not included in $n_i$ or $n_e$.
+This particle is a point charge $$q_t$$,
+whose density is simply a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(\vb{r})$$,
+and is not included in $$n_i$$ or $$n_e$$.
For a plasma in thermal equilibrium,
we have the [Boltzmann relations](/know/concept/boltzmann-relation/)
@@ -45,7 +45,7 @@ $$\begin{aligned}
\end{aligned}$$
We assume that electrical interactions are weak compared to thermal effects,
-i.e. $k_B T \gg q \phi$ in both cases.
+i.e. $$k_B T \gg q \phi$$ in both cases.
Then we Taylor-expand the Boltzmann relations to first order:
$$\begin{aligned}
@@ -57,8 +57,8 @@ $$\begin{aligned}
\end{aligned}$$
Inserting this back into Gauss' law,
-we arrive at the following equation for $\phi(\vb{r})$,
-where we have assumed quasi-neutrality such that $q_i n_{i0} = q_e n_{e0}$:
+we arrive at the following equation for $$\phi(\vb{r})$$,
+where we have assumed quasi-neutrality such that $$q_i n_{i0} = q_e n_{e0}$$:
$$\begin{aligned}
\nabla^2 \phi
@@ -70,7 +70,7 @@ $$\begin{aligned}
\end{aligned}$$
We now define the **ion** and **electron Debye lengths**
-$\lambda_{Di}$ and $\lambda_{De}$ as follows:
+$$\lambda_{Di}$$ and $$\lambda_{De}$$ as follows:
$$\begin{aligned}
\boxed{
@@ -84,7 +84,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-And then the **total Debye length** $\lambda_D$ is defined as the sum of their inverses,
+And then the **total Debye length** $$\lambda_D$$ is defined as the sum of their inverses,
and gives the rough thickness of the Debye sheath:
$$\begin{aligned}
@@ -108,7 +108,7 @@ This has the following solution,
known as the **Yukawa potential**,
which decays exponentially,
representing the plasma's **self-shielding**
-over a characteristic distance $\lambda_D$:
+over a characteristic distance $$\lambda_D$$:
$$\begin{aligned}
\boxed{
@@ -117,13 +117,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that $r$ is a scalar,
-i.e. the potential depends only on the radial distance to $q_t$.
+Note that $$r$$ is a scalar,
+i.e. the potential depends only on the radial distance to $$q_t$$.
This treatment only makes sense
if the plasma is sufficiently dense,
such that there is a large number of particles
-in a sphere with radius $\lambda_D$.
-This corresponds to a large [Coulomb logarithm](/know/concept/coulomb-logarithm/) $\ln\!(\Lambda)$:
+in a sphere with radius $$\lambda_D$$.
+This corresponds to a large [Coulomb logarithm](/know/concept/coulomb-logarithm/) $$\ln\!(\Lambda)$$:
$$\begin{aligned}
1 \ll \frac{4 \pi}{3} n_0 \lambda_D^3 = \frac{2}{9} \Lambda
@@ -138,7 +138,7 @@ $$\begin{aligned}
= \frac{A}{r} \exp(-B r)
\end{aligned}$$
-Where $A$ and $B$ are scaling constants that depend on the problem at hand.
+Where $$A$$ and $$B$$ are scaling constants that depend on the problem at hand.
diff --git a/source/know/concept/density-of-states/index.md b/source/know/concept/density-of-states/index.md
index 60b81d5..1c3b511 100644
--- a/source/know/concept/density-of-states/index.md
+++ b/source/know/concept/density-of-states/index.md
@@ -8,15 +8,15 @@ categories:
layout: "concept"
---
-The **density of states** $g(E)$ of a physical system is defined such that
-$g(E) \dd{E}$ is the number of states which could be occupied
-with an energy in the interval $[E, E + \dd{E}]$.
-In fact, $E$ need not be an energy;
+The **density of states** $$g(E)$$ of a physical system is defined such that
+$$g(E) \dd{E}$$ is the number of states which could be occupied
+with an energy in the interval $$[E, E + \dd{E}]$$.
+In fact, $$E$$ need not be an energy;
it should just be something that effectively identifies the state.
In its simplest form, the density of states is as follows,
-where $\Gamma(E)$ is the number of states with energy
-less than or equal to the argument $E$:
+where $$\Gamma(E)$$ is the number of states with energy
+less than or equal to the argument $$E$$:
$$\begin{aligned}
g(E)
@@ -25,9 +25,9 @@ $$\begin{aligned}
If the states can be treated as waves,
which is often the case,
-then we can calculate the density of states $g(k)$ in
-$k$-space, i.e. as a function of the wavenumber $k = |\vb{k}|$.
-Once we have $g(k)$, we use the dispersion relation $E(k)$ to find $g(E)$,
+then we can calculate the density of states $$g(k)$$ in
+$$k$$-space, i.e. as a function of the wavenumber $$k = |\vb{k}|$$.
+Once we have $$g(k)$$, we use the dispersion relation $$E(k)$$ to find $$g(E)$$,
by demanding that:
$$\begin{aligned}
@@ -37,14 +37,14 @@ $$\begin{aligned}
= g(k) \dv{k}{E}
\end{aligned}$$
-Inverting the dispersion relation $E(k)$ to get $k(E)$ might be difficult,
+Inverting the dispersion relation $$E(k)$$ to get $$k(E)$$ might be difficult,
in which case the left-hand equation can be satisfied numerically.
-Define $\Omega_n(k)$ as the number of states with
-a $k$-value less than or equal to the argument,
-or in other words, the volume of a hypersphere with radius $k$.
-Then the $n$-dimensional density of states $g_n(k)$
+Define $$\Omega_n(k)$$ as the number of states with
+a $$k$$-value less than or equal to the argument,
+or in other words, the volume of a hypersphere with radius $$k$$.
+Then the $$n$$-dimensional density of states $$g_n(k)$$
has the following general form:
$$\begin{aligned}
@@ -54,14 +54,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $D$ is each state's degeneracy (e.g. due to spin),
-and $k_{\mathrm{min}}$ is the smallest allowed $k$-value,
-according to the characteristic length $L$ of the system.
-We divide by $2^n$ to limit ourselves to the sector where all axes are positive,
-because we are only considering the magnitude of $k$.
+Where $$D$$ is each state's degeneracy (e.g. due to spin),
+and $$k_{\mathrm{min}}$$ is the smallest allowed $$k$$-value,
+according to the characteristic length $$L$$ of the system.
+We divide by $$2^n$$ to limit ourselves to the sector where all axes are positive,
+because we are only considering the magnitude of $$k$$.
-In one dimension $n = 1$, the number of states within a distance $k$ from the
-origin is the distance from $k$ to $-k$
+In one dimension $$n = 1$$, the number of states within a distance $$k$$ from the
+origin is the distance from $$k$$ to $$-k$$
(we let it run negative, since its meaning does not matter here), given by:
$$\begin{aligned}
@@ -69,9 +69,9 @@ $$\begin{aligned}
= 2 k
\end{aligned}$$
-To get $k_{\mathrm{min}}$, we choose to look at a rod of length $L$,
+To get $$k_{\mathrm{min}}$$, we choose to look at a rod of length $$L$$,
across which the function is a standing wave, meaning that
-the allowed values of $k$ must be as follows, where $m \in \mathbb{N}$:
+the allowed values of $$k$$ must be as follows, where $$m \in \mathbb{N}$$:
$$\begin{aligned}
\lambda = \frac{2 L}{m}
@@ -79,9 +79,9 @@ $$\begin{aligned}
k = \frac{2 \pi}{\lambda} = \frac{m \pi}{L}
\end{aligned}$$
-Take the smallest option $m = 1$,
-such that $k_{\mathrm{min}} = \pi / L$,
-the 1D density of states $g_1(k)$ is:
+Take the smallest option $$m = 1$$,
+such that $$k_{\mathrm{min}} = \pi / L$$,
+the 1D density of states $$g_1(k)$$ is:
$$\begin{aligned}
\boxed{
@@ -91,8 +91,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-In 2D, the number of states within a range $k$ of the
-origin is the area of a circle with radius $k$:
+In 2D, the number of states within a range $$k$$ of the
+origin is the area of a circle with radius $$k$$:
$$\begin{aligned}
\Omega_2(k)
@@ -100,8 +100,8 @@ $$\begin{aligned}
\end{aligned}$$
Analogously to the 1D case,
-we take the system to be a square of side $L$,
-so $k_{\mathrm{min}} = \pi / L$ again.
+we take the system to be a square of side $$L$$,
+so $$k_{\mathrm{min}} = \pi / L$$ again.
The density of states then becomes:
$$\begin{aligned}
@@ -112,14 +112,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-In 3D, the number of states is the volume of a sphere with radius $k$:
+In 3D, the number of states is the volume of a sphere with radius $$k$$:
$$\begin{aligned}
\Omega_3(k)
= \frac{4 \pi}{3} k^3
\end{aligned}$$
-For a cube with side $L$, we once again find $k_{\mathrm{min}} = \pi / L$.
+For a cube with side $$L$$, we once again find $$k_{\mathrm{min}} = \pi / L$$.
We thus get:
$$\begin{aligned}
@@ -130,17 +130,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-All these expressions contain the characteristic length/area/volume $L^n$,
+All these expressions contain the characteristic length/area/volume $$L^n$$,
and therefore give the number of states in that region only.
-Keep in mind that $L$ is free to choose;
+Keep in mind that $$L$$ is free to choose;
it need not be the physical size of the system.
In fact, we typically want the density of states
per unit length/area/volume,
-so we can just set $L = 1$ in our preferred unit of distance.
+so we can just set $$L = 1$$ in our preferred unit of distance.
If the system is infinitely large, or if it has periodic boundaries,
-then $k$ becomes a continuous variable and $k_\mathrm{min} \to 0$.
-But again, $L$ is arbitrary,
+then $$k$$ becomes a continuous variable and $$k_\mathrm{min} \to 0$$.
+But again, $$L$$ is arbitrary,
so a finite value can be chosen.
diff --git a/source/know/concept/density-operator/index.md b/source/know/concept/density-operator/index.md
index d2042ef..ffc5444 100644
--- a/source/know/concept/density-operator/index.md
+++ b/source/know/concept/density-operator/index.md
@@ -9,18 +9,18 @@ layout: "concept"
---
In quantum mechanics, the expectation value of an observable
-$\expval{\hat{L}}$ represents the average result from measuring
-$\hat{L}$ on a large number of systems (an **ensemble**)
-prepared in the same state $\Ket{\Psi}$,
+$$\expval{\hat{L}}$$ represents the average result from measuring
+$$\hat{L}$$ on a large number of systems (an **ensemble**)
+prepared in the same state $$\Ket{\Psi}$$,
known as a **pure ensemble** or (somewhat confusingly) **pure state**.
But what if the systems of the ensemble are not all in the same state?
To work with such a **mixed ensemble** or **mixed state**,
-the **density operator** $\hat{\rho}$ or **density matrix** (in a basis) is useful.
-It is defined as follows, where $p_n$ is the probability
-that the system is in state $\Ket{\Psi_n}$,
+the **density operator** $$\hat{\rho}$$ or **density matrix** (in a basis) is useful.
+It is defined as follows, where $$p_n$$ is the probability
+that the system is in state $$\Ket{\Psi_n}$$,
i.e. the proportion of systems in the ensemble that are
-in state $\Ket{\Psi_n}$:
+in state $$\Ket{\Psi_n}$$:
$$\begin{aligned}
\boxed{
@@ -29,18 +29,18 @@ $$\begin{aligned}
}
\end{aligned}$$
-Do not let is this form fool you into thinking that $\hat{\rho}$ is diagonal:
-$\Ket{\Psi_n}$ need not be basis vectors.
-Instead, the matrix elements of $\hat{\rho}$ are found as usual,
-where $\Ket{j}$ and $\Ket{k}$ are basis vectors:
+Do not let is this form fool you into thinking that $$\hat{\rho}$$ is diagonal:
+$$\Ket{\Psi_n}$$ need not be basis vectors.
+Instead, the matrix elements of $$\hat{\rho}$$ are found as usual,
+where $$\Ket{j}$$ and $$\Ket{k}$$ are basis vectors:
$$\begin{aligned}
\matrixel{j}{\hat{\rho}}{k}
= \sum_{n} p_n \Inprod{j}{\Psi_n} \Inprod{\Psi_n}{k}
\end{aligned}$$
-However, from the special case where $\Ket{\Psi_n}$ are indeed basis vectors,
-we can conclude that $\hat{\rho}$ is positive semidefinite and Hermitian,
+However, from the special case where $$\Ket{\Psi_n}$$ are indeed basis vectors,
+we can conclude that $$\hat{\rho}$$ is positive semidefinite and Hermitian,
and that its trace (i.e. the total probability) is 100%:
$$\begin{gathered}
@@ -58,26 +58,26 @@ $$\begin{gathered}
\end{gathered}$$
These properties are preserved by all changes of basis.
-If the ensemble is purely $\Ket{\Psi}$,
-then $\hat{\rho}$ is given by a single state vector:
+If the ensemble is purely $$\Ket{\Psi}$$,
+then $$\hat{\rho}$$ is given by a single state vector:
$$\begin{aligned}
\hat{\rho} = \Ket{\Psi} \Bra{\Psi}
\end{aligned}$$
-From the special case where $\Ket{\Psi}$ is a basis vector,
+From the special case where $$\Ket{\Psi}$$ is a basis vector,
we can conclude that for a pure ensemble,
-$\hat{\rho}$ is idempotent, which means that:
+$$\hat{\rho}$$ is idempotent, which means that:
$$\begin{aligned}
\hat{\rho}^2 = \hat{\rho}
\end{aligned}$$
-This can be used to find out whether a given $\hat{\rho}$
+This can be used to find out whether a given $$\hat{\rho}$$
represents a pure or mixed ensemble.
-Next, we define the ensemble average $\expval{\hat{O}}$
-as the mean of the expectation values of $\hat{O}$ for states in the ensemble.
+Next, we define the ensemble average $$\expval{\hat{O}}$$
+as the mean of the expectation values of $$\hat{O}$$ for states in the ensemble.
We use the same notation as for the pure expectation value,
since this is only a small extension of the concept to mixed ensembles.
It is calculated like so:
@@ -91,7 +91,7 @@ $$\begin{aligned}
\end{aligned}$$
To prove the latter,
-we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so:
+we write out the trace $$\mathrm{Tr}$$ as the sum of the diagonal elements, so:
$$\begin{aligned}
\mathrm{Tr}(\hat{\rho} \hat{O})
@@ -104,7 +104,7 @@ $$\begin{aligned}
\end{aligned}$$
In both the pure and mixed cases,
-if the state probabilities $p_n$ are constant with respect to time,
+if the state probabilities $$p_n$$ are constant with respect to time,
then the evolution of the ensemble obeys the **Von Neumann equation**:
$$\begin{aligned}
@@ -115,7 +115,7 @@ $$\begin{aligned}
This equivalent to the Schrödinger equation:
one can be derived from the other.
-We differentiate $\hat{\rho}$ with the product rule,
+We differentiate $$\hat{\rho}$$ with the product rule,
and then substitute the opposite side of the Schrödinger equation:
$$\begin{aligned}
diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md
index 9745959..b89d5da 100644
--- a/source/know/concept/detailed-balance/index.md
+++ b/source/know/concept/detailed-balance/index.md
@@ -24,9 +24,9 @@ since all net transition rates are zero.
We will focus on the case where both time and the state space are continuous.
Given some initial conditions,
assume that a component's trajectory can be described
-as an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
-with a time-independent drift $f$ and intensity $g$,
-and with a probability density $\phi(t, x)$ governed by the
+as an [Itō diffusion](/know/concept/ito-calculus/) $$X_t$$
+with a time-independent drift $$f$$ and intensity $$g$$,
+and with a probability density $$\phi(t, x)$$ governed by the
[forward Kolmogorov equation](/know/concept/kolmogorov-equations/)
(in 3D):
@@ -37,7 +37,7 @@ $$\begin{aligned}
We start by demanding **stationarity**,
which is a weaker condition than detailed balance.
-We want the probability $P$ of being in an arbitrary state volume $V$
+We want the probability $$P$$ of being in an arbitrary state volume $$V$$
to be constant in time:
$$\begin{aligned}
@@ -56,11 +56,11 @@ $$\begin{aligned}
= - \oint_{\partial V} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
\end{aligned}$$
-In other words, the "flow" of probability *into* the volume $V$
-is equal to the flow *out of* $V$.
+In other words, the "flow" of probability *into* the volume $$V$$
+is equal to the flow *out of* $$V$$.
If such a probability density exists,
-it is called a **stationary distribution** $\phi(t, x) = \pi(x)$.
-Because $V$ was arbitrary, $\pi$ can be found by solving:
+it is called a **stationary distribution** $$\phi(t, x) = \pi(x)$$.
+Because $$V$$ was arbitrary, $$\pi$$ can be found by solving:
$$\begin{aligned}
0
@@ -70,7 +70,7 @@ $$\begin{aligned}
Therefore, stationarity means that the state transition rates are constant.
To get detailed balance, however, we demand that
the transition rates are zero everywhere:
-the probability flux through an arbitrary surface $S$ must vanish
+the probability flux through an arbitrary surface $$S$$ must vanish
(compare to closed surface integral above):
$$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
= - \int_{S} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
\end{aligned}$$
-And since $S$ is arbitrary, this is only satisfied if the flux is trivially zero
+And since $$S$$ is arbitrary, this is only satisfied if the flux is trivially zero
(the above justification can easily be repeated in 1D, 2D, 4D, etc.):
$$\begin{aligned}
@@ -93,7 +93,7 @@ but fortunately often satisfied in practice.
The fact that a system in detailed balance appears "frozen"
implies it is **time-reversible**,
meaning its statistics are the same for both directions of time.
-Formally, given two arbitrary functions $h(x)$ and $k(x)$,
+Formally, given two arbitrary functions $$h(x)$$ and $$k(x)$$,
we have the property:
$$\begin{aligned}
@@ -109,9 +109,9 @@ $$\begin{aligned}
Consider the following weighted inner product,
-whose weight function is a stationary distribution $\pi$
+whose weight function is a stationary distribution $$\pi$$
satisfying detailed balance,
-where $\hat{L}$ is the Kolmogorov operator:
+where $$\hat{L}$$ is the Kolmogorov operator:
$$\begin{aligned}
\inprod{\hat{L} h}{k}_\pi
@@ -128,8 +128,8 @@ $$\begin{aligned}
= -\nabla \cdot (\vb{u} \pi k - D k \nabla \pi - D \pi \nabla k)
\end{aligned}$$
-Since $\pi$ is stationary by definition,
-we know that $\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$,
+Since $$\pi$$ is stationary by definition,
+we know that $$\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$$,
meaning:
$$\begin{aligned}
@@ -138,7 +138,7 @@ $$\begin{aligned}
= \nabla \pi \cdot (D \nabla k) + \pi \nabla \cdot (D \nabla k)
\end{aligned}$$
-Detailed balance demands that $\vb{u} \pi = D \nabla \pi$,
+Detailed balance demands that $$\vb{u} \pi = D \nabla \pi$$,
leading to the following:
$$\begin{aligned}
@@ -150,9 +150,9 @@ $$\begin{aligned}
= \pi \hat{L}\{k\}
\end{aligned}$$
-Where we recognized the definition of $\hat{L}$
+Where we recognized the definition of $$\hat{L}$$
from the backward Kolmogorov equation.
-Now that we have established that $\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$,
+Now that we have established that $$\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$$,
we return to the inner product:
$$\begin{aligned}
@@ -170,7 +170,7 @@ $$\begin{aligned}
Now, consider the time evolution of the
[conditional expectation](/know/concept/conditional-expectation/)
-$\mathbf{E}\big[ k(X_t) | X_0 \big]$:
+$$\mathbf{E}\big[ k(X_t) | X_0 \big]$$:
$$\begin{aligned}
\pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big]
@@ -184,10 +184,10 @@ $$\begin{aligned}
Where we used the forward Kolmogorov equation
and the definition of an adjoint operator.
-Therefore, since the expectation $\mathbf{E}$
-does not explicitly depend on $t$ (only implicitly via $X_t$),
+Therefore, since the expectation $$\mathbf{E}$$
+does not explicitly depend on $$t$$ (only implicitly via $$X_t$$),
we can naively move the differentiation inside
-(only valid within $\mathbf{E}$):
+(only valid within $$\mathbf{E}$$):
$$\begin{aligned}
\pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big]
@@ -195,9 +195,9 @@ $$\begin{aligned}
= \mathbf{E}\bigg[ \hat{L}\{k(X_0)\} \bigg| X_0 \bigg]
\end{aligned}$$
-A differential equation of the form $\ipdv{k}{t} = \hat{L}\{k(t, x)\}$
-for a time-independent operator $\hat{L}$
-has a general solution $k(t, x) = \exp(t \hat{L})\{k(0,x)\}$,
+A differential equation of the form $$\ipdv{k}{t} = \hat{L}\{k(t, x)\}$$
+for a time-independent operator $$\hat{L}$$
+has a general solution $$k(t, x) = \exp(t \hat{L})\{k(0,x)\}$$,
therefore:
$$\begin{aligned}
@@ -220,8 +220,8 @@ $$\begin{aligned}
= \mathbf{E}\big[ h(X_t) \: k(X_0) \big]
\end{aligned}$$
-Where the integral gave the expectation value at $X_0$,
-since $\pi$ does not change in time.
+Where the integral gave the expectation value at $$X_0$$,
+since $$\pi$$ does not change in time.
diff --git a/source/know/concept/deutsch-jozsa-algorithm/index.md b/source/know/concept/deutsch-jozsa-algorithm/index.md
index c2bd1e6..bbdd58d 100644
--- a/source/know/concept/deutsch-jozsa-algorithm/index.md
+++ b/source/know/concept/deutsch-jozsa-algorithm/index.md
@@ -13,40 +13,40 @@ were first to prove that quantum computers can
solve certain problems more efficiently
than any classical system.
-Given an unknown "black box" binary function $f(x)$ of one or more bits $x$,
-the goal is determine whether $f$ is
-**constant** (i.e. $f(x)$ is the same for all $x$)
-or **balanced** (i.e. exactly 50% of all $x$-values yield $f(x) = 0$,
-and the other 50% yield $f(x) = 1$).
-We can query $f$ as many times as we want with inputs of our choice,
+Given an unknown "black box" binary function $$f(x)$$ of one or more bits $$x$$,
+the goal is determine whether $$f$$ is
+**constant** (i.e. $$f(x)$$ is the same for all $$x$$)
+or **balanced** (i.e. exactly 50% of all $$x$$-values yield $$f(x) = 0$$,
+and the other 50% yield $$f(x) = 1$$).
+We can query $$f$$ as many times as we want with inputs of our choice,
but we want to solve the problem using as few queries as possible.
The problem is extremely artificial and of no practical use,
but quantum computers can solve it with a single query,
-while classical computers need up to $2^{N - 1} + 1$ queries
-for an $N$-bit $x$.
+while classical computers need up to $$2^{N - 1} + 1$$ queries
+for an $$N$$-bit $$x$$.
## Deutsch algorithm
The Deutsch algorithm handles the simplest case,
-where $x$ is only a single bit.
-Only four $f$ exist:
+where $$x$$ is only a single bit.
+Only four $$f$$ exist:
-+ **Constant**: $(f(0) = f(1) = 0)$ or $(f(0) = f(1) = 1)$.
-+ **Balanced**: $(f(0) = 0, f(1) = 1)$, or $(f(0) = 1, f(1) = 0)$.
++ **Constant**: $$(f(0) = f(1) = 0)$$ or $$(f(0) = f(1) = 1)$$.
++ **Balanced**: $$(f(0) = 0, f(1) = 1)$$, or $$(f(0) = 1, f(1) = 0)$$.
-In other words, we only need to determine if $f(0) = f(1)$ or $f(0) \neq f(1)$.
+In other words, we only need to determine if $$f(0) = f(1)$$ or $$f(0) \neq f(1)$$.
To do this, we use the following quantum circuit,
-where $U_f$ is the oracle we query:
+where $$U_f$$ is the oracle we query:
Due to unitarity constraints,
-the action of $U_f$ is defined to be as follows,
-with $\oplus$ meaning XOR:
+the action of $$U_f$$ is defined to be as follows,
+with $$\oplus$$ meaning XOR:
$$\begin{aligned}
\Ket{x} \Ket{y}
@@ -54,8 +54,8 @@ $$\begin{aligned}
\Ket{x} \Ket{y \oplus f(x)}
\end{aligned}$$
-Starting on the left from two qubits $\Ket{0}$ and $\Ket{1}$,
-we apply the [Hadamard gate](/know/concept/quantum-gate/) $H$ to both:
+Starting on the left from two qubits $$\Ket{0}$$ and $$\Ket{1}$$,
+we apply the [Hadamard gate](/know/concept/quantum-gate/) $$H$$ to both:
$$\begin{aligned}
\Ket{0} \Ket{1}
@@ -64,7 +64,7 @@ $$\begin{aligned}
= \frac{1}{2} \Big( \Ket{0} + \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
\end{aligned}$$
-Feeding this result into the oracle $U_f$ then leads us to:
+Feeding this result into the oracle $$U_f$$ then leads us to:
$$\begin{aligned}
\to \boxed{U_f} \to \quad
@@ -73,7 +73,7 @@ $$\begin{aligned}
\end{aligned}$$
The parenthesized superpositions can be reduced.
-Assuming that $f(b) = 0$, we notice:
+Assuming that $$f(b) = 0$$, we notice:
$$\begin{aligned}
\Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
@@ -81,7 +81,7 @@ $$\begin{aligned}
= \Ket{0} - \Ket{1}
\end{aligned}$$
-On the other hand, if we assume that $f(b) = 1$,
+On the other hand, if we assume that $$f(b) = 1$$,
we get the opposite result:
$$\begin{aligned}
@@ -90,7 +90,7 @@ $$\begin{aligned}
= - \big(\Ket{0} - \Ket{1}\big)
\end{aligned}$$
-We can thus combine both cases, $f(b) = 0$ or $f(b) = 1$,
+We can thus combine both cases, $$f(b) = 0$$ or $$f(b) = 1$$,
into the following single expression:
$$\begin{aligned}
@@ -106,7 +106,7 @@ $$\begin{aligned}
\frac{1}{2} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
\end{aligned}$$
-The second qubit in state $\Ket{-}$ is garbage; it is no longer of interest.
+The second qubit in state $$\Ket{-}$$ is garbage; it is no longer of interest.
The first qubit is given by:
$$\begin{aligned}
@@ -114,10 +114,10 @@ $$\begin{aligned}
= \frac{(-1)^{f(0)}}{\sqrt{2}} \Big( \Ket{0} + (-1)^{f(0) \oplus f(1)} \Ket{1} \Big)
\end{aligned}$$
-If $f$ is constant, then $f(0) \oplus f(1) = 0$,
-meaning this state is $(-1)^{f(0)} \Ket{+}$.
-On the other hand, if $f$ is balanced, then $f(0) \oplus f(1) = 1$,
-meaning this state is $(-1)^{f(0)} \Ket{-}$.
+If $$f$$ is constant, then $$f(0) \oplus f(1) = 0$$,
+meaning this state is $$(-1)^{f(0)} \Ket{+}$$.
+On the other hand, if $$f$$ is balanced, then $$f(0) \oplus f(1) = 1$$,
+meaning this state is $$(-1)^{f(0)} \Ket{-}$$.
Taking the Hadamard transform of this qubit therefore yields:
$$\begin{aligned}
@@ -125,19 +125,19 @@ $$\begin{aligned}
(-1)^{f(0)} \Ket{f(0) \oplus f(1)}
\end{aligned}$$
-Depending on whether $f$ is constant or balanced,
-the mearurement outcome of this state will be $\Ket{0}$ or $\Ket{1}$
+Depending on whether $$f$$ is constant or balanced,
+the mearurement outcome of this state will be $$\Ket{0}$$ or $$\Ket{1}$$
with 100\% probability. We have solved the problem!
-Note that we only consulted the oracle (i.e. applied $U_f$) once.
+Note that we only consulted the oracle (i.e. applied $$U_f$$) once.
A classical computer would need to query it twice,
-once with input $x = 0$, and again with $x = 1$.
+once with input $$x = 0$$, and again with $$x = 1$$.
## Full Deutsch-Jozsa algorithm
-The Deutsch-Jozsa algorithm generalizes the above to $N$-bit inputs $x$.
-We are promised that $f(x)$ is either constant or balanced;
+The Deutsch-Jozsa algorithm generalizes the above to $$N$$-bit inputs $$x$$.
+We are promised that $$f(x)$$ is either constant or balanced;
other possibilities are assumed to be impossible.
This algorithm is then implemented by the following quantum circuit:
@@ -145,8 +145,8 @@ This algorithm is then implemented by the following quantum circuit:
-There are $N$ qubits in initial state $\Ket{0}$, and one in $\Ket{1}$.
-For clarity, the oracle $U_f$ works like so:
+There are $$N$$ qubits in initial state $$\Ket{0}$$, and one in $$\Ket{1}$$.
+For clarity, the oracle $$U_f$$ works like so:
$$\begin{aligned}
\Ket{x_1} \Ket{x_2} \cdots \Ket{x_N} \Ket{y}
@@ -154,7 +154,7 @@ $$\begin{aligned}
\Ket{x_1} \cdots \Ket{x_N} \Ket{y \oplus f(x_1, ..., x_N)}
\end{aligned}$$
-Applying the $N + 1$ Hadamard gates to the initial state
+Applying the $$N + 1$$ Hadamard gates to the initial state
yields the following superposition:
$$\begin{aligned}
@@ -164,9 +164,9 @@ $$\begin{aligned}
= \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x} \Ket{-}
\end{aligned}$$
-Where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$ denotes a classical binary state.
-For example, if $x = 5 = 2^0 + 2^2$ in the summation,
-then $\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$
+Where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$ denotes a classical binary state.
+For example, if $$x = 5 = 2^0 + 2^2$$ in the summation,
+then $$\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$$
(from least to most significant).
We give this state to the oracle,
@@ -178,8 +178,8 @@ $$\begin{aligned}
\frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \Ket{x} \Ket{-}
\end{aligned}$$
-The last qubit $\Ket{-}$ is garbage.
-Next, applying the Hadamard transform to the other $N$ gives:
+The last qubit $$\Ket{-}$$ is garbage.
+Next, applying the Hadamard transform to the other $$N$$ gives:
$$\begin{aligned}
\to \boxed{H^{\otimes N}} \to \quad
@@ -187,8 +187,8 @@ $$\begin{aligned}
\bigg( \frac{1}{\sqrt{2^N}} \sum_{y = 0}^{2^N - 1} (-1)^{x \cdot y} \Ket{y} \bigg)
\end{aligned}$$
-Where $x \cdot y$ is the bitwise dot product of the binary representations of $x$ and $y$,
-so, for example, if $N = 2$, then $x \cdot y = x_1 y_1 + x_2 y_2$.
+Where $$x \cdot y$$ is the bitwise dot product of the binary representations of $$x$$ and $$y$$,
+so, for example, if $$N = 2$$, then $$x \cdot y = x_1 y_1 + x_2 y_2$$.
Note that the above expression has not been reduced at all;
it follows from the definition of the Hadamard transform.
We can rewrite it like so:
@@ -200,24 +200,24 @@ $$\begin{aligned}
\end{aligned}$$
The parenthesized expression can be interpreted as the coefficients
-of a superposition of several $y$-values.
-Therefore, the probability that a measurement yields $y = 0$,
-i.e. $\Ket{y} = \Ket{0}^{\otimes N}$, is:
+of a superposition of several $$y$$-values.
+Therefore, the probability that a measurement yields $$y = 0$$,
+i.e. $$\Ket{y} = \Ket{0}^{\otimes N}$$, is:
$$\begin{aligned}
|c_0|^2
= \bigg| \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \bigg|^2
\end{aligned}$$
-The summation always contains an even number of terms, for all values of $N$.
-Consequently, if $f$ is constant, then $|c_0|^2 = |\!\pm\! 2^N / 2^N|^2 = 1$.
-Otherwise, if $f$ is balanced, all the terms cancel out, so we are left with $|c_0|^2 = 0$.
+The summation always contains an even number of terms, for all values of $$N$$.
+Consequently, if $$f$$ is constant, then $$|c_0|^2 = |\!\pm\! 2^N / 2^N|^2 = 1$$.
+Otherwise, if $$f$$ is balanced, all the terms cancel out, so we are left with $$|c_0|^2 = 0$$.
In other words, we reach the same result as the Deutsch algorithm:
-we only need to measure the $N$ qubits once;
-$f$ is constant if and only if all are zero.
+we only need to measure the $$N$$ qubits once;
+$$f$$ is constant if and only if all are zero.
The Deutsch-Jozsa algorithm needs only one oracle query to give an error-free result,
-whereas a classical computer needs $2^{N-1} + 1$ queries in the worst case;
+whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case;
a revolutionary discovery.
diff --git a/source/know/concept/dielectric-function/index.md b/source/know/concept/dielectric-function/index.md
index 30622f4..529ce2a 100644
--- a/source/know/concept/dielectric-function/index.md
+++ b/source/know/concept/dielectric-function/index.md
@@ -9,11 +9,11 @@ categories:
layout: "concept"
---
-The **dielectric function** or **relative permittivity** $\varepsilon_r$
+The **dielectric function** or **relative permittivity** $$\varepsilon_r$$
is a measure of how strongly a given medium counteracts
[electric fields](/know/concept/electric-field/) compared to a vacuum.
-Let $\vb{D}$ be the applied external field,
-and $\vb{E}$ the effective field inside the material:
+Let $$\vb{D}$$ be the applied external field,
+and $$\vb{E}$$ the effective field inside the material:
$$\begin{aligned}
\boxed{
@@ -21,15 +21,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\varepsilon_r$ is large, then $\vb{D}$ is strongly suppressed,
+If $$\varepsilon_r$$ is large, then $$\vb{D}$$ is strongly suppressed,
because the material's electrons and nuclei move to create an opposing field.
-In order for $\varepsilon_r$ to be well defined, we only consider linear media,
-where the induced polarization $\vb{P}$ is proportional to $\vb{E}$.
+In order for $$\varepsilon_r$$ to be well defined, we only consider linear media,
+where the induced polarization $$\vb{P}$$ is proportional to $$\vb{E}$$.
-We would like to find an alternative definition of $\varepsilon_r$.
-Consider that the usual electric fields $\vb{E}$, $\vb{D}$, and $\vb{P}$
+We would like to find an alternative definition of $$\varepsilon_r$$.
+Consider that the usual electric fields $$\vb{E}$$, $$\vb{D}$$, and $$\vb{P}$$
can each be written as the gradient of an electrostatic potential like so,
-where $\Phi_\mathrm{tot}$, $\Phi_\mathrm{ext}$ and $\Phi_\mathrm{ind}$
+where $$\Phi_\mathrm{tot}$$, $$\Phi_\mathrm{ext}$$ and $$\Phi_\mathrm{ind}$$
are the total, external and induced potentials, respectively:
$$\begin{aligned}
@@ -43,8 +43,8 @@ $$\begin{aligned}
= \varepsilon_0 \nabla \Phi_\mathrm{ind}
\end{aligned}$$
-Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$.
-Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$
+Such that $$\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$$.
+Inserting this into $$\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$$
then suggests defining:
$$\begin{aligned}
@@ -57,10 +57,10 @@ $$\begin{aligned}
## From induced charge density
-A common way to calculate $\varepsilon_r$ is from
-the induced charge density $\rho_\mathrm{ind}$,
+A common way to calculate $$\varepsilon_r$$ is from
+the induced charge density $$\rho_\mathrm{ind}$$,
i.e. the offset caused by the material's particles responding to the field.
-We start from [Gauss' law](/know/concept/maxwells-equations/) for $\vb{P}$:
+We start from [Gauss' law](/know/concept/maxwells-equations/) for $$\vb{P}$$:
$$\begin{aligned}
\nabla \cdot \vb{P}
@@ -77,8 +77,8 @@ $$\begin{aligned}
= V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
\end{aligned}$$
-Where $V(\vb{q})$ represents Coulomb interactions,
-and $V(0) = 0$ to ensure overall neutrality:
+Where $$V(\vb{q})$$ represents Coulomb interactions,
+and $$V(0) = 0$$ to ensure overall neutrality:
$$\begin{aligned}
V(\vb{q})
@@ -89,7 +89,7 @@ $$\begin{aligned}
\end{aligned}$$
The [convolution theorem](/know/concept/convolution-theorem/)
-then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain:
+then gives us the solution $$\Phi_\mathrm{ind}$$ in the $$\vb{r}$$-domain:
$$\begin{aligned}
\Phi_\mathrm{ind}(\vb{r})
@@ -97,13 +97,13 @@ $$\begin{aligned}
= \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$
-To proceed, we need to find an expression for $\rho_\mathrm{ind}$
-that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$,
+To proceed, we need to find an expression for $$\rho_\mathrm{ind}$$
+that is proportional to $$\Phi_\mathrm{tot}$$ or $$\Phi_\mathrm{ext}$$,
or some linear combination thereof.
Such an expression must exist for a linear material.
-Suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$,
-for some $C_\mathrm{ext}$, which may depend on $\vb{q}$. Then:
+Suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$$,
+for some $$C_\mathrm{ext}$$, which may depend on $$\vb{q}$$. Then:
$$\begin{aligned}
\Phi_\mathrm{tot}
@@ -115,8 +115,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$,
-for some quantity $C_\mathrm{tot}$, then:
+Similarly, suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$$,
+for some quantity $$C_\mathrm{tot}$$, then:
$$\begin{aligned}
\Phi_\mathrm{ext}
diff --git a/source/know/concept/diffie-hellman-key-exchange/index.md b/source/know/concept/diffie-hellman-key-exchange/index.md
index 951d5f5..4735209 100644
--- a/source/know/concept/diffie-hellman-key-exchange/index.md
+++ b/source/know/concept/diffie-hellman-key-exchange/index.md
@@ -13,30 +13,30 @@ when they can only communicate over an insecure channel.
The fundamental assumption of the Diffie-Hellman scheme,
upon which its security rests,
-is that the following function $f(n)$ is a **trapdoor function**,
-which means that calculating $f$ is easy,
-but its inverse $f^{-1}$ is extremely hard to find:
+is that the following function $$f(n)$$ is a **trapdoor function**,
+which means that calculating $$f$$ is easy,
+but its inverse $$f^{-1}$$ is extremely hard to find:
$$\begin{aligned}
f(n) = g^n \bmod p
\end{aligned}$$
-Where $n$ is a natural number, and $p$ is a prime.
-The natural number $g$ is a so-called *primitive root modulo* $p$.
-Importantly, $g$ and $p$ have been specifically chosen
-such that $f(n)$ can take any value in $\{1, ..., p \!-\! 1\}$
-for $n$ in $\{0, ..., p \!-\! 2\}$.
-The trapdoor assumption is that, given $g$, $p$ and $f(n)$,
-there is no efficient algorithm to recover $n$.
+Where $$n$$ is a natural number, and $$p$$ is a prime.
+The natural number $$g$$ is a so-called *primitive root modulo* $$p$$.
+Importantly, $$g$$ and $$p$$ have been specifically chosen
+such that $$f(n)$$ can take any value in $$\{1, ..., p \!-\! 1\}$$
+for $$n$$ in $$\{0, ..., p \!-\! 2\}$$.
+The trapdoor assumption is that, given $$g$$, $$p$$ and $$f(n)$$,
+there is no efficient algorithm to recover $$n$$.
Suppose that Alice and Bob want to exchange encrypted data in the future,
so they need to agree on an encryption key to use.
However, they can only exchange messages with each other over
an insecure channel, which is being eavesdropped.
-After they publicly agree on the values of $g$ and $p$,
-Alice and Bob each choose a secret number from $\{0, ..., p \!-\! 2\}$, respectively $a$ and $b$,
-and then privately calculate $A$ and $B$ as follows:
+After they publicly agree on the values of $$g$$ and $$p$$,
+Alice and Bob each choose a secret number from $$\{0, ..., p \!-\! 2\}$$, respectively $$a$$ and $$b$$,
+and then privately calculate $$A$$ and $$B$$ as follows:
$$\begin{aligned}
A = g^a \bmod p
@@ -44,9 +44,9 @@ $$\begin{aligned}
B = g^b \bmod p
\end{aligned}$$
-Finally, they transmit these numbers $A$ and $B$
+Finally, they transmit these numbers $$A$$ and $$B$$
to each other over the insecure connection,
-and then each side calculates $k$, which is the desired secret key:
+and then each side calculates $$k$$, which is the desired secret key:
$$\begin{aligned}
\boxed{
@@ -54,11 +54,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-The point is that $k$ includes both $a$ *and* $b$,
-but each side only needs to know *either* $a$ *or* $b$.
+The point is that $$k$$ includes both $$a$$ *and* $$b$$,
+but each side only needs to know *either* $$a$$ *or* $$b$$.
And, due to the trapdoor assumption,
-the eavesdropper knows $A$ and $B$,
-but cannot recover $a$ or $b$.
+the eavesdropper knows $$A$$ and $$B$$,
+but cannot recover $$a$$ or $$b$$.
This assumption is just that: an assumption.
So far, nobody has been able to prove or disprove it
diff --git a/source/know/concept/dirac-delta-function/index.md b/source/know/concept/dirac-delta-function/index.md
index 88a08cb..518eba1 100644
--- a/source/know/concept/dirac-delta-function/index.md
+++ b/source/know/concept/dirac-delta-function/index.md
@@ -8,10 +8,10 @@ categories:
layout: "concept"
---
-The **Dirac delta function** $\delta(x)$, often just the **delta function**,
+The **Dirac delta function** $$\delta(x)$$, often just the **delta function**,
is a function (or, more accurately, a [Schwartz distribution](/know/concept/schwartz-distribution/))
that is commonly used in physics.
-It is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is
+It is an infinitely narrow discontinuous "spike" at $$x = 0$$ whose area is
defined to be 1:
$$\begin{aligned}
@@ -35,7 +35,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-$\delta(x)$ is thus quite an effective weapon against integrals. This may not seem very
+$$\delta(x)$$ is thus quite an effective weapon against integrals. This may not seem very
useful due to its "unnatural" definition, but in fact it appears as the
limit of several reasonable functions:
@@ -57,7 +57,7 @@ $$\begin{aligned}
\:\:\propto\:\: \hat{\mathcal{F}}\{1\}
\end{aligned}$$
-When the argument of $\delta(x)$ is scaled, the delta function is itself scaled:
+When the argument of $$\delta(x)$$ is scaled, the delta function is itself scaled:
$$\begin{aligned}
\boxed{
@@ -70,17 +70,18 @@ $$\begin{aligned}
-Because it is symmetric, $\delta(s x) = \delta(|s| x)$.
-Then by substituting $\sigma = |s| x$:
+Because it is symmetric, $$\delta(s x) = \delta(|s| x)$$.
+Then by substituting $$\sigma = |s| x$$:
$$\begin{aligned}
\int \delta(|s| x) \dd{x}
&= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
\end{aligned}$$
+
-An even more impressive property is the behaviour of the derivative of $\delta(x)$:
+An even more impressive property is the behaviour of the derivative of $$\delta(x)$$:
$$\begin{aligned}
\boxed{
@@ -94,13 +95,14 @@ $$\begin{aligned}
Note which variable is used for the
-differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:
+differentiation, and that $$\delta'(x - \xi) = - \delta'(\xi - x)$$:
$$\begin{aligned}
\int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi}
&= \dv{}{x}\int f(\xi) \: \delta(x - \xi) \dd{x}
= f'(x)
\end{aligned}$$
+
diff --git a/source/know/concept/dirac-notation/index.md b/source/know/concept/dirac-notation/index.md
index 8861505..46cc325 100644
--- a/source/know/concept/dirac-notation/index.md
+++ b/source/know/concept/dirac-notation/index.md
@@ -12,16 +12,16 @@ layout: "concept"
without needing to worry about the space's representation. It is
basically the *lingua franca* of quantum mechanics.
-In Dirac notation there are **kets** $\Ket{V}$ from the Hilbert space
-$\mathbb{H}$ and **bras** $\Bra{V}$ from a dual $\mathbb{H}'$ of the
+In Dirac notation there are **kets** $$\Ket{V}$$ from the Hilbert space
+$$\mathbb{H}$$ and **bras** $$\Bra{V}$$ from a dual $$\mathbb{H}'$$ of the
former. Crucially, the bras and kets are from different Hilbert spaces
and therefore cannot be added, but every bra has a corresponding ket and
vice versa.
Bras and kets can be combined in two ways: the **inner product**
-$\Inprod{V}{W}$, which returns a scalar, and the **outer product**
-$\Ket{V} \Bra{W}$, which returns a mapping $\hat{L}$ from kets $\Ket{V}$
-to other kets $\Ket{V'}$, i.e. a linear operator. Recall that the
+$$\Inprod{V}{W}$$, which returns a scalar, and the **outer product**
+$$\Ket{V} \Bra{W}$$, which returns a mapping $$\hat{L}$$ from kets $$\Ket{V}$$
+to other kets $$\Ket{V'}$$, i.e. a linear operator. Recall that the
Hilbert inner product must satisfy:
$$\begin{aligned}
@@ -29,7 +29,7 @@ $$\begin{aligned}
\end{aligned}$$
So far, nothing has been said about the actual representation of bras or
-kets. If we represent kets as $N$-dimensional columns vectors, the
+kets. If we represent kets as $$N$$-dimensional columns vectors, the
corresponding bras are given by the kets' adjoints, i.e. their transpose
conjugates:
@@ -45,7 +45,7 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-The inner product $\Inprod{V}{W}$ is then just the familiar dot product $V \cdot W$:
+The inner product $$\Inprod{V}{W}$$ is then just the familiar dot product $$V \cdot W$$:
$$\begin{gathered}
\Inprod{V}{W}
@@ -60,7 +60,7 @@ $$\begin{gathered}
= v_1^* w_1 + ... + v_N^* w_N
\end{gathered}$$
-Meanwhile, the outer product $\Ket{V} \Bra{W}$ creates an $N \cross N$ matrix:
+Meanwhile, the outer product $$\Ket{V} \Bra{W}$$ creates an $$N \cross N$$ matrix:
$$\begin{gathered}
\Ket{V} \Bra{W}
@@ -80,9 +80,9 @@ $$\begin{gathered}
\end{bmatrix}
\end{gathered}$$
-If the kets are instead represented by functions $f(x)$ of
-$x \in [a, b]$, then the bras represent *functionals* $F[u(x)]$ which
-take an unknown function $u(x)$ as an argument and turn it into a scalar
+If the kets are instead represented by functions $$f(x)$$ of
+$$x \in [a, b]$$, then the bras represent *functionals* $$F[u(x)]$$ which
+take an unknown function $$u(x)$$ as an argument and turn it into a scalar
using integration:
$$\begin{aligned}
diff --git a/source/know/concept/dispersive-broadening/index.md b/source/know/concept/dispersive-broadening/index.md
index 816135a..4e4cf82 100644
--- a/source/know/concept/dispersive-broadening/index.md
+++ b/source/know/concept/dispersive-broadening/index.md
@@ -12,11 +12,11 @@ layout: "concept"
In optical fibers, **dispersive broadening** is a (linear) effect
where group velocity dispersion (GVD) "smears out" a pulse in the time domain
due to the different group velocities of its frequencies,
-since pulses always have a non-zero width in the $\omega$-domain.
+since pulses always have a non-zero width in the $$\omega$$-domain.
No new frequencies are created.
-A pulse envelope $A(z, t)$ inside a fiber must obey the nonlinear Schrödinger equation,
-where the parameters $\beta_2$ and $\gamma$ respectively
+A pulse envelope $$A(z, t)$$ inside a fiber must obey the nonlinear Schrödinger equation,
+where the parameters $$\beta_2$$ and $$\gamma$$ respectively
control dispersion and nonlinearity:
$$\begin{aligned}
@@ -24,7 +24,7 @@ $$\begin{aligned}
= i \pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma |A|^2 A
\end{aligned}$$
-We set $\gamma = 0$ to ignore all nonlinear effects,
+We set $$\gamma = 0$$ to ignore all nonlinear effects,
and consider a Gaussian initial condition:
$$\begin{aligned}
@@ -32,10 +32,10 @@ $$\begin{aligned}
= \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big)
\end{aligned}$$
-By [Fourier transforming](/know/concept/fourier-transform/) in $t$,
-the full analytical solution $A(z, t)$ is found to be as follows,
+By [Fourier transforming](/know/concept/fourier-transform/) in $$t$$,
+the full analytical solution $$A(z, t)$$ is found to be as follows,
where it can be seen that the amplitude
-decreases and the width increases with $z$:
+decreases and the width increases with $$z$$:
$$\begin{aligned}
A(z,t) = \sqrt{\frac{P_0}{1 - i \beta_2 z / T_0^2}}
@@ -43,9 +43,9 @@ $$\begin{aligned}
\end{aligned}$$
To quantify the strength of dispersive effects,
-we define the dispersion length $L_D$
-as the distance over which the half-width at $1/e$ of maximum power
-(initially $T_0$) increases by a factor of $\sqrt{2}$:
+we define the dispersion length $$L_D$$
+as the distance over which the half-width at $$1/e$$ of maximum power
+(initially $$T_0$$) increases by a factor of $$\sqrt{2}$$:
$$\begin{aligned}
T_0 \sqrt{1 + \beta_2^2 L_D^2 / T_0^4} = T_0 \sqrt{2}
@@ -56,17 +56,17 @@ $$\begin{aligned}
\end{aligned}$$
This phenomenon is illustrated below for our example of a Gaussian pulse
-with parameter values $T_0 = 1\:\mathrm{ps}$, $P_0 = 1\:\mathrm{kW}$,
-$\beta_2 = -10 \:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0$:
+with parameter values $$T_0 = 1\:\mathrm{ps}$$, $$P_0 = 1\:\mathrm{kW}$$,
+$$\beta_2 = -10 \:\mathrm{ps}^2/\mathrm{m}$$ and $$\gamma = 0$$:
-The **instantaneous frequency** $\omega_\mathrm{GVD}(z, t)$,
+The **instantaneous frequency** $$\omega_\mathrm{GVD}(z, t)$$,
which describes the dominant angular frequency at a given point in the time domain,
is found to be as follows for the Gaussian pulse,
-where $\phi(z, t)$ is the phase of $A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$:
+where $$\phi(z, t)$$ is the phase of $$A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$$:
$$\begin{aligned}
\omega_{\mathrm{GVD}}(z,t)
@@ -74,20 +74,20 @@ $$\begin{aligned}
= \frac{\beta_2 z / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} \frac{t}{T_0^2}
\end{aligned}$$
-This expression is linear in time, and depending on the sign of $\beta_2$,
+This expression is linear in time, and depending on the sign of $$\beta_2$$,
frequencies on one side of the pulse arrive first,
and those on the other side arrive last.
-The effect is stronger for smaller $T_0$:
+The effect is stronger for smaller $$T_0$$:
this makes sense, since short pulses are spectrally wider.
The interaction between dispersion and [self-phase modulation](/know/concept/self-phase-modulation/)
leads to many interesting effects,
such as [modulational instability](/know/concept/modulational-instability/)
and [optical wave breaking](/know/concept/optical-wave-breaking/).
-Of great importance is the sign of $\beta_2$:
-in the **anomalous dispersion regime** ($\beta_2 < 0$),
+Of great importance is the sign of $$\beta_2$$:
+in the **anomalous dispersion regime** ($$\beta_2 < 0$$),
lower frequencies travel more slowly than higher ones,
-and vice versa in the **normal dispersion regime** ($\beta_2 > 0$).
+and vice versa in the **normal dispersion regime** ($$\beta_2 > 0$$).
diff --git a/source/know/concept/drude-model/index.md b/source/know/concept/drude-model/index.md
index 28e6dc2..b175e64 100644
--- a/source/know/concept/drude-model/index.md
+++ b/source/know/concept/drude-model/index.md
@@ -18,19 +18,19 @@ as found in metals and doped semiconductors.
An [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
has an oscillating [electric field](/know/concept/electric-field/)
-$E(t) = E_0 \exp(- i \omega t)$
+$$E(t) = E_0 \exp(- i \omega t)$$
that exerts a force on the charge carriers,
-which have mass $m$ and charge $q$.
+which have mass $$m$$ and charge $$q$$.
They thus obey the following equation of motion,
-where $\gamma$ is a frictional damping coefficient:
+where $$\gamma$$ is a frictional damping coefficient:
$$\begin{aligned}
m \dvn{2}{x}{t} + m \gamma \dv{x}{t}
= q E_0 \exp(- i \omega t)
\end{aligned}$$
-Inserting the ansatz $x(t) = x_0 \exp(- i \omega t)$
-and isolating for the displacement $x_0$ yields:
+Inserting the ansatz $$x(t) = x_0 \exp(- i \omega t)$$
+and isolating for the displacement $$x_0$$ yields:
$$\begin{aligned}
- x_0 m \omega^2 - i x_0 m \gamma \omega
@@ -40,10 +40,10 @@ $$\begin{aligned}
= - \frac{q E_0}{m (\omega^2 + i \gamma \omega)}
\end{aligned}$$
-The polarization density $P(t)$ is therefore as shown below.
-Note that the dipole moment $p$ goes from negative to positive,
-and the electric field $E$ from positive to negative.
-Let $N$ be the density of carriers in the gas, then:
+The polarization density $$P(t)$$ is therefore as shown below.
+Note that the dipole moment $$p$$ goes from negative to positive,
+and the electric field $$E$$ from positive to negative.
+Let $$N$$ be the density of carriers in the gas, then:
$$\begin{aligned}
P(t)
@@ -52,8 +52,8 @@ $$\begin{aligned}
= - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} E(t)
\end{aligned}$$
-The electric displacement field $D$ is thus as follows,
-where $\varepsilon_r$ is the unknown relative permittivity of the gas,
+The electric displacement field $$D$$ is thus as follows,
+where $$\varepsilon_r$$ is the unknown relative permittivity of the gas,
which we will find shortly:
$$\begin{aligned}
@@ -63,8 +63,8 @@ $$\begin{aligned}
= \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) E
\end{aligned}$$
-The parenthesized expression is the desired dielectric function $\varepsilon_r$,
-which depends on $\omega$:
+The parenthesized expression is the desired dielectric function $$\varepsilon_r$$,
+which depends on $$\omega$$:
$$\begin{aligned}
\boxed{
@@ -82,26 +82,26 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\gamma = 0$, then $\varepsilon_r$ is
-negative $\omega < \omega_p$,
-positive for $\omega > \omega_p$,
-and zero for $\omega = \omega_p$.
+If $$\gamma = 0$$, then $$\varepsilon_r$$ is
+negative $$\omega < \omega_p$$,
+positive for $$\omega > \omega_p$$,
+and zero for $$\omega = \omega_p$$.
Respectively, this leads to
-an imaginary index $\sqrt{\varepsilon_r}$ (high absorption),
-a real index tending to $1$ (transparency),
+an imaginary index $$\sqrt{\varepsilon_r}$$ (high absorption),
+a real index tending to $$1$$ (transparency),
and the possibility of self-sustained plasma oscillations.
-For metals, $\omega_p$ lies in the UV.
+For metals, $$\omega_p$$ lies in the UV.
-We can refine this result for $\varepsilon_r$,
-by recognizing the (mean) velocity $v = \idv{x}{t}$,
+We can refine this result for $$\varepsilon_r$$,
+by recognizing the (mean) velocity $$v = \idv{x}{t}$$,
and rewriting the equation of motion accordingly:
$$\begin{aligned}
m \dv{v}{t} + m \gamma v = q E(t)
\end{aligned}$$
-Note that $m v$ is simply the momentum $p$.
-We define the **momentum scattering time** $\tau \equiv 1 / \gamma$,
+Note that $$m v$$ is simply the momentum $$p$$.
+We define the **momentum scattering time** $$\tau \equiv 1 / \gamma$$,
which represents the average time between collisions,
where each collision resets the involved particles' momentums to zero.
Or, more formally:
@@ -111,9 +111,9 @@ $$\begin{aligned}
= - \frac{p}{\tau} + q E
\end{aligned}$$
-Returning to the equation for the mean velocity $v$,
-we insert the ansatz $v(t) = v_0 \exp(- i \omega t)$,
-for the same electric field $E(t) = E_0 \exp(-i \omega t)$ as before:
+Returning to the equation for the mean velocity $$v$$,
+we insert the ansatz $$v(t) = v_0 \exp(- i \omega t)$$,
+for the same electric field $$E(t) = E_0 \exp(-i \omega t)$$ as before:
$$\begin{aligned}
- i m \omega v_0 + \frac{m}{\tau} v_0 = q E_0
@@ -121,7 +121,7 @@ $$\begin{aligned}
v_0 = \frac{q \tau}{m (1 - i \omega \tau)} E_0
\end{aligned}$$
-From $v(t)$, we find the resulting average current density $J(t)$ to be as follows:
+From $$v(t)$$, we find the resulting average current density $$J(t)$$ to be as follows:
$$\begin{aligned}
J(t)
@@ -129,8 +129,8 @@ $$\begin{aligned}
= \sigma E(t)
\end{aligned}$$
-Where $\sigma(\omega)$ is the **AC conductivity**,
-which depends on the **DC conductivity** $\sigma_0$:
+Where $$\sigma(\omega)$$ is the **AC conductivity**,
+which depends on the **DC conductivity** $$\sigma_0$$:
$$\begin{aligned}
\boxed{
@@ -145,7 +145,7 @@ $$\begin{aligned}
\end{aligned}$$
We can use these quantities to rewrite
-the dielectric function $\varepsilon_r$ from earlier:
+the dielectric function $$\varepsilon_r$$ from earlier:
$$\begin{aligned}
\boxed{
@@ -164,12 +164,12 @@ which can be treated as free particles
moving in the bands of the material.
The Drude model can also be used in this case,
-by replacing the actual carrier mass $m$
-by the effective mass $m^*$.
+by replacing the actual carrier mass $$m$$
+by the effective mass $$m^*$$.
Furthermore, semiconductors already have
-a high intrinsic permittivity $\varepsilon_{\mathrm{int}}$
+a high intrinsic permittivity $$\varepsilon_{\mathrm{int}}$$
before the dopant is added,
-so the diplacement field $D$ is:
+so the diplacement field $$D$$ is:
$$\begin{aligned}
D
@@ -177,9 +177,9 @@ $$\begin{aligned}
= \varepsilon_{\mathrm{int}} \varepsilon_0 E - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} E
\end{aligned}$$
-Where $P_{\mathrm{int}}$ is the intrinsic undoped polarization,
-and $P_{\mathrm{free}}$ is the contribution of the free carriers.
-The dielectric function $\varepsilon_r(\omega)$ is therefore given by:
+Where $$P_{\mathrm{int}}$$ is the intrinsic undoped polarization,
+and $$P_{\mathrm{free}}$$ is the contribution of the free carriers.
+The dielectric function $$\varepsilon_r(\omega)$$ is therefore given by:
$$\begin{aligned}
\boxed{
@@ -188,8 +188,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the plasma frequency $\omega_p$ has been redefined as follows
-to include $\varepsilon_\mathrm{int}$:
+Where the plasma frequency $$\omega_p$$ has been redefined as follows
+to include $$\varepsilon_\mathrm{int}$$:
$$\begin{aligned}
\boxed{
@@ -198,14 +198,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-The meaning of $\omega_p$ is the same as for metals,
-with high absorption for $\omega < \omega_p$.
-However, due to the lower carrier density $N$ in a semiconductor,
-$\omega_p$ lies in the IR rather than UV.
+The meaning of $$\omega_p$$ is the same as for metals,
+with high absorption for $$\omega < \omega_p$$.
+However, due to the lower carrier density $$N$$ in a semiconductor,
+$$\omega_p$$ lies in the IR rather than UV.
-However, instead of asymptotically going to $1$ for $\omega > \omega_p$ like a metal,
-$\varepsilon_r$ tends to $\varepsilon_\mathrm{int}$ instead,
-and crosses $1$ along the way,
+However, instead of asymptotically going to $$1$$ for $$\omega > \omega_p$$ like a metal,
+$$\varepsilon_r$$ tends to $$\varepsilon_\mathrm{int}$$ instead,
+and crosses $$1$$ along the way,
at which point the reflectivity is zero.
This occurs at:
@@ -214,9 +214,9 @@ $$\begin{aligned}
= \frac{\varepsilon_{\mathrm{int}}}{\varepsilon_{\mathrm{int}} - 1} \omega_p^2
\end{aligned}$$
-This is used to experimentally determine the effective mass $m^*$
+This is used to experimentally determine the effective mass $$m^*$$
of the doped semiconductor,
-by finding which value of $m^*$ gives the measured $\omega$.
+by finding which value of $$m^*$$ gives the measured $$\omega$$.
diff --git a/source/know/concept/dynkins-formula/index.md b/source/know/concept/dynkins-formula/index.md
index 53da86f..c0d20c5 100644
--- a/source/know/concept/dynkins-formula/index.md
+++ b/source/know/concept/dynkins-formula/index.md
@@ -8,12 +8,12 @@ categories:
layout: "concept"
---
-Given an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
-with a time-independent drift $f$ and intensity $g$
-such that the diffusion uniquely exists on the $t$-axis.
-We define the **infinitesimal generator** $\hat{A}$
-as an operator with the following action on a given function $h(x)$,
-where $\mathbf{E}$ is a
+Given an [Itō diffusion](/know/concept/ito-calculus/) $$X_t$$
+with a time-independent drift $$f$$ and intensity $$g$$
+such that the diffusion uniquely exists on the $$t$$-axis.
+We define the **infinitesimal generator** $$\hat{A}$$
+as an operator with the following action on a given function $$h(x)$$,
+where $$\mathbf{E}$$ is a
[conditional expectation](/know/concept/conditional-expectation/):
$$\begin{aligned}
@@ -23,13 +23,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Which only makes sense for $h$ where this limit exists.
-The assumption that $X_t$ does not have any explicit time-dependence
-means that $X_0$ need not be the true initial condition;
-it can also be the state $X_s$ at any $s$ infinitesimally smaller than $t$.
+Which only makes sense for $$h$$ where this limit exists.
+The assumption that $$X_t$$ does not have any explicit time-dependence
+means that $$X_0$$ need not be the true initial condition;
+it can also be the state $$X_s$$ at any $$s$$ infinitesimally smaller than $$t$$.
-Conveniently, for a sufficiently well-behaved $h$,
-the generator $\hat{A}$ is identical to the Kolmogorov operator $\hat{L}$
+Conveniently, for a sufficiently well-behaved $$h$$,
+the generator $$\hat{A}$$ is identical to the Kolmogorov operator $$\hat{L}$$
found in the [backward Kolmogorov equation](/know/concept/kolmogorov-equations/):
$$\begin{aligned}
@@ -44,7 +44,7 @@ $$\begin{aligned}
-We define a new process $Y_t \equiv h(X_t)$, and then apply Itō's lemma, leading to:
+We define a new process $$Y_t \equiv h(X_t)$$, and then apply Itō's lemma, leading to:
$$\begin{aligned}
\dd{Y_t}
@@ -53,7 +53,7 @@ $$\begin{aligned}
&= \hat{L}\{h(X_t)\} \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
\end{aligned}$$
-Where we have recognized the definition of $\hat{L}$.
+Where we have recognized the definition of $$\hat{L}$$.
Integrating the above equation yields:
$$\begin{aligned}
@@ -63,14 +63,14 @@ $$\begin{aligned}
As always, the latter [Itō integral](/know/concept/ito-integral/)
is a [martingale](/know/concept/martingale/), so it vanishes
-when we take the expectation conditioned on the "initial" state $X_0$, leaving:
+when we take the expectation conditioned on the "initial" state $$X_0$$, leaving:
$$\begin{aligned}
\mathbf{E}[Y_t | X_0]
= Y_0 + \mathbf{E}\bigg[ \int_0^t \hat{L}\{h(X_s)\} \dd{s} \bigg| X_0 \bigg]
\end{aligned}$$
-For suffiently small $t$, the integral can be replaced by its first-order approximation:
+For suffiently small $$t$$, the integral can be replaced by its first-order approximation:
$$\begin{aligned}
\mathbf{E}[Y_t | X_0]
@@ -78,22 +78,23 @@ $$\begin{aligned}
\end{aligned}$$
Rearranging this gives the following,
-to be understood in the limit $t \to 0^+$:
+to be understood in the limit $$t \to 0^+$$:
$$\begin{aligned}
\hat{L}\{h(X_0)\}
\approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0]
\end{aligned}$$
+
The general definition of resembles that of a classical derivative,
-and indeed, the generator $\hat{A}$ can be thought of as a differential operator.
+and indeed, the generator $$\hat{A}$$ can be thought of as a differential operator.
In that case, we would like an analogue of the classical
fundamental theorem of calculus to relate it to integration.
Such an analogue is provided by **Dynkin's formula**:
-for a stopping time $\tau$ with a finite expected value $\mathbf{E}[\tau|X_0] < \infty$,
+for a stopping time $$\tau$$ with a finite expected value $$\mathbf{E}[\tau|X_0] < \infty$$,
it states that:
$$\begin{aligned}
@@ -109,7 +110,7 @@ $$\begin{aligned}
The proof is similar to the one above.
-Define $Y_t = h(X_t)$ and use Itō’s lemma:
+Define $$Y_t = h(X_t)$$ and use Itō’s lemma:
$$\begin{aligned}
\dd{Y_t}
@@ -118,7 +119,7 @@ $$\begin{aligned}
&= \hat{L} \{h(X_t)\} \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
\end{aligned}$$
-And then integrate this from $t = 0$ to the provided stopping time $t = \tau$:
+And then integrate this from $$t = 0$$ to the provided stopping time $$t = \tau$$:
$$\begin{aligned}
Y_\tau
@@ -127,7 +128,7 @@ $$\begin{aligned}
All [Itō integrals](/know/concept/ito-integral/)
are [martingales](/know/concept/martingale/),
-so the latter integral's conditional expectation is zero for the "initial" condition $X_0$.
+so the latter integral's conditional expectation is zero for the "initial" condition $$X_0$$.
The rest of the above equality is also a martingale:
$$\begin{aligned}
@@ -135,30 +136,30 @@ $$\begin{aligned}
= \mathbf{E}\bigg[ Y_\tau - Y_0 - \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
\end{aligned}$$
-Isolating this equation for $\mathbf{E}[Y_\tau | X_0]$ then gives Dynkin's formula.
+Isolating this equation for $$\mathbf{E}[Y_\tau | X_0]$$ then gives Dynkin's formula.
A common application of Dynkin's formula is predicting
-when the stopping time $\tau$ occurs, and in what state $X_\tau$ this happens.
+when the stopping time $$\tau$$ occurs, and in what state $$X_\tau$$ this happens.
Consider an example:
-for a region $\Omega$ of state space with $X_0 \in \Omega$,
-we define the exit time $\tau \equiv \inf\{ t : X_t \notin \Omega \}$,
-provided that $\mathbf{E}[\tau | X_0] < \infty$.
+for a region $$\Omega$$ of state space with $$X_0 \in \Omega$$,
+we define the exit time $$\tau \equiv \inf\{ t : X_t \notin \Omega \}$$,
+provided that $$\mathbf{E}[\tau | X_0] < \infty$$.
-To get information about when and where $X_t$ exits $\Omega$,
-we define the *general reward* $\Gamma$ as follows,
-consisting of a *running reward* $R$ for $X_t$ inside $\Omega$,
-and a *terminal reward* $T$ on the boundary $\partial \Omega$ where we stop at $X_\tau$:
+To get information about when and where $$X_t$$ exits $$\Omega$$,
+we define the *general reward* $$\Gamma$$ as follows,
+consisting of a *running reward* $$R$$ for $$X_t$$ inside $$\Omega$$,
+and a *terminal reward* $$T$$ on the boundary $$\partial \Omega$$ where we stop at $$X_\tau$$:
$$\begin{aligned}
\Gamma
= \int_0^\tau R(X_t) \dd{t} + \: T(X_\tau)
\end{aligned}$$
-For example, for $R = 1$ and $T = 0$, this becomes $\Gamma = \tau$,
-and if $R = 0$, then $T(X_\tau)$ can tell us the exit point.
-Let us now define $h(X_0) = \mathbf{E}[\Gamma | X_0]$,
+For example, for $$R = 1$$ and $$T = 0$$, this becomes $$\Gamma = \tau$$,
+and if $$R = 0$$, then $$T(X_\tau)$$ can tell us the exit point.
+Let us now define $$h(X_0) = \mathbf{E}[\Gamma | X_0]$$,
and apply Dynkin's formula:
$$\begin{aligned}
@@ -168,11 +169,11 @@ $$\begin{aligned}
&= \mathbf{E}\big[ T(X_\tau) | X_0 \big] + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} + R(X_t) \dd{t} \bigg| X_0 \bigg]
\end{aligned}$$
-The two leftmost terms depend on the exit point $X_\tau$,
-but not directly on $X_t$ for $t < \tau$,
-while the rightmost depends on the whole trajectory $X_t$.
+The two leftmost terms depend on the exit point $$X_\tau$$,
+but not directly on $$X_t$$ for $$t < \tau$$,
+while the rightmost depends on the whole trajectory $$X_t$$.
Therefore, the above formula is fulfilled
-if $h(x)$ satisfies the following equation and boundary conditions:
+if $$h(x)$$ satisfies the following equation and boundary conditions:
$$\begin{aligned}
\boxed{
@@ -183,8 +184,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-In other words, we have just turned a difficult question about a stochastic trajectory $X_t$
-into a classical differential boundary value problem for $h(x)$.
+In other words, we have just turned a difficult question about a stochastic trajectory $$X_t$$
+into a classical differential boundary value problem for $$h(x)$$.
diff --git a/source/know/concept/dyson-equation/index.md b/source/know/concept/dyson-equation/index.md
index 962b64a..ae9eb35 100644
--- a/source/know/concept/dyson-equation/index.md
+++ b/source/know/concept/dyson-equation/index.md
@@ -9,7 +9,7 @@ layout: "concept"
---
Consider the time-dependent Schrödinger equation,
-describing a wavefunction $\Psi_0(\vb{r}, t)$:
+describing a wavefunction $$\Psi_0(\vb{r}, t)$$:
$$\begin{aligned}
i \hbar \pdv{}{t}\Psi_0(\vb{r}, t)
@@ -18,50 +18,50 @@ $$\begin{aligned}
By definition, this equation's
[fundamental solution](/know/concept/fundamental-solution/)
-$G_0(\vb{r}, t; \vb{r}', t')$ satisfies the following:
+$$G_0(\vb{r}, t; \vb{r}', t')$$ satisfies the following:
$$\begin{aligned}
\Big( i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) \Big) G_0(\vb{r}, t; \vb{r}', t')
= \delta(\vb{r} - \vb{r}') \: \delta(t - t')
\end{aligned}$$
-From this, we define the inverse $\hat{G}{}_0^{-1}(\vb{r}, t)$
-as follows, so that $\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$:
+From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$
+as follows, so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:
$$\begin{aligned}
\hat{G}{}_0^{-1}(\vb{r}, t)
&\equiv i \hbar \pdv{}{t}- \hat{H}_0(\vb{r})
\end{aligned}$$
-Note that $\hat{G}{}_0^{-1}$ is an operator, while $G_0$ is a function.
+Note that $$\hat{G}{}_0^{-1}$$ is an operator, while $$G_0$$ is a function.
For the sake of consistency, we thus define
-the operator $\hat{G}_0(\vb{r}, t)$
-as a multiplication by $G_0$
-and integration over $\vb{r}'$ and $t'$:
+the operator $$\hat{G}_0(\vb{r}, t)$$
+as a multiplication by $$G_0$$
+and integration over $$\vb{r}'$$ and $$t'$$:
$$\begin{aligned}
\hat{G}_0(\vb{r}, t) \: f
\equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
\end{aligned}$$
-For an arbitrary function $f(\vb{r}, t)$,
-so that $\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$.
+For an arbitrary function $$f(\vb{r}, t)$$,
+so that $$\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$$.
Moving on, the Schrödinger equation can be rewritten like so,
-using $\hat{G}{}_0^{-1}$:
+using $$\hat{G}{}_0^{-1}$$:
$$\begin{aligned}
\hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t)
= 0
\end{aligned}$$
-Let us assume that $\hat{H}_0$ is simple,
-such that $G_0$ and $\hat{G}{}_0^{-1}$ can be found without issues
+Let us assume that $$\hat{H}_0$$ is simple,
+such that $$G_0$$ and $$\hat{G}{}_0^{-1}$$ can be found without issues
by solving the defining equation above.
Suppose we now add a more complicated and
-possibly time-dependent term $\hat{H}_1(\vb{r}, t)$,
+possibly time-dependent term $$\hat{H}_1(\vb{r}, t)$$,
in which case the corresponding fundamental solution
-$G(\vb{r}, \vb{r}', t, t')$ satisfies:
+$$G(\vb{r}, \vb{r}', t, t')$$ satisfies:
$$\begin{aligned}
\delta(\vb{r} - \vb{r}') \: \delta(t - t')
@@ -71,8 +71,8 @@ $$\begin{aligned}
\end{aligned}$$
This equation is typically too complicated to solve,
-so we would like an easier way to calculate this new $G$.
-The perturbed wavefunction $\Psi(\vb{r}, t)$
+so we would like an easier way to calculate this new $$G$$.
+The perturbed wavefunction $$\Psi(\vb{r}, t)$$
satisfies the Schrödinger equation:
$$\begin{aligned}
@@ -80,9 +80,9 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-We know that $\hat{G}{}_0^{-1} \Psi_0 = 0$,
+We know that $$\hat{G}{}_0^{-1} \Psi_0 = 0$$,
which we put on the right,
-and then we apply $\hat{G}_0$ in front:
+and then we apply $$\hat{G}_0$$ in front:
$$\begin{aligned}
\hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi
@@ -110,7 +110,7 @@ $$\begin{aligned}
\end{aligned}$$
The parenthesized expression clearly has the same recursive pattern,
-so we denote it by $\hat{G}$ and write the so-called **Dyson equation**:
+so we denote it by $$\hat{G}$$ and write the so-called **Dyson equation**:
$$\begin{aligned}
\boxed{
@@ -119,9 +119,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Such an iterative scheme is excellent for approximating $\hat{G}(\vb{r}, t)$.
+Such an iterative scheme is excellent for approximating $$\hat{G}(\vb{r}, t)$$.
Once a satisfactory accuracy is obtained,
-the perturbed wavefunction $\Psi$ can be calculated from:
+the perturbed wavefunction $$\Psi$$ can be calculated from:
$$\begin{aligned}
\boxed{
@@ -131,10 +131,10 @@ $$\begin{aligned}
\end{aligned}$$
This relation is equivalent to the Schrödinger equation.
-So now we have the operator $\hat{G}(\vb{r}, t)$,
-but what about the fundamental solution function $G(\vb{r}, t; \vb{r}', t')$?
-Let us take its definition, multiply it by an arbitrary $f(\vb{r}, t)$,
-and integrate over $G$'s second argument pair:
+So now we have the operator $$\hat{G}(\vb{r}, t)$$,
+but what about the fundamental solution function $$G(\vb{r}, t; \vb{r}', t')$$?
+Let us take its definition, multiply it by an arbitrary $$f(\vb{r}, t)$$,
+and integrate over $$G$$'s second argument pair:
$$\begin{aligned}
\iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
@@ -142,9 +142,9 @@ $$\begin{aligned}
= f
\end{aligned}$$
-Where we have hidden the arguments $(\vb{r}, t)$ for brevity.
-We now apply $\hat{G}_0(\vb{r}, t)$ to this equation
-(which contains an integral over $t''$ independent of $t'$):
+Where we have hidden the arguments $$(\vb{r}, t)$$ for brevity.
+We now apply $$\hat{G}_0(\vb{r}, t)$$ to this equation
+(which contains an integral over $$t''$$ independent of $$t'$$):
$$\begin{aligned}
\hat{G}_0 f
@@ -154,8 +154,8 @@ $$\begin{aligned}
\end{aligned}$$
Here, the shape of Dyson's equation is clearly recognizable,
-so we conclude that, as expected, the operator $\hat{G}$
-is defined as multiplication by the function $G$ followed by integration:
+so we conclude that, as expected, the operator $$\hat{G}$$
+is defined as multiplication by the function $$G$$ followed by integration:
$$\begin{aligned}
\hat{G}(\vb{r}, t) \: f(\vb{r}, t)
diff --git a/source/know/concept/ehrenfests-theorem/index.md b/source/know/concept/ehrenfests-theorem/index.md
index 4d96989..fba0192 100644
--- a/source/know/concept/ehrenfests-theorem/index.md
+++ b/source/know/concept/ehrenfests-theorem/index.md
@@ -9,10 +9,10 @@ layout: "concept"
---
In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the
-time evolution of an observable's expectation value $\expval{\hat{L}}$.
+time evolution of an observable's expectation value $$\expval{\hat{L}}$$.
The time-dependent Schrödinger equation is as follows,
-where prime denotes differentiation with respect to time $t$:
+where prime denotes differentiation with respect to time $$t$$:
$$\begin{aligned}
\Ket{\psi'} = \frac{1}{i \hbar} \hat{H} \Ket{\psi}
@@ -20,8 +20,8 @@ $$\begin{aligned}
\Bra{\psi'} = - \frac{1}{i \hbar} \Bra{\psi} \hat{H}
\end{aligned}$$
-Given an observable operator $\hat{L}$ and a state $\Ket{\psi}$,
-the time-derivative of the expectation value $\expval{\hat{L}}$ is as follows
+Given an observable operator $$\hat{L}$$ and a state $$\Ket{\psi}$$,
+the time-derivative of the expectation value $$\expval{\hat{L}}$$ is as follows
(due to the product rule of differentiation):
$$\begin{aligned}
@@ -48,12 +48,12 @@ the last term often vanishes.
As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
this relation proves itself,
-when one simply wraps all terms in $\Bra{\psi}$ and $\Ket{\psi}$.
+when one simply wraps all terms in $$\Bra{\psi}$$ and $$\Ket{\psi}$$.
-Two observables of particular interest are the position $\hat{X}$ and momentum $\hat{P}$.
-Applying the above theorem to $\hat{X}$ yields the following,
-which we reduce using the fact that $\hat{X}$ commutes
-with the potential $V(\hat{X})$,
+Two observables of particular interest are the position $$\hat{X}$$ and momentum $$\hat{P}$$.
+Applying the above theorem to $$\hat{X}$$ yields the following,
+which we reduce using the fact that $$\hat{X}$$ commutes
+with the potential $$V(\hat{X})$$,
because one is a function of the other:
$$\begin{aligned}
@@ -76,7 +76,7 @@ $$\begin{gathered}
}
\end{gathered}$$
-Next, applying the general formula to the expected momentum $\expval{\hat{P}}$
+Next, applying the general formula to the expected momentum $$\expval{\hat{P}}$$
gives us:
$$\begin{aligned}
@@ -86,8 +86,8 @@ $$\begin{aligned}
= \frac{1}{i \hbar} \Expval{[\hat{P}, V(\hat{X})]}
\end{aligned}$$
-To find the commutator, we go to the $\hat{X}$-basis and use a test
-function $f(x)$:
+To find the commutator, we go to the $$\hat{X}$$-basis and use a test
+function $$f(x)$$:
$$\begin{aligned}
\Comm{- i \hbar \dv{}{x}}{V(x)} \: f(x)
@@ -113,8 +113,8 @@ $$\begin{gathered}
\end{gathered}$$
There is an important consequence of Ehrenfest's original theorems
-for the symbolic derivatives of the Hamiltonian $\hat{H}$
-with respect to $\hat{X}$ and $\hat{P}$:
+for the symbolic derivatives of the Hamiltonian $$\hat{H}$$
+with respect to $$\hat{X}$$ and $$\hat{P}$$:
$$\begin{gathered}
\boxed{
diff --git a/source/know/concept/einstein-coefficients/index.md b/source/know/concept/einstein-coefficients/index.md
index 27d6013..179d866 100644
--- a/source/know/concept/einstein-coefficients/index.md
+++ b/source/know/concept/einstein-coefficients/index.md
@@ -19,16 +19,16 @@ in several useful situations.
## Qualitative description
-Suppose we have a ground state with energy $E_1$ containing $N_1$ electrons,
-and an excited state with energy $E_2$ containing $N_2$ electrons.
-The resonance $\omega_0 \equiv (E_2 \!-\! E_1)/\hbar$
+Suppose we have a ground state with energy $$E_1$$ containing $$N_1$$ electrons,
+and an excited state with energy $$E_2$$ containing $$N_2$$ electrons.
+The resonance $$\omega_0 \equiv (E_2 \!-\! E_1)/\hbar$$
is the frequency of the photon emitted
-when an electron falls from $E_2$ to $E_1$.
+when an electron falls from $$E_2$$ to $$E_1$$.
-The first Einstein coefficient is the **spontaneous emission rate** $A_{21}$,
+The first Einstein coefficient is the **spontaneous emission rate** $$A_{21}$$,
which gives the probability per unit time
that an excited electron falls from state 2 to 1,
-so that $N_2(t)$ obeys the following equation,
+so that $$N_2(t)$$ obeys the following equation,
which is easily solved:
$$\begin{aligned}
@@ -37,43 +37,43 @@ $$\begin{aligned}
N_2(t) = N_2(0) \exp(- t / \tau)
\end{aligned}$$
-Where $\tau = 1 / A_{21}$ is the **natural radiative lifetime** of the excited state,
+Where $$\tau = 1 / A_{21}$$ is the **natural radiative lifetime** of the excited state,
which gives the lifetime of an excited electron,
before it decays to the ground state.
-The next coefficient is the **absorption rate** $B_{12}$,
+The next coefficient is the **absorption rate** $$B_{12}$$,
which is the probability that an incoming photon excites an electron,
per unit time and per unit spectral energy density
(i.e. the rate depends on the frequency of the incoming light).
-Then $N_1(t)$ obeys the following equation:
+Then $$N_1(t)$$ obeys the following equation:
$$\begin{aligned}
\dv{N_1}{t} = - B_{12} N_1 u(\omega_0)
\end{aligned}$$
-Where $u(\omega)$ is the spectral energy density of the incoming light,
-put here to express the fact that only photons with frequency $\omega_0$ are absorbed.
+Where $$u(\omega)$$ is the spectral energy density of the incoming light,
+put here to express the fact that only photons with frequency $$\omega_0$$ are absorbed.
-There is one more Einstein coefficient: the **stimulated emission rate** $B_{21}$.
+There is one more Einstein coefficient: the **stimulated emission rate** $$B_{21}$$.
An incoming photon has an associated electromagnetic field,
which can encourage an excited electron to drop to the ground state,
-such that for $A_{21} = 0$:
+such that for $$A_{21} = 0$$:
$$\begin{aligned}
\dv{N_2}{t} = - B_{21} N_2 u(\omega_0)
\end{aligned}$$
-These three coefficients $A_{21}$, $B_{12}$ and $B_{21}$ are related to each other.
+These three coefficients $$A_{21}$$, $$B_{12}$$ and $$B_{21}$$ are related to each other.
Suppose that the system is in equilibrium,
-i.e. that $N_1$ and $N_2$ are constant.
+i.e. that $$N_1$$ and $$N_2$$ are constant.
We assume that the number of particles in the system is constant,
-implying that $N_1'(t) = - N_2'(t) = 0$, so:
+implying that $$N_1'(t) = - N_2'(t) = 0$$, so:
$$\begin{aligned}
B_{12} N_1 u(\omega_0) = A_{21} N_2 + B_{21} N_2 u(\omega_0) = 0
\end{aligned}$$
-Isolating this equation for $u(\omega_0)$,
+Isolating this equation for $$u(\omega_0)$$,
gives following expression for the radiation:
$$\begin{aligned}
@@ -84,7 +84,7 @@ $$\begin{aligned}
We assume that the system is in thermal equilibrium
with its own black-body radiation, and that there is no external light.
Then this is a [canonical ensemble](/know/concept/canonical-ensemble/),
-meaning that the relative probability that an electron has $E_2$ compared to $E_1$
+meaning that the relative probability that an electron has $$E_2$$ compared to $$E_1$$
is given by the Boltzmann distribution:
$$\begin{aligned}
@@ -93,15 +93,15 @@ $$\begin{aligned}
= \frac{g_2}{g_1} \exp(- \hbar \omega_0 \beta)
\end{aligned}$$
-Where $g_2$ and $g_1$ are the degeneracies of the energy levels.
-Inserting this back into the equation for the spectrum $u(\omega_0)$ yields:
+Where $$g_2$$ and $$g_1$$ are the degeneracies of the energy levels.
+Inserting this back into the equation for the spectrum $$u(\omega_0)$$ yields:
$$\begin{aligned}
u(\omega_0)
= \frac{A_{21}}{(g_1 / g_2) B_{12} \exp(\hbar \omega_0 \beta) - B_{21}}
\end{aligned}$$
-Since $u(\omega_0)$ represents only black-body radiation,
+Since $$u(\omega_0)$$ represents only black-body radiation,
our result must agree with [Planck's law](/know/concept/plancks-law/):
$$\begin{aligned}
@@ -120,14 +120,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that this result holds even if $E_1$ is not the ground state,
-but instead some lower excited state below $E_2$,
+Note that this result holds even if $$E_1$$ is not the ground state,
+but instead some lower excited state below $$E_2$$,
due to the principle of [detailed balance](/know/concept/detailed-balance/).
Furthermore, it turns out that these relations
also hold if the system is not in equilibrium.
A notable case is **population inversion**,
-where $B_{21} N_2 > B_{12} N_1$ such that $N_2 > (g_2 / g_1) N_1$.
+where $$B_{21} N_2 > B_{12} N_1$$ such that $$N_2 > (g_2 / g_1) N_1$$.
This situation is mandatory for lasers, where stimulated emission must dominate,
such that the light becomes stronger as it travels through the medium.
@@ -137,38 +137,38 @@ such that the light becomes stronger as it travels through the medium.
In fact, we can analytically calculate the Einstein coefficients in some cases,
by treating incoming light as a perturbation
to an electron in a two-level system,
-and then finding $B_{12}$ and $B_{21}$ from the resulting transition rate.
+and then finding $$B_{12}$$ and $$B_{21}$$ from the resulting transition rate.
We need to make the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
-in which case the perturbing Hamiltonian $\hat{H}_1(t)$ is given by:
+in which case the perturbing Hamiltonian $$\hat{H}_1(t)$$ is given by:
$$\begin{aligned}
\hat{H}_1(t)
= - q \vec{r} \cdot \vec{E}_0 \cos(\omega t)
\end{aligned}$$
-Where $q = -e$ is the electron charge,
-$\vec{r}$ is the position operator,
-and $\vec{E}_0$ is the amplitude of
+Where $$q = -e$$ is the electron charge,
+$$\vec{r}$$ is the position operator,
+and $$\vec{E}_0$$ is the amplitude of
the [electromagnetic wave](/know/concept/electromagnetic-wave-equation/).
-For simplicity, we let the amplitude be along the $z$-axis:
+For simplicity, we let the amplitude be along the $$z$$-axis:
$$\begin{aligned}
\hat{H}_1(t)
= - q E_0 z \cos(\omega t)
\end{aligned}$$
-This form of $\hat{H}_1$ is a well-known case for
+This form of $$\hat{H}_1$$ is a well-known case for
[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
-which tells us that the transition probability from $\Ket{a}$ to $\Ket{b}$ is:
+which tells us that the transition probability from $$\Ket{a}$$ to $$\Ket{b}$$ is:
$$\begin{aligned}
P_{ab}
= \frac{\big|\!\matrixel{a}{H_1}{b}\!\big|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
\end{aligned}$$
-If the nucleus is at $z = 0$,
-then generally $\Ket{1}$ and $\Ket{2}$ will be even or odd functions of $z$,
-meaning that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$
+If the nucleus is at $$z = 0$$,
+then generally $$\Ket{1}$$ and $$\Ket{2}$$ will be even or odd functions of $$z$$,
+meaning that $$\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$$
(see also [Laporte's selection rule](/know/concept/selection-rules/)),
leading to:
@@ -180,8 +180,8 @@ $$\begin{gathered}
\matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
\end{gathered}$$
-Where $d \equiv q \matrixel{2}{z}{1}$ is a constant,
-namely the $z$-component of the **transition dipole moment**.
+Where $$d \equiv q \matrixel{2}{z}{1}$$ is a constant,
+namely the $$z$$-component of the **transition dipole moment**.
The chance of an upward jump (i.e. absorption) is:
$$\begin{aligned}
@@ -190,8 +190,8 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, the transition probability for stimulated emission is as follows,
-using the fact that $P_{ab}$ is a sinc-function,
-and is therefore symmetric around $\omega_{ba}$:
+using the fact that $$P_{ab}$$ is a sinc-function,
+and is therefore symmetric around $$\omega_{ba}$$:
$$\begin{aligned}
P_{21}
@@ -202,17 +202,17 @@ Surprisingly, the probabilities of absorption and stimulated emission are the sa
In practice, however, the relative rates of these two processes depends heavily on
the availability of electrons and holes in both states.
-In theory, we could calculate the transition rate $R_{12} = \ipdv{P_{12}}{t}$,
-which would give us Einstein's absorption coefficient $B_{12}$,
+In theory, we could calculate the transition rate $$R_{12} = \ipdv{P_{12}}{t}$$,
+which would give us Einstein's absorption coefficient $$B_{12}$$,
for this specific case of coherent monochromatic light.
-However, the result would not be constant in time $t$,
+However, the result would not be constant in time $$t$$,
so is not really useful.
## Polarized light
To solve this "problem", we generalize to (incoherent) polarized polychromatic light.
-To do so, we note that the energy density $u$ of an electric field $E_0$ is given by:
+To do so, we note that the energy density $$u$$ of an electric field $$E_0$$ is given by:
$$\begin{aligned}
u = \frac{1}{2} \varepsilon_0 E_0^2
@@ -220,8 +220,8 @@ $$\begin{aligned}
E_0^2 = \frac{2 u}{\varepsilon_0}
\end{aligned}$$
-Where $\varepsilon_0$ is the vacuum permittivity.
-Putting this in the previous result for $P_{12}$ gives us:
+Where $$\varepsilon_0$$ is the vacuum permittivity.
+Putting this in the previous result for $$P_{12}$$ gives us:
$$\begin{aligned}
P_{12}
@@ -229,7 +229,7 @@ $$\begin{aligned}
\end{aligned}$$
For a continuous light spectrum,
-this $u$ turns into the spectral energy density $u(\omega)$:
+this $$u$$ turns into the spectral energy density $$u(\omega)$$:
$$\begin{aligned}
P_{12}
@@ -241,11 +241,11 @@ From here, the derivation is similar to that of
[Fermi's golden rule](/know/concept/fermis-golden-rule/),
despite the distinction that we are integrating over frequencies rather than states.
-At sufficiently large $t$, the integrand is sharply peaked at $\omega = \omega_0$
+At sufficiently large $$t$$, the integrand is sharply peaked at $$\omega = \omega_0$$
and negligible everywhere else,
-so we take $u(\omega)$ out of the integral and extend the integration limits.
+so we take $$u(\omega)$$ out of the integral and extend the integration limits.
Then we rewrite and look up the integral,
-which turns out to be $\pi t$:
+which turns out to be $$\pi t$$:
$$\begin{aligned}
P_{12}
@@ -253,7 +253,7 @@ $$\begin{aligned}
= \frac{\pi |d|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
\end{aligned}$$
-From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
+From this, the transition rate $$R_{12} = B_{12} u(\omega_0)$$
is then calculated as follows:
$$\begin{aligned}
@@ -262,7 +262,7 @@ $$\begin{aligned}
= \frac{\pi |d|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
\end{aligned}$$
-Using the relations from earlier with $g_1 = g_2$,
+Using the relations from earlier with $$g_1 = g_2$$,
the Einstein coefficients are found to be as follows
for a polarized incoming light spectrum:
@@ -278,8 +278,8 @@ $$\begin{aligned}
## Unpolarized light
We can generalize the above result even further to unpolarized light.
-Let us return to the matrix elements of the perturbation $\hat{H}_1$,
-and define the polarization unit vector $\vec{n}$:
+Let us return to the matrix elements of the perturbation $$\hat{H}_1$$,
+and define the polarization unit vector $$\vec{n}$$:
$$\begin{aligned}
\matrixel{2}{\hat{H}_1}{1}
@@ -287,22 +287,22 @@ $$\begin{aligned}
= - E_0 (\vec{d} \cdot \vec{n})
\end{aligned}$$
-Where $\vec{d} \equiv q \matrixel{2}{\vec{r}}{1}$ is
+Where $$\vec{d} \equiv q \matrixel{2}{\vec{r}}{1}$$ is
the full **transition dipole moment** vector, which is usually complex.
-The goal is to calculate the average of $|\vec{d} \cdot \vec{n}|^2$.
+The goal is to calculate the average of $$|\vec{d} \cdot \vec{n}|^2$$.
In [spherical coordinates](/know/concept/spherical-coordinates/),
-we integrate over all directions $\vec{n}$ for fixed $\vec{d}$,
-using that $\vec{d} \cdot \vec{n} = |\vec{d}| \cos(\theta)$
-with $|\vec{d}| \equiv |d_x|^2 \!+\! |d_y|^2 \!+\! |d_z|^2$:
+we integrate over all directions $$\vec{n}$$ for fixed $$\vec{d}$$,
+using that $$\vec{d} \cdot \vec{n} = |\vec{d}| \cos(\theta)$$
+with $$|\vec{d}| \equiv |d_x|^2 \!+\! |d_y|^2 \!+\! |d_z|^2$$:
$$\begin{aligned}
\Expval{|\vec{d} \cdot \vec{n}|^2}
= \frac{1}{4 \pi} \int_0^\pi \int_0^{2 \pi} |\vec{d}|^2 \cos^2(\theta) \sin(\theta) \dd{\varphi} \dd{\theta}
\end{aligned}$$
-Where we have divided by $4\pi$ (the surface area of a unit sphere) for normalization,
-and $\theta$ is the polar angle between $\vec{n}$ and $\vec{d}$.
+Where we have divided by $$4\pi$$ (the surface area of a unit sphere) for normalization,
+and $$\theta$$ is the polar angle between $$\vec{n}$$ and $$\vec{d}$$.
Evaluating the integrals yields:
$$\begin{aligned}
@@ -312,8 +312,8 @@ $$\begin{aligned}
= \frac{|\vec{d}|^2}{3}
\end{aligned}$$
-With this additional constant factor $1/3$,
-the transition rate $R_{12}$ is modified to:
+With this additional constant factor $$1/3$$,
+the transition rate $$R_{12}$$ is modified to:
$$\begin{aligned}
R_{12}
diff --git a/source/know/concept/elastic-collision/index.md b/source/know/concept/elastic-collision/index.md
index 11c3115..ac15e06 100644
--- a/source/know/concept/elastic-collision/index.md
+++ b/source/know/concept/elastic-collision/index.md
@@ -19,8 +19,8 @@ for example heat.
## One dimension
In 1D, not only the kinetic energy is conserved, but also the total momentum.
-Let $v_1$ and $v_2$ be the initial velocities of objects 1 and 2,
-and $v_1'$ and $v_2'$ their velocities afterwards:
+Let $$v_1$$ and $$v_2$$ be the initial velocities of objects 1 and 2,
+and $$v_1'$$ and $$v_2'$$ their velocities afterwards:
$$\begin{aligned}
\begin{cases}
@@ -45,8 +45,8 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-Using the first equation to replace $m_1 (v_1 \!-\! v_1')$
-with $m_2 (v_2 \!-\! v_2')$ in the second:
+Using the first equation to replace $$m_1 (v_1 \!-\! v_1')$$
+with $$m_2 (v_2 \!-\! v_2')$$ in the second:
$$\begin{aligned}
m_2 (v_1 + v_1') (v_2' - v_2)
@@ -66,10 +66,10 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-Note that the first relation is equivalent to $v_1 - v_2 = v_2' - v_1'$,
+Note that the first relation is equivalent to $$v_1 - v_2 = v_2' - v_1'$$,
meaning that the objects' relative velocity
is reversed by the collision.
-Moving on, we replace $v_1'$ in the second equation:
+Moving on, we replace $$v_1'$$ in the second equation:
$$\begin{aligned}
m_1 v_1 + m_2 v_2
@@ -79,8 +79,8 @@ $$\begin{aligned}
&= 2 m_1 v_1 + (m_2 - m_1) v_2
\end{aligned}$$
-Dividing by $m_1 + m_2$,
-and going through the same process for $v_1'$,
+Dividing by $$m_1 + m_2$$,
+and going through the same process for $$v_1'$$,
we arrive at:
$$\begin{aligned}
@@ -96,7 +96,7 @@ $$\begin{aligned}
\end{aligned}$$
To analyze this result,
-for practicality, we simplify it by setting $v_2 = 0$.
+for practicality, we simplify it by setting $$v_2 = 0$$.
In that case:
$$\begin{aligned}
@@ -108,7 +108,7 @@ $$\begin{aligned}
\end{aligned}$$
How much of its energy and momentum does object 1 transfer to object 2?
-The following ratios compare $v_1$ and $v_2'$ to quantify the transfer:
+The following ratios compare $$v_1$$ and $$v_2'$$ to quantify the transfer:
$$\begin{aligned}
\frac{m_2 v_2'}{m_1 v_1}
@@ -118,15 +118,15 @@ $$\begin{aligned}
= \frac{4 m_1 m_2}{(m_1 + m_2)^2}
\end{aligned}$$
-If $m_1 = m_2$, both ratios reduce to $1$,
+If $$m_1 = m_2$$, both ratios reduce to $$1$$,
meaning that all energy and momentum is transferred,
and object 1 is at rest after the collision.
Newton's cradle is an example of this.
-If $m_1 \ll m_2$, object 1 simply bounces off object 2,
+If $$m_1 \ll m_2$$, object 1 simply bounces off object 2,
barely transferring any energy.
Object 2 ends up with twice object 1's momentum,
-but $v_2'$ is very small and thus negligible:
+but $$v_2'$$ is very small and thus negligible:
$$\begin{aligned}
\frac{m_2 v_2'}{m_1 v_1}
@@ -136,7 +136,7 @@ $$\begin{aligned}
\approx \frac{4 m_1}{m_2}
\end{aligned}$$
-If $m_1 \gg m_2$, object 1 barely notices the collision,
+If $$m_1 \gg m_2$$, object 1 barely notices the collision,
so not much is transferred to object 2:
$$\begin{aligned}
diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md
index 66501e2..7c710ec 100644
--- a/source/know/concept/electric-dipole-approximation/index.md
+++ b/source/know/concept/electric-dipole-approximation/index.md
@@ -22,11 +22,11 @@ $$\begin{aligned}
&= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
\end{aligned}$$
-With charge $q = - e$,
-canonical momentum operator $\vu{P} = - i \hbar \nabla$,
-and magnetic vector potential $\vb{A}(\vb{x}, t)$.
-We reduce this by fixing the Coulomb gauge $\nabla \cdot \vb{A} = 0$,
-so that $\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$:
+With charge $$q = - e$$,
+canonical momentum operator $$\vu{P} = - i \hbar \nabla$$,
+and magnetic vector potential $$\vb{A}(\vb{x}, t)$$.
+We reduce this by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$,
+so that $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$:
$$\begin{aligned}
\comm{\vb{A}}{\vu{P}} \psi
@@ -36,9 +36,9 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Where $\psi$ is an arbitrary test function.
-Assuming $\vb{A}$ is so small that $\vb{A}{}^2$ is negligible, we split $\hat{H}$ as follows,
-where $\hat{H}_1$ can be regarded as a perturbation to $\hat{H}_0$:
+Where $$\psi$$ is an arbitrary test function.
+Assuming $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, we split $$\hat{H}$$ as follows,
+where $$\hat{H}_1$$ can be regarded as a perturbation to $$\hat{H}_0$$:
$$\begin{aligned}
\hat{H}
@@ -51,7 +51,7 @@ $$\begin{aligned}
\equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
\end{aligned}$$
-In an electromagnetic wave, $\vb{A}$ is oscillating sinusoidally in time and space:
+In an electromagnetic wave, $$\vb{A}$$ is oscillating sinusoidally in time and space:
$$\begin{aligned}
\vb{A}(\vb{x}, t) = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t)
@@ -64,7 +64,7 @@ $$\begin{aligned}
\vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
\end{aligned}$$
-The corresponding perturbative [electric field](/know/concept/electric-field/) $\vb{E}$ is then given by:
+The corresponding perturbative [electric field](/know/concept/electric-field/) $$\vb{E}$$ is then given by:
$$\begin{aligned}
\vb{E}(\vb{x}, t)
@@ -72,11 +72,11 @@ $$\begin{aligned}
= \vb{E}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
\end{aligned}$$
-Where $\vb{E}_0 = \omega \vb{A}_0$.
+Where $$\vb{E}_0 = \omega \vb{A}_0$$.
Let us restrict ourselves to visible light,
-whose wavelength $2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$.
-Meanwhile, an atomic orbital is several Bohr $\sim 10^{-10} \:\mathrm{m}$,
-so $\vb{k} \cdot \vb{x}$ is negligible:
+whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$.
+Meanwhile, an atomic orbital is several Bohr $$\sim 10^{-10} \:\mathrm{m}$$,
+so $$\vb{k} \cdot \vb{x}$$ is negligible:
$$\begin{aligned}
\boxed{
@@ -86,15 +86,15 @@ $$\begin{aligned}
\end{aligned}$$
This is the **electric dipole approximation**:
-we ignore all spatial variation of $\vb{E}$,
+we ignore all spatial variation of $$\vb{E}$$,
and only consider its temporal oscillation.
Also, since we have not used the word "photon",
we are implicitly treating the radiation classically,
and the electron quantum-mechanically.
-Next, we want to rewrite $\hat{H}_1$
-to use the electric field $\vb{E}$ instead of the potential $\vb{A}$.
-To do so, we use that $\vu{P} = m \: \idv{\vu{x}}{t}$
+Next, we want to rewrite $$\hat{H}_1$$
+to use the electric field $$\vb{E}$$ instead of the potential $$\vb{A}$$.
+To do so, we use that $$\vu{P} = m \: \idv{\vu{x}}{t}$$
and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
$$\begin{aligned}
@@ -105,7 +105,7 @@ $$\begin{aligned}
\end{aligned}$$
Taking the off-diagonal inner product with
-the two-level system's states $\Ket{1}$ and $\Ket{2}$ gives:
+the two-level system's states $$\Ket{1}$$ and $$\Ket{2}$$ gives:
$$\begin{aligned}
\matrixel{2}{\vu{P}}{1}
@@ -113,9 +113,9 @@ $$\begin{aligned}
= m i \omega_0 \matrixel{2}{\vu{x}}{1}
\end{aligned}$$
-Therefore, $\vu{P} / m = i \omega_0 \vu{x}$,
-where $\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$ is the resonance of the energy gap,
-close to which we assume that $\vb{A}$ and $\vb{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
+Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$,
+where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap,
+close to which we assume that $$\vb{A}$$ and $$\vb{E}$$ are oscillating, i.e. $$\omega \approx \omega_0$$.
We thus get:
$$\begin{aligned}
@@ -127,7 +127,7 @@ $$\begin{aligned}
= - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
\end{aligned}$$
-Where $\vu{d} \equiv q \vu{x} = - e \vu{x}$ is
+Where $$\vu{d} \equiv q \vu{x} = - e \vu{x}$$ is
the **transition dipole moment operator** of the electron,
hence the name **electric dipole approximation**.
Finally, we take the real part, yielding:
@@ -141,7 +141,7 @@ $$\begin{aligned}
\end{aligned}$$
If this approximation is too rough,
-$\vb{E}$ can always be Taylor-expanded in $(i \vb{k} \cdot \vb{x})$:
+$$\vb{E}$$ can always be Taylor-expanded in $$(i \vb{k} \cdot \vb{x})$$:
$$\begin{aligned}
\vb{E}(\vb{x}, t)
diff --git a/source/know/concept/electric-field/index.md b/source/know/concept/electric-field/index.md
index 38c1ff6..2433edf 100644
--- a/source/know/concept/electric-field/index.md
+++ b/source/know/concept/electric-field/index.md
@@ -8,43 +8,43 @@ categories:
layout: "concept"
---
-The **electric field** $\vb{E}$ is a vector field
+The **electric field** $$\vb{E}$$ is a vector field
that describes electric effects,
and is defined as the field that correctly predicts
the [Lorentz force](/know/concept/lorentz-force/)
-on a particle with electric charge $q$:
+on a particle with electric charge $$q$$:
$$\begin{aligned}
\vb{F}
= q \vb{E}
\end{aligned}$$
-This definition implies that the direction of $\vb{E}$
+This definition implies that the direction of $$\vb{E}$$
is from positive to negative charges,
since opposite charges attracts and like charges repel.
-If two opposite point charges with magnitude $q$
+If two opposite point charges with magnitude $$q$$
are observed from far away,
they can be treated as a single object called a **dipole**,
-which has an **electric dipole moment** $\vb{p}$ defined like so,
-where $\vb{d}$ is the vector going from
-the negative to the positive charge (opposite direction of $\vb{E}$):
+which has an **electric dipole moment** $$\vb{p}$$ defined like so,
+where $$\vb{d}$$ is the vector going from
+the negative to the positive charge (opposite direction of $$\vb{E}$$):
$$\begin{aligned}
\vb{p} = q \vb{d}
\end{aligned}$$
Alternatively, for consistency with [magnetic fields](/know/concept/magnetic-field/),
-$\vb{p}$ can be defined from the aligning torque $\vb{\tau}$
-experienced by the dipole when placed in an $\vb{E}$-field.
-In other words, $\vb{p}$ satisfies:
+$$\vb{p}$$ can be defined from the aligning torque $$\vb{\tau}$$
+experienced by the dipole when placed in an $$\vb{E}$$-field.
+In other words, $$\vb{p}$$ satisfies:
$$\begin{aligned}
\vb{\tau} = \vb{p} \times \vb{E}
\end{aligned}$$
-Where $\vb{p}$ has units of $\mathrm{C m}$.
-The **polarization density** $\vb{P}$ is defined from $\vb{p}$,
+Where $$\vb{p}$$ has units of $$\mathrm{C m}$$.
+The **polarization density** $$\vb{P}$$ is defined from $$\vb{p}$$,
and roughly speaking represents the moments per unit volume:
$$\begin{aligned}
@@ -53,16 +53,16 @@ $$\begin{aligned}
\vb{p} = \int_V \vb{P} \dd{V}
\end{aligned}$$
-If $\vb{P}$ has the same magnitude and direction throughout the body,
-then this becomes $\vb{p} = \vb{P} V$, where $V$ is the volume.
-Therefore, $\vb{P}$ has units of $\mathrm{C / m^2}$.
+If $$\vb{P}$$ has the same magnitude and direction throughout the body,
+then this becomes $$\vb{p} = \vb{P} V$$, where $$V$$ is the volume.
+Therefore, $$\vb{P}$$ has units of $$\mathrm{C / m^2}$$.
-A nonzero $\vb{P}$ complicates things,
-since it contributes to the field and hence modifies $\vb{E}$.
+A nonzero $$\vb{P}$$ complicates things,
+since it contributes to the field and hence modifies $$\vb{E}$$.
We thus define
-the "free" **displacement field** $\vb{D}$
-from the "bound" field $\vb{P}$
-and the "net" field $\vb{E}$:
+the "free" **displacement field** $$\vb{D}$$
+from the "bound" field $$\vb{P}$$
+and the "net" field $$\vb{E}$$:
$$\begin{aligned}
\vb{D} \equiv \varepsilon_0 \vb{E} + \vb{P}
@@ -70,21 +70,21 @@ $$\begin{aligned}
\vb{E} = \frac{1}{\varepsilon_0} (\vb{D} - \vb{P})
\end{aligned}$$
-Where the **electric permittivity of free space** $\varepsilon_0$ is a known constant.
+Where the **electric permittivity of free space** $$\varepsilon_0$$ is a known constant.
It is important to point out some inconsistencies here:
-$\vb{D}$ and $\vb{P}$ contain a factor of $\varepsilon_0$,
+$$\vb{D}$$ and $$\vb{P}$$ contain a factor of $$\varepsilon_0$$,
and therefore measure **flux density**,
-while $\vb{E}$ does not contain $\varepsilon_0$,
+while $$\vb{E}$$ does not contain $$\varepsilon_0$$,
and thus measures **field intensity**.
Note that this convention is the opposite
-of the magnetic analogues $\vb{B}$, $\vb{H}$ and $\vb{M}$,
-and that $\vb{M}$ has the opposite sign of $\vb{P}$.
+of the magnetic analogues $$\vb{B}$$, $$\vb{H}$$ and $$\vb{M}$$,
+and that $$\vb{M}$$ has the opposite sign of $$\vb{P}$$.
-The polarization $\vb{P}$ is a function of $\vb{E}$.
+The polarization $$\vb{P}$$ is a function of $$\vb{E}$$.
In addition to the inherent polarity
-of the material $\vb{P}_0$ (zero in most cases),
+of the material $$\vb{P}_0$$ (zero in most cases),
there is a (possibly nonlinear) response
-to the applied $\vb{E}$-field:
+to the applied $$\vb{E}$$-field:
$$\begin{aligned}
\vb{P} =
@@ -93,10 +93,10 @@ $$\begin{aligned}
+ \varepsilon_0 \chi_e^{(3)} |\vb{E}|^2 \: \vb{E} + ...
\end{aligned}$$
-Where the $\chi_e^{(n)}$ are the **electric susceptibilities** of the medium.
-For simplicity, we often assume that only the $n\!=\!1$ term is nonzero,
-which is the linear response to $\vb{E}$.
-In that case, we define the **absolute permittivity** $\varepsilon$ so that:
+Where the $$\chi_e^{(n)}$$ are the **electric susceptibilities** of the medium.
+For simplicity, we often assume that only the $$n\!=\!1$$ term is nonzero,
+which is the linear response to $$\vb{E}$$.
+In that case, we define the **absolute permittivity** $$\varepsilon$$ so that:
$$\begin{aligned}
\vb{D}
@@ -106,15 +106,15 @@ $$\begin{aligned}
= \varepsilon \vb{E}
\end{aligned}$$
-I.e. $\varepsilon \equiv \varepsilon_r \varepsilon_0$,
-where $\varepsilon_r \equiv 1 + \chi_e^{(1)}$ is
+I.e. $$\varepsilon \equiv \varepsilon_r \varepsilon_0$$,
+where $$\varepsilon_r \equiv 1 + \chi_e^{(1)}$$ is
the [**dielectric function**](/know/concept/dielectric-function/)
or **relative permittivity**,
whose calculation is of great interest in physics.
-In reality, a material cannot respond instantly to $\vb{E}$,
-meaning that $\chi_e^{(1)}$ is a function of time,
-and that $\vb{P}$ is the convolution of $\chi_e^{(1)}(t)$ and $\vb{E}(t)$:
+In reality, a material cannot respond instantly to $$\vb{E}$$,
+meaning that $$\chi_e^{(1)}$$ is a function of time,
+and that $$\vb{P}$$ is the convolution of $$\chi_e^{(1)}(t)$$ and $$\vb{E}(t)$$:
$$\begin{aligned}
\vb{P}(t)
@@ -122,6 +122,6 @@ $$\begin{aligned}
= \varepsilon_0 \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau
\end{aligned}$$
-Note that this definition requires $\chi_e^{(1)}(t) = 0$ for $t < 0$
+Note that this definition requires $$\chi_e^{(1)}(t) = 0$$ for $$t < 0$$
in order to ensure causality,
which leads to the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations/).
diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md
index c4cd9eb..a27fe6f 100644
--- a/source/know/concept/electromagnetic-wave-equation/index.md
+++ b/source/know/concept/electromagnetic-wave-equation/index.md
@@ -21,8 +21,8 @@ in order to derive the wave equation.
## Uniform medium
We will use all of Maxwell's equations,
-but we start with Ampère's circuital law for the "free" fields $\vb{H}$ and $\vb{D}$,
-in the absence of a free current $\vb{J}_\mathrm{free} = 0$:
+but we start with Ampère's circuital law for the "free" fields $$\vb{H}$$ and $$\vb{D}$$,
+in the absence of a free current $$\vb{J}_\mathrm{free} = 0$$:
$$\begin{aligned}
\nabla \cross \vb{H}
@@ -39,9 +39,9 @@ $$\begin{aligned}
\end{aligned}$$
Which, upon insertion into Ampère's law,
-yields an equation relating $\vb{B}$ and $\vb{E}$.
+yields an equation relating $$\vb{B}$$ and $$\vb{E}$$.
This may seem to contradict Ampère's "total" law,
-but keep in mind that $\vb{J}_\mathrm{bound} \neq 0$ here:
+but keep in mind that $$\vb{J}_\mathrm{bound} \neq 0$$ here:
$$\begin{aligned}
\nabla \cross \vb{B}
@@ -49,7 +49,7 @@ $$\begin{aligned}
\end{aligned}$$
Now we take the curl, rearrange,
-and substitute $\nabla \cross \vb{E}$ according to Faraday's law:
+and substitute $$\nabla \cross \vb{E}$$ according to Faraday's law:
$$\begin{aligned}
\nabla \cross (\nabla \cross \vb{B})
@@ -58,7 +58,7 @@ $$\begin{aligned}
\end{aligned}$$
Using a vector identity, we rewrite the leftmost expression,
-which can then be reduced thanks to Gauss' law for magnetism $\nabla \cdot \vb{B} = 0$:
+which can then be reduced thanks to Gauss' law for magnetism $$\nabla \cdot \vb{B} = 0$$:
$$\begin{aligned}
- \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t}
@@ -66,8 +66,8 @@ $$\begin{aligned}
= - \nabla^2 \vb{B}
\end{aligned}$$
-This describes $\vb{B}$.
-Next, we repeat the process for $\vb{E}$:
+This describes $$\vb{B}$$.
+Next, we repeat the process for $$\vb{E}$$:
taking the curl of Faraday's law yields:
$$\begin{aligned}
@@ -77,8 +77,8 @@ $$\begin{aligned}
\end{aligned}$$
Which can be rewritten using same vector identity as before,
-and then reduced by assuming that there is no net charge density $\rho = 0$
-in Gauss' law, such that $\nabla \cdot \vb{E} = 0$:
+and then reduced by assuming that there is no net charge density $$\rho = 0$$
+in Gauss' law, such that $$\nabla \cdot \vb{E} = 0$$:
$$\begin{aligned}
- \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
@@ -87,8 +87,8 @@ $$\begin{aligned}
\end{aligned}$$
We thus arrive at the following two (implicitly coupled)
-wave equations for $\vb{E}$ and $\vb{B}$,
-where we have defined the phase velocity $v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$:
+wave equations for $$\vb{E}$$ and $$\vb{B}$$,
+where we have defined the phase velocity $$v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$$:
$$\begin{aligned}
\boxed{
@@ -103,7 +103,7 @@ $$\begin{aligned}
\end{aligned}$$
Traditionally, it is said that the solutions are as follows,
-where the wavenumber $|\vb{k}| = \omega / v$:
+where the wavenumber $$|\vb{k}| = \omega / v$$:
$$\begin{aligned}
\vb{E}(\vb{r}, t)
@@ -115,9 +115,9 @@ $$\begin{aligned}
In fact, thanks to linearity, these **plane waves** can be treated as
terms in a Fourier series, meaning that virtually
-*any* function $f(\vb{k} \cdot \vb{r} - \omega t)$ is a valid solution.
+*any* function $$f(\vb{k} \cdot \vb{r} - \omega t)$$ is a valid solution.
-Keep in mind that in reality $\vb{E}$ and $\vb{B}$ are real,
+Keep in mind that in reality $$\vb{E}$$ and $$\vb{B}$$ are real,
so although it is mathematically convenient to use plane waves,
in the end you will need to take the real part.
@@ -125,8 +125,8 @@ in the end you will need to take the real part.
## Non-uniform medium
A useful generalization is to allow spatial change
-in the relative permittivity $\varepsilon_r(\vb{r})$
-and the relative permeability $\mu_r(\vb{r})$.
+in the relative permittivity $$\varepsilon_r(\vb{r})$$
+and the relative permeability $$\mu_r(\vb{r})$$.
We still assume that the medium is linear and isotropic, so:
$$\begin{aligned}
@@ -148,7 +148,7 @@ $$\begin{aligned}
= \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t}
\end{aligned}$$
-We then divide Ampère's law by $\varepsilon_r(\vb{r})$,
+We then divide Ampère's law by $$\varepsilon_r(\vb{r})$$,
take the curl, and substitute Faraday's law, giving:
$$\begin{aligned}
@@ -157,7 +157,7 @@ $$\begin{aligned}
= - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t}
\end{aligned}$$
-Next, we exploit linearity by decomposing $\vb{H}$ and $\vb{E}$
+Next, we exploit linearity by decomposing $$\vb{H}$$ and $$\vb{E}$$
into Fourier series, with terms given by:
$$\begin{aligned}
@@ -176,9 +176,9 @@ $$\begin{aligned}
= \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t)
\end{aligned}$$
-Dividing out $\exp(- i \omega t)$,
-we arrive at an eigenvalue problem for $\omega^2$,
-with $c = 1 / \sqrt{\mu_0 \varepsilon_0}$:
+Dividing out $$\exp(- i \omega t)$$,
+we arrive at an eigenvalue problem for $$\omega^2$$,
+with $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$:
$$\begin{aligned}
\boxed{
@@ -187,12 +187,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Compared to a uniform medium, $\omega$ is often not arbitrary here:
-there are discrete eigenvalues $\omega$,
-corresponding to discrete **modes** $\vb{H}(\vb{r})$.
+Compared to a uniform medium, $$\omega$$ is often not arbitrary here:
+there are discrete eigenvalues $$\omega$$,
+corresponding to discrete **modes** $$\vb{H}(\vb{r})$$.
-Next, we go through the same process to find an equation for $\vb{E}$.
-Starting from Faraday's law, we divide by $\mu_r(\vb{r})$,
+Next, we go through the same process to find an equation for $$\vb{E}$$.
+Starting from Faraday's law, we divide by $$\mu_r(\vb{r})$$,
take the curl, and insert Ampère's law:
$$\begin{aligned}
@@ -201,7 +201,7 @@ $$\begin{aligned}
= - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
\end{aligned}$$
-Then, by replacing $\vb{E}(\vb{r}, t)$ with our plane-wave ansatz,
+Then, by replacing $$\vb{E}(\vb{r}, t)$$ with our plane-wave ansatz,
we remove the time dependence:
$$\begin{aligned}
@@ -209,8 +209,8 @@ $$\begin{aligned}
= - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t)
\end{aligned}$$
-Which, after dividing out $\exp(- i \omega t)$,
-yields an analogous eigenvalue problem with $\vb{E}(r)$:
+Which, after dividing out $$\exp(- i \omega t)$$,
+yields an analogous eigenvalue problem with $$\vb{E}(r)$$:
$$\begin{aligned}
\boxed{
@@ -220,13 +220,13 @@ $$\begin{aligned}
\end{aligned}$$
Usually, it is a reasonable approximation
-to say $\mu_r(\vb{r}) = 1$,
-in which case the equation for $\vb{H}(\vb{r})$
+to say $$\mu_r(\vb{r}) = 1$$,
+in which case the equation for $$\vb{H}(\vb{r})$$
becomes a Hermitian eigenvalue problem,
-and is thus easier to solve than for $\vb{E}(\vb{r})$.
+and is thus easier to solve than for $$\vb{E}(\vb{r})$$.
Keep in mind, however, that in any case,
-the solutions $\vb{H}(\vb{r})$ and/or $\vb{E}(\vb{r})$
+the solutions $$\vb{H}(\vb{r})$$ and/or $$\vb{E}(\vb{r})$$
must satisfy the two Maxwell's equations that were not explicitly used:
$$\begin{aligned}
@@ -237,8 +237,8 @@ $$\begin{aligned}
This is equivalent to demanding that the resulting waves are *transverse*,
or in other words,
-the wavevector $\vb{k}$ must be perpendicular to
-the amplitudes $\vb{H}_0$ and $\vb{E}_0$.
+the wavevector $$\vb{k}$$ must be perpendicular to
+the amplitudes $$\vb{H}_0$$ and $$\vb{E}_0$$.
## References
diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md
index f62fb56..02ed856 100644
--- a/source/know/concept/equation-of-motion-theory/index.md
+++ b/source/know/concept/equation-of-motion-theory/index.md
@@ -13,9 +13,9 @@ is a method to calculate the time evolution of a system's properties
using [Green's functions](/know/concept/greens-functions/).
Starting from the definition of
-the retarded single-particle Green's function $G_{\nu \nu'}^R(t, t')$,
-we simply take the $t$-derivative
-(we could do the same with the advanced function $G_{\nu \nu'}^A$):
+the retarded single-particle Green's function $$G_{\nu \nu'}^R(t, t')$$,
+we simply take the $$t$$-derivative
+(we could do the same with the advanced function $$G_{\nu \nu'}^A$$):
$$\begin{aligned}
i \hbar \pdv{G^R_{\nu \nu'}(t, t')}{t}
@@ -27,22 +27,22 @@ $$\begin{aligned}
\end{aligned}$$
Where we have used that the derivative
-of a [Heaviside step function](/know/concept/heaviside-step-function/) $\Theta$
-is a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$.
+of a [Heaviside step function](/know/concept/heaviside-step-function/) $$\Theta$$
+is a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$.
Also, from the [second quantization](/know/concept/second-quantization/),
-$\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$
-for $t = t'$ is zero when $\nu \neq \nu'$.
+$$\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$$
+for $$t = t'$$ is zero when $$\nu \neq \nu'$$.
Since we are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
-we know the equation of motion of $\hat{c}_\nu(t)$:
+we know the equation of motion of $$\hat{c}_\nu(t)$$:
$$\begin{aligned}
\dv{\hat{c}_\nu(t)}{t}
= \frac{i}{\hbar} \comm{\hat{H}_0(t)}{\hat{c}_\nu(t)} + \frac{i}{\hbar} \comm{\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}
\end{aligned}$$
-Where the single-particle part of the Hamiltonian $\hat{H}_0$
-and the interaction part $\hat{H}_\mathrm{int}$
+Where the single-particle part of the Hamiltonian $$\hat{H}_0$$
+and the interaction part $$\hat{H}_\mathrm{int}$$
are assumed to be time-independent in the Schrödinger picture.
We thus get:
@@ -52,8 +52,8 @@ $$\begin{aligned}
\Expval{\Comm{\comm{\hat{H}_0}{\hat{c}_\nu} + \comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
\end{aligned}$$
-The most general form of $\hat{H}_0$, for any basis,
-is as follows, where $u_{\nu' \nu''}$ are constants:
+The most general form of $$\hat{H}_0$$, for any basis,
+is as follows, where $$u_{\nu' \nu''}$$ are constants:
$$\begin{aligned}
\hat{H}_0
@@ -68,7 +68,7 @@ $$\begin{aligned}
-Using the commutator identity for $\comm{A B}{C}$,
+Using the commutator identity for $$\comm{A B}{C}$$,
we decompose it like so:
$$\begin{aligned}
@@ -105,11 +105,12 @@ $$\begin{aligned}
- 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
= - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
\end{aligned}$$
+
-Substituting this into $G_{\nu \nu'}^R$'s equation of motion,
-we recognize another Green's function $G_{\nu'' \nu'}^R$:
+Substituting this into $$G_{\nu \nu'}^R$$'s equation of motion,
+we recognize another Green's function $$G_{\nu'' \nu'}^R$$:
$$\begin{aligned}
i \hbar \pdv{G^R_{\nu \nu'}}{t}
@@ -132,9 +133,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $D_{\nu \nu'}^R$ represents a correction due to interactions $\hat{H}_\mathrm{int}$,
+Where $$D_{\nu \nu'}^R$$ represents a correction due to interactions $$\hat{H}_\mathrm{int}$$,
and also has the form of a retarded Green's function,
-but with $\hat{c}_{\nu}$ replaced by $\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$:
+but with $$\hat{c}_{\nu}$$ replaced by $$\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$$:
$$\begin{aligned}
\boxed{
@@ -143,19 +144,19 @@ $$\begin{aligned}
}
\end{aligned}$$
-Unfortunately, calculating $D_{\nu \nu'}^R$
-might still not be doable due to $\hat{H}_\mathrm{int}$.
-The key idea of equation-of-motion theory is to either approximate $D_{\nu \nu'}^R$ now,
-or to differentiate it again $i \hbar \idv{D_{\nu \nu'}^R}{t}$,
+Unfortunately, calculating $$D_{\nu \nu'}^R$$
+might still not be doable due to $$\hat{H}_\mathrm{int}$$.
+The key idea of equation-of-motion theory is to either approximate $$D_{\nu \nu'}^R$$ now,
+or to differentiate it again $$i \hbar \idv{D_{\nu \nu'}^R}{t}$$,
and try again for the resulting corrections,
until a solvable equation is found.
There is no guarantee that that will ever happen;
if not, one of the corrections needs to be approximated.
-For non-interacting particles $\hat{H}_\mathrm{int} = 0$,
-so clearly $D_{\nu \nu'}^R$ trivially vanishes then.
-Let us assume that $\hat{H}_0$ is also time-independent,
-such that $G_{\nu'' \nu'}^R$ only depends on the difference $t - t'$:
+For non-interacting particles $$\hat{H}_\mathrm{int} = 0$$,
+so clearly $$D_{\nu \nu'}^R$$ trivially vanishes then.
+Let us assume that $$\hat{H}_0$$ is also time-independent,
+such that $$G_{\nu'' \nu'}^R$$ only depends on the difference $$t - t'$$:
$$\begin{aligned}
\sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t - t')
@@ -163,14 +164,14 @@ $$\begin{aligned}
\end{aligned}$$
We take the [Fourier transform](/know/concept/fourier-transform/)
-$(t \!-\! t') \to (\omega + i \eta)$, where $\eta \to 0^+$ ensures convergence:
+$$(t \!-\! t') \to (\omega + i \eta)$$, where $$\eta \to 0^+$$ ensures convergence:
$$\begin{aligned}
\sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
= \delta_{\nu \nu'}
\end{aligned}$$
-If we assume a diagonal basis $u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$,
+If we assume a diagonal basis $$u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$$,
this reduces to the following:
$$\begin{aligned}
diff --git a/source/know/concept/euler-bernoulli-law/index.md b/source/know/concept/euler-bernoulli-law/index.md
index c9dc7af..dad67ca 100644
--- a/source/know/concept/euler-bernoulli-law/index.md
+++ b/source/know/concept/euler-bernoulli-law/index.md
@@ -13,18 +13,18 @@ subject to certain simplifying assumptions,
which are generally valid for beams that are narrow,
i.e. longitudinally much larger than transversely.
-Consider a beam of length $L$, placed upright
-on the $z = 0$ plane, above the origin.
-If we pull the top of this beam in the postive $y$-direction,
+Consider a beam of length $$L$$, placed upright
+on the $$z = 0$$ plane, above the origin.
+If we pull the top of this beam in the postive $$y$$-direction,
we assume that it bends uniformly,
-i.e. with constant radius of [curvature](/know/concept/curvature/) $R$.
+i.e. with constant radius of [curvature](/know/concept/curvature/) $$R$$.
We also assume that the bending is **shear-free**:
if we treat the beam as a bundle of elastic strings,
then there is no friction between them.
-The central string has its length unchanged (i.e. still $L$),
-while an arbitrary non-central string is extended or compressed to $L'$.
-The [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) element $u_{zz}$ is then:
+The central string has its length unchanged (i.e. still $$L$$),
+while an arbitrary non-central string is extended or compressed to $$L'$$.
+The [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) element $$u_{zz}$$ is then:
$$\begin{aligned}
u_{zz}
@@ -32,11 +32,11 @@ $$\begin{aligned}
\end{aligned}$$
Because the bending is uniform, the central string
-is an arc with radius $R$ and central angle $\theta$,
-where $L = \theta R$.
-The non-central string has $L' = \theta R'$,
-where $R'$ is geometrically shown to be $R' = R - y$,
-with $y$ being the $y$-coordinate of that string at the beam's base.
+is an arc with radius $$R$$ and central angle $$\theta$$,
+where $$L = \theta R$$.
+The non-central string has $$L' = \theta R'$$,
+where $$R'$$ is geometrically shown to be $$R' = R - y$$,
+with $$y$$ being the $$y$$-coordinate of that string at the beam's base.
So:
$$\begin{aligned}
@@ -48,17 +48,17 @@ $$\begin{aligned}
By assumption, there are no shear stresses
and no forces acting on the beam's sides,
so the only nonzero component of the
-[Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
-is $\sigma_{zz}$, given by [Hooke's law](/know/concept/hookes-law/):
+[Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$
+is $$\sigma_{zz}$$, given by [Hooke's law](/know/concept/hookes-law/):
$$\begin{aligned}
\sigma_{zz} = E u_{zz}
\end{aligned}$$
-Where $E$ is the elastic modulus of the material.
+Where $$E$$ is the elastic modulus of the material.
By Hooke's inverse law,
the other nonzero strain components are as follows,
-where $\nu$ is Poisson's ratio:
+where $$\nu$$ is Poisson's ratio:
$$\begin{aligned}
u_{xx}
@@ -68,8 +68,8 @@ $$\begin{aligned}
= \nu \frac{y}{R}
\end{aligned}$$
-For completeness, we turn the strain tensor $\hat{u}$
-into a full displacement field $\va{u}$:
+For completeness, we turn the strain tensor $$\hat{u}$$
+into a full displacement field $$\va{u}$$:
$$\begin{aligned}
\boxed{
@@ -86,8 +86,8 @@ $$\begin{aligned}
-By integrating the above strains $u_{ii} = \ipdv{u_i}{i}$,
-we get the components of $\va{u}$:
+By integrating the above strains $$u_{ii} = \ipdv{u_i}{i}$$,
+we get the components of $$\va{u}$$:
$$\begin{aligned}
u_x
@@ -100,9 +100,9 @@ $$\begin{aligned}
= - \frac{y z}{R} + f_z(x, y)
\end{aligned}$$
-Where $f_x$, $f_y$ and $f_z$ are integration constants,
-which we find by demanding that the off-diagonal strains $u_{ij}$ are zero.
-Starting with $u_{xz} = 0$:
+Where $$f_x$$, $$f_y$$ and $$f_z$$ are integration constants,
+which we find by demanding that the off-diagonal strains $$u_{ij}$$ are zero.
+Starting with $$u_{xz} = 0$$:
$$\begin{aligned}
0
@@ -111,8 +111,8 @@ $$\begin{aligned}
= \frac{1}{2} \Big( \pdv{f_x}{z} + \pdv{f_z}{x} \Big)
\end{aligned}$$
-Here, only $f_x$ may depend on $z$,
-and only $f_z$ may depend on $x$.
+Here, only $$f_x$$ may depend on $$z$$,
+and only $$f_z$$ may depend on $$x$$.
This equation thus tell us:
$$\begin{aligned}
@@ -123,8 +123,8 @@ $$\begin{aligned}
= - x \: g(y)
\end{aligned}$$
-Where $g(y)$ is an unknown integration constant.
-Moving on to $u_{xy} = 0$:
+Where $$g(y)$$ is an unknown integration constant.
+Moving on to $$u_{xy} = 0$$:
$$\begin{aligned}
0
@@ -132,9 +132,9 @@ $$\begin{aligned}
= \frac{1}{2} \Big( \nu \frac{x}{R} + \pdv{f_x}{y} + \pdv{f_y}{x} \Big)
\end{aligned}$$
-Only $f_x$ may contain $y$,
-so its $y$-derivative must be a constant,
-so $g(y) = C y$. Therefore:
+Only $$f_x$$ may contain $$y$$,
+so its $$y$$-derivative must be a constant,
+so $$g(y) = C y$$. Therefore:
$$\begin{aligned}
f_x(y, z)
@@ -147,8 +147,8 @@ $$\begin{aligned}
= - C x y
\end{aligned}$$
-Where $h(z)$ is an unknown integration constant.
-Finally, we put everything in $u_{yz} = 0$:
+Where $$h(z)$$ is an unknown integration constant.
+Finally, we put everything in $$u_{yz} = 0$$:
$$\begin{aligned}
0
@@ -157,8 +157,8 @@ $$\begin{aligned}
= \frac{1}{2} \Big( \!-\! 2 C x + \dv{h}{z} - \frac{z}{R} \Big)
\end{aligned}$$
-Only the first term contains $x$, so to satisfy this equation, we must set $C = 0$.
-The remaining terms then tell us that $h(z) = z^2 / (2 R)$.
+Only the first term contains $$x$$, so to satisfy this equation, we must set $$C = 0$$.
+The remaining terms then tell us that $$h(z) = z^2 / (2 R)$$.
Therefore:
$$\begin{aligned}
@@ -169,12 +169,12 @@ $$\begin{aligned}
f_z = 0
\end{aligned}$$
-Inserting this into the components $u_x$, $u_y$ and $u_z$
+Inserting this into the components $$u_x$$, $$u_y$$ and $$u_z$$
then yields the full displacement field.
-The inverse FT of the forward FT of $f(\vb{x})$ must be equal to $f(\vb{x})$ again, so:
+The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so:
$$\begin{aligned}
\hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
@@ -180,14 +180,15 @@ $$\begin{aligned}
&= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
= \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
\end{aligned}$$
+
-Differentiation is more complicated for $N > 1$,
+Differentiation is more complicated for $$N > 1$$,
but the FT is still useful,
-notably for the Laplacian $\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$.
-Let $|\vb{k}|$ be the norm of $\vb{k}$,
-then for a localized $f$:
+notably for the Laplacian $$\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$$.
+Let $$|\vb{k}|$$ be the norm of $$\vb{k}$$,
+then for a localized $$f$$:
$$\begin{aligned}
\boxed{
@@ -201,9 +202,9 @@ $$\begin{aligned}
-We insert $\nabla^2 f$ into the FT,
+We insert $$\nabla^2 f$$ into the FT,
decompose the exponential and the Laplacian,
-and then integrate by parts (limits $\pm \infty$ omitted):
+and then integrate by parts (limits $$\pm \infty$$ omitted):
$$\begin{aligned}
\hat{\mathcal{F}}\{\nabla^2 f\}
@@ -216,7 +217,7 @@ $$\begin{aligned}
\end{aligned}$$
Just like in 1D, we get rid of the boundary term
-by assuming that all derivatives $\idv{f}{x_n}$ are nicely localized.
+by assuming that all derivatives $$\idv{f}{x_n}$$ are nicely localized.
To proceed, we then integrate by parts again:
$$\begin{aligned}
@@ -228,13 +229,14 @@ $$\begin{aligned}
\end{aligned}$$
Once again, we remove the boundary term
-by assuming that $f$ is localized, yielding:
+by assuming that $$f$$ is localized, yielding:
$$\begin{aligned}
\hat{\mathcal{F}}\{\nabla^2 f\}
&= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
= - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
\end{aligned}$$
+
diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md
index 5d53e79..c954272 100644
--- a/source/know/concept/fredholm-alternative/index.md
+++ b/source/know/concept/fredholm-alternative/index.md
@@ -8,51 +8,51 @@ layout: "concept"
---
The **Fredholm alternative** is a theorem regarding equations involving
-a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/hilbert-space/),
+a linear operator $$\hat{L}$$ on a [Hilbert space](/know/concept/hilbert-space/),
and is useful in the context of multiple-scale perturbation theory.
It is an *alternative* because it gives two mutually exclusive options,
given here in [Dirac notation](/know/concept/dirac-notation/):
-1. $\hat{L} \Ket{u} = \Ket{f}$ has a unique solution $\Ket{u}$ for every $\Ket{f}$.
-2. $\hat{L}^\dagger \Ket{w} = 0$ has non-zero solutions.
- Then regarding $\hat{L} \Ket{u} = \Ket{f}$:
- 1. If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then it has infinitely many solutions $\Ket{u}$.
- 2. If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then it has no solutions $\Ket{u}$.
+1. $$\hat{L} \Ket{u} = \Ket{f}$$ has a unique solution $$\Ket{u}$$ for every $$\Ket{f}$$.
+2. $$\hat{L}^\dagger \Ket{w} = 0$$ has non-zero solutions.
+ Then regarding $$\hat{L} \Ket{u} = \Ket{f}$$:
+ 1. If $$\Inprod{w}{f} = 0$$ for all $$\Ket{w}$$, then it has infinitely many solutions $$\Ket{u}$$.
+ 2. If $$\Inprod{w}{f} \neq 0$$ for any $$\Ket{w}$$, then it has no solutions $$\Ket{u}$$.
-Where $\hat{L}^\dagger$ is the adjoint of $\hat{L}$.
-In other words, $\hat{L} \Ket{u} = \Ket{f}$ has non-trivial solutions if
-and only if for all $\Ket{w}$ (including the trivial case $\Ket{w} = 0$)
-it holds that $\Inprod{w}{f} = 0$.
+Where $$\hat{L}^\dagger$$ is the adjoint of $$\hat{L}$$.
+In other words, $$\hat{L} \Ket{u} = \Ket{f}$$ has non-trivial solutions if
+and only if for all $$\Ket{w}$$ (including the trivial case $$\Ket{w} = 0$$)
+it holds that $$\Inprod{w}{f} = 0$$.
As a specific example,
-if $\hat{L}$ is a matrix and the kets are vectors,
+if $$\hat{L}$$ is a matrix and the kets are vectors,
this theorem can alternatively be stated as follows using the determinant:
-1. If $\mathrm{det}(\hat{L}) \neq 0$, then $\hat{L} \vec{u} = \vec{f}$
- has a unique solution $\vec{u}$ for every $\vec{f}$.
-2. If $\mathrm{det}(\hat{L}) = 0$,
- then $\hat{L}^\dagger \vec{w} = \vec{0}$ has non-zero solutions.
- Then regarding $\hat{L} \vec{u} = \vec{f}$:
- 1. If $\vec{w} \cdot \vec{f} = 0$ for all $\vec{w}$, then it has
- infinitely many solutions $\vec{u}$.
- 2. If $\vec{w} \cdot \vec{f} \neq 0$ for any $\vec{w}$, then it has
- no solutions $\vec{u}$.
+1. If $$\mathrm{det}(\hat{L}) \neq 0$$, then $$\hat{L} \vec{u} = \vec{f}$$
+ has a unique solution $$\vec{u}$$ for every $$\vec{f}$$.
+2. If $$\mathrm{det}(\hat{L}) = 0$$,
+ then $$\hat{L}^\dagger \vec{w} = \vec{0}$$ has non-zero solutions.
+ Then regarding $$\hat{L} \vec{u} = \vec{f}$$:
+ 1. If $$\vec{w} \cdot \vec{f} = 0$$ for all $$\vec{w}$$, then it has
+ infinitely many solutions $$\vec{u}$$.
+ 2. If $$\vec{w} \cdot \vec{f} \neq 0$$ for any $$\vec{w}$$, then it has
+ no solutions $$\vec{u}$$.
Consequently, the Fredholm alternative is also brought up
in the context of eigenvalue problems.
-Define $\hat{M} = (\hat{L} - \lambda \hat{I})$,
-where $\lambda$ is an eigenvalue of $\hat{L}$
-if and only if $\mathrm{det}(\hat{M}) = 0$.
-Then for the equation $\hat{M} \Ket{u} = \Ket{f}$, we can say that:
-
-1. If $\lambda$ is *not* an eigenvalue,
- then there is a unique solution $\Ket{u}$ for each $\Ket{f}$.
-2. If $\lambda$ is an eigenvalue, then $\hat{M}^\dagger \Ket{w} = 0$
+Define $$\hat{M} = (\hat{L} - \lambda \hat{I})$$,
+where $$\lambda$$ is an eigenvalue of $$\hat{L}$$
+if and only if $$\mathrm{det}(\hat{M}) = 0$$.
+Then for the equation $$\hat{M} \Ket{u} = \Ket{f}$$, we can say that:
+
+1. If $$\lambda$$ is *not* an eigenvalue,
+ then there is a unique solution $$\Ket{u}$$ for each $$\Ket{f}$$.
+2. If $$\lambda$$ is an eigenvalue, then $$\hat{M}^\dagger \Ket{w} = 0$$
has non-zero solutions. Then:
- 1. If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then there are
- infinitely many solutions $\Ket{u}$.
- 2. If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then there are no
- solutions $\Ket{u}$.
+ 1. If $$\Inprod{w}{f} = 0$$ for all $$\Ket{w}$$, then there are
+ infinitely many solutions $$\Ket{u}$$.
+ 2. If $$\Inprod{w}{f} \neq 0$$ for any $$\Ket{w}$$, then there are no
+ solutions $$\Ket{u}$$.
diff --git a/source/know/concept/fundamental-solution/index.md b/source/know/concept/fundamental-solution/index.md
index e8ffda6..312cc2e 100644
--- a/source/know/concept/fundamental-solution/index.md
+++ b/source/know/concept/fundamental-solution/index.md
@@ -8,10 +8,10 @@ categories:
layout: "concept"
---
-Given a linear operator $\hat{L}$ acting on $x \in [a, b]$,
-its **fundamental solution** $G(x, x')$ is defined as the response
-of $\hat{L}$ to a [Dirac delta function](/know/concept/dirac-delta-function/)
-$\delta(x - x')$ for $x \in ]a, b[$:
+Given a linear operator $$\hat{L}$$ acting on $$x \in [a, b]$$,
+its **fundamental solution** $$G(x, x')$$ is defined as the response
+of $$\hat{L}$$ to a [Dirac delta function](/know/concept/dirac-delta-function/)
+$$\delta(x - x')$$ for $$x \in ]a, b[$$:
$$\begin{aligned}
\boxed{
@@ -20,17 +20,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $A$ is a constant, usually $1$.
+Where $$A$$ is a constant, usually $$1$$.
Fundamental solutions are often called **Green's functions**,
but are distinct from the (somewhat related)
[Green's functions](/know/concept/greens-functions/)
in many-body quantum theory.
-Note that the definition of $G(x, x')$ generalizes that of
+Note that the definition of $$G(x, x')$$ generalizes that of
the [impulse response](/know/concept/impulse-response/).
And likewise, due to the superposition principle,
-once $G$ is known, $\hat{L}$'s response $u(x)$ to
-*any* forcing function $f(x)$ can easily be found as follows:
+once $$G$$ is known, $$\hat{L}$$'s response $$u(x)$$ to
+*any* forcing function $$f(x)$$ can easily be found as follows:
$$\begin{aligned}
\hat{L} \{ u(x) \}
@@ -47,10 +47,10 @@ $$\begin{aligned}
-$\hat{L}$ only acts on $x$, so $x' \in ]a, b[$ is simply a parameter,
-meaning we are free to multiply the definition of $G$
-by the constant $f(x')$ on both sides,
-and exploit $\hat{L}$'s linearity:
+$$\hat{L}$$ only acts on $$x$$, so $$x' \in ]a, b[$$ is simply a parameter,
+meaning we are free to multiply the definition of $$G$$
+by the constant $$f(x')$$ on both sides,
+and exploit $$\hat{L}$$'s linearity:
$$\begin{aligned}
A f(x') \: \delta(x - x')
@@ -58,9 +58,9 @@ $$\begin{aligned}
= \hat{L}\{ f(x') \: G(x, x') \}
\end{aligned}$$
-We then integrate both sides over $x'$ in the interval $[a, b]$,
-allowing us to consume $\delta(x \!-\! x')$.
-Note that $\int \dd{x'}$ commutes with $\hat{L}$ acting on $x$:
+We then integrate both sides over $$x'$$ in the interval $$[a, b]$$,
+allowing us to consume $$\delta(x \!-\! x')$$.
+Note that $$\int \dd{x'}$$ commutes with $$\hat{L}$$ acting on $$x$$:
$$\begin{aligned}
A \int_a^b f(x') \: \delta(x - x') \dd{x'}
@@ -70,15 +70,15 @@ $$\begin{aligned}
&= \hat{L} \int_a^b f(x') \: G(x, x') \dd{x'}
\end{aligned}$$
-By definition, $\hat{L}$'s response $u(x)$ to $f(x)$
-satisfies $\hat{L}\{ u(x) \} = f(x)$, recognizable here.
+By definition, $$\hat{L}$$'s response $$u(x)$$ to $$f(x)$$
+satisfies $$\hat{L}\{ u(x) \} = f(x)$$, recognizable here.
While the impulse response is typically used for initial value problems,
-the fundamental solution $G$ is used for boundary value problems.
+the fundamental solution $$G$$ is used for boundary value problems.
Suppose those boundary conditions are homogeneous,
-i.e. $u(x)$ or one of its derivatives is zero at the boundaries.
+i.e. $$u(x)$$ or one of its derivatives is zero at the boundaries.
Then:
$$\begin{aligned}
@@ -95,20 +95,20 @@ $$\begin{aligned}
G_x(a, x') = 0
\end{aligned}$$
-This holds for all $x'$, and analogously for the other boundary $x = b$.
-In other words, the boundary conditions are built into $G$.
+This holds for all $$x'$$, and analogously for the other boundary $$x = b$$.
+In other words, the boundary conditions are built into $$G$$.
What if the boundary conditions are inhomogeneous?
-No problem: thanks to the linearity of $\hat{L}$,
-those conditions can be given to the homogeneous solution $u_h(x)$,
-where $\hat{L}\{ u_h(x) \} = 0$,
-such that the inhomogeneous solution $u_i(x) = u(x) - u_h(x)$
+No problem: thanks to the linearity of $$\hat{L}$$,
+those conditions can be given to the homogeneous solution $$u_h(x)$$,
+where $$\hat{L}\{ u_h(x) \} = 0$$,
+such that the inhomogeneous solution $$u_i(x) = u(x) - u_h(x)$$
has homogeneous boundaries again,
-so we can use $G$ as usual to find $u_i(x)$, and then just add $u_h(x)$.
+so we can use $$G$$ as usual to find $$u_i(x)$$, and then just add $$u_h(x)$$.
-If $\hat{L}$ is self-adjoint
+If $$\hat{L}$$ is self-adjoint
(see e.g. [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)),
-then the fundamental solution $G(x, x')$
+then the fundamental solution $$G(x, x')$$
has the following **reciprocity** boundary condition:
$$\begin{aligned}
@@ -122,8 +122,8 @@ $$\begin{aligned}
-Consider two parameters $x_1'$ and $x_2'$.
-The self-adjointness of $\hat{L}$ means that:
+Consider two parameters $$x_1'$$ and $$x_2'$$.
+The self-adjointness of $$\hat{L}$$ means that:
$$\begin{aligned}
\int_a^b G^*(x, x_1') \Big( \hat{L} \{ G(x, x_2') \} \Big) \dd{x}
@@ -135,6 +135,7 @@ $$\begin{aligned}
G^*(x_2', x_1')
&= G(x_1', x_2')
\end{aligned}$$
+
diff --git a/source/know/concept/fundamental-thermodynamic-relation/index.md b/source/know/concept/fundamental-thermodynamic-relation/index.md
index 392e0b3..0d945fa 100644
--- a/source/know/concept/fundamental-thermodynamic-relation/index.md
+++ b/source/know/concept/fundamental-thermodynamic-relation/index.md
@@ -10,35 +10,35 @@ layout: "concept"
The **fundamental thermodynamic relation** combines the first two
[laws of thermodynamics](/know/concept/laws-of-thermodynamics/),
-and gives the change of the internal energy $U$,
+and gives the change of the internal energy $$U$$,
which is a [thermodynamic potential](/know/concept/thermodynamic-potential/),
in terms of the change in
-entropy $S$, volume $V$, and the number of particles $N$.
+entropy $$S$$, volume $$V$$, and the number of particles $$N$$.
Starting from the first law of thermodynamics,
-we write an infinitesimal change in energy $\dd{U}$ as follows,
-where $T$ is the temperature and $P$ is the pressure:
+we write an infinitesimal change in energy $$\dd{U}$$ as follows,
+where $$T$$ is the temperature and $$P$$ is the pressure:
$$\begin{aligned}
\dd{U} &= \dd{Q} + \dd{W} = T \dd{S} - P \dd{V}
\end{aligned}$$
-The term $T \dd{S}$ comes from the second law of thermodynamics,
+The term $$T \dd{S}$$ comes from the second law of thermodynamics,
and represents the transfer of thermal energy,
-while $P \dd{V}$ represents physical work.
+while $$P \dd{V}$$ represents physical work.
However, we are missing a term, namely matter transfer.
-If particles can enter/leave the system (i.e. the population $N$ is variable),
-then each such particle costs an amount $\mu$ of energy,
-where $\mu$ is known as the **chemical potential**:
+If particles can enter/leave the system (i.e. the population $$N$$ is variable),
+then each such particle costs an amount $$\mu$$ of energy,
+where $$\mu$$ is known as the **chemical potential**:
$$\begin{aligned}
\dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N}
\end{aligned}$$
To generalize even further, there may be multiple species of particle,
-which each have a chemical potential $\mu_i$.
-In that case, we sum over all species $i$:
+which each have a chemical potential $$\mu_i$$.
+In that case, we sum over all species $$i$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md
index e1129d4..a59ccfe 100644
--- a/source/know/concept/ghz-paradox/index.md
+++ b/source/know/concept/ghz-paradox/index.md
@@ -12,7 +12,7 @@ layout: "concept"
The **Greenberger-Horne-Zeilinger** or **GHZ paradox**
is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/)
that does not use inequalities,
-but the three-particle entangled **GHZ state** $\Ket{\mathrm{GHZ}}$ instead,
+but the three-particle entangled **GHZ state** $$\Ket{\mathrm{GHZ}}$$ instead,
$$\begin{aligned}
\boxed{
@@ -21,10 +21,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Ket{0}$ and $\Ket{1}$ are qubit states,
-for example, the eigenvalues of the Pauli matrix $\hat{\sigma}_z$.
+Where $$\Ket{0}$$ and $$\Ket{1}$$ are qubit states,
+for example, the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$.
-If we now apply certain products of the Pauli matrices $\hat{\sigma}_x$ and $\hat{\sigma}_y$
+If we now apply certain products of the Pauli matrices $$\hat{\sigma}_x$$ and $$\hat{\sigma}_y$$
to the three particles, we find:
@@ -45,10 +45,10 @@ $$\begin{aligned}
\end{aligned}$$
In other words, the GHZ state is a simultaneous eigenstate of these composite operators,
-with eigenvalues $+1$ and $-1$, respectively.
+with eigenvalues $$+1$$ and $$-1$$, respectively.
Let us introduce two other product operators,
such that we have a set of four observables,
-for which $\Ket{\mathrm{GHZ}}$ gives these eigenvalues:
+for which $$\Ket{\mathrm{GHZ}}$$ gives these eigenvalues:
$$\begin{aligned}
\hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
@@ -66,7 +66,7 @@ $$\begin{aligned}
According to any local hidden variable (LHV) theory,
the measurement outcomes of the operators are predetermined,
-and the three particles $A$, $B$ and $C$ can be measured separately,
+and the three particles $$A$$, $$B$$ and $$C$$ can be measured separately,
or in other words, the eigenvalues can be factorized:
$$\begin{aligned}
@@ -83,7 +83,7 @@ $$\begin{aligned}
\quad &\implies \quad -1 = m_y^A m_y^B m_x^C
\end{aligned}$$
-Where $m_x^A = \pm 1$ etc.
+Where $$m_x^A = \pm 1$$ etc.
Let us now multiply both sides of these four equations together:
$$\begin{aligned}
@@ -94,13 +94,13 @@ $$\begin{aligned}
&= (m_x^A)^2 (m_x^B)^2 (m_x^C)^2 (m_y^A)^2 (m_y^B)^2 (m_y^C)^2
\end{aligned}$$
-This is a contradiction: the left-hand side is $-1$,
-but all six factors on the right are $+1$.
+This is a contradiction: the left-hand side is $$-1$$,
+but all six factors on the right are $$+1$$.
This means that we must have made an incorrect assumption along the way.
Our only assumption was that we could factorize the eigenvalues,
-so that e.g. particle $A$ could be measured on its own
-without an "action-at-a-distance" effect on $B$ or $C$.
+so that e.g. particle $$A$$ could be measured on its own
+without an "action-at-a-distance" effect on $$B$$ or $$C$$.
However, because that leads us to a contradiction,
we must conclude that action-at-a-distance exists,
and that therefore all LHV-based theories are invalid.
diff --git a/source/know/concept/grad-shafranov-equation/index.md b/source/know/concept/grad-shafranov-equation/index.md
index 35d23f4..b86c032 100644
--- a/source/know/concept/grad-shafranov-equation/index.md
+++ b/source/know/concept/grad-shafranov-equation/index.md
@@ -20,7 +20,7 @@ We would like to find the equilibrium state of the plasma
in the general case of a reactor with toroidal symmetry.
Using ideal [magnetohydrodynamics](/know/concept/magnetohydrodynamics/) (MHD),
we start by assuming that the fluid is stationary,
-and that the confining field $\vb{B}$ is fixed:
+and that the confining field $$\vb{B}$$ is fixed:
$$\begin{aligned}
\vb{u}
@@ -36,7 +36,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Notice that $\vb{E} = 0$ is a result of the ideal generalized Ohm's law.
+Notice that $$\vb{E} = 0$$ is a result of the ideal generalized Ohm's law.
Under these assumptions, the relevant MHD equations to be solved are
Gauss' law for magnetism, Ampère's law, and the MHD momentum equation, respectively:
@@ -54,11 +54,11 @@ $$\begin{aligned}
The goal is to analyze them in this order,
exploiting toroidal symmetry along the way,
to arrive at a general equilibrium condition.
-[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $(r, \theta, z)$
-are a natural choice, with the $z$-axis running through the middle of the torus.
+[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $$(r, \theta, z)$$
+are a natural choice, with the $$z$$-axis running through the middle of the torus.
-As preparation, it is a good idea to write $\vb{B}$
-as the curl of a magnetic vector potential $\vb{A}$,
+As preparation, it is a good idea to write $$\vb{B}$$
+as the curl of a magnetic vector potential $$\vb{A}$$,
which looks like this in cylindrical polar coordinates:
$$\begin{aligned}
@@ -76,7 +76,7 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-Here, it is convenient to define the so-called **stream function** $\psi$ as follows:
+Here, it is convenient to define the so-called **stream function** $$\psi$$ as follows:
$$\begin{aligned}
\boxed{
@@ -85,8 +85,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Such that $\vb{B}$ can be written as below,
-where we will regard $B_\theta$ as a given quantity:
+Such that $$\vb{B}$$ can be written as below,
+where we will regard $$B_\theta$$ as a given quantity:
$$\begin{aligned}
\vb{B}
@@ -103,7 +103,7 @@ $$\begin{aligned}
Inserting this into Gauss' law,
we see that it is trivially satisfied,
-thanks to circular symmetry guaranteeing that $\ipdv{B_\theta}{\theta} = 0$:
+thanks to circular symmetry guaranteeing that $$\ipdv{B_\theta}{\theta} = 0$$:
$$\begin{aligned}
0
@@ -116,8 +116,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-What matters is that we have expressions for the components of $\vb{B}$.
-Moving on, to find the current density $\vb{J}$,
+What matters is that we have expressions for the components of $$\vb{B}$$.
+Moving on, to find the current density $$\vb{J}$$,
we use Ampère's law and symmetry to get:
@@ -138,9 +138,9 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-Where we have assumed that $B_\theta$ depends only on $r$, not $z$ or $\theta$.
+Where we have assumed that $$B_\theta$$ depends only on $$r$$, not $$z$$ or $$\theta$$.
Substituting this into the MHD momentum equation
-gives the following pressure gradient $\nabla p$:
+gives the following pressure gradient $$\nabla p$$:
$$\begin{aligned}
\nabla p
@@ -157,8 +157,8 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-Now, the idea is to focus on this $r$-component to get an equation for $\psi$,
-whose solution can then be used to calculate the $\theta$ and $z$-components of $\nabla p$.
+Now, the idea is to focus on this $$r$$-component to get an equation for $$\psi$$,
+whose solution can then be used to calculate the $$\theta$$ and $$z$$-components of $$\nabla p$$.
Therefore, we evaluate:
$$\begin{aligned}
@@ -176,8 +176,8 @@ $$\begin{aligned}
- \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
\end{aligned}$$
-By using the chain rule to rewrite $\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$,
-we get $\ipdv{\psi}{r}$ in each term:
+By using the chain rule to rewrite $$\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$$,
+we get $$\ipdv{\psi}{r}$$ in each term:
$$\begin{aligned}
\pdv{\psi}{r} \pdv{p}{\psi}
@@ -185,7 +185,7 @@ $$\begin{aligned}
- \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta
\end{aligned}$$
-Dividing out $\ipdv{\psi}{r}$ and multiplying by $\mu_0 r^2$
+Dividing out $$\ipdv{\psi}{r}$$ and multiplying by $$\mu_0 r^2$$
leads us to the **Grad-Shafranov equation**,
which gives the equilibrium condition of a plasma in a toroidal reactor:
@@ -196,13 +196,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Weirdly, $\psi$ appears both as an unknown and as a differentiation variable,
+Weirdly, $$\psi$$ appears both as an unknown and as a differentiation variable,
but this equation can still be solved analytically by
-assuming a certain $\psi$-dependence of $p$ and $r B_\theta$.
+assuming a certain $$\psi$$-dependence of $$p$$ and $$r B_\theta$$.
-Suppose that $B_\theta$ is induced by a poloidal electrical current $I_\mathrm{pol}$,
+Suppose that $$B_\theta$$ is induced by a poloidal electrical current $$I_\mathrm{pol}$$,
i.e. a current around the "tube" of the torus,
-then, assuming $I_\mathrm{pol}$ only depends on $r$, we have:
+then, assuming $$I_\mathrm{pol}$$ only depends on $$r$$, we have:
$$\begin{aligned}
B_\theta
diff --git a/source/know/concept/gram-schmidt-method/index.md b/source/know/concept/gram-schmidt-method/index.md
index 374b169..70ad512 100644
--- a/source/know/concept/gram-schmidt-method/index.md
+++ b/source/know/concept/gram-schmidt-method/index.md
@@ -9,36 +9,36 @@ layout: "concept"
---
Given a set of linearly independent non-orthonormal vectors
-$\ket{V_1}, \ket{V_2}, ...$ from a [Hilbert space](/know/concept/hilbert-space/),
+$$\ket{V_1}, \ket{V_2}, ...$$ from a [Hilbert space](/know/concept/hilbert-space/),
the **Gram-Schmidt method**
-turns them into an orthonormal set $\ket{n_1}, \ket{n_2}, ...$ as follows:
+turns them into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows:
-1. Take the first vector $\ket{V_1}$ and normalize it to get $\ket{n_1}$:
+1. Take the first vector $$\ket{V_1}$$ and normalize it to get $$\ket{n_1}$$:
$$\begin{aligned}
\ket{n_1} = \frac{\ket{V_1}}{\sqrt{\inprod{V_1}{V_1}}}
\end{aligned}$$
-2. Begin loop. Take the next non-orthonormal vector $\ket{V_j}$, and
+2. Begin loop. Take the next non-orthonormal vector $$\ket{V_j}$$, and
subtract from it its projection onto every already-processed vector:
$$\begin{aligned}
\ket{n_j'} = \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} - \ket{n_2} \inprod{n_2}{V_j} - ... - \ket{n_{j-1}} \inprod{n_{j-1}}{V_{j-1}}
\end{aligned}$$
- This leaves only the part of $\ket{V_j}$ which is orthogonal to
- $\ket{n_1}$, $\ket{n_2}$, etc. This why the input vectors must be
- linearly independent; otherwise $\Ket{n_j'}$ may become zero at some
+ This leaves only the part of $$\ket{V_j}$$ which is orthogonal to
+ $$\ket{n_1}$$, $$\ket{n_2}$$, etc. This why the input vectors must be
+ linearly independent; otherwise $$\Ket{n_j'}$$ may become zero at some
point.
-3. Normalize the resulting ortho*gonal* vector $\ket{n_j'}$ to make it
+3. Normalize the resulting ortho*gonal* vector $$\ket{n_j'}$$ to make it
ortho*normal*:
$$\begin{aligned}
\ket{n_j} = \frac{\ket{n_j'}}{\sqrt{\inprod{n_j'}{n_j'}}}
\end{aligned}$$
-4. Loop back to step 2, taking the next vector $\ket{V_{j+1}}$.
+4. Loop back to step 2, taking the next vector $$\ket{V_{j+1}}$$.
If you are unfamiliar with this notation, take a look at [Dirac notation](/know/concept/dirac-notation/).
diff --git a/source/know/concept/grand-canonical-ensemble/index.md b/source/know/concept/grand-canonical-ensemble/index.md
index 4f66fd9..62ca896 100644
--- a/source/know/concept/grand-canonical-ensemble/index.md
+++ b/source/know/concept/grand-canonical-ensemble/index.md
@@ -11,18 +11,18 @@ layout: "concept"
The **grand canonical ensemble** or **μVT ensemble**
extends the [canonical ensemble](/know/concept/canonical-ensemble/)
-by allowing the exchange of both energy $U$ and particles $N$
+by allowing the exchange of both energy $$U$$ and particles $$N$$
with an external reservoir,
so that the conserved state functions are
-the temperature $T$, the volume $V$, and the chemical potential $\mu$.
+the temperature $$T$$, the volume $$V$$, and the chemical potential $$\mu$$.
The derivation is practically identical to that of the canonical ensemble.
-We refer to the system of interest as $A$,
-and the reservoir as $B$.
-In total, $A\!+\!B$ has energy $U$ and population $N$.
+We refer to the system of interest as $$A$$,
+and the reservoir as $$B$$.
+In total, $$A\!+\!B$$ has energy $$U$$ and population $$N$$.
-Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
-Then the probability that $A$ is in a specific microstate $s_A$ is as follows:
+Let $$c_B(U_B)$$ be the number of $$B$$-microstates with energy $$U_B$$.
+Then the probability that $$A$$ is in a specific microstate $$s_A$$ is as follows:
$$\begin{aligned}
p(s)
@@ -30,11 +30,11 @@ $$\begin{aligned}
\end{aligned}$$
Then, as for the canonical ensemble,
-we assume $U_B \gg U_A$ and $N_B \gg N_A$,
-and approximate $\ln{p(s_A)}$
-by Taylor-expanding $\ln{c_B}$ around $U_B = U$ and $N_B = N$.
+we assume $$U_B \gg U_A$$ and $$N_B \gg N_A$$,
+and approximate $$\ln{p(s_A)}$$
+by Taylor-expanding $$\ln{c_B}$$ around $$U_B = U$$ and $$N_B = N$$.
The resulting probability distribution is known as the **Gibbs distribution**,
-with $\beta \equiv 1/(kT)$:
+with $$\beta \equiv 1/(kT)$$:
$$\begin{aligned}
\boxed{
@@ -42,7 +42,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the normalizing **grand partition function** $\mathcal{Z}(\mu, V, T)$ is defined as follows:
+Where the normalizing **grand partition function** $$\mathcal{Z}(\mu, V, T)$$ is defined as follows:
$$\begin{aligned}
\boxed{
@@ -52,9 +52,9 @@ $$\begin{aligned}
In contrast to the canonical ensemble,
whose [thermodynamic potential](/know/concept/thermodynamic-potential/)
-was the Helmholtz free energy $F$,
+was the Helmholtz free energy $$F$$,
the grand canonical ensemble instead
-minimizes the **grand potential** $\Omega$:
+minimizes the **grand potential** $$\Omega$$:
$$\begin{aligned}
\boxed{
@@ -64,8 +64,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-So $\mathcal{Z} = \exp(- \beta \Omega)$.
-This is proven in the same way as for $F$ in the canonical ensemble.
+So $$\mathcal{Z} = \exp(- \beta \Omega)$$.
+This is proven in the same way as for $$F$$ in the canonical ensemble.
diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md
index 0e53945..ddba2cd 100644
--- a/source/know/concept/greens-functions/index.md
+++ b/source/know/concept/greens-functions/index.md
@@ -25,12 +25,12 @@ except in a special case, see below.
If the two operators are single-particle creation/annihilation operators,
then we get the **single-particle Green's functions**,
-for which the symbol $G$ is used.
+for which the symbol $$G$$ is used.
-The **time-ordered** or **causal Green's function** $G_{\nu \nu'}$ is as follows,
-where $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/),
-$\nu$ and $\nu'$ are single-particle states,
-and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.:
+The **time-ordered** or **causal Green's function** $$G_{\nu \nu'}$$ is as follows,
+where $$\mathcal{T}$$ is the [time-ordered product](/know/concept/time-ordered-product/),
+$$\nu$$ and $$\nu'$$ are single-particle states,
+and $$\hat{c}_\nu$$ annihilates a particle from $$\nu$$, etc.:
$$\begin{aligned}
\boxed{
@@ -39,15 +39,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-The expectation value $\Expval{}$ is
+The expectation value $$\Expval{}$$ is
with respect to thermodynamic equilibrium.
This is sometimes in the [canonical ensemble](/know/concept/canonical-ensemble/)
(for some two-particle Green's functions, see below),
but usually in the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
since we are adding/removing particles.
-In the latter case, we assume that the chemical potential $\mu$
-is already included in the Hamiltonian $\hat{H}$.
-Explicitly, for a complete set of many-particle states $\Ket{\Psi_n}$, we have:
+In the latter case, we assume that the chemical potential $$\mu$$
+is already included in the Hamiltonian $$\hat{H}$$.
+Explicitly, for a complete set of many-particle states $$\Ket{\Psi_n}$$, we have:
$$\begin{aligned}
G_{\nu \nu'}(t, t')
@@ -58,8 +58,8 @@ $$\begin{aligned}
\end{aligned}$$
Arguably more prevalent are
-the **retarded Green's function** $G_{\nu \nu'}^R$
-and the **advanced Green's function** $G_{\nu \nu'}^A$
+the **retarded Green's function** $$G_{\nu \nu'}^R$$
+and the **advanced Green's function** $$G_{\nu \nu'}^A$$
which are defined like so:
$$\begin{aligned}
@@ -74,16 +74,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/),
-and $[,]_{\mp}$ is a commutator for bosons,
+Where $$\Theta$$ is a [Heaviside function](/know/concept/heaviside-step-function/),
+and $$[,]_{\mp}$$ is a commutator for bosons,
and an anticommutator for fermions.
Depending on the context,
we could either be in the [Heisenberg picture](/know/concept/heisenberg-picture/)
or in the [interaction picture](/know/concept/interaction-picture/),
-hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent.
+hence $$\hat{c}_\nu$$ and $$\hat{c}_{\nu'}^\dagger$$ are time-dependent.
-Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$
-and **lesser Green's function** $G_{\nu \nu'}^<$ are:
+Furthermore, the **greater Green's function** $$G_{\nu \nu'}^>$$
+and **lesser Green's function** $$G_{\nu \nu'}^<$$ are:
$$\begin{aligned}
\boxed{
@@ -97,7 +97,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $-$ is for bosons, and $+$ for fermions.
+Where $$-$$ is for bosons, and $$+$$ for fermions.
With this, the causal, retarded and advanced Green's functions
can thus be expressed as follows:
@@ -113,10 +113,10 @@ $$\begin{aligned}
\end{aligned}$$
If the Hamiltonian involves interactions,
-it might be more natural to use quantum field operators $\hat{\Psi}(\vb{r}, t)$
-instead of choosing a basis of single-particle states $\psi_\nu$.
-In that case, instead of a label $\nu$,
-we use the spin $s$ and position $\vb{r}$, leading to:
+it might be more natural to use quantum field operators $$\hat{\Psi}(\vb{r}, t)$$
+instead of choosing a basis of single-particle states $$\psi_\nu$$.
+In that case, instead of a label $$\nu$$,
+we use the spin $$s$$ and position $$\vb{r}$$, leading to:
$$\begin{aligned}
G_{ss'}(\vb{r}, t; \vb{r}', t')
@@ -125,14 +125,14 @@ $$\begin{aligned}
&= \sum_{\nu \nu'} \psi_\nu(\vb{r}) \: \psi^*_{\nu'}(\vb{r}') \: G_{\nu \nu'}(t, t')
\end{aligned}$$
-And analogously for $G_{ss'}^R$, $G_{ss'}^A$, $G_{ss'}^>$ and $G_{ss'}^<$.
-Note that the time-dependence is given to the old $G_{\nu \nu'}$,
-i.e. to $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$,
+And analogously for $$G_{ss'}^R$$, $$G_{ss'}^A$$, $$G_{ss'}^>$$ and $$G_{ss'}^<$$.
+Note that the time-dependence is given to the old $$G_{\nu \nu'}$$,
+i.e. to $$\hat{c}_\nu$$ and $$\hat{c}_{\nu'}^\dagger$$,
because we are in the Heisenberg picture.
If the Hamiltonian is time-independent,
then it can be shown that all the Green's functions
-only depend on the time-difference $t - t'$:
+only depend on the time-difference $$t - t'$$:
$$\begin{gathered}
G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t - t')
@@ -152,20 +152,20 @@ $$\begin{gathered}
We will prove that the thermal expectation value
-$\expval{\hat{A}(t) \hat{B}(t')}$ only depends on $t - t'$
-for arbitrary $\hat{A}$ and $\hat{B}$,
+$$\expval{\hat{A}(t) \hat{B}(t')}$$ only depends on $$t - t'$$
+for arbitrary $$\hat{A}$$ and $$\hat{B}$$,
and it trivially follows that the Green's functions do too.
In (grand) canonical equilibrium, we know that the
[density operator](/know/concept/density-operator/)
-$\hat{\rho}$ is as follows:
+$$\hat{\rho}$$ is as follows:
$$\begin{aligned}
\hat{\rho} = \frac{1}{Z} \exp(- \beta \hat{H})
\end{aligned}$$
The expected value of the product
-of the time-independent operators $\hat{A}$ and $\hat{B}$ is then:
+of the time-independent operators $$\hat{A}$$ and $$\hat{B}$$ is then:
$$\begin{aligned}
\expval{\hat{A}(t) \hat{B}(t')}
@@ -175,17 +175,17 @@ $$\begin{aligned}
e^{i t' \hat{H} / \hbar} \hat{B} e^{-i t' \hat{H} / \hbar} \Big)
\end{aligned}$$
-Using that the trace $\Tr$ is invariant
+Using that the trace $$\Tr$$ is invariant
under cyclic permutations of its argument,
-and that all functions of $\hat{H}$ commute, we find:
+and that all functions of $$\hat{H}$$ commute, we find:
$$\begin{aligned}
\expval{\hat{A}(t) \hat{B}(t')}
= \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i (t - t') \hat{H} / \hbar} \hat{A} e^{-i (t - t') \hat{H} / \hbar} \hat{B} \Big)
\end{aligned}$$
-As expected, this only depends on the time difference $t - t'$,
-because $\hat{H}$ is time-independent by assumption.
+As expected, this only depends on the time difference $$t - t'$$,
+because $$\hat{H}$$ is time-independent by assumption.
Note that thermodynamic equilibrium is crucial:
intuitively, if the system is not in equilibrium,
then it evolves in some transient time-dependent way.
@@ -193,11 +193,11 @@ then it evolves in some transient time-dependent way.
If the Hamiltonian is both time-independent and non-interacting,
-then the time-dependence of $\hat{c}_\nu$
+then the time-dependence of $$\hat{c}_\nu$$
can simply be factored out as
-$\hat{c}_\nu(t) = \hat{c}_\nu \exp(- i \varepsilon_\nu t / \hbar)$.
-Then the diagonal ($\nu = \nu'$) greater and lesser Green's functions
-can be written in the form below, where $f_\nu$ is either
+$$\hat{c}_\nu(t) = \hat{c}_\nu \exp(- i \varepsilon_\nu t / \hbar)$$.
+Then the diagonal ($$\nu = \nu'$$) greater and lesser Green's functions
+can be written in the form below, where $$f_\nu$$ is either
the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/)
or the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/).
@@ -219,7 +219,7 @@ $$\begin{aligned}
In the absence of interactions,
we know from the derivation of
[equation-of-motion theory](/know/concept/equation-of-motion-theory/)
-that the equation of motion of $G^R(\vb{r}, t; \vb{r}', t')$
+that the equation of motion of $$G^R(\vb{r}, t; \vb{r}', t')$$
is as follows (neglecting spin):
$$\begin{aligned}
@@ -228,7 +228,7 @@ $$\begin{aligned}
+ \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\comm{\hat{H}_0}{\hat{\Psi}(\vb{r}, t)}}{\hat{\Psi}^\dagger(\vb{r}', t')}}
\end{aligned}$$
-If $\hat{H}_0$ only contains kinetic energy,
+If $$\hat{H}_0$$ only contains kinetic energy,
i.e. there is no external potential,
it can be shown that:
@@ -243,7 +243,7 @@ $$\begin{aligned}
In the second quantization,
-the Hamiltonian $\hat{H}_0$ is written like so:
+the Hamiltonian $$\hat{H}_0$$ is written like so:
$$\begin{aligned}
\hat{H}_0
@@ -289,7 +289,7 @@ $$\begin{aligned}
\int \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r}) \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$
-We know that the $\psi_\nu$ form a *complete* basis,
+We know that the $$\psi_\nu$$ form a *complete* basis,
which implies (see [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)):
$$\begin{aligned}
@@ -307,11 +307,12 @@ $$\begin{aligned}
&= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
= \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$
+
After substituting this into the equation of motion,
-we recognize $G^R(\vb{r}, t; \vb{r}', t')$ itself:
+we recognize $$G^R(\vb{r}, t; \vb{r}', t')$$ itself:
$$\begin{aligned}
i \hbar \pdv{G^R}{t}
@@ -342,12 +343,12 @@ i.e. the Hamiltonian only contains kinetic energy.
## Two-particle functions
-We generalize the above to two arbitrary operators $\hat{A}$ and $\hat{B}$,
+We generalize the above to two arbitrary operators $$\hat{A}$$ and $$\hat{B}$$,
giving us the **two-particle Green's functions**,
or just **correlation functions**.
-The **causal correlation function** $C_{AB}$,
-the **retarded correlation function** $C_{AB}^R$
-and the **advanced correlation function** $C_{AB}^A$ are defined as follows
+The **causal correlation function** $$C_{AB}$$,
+the **retarded correlation function** $$C_{AB}^R$$
+and the **advanced correlation function** $$C_{AB}^A$$ are defined as follows
(in the Heisenberg picture):
$$\begin{aligned}
@@ -365,15 +366,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the expectation value $\Expval{}$ is taken of thermodynamic equilibrium.
-The name *two-particle* comes from the fact that $\hat{A}$ and $\hat{B}$
+Where the expectation value $$\Expval{}$$ is taken of thermodynamic equilibrium.
+The name *two-particle* comes from the fact that $$\hat{A}$$ and $$\hat{B}$$
will often consist of a sum of products
of two single-particle creation/annihilation operators.
Like for the single-particle Green's functions,
if the Hamiltonian is time-independent,
then it can be shown that the two-particle functions
-only depend on the time-difference $t - t'$:
+only depend on the time-difference $$t - t'$$:
$$\begin{aligned}
G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t \!-\! t')
diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md
index 417b033..8096aaf 100644
--- a/source/know/concept/gronwall-bellman-inequality/index.md
+++ b/source/know/concept/gronwall-bellman-inequality/index.md
@@ -8,16 +8,16 @@ layout: "concept"
---
Suppose we have a first-order ordinary differential equation
-for some function $u(t)$, and that it can be shown from this equation
-that the derivative $u'(t)$ is bounded as follows:
+for some function $$u(t)$$, and that it can be shown from this equation
+that the derivative $$u'(t)$$ is bounded as follows:
$$\begin{aligned}
u'(t)
\le \beta(t) \: u(t)
\end{aligned}$$
-Where $\beta(t)$ is known.
-Then **Grönwall's inequality** states that the solution $u(t)$ is bounded:
+Where $$\beta(t)$$ is known.
+Then **Grönwall's inequality** states that the solution $$u(t)$$ is bounded:
$$\begin{aligned}
\boxed{
@@ -31,8 +31,8 @@ $$\begin{aligned}
-We define $w(t)$ to equal the upper bounds above
-on both $w'(t)$ and $w(t)$ itself:
+We define $$w(t)$$ to equal the upper bounds above
+on both $$w'(t)$$ and $$w(t)$$ itself:
$$\begin{aligned}
w(t)
@@ -42,15 +42,15 @@ $$\begin{aligned}
= \beta(t) \: w(t)
\end{aligned}$$
-Where $w(0) = u(0)$.
-The goal is to show the following for all $t$:
+Where $$w(0) = u(0)$$.
+The goal is to show the following for all $$t$$:
$$\begin{aligned}
\frac{u(t)}{w(t)} \le 1
\end{aligned}$$
-For $t = 0$, this is trivial, since $w(0) = u(0)$ by definition.
-For $t > 0$, we want $w(t)$ to grow at least as fast as $u(t)$
+For $$t = 0$$, this is trivial, since $$w(0) = u(0)$$ by definition.
+For $$t > 0$$, we want $$w(t)$$ to grow at least as fast as $$u(t)$$
in order to satisfy the inequality.
We thus calculate:
@@ -61,7 +61,7 @@ $$\begin{aligned}
= \frac{u' - u \beta}{w}
\end{aligned}$$
-Since $u' \le \beta u$ as a condition,
+Since $$u' \le \beta u$$ as a condition,
the above derivative is always negative.
@@ -74,7 +74,7 @@ $$\begin{aligned}
\le \alpha(t) + \int_0^t \beta(s) \: u(s) \dd{s}
\end{aligned}$$
-Where $\alpha(t)$ and $\beta(t)$ are known.
+Where $$\alpha(t)$$ and $$\beta(t)$$ are known.
Then the **Grönwall-Bellman inequality** states that:
$$\begin{aligned}
@@ -89,7 +89,7 @@ $$\begin{aligned}
-We start by defining $w(t)$ as follows,
+We start by defining $$w(t)$$ as follows,
which will act as shorthand:
$$\begin{aligned}
@@ -97,7 +97,7 @@ $$\begin{aligned}
\equiv \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg) \bigg( \int_0^t \beta(s) \: u(s) \dd{s} \bigg)
\end{aligned}$$
-Its derivative $w'(t)$ is then straightforwardly calculated to be given by:
+Its derivative $$w'(t)$$ is then straightforwardly calculated to be given by:
$$\begin{aligned}
w'(t)
@@ -108,8 +108,8 @@ $$\begin{aligned}
\exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
\end{aligned}$$
-The parenthesized expression it bounded from above by $\alpha(t)$,
-thanks to the condition that $u(t)$ is assumed to satisfy,
+The parenthesized expression it bounded from above by $$\alpha(t)$$,
+thanks to the condition that $$u(t)$$ is assumed to satisfy,
for the Grönwall-Bellman inequality to be true:
$$\begin{aligned}
@@ -117,16 +117,16 @@ $$\begin{aligned}
\le \alpha(t) \: \beta(t) \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
\end{aligned}$$
-Integrating this to find $w(t)$ yields the following result:
+Integrating this to find $$w(t)$$ yields the following result:
$$\begin{aligned}
w(t)
\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s}
\end{aligned}$$
-In the initial definition of $w(t)$,
+In the initial definition of $$w(t)$$,
we now move the exponential to the other side,
-and rewrite it using the above inequality for $w(t)$:
+and rewrite it using the above inequality for $$w(t)$$:
$$\begin{aligned}
\int_0^t \beta(s) \: u(s) \dd{s}
@@ -141,7 +141,7 @@ Insert this into the condition under which the Grönwall-Bellman inequality hold
-In the special case where $\alpha(t)$ is non-decreasing with $t$,
+In the special case where $$\alpha(t)$$ is non-decreasing with $$t$$,
the inequality reduces to:
$$\begin{aligned}
@@ -157,8 +157,8 @@ $$\begin{aligned}
Starting from the "ordinary" Grönwall-Bellman inequality,
-the fact that $\alpha(t)$ is non-decreasing tells us that
-$\alpha(s) \le \alpha(t)$ for all $s \le t$, so:
+the fact that $$\alpha(t)$$ is non-decreasing tells us that
+$$\alpha(s) \le \alpha(t)$$ for all $$s \le t$$, so:
$$\begin{aligned}
u(t)
@@ -194,6 +194,7 @@ $$\begin{aligned}
\\
&\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
\end{aligned}$$
+
diff --git a/source/know/concept/guiding-center-theory/index.md b/source/know/concept/guiding-center-theory/index.md
index 6de54fa..5368966 100644
--- a/source/know/concept/guiding-center-theory/index.md
+++ b/source/know/concept/guiding-center-theory/index.md
@@ -11,14 +11,14 @@ layout: "concept"
When discussing the [Lorentz force](/know/concept/lorentz-force/),
we introduced the concept of *gyration*:
-a particle in a uniform [magnetic field](/know/concept/magnetic-field/) $\vb{B}$
+a particle in a uniform [magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$
*gyrates* in a circular orbit around a **guiding center**.
Here, we will generalize this result
to more complicated situations,
for example involving [electric fields](/know/concept/electric-field/).
The particle's equation of motion
-combines the Lorentz force $\vb{F}$
+combines the Lorentz force $$\vb{F}$$
with Newton's second law:
$$\begin{aligned}
@@ -28,7 +28,7 @@ $$\begin{aligned}
\end{aligned}$$
We now allow the fields vary slowly in time and space.
-We thus add deviations $\delta\vb{E}$ and $\delta\vb{B}$:
+We thus add deviations $$\delta\vb{E}$$ and $$\delta\vb{B}$$:
$$\begin{aligned}
\vb{E}
@@ -38,10 +38,10 @@ $$\begin{aligned}
\to \vb{B} + \delta\vb{B}(\vb{x}, t)
\end{aligned}$$
-Meanwhile, the velocity $\vb{u}$ can be split into
-the guiding center's motion $\vb{u}_{gc}$
-and the *known* Larmor gyration $\vb{u}_L$ around the guiding center,
-such that $\vb{u} = \vb{u}_{gc} + \vb{u}_L$.
+Meanwhile, the velocity $$\vb{u}$$ can be split into
+the guiding center's motion $$\vb{u}_{gc}$$
+and the *known* Larmor gyration $$\vb{u}_L$$ around the guiding center,
+such that $$\vb{u} = \vb{u}_{gc} + \vb{u}_L$$.
Inserting:
$$\begin{aligned}
@@ -49,7 +49,7 @@ $$\begin{aligned}
= q \big( \vb{E} + \delta\vb{E} + (\vb{u}_{gc} + \vb{u}_L) \cross (\vb{B} + \delta\vb{B}) \big)
\end{aligned}$$
-We already know that $m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$,
+We already know that $$m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$$,
which we subtract from the total to get:
$$\begin{aligned}
@@ -59,8 +59,8 @@ $$\begin{aligned}
This will be our starting point.
Before proceeding, we also define
-the average of $\Expval{f}$ of a function $f$ over a single gyroperiod,
-where $\omega_c$ is the cyclotron frequency:
+the average of $$\Expval{f}$$ of a function $$f$$ over a single gyroperiod,
+where $$\omega_c$$ is the cyclotron frequency:
$$\begin{aligned}
\Expval{f}
@@ -74,19 +74,19 @@ and focus only on the guiding center.
## Uniform electric and magnetic field
-Consider the case where $\vb{E}$ and $\vb{B}$ are both uniform,
-such that $\delta\vb{B} = 0$ and $\delta\vb{E} = 0$:
+Consider the case where $$\vb{E}$$ and $$\vb{B}$$ are both uniform,
+such that $$\delta\vb{B} = 0$$ and $$\delta\vb{E} = 0$$:
$$\begin{aligned}
m \dv{\vb{u}_{gc}}{t}
= q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B} \big)
\end{aligned}$$
-Dotting this with the unit vector $\vu{b} \equiv \vb{B} / |\vb{B}|$
-makes all components perpendicular to $\vb{B}$ vanish,
+Dotting this with the unit vector $$\vu{b} \equiv \vb{B} / |\vb{B}|$$
+makes all components perpendicular to $$\vb{B}$$ vanish,
including the cross product,
leaving only the (scalar) parallel components
-$u_{gc\parallel}$ and $E_\parallel$:
+$$u_{gc\parallel}$$ and $$E_\parallel$$:
$$\begin{aligned}
m \dv{u_{gc\parallel}}{t}
@@ -95,8 +95,8 @@ $$\begin{aligned}
This simply describes a constant acceleration,
and is easy to integrate.
-Next, the equation for $\vb{u}_{gc\perp}$ is found by
-subtracting $u_{gc\parallel}$'s equation from the original:
+Next, the equation for $$\vb{u}_{gc\perp}$$ is found by
+subtracting $$u_{gc\parallel}$$'s equation from the original:
$$\begin{aligned}
m \dv{\vb{u}_{gc\perp}}{t}
@@ -104,11 +104,11 @@ $$\begin{aligned}
= q (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B})
\end{aligned}$$
-Keep in mind that $\vb{u}_{gc\perp}$ explicitly excludes gyration.
-If we try to split $\vb{u}_{gc\perp}$ into a constant and a time-dependent part,
+Keep in mind that $$\vb{u}_{gc\perp}$$ explicitly excludes gyration.
+If we try to split $$\vb{u}_{gc\perp}$$ into a constant and a time-dependent part,
and choose the most convenient constant,
we notice that the only way to exclude gyration
-is to demand that $\vb{u}_{gc\perp}$ does not depend on time.
+is to demand that $$\vb{u}_{gc\perp}$$ does not depend on time.
Therefore:
$$\begin{aligned}
@@ -116,8 +116,8 @@ $$\begin{aligned}
= \vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B}
\end{aligned}$$
-To find $\vb{u}_{gc\perp}$, we take the cross product with $\vb{B}$,
-and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:
+To find $$\vb{u}_{gc\perp}$$, we take the cross product with $$\vb{B}$$,
+and use the fact that $$\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$$:
$$\begin{aligned}
0
@@ -125,10 +125,10 @@ $$\begin{aligned}
= \vb{B} \cross \vb{E} + \vb{u}_{gc\perp} B^2
\end{aligned}$$
-Rearranging this shows that $\vb{u}_{gc\perp}$ is constant.
+Rearranging this shows that $$\vb{u}_{gc\perp}$$ is constant.
The guiding center drifts sideways at this speed,
-hence it is called a **drift velocity** $\vb{v}_E$.
-Curiously, $\vb{v}_E$ is independent of $q$:
+hence it is called a **drift velocity** $$\vb{v}_E$$.
+Curiously, $$\vb{v}_E$$ is independent of $$q$$:
$$\begin{aligned}
\boxed{
@@ -138,8 +138,8 @@ $$\begin{aligned}
\end{aligned}$$
Drift is not specific to an electric field:
-$\vb{E}$ can be replaced by a general force $\vb{F}/q$ without issues.
-In that case, the resulting drift velocity $\vb{v}_F$ does depend on $q$:
+$$\vb{E}$$ can be replaced by a general force $$\vb{F}/q$$ without issues.
+In that case, the resulting drift velocity $$\vb{v}_F$$ does depend on $$q$$:
$$\begin{aligned}
\boxed{
@@ -151,18 +151,18 @@ $$\begin{aligned}
## Non-uniform magnetic field
-Next, consider a more general case, where $\vb{B}$ is non-uniform,
-but $\vb{E}$ is still uniform:
+Next, consider a more general case, where $$\vb{B}$$ is non-uniform,
+but $$\vb{E}$$ is still uniform:
$$\begin{aligned}
m \dv{\vb{u}_{gc}}{t}
= q \big( \vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big)
\end{aligned}$$
-Assuming the gyroradius $r_L$ is small compared to the variation of $\vb{B}$,
-we set $\delta\vb{B}$ to the first-order term
-of a Taylor expansion of $\vb{B}$ around $\vb{x}_{gc}$,
-that is, $\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}$.
+Assuming the gyroradius $$r_L$$ is small compared to the variation of $$\vb{B}$$,
+we set $$\delta\vb{B}$$ to the first-order term
+of a Taylor expansion of $$\vb{B}$$ around $$\vb{x}_{gc}$$,
+that is, $$\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}$$.
We thus have:
$$\begin{aligned}
@@ -182,7 +182,7 @@ $$\begin{aligned}
+ \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } \big)
\end{aligned}$$
-Where we have used that $\Expval{\vb{u}_{gc}} = \vb{u}_{gc}$.
+Where we have used that $$\Expval{\vb{u}_{gc}} = \vb{u}_{gc}$$.
The two averaged expressions turn out to be:
$$\begin{aligned}
@@ -198,9 +198,9 @@ $$\begin{aligned}
-We know what $\vb{x}_L$ is,
-so we can write out $(\vb{x}_L \cdot \nabla) \vb{B}$
-for $\vb{B} = (B_x, B_y, B_z)$:
+We know what $$\vb{x}_L$$ is,
+so we can write out $$(\vb{x}_L \cdot \nabla) \vb{B}$$
+for $$\vb{B} = (B_x, B_y, B_z)$$:
$$\begin{aligned}
(\vb{x}_L \cdot \nabla) \vb{B}
@@ -212,7 +212,7 @@ $$\begin{aligned}
\end{pmatrix}
\end{aligned}$$
-Integrating $\sin$ and $\cos$ over their period yields zero,
+Integrating $$\sin$$ and $$\cos$$ over their period yields zero,
so the average vanishes:
$$\begin{aligned}
@@ -220,8 +220,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Moving on, we write out $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$,
-suppressing the arguments of $\sin$ and $\cos$:
+Moving on, we write out $$\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$$,
+suppressing the arguments of $$\sin$$ and $$\cos$$:
$$\begin{aligned}
\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}
@@ -247,7 +247,7 @@ $$\begin{aligned}
\end{pmatrix}
\end{aligned}$$
-Integrating products of $\sin$ and $\cos$ over their period gives us the following:
+Integrating products of $$\sin$$ and $$\cos$$ over their period gives us the following:
$$\begin{aligned}
\Expval{\cos^2} = \Expval{\sin^2} = \frac{1}{2}
@@ -256,7 +256,7 @@ $$\begin{aligned}
\end{aligned}$$
Inserting this tells us that the average
-of $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$ is given by:
+of $$\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$$ is given by:
$$\begin{aligned}
\Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
@@ -268,11 +268,11 @@ $$\begin{aligned}
\end{pmatrix}
\end{aligned}$$
-We use [Maxwell's equation](/know/concept/maxwells-equations/) $\nabla \cdot \vb{B} = 0$
-to rewrite the $z$-component,
-and follow the convention that $\vb{B}$
-points mostly in the $z$-direction,
-such that $B \equiv |\vb{B}| \approx B_z$:
+We use [Maxwell's equation](/know/concept/maxwells-equations/) $$\nabla \cdot \vb{B} = 0$$
+to rewrite the $$z$$-component,
+and follow the convention that $$\vb{B}$$
+points mostly in the $$z$$-direction,
+such that $$B \equiv |\vb{B}| \approx B_z$$:
$$\begin{aligned}
\Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
@@ -290,6 +290,7 @@ $$\begin{aligned}
\end{pmatrix}
= - \frac{u_L^2}{2 \omega_c} \nabla B
\end{aligned}$$
+
@@ -301,11 +302,11 @@ $$\begin{aligned}
= q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
\end{aligned}$$
-Let us now split $\vb{u}_{gc}$ into
-components $\vb{u}_{gc\perp}$ and $u_{gc\parallel} \vu{b}$,
+Let us now split $$\vb{u}_{gc}$$ into
+components $$\vb{u}_{gc\perp}$$ and $$u_{gc\parallel} \vu{b}$$,
which are respectively perpendicular and parallel
-to the magnetic unit vector $\vu{b}$,
-such that $\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$.
+to the magnetic unit vector $$\vu{b}$$,
+such that $$\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$$.
Consequently:
$$\begin{aligned}
@@ -322,9 +323,9 @@ $$\begin{aligned}
= q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
\end{aligned}$$
-The derivative of $\vu{b}$ can be rewritten as follows,
-where $R_c$ is the radius of the field's [curvature](/know/concept/curvature/),
-and $\vb{R}_c$ is the corresponding vector from the center of curvature:
+The derivative of $$\vu{b}$$ can be rewritten as follows,
+where $$R_c$$ is the radius of the field's [curvature](/know/concept/curvature/),
+and $$\vb{R}_c$$ is the corresponding vector from the center of curvature:
$$\begin{aligned}
\dv{\vu{b}}{t}
@@ -336,8 +337,8 @@ $$\begin{aligned}
-Assuming that $\vu{b}$ does not explicitly depend on time,
-i.e. $\ipdv{\vu{b}}{t} = 0$,
+Assuming that $$\vu{b}$$ does not explicitly depend on time,
+i.e. $$\ipdv{\vu{b}}{t} = 0$$,
we can rewrite the derivative using the chain rule:
$$\begin{aligned}
@@ -346,16 +347,16 @@ $$\begin{aligned}
= u_{gc\parallel} \dv{\vu{b}}{s}
\end{aligned}$$
-Where $\dd{s}$ is the arc length of the magnetic field line,
-which is equal to the radius $R_c$ times the infinitesimal subtended angle $\dd{\theta}$:
+Where $$\dd{s}$$ is the arc length of the magnetic field line,
+which is equal to the radius $$R_c$$ times the infinitesimal subtended angle $$\dd{\theta}$$:
$$\begin{aligned}
\dd{s}
= R_c \dd{\theta}
\end{aligned}$$
-Meanwhile, across this arc, $\vu{b}$ rotates by $\dd{\theta}$,
-such that the tip travels a distance $|\dd{\vu{b}}|$:
+Meanwhile, across this arc, $$\vu{b}$$ rotates by $$\dd{\theta}$$,
+such that the tip travels a distance $$|\dd{\vu{b}}|$$:
$$\begin{aligned}
|\!\dd{\vu{b}}\!|
@@ -363,15 +364,15 @@ $$\begin{aligned}
= \dd{\theta}
\end{aligned}$$
-Furthermore, the direction $\dd{\vu{b}}$ is always opposite to $\vu{R}_c$,
-which is defined as the unit vector from the center of curvature to the base of $\vu{b}$:
+Furthermore, the direction $$\dd{\vu{b}}$$ is always opposite to $$\vu{R}_c$$,
+which is defined as the unit vector from the center of curvature to the base of $$\vu{b}$$:
$$\begin{aligned}
\dd{\vu{b}}
= - \vu{R}_c \dd{\theta}
\end{aligned}$$
-Combining these expressions for $\dd{s}$ and $\dd{\vu{b}}$,
+Combining these expressions for $$\dd{s}$$ and $$\dd{\vu{b}}$$,
we find the following derivative:
$$\begin{aligned}
@@ -380,6 +381,7 @@ $$\begin{aligned}
= - \frac{\vu{R}_c}{R_c}
= - \frac{\vb{R}_c}{R_c^2}
\end{aligned}$$
+
@@ -391,8 +393,8 @@ $$\begin{aligned}
= q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
\end{aligned}$$
-Since both $\vb{R}_c$ and any cross product with $\vb{B}$
-will always be perpendicular to $\vb{B}$,
+Since both $$\vb{R}_c$$ and any cross product with $$\vb{B}$$
+will always be perpendicular to $$\vb{B}$$,
we can split this equation into perpendicular and parallel components like so:
$$\begin{aligned}
@@ -405,7 +407,7 @@ $$\begin{aligned}
The parallel part simply describes an acceleration.
The perpendicular part is more interesting:
-we rewrite it as follows, defining an effective force $\vb{F}_{\!\perp}$:
+we rewrite it as follows, defining an effective force $$\vb{F}_{\!\perp}$$:
$$\begin{aligned}
m \dv{\vb{u}_{gc\perp}}{t}
@@ -416,7 +418,7 @@ $$\begin{aligned}
\end{aligned}$$
To solve this, we make a crude approximation now, and improve it later.
-We thus assume that $\vb{u}_{gc\perp}$ is constant in time,
+We thus assume that $$\vb{u}_{gc\perp}$$ is constant in time,
such that the equation reduces to:
$$\begin{aligned}
@@ -426,8 +428,8 @@ $$\begin{aligned}
\end{aligned}$$
This is analogous to the previous case of a uniform electric field,
-with $q \vb{E}$ replaced by $\vb{F}_{\!\perp}$,
-so it is also solved by crossing with $\vb{B}$ in front,
+with $$q \vb{E}$$ replaced by $$\vb{F}_{\!\perp}$$,
+so it is also solved by crossing with $$\vb{B}$$ in front,
yielding a drift:
$$\begin{aligned}
@@ -436,11 +438,11 @@ $$\begin{aligned}
\equiv \frac{\vb{F}_{\!\perp} \cross \vb{B}}{q B^2}
\end{aligned}$$
-From the definition of $\vb{F}_{\!\perp}$,
-this total $\vb{v}_F$ can be split into three drifts:
-the previously seen electric field drift $\vb{v}_E$,
-the **curvature drift** $\vb{v}_c$,
-and the **grad-$\vb{B}$ drift** $\vb{v}_{\nabla B}$:
+From the definition of $$\vb{F}_{\!\perp}$$,
+this total $$\vb{v}_F$$ can be split into three drifts:
+the previously seen electric field drift $$\vb{v}_E$$,
+the **curvature drift** $$\vb{v}_c$$,
+and the **grad-$$\vb{B}$$ drift** $$\vb{v}_{\nabla B}$$:
$$\begin{aligned}
\boxed{
@@ -454,11 +456,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Such that $\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$.
+Such that $$\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$$.
We are still missing a correction,
-since we neglected the time dependence of $\vb{u}_{gc\perp}$ earlier.
-This correction is called $\vb{v}_p$,
-where $\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$.
+since we neglected the time dependence of $$\vb{u}_{gc\perp}$$ earlier.
+This correction is called $$\vb{v}_p$$,
+where $$\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$$.
We revisit the perpendicular equation, which now reads:
$$\begin{aligned}
@@ -466,10 +468,10 @@ $$\begin{aligned}
= \vb{F}_{\!\perp} + q \big( \vb{v}_F + \vb{v}_p \big) \cross \vb{B}
\end{aligned}$$
-We assume that $\vb{v}_F$ varies much faster than $\vb{v}_p$,
-such that $\idv{}{\vb{v}p}{t}$ is negligible.
-In addition, from the derivation of $\vb{v}_F$,
-we know that $\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$,
+We assume that $$\vb{v}_F$$ varies much faster than $$\vb{v}_p$$,
+such that $$\idv{}{\vb{v}p}{t}$$ is negligible.
+In addition, from the derivation of $$\vb{v}_F$$,
+we know that $$\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$$,
leaving only:
$$\begin{aligned}
@@ -477,11 +479,11 @@ $$\begin{aligned}
= q \vb{v}_p \cross \vb{B}
\end{aligned}$$
-To isolate this for $\vb{v}_p$,
-we take the cross product with $\vb{B}$ in front,
+To isolate this for $$\vb{v}_p$$,
+we take the cross product with $$\vb{B}$$ in front,
like earlier.
We thus arrive at the following correction,
-known as the **polarization drift** $\vb{v}_p$:
+known as the **polarization drift** $$\vb{v}_p$$:
$$\begin{aligned}
\boxed{
@@ -490,8 +492,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-In many cases $\vb{v}_E$ dominates $\vb{v}_F$,
-so in some literature $\vb{v}_p$ is approximated as follows:
+In many cases $$\vb{v}_E$$ dominates $$\vb{v}_F$$,
+so in some literature $$\vb{v}_p$$ is approximated as follows:
$$\begin{aligned}
\vb{v}_p
@@ -502,7 +504,7 @@ $$\begin{aligned}
The polarization drift stands out from the others:
it has the opposite sign,
-it is proportional to $m$,
+it is proportional to $$m$$,
and it is often only temporary.
Therefore, it is also called the **inertia drift**.
diff --git a/source/know/concept/hagen-poiseuille-equation/index.md b/source/know/concept/hagen-poiseuille-equation/index.md
index d59ee9a..6484631 100644
--- a/source/know/concept/hagen-poiseuille-equation/index.md
+++ b/source/know/concept/hagen-poiseuille-equation/index.md
@@ -13,12 +13,12 @@ The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**,
describes the flow of a fluid with nonzero [viscosity](/know/concept/viscosity/)
through a cylindrical pipe.
Due to its viscosity, the fluid clings to the sides,
-limiting the amount that can pass through, for a pipe with radius $R$.
+limiting the amount that can pass through, for a pipe with radius $$R$$.
Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
-of an incompressible fluid with spatially uniform density $\rho$.
-Assuming that the flow is steady $\ipdv{\va{v}}{t} = 0$,
-and that gravity is negligible $\va{g} = 0$, we get:
+of an incompressible fluid with spatially uniform density $$\rho$$.
+Assuming that the flow is steady $$\ipdv{\va{v}}{t} = 0$$,
+and that gravity is negligible $$\va{g} = 0$$, we get:
$$\begin{aligned}
(\va{v} \cdot \nabla) \va{v}
@@ -27,26 +27,26 @@ $$\begin{aligned}
\nabla \cdot \va{v} = 0
\end{aligned}$$
-Into this, we insert the ansatz $\va{v} = \vu{e}_z \: v_z(r)$,
-where $\vu{e}_z$ is the $z$-axis' unit vector.
-In other words, we assume that the flow velocity depends only on $r$;
-not on $\phi$ or $z$.
+Into this, we insert the ansatz $$\va{v} = \vu{e}_z \: v_z(r)$$,
+where $$\vu{e}_z$$ is the $$z$$-axis' unit vector.
+In other words, we assume that the flow velocity depends only on $$r$$;
+not on $$\phi$$ or $$z$$.
Plugging this into the Navier-Stokes equations,
-$\nabla \cdot \va{v}$ is trivially zero,
-and in the other equation we multiply out $\rho$, yielding this,
-where $\eta = \rho \nu$ is the dynamic viscosity:
+$$\nabla \cdot \va{v}$$ is trivially zero,
+and in the other equation we multiply out $$\rho$$, yielding this,
+where $$\eta = \rho \nu$$ is the dynamic viscosity:
$$\begin{aligned}
\nabla p
= \vu{e}_z \: \eta \nabla^2 v_z
\end{aligned}$$
-Because only $\vu{e}_z$ appears on the right-hand side,
-only the $z$-component of $\nabla p$ can be nonzero.
-However, $v_z(r)$ is a function of $r$, not $z$!
-The left thus only depends on $z$, and the right only on $r$,
+Because only $$\vu{e}_z$$ appears on the right-hand side,
+only the $$z$$-component of $$\nabla p$$ can be nonzero.
+However, $$v_z(r)$$ is a function of $$r$$, not $$z$$!
+The left thus only depends on $$z$$, and the right only on $$r$$,
meaning that both sides must equal a constant,
-which we call $-G$:
+which we call $$-G$$:
$$\begin{aligned}
\dv{p}{z}
@@ -56,22 +56,22 @@ $$\begin{aligned}
= - G
\end{aligned}$$
-The former equation, for $p(z)$, is easy to solve.
-We get an integration constant $p(0)$:
+The former equation, for $$p(z)$$, is easy to solve.
+We get an integration constant $$p(0)$$:
$$\begin{aligned}
p(z)
= p(0) - G z
\end{aligned}$$
-This gives meaning to the **pressure gradient** $G$:
-for a pipe of length $L$,
-it describes the pressure difference $\Delta p = p(0) - p(L)$
+This gives meaning to the **pressure gradient** $$G$$:
+for a pipe of length $$L$$,
+it describes the pressure difference $$\Delta p = p(0) - p(L)$$
that is driving the fluid,
-i.e. $G = \Delta p / L$
+i.e. $$G = \Delta p / L$$
-As for the latter equation, for $v_z(r)$,
-we start by integrating it once, introducing a constant $A$:
+As for the latter equation, for $$v_z(r)$$,
+we start by integrating it once, introducing a constant $$A$$:
$$\begin{aligned}
\dv{}{r}\Big( r \dv{v_z}{r} \Big)
@@ -82,7 +82,7 @@ $$\begin{aligned}
\end{aligned}$$
Integrating this one more time,
-thereby introducing another constant $B$,
+thereby introducing another constant $$B$$,
we arrive at:
$$\begin{aligned}
@@ -90,11 +90,11 @@ $$\begin{aligned}
= - \frac{G}{4 \eta} r^2 + A \ln{r} + B
\end{aligned}$$
-The velocity must be finite at $r = 0$, so we set $A = 0$.
+The velocity must be finite at $$r = 0$$, so we set $$A = 0$$.
Furthermore, the Navier-Stokes equation's *no-slip* condition
-demands that $v_z = 0$ at the boundary $r = R$,
-so $B = G R^2 / (4 \eta)$.
-This brings us to the **Poiseuille solution** for $v_z(r)$:
+demands that $$v_z = 0$$ at the boundary $$r = R$$,
+so $$B = G R^2 / (4 \eta)$$.
+This brings us to the **Poiseuille solution** for $$v_z(r)$$:
$$\begin{aligned}
\boxed{
@@ -104,8 +104,8 @@ $$\begin{aligned}
\end{aligned}$$
How much fluid can pass through the pipe per unit time?
-This is denoted by the **volumetric flow rate** $Q$,
-which is the integral of $v_z$ over the circular cross-section:
+This is denoted by the **volumetric flow rate** $$Q$$,
+which is the integral of $$v_z$$ over the circular cross-section:
$$\begin{aligned}
Q
@@ -115,7 +115,7 @@ $$\begin{aligned}
\end{aligned}$$
We thus arrive at the main Hagen-Poiseuille equation,
-which predicts $Q$ for a given setup:
+which predicts $$Q$$ for a given setup:
$$\begin{aligned}
\boxed{
@@ -124,8 +124,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Consequently, the average flow velocity $\Expval{v_z}$
-is simply $Q$ divided by the cross-sectional area:
+Consequently, the average flow velocity $$\Expval{v_z}$$
+is simply $$Q$$ divided by the cross-sectional area:
$$\begin{aligned}
\Expval{v_z}
@@ -133,12 +133,12 @@ $$\begin{aligned}
= \frac{G R^2}{8 \eta}
\end{aligned}$$
-The fluid's viscous stickiness means it exerts a drag force $D$
-on the pipe as it flows. For a pipe of length $L$ and radius $R$,
-we calculate $D$ by multiplying the internal area $2 \pi R L$
+The fluid's viscous stickiness means it exerts a drag force $$D$$
+on the pipe as it flows. For a pipe of length $$L$$ and radius $$R$$,
+we calculate $$D$$ by multiplying the internal area $$2 \pi R L$$
by the [shear stress](/know/concept/cauchy-stress-tensor/)
-$-\sigma_{zr}$ on the wall
-(i.e. the wall applies $\sigma_{zr}$, the fluid responds with $- \sigma_{zr}$):
+$$-\sigma_{zr}$$ on the wall
+(i.e. the wall applies $$\sigma_{zr}$$, the fluid responds with $$- \sigma_{zr}$$):
$$\begin{aligned}
D
@@ -148,8 +148,8 @@ $$\begin{aligned}
= \pi R^2 L G
\end{aligned}$$
-We would like to get rid of $G$ for being impractical,
-so we substitute $R^2 G = 8 \eta \Expval{v_z}$, yielding:
+We would like to get rid of $$G$$ for being impractical,
+so we substitute $$R^2 G = 8 \eta \Expval{v_z}$$, yielding:
$$\begin{aligned}
\boxed{
@@ -158,8 +158,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Due to this drag, the pressure difference $\Delta p = p(0) - p(L)$
-does work on the fluid, at a rate $P$,
+Due to this drag, the pressure difference $$\Delta p = p(0) - p(L)$$
+does work on the fluid, at a rate $$P$$,
since power equals force (i.e. pressure times area) times velocity:
$$\begin{aligned}
@@ -167,10 +167,10 @@ $$\begin{aligned}
= 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r}
\end{aligned}$$
-Because $\Delta p$ is independent of $r$,
-we get the same integral we used to calculate $Q$.
-Then, thanks to the fact that $\Delta p = G L$
-and $Q = \pi R^2 \Expval{v_z}$, it follows that:
+Because $$\Delta p$$ is independent of $$r$$,
+we get the same integral we used to calculate $$Q$$.
+Then, thanks to the fact that $$\Delta p = G L$$
+and $$Q = \pi R^2 \Expval{v_z}$$, it follows that:
$$\begin{aligned}
P
@@ -179,8 +179,8 @@ $$\begin{aligned}
= D \Expval{v_z}
\end{aligned}$$
-In conclusion, the power $P$,
-needed to drive a fluid through the pipe at a rate $Q$,
+In conclusion, the power $$P$$,
+needed to drive a fluid through the pipe at a rate $$Q$$,
is given by:
$$\begin{aligned}
diff --git a/source/know/concept/hamiltonian-mechanics/index.md b/source/know/concept/hamiltonian-mechanics/index.md
index 466a78f..19e55b0 100644
--- a/source/know/concept/hamiltonian-mechanics/index.md
+++ b/source/know/concept/hamiltonian-mechanics/index.md
@@ -17,12 +17,12 @@ which is in turn built on [variational calculus](/know/concept/calculus-of-varia
## Definitions
-In Lagrangian mechanics, use a Lagrangian $L$,
-which depends on position $q(t)$ and velocity $\dot{q}(t)$,
-to define the momentum $p(t)$ as a derived quantity.
-Hamiltonian mechanics switches the roles of $\dot{q}$ and $p$:
-the **Hamiltonian** $H$ is a function of $q$ and $p$,
-and the velocity $\dot{q}$ is derived from it:
+In Lagrangian mechanics, use a Lagrangian $$L$$,
+which depends on position $$q(t)$$ and velocity $$\dot{q}(t)$$,
+to define the momentum $$p(t)$$ as a derived quantity.
+Hamiltonian mechanics switches the roles of $$\dot{q}$$ and $$p$$:
+the **Hamiltonian** $$H$$ is a function of $$q$$ and $$p$$,
+and the velocity $$\dot{q}$$ is derived from it:
$$\begin{aligned}
\pdv{L(q, \dot{q})}{\dot{q}} = p
@@ -32,9 +32,9 @@ $$\begin{aligned}
Conveniently, this switch turns out to be
[Legendre transformation](/know/concept/legendre-transform/):
-$H$ is the Legendre transform of $L$,
-with $p = \partial L / \partial \dot{q}$ taken as
-the coordinate to replace $\dot{q}$.
+$$H$$ is the Legendre transform of $$L$$,
+with $$p = \partial L / \partial \dot{q}$$ taken as
+the coordinate to replace $$\dot{q}$$.
Therefore:
$$\begin{aligned}
@@ -44,15 +44,15 @@ $$\begin{aligned}
\end{aligned}$$
This almost always works,
-because $L$ is usually a second-order polynomial of $\dot{q}$,
+because $$L$$ is usually a second-order polynomial of $$\dot{q}$$,
and thus convex as required for Legendre transformation.
In the above expression,
-$\dot{q}$ must be rewritten in terms of $p$ and $q$,
-which is trivial, since $p$ is proportional to $\dot{q}$ by definition.
+$$\dot{q}$$ must be rewritten in terms of $$p$$ and $$q$$,
+which is trivial, since $$p$$ is proportional to $$\dot{q}$$ by definition.
-The Hamiltonian $H$ also has a direct physical meaning:
-for a mass $m$, and for $L = T - V$,
-it is straightforward to show that $H$ represents the total energy $T + V$:
+The Hamiltonian $$H$$ also has a direct physical meaning:
+for a mass $$m$$, and for $$L = T - V$$,
+it is straightforward to show that $$H$$ represents the total energy $$T + V$$:
$$\begin{aligned}
H
@@ -64,8 +64,8 @@ $$\begin{aligned}
Just as Lagrangian mechanics,
Hamiltonian mechanics scales well for large systems.
-Its definition is generalized as follows to $N$ objects,
-where $p$ is shorthand for $p_1, ..., p_N$:
+Its definition is generalized as follows to $$N$$ objects,
+where $$p$$ is shorthand for $$p_1, ..., p_N$$:
$$\begin{aligned}
\boxed{
@@ -74,13 +74,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-The positions and momenta $(q, p)$ form a phase space,
+The positions and momenta $$(q, p)$$ form a phase space,
i.e. they fully describe the state.
An extremely useful concept in Hamiltonian mechanics
is the **Poisson bracket** (PB),
-which is a binary operation on two quantities $A(q, p)$ and $B(q, p)$,
-denoted by $\{A, B\}$:
+which is a binary operation on two quantities $$A(q, p)$$ and $$B(q, p)$$,
+denoted by $$\{A, B\}$$:
$$\begin{aligned}
\boxed{
@@ -93,8 +93,8 @@ $$\begin{aligned}
## Canonical equations
Lagrangian mechanics has a single Euler-Lagrange equation per object,
-yielding $N$ second-order equations of motion in total.
-In contrast, Hamiltonian mechanics has $2 N$ first-order equations of motion,
+yielding $$N$$ second-order equations of motion in total.
+In contrast, Hamiltonian mechanics has $$2 N$$ first-order equations of motion,
known as **Hamilton's canonical equations**:
$$\begin{aligned}
@@ -111,7 +111,7 @@ $$\begin{aligned}
For the first equation,
-we differentiate $H$ with respect to $q_n$,
+we differentiate $$H$$ with respect to $$q_n$$,
and use the chain rule:
$$\begin{aligned}
@@ -133,7 +133,7 @@ $$\begin{aligned}
\end{aligned}$$
The second equation is somewhat trivial,
-since $H$ is defined to satisfy it in the first place.
+since $$H$$ is defined to satisfy it in the first place.
Nevertheless, we can prove it by brute force,
using the same approach as above:
@@ -148,11 +148,12 @@ $$\begin{aligned}
- 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
= \dot{q}_n
\end{aligned}$$
+
-Just like in Lagrangian mechanics, if $H$ does not explicitly contain $q_n$,
-then $q_n$ is called a **cyclic coordinate**, and leads to the conservation of $p_n$:
+Just like in Lagrangian mechanics, if $$H$$ does not explicitly contain $$q_n$$,
+then $$q_n$$ is called a **cyclic coordinate**, and leads to the conservation of $$p_n$$:
$$\begin{aligned}
\dot{p}_n = - \pdv{H}{q_n} = 0
@@ -161,11 +162,11 @@ $$\begin{aligned}
\end{aligned}$$
Of course, there may be other conserved quantities.
-Generally speaking, the $t$-derivative of an arbitrary quantity $A(q, p, t)$ is as follows,
-where $\ipdv{}{t}$ is a "soft" derivative
-(only affects explicit occurrences of $t$),
-and $\idv{}{t}$ is a "hard" derivative
-(also affects implicit $t$ inside $q$ and $p$):
+Generally speaking, the $$t$$-derivative of an arbitrary quantity $$A(q, p, t)$$ is as follows,
+where $$\ipdv{}{t}$$ is a "soft" derivative
+(only affects explicit occurrences of $$t$$),
+and $$\idv{}{t}$$ is a "hard" derivative
+(also affects implicit $$t$$ inside $$q$$ and $$p$$):
$$\begin{aligned}
\boxed{
@@ -191,10 +192,11 @@ $$\begin{aligned}
\\
&= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
\end{aligned}$$
+
-Assuming that $H$ does not explicitly depend on $t$,
+Assuming that $$H$$ does not explicitly depend on $$t$$,
the above property naturally leads us to an alternative
way of writing Hamilton's canonical equations:
@@ -208,7 +210,7 @@ $$\begin{aligned}
## Canonical coordinates
-So far, we have assumed that the phase space coordinates $(q, p)$
+So far, we have assumed that the phase space coordinates $$(q, p)$$
are the *positions* and *canonical momenta*, respectively,
and that led us to Hamilton's canonical equations.
@@ -220,11 +222,11 @@ $$\begin{aligned}
p \to P(q, p)
\end{aligned}$$
-However, most choices of $(Q, P)$ would not preserve Hamilton's equations.
-Any $(Q, P)$ that do keep this form
+However, most choices of $$(Q, P)$$ would not preserve Hamilton's equations.
+Any $$(Q, P)$$ that do keep this form
are known as **canonical coordinates**,
and the corresponding transformation is a **canonical transformation**.
-That is, any $(Q, P)$ that satisfy:
+That is, any $$(Q, P)$$ that satisfy:
$$\begin{aligned}
- \pdv{H}{Q_n} = \dot{P}_n
@@ -232,10 +234,10 @@ $$\begin{aligned}
\pdv{H}{P_n} = \dot{Q}_n
\end{aligned}$$
-Then we might as well write $H(q, p)$ as $H(Q, P)$.
-So, which $(Q, P)$ fulfill this?
-It turns out that the following must be satisfied for all $n, j$,
-where $\delta_{nj}$ is the Kronecker delta:
+Then we might as well write $$H(q, p)$$ as $$H(Q, P)$$.
+So, which $$(Q, P)$$ fulfill this?
+It turns out that the following must be satisfied for all $$n, j$$,
+where $$\delta_{nj}$$ is the Kronecker delta:
$$\begin{aligned}
\boxed{
@@ -250,8 +252,8 @@ $$\begin{aligned}
-Assuming that $Q_n$, $P_n$ and $H$ do not explicitly depend on $t$,
-we use our expression for the $t$-derivative of an arbitrary quantity,
+Assuming that $$Q_n$$, $$P_n$$ and $$H$$ do not explicitly depend on $$t$$,
+we use our expression for the $$t$$-derivative of an arbitrary quantity,
and apply the multivariate chain rule to it:
$$\begin{aligned}
@@ -268,11 +270,11 @@ $$\begin{aligned}
&= \sum_{j} \bigg( \pdv{H}{Q_j} \{Q_n, Q_j\} + \pdv{H}{P_j} \{Q_n, P_j\} \bigg)
\end{aligned}$$
-This is equivalent to Hamilton's equation $\dot{Q}_n = \ipdv{H}{P_n}$
-if and only if $\{Q_n, Q_j\} = 0$ for all $n$ and $j$,
-and if $\{Q_n, P_j\} = \delta_{nj}$.
+This is equivalent to Hamilton's equation $$\dot{Q}_n = \ipdv{H}{P_n}$$
+if and only if $$\{Q_n, Q_j\} = 0$$ for all $$n$$ and $$j$$,
+and if $$\{Q_n, P_j\} = \delta_{nj}$$.
-Next, we do the exact same thing with $P_n$ instead of $Q_n$,
+Next, we do the exact same thing with $$P_n$$ instead of $$Q_n$$,
giving an analogous result:
$$\begin{aligned}
@@ -289,17 +291,17 @@ $$\begin{aligned}
&= \sum_{j} \bigg( \pdv{H}{Q_j} \{P_n, Q_j\} + \pdv{H}{P_j} \{P_n, P_j\} \bigg)
\end{aligned}$$
-Which is equivalent to Hamilton's equation $\dot{P}_n = -\ipdv{H}{Q_n}$
-if and only if $\{P_n, P_j\} = 0$,
-and $\{Q_n, P_j\} = - \delta_{nj}$.
+Which is equivalent to Hamilton's equation $$\dot{P}_n = -\ipdv{H}{Q_n}$$
+if and only if $$\{P_n, P_j\} = 0$$,
+and $$\{Q_n, P_j\} = - \delta_{nj}$$.
The PB is anticommutative,
-i.e. $\{A, B\} = - \{B, A\}$.
+i.e. $$\{A, B\} = - \{B, A\}$$.
If you have experience with quantum mechanics,
the latter equation should look suspiciously similar
-to the *canonical commutation relation* $[\hat{Q}, \hat{P}] = i \hbar$.
+to the *canonical commutation relation* $$[\hat{Q}, \hat{P}] = i \hbar$$.
diff --git a/source/know/concept/harmonic-oscillator/index.md b/source/know/concept/harmonic-oscillator/index.md
index 6729dc2..6d9eac0 100644
--- a/source/know/concept/harmonic-oscillator/index.md
+++ b/source/know/concept/harmonic-oscillator/index.md
@@ -11,34 +11,34 @@ layout: "concept"
A **harmonic oscillator** obeys
the simple 1D version of [Hooke's law](/know/concept/hookes-law/):
to displace the system away from its equilibrium,
-the needed force $F_d(x)$ scales linearly with the displacement $x(t)$:
+the needed force $$F_d(x)$$ scales linearly with the displacement $$x(t)$$:
$$\begin{aligned}
F_d(x) = k x
\end{aligned}$$
-Where $k$ is a system-specific proportionality constant,
+Where $$k$$ is a system-specific proportionality constant,
called the **spring constant**,
since a spring is a good example of a harmonic oscillator,
at least for small displacements.
Hooke's law is also often stated for
-the restoring force $F_r(x)$ instead:
+the restoring force $$F_r(x)$$ instead:
$$\begin{aligned}
F_r(x) = - k x
\end{aligned}$$
-Let a mass $m$ be attached to the end of the spring.
-After displacing it, we let it go $F_d = 0$,
-so Newton's second law for the restoring force $F_r$ demands that:
+Let a mass $$m$$ be attached to the end of the spring.
+After displacing it, we let it go $$F_d = 0$$,
+so Newton's second law for the restoring force $$F_r$$ demands that:
$$\begin{aligned}
F_r = m x''
\end{aligned}$$
-But $F_r = - k x$,
-meaning $m x'' = - k x$,
-leading to the following equation for $x(t)$:
+But $$F_r = - k x$$,
+meaning $$m x'' = - k x$$,
+leading to the following equation for $$x(t)$$:
$$\begin{aligned}
\boxed{
@@ -46,7 +46,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\omega_0 \equiv \sqrt{k / m}$ is the **natural frequency** of the system.
+Where $$\omega_0 \equiv \sqrt{k / m}$$ is the **natural frequency** of the system.
This differential equation has the following general solution:
$$\begin{aligned}
@@ -56,8 +56,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $C_1$ and $C_2$ are constants determined by the initial conditions.
-For example, for $x(0) = 1$ and $x'(0) = 0$, the solution becomes:
+Where $$C_1$$ and $$C_2$$ are constants determined by the initial conditions.
+For example, for $$x(0) = 1$$ and $$x'(0) = 0$$, the solution becomes:
$$\begin{aligned}
x(t) = \cos(\omega_0 t)
@@ -65,8 +65,8 @@ $$\begin{aligned}
When using [Lagrangian](/know/concept/lagrangian-mechanics/)
or Hamiltonian mechanics,
-we need to know the potential energy $V(x)$
-added to the system by a displacement to $x$.
+we need to know the potential energy $$V(x)$$
+added to the system by a displacement to $$x$$.
This equals the work done by the displacement,
and is therefore given by:
@@ -77,16 +77,16 @@ $$\begin{aligned}
## Damped oscillation
-If there is a **friction force** $F_f$ affecting the system,
+If there is a **friction force** $$F_f$$ affecting the system,
then the oscillation amplitude will decrease,
or it might not oscillate at all.
-We define $F_f$ using a **viscous damping coefficient** $c$:
+We define $$F_f$$ using a **viscous damping coefficient** $$c$$:
$$\begin{aligned}
F_f = - c x'
\end{aligned}$$
-Both $F_r$ and $F_f$ are acting on the system,
+Both $$F_r$$ and $$F_f$$ are acting on the system,
so Newton's second law states that:
$$\begin{aligned}
@@ -94,7 +94,7 @@ $$\begin{aligned}
\end{aligned}$$
This can be rewritten in the following conventional form
-by defining the **damping coefficient** $\zeta \equiv c / (2 \sqrt{m k})$,
+by defining the **damping coefficient** $$\zeta \equiv c / (2 \sqrt{m k})$$,
which determines the expected behaviour of the system:
$$\begin{aligned}
@@ -103,20 +103,20 @@ $$\begin{aligned}
}
\end{aligned}$$
-The general solution is found from the roots $u$ of the auxiliary quadratic equation:
+The general solution is found from the roots $$u$$ of the auxiliary quadratic equation:
$$\begin{aligned}
u^2 + 2 \zeta \omega_0 u + \omega_0^2 = 0
\end{aligned}$$
-The discriminant $D = 4 \zeta^2 \omega_0^2 - 4 \omega_0^2$
+The discriminant $$D = 4 \zeta^2 \omega_0^2 - 4 \omega_0^2$$
tells us that the behaviour changes substantially
-depending on the damping coefficient $\zeta$,
-with three possibilities: $\zeta < 1$ or $\zeta = 1$ or $\zeta > 1$.
+depending on the damping coefficient $$\zeta$$,
+with three possibilities: $$\zeta < 1$$ or $$\zeta = 1$$ or $$\zeta > 1$$.
-If $\zeta < 1$, there is **underdamping**:
+If $$\zeta < 1$$, there is **underdamping**:
the system oscillates with exponentially decaying
-amplitude and reduced frequency $\omega_1 \equiv \omega_0 \sqrt{1 - \zeta^2}$.
+amplitude and reduced frequency $$\omega_1 \equiv \omega_0 \sqrt{1 - \zeta^2}$$.
The general solution is:
$$\begin{aligned}
@@ -126,7 +126,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\zeta = 1$, there is **critical damping**:
+If $$\zeta = 1$$, there is **critical damping**:
the system returns to its equilibrium point in minimum time.
The general solution is given by:
@@ -137,10 +137,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\zeta > 1$, there is **overdamping**:
+If $$\zeta > 1$$, there is **overdamping**:
the system returns to equilibrium slowly.
The general solution is as follows,
-where $\omega_1 \equiv \omega_0 \sqrt{\zeta^2 - 1}$:
+where $$\omega_1 \equiv \omega_0 \sqrt{\zeta^2 - 1}$$:
$$\begin{aligned}
\boxed{
@@ -161,9 +161,9 @@ $$\begin{aligned}
x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = f(t)
\end{aligned}$$
-Obviously, there exist infinitely many $f(t)$ to choose from,
+Obviously, there exist infinitely many $$f(t)$$ to choose from,
and each needs a separate analysis.
-However, there is one type of $f(t)$ that deserves special mention,
+However, there is one type of $$f(t)$$ that deserves special mention,
namely sinusoids:
$$\begin{aligned}
@@ -172,25 +172,25 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $F$ is a constant force, $\chi$ is an arbitrary phase,
-and the frequency $\omega$ is not necessarily $\omega_0$.
-We solve this case for $x(t)$ in detail.
+Where $$F$$ is a constant force, $$\chi$$ is an arbitrary phase,
+and the frequency $$\omega$$ is not necessarily $$\omega_0$$.
+We solve this case for $$x(t)$$ in detail.
Consider the complex version of the equation:
$$\begin{aligned}
X'' + 2 \zeta \omega_0 X' + \omega_0^2 X = \frac{F}{m} \exp\!\big(i (\omega t + \chi)\big)
\end{aligned}$$
-Then $x(t) = \Real\{X(t)\}$.
-Inserting the ansatz $X(t) = C \exp(i \omega t)$,
-for some constant $C$:
+Then $$x(t) = \Real\{X(t)\}$$.
+Inserting the ansatz $$X(t) = C \exp(i \omega t)$$,
+for some constant $$C$$:
$$\begin{aligned}
- C \omega^2 + C 2 i \zeta \omega_0 \omega + C \omega_0^2 = \frac{F}{m} \exp(i \chi)
\end{aligned}$$
-Where $\exp(i \omega t)$ has already been divided out.
-We isolate this equation for $C$:
+Where $$\exp(i \omega t)$$ has already been divided out.
+We isolate this equation for $$C$$:
$$\begin{aligned}
C
@@ -200,7 +200,7 @@ $$\begin{aligned}
\exp(i \chi)
\end{aligned}$$
-We would like to rewrite this in polar form $C = r \exp(i \theta)$,
+We would like to rewrite this in polar form $$C = r \exp(i \theta)$$,
which turns out to be as follows:
$$\begin{aligned}
@@ -209,8 +209,8 @@ $$\begin{aligned}
\exp\!\bigg(i \chi - i \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)\bigg)
\end{aligned}$$
-For brevity, let us define the **impedance** $Z$
-and the **phase shift** $\phi$
+For brevity, let us define the **impedance** $$Z$$
+and the **phase shift** $$\phi$$
in the following way:
$$\begin{aligned}
@@ -221,8 +221,8 @@ $$\begin{aligned}
\equiv \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)
\end{aligned}$$
-Returning to the original ansatz $X(t) = C \exp(i \omega t)$,
-we take its real part to find $x(t)$:
+Returning to the original ansatz $$X(t) = C \exp(i \omega t)$$,
+we take its real part to find $$x(t)$$:
$$\begin{aligned}
\boxed{
@@ -232,18 +232,18 @@ $$\begin{aligned}
\end{aligned}$$
Two things are noteworthy here.
-Firstly, $f(t)$ and $x(t)$ are out of phase by $\phi$; there is some lag.
-This is caused by damping, because if $\zeta = 0$, it disappears $\phi = 0$.
+Firstly, $$f(t)$$ and $$x(t)$$ are out of phase by $$\phi$$; there is some lag.
+This is caused by damping, because if $$\zeta = 0$$, it disappears $$\phi = 0$$.
-Secondly, the amplitude of $x(t)$ depends on $\omega$ and $\omega_0$.
+Secondly, the amplitude of $$x(t)$$ depends on $$\omega$$ and $$\omega_0$$.
This brings us to **resonance**,
where the amplitude can become extremely large.
Actually, resonance has two subtly different definitions,
-depending on which one of $\omega$ and $\omega_0$ is a free parameter,
+depending on which one of $$\omega$$ and $$\omega_0$$ is a free parameter,
and which one is fixed.
-If the natural $\omega_0$ is fixed and the driving $\omega$ is variable,
-we find for which $\omega$ resonance occurs by minimizing the amplitude denominator $\omega Z$.
+If the natural $$\omega_0$$ is fixed and the driving $$\omega$$ is variable,
+we find for which $$\omega$$ resonance occurs by minimizing the amplitude denominator $$\omega Z$$.
We thus find:
$$\begin{aligned}
@@ -256,13 +256,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Meaning the resonant $\omega$ is lower than $\omega_0$,
-and resonance can only occur if $\zeta < 1 / \sqrt{2}$.
+Meaning the resonant $$\omega$$ is lower than $$\omega_0$$,
+and resonance can only occur if $$\zeta < 1 / \sqrt{2}$$.
-However, if the driving $\omega$ is fixed and the natural is $\omega_0$ is variable,
+However, if the driving $$\omega$$ is fixed and the natural is $$\omega_0$$ is variable,
the problem is bit more subtle:
-the damping coefficient $\zeta = c / (2 m \omega_0)$
-depends on $\omega_0$.
+the damping coefficient $$\zeta = c / (2 m \omega_0)$$
+depends on $$\omega_0$$.
This leads us to:
$$\begin{aligned}
@@ -275,10 +275,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Surprisingly, the damping does not affect $\omega_0$, if $\omega$ is given.
+Surprisingly, the damping does not affect $$\omega_0$$, if $$\omega$$ is given.
However, in both cases, the damping *does* matter for the eventual amplitude:
-$c \to 0$ leads to $x \to \infty$,
-and resonance disappears or becomes negligible for $c \to \infty$.
+$$c \to 0$$ leads to $$x \to \infty$$,
+and resonance disappears or becomes negligible for $$c \to \infty$$.
diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
index 1933037..15d1729 100644
--- a/source/know/concept/heaviside-step-function/index.md
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -8,9 +8,9 @@ categories:
layout: "concept"
---
-The **Heaviside step function** $\Theta(t)$,
+The **Heaviside step function** $$\Theta(t)$$,
is a discontinuous function used for enforcing causality
-or for representing a signal switched on at $t = 0$.
+or for representing a signal switched on at $$t = 0$$.
It is defined as:
$$\begin{aligned}
@@ -23,11 +23,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-The value of $\Theta(t \!=\! 0)$ varies between definitions;
-common choices are $0$, $1$ and $1/2$.
+The value of $$\Theta(t \!=\! 0)$$ varies between definitions;
+common choices are $$0$$, $$1$$ and $$1/2$$.
In practice, this rarely matters, and some authors even
change their definition on the fly for convenience.
-For physicists, $\Theta(0) = 1$ is generally best, such that:
+For physicists, $$\Theta(0) = 1$$ is generally best, such that:
$$\begin{aligned}
\boxed{
@@ -35,7 +35,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Unsurprisingly, the first-order derivative of $\Theta(t)$ is
+Unsurprisingly, the first-order derivative of $$\Theta(t)$$ is
the [Dirac delta function](/know/concept/dirac-delta-function/):
$$\begin{aligned}
@@ -45,10 +45,10 @@ $$\begin{aligned}
\end{aligned}$$
The [Fourier transform](/know/concept/fourier-transform/)
-of $\Theta(t)$ is as follows,
-where $\pv{}$ is the Cauchy principal value,
-$A$ and $s$ are constants from the FT's definition,
-and $\mathrm{sgn}$ is the signum function:
+of $$\Theta(t)$$ is as follows,
+where $$\pv{}$$ is the Cauchy principal value,
+$$A$$ and $$s$$ are constants from the FT's definition,
+and $$\mathrm{sgn}$$ is the signum function:
$$\begin{aligned}
\boxed{
@@ -62,16 +62,16 @@ $$\begin{aligned}
-In this case, it is easiest to use $\Theta(0) = 1/2$,
+In this case, it is easiest to use $$\Theta(0) = 1/2$$,
such that the Heaviside step function can be expressed
-using the signum function $\mathrm{sgn}(t)$:
+using the signum function $$\mathrm{sgn}(t)$$:
$$\begin{aligned}
\Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2}
\end{aligned}$$
We then take the Fourier transform,
-where $A$ and $s$ are constants from its definition:
+where $$A$$ and $$s$$ are constants from its definition:
$$\begin{aligned}
\tilde{\Theta}(\omega)
@@ -80,7 +80,7 @@ $$\begin{aligned}
\end{aligned}$$
The first term is proportional to the Dirac delta function.
-The second integral is problematic, so we take the Cauchy principal value $\pv{}$
+The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$
and look up the integral:
$$\begin{aligned}
@@ -88,10 +88,11 @@ $$\begin{aligned}
&= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
= \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
\end{aligned}$$
+
-The use of $\pv{}$ without an integral is an abuse of notation,
+The use of $$\pv{}$$ without an integral is an abuse of notation,
and means that this result only makes sense when wrapped in an integral.
-Formally, $\pv{\{1 / \omega\}}$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
+Formally, $$\pv{\{1 / \omega\}}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
index 59ed2da..359ecfe 100644
--- a/source/know/concept/heisenberg-picture/index.md
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -13,9 +13,9 @@ mechanics, and is equivalent to the traditionally-taught Schrödinger equation.
In the Schrödinger picture, the operators (observables) are fixed
(as long as they do not depend on time), while the state
-$\Ket{\psi_S(t)}$ changes according to the Schrödinger equation,
-which can be written using the generator of translations $\hat{U}(t)$ like so,
-for a time-independent $\hat{H}_S$:
+$$\Ket{\psi_S(t)}$$ changes according to the Schrödinger equation,
+which can be written using the generator of translations $$\hat{U}(t)$$ like so,
+for a time-independent $$\hat{H}_S$$:
$$\begin{aligned}
\Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
@@ -26,12 +26,12 @@ $$\begin{aligned}
\end{aligned}$$
In contrast, the Heisenberg picture reverses the roles:
-the states $\Ket{\psi_H}$ are invariant,
+the states $$\Ket{\psi_H}$$ are invariant,
and instead the operators vary with time.
An advantage of this is that the basis states remain the same.
-Given a Schrödinger-picture state $\Ket{\psi_S(t)}$, and operator
-$\hat{L}_S(t)$ which may or may not depend on time, they can be
+Given a Schrödinger-picture state $$\Ket{\psi_S(t)}$$, and operator
+$$\hat{L}_S(t)$$ which may or may not depend on time, they can be
converted to the Heisenberg picture by the following change of basis:
$$\begin{aligned}
@@ -42,7 +42,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Since $\hat{U}(t)$ is unitary, the expectation value of a given operator is unchanged:
+Since $$\hat{U}(t)$$ is unitary, the expectation value of a given operator is unchanged:
$$\begin{aligned}
\expval{\hat{L}_H}
@@ -55,7 +55,7 @@ $$\begin{aligned}
\end{aligned}$$
The Schrödinger and Heisenberg pictures therefore respectively
-correspond to active and passive transformations by $\hat{U}(t)$
+correspond to active and passive transformations by $$\hat{U}(t)$$
in [Hilbert space](/know/concept/hilbert-space/).
The two formulations are thus entirely equivalent,
and can be derived from one another,
@@ -63,15 +63,15 @@ as will be shown shortly.
In the Heisenberg picture, the states are constant,
so the time-dependent Schrödinger equation is not directly useful.
-Instead, we will use it derive a new equation for $\hat{L}_H(t)$.
-The key is that the generator $\hat{U}(t)$ is defined from the Schrödinger equation:
+Instead, we will use it derive a new equation for $$\hat{L}_H(t)$$.
+The key is that the generator $$\hat{U}(t)$$ is defined from the Schrödinger equation:
$$\begin{aligned}
\dv{}{t}\hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
\end{aligned}$$
-Where $\hat{H}_S(t)$ may depend on time. We differentiate the definition of
-$\hat{L}_H(t)$ and insert the other side of the Schrödinger equation
+Where $$\hat{H}_S(t)$$ may depend on time. We differentiate the definition of
+$$\hat{L}_H(t)$$ and insert the other side of the Schrödinger equation
when necessary:
$$\begin{aligned}
@@ -99,7 +99,7 @@ $$\begin{aligned}
\end{aligned}$$
This equation is closer to classical mechanics than the Schrödinger picture:
-inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: \idv{}{\hat{X}}$
+inserting the position $$\hat{X}$$ and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
gives the following Newton-style equations:
$$\begin{aligned}
diff --git a/source/know/concept/hellmann-feynman-theorem/index.md b/source/know/concept/hellmann-feynman-theorem/index.md
index 9ffff34..e18acc2 100644
--- a/source/know/concept/hellmann-feynman-theorem/index.md
+++ b/source/know/concept/hellmann-feynman-theorem/index.md
@@ -9,7 +9,7 @@ layout: "concept"
---
Consider the time-independent Schrödinger equation,
-where the Hamiltonian $\hat{H}$ depends on a general parameter $\lambda$,
+where the Hamiltonian $$\hat{H}$$ depends on a general parameter $$\lambda$$,
whose meaning or type we will not specify:
$$\begin{aligned}
@@ -17,7 +17,7 @@ $$\begin{aligned}
= E_n(\lambda) \Ket{\psi_n(\lambda)}
\end{aligned}$$
-Assuming all eigenstates $\Ket{\psi_n}$ are normalized,
+Assuming all eigenstates $$\Ket{\psi_n}$$ are normalized,
this gives us the following basic relation:
$$\begin{aligned}
@@ -26,7 +26,7 @@ $$\begin{aligned}
= \delta_{mn} E_n
\end{aligned}$$
-We differentiate this with respect to $\lambda$,
+We differentiate this with respect to $$\lambda$$,
which could be a scalar or a vector.
This yields:
@@ -43,9 +43,9 @@ $$\begin{aligned}
In order to simplify this,
we differentiate the orthogonality relation
-$\Inprod{\psi_m}{\psi_n} = \delta_{mn}$,
+$$\Inprod{\psi_m}{\psi_n} = \delta_{mn}$$,
which ends up telling us that
-$\Inprod{\nabla_\lambda \psi_m}{\psi_n} = - \Inprod{\psi_m}{\nabla_\lambda \psi_n}$:
+$$\Inprod{\nabla_\lambda \psi_m}{\psi_n} = - \Inprod{\psi_m}{\nabla_\lambda \psi_n}$$:
$$\begin{aligned}
0
@@ -54,7 +54,7 @@ $$\begin{aligned}
= \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \Inprod{\psi_m}{\nabla_\lambda \psi_n}
\end{aligned}$$
-Using this result to replace $\Inprod{\nabla_\lambda \psi_m}{\psi_n}$
+Using this result to replace $$\Inprod{\nabla_\lambda \psi_m}{\psi_n}$$
in the previous equation leads to:
$$\begin{aligned}
@@ -62,9 +62,9 @@ $$\begin{aligned}
&= (E_m - E_n) \Inprod{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
\end{aligned}$$
-For $m = n$, we therefore arrive at the **Hellmann-Feynman theorem**,
+For $$m = n$$, we therefore arrive at the **Hellmann-Feynman theorem**,
which is useful when doing numerical calculations
-to minimize energies with respect to $\lambda$:
+to minimize energies with respect to $$\lambda$$:
$$\begin{aligned}
\boxed{
@@ -73,7 +73,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-While for $m \neq n$, we get the **Epstein generalization**
+While for $$m \neq n$$, we get the **Epstein generalization**
of the Hellmann-Feynman theorem, which is for example relevant for
the [Berry phase](/know/concept/berry-phase/):
diff --git a/source/know/concept/hermite-polynomials/index.md b/source/know/concept/hermite-polynomials/index.md
index ce34030..2eb8e06 100644
--- a/source/know/concept/hermite-polynomials/index.md
+++ b/source/know/concept/hermite-polynomials/index.md
@@ -15,10 +15,10 @@ although slightly different definitions are used in those fields.
## Physicists' definition
-The **Hermite equation** is an eigenvalue problem for $n$,
-and the Hermite polynomials $H_n(x)$ are its eigenfunctions $u(x)$,
-subject to the boundary condition that $u$ grows at most polynomially,
-in which case the eigenvalues $n$ are non-negative integers:
+The **Hermite equation** is an eigenvalue problem for $$n$$,
+and the Hermite polynomials $$H_n(x)$$ are its eigenfunctions $$u(x)$$,
+subject to the boundary condition that $$u$$ grows at most polynomially,
+in which case the eigenvalues $$n$$ are non-negative integers:
$$\begin{aligned}
\boxed{
@@ -26,7 +26,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The $n$th-order Hermite polynomial $H_n(x)$
+The $$n$$th-order Hermite polynomial $$H_n(x)$$
is therefore as follows, according to physicists:
$$\begin{aligned}
@@ -51,7 +51,7 @@ $$\begin{gathered}
H_4(x) = 16 x^4 - 48 x^2 + 12
\end{gathered}$$
-And then more $H_n$ can be computed quickly
+And then more $$H_n$$ can be computed quickly
using the following recurrence relation:
$$\begin{aligned}
@@ -70,7 +70,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Importantly, all $H_n$ are orthogonal with respect to the weight function $w(x) \equiv \exp(- x^2)$:
+Importantly, all $$H_n$$ are orthogonal with respect to the weight function $$w(x) \equiv \exp(- x^2)$$:
$$\begin{aligned}
\boxed{
@@ -80,10 +80,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\delta_{nm}$ is the Kronecker delta.
+Where $$\delta_{nm}$$ is the Kronecker delta.
Finally, they form a basis in the [Hilbert space](/know/concept/hilbert-space/)
-of all functions $f(x)$ for which $\Inprod{f}{w f}$ is finite.
-This means that every such $f$ can be expanded in $H_n$:
+of all functions $$f(x)$$ for which $$\Inprod{f}{w f}$$ is finite.
+This means that every such $$f$$ can be expanded in $$H_n$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/hilbert-space/index.md b/source/know/concept/hilbert-space/index.md
index ef55d2b..57926ce 100644
--- a/source/know/concept/hilbert-space/index.md
+++ b/source/know/concept/hilbert-space/index.md
@@ -14,28 +14,28 @@ abstract **vector space** with a notion of length and angle.
## Vector space
-An abstract **vector space** $\mathbb{V}$ is a generalization of the
+An abstract **vector space** $$\mathbb{V}$$ is a generalization of the
traditional concept of vectors as "arrows". It consists of a set of
objects called **vectors** which support the following (familiar)
operations:
-+ **Vector addition**: the sum of two vectors $V$ and $W$, denoted $V + W$.
-+ **Scalar multiplication**: product of a vector $V$ with a scalar $a$, denoted $a V$.
++ **Vector addition**: the sum of two vectors $$V$$ and $$W$$, denoted $$V + W$$.
++ **Scalar multiplication**: product of a vector $$V$$ with a scalar $$a$$, denoted $$a V$$.
-In addition, for a given $\mathbb{V}$ to qualify as a proper vector
+In addition, for a given $$\mathbb{V}$$ to qualify as a proper vector
space, these operations must obey the following axioms:
-+ **Addition is associative**: $U + (V + W) = (U + V) + W$
-+ **Addition is commutative**: $U + V = V + U$
-+ **Addition has an identity**: there exists a $\mathbf{0}$ such that $V + 0 = V$
-+ **Addition has an inverse**: for every $V$ there exists $-V$ so that $V + (-V) = 0$
-+ **Multiplication is associative**: $a (b V) = (a b) V$
-+ **Multiplication has an identity**: There exists a $1$ such that $1 V = V$
-+ **Multiplication is distributive over scalars**: $(a + b)V = aV + bV$
-+ **Multiplication is distributive over vectors**: $a (U + V) = a U + a V$
++ **Addition is associative**: $$U + (V + W) = (U + V) + W$$
++ **Addition is commutative**: $$U + V = V + U$$
++ **Addition has an identity**: there exists a $$\mathbf{0}$$ such that $$V + 0 = V$$
++ **Addition has an inverse**: for every $$V$$ there exists $$-V$$ so that $$V + (-V) = 0$$
++ **Multiplication is associative**: $$a (b V) = (a b) V$$
++ **Multiplication has an identity**: There exists a $$1$$ such that $$1 V = V$$
++ **Multiplication is distributive over scalars**: $$(a + b)V = aV + bV$$
++ **Multiplication is distributive over vectors**: $$a (U + V) = a U + a V$$
-A set of $N$ vectors $V_1, V_2, ..., V_N$ is **linearly independent** if
-the only way to satisfy the following relation is to set all the scalar coefficients $a_n = 0$:
+A set of $$N$$ vectors $$V_1, V_2, ..., V_N$$ is **linearly independent** if
+the only way to satisfy the following relation is to set all the scalar coefficients $$a_n = 0$$:
$$\begin{aligned}
\mathbf{0} = \sum_{n = 1}^N a_n V_n
@@ -44,13 +44,13 @@ $$\begin{aligned}
In other words, these vectors cannot be expressed in terms of each
other. Otherwise, they would be **linearly dependent**.
-A vector space $\mathbb{V}$ has **dimension** $N$ if only up to $N$ of
+A vector space $$\mathbb{V}$$ has **dimension** $$N$$ if only up to $$N$$ of
its vectors can be linearly indepedent. All other vectors in
-$\mathbb{V}$ can then be written as a **linear combination** of these $N$ **basis vectors**.
+$$\mathbb{V}$$ can then be written as a **linear combination** of these $$N$$ **basis vectors**.
-Let $\vu{e}_1, ..., \vu{e}_N$ be the basis vectors, then any
-vector $V$ in the same space can be **expanded** in the basis according to
-the unique weights $v_n$, known as the **components** of $V$
+Let $$\vu{e}_1, ..., \vu{e}_N$$ be the basis vectors, then any
+vector $$V$$ in the same space can be **expanded** in the basis according to
+the unique weights $$v_n$$, known as the **components** of $$V$$
in that basis:
$$\begin{aligned}
@@ -73,25 +73,25 @@ $$\begin{gathered}
## Inner product
-A given vector space $\mathbb{V}$ can be promoted to a **Hilbert space**
-or **inner product space** if it supports an operation $\Inprod{U}{V}$
+A given vector space $$\mathbb{V}$$ can be promoted to a **Hilbert space**
+or **inner product space** if it supports an operation $$\Inprod{U}{V}$$
called the **inner product**, which takes two vectors and returns a
scalar, and has the following properties:
-+ **Skew symmetry**: $\Inprod{U}{V} = (\Inprod{V}{U})^*$, where ${}^*$ is the complex conjugate.
-+ **Positive semidefiniteness**: $\Inprod{V}{V} \ge 0$, and $\Inprod{V}{V} = 0$ if $V = \mathbf{0}$.
-+ **Linearity in second operand**: $\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$.
++ **Skew symmetry**: $$\Inprod{U}{V} = (\Inprod{V}{U})^*$$, where $${}^*$$ is the complex conjugate.
++ **Positive semidefiniteness**: $$\Inprod{V}{V} \ge 0$$, and $$\Inprod{V}{V} = 0$$ if $$V = \mathbf{0}$$.
++ **Linearity in second operand**: $$\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$$.
The inner product describes the lengths and angles of vectors, and in
Euclidean space it is implemented by the dot product.
-The **magnitude** or **norm** $|V|$ of a vector $V$ is given by
-$|V| = \sqrt{\Inprod{V}{V}}$ and represents the real positive length of $V$.
+The **magnitude** or **norm** $$|V|$$ of a vector $$V$$ is given by
+$$|V| = \sqrt{\Inprod{V}{V}}$$ and represents the real positive length of $$V$$.
A **unit vector** has a norm of 1.
-Two vectors $U$ and $V$ are **orthogonal** if their inner product
-$\Inprod{U}{V} = 0$. If in addition to being orthogonal, $|U| = 1$ and
-$|V| = 1$, then $U$ and $V$ are known as **orthonormal** vectors.
+Two vectors $$U$$ and $$V$$ are **orthogonal** if their inner product
+$$\Inprod{U}{V} = 0$$. If in addition to being orthogonal, $$|U| = 1$$ and
+$$|V| = 1$$, then $$U$$ and $$V$$ are known as **orthonormal** vectors.
Orthonormality is desirable for basis vectors, so if they are
not already like that, it is common to manually turn them into a new
@@ -108,15 +108,15 @@ $$\begin{gathered}
\Inprod{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \Inprod{\vu{e}_n}{\vu{e}_j}
\end{gathered}$$
-If the basis vectors $\vu{e}_1, ..., \vu{e}_N$ are already
+If the basis vectors $$\vu{e}_1, ..., \vu{e}_N$$ are already
orthonormal, this reduces to:
$$\begin{aligned}
\Inprod{V}{W} = \sum_{n = 1}^N v_n^* w_n
\end{aligned}$$
-As it turns out, the components $v_n$ are given by the inner product
-with $\vu{e}_n$, where $\delta_{nm}$ is the Kronecker delta:
+As it turns out, the components $$v_n$$ are given by the inner product
+with $$\vu{e}_n$$, where $$\delta_{nm}$$ is the Kronecker delta:
$$\begin{aligned}
\Inprod{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n
@@ -125,12 +125,12 @@ $$\begin{aligned}
## Infinite dimensions
-As the dimensionality $N$ tends to infinity, things may or may not
-change significantly, depending on whether $N$ is **countably** or
+As the dimensionality $$N$$ tends to infinity, things may or may not
+change significantly, depending on whether $$N$$ is **countably** or
**uncountably** infinite.
In the former case, not much changes: the infinitely many **discrete**
-basis vectors $\vu{e}_n$ can all still be made orthonormal as usual,
+basis vectors $$\vu{e}_n$$ can all still be made orthonormal as usual,
and as before:
$$\begin{aligned}
@@ -141,16 +141,16 @@ A good example of such a countably-infinitely-dimensional basis are the
solution eigenfunctions of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
However, if the dimensionality is uncountably infinite, the basis
-vectors are **continuous** and cannot be labeled by $n$. For example, all
-complex functions $f(x)$ defined for $x \in [a, b]$ which
-satisfy $f(a) = f(b) = 0$ form such a vector space.
-In this case $f(x)$ is expanded as follows, where $x$ is a basis vector:
+vectors are **continuous** and cannot be labeled by $$n$$. For example, all
+complex functions $$f(x)$$ defined for $$x \in [a, b]$$ which
+satisfy $$f(a) = f(b) = 0$$ form such a vector space.
+In this case $$f(x)$$ is expanded as follows, where $$x$$ is a basis vector:
$$\begin{aligned}
f(x) = \int_a^b \Inprod{x}{f} \dd{x}
\end{aligned}$$
-Similarly, the inner product $\Inprod{f}{g}$ must also be redefined as
+Similarly, the inner product $$\Inprod{f}{g}$$ must also be redefined as
follows:
$$\begin{aligned}
@@ -158,17 +158,17 @@ $$\begin{aligned}
\end{aligned}$$
The concept of orthonormality must be also weakened. A finite function
-$f(x)$ can be normalized as usual, but the basis vectors $x$ themselves
+$$f(x)$$ can be normalized as usual, but the basis vectors $$x$$ themselves
cannot, since each represents an infinitesimal section of the real line.
-The rationale in this case is that action of the identity operator $\hat{I}$ must
+The rationale in this case is that action of the identity operator $$\hat{I}$$ must
be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/):
$$\begin{aligned}
\hat{I} = \int_a^b \Ket{\xi} \Bra{\xi} \dd{\xi}
\end{aligned}$$
-Applying the identity operator to $f(x)$ should just give $f(x)$ again:
+Applying the identity operator to $$f(x)$$ should just give $$f(x)$$ again:
$$\begin{aligned}
f(x) = \Inprod{x}{f} = \matrixel{x}{\hat{I}}{f}
@@ -176,9 +176,9 @@ $$\begin{aligned}
= \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
\end{aligned}$$
-Since we want the latter integral to reduce to $f(x)$, it is plain to see that
-$\Inprod{x}{\xi}$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
-i.e $\Inprod{x}{\xi} = \delta(x - \xi)$:
+Since we want the latter integral to reduce to $$f(x)$$, it is plain to see that
+$$\Inprod{x}{\xi}$$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
+i.e $$\Inprod{x}{\xi} = \delta(x - \xi)$$:
$$\begin{aligned}
\int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
@@ -186,11 +186,11 @@ $$\begin{aligned}
= f(x)
\end{aligned}$$
-Consequently, $\Inprod{x}{\xi} = 0$ if $x \neq \xi$ as expected for an
-orthogonal set of vectors, but if $x = \xi$ the inner product
-$\Inprod{x}{\xi}$ is infinite, unlike earlier.
+Consequently, $$\Inprod{x}{\xi} = 0$$ if $$x \neq \xi$$ as expected for an
+orthogonal set of vectors, but if $$x = \xi$$ the inner product
+$$\Inprod{x}{\xi}$$ is infinite, unlike earlier.
-Technically, because the basis vectors $x$ cannot be normalized, they
+Technically, because the basis vectors $$x$$ cannot be normalized, they
are not members of a Hilbert space, but rather of a superset called a
**rigged Hilbert space**. Such vectors have no finite inner product with
themselves, but do have one with all vectors from the actual Hilbert
diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md
index e22799c..5dde240 100644
--- a/source/know/concept/holomorphic-function/index.md
+++ b/source/know/concept/holomorphic-function/index.md
@@ -8,7 +8,7 @@ categories:
layout: "concept"
---
-In complex analysis, a complex function $f(z)$ of a complex variable $z$
+In complex analysis, a complex function $$f(z)$$ of a complex variable $$z$$
is called **holomorphic** or **analytic** if it is complex differentiable in the
neighbourhood of every point of its domain.
This is a very strong condition.
@@ -17,9 +17,9 @@ As a result, holomorphic functions are infinitely differentiable and
equal their Taylor expansion at every point. In physicists' terms,
they are extremely "well-behaved" throughout their domain.
-More formally, a given function $f(z)$ is holomorphic in a certain region
-if the following limit exists for all $z$ in that region,
-and for all directions of $\Delta z$:
+More formally, a given function $$f(z)$$ is holomorphic in a certain region
+if the following limit exists for all $$z$$ in that region,
+and for all directions of $$\Delta z$$:
$$\begin{aligned}
\boxed{
@@ -27,13 +27,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$:
+We decompose $$f$$ into the real functions $$u$$ and $$v$$ of real variables $$x$$ and $$y$$:
$$\begin{aligned}
f(z) = f(x + i y) = u(x, y) + i v(x, y)
\end{aligned}$$
-Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$:
+Since we are free to choose the direction of $$\Delta z$$, we choose $$\Delta x$$ and $$\Delta y$$:
$$\begin{aligned}
f'(z)
@@ -44,8 +44,8 @@ $$\begin{aligned}
= \pdv{v}{y} - i \pdv{u}{y}
\end{aligned}$$
-For $f(z)$ to be holomorphic, these two results must be equivalent.
-Because $u$ and $v$ are real by definition,
+For $$f(z)$$ to be holomorphic, these two results must be equivalent.
+Because $$u$$ and $$v$$ are real by definition,
we thus arrive at the **Cauchy-Riemann equations**:
$$\begin{aligned}
@@ -56,7 +56,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Therefore, a given function $f(z)$ is holomorphic if and only if its real
+Therefore, a given function $$f(z)$$ is holomorphic if and only if its real
and imaginary parts satisfy these equations. This gives an idea of how
strict the criteria are to qualify as holomorphic.
@@ -64,8 +64,8 @@ strict the criteria are to qualify as holomorphic.
## Integration formulas
Holomorphic functions satisfy **Cauchy's integral theorem**, which states
-that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero,
-provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$:
+that the integral of $$f(z)$$ over any closed curve $$C$$ in the complex plane is zero,
+provided that $$f(z)$$ is holomorphic for all $$z$$ in the area enclosed by $$C$$:
$$\begin{aligned}
\boxed{
@@ -78,7 +78,7 @@ $$\begin{aligned}
-Just like before, we decompose $f(z)$ into its real and imaginary parts:
+Just like before, we decompose $$f(z)$$ into its real and imaginary parts:
$$\begin{aligned}
\oint_C f(z) \dd{z}
@@ -88,21 +88,21 @@ $$\begin{aligned}
&= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
\end{aligned}$$
-Using Green's theorem, we integrate over the area $A$ enclosed by $C$:
+Using Green's theorem, we integrate over the area $$A$$ enclosed by $$C$$:
$$\begin{aligned}
\oint_C f(z) \dd{z}
&= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
\end{aligned}$$
-Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+Since $$f(z)$$ is holomorphic, $$u$$ and $$v$$ satisfy the Cauchy-Riemann
equations, such that the integrands disappear and the final result is zero.
An interesting consequence is **Cauchy's integral formula**, which
-states that the value of $f(z)$ at an arbitrary point $z_0$ is
-determined by its values on an arbitrary contour $C$ around $z_0$:
+states that the value of $$f(z)$$ at an arbitrary point $$z_0$$ is
+determined by its values on an arbitrary contour $$C$$ around $$z_0$$:
$$\begin{aligned}
\boxed{
@@ -116,8 +116,8 @@ $$\begin{aligned}
Thanks to the integral theorem, we know that the shape and size
-of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$,
-such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then
+of $$C$$ is irrelevant. Therefore we choose it to be a circle with radius $$r$$,
+such that the integration variable becomes $$z = z_0 + r e^{i \theta}$$. Then
we integrate by substitution:
$$\begin{aligned}
@@ -126,13 +126,14 @@ $$\begin{aligned}
= \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
\end{aligned}$$
-We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:
+We may choose an arbitrarily small radius $$r$$, such that the contour approaches $$z_0$$:
$$\begin{aligned}
\lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
&= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
= f(z_0)
\end{aligned}$$
+
@@ -153,7 +154,7 @@ $$\begin{aligned}
-By definition, the first derivative $f'(z)$ of a
+By definition, the first derivative $$f'(z)$$ of a
holomorphic function exists and is:
$$\begin{aligned}
@@ -183,8 +184,8 @@ $$\begin{aligned}
= \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta}
\end{aligned}$$
-Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
-this proof works inductively for all higher orders $n$.
+Since the second-order derivative $$f''(z)$$ is simply the derivative of $$f'(z)$$,
+this proof works inductively for all higher orders $$n$$.
diff --git a/source/know/concept/hookes-law/index.md b/source/know/concept/hookes-law/index.md
index 87e04de..292b735 100644
--- a/source/know/concept/hookes-law/index.md
+++ b/source/know/concept/hookes-law/index.md
@@ -12,8 +12,8 @@ In its simplest form, **Hooke's law** dictates that
changing the length of an elastic object requires
a force that is proportional the desired length difference.
In its most general form, it gives a linear relationship
-between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
-to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $\hat{u}$.
+between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$
+to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $$\hat{u}$$.
Importantly, all forms of Hooke's law are only valid for small deformations,
since the stress-strain relationship becomes nonlinear otherwise.
@@ -22,8 +22,8 @@ since the stress-strain relationship becomes nonlinear otherwise.
## Simple form
The simple form of the law is traditionally quoted for springs,
-since they have a spring constant $k$ giving the ratio
-between the force $F$ and extension $x$:
+since they have a spring constant $$k$$ giving the ratio
+between the force $$F$$ and extension $$x$$:
$$\begin{aligned}
\boxed{
@@ -35,23 +35,23 @@ $$\begin{aligned}
In general, all solids are elastic for small extensions,
and therefore also obey Hooke's law.
In light of this fact, we replace the traditional spring
-with a rod of length $L$ and cross-section $A$.
+with a rod of length $$L$$ and cross-section $$A$$.
-The constant $k$ depends on, among several things,
-the spring's length $L$ and cross-section $A$,
+The constant $$k$$ depends on, among several things,
+the spring's length $$L$$ and cross-section $$A$$,
so for our generalization, we want a new parameter
to describe the proportionality independently of the rod's dimensions.
-To achieve this, we realize that the force $F$ is spread across $A$,
-and that the extension $x$ should be take relative to $L$.
+To achieve this, we realize that the force $$F$$ is spread across $$A$$,
+and that the extension $$x$$ should be take relative to $$L$$.
$$\begin{aligned}
\frac{F}{A}
= \Big( k \frac{L}{A} \Big) \frac{x}{L}
\end{aligned}$$
-The force-per-area $F/A$ on a solid is the definition of **stress**,
-and the relative elongation $x/L$ is the defintion of **strain**.
-If $F$ acts along the $x$-axis, we can then write:
+The force-per-area $$F/A$$ on a solid is the definition of **stress**,
+and the relative elongation $$x/L$$ is the defintion of **strain**.
+If $$F$$ acts along the $$x$$-axis, we can then write:
$$\begin{aligned}
\boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the proportionality constant $E$,
+Where the proportionality constant $$E$$,
known as the **elastic modulus** or **Young's modulus**,
is the general material parameter that we wanted:
@@ -70,7 +70,7 @@ $$\begin{aligned}
\end{aligned}$$
Due to the microscopic structure of some (usually crystalline) materials,
-$E$ might be dependent on the direction of the force $F$.
+$$E$$ might be dependent on the direction of the force $$F$$.
For simplicity, we only consider **isotropic** materials,
which have the same properties measured from any direction.
@@ -78,8 +78,8 @@ However, we are still missing something.
When a spring is pulled,
it becomes narrower as its coils move apart,
and this effect is also seen when stretching solids in general:
-if we pull our rod along the $x$-axis, we expect it to deform in $y$ and $z$ as well.
-This is described by **Poisson's ratio** $\nu$:
+if we pull our rod along the $$x$$-axis, we expect it to deform in $$y$$ and $$z$$ as well.
+This is described by **Poisson's ratio** $$\nu$$:
$$\begin{aligned}
\boxed{
@@ -88,12 +88,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that $u_{yy} = u_{zz}$ because the material is assumed to be isotropic.
+Note that $$u_{yy} = u_{zz}$$ because the material is assumed to be isotropic.
Intuitively, you may expect that the volume of the object is conserved,
but for most materials that is not accurate.
-In summary, for our example case with a force $F = T A$ pulling at the rod
-along the $x$-axis, the full stress and strain tensors are given by:
+In summary, for our example case with a force $$F = T A$$ pulling at the rod
+along the $$x$$-axis, the full stress and strain tensors are given by:
$$\begin{aligned}
\hat{\sigma} =
@@ -124,9 +124,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Tr{}$ is the trace.
+Where $$\Tr{}$$ is the trace.
This is often written in index notation,
-with the Kronecker delta $\delta_{ij}$:
+with the Kronecker delta $$\delta_{ij}$$:
$$\begin{aligned}
\boxed{
@@ -135,10 +135,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-The constants $\mu$ and $\lambda$ are called the **Lamé coefficients**,
-and are related to $E$ and $\nu$ in a way we can derive
-by returning to the example with a tension $T = F/A$ along $x$.
-For $\sigma_{xx}$, we have:
+The constants $$\mu$$ and $$\lambda$$ are called the **Lamé coefficients**,
+and are related to $$E$$ and $$\nu$$ in a way we can derive
+by returning to the example with a tension $$T = F/A$$ along $$x$$.
+For $$\sigma_{xx}$$, we have:
$$\begin{aligned}
T
@@ -150,7 +150,7 @@ $$\begin{aligned}
&= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big)
\end{aligned}$$
-Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$
+Meanwhile, the other diagonal stresses $$\sigma_{yy} = \sigma_{zz}$$
are expressed in terms of the strain like so:
$$\begin{aligned}
@@ -188,7 +188,7 @@ $$\begin{aligned}
\end{aligned}$$
Which can straightforwardly be inverted
-to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$:
+to express $$E$$ and $$\nu$$ as a function of $$\mu$$ and $$\lambda$$:
$$\begin{aligned}
\boxed{
@@ -212,14 +212,14 @@ $$\begin{aligned}
\end{aligned}$$
Inserting this into Hooke's law
-yields an equation that only contains one strain component $u_{ij}$:
+yields an equation that only contains one strain component $$u_{ij}$$:
$$\begin{aligned}
\sigma_{ij}
= 2 \mu u_{ij} + \frac{\lambda}{2 \mu + 3 \lambda} \delta_{ij} \sum_{k} \sigma_{kk}
\end{aligned}$$
-Which is therefore trivial to isolate for $u_{ij}$,
+Which is therefore trivial to isolate for $$u_{ij}$$,
leading us to Hooke's inverted law:
$$\begin{aligned}
diff --git a/source/know/concept/hydrostatic-pressure/index.md b/source/know/concept/hydrostatic-pressure/index.md
index 2e55246..020fc75 100644
--- a/source/know/concept/hydrostatic-pressure/index.md
+++ b/source/know/concept/hydrostatic-pressure/index.md
@@ -9,7 +9,7 @@ categories:
layout: "concept"
---
-The pressure $p$ inside a fluid at rest,
+The pressure $$p$$ inside a fluid at rest,
the so-called **hydrostatic pressure**,
is an important quantity.
Here we will properly define it,
@@ -20,8 +20,8 @@ both with and without an arbitrary gravity field.
## Without gravity
Inside the fluid, we can imagine small arbitrary partition surfaces,
-with normal vector $\vu{n}$ and area $\dd{S}$,
-yielding the following vector element $\dd{\va{S}}$:
+with normal vector $$\vu{n}$$ and area $$\dd{S}$$,
+yielding the following vector element $$\dd{\va{S}}$$:
$$\begin{aligned}
\dd{\va{S}}
@@ -29,7 +29,7 @@ $$\begin{aligned}
\end{aligned}$$
The orientation of these surfaces does not matter.
-The **pressure** $p(\va{r})$ is defined as the force-per-area
+The **pressure** $$p(\va{r})$$ is defined as the force-per-area
of these tiny surface elements:
$$\begin{aligned}
@@ -38,9 +38,9 @@ $$\begin{aligned}
\end{aligned}$$
The negative sign is there because a positive pressure is conventionally defined
-to push from the positive (normal) side of $\dd{\va{S}}$ to the negative side.
-The total force $\va{F}$ on a larger surface inside the fluid is
-then given by the surface integral over many adjacent $\dd{\va{S}}$:
+to push from the positive (normal) side of $$\dd{\va{S}}$$ to the negative side.
+The total force $$\va{F}$$ on a larger surface inside the fluid is
+then given by the surface integral over many adjacent $$\dd{\va{S}}$$:
$$\begin{aligned}
\va{F}
@@ -57,8 +57,8 @@ $$\begin{aligned}
= - \int_V \nabla p \dd{V}
\end{aligned}$$
-Since the total force on the blob is simply the sum of the forces $\dd{\va{F}}$
-on all its constituent volume elements $\dd{V}$,
+Since the total force on the blob is simply the sum of the forces $$\dd{\va{F}}$$
+on all its constituent volume elements $$\dd{V}$$,
we arrive at the following relation:
$$\begin{aligned}
@@ -71,8 +71,8 @@ $$\begin{aligned}
If the fluid is at rest, then all forces on the blob cancel out
(otherwise it would move).
Since we are currently neglecting all forces other than pressure,
-this is equivalent to demanding that $\dd{\va{F}} = 0$,
-which implies that $\nabla p = 0$, i.e. the pressure is constant.
+this is equivalent to demanding that $$\dd{\va{F}} = 0$$,
+which implies that $$\nabla p = 0$$, i.e. the pressure is constant.
$$\begin{aligned}
\boxed{
@@ -84,17 +84,17 @@ $$\begin{aligned}
## With gravity
If we include gravity, then,
-in addition to the pressure's *contact force* $\va{F}_p$ from earlier,
-there is also a *body force* $\va{F}_g$ acting on
-the arbitrary blob $V$ of fluid enclosed by $S$:
+in addition to the pressure's *contact force* $$\va{F}_p$$ from earlier,
+there is also a *body force* $$\va{F}_g$$ acting on
+the arbitrary blob $$V$$ of fluid enclosed by $$S$$:
$$\begin{aligned}
\va{F}_g
= \int_V \rho \va{g} \dd{V}
\end{aligned}$$
-Where $\rho$ is the fluid's density (which need not be constant)
-and $\va{g}$ is the gravity field given in units of force-per-mass.
+Where $$\rho$$ is the fluid's density (which need not be constant)
+and $$\va{g}$$ is the gravity field given in units of force-per-mass.
For a fluid at rest, these forces must cancel out:
$$\begin{aligned}
@@ -116,7 +116,7 @@ $$\begin{aligned}
\end{aligned}$$
On Earth (or another body with strong gravity),
-it is reasonable to treat $\va{g}$ as only pointing in the downward $z$-direction,
+it is reasonable to treat $$\va{g}$$ as only pointing in the downward $$z$$-direction,
in which case the above condition turns into:
$$\begin{aligned}
@@ -124,9 +124,9 @@ $$\begin{aligned}
= \rho g_0 z
\end{aligned}$$
-Where $g_0$ is the magnitude of the $z$-component of $\va{g}$.
+Where $$g_0$$ is the magnitude of the $$z$$-component of $$\va{g}$$.
We can generalize the equilibrium condition by treating
-the gravity field as the gradient of the gravitational potential $\Phi$:
+the gravity field as the gradient of the gravitational potential $$\Phi$$:
$$\begin{aligned}
\va{g}(\va{r})
@@ -142,13 +142,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-In practice, the density $\rho$ of the fluid
-may be a function of the pressure $p$ (compressibility)
-and/or temperature $T$ (thermal expansion).
+In practice, the density $$\rho$$ of the fluid
+may be a function of the pressure $$p$$ (compressibility)
+and/or temperature $$T$$ (thermal expansion).
We will tackle the first complication, but neglect the second,
i.e. we assume that the temperature is equal across the fluid.
-We then define the **pressure potential** $w(p)$ as
+We then define the **pressure potential** $$w(p)$$ as
the indefinite integral of the density:
$$\begin{aligned}
@@ -166,7 +166,7 @@ $$\begin{aligned}
\end{aligned}$$
From this, let us now define the
-**effective gravitational potential** $\Phi^*$ as follows:
+**effective gravitational potential** $$\Phi^*$$ as follows:
$$\begin{aligned}
\Phi^* \equiv \Phi + w(p)
@@ -181,7 +181,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-At every point in the fluid, despite $p$ being variable,
+At every point in the fluid, despite $$p$$ being variable,
the force that is applied by the pressure must have the same magnitude in all directions at that point.
This statement is known as **Pascal's law**,
and is due to the fact that all forces must cancel out
@@ -193,15 +193,15 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Let the blob be a cube with side $a$.
-Now, $\va{F}_p$ is a contact force,
-meaning it acts on the surface, and is thus proportional to $a^2$,
-however, $\va{F}_g$ is a body force,
-meaning it acts on the volume, and is thus proportional to $a^3$.
+Let the blob be a cube with side $$a$$.
+Now, $$\va{F}_p$$ is a contact force,
+meaning it acts on the surface, and is thus proportional to $$a^2$$,
+however, $$\va{F}_g$$ is a body force,
+meaning it acts on the volume, and is thus proportional to $$a^3$$.
Since we are considering a *point* in the fluid,
-$a$ is infinitesimally small,
-so that $\va{F}_p$ dominates $\va{F}_g$.
-Consequently, at equilibrium, $\va{F}_p$ must cancel out by itself,
+$$a$$ is infinitesimally small,
+so that $$\va{F}_p$$ dominates $$\va{F}_g$$.
+Consequently, at equilibrium, $$\va{F}_p$$ must cancel out by itself,
which means that the pressure is the same in all directions.
diff --git a/source/know/concept/imaginary-time/index.md b/source/know/concept/imaginary-time/index.md
index 1dbdf11..245c3c1 100644
--- a/source/know/concept/imaginary-time/index.md
+++ b/source/know/concept/imaginary-time/index.md
@@ -8,7 +8,7 @@ categories:
layout: "concept"
---
-Let $\hat{A}_S$ and $\hat{B}_S$ be time-independent in the Schrödinger picture.
+Let $$\hat{A}_S$$ and $$\hat{B}_S$$ be time-independent in the Schrödinger picture.
Then, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
consider the following expectation value
with respect to thermodynamic equilibium
@@ -19,9 +19,9 @@ $$\begin{aligned}
&= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_{0,S}(t)) \: \hat{A}_H(t) \: \hat{B}_H(t') \Big)
\end{aligned}$$
-Where the "simple" Hamiltonian $\hat{H}_{0,S}$ is time-independent.
-Suppose a (maybe time-dependent) "difficult" $\hat{H}_{1,S}$ is added,
-so that the total Hamiltonian is $\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$.
+Where the "simple" Hamiltonian $$\hat{H}_{0,S}$$ is time-independent.
+Suppose a (maybe time-dependent) "difficult" $$\hat{H}_{1,S}$$ is added,
+so that the total Hamiltonian is $$\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$$.
Then it is easier to consider the expectation value
in the [interaction picture](/know/concept/interaction-picture/):
@@ -30,16 +30,16 @@ $$\begin{aligned}
&= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_S(t)) \: \hat{K}_I(0, t) \hat{A}_I(t) \hat{K}_I(t, t') \hat{B}_I(t') \hat{K}_I(t', 0) \Big)
\end{aligned}$$
-Where $\hat{K}_I(t, t_0)$ is the time evolution operator of $\hat{H}_{1,S}$.
-In front, we have $\exp(-\beta \hat{H}_S(t))$,
-while $\hat{K}_I$ is an exponential of an integral of $\hat{H}_{1,I}$, so we are stuck.
+Where $$\hat{K}_I(t, t_0)$$ is the time evolution operator of $$\hat{H}_{1,S}$$.
+In front, we have $$\exp(-\beta \hat{H}_S(t))$$,
+while $$\hat{K}_I$$ is an exponential of an integral of $$\hat{H}_{1,I}$$, so we are stuck.
Keep in mind that exponentials of operators
cannot just be factorized, i.e. in general
-$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$
+$$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$$
To get around this, a useful mathematical trick is
-to use an **imaginary time** variable $\tau$ instead of the real time $t$.
-Fixing a $t$, we "redefine" the interaction picture along the imaginary axis:
+to use an **imaginary time** variable $$\tau$$ instead of the real time $$t$$.
+Fixing a $$t$$, we "redefine" the interaction picture along the imaginary axis:
$$\begin{aligned}
\boxed{
@@ -48,13 +48,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Ironically, $\tau$ is real; the point is that this formula
-comes from the real-time definition by replacing $t \to -i \tau$.
+Ironically, $$\tau$$ is real; the point is that this formula
+comes from the real-time definition by replacing $$t \to -i \tau$$.
The Heisenberg and Schrödinger pictures can be redefined in the same way.
-In fact, by substituting $t \to -i \tau$,
+In fact, by substituting $$t \to -i \tau$$,
all the key results of the interaction picture can be updated,
-for example the Schrödinger equation for $\Ket{\psi_S(\tau)}$ becomes:
+for example the Schrödinger equation for $$\Ket{\psi_S(\tau)}$$ becomes:
$$\begin{aligned}
\hbar \dv{}{t}\Ket{\psi_S(\tau)}
@@ -64,7 +64,7 @@ $$\begin{aligned}
= \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar} \bigg) \Ket{\psi_H}
\end{aligned}$$
-And the interaction picture's time evolution operator $\hat{K}_I$
+And the interaction picture's time evolution operator $$\hat{K}_I$$
turns out to be given by:
$$\begin{aligned}
@@ -74,9 +74,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\mathcal{T}$ is the
+Where $$\mathcal{T}$$ is the
[time-ordered product](/know/concept/time-ordered-product/)
-with respect to $\tau$.
+with respect to $$\tau$$.
This operator works as expected:
$$\begin{aligned}
@@ -84,7 +84,7 @@ $$\begin{aligned}
= \hat{K}_I(\tau, \tau_0) \Ket{\psi_I(\tau_0)}
\end{aligned}$$
-Where $\Ket{\psi_I(\tau)}$ is related to
+Where $$\Ket{\psi_I(\tau)}$$ is related to
the Schrödinger and Heisenberg pictures as follows:
$$\begin{aligned}
@@ -94,7 +94,7 @@ $$\begin{aligned}
\end{aligned}$$
It is interesting to combine this definition
-with the action of time evolution $\hat{K}_I(\tau, \tau_0)$:
+with the action of time evolution $$\hat{K}_I(\tau, \tau_0)$$:
$$\begin{aligned}
\Ket{\psi_I(\tau)}
@@ -105,7 +105,7 @@ $$\begin{aligned}
\end{aligned}$$
Rearranging this leads to the following useful
-alternative expression for $\hat{K}_I(\tau, \tau_0)$:
+alternative expression for $$\hat{K}_I(\tau, \tau_0)$$:
$$\begin{aligned}
\boxed{
@@ -117,8 +117,8 @@ $$\begin{aligned}
\end{aligned}$$
Returning to our initial example,
-we can set $\tau = \hbar \beta$ and $\tau_0 = 0$,
-so $\hat{K}_I(\tau, \tau_0)$ becomes:
+we can set $$\tau = \hbar \beta$$ and $$\tau_0 = 0$$,
+so $$\hat{K}_I(\tau, \tau_0)$$ becomes:
$$\begin{aligned}
\hat{K}_I(\hbar \beta, 0)
@@ -130,7 +130,7 @@ $$\begin{aligned}
\end{aligned}$$
Using the easily-shown fact that
-$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$,
+$$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$$,
we can therefore rewrite the thermodynamic expectation value like so:
$$\begin{aligned}
@@ -139,8 +139,8 @@ $$\begin{aligned}
\hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big)
\end{aligned}$$
-We now introduce a time-ordering $\mathcal{T}$,
-letting us reorder the (bosonic) $\hat{K}_I$-operators inside,
+We now introduce a time-ordering $$\mathcal{T}$$,
+letting us reorder the (bosonic) $$\hat{K}_I$$-operators inside,
and thereby reduce the expression considerably:
$$\begin{aligned}
@@ -151,9 +151,9 @@ $$\begin{aligned}
&= \frac{1}{Z} \Tr\!\Big( \mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp(-\beta \hat{H}_{0,S}) \Big)
\end{aligned}$$
-Where $Z = \Tr\!\big(\exp(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp(-\beta \hat{H}_{0,S})\big)$.
-If we now define $\Expval{}_0$ as the expectation value with respect
-to the unperturbed equilibrium involving only $\hat{H}_{0,S}$,
+Where $$Z = \Tr\!\big(\exp(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp(-\beta \hat{H}_{0,S})\big)$$.
+If we now define $$\Expval{}_0$$ as the expectation value with respect
+to the unperturbed equilibrium involving only $$\hat{H}_{0,S}$$,
we arrive at the following way of writing this time-ordered expectation:
$$\begin{aligned}
diff --git a/source/know/concept/impulse-response/index.md b/source/know/concept/impulse-response/index.md
index 533580d..397ac2d 100644
--- a/source/know/concept/impulse-response/index.md
+++ b/source/know/concept/impulse-response/index.md
@@ -8,9 +8,9 @@ categories:
layout: "concept"
---
-The **impulse response** $u_p(t)$ of a system whose behaviour is described
-by a linear operator $\hat{L}$, is defined as the reponse of the system
-when forced by the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(t)$:
+The **impulse response** $$u_p(t)$$ of a system whose behaviour is described
+by a linear operator $$\hat{L}$$, is defined as the reponse of the system
+when forced by the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(t)$$:
$$\begin{aligned}
\boxed{
@@ -18,9 +18,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-This can be used to find the response $u(t)$ of $\hat{L}$ to
-*any* forcing function $f(t)$, i.e. not only $\delta(t)$,
-by simply taking the convolution with $u_p(t)$:
+This can be used to find the response $$u(t)$$ of $$\hat{L}$$ to
+*any* forcing function $$f(t)$$, i.e. not only $$\delta(t)$$,
+by simply taking the convolution with $$u_p(t)$$:
$$\begin{aligned}
\hat{L} \{ u(t) \} = f(t)
@@ -35,9 +35,9 @@ $$\begin{aligned}
-Starting from the definition of $u_p(t)$,
-we shift the argument by some constant $\tau$,
-and multiply both sides by the constant $f(\tau)$:
+Starting from the definition of $$u_p(t)$$,
+we shift the argument by some constant $$\tau$$,
+and multiply both sides by the constant $$f(\tau)$$:
$$\begin{aligned}
\hat{L} \{ u_p(t - \tau) \} &= \delta(t - \tau)
@@ -45,8 +45,8 @@ $$\begin{aligned}
\hat{L} \{ f(\tau) \: u_p(t - \tau) \} &= f(\tau) \: \delta(t - \tau)
\end{aligned}$$
-Where $f(\tau)$ can be moved inside using the
-linearity of $\hat{L}$. Integrating over $\tau$ then gives us:
+Where $$f(\tau)$$ can be moved inside using the
+linearity of $$\hat{L}$$. Integrating over $$\tau$$ then gives us:
$$\begin{aligned}
\int_0^\infty \hat{L} \{ f(\tau) \: u_p(t - \tau) \} \dd{\tau}
@@ -54,20 +54,21 @@ $$\begin{aligned}
= f(t)
\end{aligned}$$
-The integral and $\hat{L}$ are operators of different variables, so we reorder them:
+The integral and $$\hat{L}$$ are operators of different variables, so we reorder them:
$$\begin{aligned}
\hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
&= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
\end{aligned}$$
+
This is useful for solving initial value problems,
because any initial condition can be satisfied
-due to the linearity of $\hat{L}$,
-by choosing the initial values of the homogeneous solution $\hat{L}\{ u_h(t) \} = 0$
-such that the total solution $(f * u_p)(t) + u_h(t)$
+due to the linearity of $$\hat{L}$$,
+by choosing the initial values of the homogeneous solution $$\hat{L}\{ u_h(t) \} = 0$$
+such that the total solution $$(f * u_p)(t) + u_h(t)$$
has the desired values.
Meanwhile, for boundary value problems,
diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md
index 05f3ad0..de469fa 100644
--- a/source/know/concept/interaction-picture/index.md
+++ b/source/know/concept/interaction-picture/index.md
@@ -13,25 +13,25 @@ is an alternative formulation of quantum mechanics,
equivalent to both the Schrödinger picture
and the [Heisenberg picture](/know/concept/heisenberg-picture/).
-Recall that Schrödinger lets states $\Ket{\psi_S(t)}$ evolve in time,
-but keeps operators $\hat{L}_S$ fixed (except for explicit time dependence).
-Meanwhile, Heisenberg keeps states $\Ket{\psi_H}$ fixed,
-and puts all time dependence on the operators $\hat{L}_H(t)$.
+Recall that Schrödinger lets states $$\Ket{\psi_S(t)}$$ evolve in time,
+but keeps operators $$\hat{L}_S$$ fixed (except for explicit time dependence).
+Meanwhile, Heisenberg keeps states $$\Ket{\psi_H}$$ fixed,
+and puts all time dependence on the operators $$\hat{L}_H(t)$$.
However, in the interaction picture,
-both the states $\Ket{\psi_I(t)}$ and the operators $\hat{L}_I(t)$
-evolve in $t$.
+both the states $$\Ket{\psi_I(t)}$$ and the operators $$\hat{L}_I(t)$$
+evolve in $$t$$.
This might seem unnecessarily complicated,
but it turns out be convenient when considering
-a time-dependent "perturbation" $\hat{H}_{1,S}$
-to a time-independent Hamiltonian $\hat{H}_{0,S}$:
+a time-dependent "perturbation" $$\hat{H}_{1,S}$$
+to a time-independent Hamiltonian $$\hat{H}_{0,S}$$:
$$\begin{aligned}
\hat{H}_S(t)
= \hat{H}_{0,S} + \hat{H}_{1,S}(t)
\end{aligned}$$
-With $\hat{H}_S(t)$ the full Schrödinger Hamiltonian.
+With $$\hat{H}_S(t)$$ the full Schrödinger Hamiltonian.
We define the unitary conversion operator:
$$\begin{aligned}
@@ -41,7 +41,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The interaction-picture states $\Ket{\psi_I(t)}$ and operators $\hat{L}_I(t)$
+The interaction-picture states $$\Ket{\psi_I(t)}$$ and operators $$\hat{L}_I(t)$$
are then defined to be:
$$\begin{aligned}
@@ -57,8 +57,8 @@ $$\begin{aligned}
## Equations of motion
-To find the equation of motion for $\Ket{\psi_I(t)}$,
-we differentiate it and multiply by $i \hbar$:
+To find the equation of motion for $$\Ket{\psi_I(t)}$$,
+we differentiate it and multiply by $$i \hbar$$:
$$\begin{aligned}
i \hbar \dv{}{t}\Ket{\psi_I}
@@ -68,7 +68,7 @@ $$\begin{aligned}
\end{aligned}$$
We insert the Schrödinger equation into the second term,
-and use $\comm{\hat{U}}{\hat{H}_{0,S}} = 0$:
+and use $$\comm{\hat{U}}{\hat{H}_{0,S}} = 0$$:
$$\begin{aligned}
i \hbar \dv{}{t}\Ket{\psi_I}
@@ -80,7 +80,7 @@ $$\begin{aligned}
\end{aligned}$$
Which leads to an analogue of the Schrödinger equation,
-with $\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$:
+with $$\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$$:
$$\begin{aligned}
\boxed{
@@ -89,7 +89,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Next, we do the same with an operator $\hat{L}_I$
+Next, we do the same with an operator $$\hat{L}_I$$
to find a description of its evolution in time:
$$\begin{aligned}
@@ -120,8 +120,8 @@ $$\begin{aligned}
Recall that an alternative form of the Schrödinger equation is as follows,
where a **time evolution operator** or
-**generator of translations in time** $K_S(t, t_0)$
-brings $\Ket{\psi_S}$ from time $t_0$ to $t$:
+**generator of translations in time** $$K_S(t, t_0)$$
+brings $$\Ket{\psi_S}$$ from time $$t_0$$ to $$t$$:
$$\begin{aligned}
\Ket{\psi_S(t)}
@@ -138,8 +138,8 @@ $$\begin{aligned}
\equiv \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)}
\end{aligned}$$
-Inserting this definition into the equation of motion for $\Ket{\psi_I}$ yields
-an equation for $\hat{K}_I$, with the logical boundary condition $\hat{K}_I(t_0, t_0) = 1$:
+Inserting this definition into the equation of motion for $$\Ket{\psi_I}$$ yields
+an equation for $$\hat{K}_I$$, with the logical boundary condition $$\hat{K}_I(t_0, t_0) = 1$$:
$$\begin{aligned}
i \hbar \dv{}{t}\Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big)
@@ -150,7 +150,7 @@ $$\begin{aligned}
\end{aligned}$$
We turn this into an integral equation
-by integrating both sides from $t_0$ to $t$:
+by integrating both sides from $$t_0$$ to $$t$$:
$$\begin{aligned}
i \hbar \int_{t_0}^t \dv{}{t'}K_I(t', t_0) \dd{t'}
@@ -158,15 +158,15 @@ $$\begin{aligned}
\end{aligned}$$
After evaluating the left integral,
-we see an expression for $\hat{K}_I$ as a function of $\hat{K}_I$ itself:
+we see an expression for $$\hat{K}_I$$ as a function of $$\hat{K}_I$$ itself:
$$\begin{aligned}
K_I(t, t_0)
= 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
\end{aligned}$$
-By recursively inserting $\hat{K}_I$ once, we get a longer expression,
-still with $\hat{K}_I$ on both sides:
+By recursively inserting $$\hat{K}_I$$ once, we get a longer expression,
+still with $$\hat{K}_I$$ on both sides:
$$\begin{aligned}
K_I(t, t_0)
@@ -175,9 +175,9 @@ $$\begin{aligned}
\end{aligned}$$
And so on. Note the ordering of the integrals and integrands:
-upon closer inspection, we see that the $n$th term is
-a [time-ordered product](/know/concept/time-ordered-product/) $\mathcal{T}$
-of $n$ factors $\hat{H}_{1,I}$:
+upon closer inspection, we see that the $$n$$th term is
+a [time-ordered product](/know/concept/time-ordered-product/) $$\mathcal{T}$$
+of $$n$$ factors $$\hat{H}_{1,I}$$:
$$\begin{aligned}
\hat{K}_I(t, t_0)
@@ -193,8 +193,8 @@ $$\begin{aligned}
\end{aligned}$$
This construction is occasionally called the **Dyson series**.
-We recognize the well-known Taylor expansion of $\exp(x)$,
-leading us to a final expression for $\hat{K}_I$:
+We recognize the well-known Taylor expansion of $$\exp(x)$$,
+leading us to a final expression for $$\hat{K}_I$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/ion-sound-wave/index.md b/source/know/concept/ion-sound-wave/index.md
index 622605c..cb86c04 100644
--- a/source/know/concept/ion-sound-wave/index.md
+++ b/source/know/concept/ion-sound-wave/index.md
@@ -16,9 +16,9 @@ at lower temperatures and pressures
than would be possible in a neutral gas.
We start from the [two-fluid model's](/know/concept/two-fluid-equations/) momentum equations,
-rewriting the [electric field](/know/concept/electric-field/) $\vb{E} = - \nabla \phi$
-and the pressure gradient $\nabla p = \gamma k_B T \nabla n$,
-and arguing that $m_e \approx 0$ because $m_e \ll m_i$:
+rewriting the [electric field](/know/concept/electric-field/) $$\vb{E} = - \nabla \phi$$
+and the pressure gradient $$\nabla p = \gamma k_B T \nabla n$$,
+and arguing that $$m_e \approx 0$$ because $$m_e \ll m_i$$:
$$\begin{aligned}
m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
@@ -29,9 +29,9 @@ $$\begin{aligned}
\end{aligned}$$
Note that we neglect ion-electron collisions,
-and allow for separate values of $\gamma$.
-We split $n_i$, $n_e$, $\vb{u}_i$ and $\phi$ into an equilibrium
-(subscript $0$) and a perturbation (subscript $1$):
+and allow for separate values of $$\gamma$$.
+We split $$n_i$$, $$n_e$$, $$\vb{u}_i$$ and $$\phi$$ into an equilibrium
+(subscript $$0$$) and a perturbation (subscript $$1$$):
$$\begin{aligned}
n_i
@@ -47,8 +47,8 @@ $$\begin{aligned}
= \phi_0 + \phi_1
\end{aligned}$$
-Where the perturbations $n_{i1}$, $n_{e1}$, $\vb{u}_{i1}$ and $\phi_1$ are tiny,
-and the equilibrium components $n_{i0}$, $n_{e0}$, $\vb{u}_{i0}$ and $\phi_0$
+Where the perturbations $$n_{i1}$$, $$n_{e1}$$, $$\vb{u}_{i1}$$ and $$\phi_1$$ are tiny,
+and the equilibrium components $$n_{i0}$$, $$n_{e0}$$, $$\vb{u}_{i0}$$ and $$\phi_0$$
by definition satisfy:
$$\begin{aligned}
@@ -65,9 +65,9 @@ $$\begin{aligned}
Inserting this decomposition into the momentum equations
yields new equations.
-Note that we will implicitly use $\vb{u}_{i0} = 0$
+Note that we will implicitly use $$\vb{u}_{i0} = 0$$
to pretend that the [material derivative](/know/concept/material-derivative/)
-$\mathrm{D}/\mathrm{D} t$ is linear:
+$$\mathrm{D}/\mathrm{D} t$$ is linear:
$$\begin{aligned}
m_i (n_{i0} \!+\! n_{i1}) \frac{\mathrm{D} (\vb{u}_{i0} \!+\! \vb{u}_{i1})}{\mathrm{D} t}
@@ -78,7 +78,7 @@ $$\begin{aligned}
\end{aligned}$$
Using the defined properties of the equilibrium components
-$n_{i0}$, $n_{e0}$, $\vb{u}_{i0}$ and $\phi_0$,
+$$n_{i0}$$, $$n_{e0}$$, $$\vb{u}_{i0}$$ and $$\phi_0$$,
and neglecting all products of perturbations for being small,
this reduces to:
@@ -118,7 +118,7 @@ $$\begin{aligned}
\end{aligned}$$
The electron equation can easily be rearranged
-to get a relation between $n_{e1}$ and $n_{e0}$:
+to get a relation between $$n_{e1}$$ and $$n_{e0}$$:
$$\begin{aligned}
i \vb{k} \gamma_e k_B T_e n_{e1}
@@ -131,12 +131,12 @@ $$\begin{aligned}
Due to their low mass, the electrons' heat conductivity
can be regarded as infinite compared to the ions'.
In that case, all electron gas compression is isothermal,
-meaning it obeys the ideal gas law $p_e = n_e k_B T_e$, so that $\gamma_e = 1$.
+meaning it obeys the ideal gas law $$p_e = n_e k_B T_e$$, so that $$\gamma_e = 1$$.
Note that this yields the first-order term of a Taylor expansion
of the [Boltzmann relation](/know/concept/boltzmann-relation/).
-At equilibrium, quasi-neutrality demands that $n_{i0} = n_{e0} = n_0$,
-so we can rearrange the above relation to $n_0 = - k_B T_e n_{e1} / (q_e \phi_1)$,
+At equilibrium, quasi-neutrality demands that $$n_{i0} = n_{e0} = n_0$$,
+so we can rearrange the above relation to $$n_0 = - k_B T_e n_{e1} / (q_e \phi_1)$$,
which we insert into the ion equation to get:
$$\begin{gathered}
@@ -148,8 +148,8 @@ $$\begin{gathered}
= T_e n_{e1} |\vb{k}|^2 - \gamma_i T_i n_{i1} |\vb{k}|^2
\end{gathered}$$
-Where we have taken the dot product with $\vb{k}$,
-and used that $q_i / q_e = -1$.
+Where we have taken the dot product with $$\vb{k}$$,
+and used that $$q_i / q_e = -1$$.
In order to simplify this equation,
we turn to the two-fluid ion continuity relation:
@@ -160,7 +160,7 @@ $$\begin{aligned}
\end{aligned}$$
Then we insert our plane-wave ansatz,
-and substitute $n_{i0} = n_0$ as before, yielding:
+and substitute $$n_{i0} = n_0$$ as before, yielding:
$$\begin{aligned}
0
@@ -172,7 +172,7 @@ $$\begin{aligned}
\end{aligned}$$
Substituting this in the ion momentum equation
-leads us to a dispersion relation $\omega(\vb{k})$:
+leads us to a dispersion relation $$\omega(\vb{k})$$:
$$\begin{gathered}
\omega^2 m_i \frac{T_e n_{e1}}{q_e \phi_1} \frac{q_e n_{i1} \phi_1}{k_B T_e n_{e1}}
@@ -184,8 +184,8 @@ $$\begin{gathered}
= \frac{|\vb{k}|^2}{m_i} \Big( k_B T_e \frac{n_{e1}}{n_{i1}} - \gamma_i k_B T_i \Big)
\end{gathered}$$
-Finally, we would like to find an expression for $n_{e1} / n_{i1}$.
-It cannot be $1$, because then $\phi_1$ could not be nonzero,
+Finally, we would like to find an expression for $$n_{e1} / n_{i1}$$.
+It cannot be $$1$$, because then $$\phi_1$$ could not be nonzero,
according to [Gauss' law](/know/concept/maxwells-equations/).
Nevertheless, authors often ignore this fact,
thereby making the so-called **plasma approximation**.
@@ -199,8 +199,8 @@ $$\begin{aligned}
\end{aligned}$$
One final time, we insert our plane-wave ansatz,
-and use our Boltzmann-like relation between $n_{e1}$ and $n_{e0}$
-to substitute $\phi_1 = - k_B T_e n_{e1} / (q_e n_{e0})$:
+and use our Boltzmann-like relation between $$n_{e1}$$ and $$n_{e0}$$
+to substitute $$\phi_1 = - k_B T_e n_{e1} / (q_e n_{e0})$$:
$$\begin{gathered}
q_e (n_{e1} - n_{i1})
@@ -213,7 +213,7 @@ $$\begin{gathered}
= n_{e1} \big( 1 + |\vb{k}|^2 \lambda_{De}^2 \big)
\end{gathered}$$
-Where $\lambda_{De}$ is the electron [Debye length](/know/concept/debye-length/).
+Where $$\lambda_{De}$$ is the electron [Debye length](/know/concept/debye-length/).
We thus reach the following dispersion relation,
which governs **ion sound waves** or **ion acoustic waves**:
@@ -224,7 +224,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The aforementioned plasma approximation is valid if $|\vb{k}| \lambda_{De} \ll 1$,
+The aforementioned plasma approximation is valid if $$|\vb{k}| \lambda_{De} \ll 1$$,
which is often reasonable,
in which case this dispersion relation reduces to:
@@ -233,7 +233,7 @@ $$\begin{aligned}
= \frac{|\vb{k}|^2}{m_i} \bigg( k_B T_e + \gamma_i k_B T_i \bigg)
\end{aligned}$$
-The phase velocity $v_s$ of these waves,
+The phase velocity $$v_s$$ of these waves,
i.e. the speed of sound, is then given by:
$$\begin{aligned}
@@ -245,11 +245,11 @@ $$\begin{aligned}
\end{aligned}$$
Curiously, unlike a neutral gas,
-this velocity is nonzero even if $T_i = 0$,
+this velocity is nonzero even if $$T_i = 0$$,
meaning that the waves still exist then.
-In fact, usually the electron temperature $T_e$ dominates $T_e \gg T_i$,
+In fact, usually the electron temperature $$T_e$$ dominates $$T_e \gg T_i$$,
even though the main feature of these waves
-is that they involve ion density fluctuations $n_{i1}$.
+is that they involve ion density fluctuations $$n_{i1}$$.
diff --git a/source/know/concept/ito-integral/index.md b/source/know/concept/ito-integral/index.md
index 3f17a9a..f087f97 100644
--- a/source/know/concept/ito-integral/index.md
+++ b/source/know/concept/ito-integral/index.md
@@ -9,10 +9,10 @@ layout: "concept"
---
The **Itō integral** offers a way to integrate
-a given [stochastic process](/know/concept/stochastic-process/) $G_t$
-with respect to a [Wiener process](/know/concept/wiener-process/) $B_t$,
+a given [stochastic process](/know/concept/stochastic-process/) $$G_t$$
+with respect to a [Wiener process](/know/concept/wiener-process/) $$B_t$$,
which is also a stochastic process.
-The Itō integral $I_t$ of $G_t$ is defined as follows:
+The Itō integral $$I_t$$ of $$G_t$$ is defined as follows:
$$\begin{aligned}
\boxed{
@@ -22,17 +22,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where have partitioned the time interval $[a, b]$ into steps of size $h$.
-The above integral exists if $G_t$ and $B_t$ are adapted
-to a common filtration $\mathcal{F}_t$,
-and $\mathbf{E}[G_t^2]$ is integrable for $t \in [a, b]$.
-If $I_t$ exists, $G_t$ is said to be **Itō-integrable** with respect to $B_t$.
+Where have partitioned the time interval $$[a, b]$$ into steps of size $$h$$.
+The above integral exists if $$G_t$$ and $$B_t$$ are adapted
+to a common filtration $$\mathcal{F}_t$$,
+and $$\mathbf{E}[G_t^2]$$ is integrable for $$t \in [a, b]$$.
+If $$I_t$$ exists, $$G_t$$ is said to be **Itō-integrable** with respect to $$B_t$$.
## Motivation
-Consider the following simple first-order differential equation for $X_t$,
-for some function $f$:
+Consider the following simple first-order differential equation for $$X_t$$,
+for some function $$f$$:
$$\begin{aligned}
\dv{X_t}{t}
@@ -40,7 +40,7 @@ $$\begin{aligned}
\end{aligned}$$
This can be solved numerically using the explicit Euler scheme
-by discretizing it with step size $h$,
+by discretizing it with step size $$h$$,
which can be applied recursively, leading to:
$$\begin{aligned}
@@ -51,7 +51,7 @@ $$\begin{aligned}
\approx X_0 + \sum_{s = 0}^{s = t} f(X_s) \: h
\end{aligned}$$
-In the limit $h \to 0$, this leads to the following unsurprising integral for $X_t$:
+In the limit $$h \to 0$$, this leads to the following unsurprising integral for $$X_t$$:
$$\begin{aligned}
\int_0^t f(X_s) \dd{s}
@@ -59,18 +59,18 @@ $$\begin{aligned}
\end{aligned}$$
In contrast, consider the *stochastic differential equation* below,
-where $\xi_t$ represents white noise,
-which is informally the $t$-derivative
-of the Wiener process $\xi_t = \idv{B_t}{t}$:
+where $$\xi_t$$ represents white noise,
+which is informally the $$t$$-derivative
+of the Wiener process $$\xi_t = \idv{B_t}{t}$$:
$$\begin{aligned}
\dv{X_t}{t}
= g(X_t) \: \xi_t
\end{aligned}$$
-Now $X_t$ is not deterministic,
-since $\xi_t$ is derived from a random variable $B_t$.
-If $g = 1$, we expect $X_t = X_0 + B_t$.
+Now $$X_t$$ is not deterministic,
+since $$\xi_t$$ is derived from a random variable $$B_t$$.
+If $$g = 1$$, we expect $$X_t = X_0 + B_t$$.
With this in mind, we introduce the **Euler-Maruyama scheme**:
$$\begin{aligned}
@@ -80,7 +80,7 @@ $$\begin{aligned}
&= X_t + g(X_t) \: (B_{t+h} - B_t)
\end{aligned}$$
-We would like to turn this into an integral for $X_t$, as we did above.
+We would like to turn this into an integral for $$X_t$$, as we did above.
Therefore, we state:
$$\begin{aligned}
@@ -89,8 +89,8 @@ $$\begin{aligned}
\end{aligned}$$
This integral is *defined* as below,
-analogously to the first, but with $h$ replaced by
-the increment $B_{t+h} \!-\! B_t$ of a Wiener process.
+analogously to the first, but with $$h$$ replaced by
+the increment $$B_{t+h} \!-\! B_t$$ of a Wiener process.
This is an Itō integral:
$$\begin{aligned}
@@ -104,14 +104,14 @@ see the [Itō calculus](/know/concept/ito-calculus/).
## Properties
-Since $G_t$ and $B_t$ must be known (i.e. $\mathcal{F}_t$-adapted)
-in order to evaluate the Itō integral $I_t$ at any given $t$,
-it logically follows that $I_t$ is also $\mathcal{F}_t$-adapted.
+Since $$G_t$$ and $$B_t$$ must be known (i.e. $$\mathcal{F}_t$$-adapted)
+in order to evaluate the Itō integral $$I_t$$ at any given $$t$$,
+it logically follows that $$I_t$$ is also $$\mathcal{F}_t$$-adapted.
Because the Itō integral is defined as the limit of a sum of linear terms,
it inherits this linearity.
-Consider two Itō-integrable processes $G_t$ and $H_t$,
-and two constants $v, w \in \mathbb{R}$:
+Consider two Itō-integrable processes $$G_t$$ and $$H_t$$,
+and two constants $$v, w \in \mathbb{R}$$:
$$\begin{aligned}
\int_a^b v G_t + w H_t \dd{B_t}
@@ -119,7 +119,7 @@ $$\begin{aligned}
\end{aligned}$$
By adding multiple summations,
-the Itō integral clearly satisfies, for $a < b < c$:
+the Itō integral clearly satisfies, for $$a < b < c$$:
$$\begin{aligned}
\int_a^c G_t \dd{B_t}
@@ -127,8 +127,8 @@ $$\begin{aligned}
\end{aligned}$$
A more interesting property is the **Itō isometry**,
-which expresses the expectation of the square of an Itō integral of $G_t$
-as a simpler "ordinary" integral of the expectation of $G_t^2$
+which expresses the expectation of the square of an Itō integral of $$G_t$$
+as a simpler "ordinary" integral of the expectation of $$G_t^2$$
(which exists by the definition of Itō-integrability):
$$\begin{aligned}
@@ -144,14 +144,14 @@ $$\begin{aligned}
We write out the left-hand side of the Itō isometry,
-where eventually $h \to 0$:
+where eventually $$h \to 0$$:
$$\begin{aligned}
\mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
&= \sum_{t = a}^{t = b} \sum_{s = a}^{s = b} \mathbf{E} \bigg[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \bigg]
\end{aligned}$$
-In the particular case $t \ge s \!+\! h$,
+In the particular case $$t \ge s \!+\! h$$,
a given term of this summation can be rewritten
as follows using the *law of total expectation*
(see [conditional expectation](/know/concept/conditional-expectation/)):
@@ -161,18 +161,18 @@ $$\begin{aligned}
= \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big| \mathcal{F}_t \Big] \bigg]
\end{aligned}$$
-Recall that $G_t$ and $B_t$ are adapted to $\mathcal{F}_t$:
-at time $t$, we have information $\mathcal{F}_t$,
-which includes knowledge of the realized values $G_t$ and $B_t$.
-Since $t \ge s \!+\! h$ by assumption, we can simply factor out the known quantities:
+Recall that $$G_t$$ and $$B_t$$ are adapted to $$\mathcal{F}_t$$:
+at time $$t$$, we have information $$\mathcal{F}_t$$,
+which includes knowledge of the realized values $$G_t$$ and $$B_t$$.
+Since $$t \ge s \!+\! h$$ by assumption, we can simply factor out the known quantities:
$$\begin{aligned}
\mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
= \mathbf{E} \bigg[ G_t G_s (B_{s + h} \!-\! B_s) \: \mathbf{E} \Big[ (B_{t + h} \!-\! B_t) \Big| \mathcal{F}_t \Big] \bigg]
\end{aligned}$$
-However, $\mathcal{F}_t$ says nothing about
-the increment $(B_{t + h} \!-\! B_t) \sim \mathcal{N}(0, h)$,
+However, $$\mathcal{F}_t$$ says nothing about
+the increment $$(B_{t + h} \!-\! B_t) \sim \mathcal{N}(0, h)$$,
meaning that the conditional expectation is zero:
$$\begin{aligned}
@@ -181,7 +181,7 @@ $$\begin{aligned}
\qquad \mathrm{for}\; t \ge s + h
\end{aligned}$$
-By swapping $s$ and $t$, the exact same result can be obtained for $s \ge t \!+\! h$:
+By swapping $$s$$ and $$t$$, the exact same result can be obtained for $$s \ge t \!+\! h$$:
$$\begin{aligned}
\mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
@@ -189,7 +189,7 @@ $$\begin{aligned}
\qquad \mathrm{for}\; s \ge t + h
\end{aligned}$$
-This leaves only one case which can be nonzero: $[t, t\!+\!h] = [s, s\!+\!h]$.
+This leaves only one case which can be nonzero: $$[t, t\!+\!h] = [s, s\!+\!h]$$.
Applying the law of total expectation again yields:
$$\begin{aligned}
@@ -199,8 +199,8 @@ $$\begin{aligned}
&= \sum_{t = a}^{t = b} \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big| \mathcal{F}_t \Big] \bigg]
\end{aligned}$$
-We know $G_t$, and the expectation value of $(B_{t+h} \!-\! B_t)^2$,
-since the increment is normally distributed, is simply the variance $h$:
+We know $$G_t$$, and the expectation value of $$(B_{t+h} \!-\! B_t)^2$$,
+since the increment is normally distributed, is simply the variance $$h$$:
$$\begin{aligned}
\mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
@@ -208,6 +208,7 @@ $$\begin{aligned}
\longrightarrow
\int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
\end{aligned}$$
+
@@ -221,9 +222,9 @@ since true white noise cannot be biased.
-We will prove that an arbitrary Itō integral $I_t$ is a martingale.
-Using additivity, we know that the increment $I_t \!-\! I_s$
-is as follows, given information $\mathcal{F}_s$:
+We will prove that an arbitrary Itō integral $$I_t$$ is a martingale.
+Using additivity, we know that the increment $$I_t \!-\! I_s$$
+is as follows, given information $$\mathcal{F}_s$$:
$$\begin{aligned}
\mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
@@ -232,8 +233,8 @@ $$\begin{aligned}
\end{aligned}$$
We rewrite this [conditional expectation](/know/concept/conditional-expectation/)
-using the *tower property* for some $\mathcal{F}_u \supset \mathcal{F}_s$,
-such that $G_u$ and $B_u$ are known, but $B_{u+h} \!-\! B_u$ is not:
+using the *tower property* for some $$\mathcal{F}_u \supset \mathcal{F}_s$$,
+such that $$G_u$$ and $$B_u$$ are known, but $$B_{u+h} \!-\! B_u$$ is not:
$$\begin{aligned}
\mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
@@ -242,7 +243,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-We now have everything we need to calculate $\mathbf{E} [ I_t | \mathcal{F_s} ]$,
+We now have everything we need to calculate $$\mathbf{E} [ I_t | \mathcal{F_s} ]$$,
giving the martingale property:
$$\begin{aligned}
@@ -252,12 +253,12 @@ $$\begin{aligned}
= I_s
\end{aligned}$$
-For the existence of $I_t$,
-we need $\mathbf{E}[G_t^2]$ to be integrable over the target interval,
-so from the Itō isometry we have $\mathbf{E}[I]^2 < \infty$,
-and therefore $\mathbf{E}[I] < \infty$,
-so $I_t$ has all the properties of a Martingale,
-since it is trivially $\mathcal{F}_t$-adapted.
+For the existence of $$I_t$$,
+we need $$\mathbf{E}[G_t^2]$$ to be integrable over the target interval,
+so from the Itō isometry we have $$\mathbf{E}[I]^2 < \infty$$,
+and therefore $$\mathbf{E}[I] < \infty$$,
+so $$I_t$$ has all the properties of a Martingale,
+since it is trivially $$\mathcal{F}_t$$-adapted.
diff --git a/source/know/concept/ito-process/index.md b/source/know/concept/ito-process/index.md
index b82835d..f192e28 100644
--- a/source/know/concept/ito-process/index.md
+++ b/source/know/concept/ito-process/index.md
@@ -9,8 +9,8 @@ layout: "concept"
---
Given two [stochastic processes](/know/concept/stochastic-process/)
-$F_t$ and $G_t$, consider the following random variable $X_t$,
-where $B_t$ is the [Wiener process](/know/concept/wiener-process/),
+$$F_t$$ and $$G_t$$, consider the following random variable $$X_t$$,
+where $$B_t$$ is the [Wiener process](/know/concept/wiener-process/),
i.e. Brownian motion:
$$\begin{aligned}
@@ -19,25 +19,25 @@ $$\begin{aligned}
\end{aligned}$$
Where the latter is an [Itō integral](/know/concept/ito-integral/),
-assuming $G_t$ is Itō-integrable.
-We call $X_t$ an **Itō process** if $F_t$ is locally integrable,
-and the initial condition $X_0$ is known,
-i.e. $X_0$ is $\mathcal{F}_0$-measurable,
-where $\mathcal{F}_t$ is the filtration
-to which $F_t$, $G_t$ and $B_t$ are adapted.
-The above definition of $X_t$ is often abbreviated as follows,
-where $X_0$ is implicit:
+assuming $$G_t$$ is Itō-integrable.
+We call $$X_t$$ an **Itō process** if $$F_t$$ is locally integrable,
+and the initial condition $$X_0$$ is known,
+i.e. $$X_0$$ is $$\mathcal{F}_0$$-measurable,
+where $$\mathcal{F}_t$$ is the filtration
+to which $$F_t$$, $$G_t$$ and $$B_t$$ are adapted.
+The above definition of $$X_t$$ is often abbreviated as follows,
+where $$X_0$$ is implicit:
$$\begin{aligned}
\dd{X_t}
= F_t \dd{t} + G_t \dd{B_t}
\end{aligned}$$
-Typically, $F_t$ is referred to as the **drift** of $X_t$,
-and $G_t$ as its **intensity**.
-Because the Itō integral of $G_t$ is a
+Typically, $$F_t$$ is referred to as the **drift** of $$X_t$$,
+and $$G_t$$ as its **intensity**.
+Because the Itō integral of $$G_t$$ is a
[martingale](/know/concept/martingale/),
-it does not contribute to the mean of $X_t$:
+it does not contribute to the mean of $$X_t$$:
$$\begin{aligned}
\mathbf{E}[X_t]
@@ -45,25 +45,25 @@ $$\begin{aligned}
\end{aligned}$$
Now, consider the following **Itō stochastic differential equation** (SDE),
-where $\xi_t = \idv{B_t}{t}$ is white noise,
-informally treated as the $t$-derivative of $B_t$:
+where $$\xi_t = \idv{B_t}{t}$$ is white noise,
+informally treated as the $$t$$-derivative of $$B_t$$:
$$\begin{aligned}
\dv{X_t}{t}
= f(X_t, t) + g(X_t, t) \: \xi_t
\end{aligned}$$
-An Itō process $X_t$ is said to satisfy this equation
-if $f(X_t, t) = F_t$ and $g(X_t, t) = G_t$,
-in which case $X_t$ is also called an **Itō diffusion**.
+An Itō process $$X_t$$ is said to satisfy this equation
+if $$f(X_t, t) = F_t$$ and $$g(X_t, t) = G_t$$,
+in which case $$X_t$$ is also called an **Itō diffusion**.
All Itō diffusions are [Markov processes](/know/concept/markov-process/),
-since only the current value of $X_t$ determines the future,
-and $B_t$ is also a Markov process.
+since only the current value of $$X_t$$ determines the future,
+and $$B_t$$ is also a Markov process.
## Itō's lemma
-Classically, given $y \equiv h(x(t), t)$,
+Classically, given $$y \equiv h(x(t), t)$$,
the chain rule of differentiation states that:
$$\begin{aligned}
@@ -71,8 +71,8 @@ $$\begin{aligned}
= \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{x}
\end{aligned}$$
-However, for a stochastic process $Y_t \equiv h(X_t, t)$,
-where $X_t$ is an Itō process,
+However, for a stochastic process $$Y_t \equiv h(X_t, t)$$,
+where $$X_t$$ is an Itō process,
the chain rule is modified to the following,
known as **Itō's lemma**:
@@ -89,7 +89,7 @@ $$\begin{aligned}
We start by applying the classical chain rule,
-but we go to second order in $x$.
+but we go to second order in $$x$$.
This is also valid classically,
but there we would neglect all higher-order infinitesimals:
@@ -98,7 +98,7 @@ $$\begin{aligned}
= \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{X_t} + \frac{1}{2} \pdvn{2}{h}{x} \dd{X_t}^2
\end{aligned}$$
-But here we cannot neglect $\dd{X_t}^2$.
+But here we cannot neglect $$\dd{X_t}^2$$.
We insert the definition of an Itō process:
$$\begin{aligned}
@@ -109,8 +109,8 @@ $$\begin{aligned}
+ \frac{1}{2} \pdvn{2}{h}{x} \Big( F_t^2 \dd{t}^2 + 2 F_t G_t \dd{t} \dd{B_t} + G_t^2 \dd{B_t}^2 \Big)
\end{aligned}$$
-In the limit of small $\dd{t}$, we can neglect $\dd{t}^2$,
-and as it turns out, $\dd{t} \dd{B_t}$ too:
+In the limit of small $$\dd{t}$$, we can neglect $$\dd{t}^2$$,
+and as it turns out, $$\dd{t} \dd{B_t}$$ too:
$$\begin{aligned}
\dd{t} \dd{B_t}
@@ -120,8 +120,8 @@ $$\begin{aligned}
\longrightarrow 0
\end{aligned}$$
-However, due to the scaling property of $B_t$,
-we cannot ignore $\dd{B_t}^2$, which has order $\dd{t}$:
+However, due to the scaling property of $$B_t$$,
+we cannot ignore $$\dd{B_t}^2$$, which has order $$\dd{t}$$:
$$\begin{aligned}
\dd{B_t}^2
@@ -131,8 +131,8 @@ $$\begin{aligned}
\longrightarrow \dd{t}
\end{aligned}$$
-Where $\chi_1^2(\dd{t})$ is the generalized chi-squared distribution
-with one term of variance $\dd{t}$.
+Where $$\chi_1^2(\dd{t})$$ is the generalized chi-squared distribution
+with one term of variance $$\dd{t}$$.
@@ -144,9 +144,9 @@ to make the solution of a given Itō SDE easier.
## Coordinate transformations
The simplest coordinate transformation is a scaling of the time axis.
-Defining $s \equiv \alpha t$, the goal is to keep the Itō process.
-We know how to scale $B_t$, be setting $W_s \equiv \sqrt{\alpha} B_{s / \alpha}$.
-Let $Y_s \equiv X_t$ be the new variable on the rescaled axis, then:
+Defining $$s \equiv \alpha t$$, the goal is to keep the Itō process.
+We know how to scale $$B_t$$, be setting $$W_s \equiv \sqrt{\alpha} B_{s / \alpha}$$.
+Let $$Y_s \equiv X_t$$ be the new variable on the rescaled axis, then:
$$\begin{aligned}
\dd{Y_s}
@@ -156,14 +156,14 @@ $$\begin{aligned}
&= \frac{1}{\alpha} f(Y_s) \dd{s} + \frac{1}{\sqrt{\alpha}} g(Y_s) \dd{W_s}
\end{aligned}$$
-$W_s$ is a valid Wiener process,
+$$W_s$$ is a valid Wiener process,
and the other changes are small,
so this is still an Itō process.
To solve SDEs analytically, it is usually best
-to have additive noise, i.e. $g = 1$.
+to have additive noise, i.e. $$g = 1$$.
This can be achieved using the **Lamperti transform**:
-define $Y_t \equiv h(X_t)$, where $h$ is given by:
+define $$Y_t \equiv h(X_t)$$, where $$h$$ is given by:
$$\begin{aligned}
\boxed{
@@ -173,8 +173,8 @@ $$\begin{aligned}
\end{aligned}$$
Then, using Itō's lemma, it is straightforward
-to show that the intensity becomes $1$.
-Note that the lower integration limit $x_0$ does not enter:
+to show that the intensity becomes $$1$$.
+Note that the lower integration limit $$x_0$$ does not enter:
$$\begin{aligned}
\dd{Y_t}
@@ -185,9 +185,9 @@ $$\begin{aligned}
&= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g'(X_t) \bigg) \dd{t} + \dd{B_t}
\end{aligned}$$
-Similarly, we can eliminate the drift $f = 0$,
+Similarly, we can eliminate the drift $$f = 0$$,
thereby making the Itō process a martingale.
-This is done by defining $Y_t \equiv h(X_t)$, with $h(x)$ given by:
+This is done by defining $$Y_t \equiv h(X_t)$$, with $$h(x)$$ given by:
$$\begin{aligned}
\boxed{
@@ -197,8 +197,8 @@ $$\begin{aligned}
\end{aligned}$$
The goal is to make the parenthesized first term (see above)
-of Itō's lemma disappear, which this $h(x)$ does indeed do.
-Note that $x_0$ and $x_1$ do not enter:
+of Itō's lemma disappear, which this $$h(x)$$ does indeed do.
+Note that $$x_0$$ and $$x_1$$ do not enter:
$$\begin{aligned}
0
@@ -212,8 +212,8 @@ $$\begin{aligned}
It is worth knowing under what condition a solution to a given SDE exists,
in the sense that it is finite on the entire time axis.
-Suppose the drift $f$ and intensity $g$ satisfy these inequalities,
-for some known constant $K$ and for all $x$:
+Suppose the drift $$f$$ and intensity $$g$$ satisfy these inequalities,
+for some known constant $$K$$ and for all $$x$$:
$$\begin{aligned}
x f(x) \le K (1 + x^2)
@@ -222,8 +222,8 @@ $$\begin{aligned}
\end{aligned}$$
When this is satisfied, we can find the following upper bound
-on an Itō process $X_t$,
-which clearly implies that $X_t$ is finite for all $t$:
+on an Itō process $$X_t$$,
+which clearly implies that $$X_t$$ is finite for all $$t$$:
$$\begin{aligned}
\boxed{
@@ -237,7 +237,7 @@ $$\begin{aligned}
-If we define $Y_t \equiv X_t^2$,
+If we define $$Y_t \equiv X_t^2$$,
then Itō's lemma tells us that the following holds:
$$\begin{aligned}
@@ -253,7 +253,7 @@ $$\begin{aligned}
= Y_0 + \mathbf{E}\! \int_0^t 2 X_s f(X_s) + g^2(X_s) \dd{s}
\end{aligned}$$
-Given that $K (1 \!+\! x^2)$ is an upper bound of $x f(x)$ and $g^2(x)$,
+Given that $$K (1 \!+\! x^2)$$ is an upper bound of $$x f(x)$$ and $$g^2(x)$$,
we get an inequality:
$$\begin{aligned}
@@ -267,7 +267,7 @@ $$\begin{aligned}
We then apply the
[Grönwall-Bellman inequality](/know/concept/gronwall-bellman-inequality/),
-noting that $(Y_0 \!+\! 3 K t)$ does not decrease with time, leading us to:
+noting that $$(Y_0 \!+\! 3 K t)$$ does not decrease with time, leading us to:
$$\begin{aligned}
\mathbf{E}[Y_t]
@@ -275,12 +275,13 @@ $$\begin{aligned}
\\
&\le (Y_0 + 3 K t) \exp\!\big(3 K t\big)
\end{aligned}$$
+
If a solution exists, it is also worth knowing whether it is unique.
-Suppose that $f$ and $g$ satisfy the following inequalities,
-for some constant $K$ and for all $x$ and $y$:
+Suppose that $$f$$ and $$g$$ satisfy the following inequalities,
+for some constant $$K$$ and for all $$x$$ and $$y$$:
$$\begin{aligned}
\big| f(x) - f(y) \big| \le K \big| x - y \big|
@@ -288,10 +289,10 @@ $$\begin{aligned}
\big| g(x) - g(y) \big| \le K \big| x - y \big|
\end{aligned}$$
-Let $X_t$ and $Y_t$ both be solutions to a given SDE,
+Let $$X_t$$ and $$Y_t$$ both be solutions to a given SDE,
but the initial conditions need not be the same,
-such that the difference is initially $X_0 \!-\! Y_0$.
-Then the difference $X_t \!-\! Y_t$ is bounded by:
+such that the difference is initially $$X_0 \!-\! Y_0$$.
+Then the difference $$X_t \!-\! Y_t$$ is bounded by:
$$\begin{aligned}
\boxed{
@@ -305,8 +306,8 @@ $$\begin{aligned}
-We define $D_t \equiv X_t \!-\! Y_t$ and $Z_t \equiv D_t^2 \ge 0$,
-together with $F_t \equiv f(X_t) \!-\! f(Y_t)$ and $G_t \equiv g(X_t) \!-\! g(Y_t)$,
+We define $$D_t \equiv X_t \!-\! Y_t$$ and $$Z_t \equiv D_t^2 \ge 0$$,
+together with $$F_t \equiv f(X_t) \!-\! f(Y_t)$$ and $$G_t \equiv g(X_t) \!-\! g(Y_t)$$,
such that Itō's lemma states:
$$\begin{aligned}
@@ -322,9 +323,9 @@ $$\begin{aligned}
= Z_0 + \mathbf{E}\! \int_0^t 2 D_s F_s + G_s^2 \dd{s}
\end{aligned}$$
-The *Cauchy-Schwarz inequality* states that $|D_s F_s| \le |D_s| |F_s|$,
-and then the given fact that $F_s$ and $G_s$ satisfy
-$|F_s| \le K |D_s|$ and $|G_s| \le K |D_s|$ gives:
+The *Cauchy-Schwarz inequality* states that $$|D_s F_s| \le |D_s| |F_s|$$,
+and then the given fact that $$F_s$$ and $$G_s$$ satisfy
+$$|F_s| \le K |D_s|$$ and $$|G_s| \le K |D_s|$$ gives:
$$\begin{aligned}
\mathbf{E}[Z_t]
@@ -333,12 +334,12 @@ $$\begin{aligned}
&\le Z_0 + \int_0^t (2 K \!+\! K^2) \: \mathbf{E}[Z_s] \dd{s}
\end{aligned}$$
-Where we have implicitly used that $D_s F_s = |D_s F_s|$
-because $Z_t$ is positive for all $G_s^2$,
-and that $|D_s|^2 = D_s^2$ because $D_s$ is real.
+Where we have implicitly used that $$D_s F_s = |D_s F_s|$$
+because $$Z_t$$ is positive for all $$G_s^2$$,
+and that $$|D_s|^2 = D_s^2$$ because $$D_s$$ is real.
We then apply the
[Grönwall-Bellman inequality](/know/concept/gronwall-bellman-inequality/),
-recognizing that $Z_0$ does not decrease with time (since it is constant):
+recognizing that $$Z_0$$ does not decrease with time (since it is constant):
$$\begin{aligned}
\mathbf{E}[Z_t]
@@ -346,13 +347,14 @@ $$\begin{aligned}
\\
&\le Z_0 \exp\!\Big( \big( 2 K \!+\! K^2 \big) t \Big)
\end{aligned}$$
+
Using these properties, it can then be shown
that if all of the above conditions are satisfied,
then the SDE has a unique solution,
-which is $\mathcal{F}_t$-adapted, continuous, and exists for all times.
+which is $$\mathcal{F}_t$$-adapted, continuous, and exists for all times.
diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md
index c951fd7..5c50f80 100644
--- a/source/know/concept/jellium/index.md
+++ b/source/know/concept/jellium/index.md
@@ -21,7 +21,7 @@ Let us start by neglecting electron-electron interactions.
This is clearly a dubious assumption, but we will stick with it for now.
For an infinitely large sample of jellium,
the single-electron states are simply plane waves.
-We consider an arbitrary cube of volume $V$,
+We consider an arbitrary cube of volume $$V$$,
and impose periodic boundary conditions on it,
such that the single-particle orbitals are (suppressing spin):
@@ -33,22 +33,22 @@ $$\begin{aligned}
\vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z)
\end{aligned}$$
-Where $n_x, n_y, n_z \in \mathbb{Z}$.
+Where $$n_x, n_y, n_z \in \mathbb{Z}$$.
This is a discrete (but infinite) set of independent orbitals,
so it is natural to use the
[second quantization](/know/concept/second-quantization/)
-to write the non-interacting Hamiltonian $\hat{H}_0$,
-where $\hbar^2 |\vb{k}|^2 / (2 m)$ is the kinetic energy
-of the orbital with wavevector $\vb{k}$, and $s$ is the spin:
+to write the non-interacting Hamiltonian $$\hat{H}_0$$,
+where $$\hbar^2 |\vb{k}|^2 / (2 m)$$ is the kinetic energy
+of the orbital with wavevector $$\vb{k}$$, and $$s$$ is the spin:
$$\begin{aligned}
\hat{H}_0
= \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}
\end{aligned}$$
-Assuming that the temperature $T = 0$,
-the $N$-electron ground state of this Hamiltonian
-is known as the **Fermi sea** or **Fermi sphere** $\Ket{\mathrm{FS}}$,
+Assuming that the temperature $$T = 0$$,
+the $$N$$-electron ground state of this Hamiltonian
+is known as the **Fermi sea** or **Fermi sphere** $$\Ket{\mathrm{FS}}$$,
and is constructed by filling up the single-electron states
starting from the lowest energy:
@@ -57,9 +57,9 @@ $$\begin{aligned}
= \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0}
\end{aligned}$$
-Because $T = 0$, all the electrons stay in their assigned state.
-The energy and wavenumber $|\vb{k}|$ of the highest filled orbital
-are called the **Fermi energy** $\epsilon_F$ and **Fermi wavenumber** $k_F$,
+Because $$T = 0$$, all the electrons stay in their assigned state.
+The energy and wavenumber $$|\vb{k}|$$ of the highest filled orbital
+are called the **Fermi energy** $$\epsilon_F$$ and **Fermi wavenumber** $$k_F$$,
and obey the expected kinetic energy relation:
$$\begin{aligned}
@@ -69,13 +69,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-The Fermi sea can be visualized in $\vb{k}$-space as a sphere with radius $k_F$.
-Because $\vb{k}$ is discrete, the sphere's surface is not smooth,
-but in the limit $V \to \infty$ it becomes perfect.
+The Fermi sea can be visualized in $$\vb{k}$$-space as a sphere with radius $$k_F$$.
+Because $$\vb{k}$$ is discrete, the sphere's surface is not smooth,
+but in the limit $$V \to \infty$$ it becomes perfect.
Now, we would like a relation between the system's parameters,
-e.g. $N$ and $V$, and the resulting values of $\epsilon_F$ or $k_F$.
-The total population $N$ must be given by:
+e.g. $$N$$ and $$V$$, and the resulting values of $$\epsilon_F$$ or $$k_F$$.
+The total population $$N$$ must be given by:
$$\begin{aligned}
N
@@ -83,11 +83,11 @@ $$\begin{aligned}
= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
\end{aligned}$$
-Where we have turned the sum over $\vb{k}$ into an integral with a constant factor,
-by using that each orbital exclusively occupies a volume $(2 \pi)^3 / V$ in $\vb{k}$-space.
+Where we have turned the sum over $$\vb{k}$$ into an integral with a constant factor,
+by using that each orbital exclusively occupies a volume $$(2 \pi)^3 / V$$ in $$\vb{k}$$-space.
-At zero temperature, this inner product can only be $0$ or $1$,
-depending on whether $\vb{k}$ is outside or inside the Fermi sphere.
+At zero temperature, this inner product can only be $$0$$ or $$1$$,
+depending on whether $$\vb{k}$$ is outside or inside the Fermi sphere.
We can therefore rewrite using a
[Heaviside step function](/know/concept/heaviside-step-function/):
@@ -98,10 +98,10 @@ $$\begin{aligned}
\end{aligned}$$
Where we realized that spin does not matter,
-and replaced the sum over $s$ by a factor $2$.
+and replaced the sum over $$s$$ by a factor $$2$$.
In order to evaluate this 3D integral,
we go to [spherical coordinates](/know/concept/spherical-coordinates/)
-$(|\vb{k}|, \theta, \varphi)$:
+$$(|\vb{k}|, \theta, \varphi)$$:
$$\begin{aligned}
N
@@ -112,7 +112,7 @@ $$\begin{aligned}
= \frac{V}{3 \pi^2} k_F^3
\end{aligned}$$
-Using that the electron density $n = N/V$,
+Using that the electron density $$n = N/V$$,
we thus arrive at the following relation:
$$\begin{aligned}
@@ -122,15 +122,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-This result also justifies our assumption that $T = 0$:
-we can accurately calculate the density $n$ for many conducting materials,
-and this relation then gives $k_F$ and $\epsilon_F$.
-It turns out that $\epsilon_F$ is usually very large
-compared to the thermal energy $k_B T$ at reasonable temperatures,
+This result also justifies our assumption that $$T = 0$$:
+we can accurately calculate the density $$n$$ for many conducting materials,
+and this relation then gives $$k_F$$ and $$\epsilon_F$$.
+It turns out that $$\epsilon_F$$ is usually very large
+compared to the thermal energy $$k_B T$$ at reasonable temperatures,
so we can conclude that thermal fluctuations are negligible.
-Now, $\epsilon_F$ is the highest single-electron energy,
-but about the total $N$-particle energy $E^{(0)}$?
+Now, $$\epsilon_F$$ is the highest single-electron energy,
+but about the total $$N$$-particle energy $$E^{(0)}$$?
$$\begin{aligned}
E^{(0)}
@@ -138,7 +138,7 @@ $$\begin{aligned}
= \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
\end{aligned}$$
-Once again, we turn the sum over $\vb{k}$ into an integral,
+Once again, we turn the sum over $$\vb{k}$$ into an integral,
and recognize the spin's irrelevance:
$$\begin{aligned}
@@ -150,7 +150,7 @@ $$\begin{aligned}
\end{aligned}$$
In spherical coordinates,
-we evaluate the integral and find that $E^{(0)}$ is proportional to $k_F^5$:
+we evaluate the integral and find that $$E^{(0)}$$ is proportional to $$k_F^5$$:
$$\begin{aligned}
E^{(0)}
@@ -163,8 +163,8 @@ $$\begin{aligned}
\end{aligned}$$
In general, it is more useful to consider
-the average kinetic energy per electron $E^{(0)} / N$,
-which we find to be as follows, using that $k_F^3 = 3 \pi^2 n$:
+the average kinetic energy per electron $$E^{(0)} / N$$,
+which we find to be as follows, using that $$k_F^3 = 3 \pi^2 n$$:
$$\begin{aligned}
\boxed{
@@ -175,9 +175,9 @@ $$\begin{aligned}
\:\sim\: n^{2/3}
\end{aligned}$$
-Traditionally, this is expressed using a dimensionless parameter $r_s$,
+Traditionally, this is expressed using a dimensionless parameter $$r_s$$,
defined as the radius of a sphere containing a single electron,
-measured in Bohr radii $a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$:
+measured in Bohr radii $$a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$$:
$$\begin{aligned}
\frac{4 \pi}{3} (a_0 r_s)^3
@@ -206,14 +206,14 @@ To include Coulomb interactions, let us try
Clearly, this will give better results when the interaction is relatively weak, if ever.
The Coulomb potential is proportional to the inverse distance,
-and the average electron spacing is roughly $n^{-1/3}$,
-so the interaction energy $E_\mathrm{int}$ should scale as $n^{1/3}$.
-We already know that the kinetic energy $E_\mathrm{kin} = E^{(0)}$ scales as $n^{2/3}$,
+and the average electron spacing is roughly $$n^{-1/3}$$,
+so the interaction energy $$E_\mathrm{int}$$ should scale as $$n^{1/3}$$.
+We already know that the kinetic energy $$E_\mathrm{kin} = E^{(0)}$$ scales as $$n^{2/3}$$,
meaning perturbation theory should be reasonable
-if $1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$,
-so in the limit of high density $n \to \infty$.
+if $$1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$$,
+so in the limit of high density $$n \to \infty$$.
-The two-body Coulomb interaction operator $\hat{W}$
+The two-body Coulomb interaction operator $$\hat{W}$$
is as follows in second-quantized form:
$$\begin{aligned}
@@ -222,7 +222,7 @@ $$\begin{aligned}
\hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
\end{aligned}$$
-The first-order correction $E^{(1)}$ to the ground state (i.e. Fermi sea) energy
+The first-order correction $$E^{(1)}$$ to the ground state (i.e. Fermi sea) energy
is then given by:
$$\begin{aligned}
@@ -235,14 +235,14 @@ $$\begin{aligned}
\end{aligned}$$
This inner product can only be nonzero
-if the two creation operators $\hat{c}^\dagger$
-are for the same orbitals as the two annihilation operators $\hat{c}$.
-Since $\vb{q} \neq 0$, this means that $s_1 = s_2$,
-and that momentum is conserved: $\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$.
-And of course both $\vb{k}_1$ and $\vb{k}_1 \!+\! \vb{q}$
+if the two creation operators $$\hat{c}^\dagger$$
+are for the same orbitals as the two annihilation operators $$\hat{c}$$.
+Since $$\vb{q} \neq 0$$, this means that $$s_1 = s_2$$,
+and that momentum is conserved: $$\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$$.
+And of course both $$\vb{k}_1$$ and $$\vb{k}_1 \!+\! \vb{q}$$
must be inside the Fermi sphere,
to avoid annihilating an empty orbital.
-Let $s = s_1$ and $\vb{k} = \vb{k}_1$:
+Let $$s = s_1$$ and $$\vb{k} = \vb{k}_1$$:
$$\begin{aligned}
E^{(1)}
@@ -260,11 +260,11 @@ $$\begin{aligned}
\Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|)
\end{aligned}$$
-Next, we convert the sum over $\vb{q}$ into an integral in spherical coordinates.
-Clearly, $\vb{q}$ is the "jump" made by an electron from one orbital to another,
+Next, we convert the sum over $$\vb{q}$$ into an integral in spherical coordinates.
+Clearly, $$\vb{q}$$ is the "jump" made by an electron from one orbital to another,
so the largest possible jump
goes from a point on the Fermi surface to the opposite point,
-and thus has length $2 k_F$.
+and thus has length $$2 k_F$$.
This yields the integration limit, and therefore leads to:
$$\begin{aligned}
@@ -277,11 +277,11 @@ $$\begin{aligned}
\int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|}
\end{aligned}$$
-Where we have used that the direction of $\vb{q}$,
-i.e. $(\theta_q,\varphi_q)$, is irrelevant,
-as long as we define $\theta_k$ as
-the angle between $\vb{q}$ and $\vb{k} \!+\! \vb{q}$
-when we go to spherical coordinates $(|\vb{k}|, \theta_k, \varphi_k)$ for $\vb{k}$:
+Where we have used that the direction of $$\vb{q}$$,
+i.e. $$(\theta_q,\varphi_q)$$, is irrelevant,
+as long as we define $$\theta_k$$ as
+the angle between $$\vb{q}$$ and $$\vb{k} \!+\! \vb{q}$$
+when we go to spherical coordinates $$(|\vb{k}|, \theta_k, \varphi_k)$$ for $$\vb{k}$$:
$$\begin{aligned}
E^{(1)}
@@ -296,10 +296,10 @@ $$\begin{aligned}
Unfortunately, this last step function is less easy to translate into integration limits.
In effect, we are trying to calculate the intersection volume of two spheres,
-both with radius $k_F$, one centered on the origin (for $\vb{k}$),
-and the other centered on $\vb{q}$ (for $\vb{k} \!+\! \vb{q}$).
-Imagine a triangle with side lengths $|\vb{k}|$, $|\vb{q}|$ and $|\vb{k} \!+\! \vb{q}|^2$,
-where $\theta_k$ is the angle between $|\vb{k}|$ and $|\vb{k} \!+\! \vb{q}|$.
+both with radius $$k_F$$, one centered on the origin (for $$\vb{k}$$),
+and the other centered on $$\vb{q}$$ (for $$\vb{k} \!+\! \vb{q}$$).
+Imagine a triangle with side lengths $$|\vb{k}|$$, $$|\vb{q}|$$ and $$|\vb{k} \!+\! \vb{q}|^2$$,
+where $$\theta_k$$ is the angle between $$|\vb{k}|$$ and $$|\vb{k} \!+\! \vb{q}|$$.
The *law of cosines* then gives the following relation:
$$\begin{aligned}
@@ -307,10 +307,10 @@ $$\begin{aligned}
= |\vb{q}|^2 + |\vb{k} \!+\! \vb{q}|^2 - 2 |\vb{q}| |\vb{k} \!+\! \vb{q}| \cos(\theta_k)
\end{aligned}$$
-We already know that $|\vb{k}| < k_F$ and $0 < |\vb{q}| < 2 k_F$,
-so by isolating for $\cos(\theta_k)$,
-we can obtain bounds on $\theta_k$ and $|\vb{k}|$.
-Let $|\vb{k}| \to k_F$ in both cases, then:
+We already know that $$|\vb{k}| < k_F$$ and $$0 < |\vb{q}| < 2 k_F$$,
+so by isolating for $$\cos(\theta_k)$$,
+we can obtain bounds on $$\theta_k$$ and $$|\vb{k}|$$.
+Let $$|\vb{k}| \to k_F$$ in both cases, then:
$$\begin{aligned}
\cos(\theta_k)
@@ -322,16 +322,16 @@ $$\begin{aligned}
= 1
\end{aligned}$$
-Meaning that $0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$.
-To get a lower limit for $|\vb{k}|$, we "cheat" by artificially demanding
-that $\vb{k}$ does not cross the halfway point between the spheres,
-with the result that $|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$.
+Meaning that $$0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$$.
+To get a lower limit for $$|\vb{k}|$$, we "cheat" by artificially demanding
+that $$\vb{k}$$ does not cross the halfway point between the spheres,
+with the result that $$|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$$.
Then, thanks to symmetry (both spheres have the same radius),
-we just multiply the integral by $2$,
-for $\vb{k}$ on the other side of the halfway point.
+we just multiply the integral by $$2$$,
+for $$\vb{k}$$ on the other side of the halfway point.
-Armed with these integration limits, we return to calculating $E^{(1)}$,
-substituting $\xi \equiv \cos(\theta_k)$:
+Armed with these integration limits, we return to calculating $$E^{(1)}$$,
+substituting $$\xi \equiv \cos(\theta_k)$$:
$$\begin{aligned}
E^{(1)}
@@ -345,7 +345,7 @@ $$\begin{aligned}
\!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
\end{aligned}$$
-Where we have used that $\varphi_k$ does not appear in the integrand.
+Where we have used that $$\varphi_k$$ does not appear in the integrand.
Evaluating these integrals:
$$\begin{aligned}
@@ -369,7 +369,7 @@ $$\begin{aligned}
= -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F
\end{aligned}$$
-Per particle, the first-order energy correction $E^{(1)}$
+Per particle, the first-order energy correction $$E^{(1)}$$
is therefore found to be as follows:
$$\begin{aligned}
@@ -379,7 +379,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-This can also be written using the parameter $r_s$ introduced above, leading to:
+This can also be written using the parameter $$r_s$$ introduced above, leading to:
$$\begin{aligned}
\frac{E^{(1)}}{N}
@@ -387,8 +387,8 @@ $$\begin{aligned}
= -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
\end{aligned}$$
-Consequently, for sufficiently high densities $n$,
-the total energy $E$ per particle is given by:
+Consequently, for sufficiently high densities $$n$$,
+the total energy $$E$$ per particle is given by:
$$\begin{aligned}
\boxed{
@@ -398,7 +398,7 @@ $$\begin{aligned}
\end{aligned}$$
Unfortunately, this is as far as we can go.
-In theory, the second-order energy correction $E^{(2)}$ is as shown below,
+In theory, the second-order energy correction $$E^{(2)}$$ is as shown below,
but it turns out that it (and all higher orders) diverge:
$$\begin{aligned}
diff --git a/source/know/concept/kolmogorov-equations/index.md b/source/know/concept/kolmogorov-equations/index.md
index 47820ee..1ca2df6 100644
--- a/source/know/concept/kolmogorov-equations/index.md
+++ b/source/know/concept/kolmogorov-equations/index.md
@@ -10,7 +10,7 @@ layout: "concept"
---
Consider the following general [Itō diffusion](/know/concept/ito-calculus/)
-$X_t \in \mathbb{R}$, which is assumed to satisfy
+$$X_t \in \mathbb{R}$$, which is assumed to satisfy
the conditions for unique existence on the entire time axis:
$$\begin{aligned}
@@ -18,14 +18,14 @@ $$\begin{aligned}
= f(X_t, t) \dd{t} + g(X_t, t) \dd{B_t}
\end{aligned}$$
-Let $\mathcal{F}_t$ be the filtration to which $X_t$ is adapted,
-then we define $Y_s$ as shown below,
+Let $$\mathcal{F}_t$$ be the filtration to which $$X_t$$ is adapted,
+then we define $$Y_s$$ as shown below,
namely as the [conditional expectation](/know/concept/conditional-expectation/)
-of $h(X_t)$, for an arbitrary bounded function $h(x)$,
-given the information $\mathcal{F}_s$ available at time $s \le t$.
-Because $X_t$ is a [Markov process](/know/concept/markov-process/),
-$Y_s$ must be $X_s$-measurable,
-so it is a function $k$ of $X_s$ and $s$:
+of $$h(X_t)$$, for an arbitrary bounded function $$h(x)$$,
+given the information $$\mathcal{F}_s$$ available at time $$s \le t$$.
+Because $$X_t$$ is a [Markov process](/know/concept/markov-process/),
+$$Y_s$$ must be $$X_s$$-measurable,
+so it is a function $$k$$ of $$X_s$$ and $$s$$:
$$\begin{aligned}
Y_s
@@ -34,8 +34,8 @@ $$\begin{aligned}
= k(X_s, s)
\end{aligned}$$
-Consequently, we can apply Itō's lemma to find $\dd{Y_s}$
-in terms of $k$, $f$ and $g$:
+Consequently, we can apply Itō's lemma to find $$\dd{Y_s}$$
+in terms of $$k$$, $$f$$ and $$g$$:
$$\begin{aligned}
\dd{Y_s}
@@ -44,19 +44,19 @@ $$\begin{aligned}
&= \bigg( \pdv{k}{s} + \hat{L} k \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s}
\end{aligned}$$
-Where we have defined the linear operator $\hat{L}$
-to have the following action on $k$:
+Where we have defined the linear operator $$\hat{L}$$
+to have the following action on $$k$$:
$$\begin{aligned}
\hat{L} k
\equiv \pdv{k}{x} f + \frac{1}{2} \pdvn{2}{k}{x} g^2
\end{aligned}$$
-At this point, we need to realize that $Y_s$ is
-a [martingale](/know/concept/martingale/) with respect to $\mathcal{F}_s$,
-since $Y_s$ is $\mathcal{F}_s$-adapted and finite,
+At this point, we need to realize that $$Y_s$$ is
+a [martingale](/know/concept/martingale/) with respect to $$\mathcal{F}_s$$,
+since $$Y_s$$ is $$\mathcal{F}_s$$-adapted and finite,
and it satisfies the martingale property,
-for $r \le s \le t$:
+for $$r \le s \le t$$:
$$\begin{aligned}
\mathbf{E}[Y_s | \mathcal{F}_r]
@@ -66,20 +66,20 @@ $$\begin{aligned}
\end{aligned}$$
Where we used the tower property of conditional expectations,
-because $\mathcal{F}_r \subset \mathcal{F}_s$.
+because $$\mathcal{F}_r \subset \mathcal{F}_s$$.
However, an Itō diffusion can only be a martingale
-if its drift term (the one containing $\dd{s}$) vanishes,
-so, looking at $\dd{Y_s}$, we must demand that:
+if its drift term (the one containing $$\dd{s}$$) vanishes,
+so, looking at $$\dd{Y_s}$$, we must demand that:
$$\begin{aligned}
\pdv{k}{s} + \hat{L} k
= 0
\end{aligned}$$
-Because $k(X_s, s)$ is a Markov process,
-we can write it with a transition density $p(s, X_s; t, X_t)$,
-where in this case $s$ and $X_s$ are given initial conditions,
-$t$ is a parameter, and the terminal state $X_t$ is a random variable.
+Because $$k(X_s, s)$$ is a Markov process,
+we can write it with a transition density $$p(s, X_s; t, X_t)$$,
+where in this case $$s$$ and $$X_s$$ are given initial conditions,
+$$t$$ is a parameter, and the terminal state $$X_t$$ is a random variable.
We thus have:
$$\begin{aligned}
@@ -87,26 +87,26 @@ $$\begin{aligned}
= \int_{-\infty}^\infty p(s, x; t, y) \: h(y) \dd{y}
\end{aligned}$$
-We insert this into the equation that we just derived for $k$, yielding:
+We insert this into the equation that we just derived for $$k$$, yielding:
$$\begin{aligned}
0
= \int_{-\infty}^\infty \!\! \Big( \pdv{}{s}p(s, x; t, y) + \hat{L} p(s, x; t, y) \Big) h(y) \dd{y}
\end{aligned}$$
-Because $h$ is arbitrary, and this must be satisfied for all $h$,
-the transition density $p$ fulfills:
+Because $$h$$ is arbitrary, and this must be satisfied for all $$h$$,
+the transition density $$p$$ fulfills:
$$\begin{aligned}
0
= \pdv{}{s}p(s, x; t, y) + \hat{L} p(s, x; t, y)
\end{aligned}$$
-Here, $t$ is a known parameter and $y$ is a "known" integration variable,
-leaving only $s$ and $x$ as free variables for us to choose.
-We therefore define the **likelihood function** $\psi(s, x)$,
-which gives the likelihood of an initial condition $(s, x)$
-given that the terminal condition is $(t, y)$:
+Here, $$t$$ is a known parameter and $$y$$ is a "known" integration variable,
+leaving only $$s$$ and $$x$$ as free variables for us to choose.
+We therefore define the **likelihood function** $$\psi(s, x)$$,
+which gives the likelihood of an initial condition $$(s, x)$$
+given that the terminal condition is $$(t, y)$$:
$$\begin{aligned}
\boxed{
@@ -116,7 +116,7 @@ $$\begin{aligned}
\end{aligned}$$
And from the above derivation,
-we conclude that $\psi$ satisfies the following PDE,
+we conclude that $$\psi$$ satisfies the following PDE,
known as the **backward Kolmogorov equation**:
$$\begin{aligned}
@@ -128,9 +128,9 @@ $$\begin{aligned}
\end{aligned}$$
Moving on, we can define the traditional
-**probability density function** $\phi(t, y)$ from the transition density $p$,
-by fixing the initial $(s, x)$
-and leaving the terminal $(t, y)$ free:
+**probability density function** $$\phi(t, y)$$ from the transition density $$p$$,
+by fixing the initial $$(s, x)$$
+and leaving the terminal $$(t, y)$$ free:
$$\begin{aligned}
\boxed{
@@ -139,10 +139,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-With this in mind, for $(s, x) = (0, X_0)$,
-the unconditional expectation $\mathbf{E}[Y_t]$
+With this in mind, for $$(s, x) = (0, X_0)$$,
+the unconditional expectation $$\mathbf{E}[Y_t]$$
(i.e. the conditional expectation without information)
-will be constant in time, because $Y_t$ is a martingale:
+will be constant in time, because $$Y_t$$ is a martingale:
$$\begin{aligned}
\mathbf{E}[Y_t]
@@ -154,8 +154,8 @@ $$\begin{aligned}
This integral has the form of an inner product,
so we switch to [Dirac notation](/know/concept/dirac-notation/).
-We differentiate with respect to $t$,
-and use the backward equation $\ipdv{k}{t} + \hat{L} k = 0$:
+We differentiate with respect to $$t$$,
+and use the backward equation $$\ipdv{k}{t} + \hat{L} k = 0$$:
$$\begin{aligned}
0
@@ -165,11 +165,11 @@ $$\begin{aligned}
= \Inprod{k}{\pdv{\phi}{t} - \hat{L}{}^\dagger \phi}
\end{aligned}$$
-Where $\hat{L}{}^\dagger$ is by definition the adjoint operator of $\hat{L}$,
+Where $$\hat{L}{}^\dagger$$ is by definition the adjoint operator of $$\hat{L}$$,
which we calculate using partial integration,
-where all boundary terms vanish thanks to the *existence* of $X_t$;
-in other words, $X_t$ cannot reach infinity at any finite $t$,
-so the integrand must decay to zero for $|y| \to \infty$:
+where all boundary terms vanish thanks to the *existence* of $$X_t$$;
+in other words, $$X_t$$ cannot reach infinity at any finite $$t$$,
+so the integrand must decay to zero for $$|y| \to \infty$$:
$$\begin{aligned}
\Inprod{\hat{L} k}{\phi}
@@ -185,9 +185,9 @@ $$\begin{aligned}
= \Inprod{k}{\hat{L}{}^\dagger \phi}
\end{aligned}$$
-Since $k$ is arbitrary, and $\ipdv{\Inprod{k}{\phi}}{t} = 0$ for all $k$,
+Since $$k$$ is arbitrary, and $$\ipdv{\Inprod{k}{\phi}}{t} = 0$$ for all $$k$$,
we thus arrive at the **forward Kolmogorov equation**,
-describing the evolution of the probability density $\phi(t, y)$:
+describing the evolution of the probability density $$\phi(t, y)$$:
$$\begin{aligned}
\boxed{
@@ -199,7 +199,7 @@ $$\begin{aligned}
This can be rewritten in a way
that highlights the connection between Itō diffusions and physical diffusion,
-if we define the **diffusivity** $D$, **advection** $u$, and **probability flux** $J$:
+if we define the **diffusivity** $$D$$, **advection** $$u$$, and **probability flux** $$J$$:
$$\begin{aligned}
D
@@ -223,7 +223,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that if $u = 0$, then this reduces to
+Note that if $$u = 0$$, then this reduces to
[Fick's second law](/know/concept/ficks-laws/).
The backward Kolmogorov equation can also be rewritten analogously,
although it is less noteworthy:
diff --git a/source/know/concept/kramers-kronig-relations/index.md b/source/know/concept/kramers-kronig-relations/index.md
index bb9b08c..3880113 100644
--- a/source/know/concept/kramers-kronig-relations/index.md
+++ b/source/know/concept/kramers-kronig-relations/index.md
@@ -10,22 +10,22 @@ categories:
layout: "concept"
---
-Let $\chi(t)$ be a complex function describing
-the response of a system to an impulse $f(t)$ starting at $t = 0$.
-The **Kramers-Kronig relations** connect the real and imaginary parts of $\chi(t)$,
+Let $$\chi(t)$$ be a complex function describing
+the response of a system to an impulse $$f(t)$$ starting at $$t = 0$$.
+The **Kramers-Kronig relations** connect the real and imaginary parts of $$\chi(t)$$,
such that one can be reconstructed from the other.
-Suppose we can only measure $\chi_r(t)$ or $\chi_i(t)$:
+Suppose we can only measure $$\chi_r(t)$$ or $$\chi_i(t)$$:
$$\begin{aligned}
\chi(t) = \chi_r(t) + i \chi_i(t)
\end{aligned}$$
-Assuming that the system was at rest until $t = 0$,
-the response $\chi(t)$ cannot depend on anything from $t < 0$,
-since the known impulse $f(t)$ had not started yet,
+Assuming that the system was at rest until $$t = 0$$,
+the response $$\chi(t)$$ cannot depend on anything from $$t < 0$$,
+since the known impulse $$f(t)$$ had not started yet,
This principle is called **causality**, and to enforce it,
we use the [Heaviside step function](/know/concept/heaviside-step-function/)
-$\Theta(t)$ to create a **causality test** for $\chi(t)$:
+$$\Theta(t)$$ to create a **causality test** for $$\chi(t)$$:
$$\begin{aligned}
\chi(t) = \chi(t) \: \Theta(t)
@@ -34,7 +34,7 @@ $$\begin{aligned}
If we [Fourier transform](/know/concept/fourier-transform/) this equation,
then it will become a convolution in the frequency domain
thanks to the [convolution theorem](/know/concept/convolution-theorem/),
-where $A$, $B$ and $s$ are constants from the FT definition:
+where $$A$$, $$B$$ and $$s$$ are constants from the FT definition:
$$\begin{aligned}
\tilde{\chi}(\omega)
@@ -42,10 +42,10 @@ $$\begin{aligned}
= B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'}
\end{aligned}$$
-We look up the FT of the step function $\tilde{\Theta}(\omega)$,
-which involves the signum function $\mathrm{sgn}(t)$,
-the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$,
-and the Cauchy principal value $\pv{}$.
+We look up the FT of the step function $$\tilde{\Theta}(\omega)$$,
+which involves the signum function $$\mathrm{sgn}(t)$$,
+the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$,
+and the Cauchy principal value $$\pv{}$$.
We arrive at:
$$\begin{aligned}
@@ -58,7 +58,7 @@ $$\begin{aligned}
\pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}}
\end{aligned}$$
-From the definition of the Fourier transform we know that $2 \pi A B / |s| = 1$:
+From the definition of the Fourier transform we know that $$2 \pi A B / |s| = 1$$:
$$\begin{aligned}
\tilde{\chi}(\omega)
@@ -66,7 +66,7 @@ $$\begin{aligned}
+ \mathrm{sgn}(s) \frac{i}{2 \pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}}
\end{aligned}$$
-We isolate this equation for $\tilde{\chi}(\omega)$
+We isolate this equation for $$\tilde{\chi}(\omega)$$
to get the final version of the causality test:
$$\begin{aligned}
@@ -76,7 +76,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-By inserting $\tilde{\chi}(\omega) = \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega)$
+By inserting $$\tilde{\chi}(\omega) = \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega)$$
and splitting the equation into real and imaginary parts,
we get the Kramers-Kronig relations:
@@ -92,13 +92,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-If the time-domain response function $\chi(t)$ is real
+If the time-domain response function $$\chi(t)$$ is real
(so far we have assumed it to be complex),
then we can take advantage of the fact that
the FT of a real function satisfies
-$\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega)$, i.e. $\tilde{\chi}_r(\omega)$
-is even and $\tilde{\chi}_i(\omega)$ is odd. We multiply the fractions by
-$(\omega' + \omega)$ above and below:
+$$\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega)$$, i.e. $$\tilde{\chi}_r(\omega)$$
+is even and $$\tilde{\chi}_i(\omega)$$ is odd. We multiply the fractions by
+$$(\omega' + \omega)$$ above and below:
$$\begin{aligned}
\tilde{\chi}_r(\omega)
@@ -110,8 +110,8 @@ $$\begin{aligned}
+ \frac{\omega}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} \bigg)
\end{aligned}$$
-For $\tilde{\chi}_r(\omega)$, the second integrand is odd, so we can drop it.
-Similarly, for $\tilde{\chi}_i(\omega)$, the first integrand is odd.
+For $$\tilde{\chi}_r(\omega)$$, the second integrand is odd, so we can drop it.
+Similarly, for $$\tilde{\chi}_i(\omega)$$, the first integrand is odd.
We therefore find the following variant of the Kramers-Kronig relations:
$$\begin{aligned}
@@ -126,7 +126,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-To reiterate: this version is only valid if $\chi(t)$ is real in the time domain.
+To reiterate: this version is only valid if $$\chi(t)$$ is real in the time domain.
diff --git a/source/know/concept/kubo-formula/index.md b/source/know/concept/kubo-formula/index.md
index 80309c5..4cb39ac 100644
--- a/source/know/concept/kubo-formula/index.md
+++ b/source/know/concept/kubo-formula/index.md
@@ -10,19 +10,19 @@ layout: "concept"
---
Consider the following quantum Hamiltonian,
-split into a main time-independent term $\hat{H}_{0,S}$
-and a small time-dependent perturbation $\hat{H}_{1,S}$,
-which is turned on at $t = t_0$:
+split into a main time-independent term $$\hat{H}_{0,S}$$
+and a small time-dependent perturbation $$\hat{H}_{1,S}$$,
+which is turned on at $$t = t_0$$:
$$\begin{aligned}
\hat{H}_S(t)
= \hat{H}_{0,S} + \hat{H}_{1,S}(t)
\end{aligned}$$
-And let $\Ket{\psi_S(t)}$ be the corresponding solutions to the Schrödinger equation.
-Then, given a time-independent observable $\hat{A}$,
-its expectation value $\expval{\hat{A}}$ evolves like so,
-where the subscripts $S$ and $I$
+And let $$\Ket{\psi_S(t)}$$ be the corresponding solutions to the Schrödinger equation.
+Then, given a time-independent observable $$\hat{A}$$,
+its expectation value $$\expval{\hat{A}}$$ evolves like so,
+where the subscripts $$S$$ and $$I$$
respectively refer to the Schrödinger
and [interaction pictures](/know/concept/interaction-picture/):
@@ -34,7 +34,7 @@ $$\begin{aligned}
&= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)}
\end{aligned}$$
-Where the time evolution operator $\hat{K}_I(t, t_0)$ is as follows,
+Where the time evolution operator $$\hat{K}_I(t, t_0)$$ is as follows,
which we Taylor-expand:
$$\begin{aligned}
@@ -56,7 +56,7 @@ $$\begin{aligned}
\end{aligned}$$
Where we have dropped the last term,
-because $\hat{H}_{1}$ is assumed to be so small
+because $$\hat{H}_{1}$$ is assumed to be so small
that it only matters to first order.
Here, we notice a commutator, so we can rewrite:
@@ -65,10 +65,10 @@ $$\begin{aligned}
&= \hat{A}_I(t) - \frac{i}{\hbar} \int_{t_0}^t \Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')} \dd{t'}
\end{aligned}$$
-Returning to $\expval{\hat{A}}$,
+Returning to $$\expval{\hat{A}}$$,
we have the following formula,
-where $\Expval{}$ is the expectation value for $\Ket{\psi(t)}$,
-and $\Expval{}_0$ is the expectation value for $\Ket{\psi_I(t_0)}$:
+where $$\Expval{}$$ is the expectation value for $$\Ket{\psi(t)}$$,
+and $$\Expval{}_0$$ is the expectation value for $$\Ket{\psi_I(t_0)}$$:
$$\begin{aligned}
\expval{\hat{A}}(t)
@@ -76,9 +76,9 @@ $$\begin{aligned}
= \expval{\hat{A}_I(t)}_0 - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
\end{aligned}$$
-Now we define $\delta\!\expval{\hat{A}}\!(t)$
-as the change of $\expval{\hat{A}}$ due to the perturbation $\hat{H}_1$,
-and insert $\expval{\hat{A}}(t)$:
+Now we define $$\delta\!\expval{\hat{A}}\!(t)$$
+as the change of $$\expval{\hat{A}}$$ due to the perturbation $$\hat{H}_1$$,
+and insert $$\expval{\hat{A}}(t)$$:
$$\begin{aligned}
\delta\!\expval{\hat{A}}\!(t)
@@ -87,10 +87,10 @@ $$\begin{aligned}
\end{aligned}$$
Finally, we introduce
-a [Heaviside step function](/know/concept/heaviside-step-function) $\Theta$
+a [Heaviside step function](/know/concept/heaviside-step-function) $$\Theta$$
and change the integration limit accordingly,
leading to the **Kubo formula**
-describing the response of $\expval{\hat{A}}$ to first order in $\hat{H}_1$:
+describing the response of $$\expval{\hat{A}}$$ to first order in $$\hat{H}_1$$:
$$\begin{aligned}
\boxed{
@@ -99,7 +99,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined the **retarded correlation function** $C^R_{A H_1}(t, t')$ as follows:
+Where we have defined the **retarded correlation function** $$C^R_{A H_1}(t, t')$$ as follows:
$$\begin{aligned}
\boxed{
@@ -115,10 +115,10 @@ of particle creation/annihiliation operators.
Therefore, this correlation function
is a two-particle [Green's function](/know/concept/greens-functions/).
-A common situation is that $\hat{H}_1$ consists of
-a time-independent operator $\hat{B}$
-and a time-dependent function $f(t)$,
-allowing us to split $C^R_{A H_1}$ as follows:
+A common situation is that $$\hat{H}_1$$ consists of
+a time-independent operator $$\hat{B}$$
+and a time-dependent function $$f(t)$$,
+allowing us to split $$C^R_{A H_1}$$ as follows:
$$\begin{aligned}
\hat{H}_{1,S}(t)
@@ -128,10 +128,10 @@ $$\begin{aligned}
= C^R_{A B}(t, t') f(t')
\end{aligned}$$
-Since $C_{AB}^R$ is a Green's function,
-we know that it only depends on the difference $t - t'$,
+Since $$C_{AB}^R$$ is a Green's function,
+we know that it only depends on the difference $$t - t'$$,
as long as the system was initially in thermodynamic equilibrium,
-and $\hat{H}_{0,S}$ is time-independent:
+and $$\hat{H}_{0,S}$$ is time-independent:
$$\begin{aligned}
C^R_{A B}(t, t')
@@ -139,7 +139,7 @@ $$\begin{aligned}
\end{aligned}$$
With this, the Kubo formula can be written as follows,
-where we have set $t_0 = - \infty$:
+where we have set $$t_0 = - \infty$$:
$$\begin{aligned}
\delta\!\expval{A}\!(t)
@@ -150,8 +150,8 @@ $$\begin{aligned}
This is a convolution,
so the [convolution theorem](/know/concept/convolution-theorem/)
states that the [Fourier transform](/know/concept/fourier-transform/)
-of $\delta\!\expval{\hat{A}}\!(t)$ is simply the product
-of the transforms of $C^R_{AB}$ and $f$:
+of $$\delta\!\expval{\hat{A}}\!(t)$$ is simply the product
+of the transforms of $$C^R_{AB}$$ and $$f$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/lagrange-multiplier/index.md b/source/know/concept/lagrange-multiplier/index.md
index 9761e75..8ee1054 100644
--- a/source/know/concept/lagrange-multiplier/index.md
+++ b/source/know/concept/lagrange-multiplier/index.md
@@ -10,27 +10,27 @@ layout: "concept"
The method of **Lagrange multipliers** or **undetermined multipliers**
is a technique for optimizing (i.e. finding the extrema of)
-a function $f(x, y, z)$,
-subject to a given constraint $\phi(x, y, z) = C$,
-where $C$ is a constant.
+a function $$f(x, y, z)$$,
+subject to a given constraint $$\phi(x, y, z) = C$$,
+where $$C$$ is a constant.
-If we ignore the constraint $\phi$,
-optimizing $f$ simply comes down to finding stationary points:
+If we ignore the constraint $$\phi$$,
+optimizing $$f$$ simply comes down to finding stationary points:
$$\begin{aligned}
0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z}
\end{aligned}$$
This problem is easy:
-$\dd{x}$, $\dd{y}$, and $\dd{z}$ are independent and arbitrary,
+$$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$ are independent and arbitrary,
so all we need to do is find the roots of
-the partial derivatives $f_x$, $f_y$ and $f_z$,
-which we respectively call $x_0$, $y_0$ and $z_0$,
-and then the extremum is simply $(x_0, y_0, z_0)$.
+the partial derivatives $$f_x$$, $$f_y$$ and $$f_z$$,
+which we respectively call $$x_0$$, $$y_0$$ and $$z_0$$,
+and then the extremum is simply $$(x_0, y_0, z_0)$$.
-But the constraint $\phi$, over which we have no control,
-adds a relation between $\dd{x}$, $\dd{y}$, and $\dd{z}$,
-so if two are known, the third is given by $\phi = C$.
+But the constraint $$\phi$$, over which we have no control,
+adds a relation between $$\dd{x}$$, $$\dd{y}$$, and $$\dd{z}$$,
+so if two are known, the third is given by $$\phi = C$$.
The problem is then a system of equations:
$$\begin{aligned}
@@ -42,15 +42,15 @@ $$\begin{aligned}
Solving this directly would be a delicate balancing act
of all the partial derivatives.
-To help us solve this, we introduce a "dummy" parameter $\lambda$,
+To help us solve this, we introduce a "dummy" parameter $$\lambda$$,
the so-called **Lagrange multiplier**,
-and contruct a new function $L$ given by:
+and contruct a new function $$L$$ given by:
$$\begin{aligned}
L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z)
\end{aligned}$$
-At the extremum, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$,
+At the extremum, $$\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$$,
so now the problem is a "single" equation again:
$$\begin{aligned}
@@ -58,13 +58,13 @@ $$\begin{aligned}
= (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z}
\end{aligned}$$
-Assuming $\phi_z \neq 0$, we now choose $\lambda$ such that $f_z + \lambda \phi_z = 0$.
+Assuming $$\phi_z \neq 0$$, we now choose $$\lambda$$ such that $$f_z + \lambda \phi_z = 0$$.
This choice represents satisfying the constraint,
-so now the remaining $\dd{x}$ and $\dd{y}$ are independent again,
-and we simply have to find the roots of $f_x + \lambda \phi_x$ and $f_y + \lambda \phi_y$.
+so now the remaining $$\dd{x}$$ and $$\dd{y}$$ are independent again,
+and we simply have to find the roots of $$f_x + \lambda \phi_x$$ and $$f_y + \lambda \phi_y$$.
-In effect, after introducing $\lambda$,
-we have four unknowns $(x, y, z, \lambda)$,
+In effect, after introducing $$\lambda$$,
+we have four unknowns $$(x, y, z, \lambda)$$,
but also four equations:
$$\begin{aligned}
@@ -73,19 +73,19 @@ $$\begin{aligned}
\phi = C
\end{aligned}$$
-We are only really interested in the first three unknowns $(x, y, z)$,
-so $\lambda$ is sometimes called the **undetermined multiplier**,
+We are only really interested in the first three unknowns $$(x, y, z)$$,
+so $$\lambda$$ is sometimes called the **undetermined multiplier**,
since it is just an algebraic helper whose value is irrelevant.
This method generalizes nicely to multiple constraints or more variables:
-suppose that we want to find the extrema of $f(x_1, ..., x_N)$
-subject to $M < N$ conditions:
+suppose that we want to find the extrema of $$f(x_1, ..., x_N)$$
+subject to $$M < N$$ conditions:
$$\begin{aligned}
\phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M
\end{aligned}$$
-This once again turns into a delicate system of $M+1$ equations to solve:
+This once again turns into a delicate system of $$M+1$$ equations to solve:
$$\begin{aligned}
0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N}
@@ -97,15 +97,15 @@ $$\begin{aligned}
0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N}
\end{aligned}$$
-Then we introduce $M$ Lagrange multipliers $\lambda_1, ..., \lambda_M$
-and define $L(x_1, ..., x_N)$:
+Then we introduce $$M$$ Lagrange multipliers $$\lambda_1, ..., \lambda_M$$
+and define $$L(x_1, ..., x_N)$$:
$$\begin{aligned}
L = f + \sum_{m = 1}^M \lambda_m \phi_m
\end{aligned}$$
-As before, we set $\dd{L} = 0$ and choose the multipliers $\lambda_1, ..., \lambda_M$
-to eliminate $M$ of its $N$ terms:
+As before, we set $$\dd{L} = 0$$ and choose the multipliers $$\lambda_1, ..., \lambda_M$$
+to eliminate $$M$$ of its $$N$$ terms:
$$\begin{aligned}
0 = \dd{L}
diff --git a/source/know/concept/lagrangian-mechanics/index.md b/source/know/concept/lagrangian-mechanics/index.md
index 7b520a2..0a7066a 100644
--- a/source/know/concept/lagrangian-mechanics/index.md
+++ b/source/know/concept/lagrangian-mechanics/index.md
@@ -17,8 +17,8 @@ and hence it is built on the **principle of least action**,
which states that the path taken by a system
will be a minimum of the **action** (i.e. energy cost) of that path.
-For a moving object with position $x(t)$ and velocity $\dot{x}(t)$,
-we define the Lagrangian $L$ as the difference
+For a moving object with position $$x(t)$$ and velocity $$\dot{x}(t)$$,
+we define the Lagrangian $$L$$ as the difference
between its kinetic and potential energies:
$$\begin{aligned}
@@ -38,21 +38,21 @@ $$\begin{aligned}
But compared to Newtonian mechanics,
Lagrangian mechanics scales better for large systems.
-For example, to describe the dynamics of $N$ objects $x_1(t), ..., x_N(t)$,
-we only need a single $L$
+For example, to describe the dynamics of $$N$$ objects $$x_1(t), ..., x_N(t)$$,
+we only need a single $$L$$
from which the equations of motion can easily be derived.
Getting these equations directly from Newton's laws could get messy.
At no point have we assumed Cartesian coordinates:
the Euler-Lagrange equations keep their form
-for any independent coordinates $q_1(t), ..., q_N(t)$:
+for any independent coordinates $$q_1(t), ..., q_N(t)$$:
$$\begin{aligned}
\dv{}{t}\Big( \pdv{L}{\dot{q}_n} \Big) = \pdv{L}{q_n}
\end{aligned}$$
-We define the **canonical momentum conjugate** $p_n(t)$
-and the **generalized force conjugate** $F_n(t)$ as follows,
+We define the **canonical momentum conjugate** $$p_n(t)$$
+and the **generalized force conjugate** $$F_n(t)$$ as follows,
such that we can always get Newton's second law:
$$\begin{aligned}
@@ -64,15 +64,15 @@ $$\begin{aligned}
\end{aligned}$$
But this is actually a bit misleading,
-since $p_n$ need not be a momentum, nor $F_n$ a force,
+since $$p_n$$ need not be a momentum, nor $$F_n$$ a force,
although often they are.
-For example, $p_n$ could be angular momentum, and $F_n$ torque.
+For example, $$p_n$$ could be angular momentum, and $$F_n$$ torque.
Another advantage of Lagrangian mechanics is that
-the conserved quantities can be extracted from $L$ using Noether's theorem.
-In the simplest case, if $L$ does not depend on $q_n$
+the conserved quantities can be extracted from $$L$$ using Noether's theorem.
+In the simplest case, if $$L$$ does not depend on $$q_n$$
(then known as a **cyclic coordinate**),
-then we know that the "momentum" $p_n$ is a conserved quantity:
+then we know that the "momentum" $$p_n$$ is a conserved quantity:
$$\begin{aligned}
F_n = \pdv{L}{q_n} = 0
@@ -80,23 +80,23 @@ $$\begin{aligned}
\dv{p_n}{t} = 0
\end{aligned}$$
-Now, as the number of particles $N$ increases to infinity,
+Now, as the number of particles $$N$$ increases to infinity,
variational calculus will give infinitely many coupled equations,
which is obviously impractical.
-Such a system can be regarded as continuous, so the $N$ functions $q_n$
-can be replaced by a single density function $u(x,t)$.
+Such a system can be regarded as continuous, so the $$N$$ functions $$q_n$$
+can be replaced by a single density function $$u(x,t)$$.
This approach can also be used for continuous fields,
-in which case the complex conjugate $u^*$ is often included.
-The Lagrangian $L$ then becomes:
+in which case the complex conjugate $$u^*$$ is often included.
+The Lagrangian $$L$$ then becomes:
$$\begin{aligned}
L(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t)
= \int_{-\infty}^\infty \mathcal{L}(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t) \dd{x}
\end{aligned}$$
-Where $\mathcal{L}$ is known as the **Lagrangian density**.
-By inserting this into the functional $J$
+Where $$\mathcal{L}$$ is known as the **Lagrangian density**.
+By inserting this into the functional $$J$$
used for the derivation of the Euler-Lagrange equations, we get:
$$\begin{aligned}
@@ -114,11 +114,11 @@ $$\begin{aligned}
0 &= \pdv{\mathcal{L}}{u^*} - \pdv{}{x}\Big( \pdv{\mathcal{L}}{u_x^*} \Big) - \pdv{}{t}\Big( \pdv{\mathcal{L}}{u_t^*} \Big)
\end{aligned}$$
-If $\mathcal{L}$ is real,
+If $$\mathcal{L}$$ is real,
then these two Euler-Lagrange equations will in fact be identical.
Finally, note that for abstract fields,
-the Lagrangian density $\mathcal{L}$ rarely has
+the Lagrangian density $$\mathcal{L}$$ rarely has
a physical interpretation, and is not unique.
Instead, it must be reverse-engineered from a relevant equation.
diff --git a/source/know/concept/laguerre-polynomials/index.md b/source/know/concept/laguerre-polynomials/index.md
index fd3deb6..ba68343 100644
--- a/source/know/concept/laguerre-polynomials/index.md
+++ b/source/know/concept/laguerre-polynomials/index.md
@@ -8,8 +8,8 @@ layout: "concept"
---
The **Laguerre polynomials** are a set of useful functions that arise in physics.
-They are the non-singular eigenfunctions $u(x)$ of **Laguerre's equation**,
-with the corresponding eigenvalues $n$ being non-negative integers:
+They are the non-singular eigenfunctions $$u(x)$$ of **Laguerre's equation**,
+with the corresponding eigenvalues $$n$$ being non-negative integers:
$$\begin{aligned}
\boxed{
@@ -17,7 +17,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The $n$th-order Laguerre polynomial $L_n(x)$
+The $$n$$th-order Laguerre polynomial $$L_n(x)$$
is given in the form of a *Rodrigues' formula* by:
$$\begin{aligned}
@@ -27,7 +27,7 @@ $$\begin{aligned}
&= \frac{1}{n!} \Big( \dv{}{x}- 1 \Big)^n x^n
\end{aligned}$$
-The first couple of Laguerre polynomials $L_n(x)$ are therefore as follows:
+The first couple of Laguerre polynomials $$L_n(x)$$ are therefore as follows:
$$\begin{gathered}
L_0(x) = 1
@@ -39,8 +39,8 @@ $$\begin{gathered}
Based on Laguerre's equation,
**Laguerre's generalized equation** is as follows,
-with an arbitrary real (but usually integer) parameter $\alpha$,
-and $n$ still a non-negative integer:
+with an arbitrary real (but usually integer) parameter $$\alpha$$,
+and $$n$$ still a non-negative integer:
$$\begin{aligned}
\boxed{
@@ -48,10 +48,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Its solutions, denoted by $L_n^\alpha(x)$,
+Its solutions, denoted by $$L_n^\alpha(x)$$,
are the **generalized** or **associated Laguerre polynomials**,
which also have a Rodrigues' formula.
-Note that if $\alpha = 0$ then $L_n^\alpha = L_n$:
+Note that if $$\alpha = 0$$ then $$L_n^\alpha = L_n$$:
$$\begin{aligned}
L_n^\alpha(x)
@@ -60,7 +60,7 @@ $$\begin{aligned}
&= \frac{x^{-\alpha}}{n!} \Big( \dv{}{x}- 1 \Big)^n x^{n + \alpha}
\end{aligned}$$
-The first couple of associated Laguerre polynomials $L_n^\alpha(x)$ are therefore as follows:
+The first couple of associated Laguerre polynomials $$L_n^\alpha(x)$$ are therefore as follows:
$$\begin{aligned}
L_0^\alpha(x) = 1
@@ -70,7 +70,7 @@ $$\begin{aligned}
L_2^\alpha(x) = \frac{1}{2} (x^2 - 2 \alpha x - 4 x + \alpha^2 + 3 \alpha + 2)
\end{aligned}$$
-And then more $L_n^\alpha$ can be computed quickly
+And then more $$L_n^\alpha$$ can be computed quickly
using the following recurrence relation:
$$\begin{aligned}
@@ -91,8 +91,8 @@ $$\begin{aligned}
\end{aligned}$$
Noteworthy is that these polynomials (both normal and associated)
-are all mutually orthogonal for $x \in [0, \infty[$,
-with respect to the weight function $w(x) \equiv x^\alpha \exp(-x)$:
+are all mutually orthogonal for $$x \in [0, \infty[$$,
+with respect to the weight function $$w(x) \equiv x^\alpha \exp(-x)$$:
$$\begin{aligned}
\boxed{
@@ -102,11 +102,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\delta_{nm}$ is the Kronecker delta.
+Where $$\delta_{nm}$$ is the Kronecker delta.
Moreover, they form a basis in
the [Hilbert space](/know/concept/hilbert-space/)
-of all functions $f(x)$ for which $\Inprod{f}{w f}$ is finite.
-Any such $f$ can thus be expanded as follows:
+of all functions $$f(x)$$ for which $$\Inprod{f}{w f}$$ is finite.
+Any such $$f$$ can thus be expanded as follows:
$$\begin{aligned}
\boxed{
@@ -116,8 +116,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, the $L_n^\alpha(x)$ are related to
-the [Hermite polynomials](/know/concept/hermite-polynomials/) $H_n(x)$ like so:
+Finally, the $$L_n^\alpha(x)$$ are related to
+the [Hermite polynomials](/know/concept/hermite-polynomials/) $$H_n(x)$$ like so:
$$\begin{aligned}
H_{2n(x)} &= (-1)^n 2^{2n} n! \: L_n^{-1/2}(x^2)
diff --git a/source/know/concept/landau-quantization/index.md b/source/know/concept/landau-quantization/index.md
index 82ea86e..592d266 100644
--- a/source/know/concept/landau-quantization/index.md
+++ b/source/know/concept/landau-quantization/index.md
@@ -8,23 +8,23 @@ categories:
layout: "concept"
---
-When a particle with charge $q$ is moving in a homogeneous
+When a particle with charge $$q$$ is moving in a homogeneous
[magnetic field](/know/concept/magnetic-field/),
quantum mechanics decrees that its allowed energies split
into degenerate discrete **Landau levels**,
a phenomenon known as **Landau quantization**.
-Starting from the Hamiltonian $\hat{H}$ for a particle with mass $m$
-in a vector potential $\vec{A}(\hat{Q})$:
+Starting from the Hamiltonian $$\hat{H}$$ for a particle with mass $$m$$
+in a vector potential $$\vec{A}(\hat{Q})$$:
$$\begin{aligned}
\hat{H}
&= \frac{1}{2 m} \big( \hat{p} - q \vec{A} \big)^2
\end{aligned}$$
-We choose $\vec{A} = (- \hat{y} B, 0, 0)$,
-yielding a magnetic field $\vec{B} = \nabla \times \vec{A}$
-pointing in the $z$-direction with strength $B$.
+We choose $$\vec{A} = (- \hat{y} B, 0, 0)$$,
+yielding a magnetic field $$\vec{B} = \nabla \times \vec{A}$$
+pointing in the $$z$$-direction with strength $$B$$.
The Hamiltonian becomes:
$$\begin{aligned}
@@ -32,18 +32,18 @@ $$\begin{aligned}
&= \frac{\big( \hat{p}_x - q B \hat{y} \big)^2}{2 m} + \frac{\hat{p}_y^2}{2 m} + \frac{\hat{p}_z^2}{2 m}
\end{aligned}$$
-The only position operator occurring in $\hat{H}$ is $\hat{y}$,
-so $[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$.
-Because $\hat{p}_z$ appears in an unmodified kinetic energy term,
-and the corresponding $\hat{z}$ does not occur at all,
-the particle has completely free motion in the $z$-direction.
-Likewise, because $\hat{x}$ does not occur in $\hat{H}$,
-we can replace $\hat{p}_x$ by its eigenvalue $\hbar k_x$,
-although the motion is not free, due to $q B \hat{y}$.
+The only position operator occurring in $$\hat{H}$$ is $$\hat{y}$$,
+so $$[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$$.
+Because $$\hat{p}_z$$ appears in an unmodified kinetic energy term,
+and the corresponding $$\hat{z}$$ does not occur at all,
+the particle has completely free motion in the $$z$$-direction.
+Likewise, because $$\hat{x}$$ does not occur in $$\hat{H}$$,
+we can replace $$\hat{p}_x$$ by its eigenvalue $$\hbar k_x$$,
+although the motion is not free, due to $$q B \hat{y}$$.
-Based on the absence of $\hat{x}$ and $\hat{z}$,
-we make the following ansatz for the wavefunction $\Psi$:
-a plane wave in the $x$ and $z$ directions, multiplied by an unknown $\phi(y)$:
+Based on the absence of $$\hat{x}$$ and $$\hat{z}$$,
+we make the following ansatz for the wavefunction $$\Psi$$:
+a plane wave in the $$x$$ and $$z$$ directions, multiplied by an unknown $$\phi(y)$$:
$$\begin{aligned}
\Psi(x, y, z)
@@ -51,15 +51,15 @@ $$\begin{aligned}
\end{aligned}$$
Inserting this into the time-independent Schrödinger equation gives,
-after dividing out the plane wave exponential $\exp(i k_x x + i k_z z)$:
+after dividing out the plane wave exponential $$\exp(i k_x x + i k_z z)$$:
$$\begin{aligned}
E \phi
&= \frac{1}{2 m} \Big( (\hbar k_x - q B y)^2 + \hat{p}_y^2 + \hbar^2 k_z^2 \Big) \phi
\end{aligned}$$
-By defining the cyclotron frequency $\omega_c \equiv q B / m$ and rearranging,
-we can turn this into a 1D quantum harmonic oscillator in $y$,
+By defining the cyclotron frequency $$\omega_c \equiv q B / m$$ and rearranging,
+we can turn this into a 1D quantum harmonic oscillator in $$y$$,
with a couple of extra terms:
$$\begin{aligned}
@@ -67,8 +67,8 @@ $$\begin{aligned}
&= \bigg( \frac{1}{2} m \omega_c^2 \Big( y - \frac{\hbar k_x}{m \omega_c} \Big)^2 + \frac{\hat{p}_y^2}{2 m} \bigg) \phi
\end{aligned}$$
-The potential minimum is shifted by $y_0 = \hbar k_x / (m \omega_c)$,
-and a plane wave in $z$ contributes to the energy $E$.
+The potential minimum is shifted by $$y_0 = \hbar k_x / (m \omega_c)$$,
+and a plane wave in $$z$$ contributes to the energy $$E$$.
In any case, the energy levels of this type of system are well-known:
$$\begin{aligned}
@@ -77,29 +77,29 @@ $$\begin{aligned}
}
\end{aligned}$$
-And $\Psi_n$ is then as follows,
-where $\phi$ is the known quantum harmonic oscillator solution:
+And $$\Psi_n$$ is then as follows,
+where $$\phi$$ is the known quantum harmonic oscillator solution:
$$\begin{aligned}
\Psi_n(x, y, z)
= \phi_n(y - y_0) \exp(i k_x x + i k_z z)
\end{aligned}$$
-Note that this wave function contains $k_x$ (also inside $y_0$),
-but $k_x$ is absent from the energy $E_n$.
+Note that this wave function contains $$k_x$$ (also inside $$y_0$$),
+but $$k_x$$ is absent from the energy $$E_n$$.
This implies degeneracy:
-assuming periodic boundary conditions $\Psi(x\!+\!L_x) = \Psi(x)$,
-then $k_x$ can take values of the form $2 \pi n / L_x$, for $n \in \mathbb{Z}$.
+assuming periodic boundary conditions $$\Psi(x\!+\!L_x) = \Psi(x)$$,
+then $$k_x$$ can take values of the form $$2 \pi n / L_x$$, for $$n \in \mathbb{Z}$$.
-However, $k_x$ also occurs in the definition of $y_0$, so the degeneracy
-is finite, since $y_0$ must still lie inside the system,
-or, more formally, $y_0 \in [0, L_y]$:
+However, $$k_x$$ also occurs in the definition of $$y_0$$, so the degeneracy
+is finite, since $$y_0$$ must still lie inside the system,
+or, more formally, $$y_0 \in [0, L_y]$$:
$$\begin{aligned}
0 \le y_0 = \frac{\hbar k_x}{m \omega_c} = \frac{\hbar 2 \pi n}{q B L_x} \le L_y
\end{aligned}$$
-Isolating this for $n$, we find the following upper bound of the degeneracy:
+Isolating this for $$n$$, we find the following upper bound of the degeneracy:
$$\begin{aligned}
\boxed{
@@ -108,12 +108,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $A \equiv L_x L_y$ is the area of the confinement in the $(x,y)$-plane.
-Evidently, the degeneracy of each level increases with larger $B$,
-but since $\omega_c = q B / m$, the energy gap between each level increases too.
+Where $$A \equiv L_x L_y$$ is the area of the confinement in the $$(x,y)$$-plane.
+Evidently, the degeneracy of each level increases with larger $$B$$,
+but since $$\omega_c = q B / m$$, the energy gap between each level increases too.
In other words: the [density of states](/know/concept/density-of-states/)
is a constant with respect to the energy,
-but the states get distributed across the $E_n$ differently depending on $B$.
+but the states get distributed across the $$E_n$$ differently depending on $$B$$.
diff --git a/source/know/concept/langmuir-waves/index.md b/source/know/concept/langmuir-waves/index.md
index 0b7f2d7..7dc5dbf 100644
--- a/source/know/concept/langmuir-waves/index.md
+++ b/source/know/concept/langmuir-waves/index.md
@@ -13,8 +13,8 @@ layout: "concept"
In plasma physics, **Langmuir waves** are oscillations in the electron density,
which may or may not propagate, depending on the temperature.
-Assuming no [magnetic field](/know/concept/magnetic-field/) $\vb{B} = 0$,
-no ion motion $\vb{u}_i = 0$ (since $m_i \gg m_e$),
+Assuming no [magnetic field](/know/concept/magnetic-field/) $$\vb{B} = 0$$,
+no ion motion $$\vb{u}_i = 0$$ (since $$m_i \gg m_e$$),
and therefore no ion-electron momentum transfer,
the [two-fluid equations](/know/concept/two-fluid-equations/)
tell us that:
@@ -34,8 +34,8 @@ $$\begin{aligned}
= q_e (n_e - n_i)
\end{aligned}$$
-We split $n_e$, $\vb{u}_e$ and $\vb{E}$ into a base component
-(subscript $0$) and a perturbation (subscript $1$):
+We split $$n_e$$, $$\vb{u}_e$$ and $$\vb{E}$$ into a base component
+(subscript $$0$$) and a perturbation (subscript $$1$$):
$$\begin{aligned}
n_e
@@ -48,8 +48,8 @@ $$\begin{aligned}
= \vb{E}_0 + \vb{E}_1
\end{aligned}$$
-Where the perturbations $n_{e1}$, $\vb{u}_{e1}$ and $\vb{E}_1$ are very small,
-and the equilibrium components $n_{e0}$, $\vb{u}_{e0}$ and $\vb{E}_0$
+Where the perturbations $$n_{e1}$$, $$\vb{u}_{e1}$$ and $$\vb{E}_1$$ are very small,
+and the equilibrium components $$n_{e0}$$, $$\vb{u}_{e0}$$ and $$\vb{E}_0$$
by definition satisfy:
$$\begin{aligned}
@@ -65,7 +65,7 @@ $$\begin{aligned}
\end{aligned}$$
We insert this decomposistion into the electron continuity equation,
-arguing that $n_{e1} \vb{u}_{e1}$ is small enough to neglect, leading to:
+arguing that $$n_{e1} \vb{u}_{e1}$$ is small enough to neglect, leading to:
$$\begin{aligned}
0
@@ -78,7 +78,7 @@ $$\begin{aligned}
\end{aligned}$$
Likewise, we insert it into Gauss' law,
-and use the plasma's quasi-neutrality $n_i = n_{e0}$ to get:
+and use the plasma's quasi-neutrality $$n_i = n_{e0}$$ to get:
$$\begin{aligned}
\varepsilon_0 \nabla \cdot \big( \vb{E}_0 \!+\! \vb{E}_1 \big)
@@ -110,7 +110,7 @@ $$\begin{aligned}
-\! i \varepsilon_0 \vb{k} \cdot \vb{E}_1 = q_e n_{e1}
\end{aligned}$$
-However, there are three unknowns $n_{e1}$, $\vb{u}_{e1}$ and $\vb{E}_1$,
+However, there are three unknowns $$n_{e1}$$, $$\vb{u}_{e1}$$ and $$\vb{E}_1$$,
so one more equation is needed.
@@ -118,7 +118,7 @@ so one more equation is needed.
We therefore turn to the electron momentum equation.
For now, let us assume that the electrons have no thermal motion,
-i.e. the electron temperature $T_e = 0$, so that $p_e = 0$, leaving:
+i.e. the electron temperature $$T_e = 0$$, so that $$p_e = 0$$, leaving:
$$\begin{aligned}
m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
@@ -126,8 +126,8 @@ $$\begin{aligned}
\end{aligned}$$
Inserting the decomposition then gives the following,
-where we neglect $(\vb{u}_{e1} \cdot \nabla) \vb{u}_{e1}$
-because $\vb{u}_{e1}$ is so small by assumption:
+where we neglect $$(\vb{u}_{e1} \cdot \nabla) \vb{u}_{e1}$$
+because $$\vb{u}_{e1}$$ is so small by assumption:
$$\begin{gathered}
m_e (n_{e0} \!+\! n_{e1}) \Big( \pdv{(\vb{u}_{e0} \!+\! \vb{u}_{e1})}{t}
@@ -147,7 +147,7 @@ $$\begin{aligned}
-i \omega m_e \vb{u}_{e1} = q_e \vb{E}_1
\end{aligned}$$
-Solving this system of three equations for $\omega^2$
+Solving this system of three equations for $$\omega^2$$
gives the following dispersion relation:
$$\begin{aligned}
@@ -158,7 +158,7 @@ $$\begin{aligned}
= \frac{n_{e0} q_e^2}{\varepsilon_0 m_e}
\end{aligned}$$
-This result is known as the **plasma frequency** $\omega_p$,
+This result is known as the **plasma frequency** $$\omega_p$$,
and describes the frequency of **cold Langmuir waves**,
otherwise known as **plasma oscillations**:
@@ -169,16 +169,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that this is a dispersion relation $\omega(k) = \omega_p$,
-but that $\omega_p$ does not contain $k$.
+Note that this is a dispersion relation $$\omega(k) = \omega_p$$,
+but that $$\omega_p$$ does not contain $$k$$.
This means that cold Langmuir waves do not propagate:
the oscillation is "stationary".
## Warm Langmuir waves
-Next, we generalize this result to nonzero $T_e$,
-in which case the pressure $p_e$ is involved:
+Next, we generalize this result to nonzero $$T_e$$,
+in which case the pressure $$p_e$$ is involved:
$$\begin{aligned}
m_e n_{e0} \pdv{}{\vb{u}{e1}}{t}
@@ -186,7 +186,7 @@ $$\begin{aligned}
\end{aligned}$$
From the two-fluid thermodynamic equation of state,
-we know that $\nabla p_e$ can be written as:
+we know that $$\nabla p_e$$ can be written as:
$$\begin{aligned}
\nabla p_e
@@ -203,7 +203,7 @@ $$\begin{aligned}
\end{aligned}$$
Which once again closes the system of three equations.
-Solving for $\omega^2$ then gives:
+Solving for $$\omega^2$$ then gives:
$$\begin{aligned}
\omega^2
@@ -213,8 +213,8 @@ $$\begin{aligned}
&= \frac{n_{e0} q_e^2}{\varepsilon_0 m_e} - \frac{i \omega}{\omega m_e n_{e1}} i \gamma k_B T_e n_{e1} \big(\vb{k} \cdot \vb{k}\big)
\end{aligned}$$
-Recognizing the first term as the plasma frequency $\omega_p^2$,
-we therefore arrive at the **Bohm-Gross dispersion relation** $\omega(\vb{k})$
+Recognizing the first term as the plasma frequency $$\omega_p^2$$,
+we therefore arrive at the **Bohm-Gross dispersion relation** $$\omega(\vb{k})$$
for **warm Langmuir waves**:
$$\begin{aligned}
@@ -225,16 +225,16 @@ $$\begin{aligned}
\end{aligned}$$
This expression is typically quoted for 1D oscillations,
-in which case $\gamma = 3$ and $k = |\vb{k}|$:
+in which case $$\gamma = 3$$ and $$k = |\vb{k}|$$:
$$\begin{aligned}
\omega^2
= \omega_p^2 + \frac{3 k_B T_e}{m_e} k^2
\end{aligned}$$
-Unlike for $T_e = 0$, these "warm" waves do propagate,
-carrying information at group velocity $v_g$,
-which, in the limit of large $k$, is given by:
+Unlike for $$T_e = 0$$, these "warm" waves do propagate,
+carrying information at group velocity $$v_g$$,
+which, in the limit of large $$k$$, is given by:
$$\begin{aligned}
v_g
@@ -244,7 +244,7 @@ $$\begin{aligned}
This is the root-mean-square velocity of the
[Maxwell-Boltzmann speed distribution](/know/concept/maxwell-boltzmann-distribution/),
-meaning that information travels at the thermal velocity for large $k$.
+meaning that information travels at the thermal velocity for large $$k$$.
diff --git a/source/know/concept/laplace-transform/index.md b/source/know/concept/laplace-transform/index.md
index 5b834c3..c7f352a 100644
--- a/source/know/concept/laplace-transform/index.md
+++ b/source/know/concept/laplace-transform/index.md
@@ -9,9 +9,9 @@ layout: "concept"
---
The **Laplace transform** is an integral transform
-that losslessly converts a function $f(t)$ of a real variable $t$,
-into a function $\tilde{f}(s)$ of a complex variable $s$,
-where $s$ is sometimes called the **complex frequency**,
+that losslessly converts a function $$f(t)$$ of a real variable $$t$$,
+into a function $$\tilde{f}(s)$$ of a complex variable $$s$$,
+where $$s$$ is sometimes called the **complex frequency**,
analogously to the [Fourier transform](/know/concept/fourier-transform/).
The transform is defined as follows:
@@ -23,14 +23,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Depending on $f(t)$, this integral may diverge.
-This is solved by restricting the domain of $\tilde{f}(s)$
-to $s$ where $\mathrm{Re}\{s\} > s_0$,
-for an $s_0$ large enough to compensate for the growth of $f(t)$.
+Depending on $$f(t)$$, this integral may diverge.
+This is solved by restricting the domain of $$\tilde{f}(s)$$
+to $$s$$ where $$\mathrm{Re}\{s\} > s_0$$,
+for an $$s_0$$ large enough to compensate for the growth of $$f(t)$$.
-The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration,
+The **inverse Laplace transform** $$\hat{\mathcal{L}}{}^{-1}$$ involves complex integration,
and is therefore a lot more difficult to calculate.
-Fortunately, it is usually avoidable by rewriting a given $s$-space expression
+Fortunately, it is usually avoidable by rewriting a given $$s$$-space expression
using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/),
and then looking up the individual terms.
@@ -47,7 +47,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-This property generalizes nicely to higher-order derivatives of $s$, so:
+This property generalizes nicely to higher-order derivatives of $$s$$, so:
$$\begin{aligned}
\boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
-The exponential $\exp(- s t)$ is the only thing that depends on $s$ here:
+The exponential $$\exp(- s t)$$ is the only thing that depends on $$s$$ here:
$$\begin{aligned}
\dvn{n}{\tilde{f}}{s}
@@ -69,11 +69,12 @@ $$\begin{aligned}
&= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t}
= (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
\end{aligned}$$
+
The Laplace transform of a derivative introduces the initial conditions into the result.
-Notice that $f(0)$ is the initial value in the original $t$-domain:
+Notice that $$f(0)$$ is the initial value in the original $$t$$-domain:
$$\begin{aligned}
\boxed{
@@ -92,17 +93,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $f^{(n)}(t)$ is shorthand for the $n$th derivative of $f(t)$,
-and $f^{(0)}(t) = f(t)$.
-As an example, $\hat{\mathcal{L}}\{f'''(t)\}$ becomes
-$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$.
+Where $$f^{(n)}(t)$$ is shorthand for the $$n$$th derivative of $$f(t)$$,
+and $$f^{(0)}(t) = f(t)$$.
+As an example, $$\hat{\mathcal{L}}\{f'''(t)\}$$ becomes
+$$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$$.
-We integrate by parts and use the fact that $\lim_{x \to \infty} \exp(-x) = 0$:
+We integrate by parts and use the fact that $$\lim_{x \to \infty} \exp(-x) = 0$$:
$$\begin{aligned}
\hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
@@ -114,7 +115,7 @@ $$\begin{aligned}
\end{aligned}$$
And so on.
-By partially integrating $n$ times in total we arrive at the conclusion.
+By partially integrating $$n$$ times in total we arrive at the conclusion.
diff --git a/source/know/concept/larmor-precession/index.md b/source/know/concept/larmor-precession/index.md
index 6b101e0..774af7b 100644
--- a/source/know/concept/larmor-precession/index.md
+++ b/source/know/concept/larmor-precession/index.md
@@ -10,18 +10,18 @@ layout: "concept"
Consider a stationary spin-1/2 particle,
placed in a [magnetic field](/know/concept/magnetic-field/)
-with magnitude $B$ pointing in the $z$-direction.
-In that case, its Hamiltonian $\hat{H}$ is given by:
+with magnitude $$B$$ pointing in the $$z$$-direction.
+In that case, its Hamiltonian $$\hat{H}$$ is given by:
$$\begin{aligned}
\hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z}
\end{aligned}$$
-Where $\gamma = - q / m$ is the gyromagnetic ratio,
-and $\hat{\sigma}_z$ is the Pauli spin matrix for the $z$-direction.
-Since $\hat{H}$ is proportional to $\hat{\sigma}_z$,
-they share eigenstates $\Ket{\downarrow}$ and $\Ket{\uparrow}$.
-The respective eigenenergies $E_{\downarrow}$ and $E_{\uparrow}$ are as follows:
+Where $$\gamma = - q / m$$ is the gyromagnetic ratio,
+and $$\hat{\sigma}_z$$ is the Pauli spin matrix for the $$z$$-direction.
+Since $$\hat{H}$$ is proportional to $$\hat{\sigma}_z$$,
+they share eigenstates $$\Ket{\downarrow}$$ and $$\Ket{\uparrow}$$.
+The respective eigenenergies $$E_{\downarrow}$$ and $$E_{\uparrow}$$ are as follows:
$$\begin{aligned}
E_{\downarrow} = \frac{\hbar}{2} \gamma B
@@ -29,9 +29,9 @@ $$\begin{aligned}
E_{\uparrow} = - \frac{\hbar}{2} \gamma B
\end{aligned}$$
-Because $\hat{H}$ is time-independent,
-the general time-dependent solution $\Ket{\chi(t)}$ is of the following form,
-where $a$ and $b$ are constants,
+Because $$\hat{H}$$ is time-independent,
+the general time-dependent solution $$\Ket{\chi(t)}$$ is of the following form,
+where $$a$$ and $$b$$ are constants,
and the exponentials are "twiddle factors":
$$\begin{aligned}
@@ -40,9 +40,9 @@ $$\begin{aligned}
\:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow}
\end{aligned}$$
-For our purposes, we can safely assume that $a$ and $b$ are real,
-and then say that there exists an angle $\theta$
-satisfying $a = \sin(\theta / 2)$ and $b = \cos(\theta / 2)$, such that:
+For our purposes, we can safely assume that $$a$$ and $$b$$ are real,
+and then say that there exists an angle $$\theta$$
+satisfying $$a = \sin(\theta / 2)$$ and $$b = \cos(\theta / 2)$$, such that:
$$\begin{aligned}
\Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow}
@@ -50,7 +50,7 @@ $$\begin{aligned}
\end{aligned}$$
Now, we find the expectation values of the spin operators
-$\expval{\hat{S}_x}$, $\expval{\hat{S}_y}$, and $\expval{\hat{S}_z}$.
+$$\expval{\hat{S}_x}$$, $$\expval{\hat{S}_y}$$, and $$\expval{\hat{S}_z}$$.
The first is:
$$\begin{aligned}
@@ -86,8 +86,8 @@ $$\begin{aligned}
\matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos(\theta)
\end{aligned}$$
-The result is that the spin axis is off by $\theta$ from the $z$-direction,
-and is rotating (or **precessing**) around the $z$-axis at the **Larmor frequency** $\omega$:
+The result is that the spin axis is off by $$\theta$$ from the $$z$$-direction,
+and is rotating (or **precessing**) around the $$z$$-axis at the **Larmor frequency** $$\omega$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/laser-rate-equations/index.md b/source/know/concept/laser-rate-equations/index.md
index 939a1a0..a84d274 100644
--- a/source/know/concept/laser-rate-equations/index.md
+++ b/source/know/concept/laser-rate-equations/index.md
@@ -12,9 +12,9 @@ layout: "concept"
The [Maxwell-Bloch equations](/know/concept/maxwell-bloch-equations/) (MBEs)
give a fundamental description of light-matter interaction
for a two-level quantum system for the purposes of laser theory.
-They govern the [electric field](/know/concept/electric-field/) $\vb{E}^{+}$,
-the induced polarization $\vb{P}^{+}$,
-and the total population inversion $D$:
+They govern the [electric field](/know/concept/electric-field/) $$\vb{E}^{+}$$,
+the induced polarization $$\vb{P}^{+}$$,
+and the total population inversion $$D$$:
$$\begin{aligned}
- \mu_0 \pdvn{2}{\vb{P}^{+}}{t}
@@ -28,16 +28,16 @@ $$\begin{aligned}
&= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
\end{aligned}$$
-Where $n$ is the background medium's refractive index,
-$\omega_0$ the two-level system's gap resonance frequency,
-$|g| \equiv |\matrixel{e}{\vu{x}}{g}|$ the transition dipole moment,
-$\gamma_\perp$ and $\gamma_\parallel$ empirical decay rates,
-and $D_0$ the equilibrium inversion.
-Note that $\vb{E}^{-} = (\vb{E}^{+})^*$.
+Where $$n$$ is the background medium's refractive index,
+$$\omega_0$$ the two-level system's gap resonance frequency,
+$$|g| \equiv |\matrixel{e}{\vu{x}}{g}|$$ the transition dipole moment,
+$$\gamma_\perp$$ and $$\gamma_\parallel$$ empirical decay rates,
+and $$D_0$$ the equilibrium inversion.
+Note that $$\vb{E}^{-} = (\vb{E}^{+})^*$$.
Let us make the following ansatz,
-where $\vb{E}_0^{+}$ and $\vb{P}_0^{+}$ are slowly-varying envelopes
-of a plane wave with angular frequency $\omega \approx \omega_0$:
+where $$\vb{E}_0^{+}$$ and $$\vb{P}_0^{+}$$ are slowly-varying envelopes
+of a plane wave with angular frequency $$\omega \approx \omega_0$$:
$$\begin{aligned}
\vb{E}^{+}(\vb{r}, t)
@@ -48,9 +48,9 @@ $$\begin{aligned}
\end{aligned}$$
We insert this into the first MBE,
-and assume that $\vb{E}_0^{+}$ and $\vb{P}_0^{+}$
+and assume that $$\vb{E}_0^{+}$$ and $$\vb{P}_0^{+}$$
vary so slowly that their second-order derivatives are negligible,
-i.e. $\ipdvn{2}{\vb{E}_0^{+}\!}{t} \approx 0$ and $\ipdvn{2}{\vb{P}_0^{+}\!}{t} \approx 0$,
+i.e. $$\ipdvn{2}{\vb{E}_0^{+}\!}{t} \approx 0$$ and $$\ipdvn{2}{\vb{P}_0^{+}\!}{t} \approx 0$$,
giving:
$$\begin{aligned}
@@ -62,7 +62,7 @@ $$\begin{aligned}
To get rid of the double curl,
consider the time-independent
[electromagnetic wave equation](/know/concept/electromagnetic-wave-equation/),
-where $\Omega$ is an eigenfrequency of the optical cavity
+where $$\Omega$$ is an eigenfrequency of the optical cavity
in which lasing will occur:
$$\begin{aligned}
@@ -71,7 +71,7 @@ $$\begin{aligned}
\end{aligned}$$
For simplicity, we restrict ourselves to a single-mode laser,
-where there is only one $\Omega$ and $\vb{E}_0^{+}$ to care about.
+where there is only one $$\Omega$$ and $$\vb{E}_0^{+}$$ to care about.
Substituting the above equation into the first MBE yields:
$$\begin{aligned}
@@ -79,9 +79,9 @@ $$\begin{aligned}
= \varepsilon_0 n^2 \bigg( (\Omega^2 - \omega^2) \vb{E}_0^{+} - i 2 \omega \pdv{\vb{E}_0^{+}}{t} \bigg)
\end{aligned}$$
-Where we used $1 / c^2 = \mu_0 \varepsilon_0$.
-Assuming the light is more or less on-resonance $\omega \approx \Omega$,
-we can approximate $\Omega^2 \!-\! \omega^2 \approx 2 \omega (\Omega \!-\! \omega)$, so:
+Where we used $$1 / c^2 = \mu_0 \varepsilon_0$$.
+Assuming the light is more or less on-resonance $$\omega \approx \Omega$$,
+we can approximate $$\Omega^2 \!-\! \omega^2 \approx 2 \omega (\Omega \!-\! \omega)$$, so:
$$\begin{aligned}
i 2 \pdv{\vb{P}_0^{+}}{t} + \omega \vb{P}_0^{+}
@@ -89,18 +89,18 @@ $$\begin{aligned}
\end{aligned}$$
Moving on to the second MBE,
-inserting the ansatz $\vb{P}^{+} = \vb{P}_0^{+} e^{-i \omega t} / 2$ leads to:
+inserting the ansatz $$\vb{P}^{+} = \vb{P}_0^{+} e^{-i \omega t} / 2$$ leads to:
$$\begin{aligned}
\pdv{\vb{P}_0^{+}}{t}
= - \Big( \gamma_\perp + i (\omega_0 - \omega) \Big) \vb{P}_0^{+} - \frac{i |g|^2}{\hbar} \vb{E}_0^{+} D
\end{aligned}$$
-Typically, $\gamma_\perp$ is much larger than the rate of any other decay process,
-in which case $\ipdv{}{\vb{P}0^{+}\!}{t}$ is negligible compared to $\gamma_\perp \vb{P}_0^{+}$.
-Effectively, this means that the polarization $\vb{P}_0^{+}$
-near-instantly follows the electric field $\vb{E}^{+}\!$.
-Setting $\ipdv{}{\vb{P}0^{+}\!}{t} \approx 0$, the second MBE becomes:
+Typically, $$\gamma_\perp$$ is much larger than the rate of any other decay process,
+in which case $$\ipdv{}{\vb{P}0^{+}\!}{t}$$ is negligible compared to $$\gamma_\perp \vb{P}_0^{+}$$.
+Effectively, this means that the polarization $$\vb{P}_0^{+}$$
+near-instantly follows the electric field $$\vb{E}^{+}\!$$.
+Setting $$\ipdv{}{\vb{P}0^{+}\!}{t} \approx 0$$, the second MBE becomes:
$$\begin{aligned}
\vb{P}^{+}
@@ -108,7 +108,7 @@ $$\begin{aligned}
= \frac{|g|^2 \gamma(\omega)}{\hbar \gamma_\perp} \vb{E}^{+} D
\end{aligned}$$
-Where the Lorentzian gain curve $\gamma(\omega)$
+Where the Lorentzian gain curve $$\gamma(\omega)$$
(which also appears in the [SALT equation](/know/concept/salt-equation/))
represents a laser's preferred spectrum for amplification,
and is defined like so:
@@ -118,7 +118,7 @@ $$\begin{aligned}
\equiv \frac{\gamma_\perp}{(\omega - \omega_0) + i \gamma_\perp}
\end{aligned}$$
-Note that $\gamma(\omega)$ satisfies the following relation,
+Note that $$\gamma(\omega)$$ satisfies the following relation,
which will be useful to us later:
$$\begin{aligned}
@@ -127,8 +127,8 @@ $$\begin{aligned}
= i 2 |\gamma(\omega)|^2
\end{aligned}$$
-Returning to the first MBE with $\ipdv{\vb{P}_0^{+}}{t} \approx 0$,
-we substitute the above expression for $\vb{P}_0^{+}$:
+Returning to the first MBE with $$\ipdv{\vb{P}_0^{+}}{t} \approx 0$$,
+we substitute the above expression for $$\vb{P}_0^{+}$$:
$$\begin{aligned}
\pdv{\vb{E}_0^{+}}{t}
@@ -137,10 +137,10 @@ $$\begin{aligned}
&= i (\omega - \Omega) \vb{E}_0^{+} + i \frac{|g|^2 \omega \gamma(\omega)}{2 \hbar \varepsilon_0 \gamma_\perp n^2} \vb{E}_0^{+} D
\end{aligned}$$
-Next, we insert our ansatz for $\vb{E}^{+}\!$ and $\vb{P}^{+}\!$
-into the third MBE, and rewrite $\vb{P}_0^{+}$ as above.
-Using our identity for $\gamma(\omega)$,
-and the fact that $\vb{E}_0^{+} \cdot \vb{E}_0^{-} = |\vb{E}|^2$, we find:
+Next, we insert our ansatz for $$\vb{E}^{+}\!$$ and $$\vb{P}^{+}\!$$
+into the third MBE, and rewrite $$\vb{P}_0^{+}$$ as above.
+Using our identity for $$\gamma(\omega)$$,
+and the fact that $$\vb{E}_0^{+} \cdot \vb{E}_0^{-} = |\vb{E}|^2$$, we find:
$$\begin{aligned}
\pdv{D}{t}
@@ -155,7 +155,7 @@ $$\begin{aligned}
This is the prototype of the first laser rate equation.
However, in order to have a practical set,
-we need an equation for $|\vb{E}|^2$,
+we need an equation for $$|\vb{E}|^2$$,
which we can obtain using the first MBE:
$$\begin{aligned}
@@ -171,7 +171,7 @@ $$\begin{aligned}
&= 2 \Imag(\Omega) |\vb{E}|^2 + \frac{|g|^2 \omega}{\hbar \varepsilon_0 \gamma_\perp n^2} |\gamma(\omega)|^2 |\vb{E}|^2 D
\end{aligned}$$
-Where $\Imag(\Omega) < 0$ represents the fact that the laser cavity is leaky.
+Where $$\Imag(\Omega) < 0$$ represents the fact that the laser cavity is leaky.
We now have the **laser rate equations**,
although they are still in an unidiomatic form:
@@ -187,8 +187,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-To rewrite this, we replace $|\vb{E}|^2$ with the photon number $N_p$ as follows,
-with $U = \varepsilon_0 n^2 |\vb{E}|^2 / 2$ being the energy density of the light:
+To rewrite this, we replace $$|\vb{E}|^2$$ with the photon number $$N_p$$ as follows,
+with $$U = \varepsilon_0 n^2 |\vb{E}|^2 / 2$$ being the energy density of the light:
$$\begin{aligned}
N_{p}
@@ -196,12 +196,12 @@ $$\begin{aligned}
= \frac{\varepsilon_0 n^2}{2 \hbar \omega} |\vb{E}|^2
\end{aligned}$$
-Furthermore, consider the definition of the inversion $D$:
+Furthermore, consider the definition of the inversion $$D$$:
because a photon emission annihilates an electron-hole pair,
-it reduces $D$ by $2$.
-Since lasing is only possible for $D > 0$,
-we can replace $D$ with the conduction band's electron population $N_e$,
-which is reduced by $1$ whenever a photon is emitted.
+it reduces $$D$$ by $$2$$.
+Since lasing is only possible for $$D > 0$$,
+we can replace $$D$$ with the conduction band's electron population $$N_e$$,
+which is reduced by $$1$$ whenever a photon is emitted.
The laser rate equations then take the following standard form:
$$\begin{aligned}
@@ -216,10 +216,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\gamma_e$ is a redefinition of $\gamma_\parallel$
+Where $$\gamma_e$$ is a redefinition of $$\gamma_\parallel$$
depending on the electron decay processes,
-and the photon loss rate $\gamma_p$, the gain $G$,
-and the carrier supply rate $R_\mathrm{pump}$
+and the photon loss rate $$\gamma_p$$, the gain $$G$$,
+and the carrier supply rate $$R_\mathrm{pump}$$
are defined like so:
$$\begin{aligned}
@@ -234,14 +234,14 @@ $$\begin{aligned}
\equiv \gamma_\parallel D_0
\end{aligned}$$
-With $Q$ being the cavity mode's quality factor.
-The nonlinear coupling term $G N_p N_e$ represents
+With $$Q$$ being the cavity mode's quality factor.
+The nonlinear coupling term $$G N_p N_e$$ represents
[stimulated emission](/know/concept/einstein-coefficients/),
which is the key to lasing.
To understand the behaviour of a laser,
consider these equations in a steady state,
-i.e. where $N_p$ and $N_e$ are constant in $t$:
+i.e. where $$N_p$$ and $$N_e$$ are constant in $$t$$:
$$\begin{aligned}
0
@@ -251,9 +251,9 @@ $$\begin{aligned}
&= R_\mathrm{pump} - \gamma_e N_e - G N_p N_e
\end{aligned}$$
-In addition to the trivial solution $N_p = 0$,
-we can also have $N_p > 0$.
-Isolating $N_p$'s equation for $N_e$ and inserting that into $N_e$'s equation, we find:
+In addition to the trivial solution $$N_p = 0$$,
+we can also have $$N_p > 0$$.
+Isolating $$N_p$$'s equation for $$N_e$$ and inserting that into $$N_e$$'s equation, we find:
$$\begin{aligned}
N_e
@@ -265,27 +265,27 @@ $$\begin{aligned}
}
\end{aligned}$$
-The quantity $R_\mathrm{thr} \equiv \gamma_e \gamma_p / G$ is called the **lasing threshold**:
-if $R_\mathrm{pump} \ge R_\mathrm{thr}$, the laser is active,
-meaning that $N_p$ is big enough to cause
+The quantity $$R_\mathrm{thr} \equiv \gamma_e \gamma_p / G$$ is called the **lasing threshold**:
+if $$R_\mathrm{pump} \ge R_\mathrm{thr}$$, the laser is active,
+meaning that $$N_p$$ is big enough to cause
a "chain reaction" of stimulated emission
that consumes all surplus carriers to maintain a steady state.
-The point is that $N_e$ is independent of the electron supply $R_\mathrm{pump}$,
+The point is that $$N_e$$ is independent of the electron supply $$R_\mathrm{pump}$$,
because all additional electrons are almost immediately
annihilated by stimulated emission.
-Consequently $N_p$ increases linearly as $R_\mathrm{pump}$ is raised,
+Consequently $$N_p$$ increases linearly as $$R_\mathrm{pump}$$ is raised,
at a much steeper slope than would be possible below threshold.
-The output of the cavity is proportional to $N_p$,
+The output of the cavity is proportional to $$N_p$$,
so the brightness is also linear.
Unfortunately, by deriving the laser rate equations from the MBEs,
we lost some interesting and important effects,
most notably spontaneous emission,
-which is needed for $N_p$ to grow if $R_\mathrm{pump}$ is below threshold.
+which is needed for $$N_p$$ to grow if $$R_\mathrm{pump}$$ is below threshold.
For this reason, the laser rate equations are typically presented
-in a more empirical form, which "bookkeeps" the processes affecting $N_p$ and $N_e$.
+in a more empirical form, which "bookkeeps" the processes affecting $$N_p$$ and $$N_e$$.
Consider the following example:
$$\begin{aligned}
@@ -301,17 +301,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\gamma_\mathrm{out}$ represents the cavity's usable output,
-$\gamma_\mathrm{abs}$ the medium's absorption,
-$\gamma_\mathrm{loss}$ scattering losses,
-$\gamma_\mathrm{spon}$ spontaneous emission,
-$\gamma_\mathrm{n.r.}$ non-radiative electron-hole recombination,
-and $\gamma_\mathrm{leak}$ the fact that
+Where $$\gamma_\mathrm{out}$$ represents the cavity's usable output,
+$$\gamma_\mathrm{abs}$$ the medium's absorption,
+$$\gamma_\mathrm{loss}$$ scattering losses,
+$$\gamma_\mathrm{spon}$$ spontaneous emission,
+$$\gamma_\mathrm{n.r.}$$ non-radiative electron-hole recombination,
+and $$\gamma_\mathrm{leak}$$ the fact that
some carriers leak away before they can be used for emission.
Unsurprisingly, this form is much harder to analyze,
but more accurately describes the dynamics inside a laser.
-To make matters even worse, many of these decay rates depend on $N_p$ or $N_e$,
+To make matters even worse, many of these decay rates depend on $$N_p$$ or $$N_e$$,
so solutions can only be obtained numerically.
diff --git a/source/know/concept/laws-of-thermodynamics/index.md b/source/know/concept/laws-of-thermodynamics/index.md
index 7758446..3605a0e 100644
--- a/source/know/concept/laws-of-thermodynamics/index.md
+++ b/source/know/concept/laws-of-thermodynamics/index.md
@@ -19,9 +19,9 @@ is a consequence of these laws.
The **first law of thermodynamics** states that energy is conserved.
When a system goes from one equilibrium to another,
-the change $\Delta U$ of its energy $U$ is equal to
-the work $\Delta W$ done by external forces,
-plus the energy transferred by heating ($\Delta Q > 0$) or cooling ($\Delta Q < 0$):
+the change $$\Delta U$$ of its energy $$U$$ is equal to
+the work $$\Delta W$$ done by external forces,
+plus the energy transferred by heating ($$\Delta Q > 0$$) or cooling ($$\Delta Q < 0$$):
$$\begin{aligned}
\boxed{
@@ -29,9 +29,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-The internal energy $U$ is a state variable,
+The internal energy $$U$$ is a state variable,
so is independent of the path taken between equilibria.
-However, the work $\Delta W$ and heating $\Delta Q$ do depend on the path,
+However, the work $$\Delta W$$ and heating $$\Delta Q$$ do depend on the path,
so the first law means that
the act of transferring energy is path-dependent,
but the result has no "memory" of that path.
@@ -44,9 +44,9 @@ the total entropy never decreases.
An important consequence is that
no machine can convert energy into work with 100% efficiency.
-It is possible for the local entropy $S_{\mathrm{loc}}$
+It is possible for the local entropy $$S_{\mathrm{loc}}$$
of a system to decrease, but doing so requires work,
-and therefore the entropy of the surroundings $S_{\mathrm{sur}}$
+and therefore the entropy of the surroundings $$S_{\mathrm{sur}}$$
must increase accordingly, such that:
$$\begin{aligned}
@@ -57,19 +57,19 @@ $$\begin{aligned}
Since the total entropy never decreases,
the equilibrium state of a system must be a maximum
-of its entropy $S$, and therefore $S$ can be used as
+of its entropy $$S$$, and therefore $$S$$ can be used as
a [thermodynamic "potential"](/know/concept/thermodynamic-potential/).
-The only situation where $\Delta S = 0$ is a reversible process,
+The only situation where $$\Delta S = 0$$ is a reversible process,
since then it must be possible to return to
the previous equilibrium state by doing the same work in the opposite direction.
According to the first law,
if a process is reversible, or if it is only heating/cooling,
then (after one reversible cycle) the energy change
-is simply the heat transfer $\dd{U} = \dd{Q}$.
-An entropy change $\dd{S}$ is then expressed as follows
-(since $\ipdv{S}{U} = 1 / T$ by definition):
+is simply the heat transfer $$\dd{U} = \dd{Q}$$.
+An entropy change $$\dd{S}$$ is then expressed as follows
+(since $$\ipdv{S}{U} = 1 / T$$ by definition):
$$\begin{aligned}
\boxed{
@@ -85,7 +85,7 @@ Confusingly, this equation is sometimes also called the second law of thermodyna
## Third law
The **third law of thermodynamics** states that
-the entropy $S$ of a system goes to zero when the temperature reaches absolute zero:
+the entropy $$S$$ of a system goes to zero when the temperature reaches absolute zero:
$$\begin{aligned}
\boxed{
@@ -93,8 +93,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-From this, the absolute quantity of $S$ is defined, otherwise we would
-only be able to speak of entropy differences $\Delta S$.
+From this, the absolute quantity of $$S$$ is defined, otherwise we would
+only be able to speak of entropy differences $$\Delta S$$.
diff --git a/source/know/concept/lawson-criterion/index.md b/source/know/concept/lawson-criterion/index.md
index ff9594b..f2f2fe0 100644
--- a/source/know/concept/lawson-criterion/index.md
+++ b/source/know/concept/lawson-criterion/index.md
@@ -13,9 +13,9 @@ the **Lawson criterion** must be met,
from which some required properties
of the plasma and the reactor chamber can be deduced.
-Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion,
-but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way.
-If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction,
+Suppose that a reactor generates a given power $$P_\mathrm{fus}$$ by nuclear fusion,
+but that it leaks energy at a rate $$P_\mathrm{loss}$$ in an unusable way.
+If an auxiliary input power $$P_\mathrm{aux}$$ sustains the fusion reaction,
then the following inequality must be satisfied
in order to have harvestable energy:
@@ -24,8 +24,8 @@ $$\begin{aligned}
\le P_\mathrm{fus} + P_\mathrm{aux}
\end{aligned}$$
-We can rewrite $P_\mathrm{aux}$ using the definition
-of the **energy gain factor** $Q$,
+We can rewrite $$P_\mathrm{aux}$$ using the definition
+of the **energy gain factor** $$Q$$,
which is the ratio of the output and input powers of the fusion reaction:
$$\begin{aligned}
@@ -45,12 +45,12 @@ $$\begin{aligned}
= P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
\end{aligned}$$
-We assume that the plasma has equal species densities $n_i = n_e$,
-so its total density $n = 2 n_i$.
-Then $P_\mathrm{fus}$ is as follows,
-where $f_{ii}$ is the frequency
+We assume that the plasma has equal species densities $$n_i = n_e$$,
+so its total density $$n = 2 n_i$$.
+Then $$P_\mathrm{fus}$$ is as follows,
+where $$f_{ii}$$ is the frequency
with which a given ion collides with other ions,
-and $E_\mathrm{fus}$ is the energy released by a single fusion reaction:
+and $$E_\mathrm{fus}$$ is the energy released by a single fusion reaction:
$$\begin{aligned}
P_\mathrm{fus}
@@ -59,11 +59,11 @@ $$\begin{aligned}
= \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus}
\end{aligned}$$
-Where $\Expval{\sigma v}$ is the mean product
-of the velocity $v$ and the collision cross-section $\sigma$.
+Where $$\Expval{\sigma v}$$ is the mean product
+of the velocity $$v$$ and the collision cross-section $$\sigma$$.
-Furthermore, assuming that both species have the same temperature $T_i = T_e = T$,
-the total energy density $W$ of the plasma is given by:
+Furthermore, assuming that both species have the same temperature $$T_i = T_e = T$$,
+the total energy density $$W$$ of the plasma is given by:
$$\begin{aligned}
W
@@ -71,8 +71,8 @@ $$\begin{aligned}
= 3 k_B T n
\end{aligned}$$
-Where $k_B$ is Boltzmann's constant.
-From this, we can define the **confinement time** $\tau_E$
+Where $$k_B$$ is Boltzmann's constant.
+From this, we can define the **confinement time** $$\tau_E$$
as the characteristic lifetime of energy in the reactor, before leakage.
Therefore:
@@ -84,7 +84,7 @@ $$\begin{aligned}
= \frac{3 n k_B T}{\tau_E}
\end{aligned}$$
-Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$
+Inserting these new expressions for $$P_\mathrm{fus}$$ and $$P_\mathrm{loss}$$
into the inequality, we arrive at:
$$\begin{aligned}
@@ -101,8 +101,8 @@ $$\begin{aligned}
\end{aligned}$$
However, it turns out that the highest fusion power density
-is reached when $T$ is at the minimum of $T^2 / \Expval{\sigma v}$.
-Therefore, we multiply by $T$ to get the Lawson triple product:
+is reached when $$T$$ is at the minimum of $$T^2 / \Expval{\sigma v}$$.
+Therefore, we multiply by $$T$$ to get the Lawson triple product:
$$\begin{aligned}
\boxed{
@@ -112,7 +112,7 @@ $$\begin{aligned}
\end{aligned}$$
For some reason,
-it is often assumed that the fusion is infinitely profitable $Q \to \infty$,
+it is often assumed that the fusion is infinitely profitable $$Q \to \infty$$,
in which case the criterion reduces to:
$$\begin{aligned}
diff --git a/source/know/concept/legendre-polynomials/index.md b/source/know/concept/legendre-polynomials/index.md
index 74543a3..f223cd3 100644
--- a/source/know/concept/legendre-polynomials/index.md
+++ b/source/know/concept/legendre-polynomials/index.md
@@ -8,10 +8,10 @@ layout: "concept"
---
The **Legendre polynomials** are a set of functions that sometimes arise in physics.
-They are the eigenfunctions $u(x)$ of **Legendre's differential equation**,
+They are the eigenfunctions $$u(x)$$ of **Legendre's differential equation**,
which is a ([Sturm-Liouville](/know/concept/sturm-liouville-theory/))
-eigenvalue problem for $\ell (\ell + 1)$,
-where $\ell$ turns out to be a non-negative integer:
+eigenvalue problem for $$\ell (\ell + 1)$$,
+where $$\ell$$ turns out to be a non-negative integer:
$$\begin{aligned}
\boxed{
@@ -19,7 +19,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The $\ell$th-degree Legendre polynomial $P_\ell(x)$
+The $$\ell$$th-degree Legendre polynomial $$P_\ell(x)$$
is given in the form of a *Rodrigues' formula* by:
$$\begin{aligned}
@@ -27,7 +27,7 @@ $$\begin{aligned}
&= \frac{1}{2^\ell \ell!} \dvn{\ell}{}{x}(x^2 - 1)^\ell
\end{aligned}$$
-The first handful of Legendre polynomials $P_\ell(x)$ are therefore as follows:
+The first handful of Legendre polynomials $$P_\ell(x)$$ are therefore as follows:
$$\begin{gathered}
P_0(x) = 1
@@ -41,7 +41,7 @@ $$\begin{gathered}
P_4(x) = \frac{1}{8} (35 x^4 - 30 x^2 + 3)
\end{gathered}$$
-And then more $P_\ell$ can be computed quickly
+And then more $$P_\ell$$ can be computed quickly
using **Bonnet's recursion formula**:
$$\begin{aligned}
@@ -50,7 +50,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The derivative of a given $P_\ell$ can be calculated recursively
+The derivative of a given $$P_\ell$$ can be calculated recursively
using the following relation:
$$\begin{aligned}
@@ -61,7 +61,7 @@ $$\begin{aligned}
\end{aligned}$$
Noteworthy is that the Legendre polynomials
-are mutually orthogonal for $x \in [-1, 1]$:
+are mutually orthogonal for $$x \in [-1, 1]$$:
$$\begin{aligned}
\boxed{
@@ -74,7 +74,7 @@ $$\begin{aligned}
As was to be expected from Sturm-Liouville theory.
Likewise, they form a complete basis in the
[Hilbert space](/know/concept/hilbert-space/)
-of piecewise continuous functions $f(x)$ on $x \in [-1, 1]$,
+of piecewise continuous functions $$f(x)$$ on $$x \in [-1, 1]$$,
meaning:
$$\begin{aligned}
@@ -85,11 +85,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Each Legendre polynomial $P_\ell$ comes with
-a set of **associated Legendre polynomials** $P_\ell^m(x)$
-of order $m$ and degree $\ell$.
+Each Legendre polynomial $$P_\ell$$ comes with
+a set of **associated Legendre polynomials** $$P_\ell^m(x)$$
+of order $$m$$ and degree $$\ell$$.
These are the non-singular solutions of the **general Legendre equation**,
-where $m$ and $\ell$ are integers satisfying $-\ell \le m \le \ell$:
+where $$m$$ and $$\ell$$ are integers satisfying $$-\ell \le m \le \ell$$:
$$\begin{aligned}
\boxed{
@@ -97,17 +97,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-The $\ell$th-degree $m$th-order associated Legendre polynomial $P_\ell^m$
-is as follows for $m \ge 0$:
+The $$\ell$$th-degree $$m$$th-order associated Legendre polynomial $$P_\ell^m$$
+is as follows for $$m \ge 0$$:
$$\begin{aligned}
P_\ell^m(x)
= (-1)^m (1 - x^2)^{m/2} \dvn{m}{}{x}P_\ell(x)
\end{aligned}$$
-Here, the $(-1)^m$ in front is called the **Condon-Shortley phase**,
+Here, the $$(-1)^m$$ in front is called the **Condon-Shortley phase**,
and is omitted by some authors.
-For negative orders $m$,
+For negative orders $$m$$,
an additional constant factor is necessary:
$$\begin{aligned}
@@ -115,6 +115,6 @@ $$\begin{aligned}
\end{aligned}$$
Beware, the name is misleading:
-if $m$ is odd, then $P_\ell^m$ is actually not a polynomial.
-Moreover, not all $P_\ell^m$ are mutually orthogonal
+if $$m$$ is odd, then $$P_\ell^m$$ is actually not a polynomial.
+Moreover, not all $$P_\ell^m$$ are mutually orthogonal
(but some are).
diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
index 354b7ae..51c003e 100644
--- a/source/know/concept/legendre-transform/index.md
+++ b/source/know/concept/legendre-transform/index.md
@@ -8,32 +8,32 @@ categories:
layout: "concept"
---
-The **Legendre transform** of a function $f(x)$ is a new function $L(f')$,
-which depends only on the derivative $f'(x)$ of $f(x)$, and from which
-the original function $f(x)$ can be reconstructed. The point is,
+The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$,
+which depends only on the derivative $$f'(x)$$ of $$f(x)$$, and from which
+the original function $$f(x)$$ can be reconstructed. The point is,
analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)),
-that $L(f')$ contains the same information as $f(x)$, just in a different form.
+that $$L(f')$$ contains the same information as $$f(x)$$, just in a different form.
-Let us choose an arbitrary point $x_0 \in [a, b]$ in the domain of
-$f(x)$. Consider a line $y(x)$ tangent to $f(x)$ at $x = x_0$, which has
-a slope $f'(x_0)$ and intersects the $y$-axis at $-C$:
+Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of
+$$f(x)$$. Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$, which has
+a slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$-C$$:
$$\begin{aligned}
y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
\end{aligned}$$
-The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$
-(or sometimes $-C$) for all $x_0 \in [a, b]$,
-where $C$ corresponds to the tangent line at $x = x_0$. This yields:
+The Legendre transform $$L(f')$$ is defined such that $$L(f'(x_0)) = C$$
+(or sometimes $$-C$$) for all $$x_0 \in [a, b]$$,
+where $$C$$ corresponds to the tangent line at $$x = x_0$$. This yields:
$$\begin{aligned}
L(f'(x)) = f'(x) \: x - f(x)
\end{aligned}$$
-We want this function to depend only on the derivative $f'$, but
-currently $x$ still appears here as a variable. We fix that problem in
-the easiest possible way: by assuming that $f'(x)$ is invertible for all
-$x \in [a, b]$. If $x(f')$ is the inverse of $f'(x)$, then $L(f')$ is
+We want this function to depend only on the derivative $$f'$$, but
+currently $$x$$ still appears here as a variable. We fix that problem in
+the easiest possible way: by assuming that $$f'(x)$$ is invertible for all
+$$x \in [a, b]$$. If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is
given by:
$$\begin{aligned}
@@ -43,12 +43,12 @@ $$\begin{aligned}
\end{aligned}$$
The only requirement for the existence of the Legendre transform is thus
-the invertibility of $f'(x)$ in the target interval $[a,b]$, which can
-only be true if $f(x)$ is either convex or concave, i.e. its derivative
-$f'(x)$ is monotonic.
+the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, which can
+only be true if $$f(x)$$ is either convex or concave, i.e. its derivative
+$$f'(x)$$ is monotonic.
-Crucially, the derivative of $L(f')$ with respect to $f'$ is simply
-$x(f')$. In other words, the roles of $f'$ and $x$ are switched by the
+Crucially, the derivative of $$L(f')$$ with respect to $$f'$$ is simply
+$$x(f')$$. In other words, the roles of $$f'$$ and $$x$$ are switched by the
transformation: the coordinate becomes the derivative and vice versa.
This is demonstrated here:
@@ -59,7 +59,7 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, Legendre transformation is an *involution*, meaning it is
-its own inverse. Let $g(L')$ be the Legendre transform of $L(f')$:
+its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$:
$$\begin{aligned}
g(L') = L' \: f'(L') - L(f'(L'))
@@ -68,7 +68,7 @@ $$\begin{aligned}
Moreover, the inverse of a (forward) transform always exists, because
the Legendre transform of a convex function is itself convex. Convexity
-of $f(x)$ means that $f''(x) > 0$ for all $x \in [a, b]$, which yields
+of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$, which yields
the following proof:
$$\begin{aligned}
diff --git a/source/know/concept/lehmann-representation/index.md b/source/know/concept/lehmann-representation/index.md
index cfc6838..74bd457 100644
--- a/source/know/concept/lehmann-representation/index.md
+++ b/source/know/concept/lehmann-representation/index.md
@@ -11,11 +11,11 @@ layout: "concept"
In many-body quantum theory, the **Lehmann representation**
is an alternative way to write the [Green's functions](/know/concept/greens-functions/),
obtained by expanding in the many-particle eigenstates
-under the assumption of a time-independent Hamiltonian $\hat{H}$.
+under the assumption of a time-independent Hamiltonian $$\hat{H}$$.
-First, we write out the greater Green's function $G_{\nu \nu'}^>(t, t')$,
-and then expand its expected value $\Expval{}$ (at thermodynamic equilibrium)
-into a sum of many-particle basis states $\Ket{n}$:
+First, we write out the greater Green's function $$G_{\nu \nu'}^>(t, t')$$,
+and then expand its expected value $$\Expval{}$$ (at thermodynamic equilibrium)
+into a sum of many-particle basis states $$\Ket{n}$$:
$$\begin{aligned}
G_{\nu \nu'}^>(t, t')
@@ -23,13 +23,13 @@ $$\begin{aligned}
&= - \frac{i}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{c}_\nu(t) \hat{c}_{\nu'}^\dagger(t') e^{-\beta \hat{H}}}{n}
\end{aligned}$$
-Where $\beta = 1 / (k_B T)$, and $Z$ is the grand partition function
+Where $$\beta = 1 / (k_B T)$$, and $$Z$$ is the grand partition function
(see [grand canonical ensemble](/know/concept/grand-canonical-ensemble/));
-the operator $e^{\beta \hat{H}}$ gives the weight of each term at equilibrium.
-Since $\Ket{n}$ is an eigenstate of $\hat{H}$ with energy $E_n$,
-this gives us a factor of $e^{\beta E_n}$.
+the operator $$e^{\beta \hat{H}}$$ gives the weight of each term at equilibrium.
+Since $$\Ket{n}$$ is an eigenstate of $$\hat{H}$$ with energy $$E_n$$,
+this gives us a factor of $$e^{\beta E_n}$$.
Furthermore, we are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
-so we write out the time-dependence of $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$:
+so we write out the time-dependence of $$\hat{c}_\nu$$ and $$\hat{c}_{\nu'}^\dagger$$:
$$\begin{aligned}
G_{\nu \nu'}^>(t, t')
@@ -40,10 +40,10 @@ $$\begin{aligned}
\Matrixel{n}{e^{i \hat{H} (t - t') / \hbar} \hat{c}_\nu e^{- i \hat{H} (t - t') / \hbar} \hat{c}_{\nu'}^\dagger}{n}
\end{aligned}$$
-Where we used that the trace $\Tr\!(x) = \sum_{n} \matrixel{n}{x}{n}$
-is invariant under cyclic permutations of $x$.
-The $\Ket{n}$ form a basis of eigenstates of $\hat{H}$,
-so we insert an identity operator $\sum_{n'} \Ket{n'} \Bra{n'}$:
+Where we used that the trace $$\Tr\!(x) = \sum_{n} \matrixel{n}{x}{n}$$
+is invariant under cyclic permutations of $$x$$.
+The $$\Ket{n}$$ form a basis of eigenstates of $$\hat{H}$$,
+so we insert an identity operator $$\sum_{n'} \Ket{n'} \Bra{n'}$$:
$$\begin{aligned}
G_{\nu \nu'}^>(t - t')
@@ -54,10 +54,10 @@ $$\begin{aligned}
\matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} e^{i (E_n - E_{n'}) (t - t') / \hbar}
\end{aligned}$$
-Note that $G_{\nu \nu'}^>$ now only depends on the time difference $t - t'$,
-because $\hat{H}$ is time-independent.
+Note that $$G_{\nu \nu'}^>$$ now only depends on the time difference $$t - t'$$,
+because $$\hat{H}$$ is time-independent.
Next, we take the [Fourier transform](/know/concept/fourier-transform/)
-$t \to \omega$ (with $t' = 0$):
+$$t \to \omega$$ (with $$t' = 0$$):
$$\begin{aligned}
G_{\nu \nu'}^>(\omega)
@@ -66,9 +66,9 @@ $$\begin{aligned}
\end{aligned}$$
Here, we recognize the integral
-as a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$,
-thereby introducing a factor of $2 \pi$,
-and arriving at the Lehmann representation of $G_{\nu \nu'}^>$:
+as a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$,
+thereby introducing a factor of $$2 \pi$$,
+and arriving at the Lehmann representation of $$G_{\nu \nu'}^>$$:
$$\begin{aligned}
\boxed{
@@ -78,7 +78,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-We now go through the same process for the lesser Green's function $G_{\nu \nu'}^<(t, t')$:
+We now go through the same process for the lesser Green's function $$G_{\nu \nu'}^<(t, t')$$:
$$\begin{aligned}
G_{\nu \nu'}^<(t - t')
@@ -88,7 +88,7 @@ $$\begin{aligned}
e^{i (E_{n'} - E_n) (t - t') / \hbar}
\end{aligned}$$
-Where $-$ is for bosons, and $+$ for fermions.
+Where $$-$$ is for bosons, and $$+$$ for fermions.
Fourier transforming yields the following:
$$\begin{aligned}
@@ -97,8 +97,8 @@ $$\begin{aligned}
\: \delta(E_{n'} - E_n + \hbar \omega)
\end{aligned}$$
-We swap $n$ and $n'$, leading to the following
-Lehmann representation of $G_{\nu \nu'}^<$:
+We swap $$n$$ and $$n'$$, leading to the following
+Lehmann representation of $$G_{\nu \nu'}^<$$:
$$\begin{aligned}
\boxed{
@@ -108,8 +108,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Due to the delta function $\delta$,
-each term is only nonzero for $E_n' = E_n + \hbar \omega$,
+Due to the delta function $$\delta$$,
+each term is only nonzero for $$E_n' = E_n + \hbar \omega$$,
so we write:
$$\begin{aligned}
@@ -119,7 +119,7 @@ $$\begin{aligned}
\end{aligned}$$
Therefore, we arrive at the following useful relation
-between $G_{\nu \nu'}^<$ and $G_{\nu \nu'}^>$:
+between $$G_{\nu \nu'}^<$$ and $$G_{\nu \nu'}^>$$:
$$\begin{aligned}
\boxed{
@@ -129,7 +129,7 @@ $$\begin{aligned}
\end{aligned}$$
Moving on, let us do the same for
-the retarded Green's function $G_{\nu \nu'}^R(t, t')$, given by:
+the retarded Green's function $$G_{\nu \nu'}^R(t, t')$$, given by:
$$\begin{aligned}
G_{\nu \nu'}^R(t \!-\! t')
@@ -141,7 +141,7 @@ $$\begin{aligned}
\end{aligned}$$
We take the Fourier transform, but to ensure convergence,
-we must introduce an infinitesimal positive $\eta \to 0^+$ to the exponent
+we must introduce an infinitesimal positive $$\eta \to 0^+$$ to the exponent
(and eventually take the limit):
$$\begin{aligned}
@@ -155,7 +155,7 @@ $$\begin{aligned}
\end{aligned}$$
Leading us to the following Lehmann representation
-of the retarded Green's function $G_{\nu \nu'}^R$:
+of the retarded Green's function $$G_{\nu \nu'}^R$$:
$$\begin{aligned}
\boxed{
@@ -166,7 +166,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, we go through the same steps for the advanced Green's function $G_{\nu \nu'}^A(t, t')$:
+Finally, we go through the same steps for the advanced Green's function $$G_{\nu \nu'}^A(t, t')$$:
$$\begin{aligned}
G_{\nu \nu'}^A(t \!-\! t')
@@ -177,7 +177,7 @@ $$\begin{aligned}
\Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
\end{aligned}$$
-For the Fourier transform, we must again introduce $\eta \to 0^+$
+For the Fourier transform, we must again introduce $$\eta \to 0^+$$
(although note the sign):
$$\begin{aligned}
@@ -191,7 +191,7 @@ $$\begin{aligned}
\end{aligned}$$
Therefore, the Lehmann representation of
-the advanced Green's function $G_{\nu \nu'}^A$ is as follows:
+the advanced Green's function $$G_{\nu \nu'}^A$$ is as follows:
$$\begin{aligned}
\boxed{
@@ -211,8 +211,8 @@ $$\begin{aligned}
\Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
\end{aligned}$$
-Note the subscripts $\nu$ and $\nu'$.
-Comparing this to $G_{\nu \nu'}^R$ gives us another useful relation:
+Note the subscripts $$\nu$$ and $$\nu'$$.
+Comparing this to $$G_{\nu \nu'}^R$$ gives us another useful relation:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index 4252bd4..4033148 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -14,28 +14,28 @@ to an external perturbation, and is a quantum-mechanical
alternative to the [Drude model](/know/concept/drude-model/).
We start from the [Kubo formula](/know/concept/kubo-formula/)
-for the electron density operator $\hat{n}$,
-which describes the change in $\Expval{\hat{n}}$
-due to a time-dependent perturbation $\hat{H}_1$:
+for the electron density operator $$\hat{n}$$,
+which describes the change in $$\Expval{\hat{n}}$$
+due to a time-dependent perturbation $$\hat{H}_1$$:
$$\begin{aligned}
\delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
= -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
\end{aligned}$$
-Where the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/),
-and the expectation $\Expval{}_0$ is for
+Where the subscript $$I$$ refers to the [interaction picture](/know/concept/interaction-picture/),
+and the expectation $$\Expval{}_0$$ is for
a thermal equilibrium before the perturbation was applied.
-Now consider a harmonic $\hat{H}_1$:
+Now consider a harmonic $$\hat{H}_1$$:
$$\begin{aligned}
\hat{H}_{1,S}(t)
= e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}_S(\vb{r}) \dd{\vb{r}}
\end{aligned}$$
-Where $S$ is the Schrödinger picture,
-$\eta$ is a positive infinitesimal to ensure convergence later,
-and $U(\vb{r})$ is an arbitrary potential function.
+Where $$S$$ is the Schrödinger picture,
+$$\eta$$ is a positive infinitesimal to ensure convergence later,
+and $$U(\vb{r})$$ is an arbitrary potential function.
The Kubo formula becomes:
$$\begin{aligned}
@@ -43,7 +43,7 @@ $$\begin{aligned}
= \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
\end{aligned}$$
-Here, $\chi$ is the density-density correlation function,
+Here, $$\chi$$ is the density-density correlation function,
i.e. a two-particle [Green's function](/know/concept/greens-functions/):
$$\begin{aligned}
@@ -51,10 +51,10 @@ $$\begin{aligned}
\equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0
\end{aligned}$$
-Let us assume that the unperturbed system (i.e. without $U$) is spatially uniform,
-so that $\chi$ only depends on the difference $\vb{r} - \vb{r}'$.
+Let us assume that the unperturbed system (i.e. without $$U$$) is spatially uniform,
+so that $$\chi$$ only depends on the difference $$\vb{r} - \vb{r}'$$.
We then take its [Fourier transform](/know/concept/fourier-transform/)
-$\vb{r}\!-\!\vb{r}' \to \vb{q}$:
+$$\vb{r}\!-\!\vb{r}' \to \vb{q}$$:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -65,9 +65,9 @@ $$\begin{aligned}
\: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
\end{aligned}$$
-Where both $\hat{n}_I$ have been written as inverse Fourier transforms,
-giving a factor $(2 \pi)^{-2 D}$, with $D$ being the number of spatial dimensions.
-We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
+Where both $$\hat{n}_I$$ have been written as inverse Fourier transforms,
+giving a factor $$(2 \pi)^{-2 D}$$, with $$D$$ being the number of spatial dimensions.
+We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -84,9 +84,9 @@ $$\begin{aligned}
\: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2}
\end{aligned}$$
-On the left, $\vb{r}'$ does not appear, so it must also disappear on the right.
-If we choose an arbitrary (hyper)cube of volume $V$ in real space,
-then clearly $\int_V \dd{\vb{r}'} = V$. Therefore:
+On the left, $$\vb{r}'$$ does not appear, so it must also disappear on the right.
+If we choose an arbitrary (hyper)cube of volume $$V$$ in real space,
+then clearly $$\int_V \dd{\vb{r}'} = V$$. Therefore:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -95,8 +95,8 @@ $$\begin{aligned}
\: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
\end{aligned}$$
-For $V \to \infty$ we get a Dirac delta function,
-but in fact the conclusion holds for finite $V$ too:
+For $$V \to \infty$$ we get a Dirac delta function,
+but in fact the conclusion holds for finite $$V$$ too:
$$\begin{aligned}
\chi(\vb{q}; t, t')
@@ -106,10 +106,10 @@ $$\begin{aligned}
&= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
\end{aligned}$$
-Similarly, if the unperturbed Hamiltonian $\hat{H}_0$ is time-independent,
-$\chi$ only depends on the time difference $t - t'$.
-Note that $\delta{\Expval{\hat{n}}}$ already has the form of a Fourier transform,
-which gives us an opportunity to rewrite $\chi$
+Similarly, if the unperturbed Hamiltonian $$\hat{H}_0$$ is time-independent,
+$$\chi$$ only depends on the time difference $$t - t'$$.
+Note that $$\delta{\Expval{\hat{n}}}$$ already has the form of a Fourier transform,
+which gives us an opportunity to rewrite $$\chi$$
in the [Lehmann representation](/know/concept/lehmann-representation/):
$$\begin{aligned}
@@ -119,11 +119,11 @@ $$\begin{aligned}
\Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big)
\end{aligned}$$
-Where $\Ket{\nu}$ and $\Ket{\nu'}$ are many-electron eigenstates of $\hat{H}_0$,
-and $Z$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
+Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-electron eigenstates of $$\hat{H}_0$$,
+and $$Z$$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
According to the [convolution theorem](/know/concept/convolution-theorem/)
-$\delta{\Expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$.
-In anticipation, we swap $\nu$ and $\nu''$ in the second term,
+$$\delta{\Expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$$.
+In anticipation, we swap $$\nu$$ and $$\nu''$$ in the second term,
so the general response function is written as:
$$\begin{aligned}
@@ -135,9 +135,9 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$.
+All operators are in the Schrödinger picture from now on, hence we dropped the subscript $$S$$.
-To proceed, we need to rewrite $\hat{n}(\vb{q})$ somehow.
+To proceed, we need to rewrite $$\hat{n}(\vb{q})$$ somehow.
If we neglect electron-electron interactions,
the single-particle states are simply plane waves, in which case:
@@ -154,10 +154,10 @@ $$\begin{aligned}
-Starting from the general definition of $\hat{n}$,
-we write out the field operators $\hat{\Psi}(\vb{r})$,
+Starting from the general definition of $$\hat{n}$$,
+we write out the field operators $$\hat{\Psi}(\vb{r})$$,
and insert the known non-interacting single-electron orbitals
-$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$:
+$$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$$:
$$\begin{aligned}
\hat{n}(\vb{r})
@@ -166,7 +166,7 @@ $$\begin{aligned}
= \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'}
\end{aligned}$$
-Taking the Fourier transfom yields a Dirac delta function $\delta$:
+Taking the Fourier transfom yields a Dirac delta function $$\delta$$:
$$\begin{aligned}
\hat{n}(\vb{q})
@@ -176,19 +176,19 @@ $$\begin{aligned}
\end{aligned}$$
If we impose periodic boundary conditions
-on our $D$-dimensional hypercube of volume $V$,
-then $\vb{k}$ becomes discrete,
-with per-value spacing $2 \pi / V^{1/D}$ along each axis.
-
-Consequently, each orbital $\psi_\vb{k}$ uniquely occupies
-a volume $(2 \pi)^D / V$ in $\vb{k}$-space, so we make the approximation
-$\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}$.
-This becomes exact for $V \to \infty$,
-in which case $\vb{k}$ also becomes continuous again,
+on our $$D$$-dimensional hypercube of volume $$V$$,
+then $$\vb{k}$$ becomes discrete,
+with per-value spacing $$2 \pi / V^{1/D}$$ along each axis.
+
+Consequently, each orbital $$\psi_\vb{k}$$ uniquely occupies
+a volume $$(2 \pi)^D / V$$ in $$\vb{k}$$-space, so we make the approximation
+$$\sum_{\vb{k}} \approx V / (2 \pi)^D \int_{-\infty}^\infty \dd{\vb{k}}$$.
+This becomes exact for $$V \to \infty$$,
+in which case $$\vb{k}$$ also becomes continuous again,
which is what we want for jellium.
-We apply this standard trick from condensed matter physics to $\hat{n}$,
-and $V$ cancels out:
+We apply this standard trick from condensed matter physics to $$\hat{n}$$,
+and $$V$$ cancels out:
$$\begin{aligned}
\hat{n}(\vb{q})
@@ -197,9 +197,9 @@ $$\begin{aligned}
= \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
\end{aligned}$$
-For negated arguments, we simply define $\vb{k}' \equiv \vb{k} - \vb{q}$
-to show that $\hat{n}(-\vb{q}) = \hat{n}{}^\dagger(\vb{q})$,
-which can also be understood as a consequence of $\hat{n}(\vb{r})$ being real:
+For negated arguments, we simply define $$\vb{k}' \equiv \vb{k} - \vb{q}$$
+to show that $$\hat{n}(-\vb{q}) = \hat{n}{}^\dagger(\vb{q})$$,
+which can also be understood as a consequence of $$\hat{n}(\vb{r})$$ being real:
$$\begin{aligned}
\hat{n}(-\vb{q})
@@ -208,13 +208,13 @@ $$\begin{aligned}
= \hat{n}^\dagger(\vb{q})
\end{aligned}$$
-The summation variable $\vb{k}$ has an associated spin $\sigma$,
-and $\hat{n}$ does not carry any spin.
+The summation variable $$\vb{k}$$ has an associated spin $$\sigma$$,
+and $$\hat{n}$$ does not carry any spin.
-When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$.
-We insert $\hat{n}$, suppressing spin:
+When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$.
+We insert $$\hat{n}$$, suppressing spin:
$$\begin{aligned}
\chi_0
@@ -227,16 +227,16 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-Here, $\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}$
-is only nonzero if $\Ket{\nu'}$ is contructed from $\Ket{\nu}$
-by moving an electron from $\vb{k}$ to $\vb{k} \!+\! \vb{q}$,
+Here, $$\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}$$
+is only nonzero if $$\Ket{\nu'}$$ is contructed from $$\Ket{\nu}$$
+by moving an electron from $$\vb{k}$$ to $$\vb{k} \!+\! \vb{q}$$,
and analogously for the other inner products.
-As a result, $\vb{k} = \vb{k}'$ (and $\sigma = \sigma'$).
+As a result, $$\vb{k} = \vb{k}'$$ (and $$\sigma = \sigma'$$).
-For the same reason, the energy difference $E_\nu \!-\! E_{\nu'}$
+For the same reason, the energy difference $$E_\nu \!-\! E_{\nu'}$$
can simply be replaced by the cost of the single-particle excitation
-$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$,
-where $\xi_{\vb{k}}$ is the energy of a $\vb{k}$-orbital.
+$$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$$,
+where $$\xi_{\vb{k}}$$ is the energy of a $$\vb{k}$$-orbital.
Therefore:
$$\begin{aligned}
@@ -250,8 +250,8 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu}
\end{aligned}$$
-Notice that we have eliminated all dependence on $\Ket{\nu'}$,
-so we remove it by $\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$:
+Notice that we have eliminated all dependence on $$\Ket{\nu'}$$,
+so we remove it by $$\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$$:
$$\begin{aligned}
\chi_0
@@ -268,9 +268,9 @@ $$\begin{aligned}
\end{aligned}$$
Where we recognized the commutator,
-and eliminated $E_\nu$ using $\hat{H}_0 \Ket{n} = E_\nu \Ket{\nu}$.
-The resulting expression has the form of a matrix trace $\Tr$
-and a thermal expectation $\Expval{}_0$:
+and eliminated $$E_\nu$$ using $$\hat{H}_0 \Ket{n} = E_\nu \Ket{\nu}$$.
+The resulting expression has the form of a matrix trace $$\Tr$$
+and a thermal expectation $$\Expval{}_0$$:
$$\begin{aligned}
\chi_0
@@ -295,7 +295,7 @@ $$\begin{aligned}
-In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have:
+In general, for any single-particle states labeled by $$m$$, $$n$$, $$o$$ and $$p$$, we have:
$$\begin{aligned}
\comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
&= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n
@@ -317,13 +317,13 @@ $$\begin{aligned}
&= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm}
\end{aligned}$$
-In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$,
+In this case, $$m = p = \vb{k}$$ and $$n = o = \vb{k} \!+\! \vb{q}$$,
so the Kronecker deltas are unnecessary.
-We substitute this result into $\chi_0$,
-and reintroduce the spin index $\sigma$ associated with $\vb{k}$:
+We substitute this result into $$\chi_0$$,
+and reintroduce the spin index $$\sigma$$ associated with $$\vb{k}$$:
$$\begin{aligned}
\chi_0(\vb{q}, \omega)
@@ -332,9 +332,9 @@ $$\begin{aligned}
{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
\end{aligned}$$
-The operator $\hat{c}_{\sigma.\vb{k}}^\dagger \hat{c}_{\sigma.\vb{k}}$
-simply counts the number of electrons in state $(\sigma, \vb{k})$,
-which is given by the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) $n_F$.
+The operator $$\hat{c}_{\sigma.\vb{k}}^\dagger \hat{c}_{\sigma.\vb{k}}$$
+simply counts the number of electrons in state $$(\sigma, \vb{k})$$,
+which is given by the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) $$n_F$$.
This gives us the **Lindhard response function**:
$$\begin{aligned}
@@ -347,8 +347,8 @@ $$\begin{aligned}
\end{aligned}$$
From this, we would like to get the
-[dielectric function](/know/concept/dielectric-function/) $\varepsilon_r$.
-Recall its definition, where $U_\mathrm{tot}$, $U_\mathrm{ext}$, and $U_\mathrm{ind}$
+[dielectric function](/know/concept/dielectric-function/) $$\varepsilon_r$$.
+Recall its definition, where $$U_\mathrm{tot}$$, $$U_\mathrm{ext}$$, and $$U_\mathrm{ind}$$
are the total, external and induced potentials, respectively:
$$\begin{aligned}
@@ -361,17 +361,17 @@ Note that these are all *energy* potentials:
this choice is justified because all energy potentials
are caused by electric fields in this case.
The *electric* potential is recoverable as
-$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$,
-where $q_e < 0$ is the charge of an electron.
-
-From the Lindhard response function $\chi_0$,
-we get the induced particle density offset $\delta{\Expval{\hat{n}}}$
-caused by a potential $U$.
-The density $\delta{\Expval{\hat{n}}}$ should be self-consistent,
-implying $U = U_\mathrm{tot}$.
+$$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$$,
+where $$q_e < 0$$ is the charge of an electron.
+
+From the Lindhard response function $$\chi_0$$,
+we get the induced particle density offset $$\delta{\Expval{\hat{n}}}$$
+caused by a potential $$U$$.
+The density $$\delta{\Expval{\hat{n}}}$$ should be self-consistent,
+implying $$U = U_\mathrm{tot}$$.
In other words, we have a linear relation
-$\delta{\Expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$,
-so the standard formula for $\varepsilon_r$ gives:
+$$\delta{\Expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$$,
+so the standard formula for $$\varepsilon_r$$ gives:
$$\begin{aligned}
\boxed{
@@ -381,7 +381,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$
+Where $$U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$$
is Coulomb repulsion.
This is the **Lindhard dielectric function** of a free
non-interacting electron gas,
diff --git a/source/know/concept/lorentz-force/index.md b/source/know/concept/lorentz-force/index.md
index d1b216d..03178f6 100644
--- a/source/know/concept/lorentz-force/index.md
+++ b/source/know/concept/lorentz-force/index.md
@@ -10,10 +10,10 @@ layout: "concept"
---
The **Lorentz force** is an empirical force used to define
-the [electric field](/know/concept/electric-field/) $\vb{E}$
-and [magnetic field](/know/concept/magnetic-field/) $\vb{B}$.
-For a particle with charge $q$ moving with velocity $\vb{u}$,
-the Lorentz force $\vb{F}$ is given by:
+the [electric field](/know/concept/electric-field/) $$\vb{E}$$
+and [magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$.
+For a particle with charge $$q$$ moving with velocity $$\vb{u}$$,
+the Lorentz force $$\vb{F}$$ is given by:
$$\begin{aligned}
\boxed{
@@ -25,9 +25,9 @@ $$\begin{aligned}
## Uniform electric field
-Consider the simple case of an electric field $\vb{E}$
+Consider the simple case of an electric field $$\vb{E}$$
that is uniform in all of space.
-In the absence of a magnetic field $\vb{B} = 0$
+In the absence of a magnetic field $$\vb{B} = 0$$
and any other forces,
Newton's second law states:
@@ -38,16 +38,16 @@ $$\begin{aligned}
\end{aligned}$$
This is straightforward to integrate in time,
-for a given initial velocity vector $\vb{u}_0$:
+for a given initial velocity vector $$\vb{u}_0$$:
$$\begin{aligned}
\vb{u}(t)
= \frac{q}{m} \vb{E} t + \vb{u}_0
\end{aligned}$$
-And then the particle's position $\vb{x}(t)$
+And then the particle's position $$\vb{x}(t)$$
is found be integrating once more,
-with $\vb{x}(0) = \vb{x}_0$:
+with $$\vb{x}(0) = \vb{x}_0$$:
$$\begin{aligned}
\boxed{
@@ -56,16 +56,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-In summary, unsurprisingly, a uniform electric field $\vb{E}$
-accelerates the particle with a constant force $\vb{F} = q \vb{E}$.
-Note that the direction depends on the sign of $q$.
+In summary, unsurprisingly, a uniform electric field $$\vb{E}$$
+accelerates the particle with a constant force $$\vb{F} = q \vb{E}$$.
+Note that the direction depends on the sign of $$q$$.
## Uniform magnetic field
Consider the simple case of a uniform magnetic field
-$\vb{B} = (0, 0, B)$ in the $z$-direction,
-without an electric field $\vb{E} = 0$.
+$$\vb{B} = (0, 0, B)$$ in the $$z$$-direction,
+without an electric field $$\vb{E} = 0$$.
If there are no other forces,
Newton's second law states:
@@ -76,7 +76,7 @@ $$\begin{aligned}
\end{aligned}$$
Evaluating the cross product yields
-three coupled equations for the components of $\vb{u}$:
+three coupled equations for the components of $$\vb{u}$$:
$$\begin{aligned}
\dv{u_x}{t}
@@ -89,15 +89,15 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Differentiating the first equation with respect to $t$,
-and substituting $\idv{u_y}{t}$ from the second,
+Differentiating the first equation with respect to $$t$$,
+and substituting $$\idv{u_y}{t}$$ from the second,
we arrive at the following harmonic oscillator:
$$\begin{aligned}
\dvn{2}{u_x}{t} = - \omega_c^2 u_x
\end{aligned}$$
-Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
+Where we have defined the **cyclotron frequency** $$\omega_c$$ as follows,
which may be negative:
$$\begin{aligned}
@@ -108,15 +108,15 @@ $$\begin{aligned}
\end{aligned}$$
Suppose we choose our initial conditions so that
-the solution for $u_x(t)$ is given by:
+the solution for $$u_x(t)$$ is given by:
$$\begin{aligned}
u_x(t)
= u_\perp \cos(\omega_c t)
\end{aligned}$$
-Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity.
-Then $u_y(t)$ is found to be:
+Where $$u_\perp \equiv \sqrt{u_x^2 + u_y^2}$$ is the constant total transverse velocity.
+Then $$u_y(t)$$ is found to be:
$$\begin{aligned}
u_y(t)
@@ -126,10 +126,10 @@ $$\begin{aligned}
\end{aligned}$$
This means that the particle moves in a circle,
-in a direction determined by the sign of $\omega_c$.
+in a direction determined by the sign of $$\omega_c$$.
Integrating the velocity yields the position,
-where we refer to the integration constants $x_{gc}$ and $y_{gc}$
+where we refer to the integration constants $$x_{gc}$$ and $$y_{gc}$$
as the **guiding center**, around which the particle orbits or **gyrates**:
$$\begin{aligned}
@@ -140,7 +140,7 @@ $$\begin{aligned}
= \frac{u_\perp}{\omega_c} \cos(\omega_c t) + y_{gc}
\end{aligned}$$
-The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:
+The radius of this orbit is known as the **Larmor radius** or **gyroradius** $$r_L$$, given by:
$$\begin{aligned}
\boxed{
@@ -151,7 +151,7 @@ $$\begin{aligned}
\end{aligned}$$
Finally, it is easy to integrate the equation
-for the $z$-axis velocity $u_z$, which is conserved:
+for the $$z$$-axis velocity $$u_z$$, which is conserved:
$$\begin{aligned}
z(t)
@@ -159,9 +159,9 @@ $$\begin{aligned}
= u_z t + z_0
\end{aligned}$$
-In conclusion, the particle's motion parallel to $\vb{B}$
+In conclusion, the particle's motion parallel to $$\vb{B}$$
is not affected by the magnetic field,
-while its motion perpendicular to $\vb{B}$
+while its motion perpendicular to $$\vb{B}$$
is circular around an imaginary guiding center.
The end result is that particles follow a helical path
when moving through a uniform magnetic field:
@@ -177,9 +177,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$
+Where $$\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$$
is the position of the guiding center.
-For a detailed look at how $\vb{B}$ and $\vb{E}$
+For a detailed look at how $$\vb{B}$$ and $$\vb{E}$$
can affect the guiding center's motion,
see [guiding center theory](/know/concept/guiding-center-theory/).
diff --git a/source/know/concept/lubrication-theory/index.md b/source/know/concept/lubrication-theory/index.md
index d2a615e..4015526 100644
--- a/source/know/concept/lubrication-theory/index.md
+++ b/source/know/concept/lubrication-theory/index.md
@@ -16,15 +16,15 @@ is the study of fluids that are tightly constrained in one dimension,
especially those in small gaps between moving surfaces.
For simplicity, we limit ourselves to 2D
-by assuming that everything is constant along the $z$-axis.
-Consider a gap of width $d$ (along $y$) and length $L$ (along $x$),
-where $d \ll L$, containing the fluid.
+by assuming that everything is constant along the $$z$$-axis.
+Consider a gap of width $$d$$ (along $$y$$) and length $$L$$ (along $$x$$),
+where $$d \ll L$$, containing the fluid.
Outside the gap, the lubricant has a
-[Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \approx U L / \nu$.
+[Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re} \approx U L / \nu$$.
-Inside the gap, the Reynolds number $\mathrm{Re}_\mathrm{gap}$ is different.
-This is because advection will dominate along the $x$-axis (gap length),
-and viscosity along the $y$-axis (gap width).
+Inside the gap, the Reynolds number $$\mathrm{Re}_\mathrm{gap}$$ is different.
+This is because advection will dominate along the $$x$$-axis (gap length),
+and viscosity along the $$y$$-axis (gap width).
Therefore:
$$\begin{aligned}
@@ -34,13 +34,13 @@ $$\begin{aligned}
\approx \frac{d^2}{L^2} \mathrm{Re}
\end{aligned}$$
-If $d$ is small enough compared to $L$,
-then $\mathrm{Re}_\mathrm{gap} \ll 1$.
-More formally, we need $d \ll L / \sqrt{\mathrm{Re}}$,
+If $$d$$ is small enough compared to $$L$$,
+then $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
+More formally, we need $$d \ll L / \sqrt{\mathrm{Re}}$$,
so we are inside the boundary layer,
in the realm of the [Prandtl equations](/know/concept/prandtl-equations/).
-Let $\mathrm{Re}_\mathrm{gap} \ll 1$.
+Let $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
We are thus dealing with *Stokes flow*, in which case
the [Navier-Stokes equations](/know/concept/navier-stokes/equations/)
can be reduced to the following *Stokes equations*:
@@ -53,16 +53,16 @@ $$\begin{aligned}
= \eta \: \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big)
\end{aligned}$$
-Let the $y = 0$ plane be an infinite flat surface,
-sliding in the positive $x$-direction at a constant velocity $U$.
+Let the $$y = 0$$ plane be an infinite flat surface,
+sliding in the positive $$x$$-direction at a constant velocity $$U$$.
On the other side of the gap,
-an arbitrary surface is described by $h(x)$.
+an arbitrary surface is described by $$h(x)$$.
Since the gap is so narrow,
and the surfaces' movements cause large shear stresses inside,
-$v_y$ is negligible compared to $v_x$.
+$$v_y$$ is negligible compared to $$v_x$$.
Furthermore, because the gap is so long,
-we assume that $\ipdv{v_x}{x}$ is negligible compared to $\ipdv{v_x}{y}$.
+we assume that $$\ipdv{v_x}{x}$$ is negligible compared to $$\ipdv{v_x}{y}$$.
This reduces the Stokes equations to:
$$\begin{aligned}
@@ -74,7 +74,7 @@ $$\begin{aligned}
\end{aligned}$$
This result could also be derived from the Prandtl equations.
-In any case, it tells us that $p$ only depends on $x$,
+In any case, it tells us that $$p$$ only depends on $$x$$,
allowing us to integrate the former equation:
$$\begin{aligned}
@@ -82,17 +82,17 @@ $$\begin{aligned}
= \frac{p'}{2 \eta} y^2 + C_1 y + C_2
\end{aligned}$$
-Where $C_1$ and $C_2$ are integration constants.
-At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$.
-Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to:
+Where $$C_1$$ and $$C_2$$ are integration constants.
+At $$y = 0$$, the viscous *no-slip* condition demands that $$v_x = U$$, so $$C_2 = U$$.
+Likewise, at $$y = h(x)$$, we need $$v_x = 0$$, leading us to:
$$\begin{aligned}
v_x
= \frac{p'}{2 \eta} y^2 - \Big( \frac{p'}{2 \eta} h + \frac{U}{h} \Big) y + U
\end{aligned}$$
-The moving bottom surface drags fluid in the $x$-direction
-at a volumetric rate $Q$, given by:
+The moving bottom surface drags fluid in the $$x$$-direction
+at a volumetric rate $$Q$$, given by:
$$\begin{aligned}
Q
@@ -103,9 +103,9 @@ $$\begin{aligned}
Assuming that the lubricant is incompressible,
meaning that the same volume of fluid must be leaving a point as is entering it.
-In other words, $Q$ is independent of $x$,
-which allows us to write $p'(x)$ in terms of
-measurable constants and the known function $h(x)$:
+In other words, $$Q$$ is independent of $$x$$,
+which allows us to write $$p'(x)$$ in terms of
+measurable constants and the known function $$h(x)$$:
$$\begin{aligned}
\boxed{
@@ -114,7 +114,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Then we insert this into our earlier expression for $v_x$, yielding:
+Then we insert this into our earlier expression for $$v_x$$, yielding:
$$\begin{aligned}
v_x
@@ -130,7 +130,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-With this, we can find $v_y$ by exploiting incompressibility,
+With this, we can find $$v_y$$ by exploiting incompressibility,
i.e. the continuity equation states:
$$\begin{aligned}
@@ -139,7 +139,7 @@ $$\begin{aligned}
= - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big)
\end{aligned}$$
-Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$:
+Integrating with respect to $$y$$ thus leads to the following transverse velocity $$v_y$$:
$$\begin{aligned}
\boxed{
@@ -150,7 +150,7 @@ $$\begin{aligned}
Typically, the lubricant is not in a preexisting pressure differential,
i.e it is not getting pumped through the system.
-Although the pressure gradient $p'$ need not be zero,
+Although the pressure gradient $$p'$$ need not be zero,
we therefore expect that its integral vanishes:
$$\begin{aligned}
@@ -159,7 +159,7 @@ $$\begin{aligned}
= 6 \eta U \int_L \frac{1}{h(x)^2} \dd{x} - 12 \eta Q \int_L \frac{1}{h(x)^3} \dd{x}
\end{aligned}$$
-Isolating this for $Q$, and defining $q$ as below, yields a simple equation:
+Isolating this for $$Q$$, and defining $$q$$ as below, yields a simple equation:
$$\begin{aligned}
Q
@@ -169,7 +169,7 @@ $$\begin{aligned}
\equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}}
\end{aligned}$$
-We substitute this into $v_x$ and rearrange to get an interesting expression:
+We substitute this into $$v_x$$ and rearrange to get an interesting expression:
$$\begin{aligned}
v_x
@@ -180,7 +180,7 @@ $$\begin{aligned}
The first factor is always positive,
but the second can be negative,
-if for some $y$-values:
+if for some $$y$$-values:
$$\begin{aligned}
h^2 < 3 y (h - q)
@@ -188,8 +188,8 @@ $$\begin{aligned}
y > \frac{h^2}{3 (h - q)}
\end{aligned}$$
-Since $h > y$, such $y$-values will only exist
-if $h$ is larger than some threshold:
+Since $$h > y$$, such $$y$$-values will only exist
+if $$h$$ is larger than some threshold:
$$\begin{aligned}
3 (h - q) > h
@@ -201,7 +201,7 @@ If this condition is satisfied,
there will be some flow reversal:
rather than just getting dragged by the shearing motion,
the lubricant instead "rolls" inside the gap.
-This is confirmed by $v_y$:
+This is confirmed by $$v_y$$:
$$\begin{aligned}
v_y
diff --git a/source/know/concept/magnetic-field/index.md b/source/know/concept/magnetic-field/index.md
index 369c8a6..8d215fb 100644
--- a/source/know/concept/magnetic-field/index.md
+++ b/source/know/concept/magnetic-field/index.md
@@ -8,28 +8,28 @@ categories:
layout: "concept"
---
-The **magnetic field** $\vb{B}$ is a vector field
+The **magnetic field** $$\vb{B}$$ is a vector field
that describes magnetic effects,
and is defined as the field that correctly predicts
the [Lorentz force](/know/concept/lorentz-force/)
-on a particle with electric charge $q$:
+on a particle with electric charge $$q$$:
$$\begin{aligned}
\vb{F}
= q \vb{v} \cross \vb{B}
\end{aligned}$$
-If an object is placed in a magnetic field $\vb{B}$,
+If an object is placed in a magnetic field $$\vb{B}$$,
and wants to rotate to align itself with the field,
-then its **magnetic dipole moment** $\vb{m}$
-is defined from the aligning torque $\vb{\tau}$:
+then its **magnetic dipole moment** $$\vb{m}$$
+is defined from the aligning torque $$\vb{\tau}$$:
$$\begin{aligned}
\vb{\tau} = \vb{m} \times \vb{B}
\end{aligned}$$
-Where $\vb{m}$ has units of $\mathrm{J / T}$.
-From this, the **magnetization** $\vb{M}$ is defined as follows,
+Where $$\vb{m}$$ has units of $$\mathrm{J / T}$$.
+From this, the **magnetization** $$\vb{M}$$ is defined as follows,
and roughly represents the moments per unit volume:
$$\begin{aligned}
@@ -38,17 +38,17 @@ $$\begin{aligned}
\vb{m} = \int_V \vb{M} \dd{V}
\end{aligned}$$
-If $\vb{M}$ has the same magnitude and orientation throughout the body,
-then $\vb{m} = \vb{M} V$, where $V$ is the volume.
-Therefore, $\vb{M}$ has units of $\mathrm{A / m}$.
+If $$\vb{M}$$ has the same magnitude and orientation throughout the body,
+then $$\vb{m} = \vb{M} V$$, where $$V$$ is the volume.
+Therefore, $$\vb{M}$$ has units of $$\mathrm{A / m}$$.
-A nonzero $\vb{M}$ complicates things,
+A nonzero $$\vb{M}$$ complicates things,
since it contributes to the field
-and hence modifies $\vb{B}$.
+and hence modifies $$\vb{B}$$.
We thus define
-the "free" **auxiliary field** $\vb{H}$
-from the "bound" field $\vb{M}$
-and the "net" field $\vb{B}$:
+the "free" **auxiliary field** $$\vb{H}$$
+from the "bound" field $$\vb{M}$$
+and the "net" field $$\vb{B}$$:
$$\begin{aligned}
\vb{H} \equiv \frac{1}{\mu_0} \vb{B} - \vb{M}
@@ -56,24 +56,24 @@ $$\begin{aligned}
\vb{B} = \mu_0 (\vb{H} + \vb{M})
\end{aligned}$$
-Where the **magnetic permeability of free space** $\mu_0$ is a known constant.
+Where the **magnetic permeability of free space** $$\mu_0$$ is a known constant.
It is important to point out some inconsistencies here:
-$\vb{B}$ contains a factor of $\mu_0$, and thus measures **flux density**,
-while $\vb{H}$ and $\vb{M}$ do not contain $\mu_0$,
+$$\vb{B}$$ contains a factor of $$\mu_0$$, and thus measures **flux density**,
+while $$\vb{H}$$ and $$\vb{M}$$ do not contain $$\mu_0$$,
and therefore measure **field intensity**.
Note that this convention is the opposite of the analogous
[electric fields](/know/concept/electric-field/)
-$\vb{E}$, $\vb{D}$ and $\vb{P}$.
-Also note that $\vb{P}$ has the opposite sign convention of $\vb{M}$.
+$$\vb{E}$$, $$\vb{D}$$ and $$\vb{P}$$.
+Also note that $$\vb{P}$$ has the opposite sign convention of $$\vb{M}$$.
Some objects, called **ferromagnets** or **permanent magnets**,
-have an inherently nonzero $\vb{M}$.
-Others objects, when placed in a $\vb{B}$-field,
-may instead gain an induced $\vb{M}$.
+have an inherently nonzero $$\vb{M}$$.
+Others objects, when placed in a $$\vb{B}$$-field,
+may instead gain an induced $$\vb{M}$$.
-When $\vb{M}$ is induced,
+When $$\vb{M}$$ is induced,
its magnitude is usually proportional
-to the applied field strength $\vb{H}$:
+to the applied field strength $$\vb{H}$$:
$$\begin{aligned}
\vb{B}
@@ -83,20 +83,20 @@ $$\begin{aligned}
= \mu \vb{H}
\end{aligned}$$
-Where $\chi_m$ is the **volume magnetic susceptibility**,
-and $\mu_r \equiv 1 + \chi_m$ and $\mu \equiv \mu_r \mu_0$ are
+Where $$\chi_m$$ is the **volume magnetic susceptibility**,
+and $$\mu_r \equiv 1 + \chi_m$$ and $$\mu \equiv \mu_r \mu_0$$ are
the **relative permeability** and **absolute permeability**
of the medium, respectively.
Materials with intrinsic magnetization, i.e. ferromagnets,
-do not have a well-defined $\chi_m$.
+do not have a well-defined $$\chi_m$$.
-If $\chi_m > 0$, the medium is **paramagnetic**,
-meaning it strengthens the net field $\vb{B}$.
-Otherwise, if $\chi_m < 0$, the medium is **diamagnetic**,
-meaning it counteracts the applied field $\vb{H}$.
+If $$\chi_m > 0$$, the medium is **paramagnetic**,
+meaning it strengthens the net field $$\vb{B}$$.
+Otherwise, if $$\chi_m < 0$$, the medium is **diamagnetic**,
+meaning it counteracts the applied field $$\vb{H}$$.
-For $|\chi_m| \ll 1$, as is often the case,
-the magnetization $\vb{M}$ can be approximated by:
+For $$|\chi_m| \ll 1$$, as is often the case,
+the magnetization $$\vb{M}$$ can be approximated by:
$$\begin{aligned}
\vb{M}
diff --git a/source/know/concept/magnetohydrodynamics/index.md b/source/know/concept/magnetohydrodynamics/index.md
index 115ce04..bcc23f3 100644
--- a/source/know/concept/magnetohydrodynamics/index.md
+++ b/source/know/concept/magnetohydrodynamics/index.md
@@ -19,8 +19,8 @@ but the results are not specific to plasmas.
In the two-fluid model, we described the plasma as two separate fluids,
but in MHD we treat it as a single conductive fluid.
-The macroscopic pressure $p$
-and electric current density $\vb{J}$ are:
+The macroscopic pressure $$p$$
+and electric current density $$\vb{J}$$ are:
$$\begin{aligned}
p
@@ -30,8 +30,8 @@ $$\begin{aligned}
= q_i n_i \vb{u}_i + q_e n_e \vb{u}_e
\end{aligned}$$
-Meanwhile, the macroscopic mass density $\rho$
-and center-of-mass flow velocity $\vb{u}$
+Meanwhile, the macroscopic mass density $$\rho$$
+and center-of-mass flow velocity $$\vb{u}$$
are as follows, although the ions dominate due to their large mass:
$$\begin{aligned}
@@ -76,8 +76,8 @@ $$\begin{aligned}
We will assume that electrons' inertia
is negligible compared to the [Lorentz force](/know/concept/lorentz-force/).
-Let $\tau_\mathrm{char}$ be the characteristic timescale of the plasma's dynamics,
-i.e. nothing noticable happens in times shorter than $\tau_\mathrm{char}$,
+Let $$\tau_\mathrm{char}$$ be the characteristic timescale of the plasma's dynamics,
+i.e. nothing noticable happens in times shorter than $$\tau_\mathrm{char}$$,
then this assumption can be written as:
$$\begin{aligned}
@@ -89,10 +89,10 @@ $$\begin{aligned}
\ll 1
\end{aligned}$$
-Where we have recognized the cyclotron frequency $\omega_c$ (see Lorentz force article).
+Where we have recognized the cyclotron frequency $$\omega_c$$ (see Lorentz force article).
In other words, our assumption is equivalent to
-the electron gyration period $2 \pi / \omega_{ce}$
-being small compared to the macroscopic dynamics' timescale $\tau_\mathrm{char}$.
+the electron gyration period $$2 \pi / \omega_{ce}$$
+being small compared to the macroscopic dynamics' timescale $$\tau_\mathrm{char}$$.
By construction, we can thus ignore the left-hand side
of the electron momentum equation, leaving:
@@ -105,7 +105,7 @@ $$\begin{aligned}
\end{aligned}$$
We add up these momentum equations,
-recognizing the pressure $p$ and current $\vb{J}$:
+recognizing the pressure $$p$$ and current $$\vb{J}$$:
$$\begin{aligned}
m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
@@ -115,7 +115,7 @@ $$\begin{aligned}
&= (q_i n_i + q_e n_e) \vb{E} + \vb{J} \cross \vb{B} - \nabla p
\end{aligned}$$
-Where we have used $f_{ie} m_i n_i = f_{ei} m_e n_e$
+Where we have used $$f_{ie} m_i n_i = f_{ei} m_e n_e$$
because momentum is conserved by the underlying
[Rutherford scattering](/know/concept/rutherford-scattering/) process,
which is [elastic](/know/concept/elastic-collision/).
@@ -123,10 +123,10 @@ In other words, the momentum given by ions to electrons
is equal to the momentum received by electrons from ions.
Since the two-fluid model assumes that
-the [Debye length](/know/concept/debye-length/) $\lambda_D$
-is small compared to a "blob" $\dd{V}$ of the fluid,
-we can invoke quasi-neutrality $q_i n_i + q_e n_e = 0$.
-Using that $\rho \approx m_i n_i$ and $\vb{u} \approx \vb{u}_i$,
+the [Debye length](/know/concept/debye-length/) $$\lambda_D$$
+is small compared to a "blob" $$\dd{V}$$ of the fluid,
+we can invoke quasi-neutrality $$q_i n_i + q_e n_e = 0$$.
+Using that $$\rho \approx m_i n_i$$ and $$\vb{u} \approx \vb{u}_i$$,
we thus arrive at the **momentum equation**:
$$\begin{aligned}
@@ -147,8 +147,8 @@ $$\begin{aligned}
= \frac{f_{ei} m_e}{q_e} (\vb{u}_e - \vb{u}_i)
\end{aligned}$$
-Again using quasi-neutrality $q_i n_i = - q_e n_e$,
-the current density $\vb{J} = q_e n_e (\vb{u}_e \!-\! \vb{u}_i)$,
+Again using quasi-neutrality $$q_i n_i = - q_e n_e$$,
+the current density $$\vb{J} = q_e n_e (\vb{u}_e \!-\! \vb{u}_i)$$,
so:
$$\begin{aligned}
@@ -159,13 +159,13 @@ $$\begin{aligned}
\equiv \frac{f_{ei} m_e}{n_e q_e^2}
\end{aligned}$$
-Where $\eta$ is the electrical resistivity of the plasma,
+Where $$\eta$$ is the electrical resistivity of the plasma,
see [Spitzer resistivity](/know/concept/spitzer-resistivity/)
for more information, and a rough estimate of this quantity for a plasma.
-Now, using that $\vb{u} \approx \vb{u}_i$,
-we add $(\vb{u} \!-\! \vb{u}_i) \cross \vb{B} \approx 0$ to the equation,
-and insert $\vb{J}$ again:
+Now, using that $$\vb{u} \approx \vb{u}_i$$,
+we add $$(\vb{u} \!-\! \vb{u}_i) \cross \vb{B} \approx 0$$ to the equation,
+and insert $$\vb{J}$$ again:
$$\begin{aligned}
\eta \vb{J}
@@ -185,8 +185,8 @@ $$\begin{aligned}
Where we have used Faraday's law.
This is the **induction equation**,
-and is used to compute $\vb{B}$.
-The pressure term can be rewritten using the ideal gas law $p_e = k_B T_e n_e$:
+and is used to compute $$\vb{B}$$.
+The pressure term can be rewritten using the ideal gas law $$p_e = k_B T_e n_e$$:
$$\begin{aligned}
\nabla \cross \frac{\nabla p_e}{q_e n_e}
@@ -195,8 +195,8 @@ $$\begin{aligned}
\end{aligned}$$
The curl of a gradient is always zero,
-and we notice that $\nabla n_e / n_e = \nabla\! \ln(n_e)$.
-Then we use the vector identity $\nabla \cross (f \nabla g) = \nabla f \cross \nabla g$,
+and we notice that $$\nabla n_e / n_e = \nabla\! \ln(n_e)$$.
+Then we use the vector identity $$\nabla \cross (f \nabla g) = \nabla f \cross \nabla g$$,
leading to:
$$\begin{aligned}
@@ -206,10 +206,10 @@ $$\begin{aligned}
= \frac{k_B}{q_e n_e} \big( \nabla T_e \cross \nabla n_e \big)
\end{aligned}$$
-It is reasonable to assume that $\nabla T_e$ and $\nabla n_e$
+It is reasonable to assume that $$\nabla T_e$$ and $$\nabla n_e$$
point in roughly the same direction,
in which case the pressure term can be neglected.
-Consequently, $p_e$ has no effect on the dynamics of $\vb{B}$,
+Consequently, $$p_e$$ has no effect on the dynamics of $$\vb{B}$$,
so we argue that it can be dropped from the original (non-curled) equation too, leaving:
$$\begin{aligned}
@@ -220,7 +220,7 @@ $$\begin{aligned}
\end{aligned}$$
This is known as the **generalized Ohm's law**,
-since it contains the relation $\vb{E} = \eta \vb{J}$.
+since it contains the relation $$\vb{E} = \eta \vb{J}$$.
Next, consider [Ampère's law](/know/concept/maxwells-equations/),
where we would like to neglect the last term:
@@ -230,9 +230,9 @@ $$\begin{aligned}
= \mu_0 \vb{J} + \frac{1}{c^2} \pdv{\vb{E}}{t}
\end{aligned}$$
-From Faraday's law, we can obtain a scale estimate for $\vb{E}$.
-Recall that $\tau_\mathrm{char}$ is the characteristic timescale of the plasma,
-and let $\lambda_\mathrm{char} \gg \lambda_D$ be its characteristic lengthscale:
+From Faraday's law, we can obtain a scale estimate for $$\vb{E}$$.
+Recall that $$\tau_\mathrm{char}$$ is the characteristic timescale of the plasma,
+and let $$\lambda_\mathrm{char} \gg \lambda_D$$ be its characteristic lengthscale:
$$\begin{aligned}
\nabla \cross \vb{E}
@@ -244,8 +244,8 @@ $$\begin{aligned}
From this, we find when we can neglect
the last term in Ampère's law:
-the characteristic velocity $v_\mathrm{char}$
-must be tiny compared to $c$,
+the characteristic velocity $$v_\mathrm{char}$$
+must be tiny compared to $$c$$,
i.e. the plasma must be non-relativistic:
$$\begin{aligned}
@@ -284,7 +284,7 @@ $$\begin{aligned}
The continuity equation allows us to rewrite
the [material derivative](/know/concept/material-derivative/)
-$\mathrm{D} \rho / \mathrm{D} t$ as follows:
+$$\mathrm{D} \rho / \mathrm{D} t$$ as follows:
$$\begin{aligned}
\pdv{\rho}{t} + \nabla \cdot (\rho \vb{u})
@@ -294,7 +294,7 @@ $$\begin{aligned}
\end{aligned}$$
Inserting this into the equation of state
-leads us to a differential equation for $p$:
+leads us to a differential equation for $$p$$:
$$\begin{aligned}
0
@@ -307,8 +307,8 @@ $$\begin{aligned}
This closes the set of 14 MHD equations for 14 unknowns.
Originally, the two-fluid model had 16 of each,
-but we have merged $n_i$ and $n_e$ into $\rho$,
-and $p_i$ and $p_i$ into $p$.
+but we have merged $$n_i$$ and $$n_e$$ into $$\rho$$,
+and $$p_i$$ and $$p_i$$ into $$p$$.
## Ohm's law variants
@@ -323,7 +323,7 @@ $$\begin{aligned}
However, most authors neglect some of its terms:
this form is used for **Hall MHD**,
-where $\vb{J} \cross \vb{B}$ is called the *Hall term*.
+where $$\vb{J} \cross \vb{B}$$ is called the *Hall term*.
This term can be dropped in any of the following cases:
$$\begin{gathered}
@@ -342,11 +342,11 @@ $$\begin{gathered}
\ll 1
\end{gathered}$$
-Where we have used the MHD momentum equation with $\nabla p \approx 0$
-to obtain the scale estimate $\vb{J} \cross \vb{B} \sim \rho v_\mathrm{char} / \tau_\mathrm{char}$.
-In other words, if the ion gyration period is short $\tau_\mathrm{char} \gg \omega_{ci}$,
+Where we have used the MHD momentum equation with $$\nabla p \approx 0$$
+to obtain the scale estimate $$\vb{J} \cross \vb{B} \sim \rho v_\mathrm{char} / \tau_\mathrm{char}$$.
+In other words, if the ion gyration period is short $$\tau_\mathrm{char} \gg \omega_{ci}$$,
and/or if the electron gyration period is long
-compared to the electron-ion collision period $\omega_{ce} \ll f_{ei}$,
+compared to the electron-ion collision period $$\omega_{ce} \ll f_{ei}$$,
then we are left with this form of Ohm's law, used in **resistive MHD**:
$$\begin{aligned}
@@ -354,10 +354,10 @@ $$\begin{aligned}
= \eta \vb{J}
\end{aligned}$$
-Finally, we can neglect the resisitive term $\eta \vb{J}$
+Finally, we can neglect the resisitive term $$\eta \vb{J}$$
if the Lorentz force is much larger.
We formalize this condition as follows,
-where we have used Ampère's law to find $\vb{J} \sim \vb{B} / \mu_0 \lambda_\mathrm{char}$:
+where we have used Ampère's law to find $$\vb{J} \sim \vb{B} / \mu_0 \lambda_\mathrm{char}$$:
$$\begin{aligned}
1
@@ -368,8 +368,8 @@ $$\begin{aligned}
\gg 1
\end{aligned}$$
-Where we have defined the **magnetic Reynolds number** $\mathrm{R_m}$ as follows,
-which is analogous to the fluid [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re}$:
+Where we have defined the **magnetic Reynolds number** $$\mathrm{R_m}$$ as follows,
+which is analogous to the fluid [Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re}$$:
$$\begin{aligned}
\boxed{
@@ -378,9 +378,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\mathrm{R_m} \ll 1$, the plasma is "electrically viscous",
+If $$\mathrm{R_m} \ll 1$$, the plasma is "electrically viscous",
such that resistivity needs to be accounted for,
-whereas if $\mathrm{R_m} \gg 1$, the resistivity is negligible,
+whereas if $$\mathrm{R_m} \gg 1$$, the resistivity is negligible,
in which case we have **ideal MHD**:
$$\begin{aligned}
diff --git a/source/know/concept/markov-process/index.md b/source/know/concept/markov-process/index.md
index fd6b076..c938866 100644
--- a/source/know/concept/markov-process/index.md
+++ b/source/know/concept/markov-process/index.md
@@ -9,37 +9,37 @@ layout: "concept"
---
Given a [stochastic process](/know/concept/stochastic-process/)
-$\{X_t : t \ge 0\}$ on a filtered probability space
-$(\Omega, \mathcal{F}, \{\mathcal{F}_t\}, P)$,
+$$\{X_t : t \ge 0\}$$ on a filtered probability space
+$$(\Omega, \mathcal{F}, \{\mathcal{F}_t\}, P)$$,
it is said to be a **Markov process**
if it satisfies the following requirements:
-1. $X_t$ is $\mathcal{F}_t$-adapted,
- meaning that the current and all past values of $X_t$
- can be reconstructed from the filtration $\mathcal{F}_t$.
-2. For some function $h(x)$,
+1. $$X_t$$ is $$\mathcal{F}_t$$-adapted,
+ meaning that the current and all past values of $$X_t$$
+ can be reconstructed from the filtration $$\mathcal{F}_t$$.
+2. For some function $$h(x)$$,
the [conditional expectation](/know/concept/conditional-expectation/)
- $\mathbf{E}[h(X_t) | \mathcal{F}_s] = \mathbf{E}[h(X_t) | X_s]$,
- i.e. at time $s \le t$, the expectation of $h(X_t)$ depends only on the current $X_s$.
- Note that $h$ must be bounded and *Borel-measurable*,
- meaning $\sigma(h(X_t)) \subseteq \mathcal{F}_t$.
+ $$\mathbf{E}[h(X_t) | \mathcal{F}_s] = \mathbf{E}[h(X_t) | X_s]$$,
+ i.e. at time $$s \le t$$, the expectation of $$h(X_t)$$ depends only on the current $$X_s$$.
+ Note that $$h$$ must be bounded and *Borel-measurable*,
+ meaning $$\sigma(h(X_t)) \subseteq \mathcal{F}_t$$.
This last condition is called the **Markov property**,
-and demands that the future of $X_t$ does not depend on the past,
-but only on the present $X_s$.
+and demands that the future of $$X_t$$ does not depend on the past,
+but only on the present $$X_s$$.
-If both $t$ and $X_t$ are taken to be discrete,
-then $X_t$ is known as a **Markov chain**.
+If both $$t$$ and $$X_t$$ are taken to be discrete,
+then $$X_t$$ is known as a **Markov chain**.
This brings us to the concept of the **transition probability**
-$P(X_t \in A | X_s = x)$, which describes the probability that
-$X_t$ will be in a given set $A$, if we know that currently $X_s = x$.
-
-If $t$ and $X_t$ are continuous, we can often (but not always) express $P$
-using a **transition density** $p(s, x; t, y)$,
-which gives the probability density that the initial condition $X_s = x$
-will evolve into the terminal condition $X_t = y$.
-Then the transition probability $P$ can be calculated like so,
-where $B$ is a given Borel set (see [$\sigma$-algebra](/know/concept/sigma-algebra/)):
+$$P(X_t \in A | X_s = x)$$, which describes the probability that
+$$X_t$$ will be in a given set $$A$$, if we know that currently $$X_s = x$$.
+
+If $$t$$ and $$X_t$$ are continuous, we can often (but not always) express $$P$$
+using a **transition density** $$p(s, x; t, y)$$,
+which gives the probability density that the initial condition $$X_s = x$$
+will evolve into the terminal condition $$X_t = y$$.
+Then the transition probability $$P$$ can be calculated like so,
+where $$B$$ is a given Borel set (see [$$\sigma$$-algebra](/know/concept/sigma-algebra/)):
$$\begin{aligned}
P(X_t \in B | X_s = x)
diff --git a/source/know/concept/martingale/index.md b/source/know/concept/martingale/index.md
index a54320f..9d3c6b4 100644
--- a/source/know/concept/martingale/index.md
+++ b/source/know/concept/martingale/index.md
@@ -13,25 +13,25 @@ A **martingale** is a type of
with important and useful properties,
especially for stochastic calculus.
-For a stochastic process $\{ M_t : t \ge 0 \}$
-on a probability filtered space $(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$,
-then $M_t$ is a martingale if it satisfies all of the following:
-
-1. $M_t$ is $\mathcal{F}_t$-adapted, meaning
- the filtration $\mathcal{F}_t$ contains enough information
- to reconstruct the current and all past values of $M_t$.
-2. For all times $t \ge 0$, the expectation value exists $\mathbf{E}(M_t) < \infty$.
-3. For all $s, t$ satisfying $0 \le s \le t$,
+For a stochastic process $$\{ M_t : t \ge 0 \}$$
+on a probability filtered space $$(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$$,
+then $$M_t$$ is a martingale if it satisfies all of the following:
+
+1. $$M_t$$ is $$\mathcal{F}_t$$-adapted, meaning
+ the filtration $$\mathcal{F}_t$$ contains enough information
+ to reconstruct the current and all past values of $$M_t$$.
+2. For all times $$t \ge 0$$, the expectation value exists $$\mathbf{E}(M_t) < \infty$$.
+3. For all $$s, t$$ satisfying $$0 \le s \le t$$,
the [conditional expectation](/know/concept/conditional-expectation/)
- $\mathbf{E}(M_t | \mathcal{F}_s) = M_s$,
- meaning the increment $M_t \!-\! M_s$ is always expected
- to be zero $\mathbf{E}(M_t \!-\! M_s | \mathcal{F}_s) = 0$.
+ $$\mathbf{E}(M_t | \mathcal{F}_s) = M_s$$,
+ meaning the increment $$M_t \!-\! M_s$$ is always expected
+ to be zero $$\mathbf{E}(M_t \!-\! M_s | \mathcal{F}_s) = 0$$.
The last condition is called the **martingale property**,
and basically means that a martingale is an unbiased random walk.
-Accordingly, the [Wiener process](/know/concept/wiener-process/) $B_t$
+Accordingly, the [Wiener process](/know/concept/wiener-process/) $$B_t$$
(Brownian motion) is an example of a martingale,
-since each of its increments $B_t \!-\! B_s$ has mean $0$ by definition.
+since each of its increments $$B_t \!-\! B_s$$ has mean $$0$$ by definition.
Martingales are easily confused with
[Markov processes](/know/concept/markov-process/),
@@ -42,15 +42,15 @@ the martingale property says nothing about history-dependence,
and the Markov property does not say *what* the future expectation should be.
Modifying property (3) leads to two common generalizations.
-The stochastic process $M_t$ above is a **submartingale**
+The stochastic process $$M_t$$ above is a **submartingale**
if the current value is a lower bound for the expectation:
-3. For $0 \le s \le t$, the conditional expectation $\mathbf{E}(M_t | \mathcal{F}_s) \ge M_s$.
+3. For $$0 \le s \le t$$, the conditional expectation $$\mathbf{E}(M_t | \mathcal{F}_s) \ge M_s$$.
-Analogouly, $M_t$ is a **supermartingale**
+Analogouly, $$M_t$$ is a **supermartingale**
if the current value is an upper bound instead:
-3. For $0 \le s \le t$, the conditional expectation $\mathbf{E}(M_t | \mathcal{F}_s) \le M_s$.
+3. For $$0 \le s \le t$$, the conditional expectation $$\mathbf{E}(M_t | \mathcal{F}_s) \le M_s$$.
Clearly, submartingales and supermartingales are *biased* random walks,
since they will tend to increase and decrease with time, respectively.
diff --git a/source/know/concept/material-derivative/index.md b/source/know/concept/material-derivative/index.md
index 16c2c66..93e8ad0 100644
--- a/source/know/concept/material-derivative/index.md
+++ b/source/know/concept/material-derivative/index.md
@@ -11,22 +11,22 @@ layout: "concept"
---
Inside a fluid (or any other continuum), we might be interested in
-the time evolution of a certain intensive quantity $f$,
+the time evolution of a certain intensive quantity $$f$$,
e.g. the temperature or pressure,
-represented by a scalar field $f(\va{r}, t)$.
+represented by a scalar field $$f(\va{r}, t)$$.
-If the fluid is static, the evolution of $f$ is simply $\ipdv{f}{t}$,
+If the fluid is static, the evolution of $$f$$ is simply $$\ipdv{f}{t}$$,
since each point of the fluid is motionless.
However, if the fluid is moving, we have a problem:
-the fluid molecules at position $\va{r} = \va{r}_0$ are not necessarily
-the same ones at time $t = t_0$ and $t = t_1$.
-Those molecules take $f$ with them as they move,
+the fluid molecules at position $$\va{r} = \va{r}_0$$ are not necessarily
+the same ones at time $$t = t_0$$ and $$t = t_1$$.
+Those molecules take $$f$$ with them as they move,
so we need to account for this transport somehow.
To do so, we choose an infinitesimal "blob" or **parcel** of the fluid,
which always contains the same specific molecules,
-and track its position $\va{r}(t)$ through time as it moves and deforms.
-The value of $f$ for this parcel is then given by:
+and track its position $$\va{r}(t)$$ through time as it moves and deforms.
+The value of $$f$$ for this parcel is then given by:
$$\begin{aligned}
f(\va{r}, t)
@@ -34,9 +34,9 @@ $$\begin{aligned}
= f\big(x(t), y(t), z(t), t\big)
\end{aligned}$$
-In effect, we have simply made the coordinate $\va{r}$ dependent on time,
+In effect, we have simply made the coordinate $$\va{r}$$ dependent on time,
and have specifically chosen the time-dependence to track the parcel.
-The net evolution of $f$ is then its "true" (i.e. non-partial) derivative with respect to $t$,
+The net evolution of $$f$$ is then its "true" (i.e. non-partial) derivative with respect to $$t$$,
allowing us to apply the chain rule:
$$\begin{aligned}
@@ -46,8 +46,8 @@ $$\begin{aligned}
&= \pdv{f}{t} + v_x \pdv{f}{x} + v_y \pdv{f}{y} + v_z \pdv{f}{z}
\end{aligned}$$
-Where $v_x$, $v_y$ and $v_z$ are the parcel's velocity components.
-Let $\va{v} = (v_x, v_y, v_z)$ be the velocity vector field,
+Where $$v_x$$, $$v_y$$ and $$v_z$$ are the parcel's velocity components.
+Let $$\va{v} = (v_x, v_y, v_z)$$ be the velocity vector field,
then we can rewrite this expression like so:
$$\begin{aligned}
@@ -55,12 +55,12 @@ $$\begin{aligned}
&= \pdv{f}{t} + (\va{v} \cdot \nabla) f
\end{aligned}$$
-Note that $\va{v} = \va{v}(\va{r}, t)$,
-that is, the velocity can change with time ($t$-dependence),
-and depends on which parcel we track ($\va{r}$-dependence).
+Note that $$\va{v} = \va{v}(\va{r}, t)$$,
+that is, the velocity can change with time ($$t$$-dependence),
+and depends on which parcel we track ($$\va{r}$$-dependence).
Of course, the parcel is in our imagination:
-$\va{r}$ does not really depend on $t$;
+$$\va{r}$$ does not really depend on $$t$$;
after all, we are dealing with a continuum.
Nevertheless, the right-hand side of the equation is very useful,
and is known as the **material derivative** or **comoving derivative**:
@@ -75,11 +75,11 @@ $$\begin{aligned}
The first term is called the **local rate of change**,
and the second is the **advective rate of change**.
In effect, the latter moves the frame of reference along with the material,
-so that we can find the evolution of $f$
+so that we can find the evolution of $$f$$
without needing to worry about the continuum's motion.
-That was for a scalar field $f(\va{r}, t)$,
-but in fact the definition also works for vector fields $\va{U}(\va{r}, t)$:
+That was for a scalar field $$f(\va{r}, t)$$,
+but in fact the definition also works for vector fields $$\va{U}(\va{r}, t)$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/matsubara-greens-function/index.md b/source/know/concept/matsubara-greens-function/index.md
index 81dd360..fdcadb3 100644
--- a/source/know/concept/matsubara-greens-function/index.md
+++ b/source/know/concept/matsubara-greens-function/index.md
@@ -21,10 +21,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the expectation value $\Expval{}$ is with respect to thermodynamic equilibrium,
-and $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator.
-Because the Hamiltonian $\hat{H}$ cannot depend on the imaginary time,
-$C_{AB}$ is a function of the difference $\tau \!-\! \tau'$ only:
+Where the expectation value $$\Expval{}$$ is with respect to thermodynamic equilibrium,
+and $$\mathcal{T}$$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator.
+Because the Hamiltonian $$\hat{H}$$ cannot depend on the imaginary time,
+$$C_{AB}$$ is a function of the difference $$\tau \!-\! \tau'$$ only:
$$\begin{aligned}
C_{AB}(\tau, \tau')
@@ -36,9 +36,9 @@ $$\begin{aligned}
&= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} \Big)
\end{aligned}$$
-For $\tau > \tau'$, we see by expanding in the many-particle eigenstates $\Ket{n}$
-that we need to demand $\hbar \beta > \tau \!-\! \tau'$ to prevent
-$C_{AB}$ from diverging for increasing temperatures:
+For $$\tau > \tau'$$, we see by expanding in the many-particle eigenstates $$\Ket{n}$$
+that we need to demand $$\hbar \beta > \tau \!-\! \tau'$$ to prevent
+$$C_{AB}$$ from diverging for increasing temperatures:
$$\begin{aligned}
C_{AB}(\tau \!-\! \tau')
@@ -48,8 +48,8 @@ $$\begin{aligned}
&= - \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} e^{-\beta E_n} e^{(\tau - \tau') E_n / \hbar}
\end{aligned}$$
-And likewise, for $\tau < \tau'$,
-we must demand that $\tau \!-\! \tau' > -\hbar \beta$
+And likewise, for $$\tau < \tau'$$,
+we must demand that $$\tau \!-\! \tau' > -\hbar \beta$$
for the same reason:
$$\begin{aligned}
@@ -61,14 +61,14 @@ $$\begin{aligned}
&= \mp \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} e^{-\beta E_n} e^{- (\tau - \tau') E_n / \hbar}
\end{aligned}$$
-With $-$ for bosons, and $+$ for fermions,
-due to the time-ordered product for $\tau > \tau'$.
+With $$-$$ for bosons, and $$+$$ for fermions,
+due to the time-ordered product for $$\tau > \tau'$$.
-On this domain $[-\hbar \beta, \hbar \beta]$,
-the Matsubara Green's function $C_{AB}$
+On this domain $$[-\hbar \beta, \hbar \beta]$$,
+the Matsubara Green's function $$C_{AB}$$
obeys a useful shift relation:
-it is $\hbar \beta$-periodic for bosons,
-and $\hbar \beta$-antiperiodic for fermions:
+it is $$\hbar \beta$$-periodic for bosons,
+and $$\hbar \beta$$-antiperiodic for fermions:
$$\begin{aligned}
\boxed{
@@ -88,8 +88,8 @@ $$\begin{aligned}
-First $\tau \!-\! \tau' < 0$.
-We insert the argument $\tau \!-\! \tau' \!+\! \hbar \beta$,
+First $$\tau \!-\! \tau' < 0$$.
+We insert the argument $$\tau \!-\! \tau' \!+\! \hbar \beta$$,
and use the cyclic property:
$$\begin{aligned}
@@ -105,8 +105,8 @@ $$\begin{aligned}
&= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big)
\end{aligned}$$
-Since $\tau < \tau'$ by assumption,
-we can bring back the time-ordered product $\mathcal{T}$:
+Since $$\tau < \tau'$$ by assumption,
+we can bring back the time-ordered product $$\mathcal{T}$$:
$$\begin{aligned}
C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
@@ -115,7 +115,7 @@ $$\begin{aligned}
&= \pm C_{AB}(\tau \!-\! \tau')
\end{aligned}$$
-Moving on to $\tau \!-\! \tau' > 0$, the proof is perfectly analogous:
+Moving on to $$\tau \!-\! \tau' > 0$$, the proof is perfectly analogous:
$$\begin{aligned}
C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta)
@@ -133,16 +133,17 @@ $$\begin{aligned}
\\
&= \pm C_{AB}(\tau \!-\! \tau')
\end{aligned}$$
+
-Due to this limited domain $\tau \in [-\hbar \beta, \hbar \beta]$,
+Due to this limited domain $$\tau \in [-\hbar \beta, \hbar \beta]$$,
the [Fourier transform](/know/concept/fourier-transform/)
-of $C_{AB}(\tau)$ consists of discrete frequencies
-$k_n \equiv n \pi / (\hbar \beta)$.
+of $$C_{AB}(\tau)$$ consists of discrete frequencies
+$$k_n \equiv n \pi / (\hbar \beta)$$.
The forward and inverse Fourier transforms
-are therefore defined as given below (with $\tau' = 0$).
-It is convention to write $C_{AB}(i k_n)$ instead of $C_{AB}(k_n)$:
+are therefore defined as given below (with $$\tau' = 0$$).
+It is convention to write $$C_{AB}(i k_n)$$ instead of $$C_{AB}(k_n)$$:
$$\begin{aligned}
\boxed{
@@ -162,8 +163,8 @@ $$\begin{aligned}
We will prove that one is indeed the inverse of the other.
-We demand that the inverse FT of the forward FT of $C_{AB}(\tau)$
-is simply $C_{AB}(\tau)$ again:
+We demand that the inverse FT of the forward FT of $$C_{AB}(\tau)$$
+is simply $$C_{AB}(\tau)$$ again:
$$\begin{aligned}
C_{AB}(\tau)
@@ -197,11 +198,12 @@ $$\begin{aligned}
\\
&= C_{AB}(\tau)
\end{aligned}$$
+
-Let us now define the **Matsubara frequencies** $\omega_n$
-as a species-dependent subset of $k_n$:
+Let us now define the **Matsubara frequencies** $$\omega_n$$
+as a species-dependent subset of $$k_n$$:
$$\begin{aligned}
\boxed{
@@ -232,7 +234,7 @@ $$\begin{aligned}
We split the integral, shift its limits,
-and use the (anti)periodicity of $C_{AB}$:
+and use the (anti)periodicity of $$C_{AB}$$:
$$\begin{aligned}
C_{AB}(i k_n)
@@ -247,9 +249,9 @@ $$\begin{aligned}
&= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
\end{aligned}$$
-With $+$ for bosons, and $-$ for fermions. Since $k_n \equiv n \pi / (\hbar \beta)$,
-we know $e^{-i k_n \hbar \beta} \in \{-1, 1\}$,
-so for bosons all odd $n$ vanish, and for fermions all even $n$,
+With $$+$$ for bosons, and $$-$$ for fermions. Since $$k_n \equiv n \pi / (\hbar \beta)$$,
+we know $$e^{-i k_n \hbar \beta} \in \{-1, 1\}$$,
+so for bosons all odd $$n$$ vanish, and for fermions all even $$n$$,
yielding the desired result.
For the other case, we simply shift the first integral's limits instead of the seconds':
@@ -263,11 +265,12 @@ $$\begin{aligned}
\\
&= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
\end{aligned}$$
+
If we actually evaluate this,
-we obtain the following form of $C_{AB}$,
+we obtain the following form of $$C_{AB}$$,
which is almost identical to the
[Lehmann representation](/know/concept/lehmann-representation/)
of the "ordinary" retarded and advanced Green's functions:
@@ -285,8 +288,8 @@ $$\begin{aligned}
-For $\tau \!-\! \tau' > 0$, we start by expanding
-in the many-particle eigenstates $\Ket{n}$:
+For $$\tau \!-\! \tau' > 0$$, we start by expanding
+in the many-particle eigenstates $$\Ket{n}$$:
$$\begin{aligned}
C_{AB}(\tau \!-\! \tau')
@@ -300,7 +303,7 @@ $$\begin{aligned}
\matrixel{n'}{\hat{B}}{n} e^{(E_n - E_{n'})(\tau - \tau') / \hbar}
\end{aligned}$$
-We take the Fourier transform by integrating over $[0, \hbar \beta]$:
+We take the Fourier transform by integrating over $$[0, \hbar \beta]$$:
$$\begin{aligned}
C_{AB}(i \omega_m)
@@ -320,8 +323,8 @@ $$\begin{aligned}
\Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
\end{aligned}$$
-Moving on to $\tau \!-\! \tau' < 0$,
-we again expand in the many-particle eigenstates $\Ket{n}$:
+Moving on to $$\tau \!-\! \tau' < 0$$,
+we again expand in the many-particle eigenstates $$\Ket{n}$$:
$$\begin{aligned}
C_{AB}(\tau \!-\! \tau')
@@ -335,8 +338,8 @@ $$\begin{aligned}
\matrixel{n'}{\hat{A}}{n} e^{-(E_n - E_{n'})(\tau - \tau') / \hbar}
\end{aligned}$$
-Since $\tau \!-\! \tau' < 0$ this time,
-we take the Fourier transform over $[-\hbar \beta, 0]$:
+Since $$\tau \!-\! \tau' < 0$$ this time,
+we take the Fourier transform over $$[-\hbar \beta, 0]$$:
$$\begin{aligned}
C_{AB}(i \omega_m)
@@ -359,17 +362,17 @@ $$\begin{aligned}
\Big( e^{- \beta E_{n'}} \mp e^{-\beta E_n} \Big)
\end{aligned}$$
-Where swapping $n$ and $n'$ gives the desired result.
+Where swapping $$n$$ and $$n'$$ gives the desired result.
-This gives us the primary use of the Matsubara Green's function $C_{AB}$:
-calculating the retarded $C_{AB}^R$ and advanced $C_{AB}^A$.
-Once we have an expression for Matsubara's $C_{AB}$,
-we can recover $C_{AB}^R$ and $C_{AB}^A$ by substituting
-$i \omega_m \to \omega \!+\! i \eta$ and $i \omega_m \to \omega \!-\! i \eta$ respectively.
+This gives us the primary use of the Matsubara Green's function $$C_{AB}$$:
+calculating the retarded $$C_{AB}^R$$ and advanced $$C_{AB}^A$$.
+Once we have an expression for Matsubara's $$C_{AB}$$,
+we can recover $$C_{AB}^R$$ and $$C_{AB}^A$$ by substituting
+$$i \omega_m \to \omega \!+\! i \eta$$ and $$i \omega_m \to \omega \!-\! i \eta$$ respectively.
-In general, we can define the **canonical Green's function** $C_{AB}(z)$
+In general, we can define the **canonical Green's function** $$C_{AB}(z)$$
on the complex plane:
$$\begin{aligned}
@@ -380,8 +383,8 @@ $$\begin{aligned}
This is a [holomorphic function](/know/concept/holomorphic-function/),
except for poles on the real axis.
-It turns out that $C_{AB}(z)$ must have these properties
-for the substitution $i \omega_n \to \omega \!\pm\! i \eta$ to be valid.
+It turns out that $$C_{AB}(z)$$ must have these properties
+for the substitution $$i \omega_n \to \omega \!\pm\! i \eta$$ to be valid.
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md
index 45381ba..8b903d4 100644
--- a/source/know/concept/matsubara-sum/index.md
+++ b/source/know/concept/matsubara-sum/index.md
@@ -18,18 +18,18 @@ $$\begin{aligned}
= \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
\end{aligned}$$
-Where $i \omega_n$ are the Matsubara frequencies
-for bosons ($B$) or fermions ($F$),
-and $g(z)$ is a function on the complex plane
+Where $$i \omega_n$$ are the Matsubara frequencies
+for bosons ($$B$$) or fermions ($$F$$),
+and $$g(z)$$ is a function on the complex plane
that is [holomorphic](/know/concept/holomorphic-function/)
except for a known set of simple poles,
-and $\tau$ is a real parameter
+and $$\tau$$ is a real parameter
(e.g. the [imaginary time](/know/concept/imaginary-time/))
-satisfying $-\hbar \beta < \tau < \hbar \beta$.
+satisfying $$-\hbar \beta < \tau < \hbar \beta$$.
Now, consider the following integral
-over a (for now) unspecified counter-clockwise contour $C$,
-with a (for now) unspecified weighting function $h(z)$:
+over a (for now) unspecified counter-clockwise contour $$C$$,
+with a (for now) unspecified weighting function $$h(z)$$:
$$\begin{aligned}
\oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
@@ -37,14 +37,14 @@ $$\begin{aligned}
\end{aligned}$$
Where we have applied the residue theorem
-to get a sum over all simple poles $z_p$
-of either $g$ or $h$ (but not both) enclosed by $C$.
+to get a sum over all simple poles $$z_p$$
+of either $$g$$ or $$h$$ (but not both) enclosed by $$C$$.
Clearly, we could make this look like a Matsubara sum,
-if we choose $h$ such that it has poles at $i \omega_n$.
+if we choose $$h$$ such that it has poles at $$i \omega_n$$.
-Therefore, we choose the weighting function $h(z)$ as follows,
-where $n_B(z)$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
-and $n_F(z)$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/):
+Therefore, we choose the weighting function $$h(z)$$ as follows,
+where $$n_B(z)$$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
+and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/):
$$\begin{aligned}
h(z)
@@ -59,11 +59,11 @@ $$\begin{aligned}
= \frac{1}{e^{\hbar \beta z} \mp 1}
\end{aligned}$$
-The distinction between the signs of $\tau$ is needed
-to ensure that the integrand $h(z) e^{z \tau}$ decays for $|z| \to \infty$,
-both for $\Real(z) > 0$ and $\Real(z) < 0$.
-This choice of $h$ indeed has poles at the respective
-Matsubara frequencies $i \omega_n$ of bosons and fermions,
+The distinction between the signs of $$\tau$$ is needed
+to ensure that the integrand $$h(z) e^{z \tau}$$ decays for $$|z| \to \infty$$,
+both for $$\Real(z) > 0$$ and $$\Real(z) < 0$$.
+This choice of $$h$$ indeed has poles at the respective
+Matsubara frequencies $$i \omega_n$$ of bosons and fermions,
and the residues are:
$$\begin{aligned}
@@ -84,10 +84,10 @@ $$\begin{aligned}
= - \frac{1}{\hbar \beta}
\end{aligned}$$
-In the definition of $h$, the sign flip for $\tau \le 0$
+In the definition of $$h$$, the sign flip for $$\tau \le 0$$
is introduced because negating the argument also negates the residues,
-i.e. $\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$.
-With this $h$, our contour integral can be rewritten as follows:
+i.e. $$\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$$.
+With this $$h$$, our contour integral can be rewritten as follows:
$$\begin{aligned}
\oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
@@ -98,8 +98,8 @@ $$\begin{aligned}
\pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
\end{aligned}$$
-Where $+$ is for bosons, and $-$ for fermions.
-Here, we recognize the last term as the Matsubara sum $S_{F,B}$,
+Where $$+$$ is for bosons, and $$-$$ for fermions.
+Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$,
for which we isolate, yielding:
$$\begin{aligned}
@@ -108,10 +108,10 @@ $$\begin{aligned}
\pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
\end{aligned}$$
-Now we must choose $C$. Assuming $g(z)$ does not interfere,
-we know that $h(z) e^{z \tau}$ decays to zero
-for $|z| \to \infty$, so a useful choice would be a circle of radius $R$.
-If we then let $R \to \infty$, the contour encloses
+Now we must choose $$C$$. Assuming $$g(z)$$ does not interfere,
+we know that $$h(z) e^{z \tau}$$ decays to zero
+for $$|z| \to \infty$$, so a useful choice would be a circle of radius $$R$$.
+If we then let $$R \to \infty$$, the contour encloses
the whole complex plane, including all of the integrand's poles.
However, thanks to the integrand's decay,
the resulting contour integral must vanish:
@@ -126,7 +126,7 @@ $$\begin{aligned}
\end{aligned}$$
We thus arrive at the following results
-for bosonic and fermionic Matsubara sums $S_{B,F}$:
+for bosonic and fermionic Matsubara sums $$S_{B,F}$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/maxwell-bloch-equations/index.md b/source/know/concept/maxwell-bloch-equations/index.md
index 8ba9a5b..b306c7d 100644
--- a/source/know/concept/maxwell-bloch-equations/index.md
+++ b/source/know/concept/maxwell-bloch-equations/index.md
@@ -12,7 +12,7 @@ layout: "concept"
---
For an electron in a two-level system with time-independent states
-$\Ket{g}$ (ground) and $\Ket{e}$ (excited),
+$$\Ket{g}$$ (ground) and $$\Ket{e}$$ (excited),
consider the following general solution
to the full Schrödinger equation:
@@ -23,9 +23,9 @@ $$\begin{aligned}
Perturbing this system with
an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
-introduces a time-dependent sinusoidal term $\hat{H}_1$ to the Hamiltonian.
+introduces a time-dependent sinusoidal term $$\hat{H}_1$$ to the Hamiltonian.
In the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
-$\hat{H}_1$ is given by:
+$$\hat{H}_1$$ is given by:
$$\begin{aligned}
\hat{H}_1(t)
@@ -38,10 +38,10 @@ $$\begin{aligned}
= \vb{E}_0 \cos(\omega t)
\end{aligned}$$
-Where $\vb{E}$ is an [electric field](/know/concept/electric-field/),
-and $\hat{\vb{p}}$ is the dipole moment operator.
+Where $$\vb{E}$$ is an [electric field](/know/concept/electric-field/),
+and $$\hat{\vb{p}}$$ is the dipole moment operator.
From [Rabi oscillation](/know/concept/rabi-oscillation/),
-we know that the time-varying coefficients $c_g$ and $c_e$
+we know that the time-varying coefficients $$c_g$$ and $$c_e$$
can then be described by:
$$\begin{aligned}
@@ -53,8 +53,8 @@ $$\begin{aligned}
\end{aligned}$$
We want to rearrange these equations a bit.
-Therefore, we split the electric field $\vb{E}$ like so,
-where the amplitudes $\vb{E}_0^{-}$ and $\vb{E}_0^{+}$ may be slowly varying:
+Therefore, we split the electric field $$\vb{E}$$ like so,
+where the amplitudes $$\vb{E}_0^{-}$$ and $$\vb{E}_0^{+}$$ may be slowly varying:
$$\begin{aligned}
\vb{E}(t)
@@ -62,8 +62,8 @@ $$\begin{aligned}
= \vb{E}_0^{-} \exp(i \omega t) + \vb{E}_0^{+} \exp(-i \omega t)
\end{aligned}$$
-Since $\vb{E}$ is real, $\vb{E}_0^{+} = (\vb{E}_0^{-})^*$.
-Similarly, we define the transition dipole moment $\vb{p}_0^{-}$:
+Since $$\vb{E}$$ is real, $$\vb{E}_0^{+} = (\vb{E}_0^{-})^*$$.
+Similarly, we define the transition dipole moment $$\vb{p}_0^{-}$$:
$$\begin{aligned}
\vb{p}_0^{-}
@@ -74,8 +74,8 @@ $$\begin{aligned}
= q \matrixel{g}{\vu{x}}{e}
\end{aligned}$$
-With these, the equations for $c_g$ and $c_e$ can be rewritten as shown below.
-Note that $\vb{E}^{-}$ and $\vb{E}^{+}$ include the driving plane wave, and the
+With these, the equations for $$c_g$$ and $$c_e$$ can be rewritten as shown below.
+Note that $$\vb{E}^{-}$$ and $$\vb{E}^{+}$$ include the driving plane wave, and the
[rotating wave approximation](/know/concept/rotating-wave-approximation/) is still made:
$$\begin{aligned}
@@ -89,9 +89,9 @@ $$\begin{aligned}
## Optical Bloch equations
-For $\Ket{\Psi}$ as defined above,
+For $$\Ket{\Psi}$$ as defined above,
the corresponding pure [density operator](/know/concept/density-operator/)
-$\hat{\rho}$ is as follows:
+$$\hat{\rho}$$ is as follows:
$$\begin{aligned}
\hat{\rho}
@@ -108,9 +108,9 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-Where $\omega_0 \equiv (E_e \!-\! E_g) / \hbar$ is the resonance frequency.
-We take the $t$-derivative of the matrix elements,
-and insert the equations for $c_g$ and $c_e$:
+Where $$\omega_0 \equiv (E_e \!-\! E_g) / \hbar$$ is the resonance frequency.
+We take the $$t$$-derivative of the matrix elements,
+and insert the equations for $$c_g$$ and $$c_e$$:
$$\begin{aligned}
\dv{\rho_{gg}}{t}
@@ -157,12 +157,12 @@ $$\begin{aligned}
&= - i \omega_0 \rho_{eg} + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \big( \rho_{gg} - \rho_{ee} \big)
\end{aligned}$$
-These equations are correct if nothing else is affecting $\hat{\rho}$.
+These equations are correct if nothing else is affecting $$\hat{\rho}$$.
But in practice, these quantities decay due to various processes,
e.g. spontaneous emission (see [Einstein coefficients](/know/concept/einstein-coefficients/)).
-Let $\rho_{ee}$ decays with rate $\gamma_e$.
-Since the total probability $\rho_{ee} + \rho_{gg} = 1$,
+Let $$\rho_{ee}$$ decays with rate $$\gamma_e$$.
+Since the total probability $$\rho_{ee} + \rho_{gg} = 1$$,
we thus have:
$$\begin{aligned}
@@ -174,7 +174,7 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, for whatever reason,
-let $\rho_{gg}$ decay into $\rho_{ee}$ with rate $\gamma_g$:
+let $$\rho_{gg}$$ decay into $$\rho_{ee}$$ with rate $$\gamma_g$$:
$$\begin{aligned}
\Big( \dv{\rho_{gg}}{t} \Big)_{g}
@@ -185,7 +185,7 @@ $$\begin{aligned}
\end{aligned}$$
And finally, let the diagonal (perpendicular) matrix elements
-both decay with rate $\gamma_\perp$:
+both decay with rate $$\gamma_\perp$$:
$$\begin{aligned}
\Big( \dv{\rho_{eg}}{t} \Big)_{\perp}
@@ -196,7 +196,7 @@ $$\begin{aligned}
\end{aligned}$$
Putting everything together,
-we arrive at the **optical Bloch equations** governing $\hat{\rho}$:
+we arrive at the **optical Bloch equations** governing $$\hat{\rho}$$:
$$\begin{aligned}
\boxed{
@@ -221,12 +221,12 @@ $$\begin{aligned}
\end{aligned}$$
Many authors simplify these equations a bit by choosing
-$\gamma_g = 0$ and $\gamma_\perp = \gamma_e / 2$.
+$$\gamma_g = 0$$ and $$\gamma_\perp = \gamma_e / 2$$.
## Including Maxwell's equations
-This two-level system has a dipole moment $\vb{p}$ as follows,
+This two-level system has a dipole moment $$\vb{p}$$ as follows,
where we use [Laporte's selection rule](/know/concept/selection-rules/)
to remove diagonal terms, by assuming that
the electron's orbitals are odd or even:
@@ -243,8 +243,8 @@ $$\begin{aligned}
\equiv \vb{p}^{-}(t) + \vb{p}^{+}(t)
\end{aligned}$$
-Where we have split $\vb{p}$ analogously to $\vb{E}$
-by defining $\vb{p}^{+} \equiv \vb{p}_0^{+} \rho_{eg}$.
+Where we have split $$\vb{p}$$ analogously to $$\vb{E}$$
+by defining $$\vb{p}^{+} \equiv \vb{p}_0^{+} \rho_{eg}$$.
Its equation of motion can then be found from the optical Bloch equations:
$$\begin{aligned}
@@ -254,7 +254,7 @@ $$\begin{aligned}
+ \frac{i}{\hbar} \vb{p}_0^{+} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \Big( \rho_{gg} - \rho_{ee} \Big)
\end{aligned}$$
-Some authors do not bother multiplying $\rho_{ge}$ by $\vb{p}_0^{+}$.
+Some authors do not bother multiplying $$\rho_{ge}$$ by $$\vb{p}_0^{+}$$.
In any case, we arrive at:
$$\begin{aligned}
@@ -265,7 +265,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined the **population inversion** $d \in [-1, 1]$ as follows,
+Where we have defined the **population inversion** $$d \in [-1, 1]$$ as follows,
which quantifies the electron's excitedness:
$$\begin{aligned}
@@ -285,8 +285,8 @@ $$\begin{aligned}
We can rewrite the first two terms in the following intuitive form,
which describes a decay with
-rate $\gamma_\parallel \equiv \gamma_g + \gamma_e$
-towards an equilbrium $d_0$:
+rate $$\gamma_\parallel \equiv \gamma_g + \gamma_e$$
+towards an equilbrium $$d_0$$:
$$\begin{aligned}
2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
@@ -313,8 +313,8 @@ $$\begin{aligned}
+ \gamma_g (\rho_{gg} - \rho_{ee}) + \gamma_e (\rho_{gg} - \rho_{ee})
\end{aligned}$$
-Since the total probability $\rho_{gg} + \rho_{ee} = 1$,
-and $d \equiv \rho_{ee} - \rho_{gg}$, this reduces to:
+Since the total probability $$\rho_{gg} + \rho_{ee} = 1$$,
+and $$d \equiv \rho_{ee} - \rho_{gg}$$, this reduces to:
$$\begin{aligned}
2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
@@ -324,10 +324,11 @@ $$\begin{aligned}
\\
&= \gamma_\parallel ( d_0 - d )
\end{aligned}$$
+
-With this, the equation for the population inversion $d$
+With this, the equation for the population inversion $$d$$
takes the following final form:
$$\begin{aligned}
@@ -338,10 +339,10 @@ $$\begin{aligned}
\end{aligned}$$
Finally, we would like a relation between the polarization
-and the electric field $\vb{E}$,
+and the electric field $$\vb{E}$$,
for which we turn to [Maxwell's equations](/know/concept/maxwells-equations/).
We start from Faraday's law,
-and split $\vb{B} = \mu_0 (\vb{H} + \vb{M})$:
+and split $$\vb{B} = \mu_0 (\vb{H} + \vb{M})$$:
$$\begin{aligned}
\nabla \cross \vb{E}
@@ -349,9 +350,9 @@ $$\begin{aligned}
= - \mu_0 \pdv{\vb{H}}{t} - \mu_0 \pdv{\vb{M}}{t}
\end{aligned}$$
-We assume that there is no magnetization $\vb{M} = 0$.
+We assume that there is no magnetization $$\vb{M} = 0$$.
Then we we take the curl of both sides,
-and replace $\nabla \cross \vb{H}$ with Ampère's circuital law:
+and replace $$\nabla \cross \vb{H}$$ with Ampère's circuital law:
$$\begin{aligned}
\nabla \cross \big( \nabla \cross \vb{E} \big)
@@ -359,23 +360,23 @@ $$\begin{aligned}
= - \mu_0 \pdv{}{t} \Big( \vb{J}_\mathrm{free} + \pdv{\vb{D}}{t} \Big)
\end{aligned}$$
-Inserting the definition $\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$
-together with Ohm's law $\vb{J}_\mathrm{free} = \sigma \vb{E}$ yields:
+Inserting the definition $$\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$$
+together with Ohm's law $$\vb{J}_\mathrm{free} = \sigma \vb{E}$$ yields:
$$\begin{aligned}
\nabla \cross \big( \nabla \cross \vb{E} \big)
= - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdvn{2}{\vb{E}}{t} - \mu_0 \pdvn{2}{\vb{P}}{t}
\end{aligned}$$
-Where $\sigma$ is the active material's conductivity, if any;
-almost all authors assume $\sigma = 0$.
+Where $$\sigma$$ is the active material's conductivity, if any;
+almost all authors assume $$\sigma = 0$$.
Recall that we are describing the dynamics of a two-level system.
In reality, such a system (e.g. a quantum dot)
is suspended in a passive background medium,
-which reacts with a polarization $\vb{P}_\mathrm{med}$
-to the electric field $\vb{E}$.
-If the medium is linear, i.e. $\vb{P}_\mathrm{med} = \varepsilon_0 \chi \vb{E}$,
+which reacts with a polarization $$\vb{P}_\mathrm{med}$$
+to the electric field $$\vb{E}$$.
+If the medium is linear, i.e. $$\vb{P}_\mathrm{med} = \varepsilon_0 \chi \vb{E}$$,
then:
$$\begin{aligned}
@@ -390,10 +391,10 @@ $$\begin{aligned}
- \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
\end{aligned}$$
-Where $\varepsilon_r \equiv 1 + \chi_e$ is the medium's relative permittivity.
-The speed of light $c^2 = 1 / (\mu_0 \varepsilon_0)$,
-and the refractive index $n^2 = \mu_r \varepsilon_r$,
-where $\mu_r = 1$ due to our assumption that $\vb{M} = 0$, so:
+Where $$\varepsilon_r \equiv 1 + \chi_e$$ is the medium's relative permittivity.
+The speed of light $$c^2 = 1 / (\mu_0 \varepsilon_0)$$,
+and the refractive index $$n^2 = \mu_r \varepsilon_r$$,
+where $$\mu_r = 1$$ due to our assumption that $$\vb{M} = 0$$, so:
$$\begin{aligned}
\boxed{
@@ -402,9 +403,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-$\vb{E}$ and $\vb{P}$ can trivially be replaced by $\vb{E}^{+}$ and $\vb{P}^{+}$.
-It is also simple to convert $\vb{p}^{+}$ and $d$
-into the macroscopic $\vb{P}^{+}$ and total $D$
+$$\vb{E}$$ and $$\vb{P}$$ can trivially be replaced by $$\vb{E}^{+}$$ and $$\vb{P}^{+}$$.
+It is also simple to convert $$\vb{p}^{+}$$ and $$d$$
+into the macroscopic $$\vb{P}^{+}$$ and total $$D$$
by summing over all two-level systems in the medium:
$$\begin{aligned}
diff --git a/source/know/concept/maxwell-boltzmann-distribution/index.md b/source/know/concept/maxwell-boltzmann-distribution/index.md
index 65e983c..ebb2460 100644
--- a/source/know/concept/maxwell-boltzmann-distribution/index.md
+++ b/source/know/concept/maxwell-boltzmann-distribution/index.md
@@ -17,23 +17,23 @@ probability distributions with applications in classical statistical physics.
In the [canonical ensemble](/know/concept/canonical-ensemble/)
(where a fixed-size system can exchange energy with its environment),
-the probability of a microstate with energy $E$ is given by the Boltzmann distribution:
+the probability of a microstate with energy $$E$$ is given by the Boltzmann distribution:
$$\begin{aligned}
f(E)
\:\propto\: \exp\!\big(\!-\! \beta E\big)
\end{aligned}$$
-Where $\beta = 1 / k_B T$.
-We split $E = K + U$,
-with $K$ and $U$ the total kinetic and potential energy contributions.
-If there are $N$ particles in the system,
-with positions $\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$
-and momenta $\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$,
-then $K$ only depends on $\tilde{p}$,
-and $U$ only depends on $\tilde{r}$,
+Where $$\beta = 1 / k_B T$$.
+We split $$E = K + U$$,
+with $$K$$ and $$U$$ the total kinetic and potential energy contributions.
+If there are $$N$$ particles in the system,
+with positions $$\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$$
+and momenta $$\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$$,
+then $$K$$ only depends on $$\tilde{p}$$,
+and $$U$$ only depends on $$\tilde{r}$$,
so the probability of a specific microstate
-$(\tilde{r}, \tilde{p})$ is as follows:
+$$(\tilde{r}, \tilde{p})$$ is as follows:
$$\begin{aligned}
f(\tilde{r}, \tilde{p})
@@ -62,10 +62,10 @@ $$\begin{aligned}
\:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
\end{aligned}$$
-We cannot evaluate $f_U(\tilde{r})$ further without knowing $U(\tilde{r})$ for a system.
-We thus turn to $f_K(\tilde{p})$, and see that the total kinetic
-energy $K(\tilde{p})$ is simply the sum of the particles' individual
-kinetic energies $K_n(\vec{p}_n)$, which are well-known:
+We cannot evaluate $$f_U(\tilde{r})$$ further without knowing $$U(\tilde{r})$$ for a system.
+We thus turn to $$f_K(\tilde{p})$$, and see that the total kinetic
+energy $$K(\tilde{p})$$ is simply the sum of the particles' individual
+kinetic energies $$K_n(\vec{p}_n)$$, which are well-known:
$$\begin{aligned}
K(\tilde{p})
@@ -75,7 +75,7 @@ $$\begin{aligned}
= \frac{|\vec{p}_n|^2}{2 m}
\end{aligned}$$
-Consequently, the probability distribution $f(p_x, p_y, p_z)$ for the
+Consequently, the probability distribution $$f(p_x, p_y, p_z)$$ for the
momentum vector of a single particle is as follows,
after normalization:
@@ -84,7 +84,7 @@ $$\begin{aligned}
= \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big)
\end{aligned}$$
-We now rewrite this using the velocities $v_x = p_x / m$,
+We now rewrite this using the velocities $$v_x = p_x / m$$,
and update the normalization, giving:
$$\begin{aligned}
@@ -114,10 +114,10 @@ $$\begin{aligned}
## Speed distribution
We know the distribution of the velocities along each axis,
-but what about the speed $v = |\vec{v}|$?
-Because we do not care about the direction of $\vec{v}$, only its magnitude,
-the [density of states](/know/concept/density-of-states/) $g(v)$ is not constant:
-it is the rate-of-change of the volume of a sphere of radius $v$:
+but what about the speed $$v = |\vec{v}|$$?
+Because we do not care about the direction of $$\vec{v}$$, only its magnitude,
+the [density of states](/know/concept/density-of-states/) $$g(v)$$ is not constant:
+it is the rate-of-change of the volume of a sphere of radius $$v$$:
$$\begin{aligned}
g(v)
@@ -125,8 +125,8 @@ $$\begin{aligned}
= 4 \pi v^2
\end{aligned}$$
-Multiplying the velocity vector distribution by $g(v)$
-and substituting $v^2 = v_x^2 + v_y^2 + v_z^2$
+Multiplying the velocity vector distribution by $$g(v)$$
+and substituting $$v^2 = v_x^2 + v_y^2 + v_z^2$$
then gives us the **Maxwell-Boltzmann speed distribution**:
$$\begin{aligned}
@@ -137,9 +137,9 @@ $$\begin{aligned}
\end{aligned}$$
Some notable points on this distribution are
-the most probable speed $v_{\mathrm{mode}}$,
-the mean average speed $v_{\mathrm{mean}}$,
-and the root-mean-square speed $v_{\mathrm{rms}}$:
+the most probable speed $$v_{\mathrm{mode}}$$,
+the mean average speed $$v_{\mathrm{mean}}$$,
+and the root-mean-square speed $$v_{\mathrm{rms}}$$:
$$\begin{aligned}
f'(v_\mathrm{mode})
@@ -176,7 +176,7 @@ $$\begin{aligned}
Using the speed distribution,
we can work out the kinetic energy distribution.
-Because $K$ is not proportional to $v$,
+Because $$K$$ is not proportional to $$v$$,
we must do this by demanding that:
$$\begin{aligned}
@@ -187,9 +187,9 @@ $$\begin{aligned}
= f(v) \dv{v}{K}
\end{aligned}$$
-We know that $K = m v^2 / 2$,
-meaning $\dd{K} = m v \dd{v}$
-so the energy distribution $f(K)$ is:
+We know that $$K = m v^2 / 2$$,
+meaning $$\dd{K} = m v \dd{v}$$
+so the energy distribution $$f(K)$$ is:
$$\begin{aligned}
f(K)
@@ -197,7 +197,7 @@ $$\begin{aligned}
= \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
\end{aligned}$$
-Substituting $v = \sqrt{2 K/m}$ leads to
+Substituting $$v = \sqrt{2 K/m}$$ leads to
the **Maxwell-Boltzmann kinetic energy distribution**:
$$\begin{aligned}
diff --git a/source/know/concept/maxwell-relations/index.md b/source/know/concept/maxwell-relations/index.md
index aa51b06..892ced1 100644
--- a/source/know/concept/maxwell-relations/index.md
+++ b/source/know/concept/maxwell-relations/index.md
@@ -14,15 +14,15 @@ for well-behaved functions (sometimes known as the *Schwarz theorem*),
applied to the [thermodynamic potentials](/know/concept/thermodynamic-potential/).
We start by proving the general "recipe".
-Given that the differential element of some $z$ is defined in terms of
-two constant quantities $A$ and $B$ and two independent variables $x$ and $y$:
+Given that the differential element of some $$z$$ is defined in terms of
+two constant quantities $$A$$ and $$B$$ and two independent variables $$x$$ and $$y$$:
$$\begin{aligned}
\dd{z} \equiv A \dd{x} + B \dd{y}
\end{aligned}$$
-Then the quantities $A$ and $B$ can be extracted
-by dividing by $\dd{x}$ and $\dd{y}$ respectively:
+Then the quantities $$A$$ and $$B$$ can be extracted
+by dividing by $$\dd{x}$$ and $$\dd{y}$$ respectively:
$$\begin{aligned}
A = \Big( \pdv{z}{x} \Big)_y
@@ -30,7 +30,7 @@ $$\begin{aligned}
B = \Big( \pdv{z}{y} \Big)_x
\end{aligned}$$
-By differentiating $A$ and $B$,
+By differentiating $$A$$ and $$B$$,
and using that the order of differentiation is irrelevant, we find:
$$\begin{aligned}
@@ -55,11 +55,11 @@ $$\begin{aligned}
\end{aligned}$$
The following quantities are useful to rewrite some of the Maxwell relations:
-the iso-$P$ thermal expansion coefficient $\alpha$,
-the iso-$T$ combressibility $\kappa_T$,
-the iso-$S$ combressibility $\kappa_S$,
-the iso-$V$ heat capacity $C_V$,
-and the iso-$P$ heat capacity $C_P$:
+the iso-$$P$$ thermal expansion coefficient $$\alpha$$,
+the iso-$$T$$ combressibility $$\kappa_T$$,
+the iso-$$S$$ combressibility $$\kappa_S$$,
+the iso-$$V$$ heat capacity $$C_V$$,
+and the iso-$$P$$ heat capacity $$C_P$$:
$$\begin{gathered}
\alpha \equiv \frac{1}{V} \Big( \pdv{V}{T} \Big)_{P,N}
@@ -77,7 +77,7 @@ $$\begin{gathered}
## Internal energy
The following Maxwell relations can be derived
-from the internal energy $U(S, V, N)$:
+from the internal energy $$U(S, V, N)$$:
$$\begin{gathered}
\mpdv{U}{V}{S} =
@@ -119,7 +119,7 @@ $$\begin{gathered}
## Enthalpy
The following Maxwell relations can be derived
-from the enthalpy $H(S, P, N)$:
+from the enthalpy $$H(S, P, N)$$:
$$\begin{gathered}
\mpdv{H}{P}{S} =
@@ -161,7 +161,7 @@ $$\begin{gathered}
## Helmholtz free energy
The following Maxwell relations can be derived
-from the Helmholtz free energy $F(T, V, N)$:
+from the Helmholtz free energy $$F(T, V, N)$$:
$$\begin{gathered}
- \mpdv{F}{V}{T} =
@@ -203,7 +203,7 @@ $$\begin{gathered}
## Gibbs free energy
The following Maxwell relations can be derived
-from the Gibbs free energy $G(T, P, N)$:
+from the Gibbs free energy $$G(T, P, N)$$:
$$\begin{gathered}
\mpdv{G}{T}{P} =
@@ -245,7 +245,7 @@ $$\begin{gathered}
## Landau potential
The following Maxwell relations can be derived
-from the Gibbs free energy $\Omega(T, V, \mu)$:
+from the Gibbs free energy $$\Omega(T, V, \mu)$$:
$$\begin{gathered}
- \mpdv{\Omega}{V}{T} =
diff --git a/source/know/concept/maxwells-equations/index.md b/source/know/concept/maxwells-equations/index.md
index ea052ce..36eb7b2 100644
--- a/source/know/concept/maxwells-equations/index.md
+++ b/source/know/concept/maxwells-equations/index.md
@@ -17,10 +17,10 @@ which describes the existence of light.
## Gauss' law
-**Gauss' law** states that the electric flux $\Phi_E$ through
-a closed surface $S(V)$ is equal to the total charge $Q$
-contained in the enclosed volume $V$,
-divided by the vacuum permittivity $\varepsilon_0$:
+**Gauss' law** states that the electric flux $$\Phi_E$$ through
+a closed surface $$S(V)$$ is equal to the total charge $$Q$$
+contained in the enclosed volume $$V$$,
+divided by the vacuum permittivity $$\varepsilon_0$$:
$$\begin{aligned}
\Phi_E
@@ -29,12 +29,12 @@ $$\begin{aligned}
= \frac{Q}{\varepsilon_0}
\end{aligned}$$
-Where $\vb{E}$ is the [electric field](/know/concept/electric-field/),
-and $\rho$ is the charge density in $V$.
+Where $$\vb{E}$$ is the [electric field](/know/concept/electric-field/),
+and $$\rho$$ is the charge density in $$V$$.
Gauss' law is usually more useful when written in its vector form,
which can be found by applying the divergence theorem
to the surface integral above.
-It states that the divergence of $\vb{E}$ is proportional to $\rho$:
+It states that the divergence of $$\vb{E}$$ is proportional to $$\rho$$:
$$\begin{aligned}
\boxed{
@@ -43,10 +43,10 @@ $$\begin{aligned}
\end{aligned}$$
This law can just as well be expressed for
-the displacement field $\vb{D}$
-and polarization density $\vb{P}$.
-We insert $\vb{E} = (\vb{D} - \vb{P}) / \varepsilon_0$
-into Gauss' law for $\vb{E}$, multiplied by $\varepsilon_0$:
+the displacement field $$\vb{D}$$
+and polarization density $$\vb{P}$$.
+We insert $$\vb{E} = (\vb{D} - \vb{P}) / \varepsilon_0$$
+into Gauss' law for $$\vb{E}$$, multiplied by $$\varepsilon_0$$:
$$\begin{aligned}
\rho
@@ -54,11 +54,11 @@ $$\begin{aligned}
= \nabla \cdot \vb{D} - \nabla \cdot \vb{P}
\end{aligned}$$
-To proceed, we split the net charge density $\rho$
-into a "free" part $\rho_\mathrm{free}$
-and a "bound" part $\rho_\mathrm{bound}$,
-respectively corresponding to $\vb{D}$ and $\vb{P}$,
-such that $\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}$.
+To proceed, we split the net charge density $$\rho$$
+into a "free" part $$\rho_\mathrm{free}$$
+and a "bound" part $$\rho_\mathrm{bound}$$,
+respectively corresponding to $$\vb{D}$$ and $$\vb{P}$$,
+such that $$\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}$$.
This yields:
$$\begin{aligned}
@@ -71,7 +71,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-By integrating over an arbitrary volume $V$
+By integrating over an arbitrary volume $$V$$
we can get integral forms of these equations:
$$\begin{aligned}
@@ -89,10 +89,10 @@ $$\begin{aligned}
## Gauss' law for magnetism
-**Gauss' law for magnetism** states that magnetic flux $\Phi_B$
-through a closed surface $S(V)$ is zero.
+**Gauss' law for magnetism** states that magnetic flux $$\Phi_B$$
+through a closed surface $$S(V)$$ is zero.
In other words, all magnetic field lines entering
-the volume $V$ must leave it too:
+the volume $$V$$ must leave it too:
$$\begin{aligned}
\Phi_B
@@ -100,7 +100,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Where $\vb{B}$ is the [magnetic field](/know/concept/magnetic-field/).
+Where $$\vb{B}$$ is the [magnetic field](/know/concept/magnetic-field/).
Thanks to the divergence theorem,
this can equivalently be stated in vector form as follows:
@@ -117,10 +117,10 @@ in contrast to electric charge.
## Faraday's law of induction
-**Faraday's law of induction** states that a magnetic field $\vb{B}$
-that changes with time will induce an electric field $E$.
-Specifically, the change in magnetic flux through a non-closed surface $S$
-creates an electromotive force around the contour $C(S)$.
+**Faraday's law of induction** states that a magnetic field $$\vb{B}$$
+that changes with time will induce an electric field $$E$$.
+Specifically, the change in magnetic flux through a non-closed surface $$S$$
+creates an electromotive force around the contour $$C(S)$$.
This is written as:
$$\begin{aligned}
@@ -141,37 +141,38 @@ $$\begin{aligned}
## Ampère's circuital law
**Ampère's circuital law**, with Maxwell's correction,
-states that a magnetic field $\vb{B}$
-can be induced along a contour $C(S)$ by two things:
-a current density $\vb{J}$ through the enclosed surface $S$,
-and a change of the electric field flux $\Phi_E$ through $S$:
+states that a magnetic field $$\vb{B}$$
+can be induced along a contour $$C(S)$$ by two things:
+a current density $$\vb{J}$$ through the enclosed surface $$S$$,
+and a change of the electric field flux $$\Phi_E$$ through $$S$$:
$$\begin{aligned}
\oint_{C(S)} \vb{B} \cdot d\vb{l}
= \mu_0 \Big( \int_S \vb{J} \cdot d\vb{A} + \varepsilon_0 \dv{}{t}\int_S \vb{E} \cdot d\vb{A} \Big)
\end{aligned}$$
+
$$\begin{aligned}
\boxed{
\nabla \times \vb{B} = \mu_0 \Big( \vb{J} + \varepsilon_0 \pdv{\vb{E}}{t} \Big)
}
\end{aligned}$$
-Where $\mu_0$ is the vacuum permeability.
-This relation also exists for the "bound" fields $\vb{H}$ and $\vb{D}$,
-and for $\vb{M}$ and $\vb{P}$.
-We insert $\vb{B} = \mu_0 (\vb{H} + \vb{M})$
-and $\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0$
-into Ampère's law, after dividing it by $\mu_0$ for simplicity:
+Where $$\mu_0$$ is the vacuum permeability.
+This relation also exists for the "bound" fields $$\vb{H}$$ and $$\vb{D}$$,
+and for $$\vb{M}$$ and $$\vb{P}$$.
+We insert $$\vb{B} = \mu_0 (\vb{H} + \vb{M})$$
+and $$\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0$$
+into Ampère's law, after dividing it by $$\mu_0$$ for simplicity:
$$\begin{aligned}
\nabla \cross \big( \vb{H} + \vb{M} \big)
&= \vb{J} + \pdv{}{t}\big( \vb{D} - \vb{P} \big)
\end{aligned}$$
-To proceed, we split the net current density $\vb{J}$
-into a "free" part $\vb{J}_\mathrm{free}$
-and a "bound" part $\vb{J}_\mathrm{bound}$,
-such that $\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}$.
+To proceed, we split the net current density $$\vb{J}$$
+into a "free" part $$\vb{J}_\mathrm{free}$$
+and a "bound" part $$\vb{J}_\mathrm{bound}$$,
+such that $$\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}$$.
This leads us to:
$$\begin{aligned}
@@ -184,7 +185,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-By integrating over an arbitrary surface $S$
+By integrating over an arbitrary surface $$S$$
we can get integral forms of these equations:
$$\begin{aligned}
@@ -195,9 +196,9 @@ $$\begin{aligned}
&= \int_S \vb{J}_{\mathrm{bound}} \cdot \dd{\vb{A}} - \dv{}{t}\int_S \vb{P} \cdot \dd{\vb{A}}
\end{aligned}$$
-Note that $\vb{J}_\mathrm{bound}$ can be split into
-the **magnetization current density** $\vb{J}_M = \nabla \cross \vb{M}$
-and the **polarization current density** $\vb{J}_P = \ipdv{\vb{P}}{t}$:
+Note that $$\vb{J}_\mathrm{bound}$$ can be split into
+the **magnetization current density** $$\vb{J}_M = \nabla \cross \vb{M}$$
+and the **polarization current density** $$\vb{J}_P = \ipdv{\vb{P}}{t}$$:
$$\begin{aligned}
\vb{J}_\mathrm{bound}
@@ -221,8 +222,8 @@ $$\begin{aligned}
Since the divergence of a curl is always zero,
the right-hand side must vanish.
-We know that $\vb{B}$ can vary in time,
-so our only option to satisfy this is to demand that $\nabla \cdot \vb{B} = 0$.
+We know that $$\vb{B}$$ can vary in time,
+so our only option to satisfy this is to demand that $$\nabla \cdot \vb{B} = 0$$.
We thus arrive arrive at Gauss' law for magnetism from Faraday's law.
The same technique works for Ampère's law.
@@ -234,7 +235,7 @@ $$\begin{aligned}
= \nabla \cdot \vb{J} + \varepsilon_0 \pdv{}{t}(\nabla \cdot \vb{E})
\end{aligned}$$
-We integrate this over an arbitrary volume $V$,
+We integrate this over an arbitrary volume $$V$$,
and apply the divergence theorem:
$$\begin{aligned}
@@ -245,9 +246,9 @@ $$\begin{aligned}
\end{aligned}$$
The first integral represents the current (charge flux)
-through the surface of $V$.
+through the surface of $$V$$.
Electric charge is not created or destroyed,
-so the second integral *must* be the total charge in $V$:
+so the second integral *must* be the total charge in $$V$$:
$$\begin{aligned}
Q
diff --git a/source/know/concept/meniscus/index.md b/source/know/concept/meniscus/index.md
index e032a4e..a53b5e9 100644
--- a/source/know/concept/meniscus/index.md
+++ b/source/know/concept/meniscus/index.md
@@ -15,19 +15,19 @@ touches a flat solid wall, it will curve to meet it.
This small rise or fall is called a **meniscus**,
and is caused by surface tension and gravity.
-In 2D, let the vertical $y$-axis be a flat wall,
-and the fluid tend to $y = 0$ when $x \to \infty$.
-Close to the wall, i.e. for small $x$, the liquid curves up or down
-to touch the wall at a height $y = d$.
+In 2D, let the vertical $$y$$-axis be a flat wall,
+and the fluid tend to $$y = 0$$ when $$x \to \infty$$.
+Close to the wall, i.e. for small $$x$$, the liquid curves up or down
+to touch the wall at a height $$y = d$$.
Three forces are at work here:
-the first two are the surface tension $\alpha$ of the fluid surface,
-and the counter-pull $\alpha \sin\phi$ of the wall against the tension,
-where $\phi$ is the contact angle.
+the first two are the surface tension $$\alpha$$ of the fluid surface,
+and the counter-pull $$\alpha \sin\phi$$ of the wall against the tension,
+where $$\phi$$ is the contact angle.
The third is the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) gradient
inside the small portion of the fluid above/below the ambient level,
which exerts a total force on the wall given by
-(for $\phi < \pi/2$ so that $d > 0$):
+(for $$\phi < \pi/2$$ so that $$d > 0$$):
$$\begin{aligned}
\int_0^d \rho g y \dd{y}
@@ -35,7 +35,7 @@ $$\begin{aligned}
\end{aligned}$$
If you were wondering about the units,
-keep in mind that there is an implicit $z$-direction here too.
+keep in mind that there is an implicit $$z$$-direction here too.
This results in the following balance equation for the forces at the wall:
$$\begin{aligned}
@@ -43,7 +43,7 @@ $$\begin{aligned}
= \alpha \sin\phi + \frac{1}{2} \rho g d^2
\end{aligned}$$
-We isolate this relation for $d$
+We isolate this relation for $$d$$
and use some trigonometric magic to rewrite it:
$$\begin{aligned}
@@ -52,10 +52,10 @@ $$\begin{aligned}
= \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)}
\end{aligned}$$
-Here, we recognize the definition of the capillary length $L_c = \sqrt{\alpha / (\rho g)}$,
-yielding an expression for $d$
-that is valid both for $\phi < \pi/2$ (where $d > 0$)
-and $\phi > \pi/2$ (where $d < 0$):
+Here, we recognize the definition of the capillary length $$L_c = \sqrt{\alpha / (\rho g)}$$,
+yielding an expression for $$d$$
+that is valid both for $$\phi < \pi/2$$ (where $$d > 0$$)
+and $$\phi > \pi/2$$ (where $$d < 0$$):
$$\begin{aligned}
\boxed{
@@ -66,9 +66,9 @@ $$\begin{aligned}
Next, we would like to know the exact shape of the meniscus.
To do this, we need to describe the liquid surface differently,
-using the elevation angle $\theta$ relative to the $y = 0$ plane.
-The curve $\theta(s)$ is a function of the arc length $s$,
-where $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$,
+using the elevation angle $$\theta$$ relative to the $$y = 0$$ plane.
+The curve $$\theta(s)$$ is a function of the arc length $$s$$,
+where $$\dd{s}^2 = \dd{x}^2 + \dd{y}^2$$,
and is governed by:
$$\begin{aligned}
@@ -82,23 +82,23 @@ $$\begin{aligned}
= \frac{1}{R}
\end{aligned}$$
-The last equation describes the curvature radius $R$
-of the surface along the $x$-axis.
+The last equation describes the curvature radius $$R$$
+of the surface along the $$x$$-axis.
Since we are considering a flat wall,
there is no curvature in the orthogonal principal direction.
Just below the liquid surface in the meniscus,
we expect the hydrostatic pressure
and the [Young-Laplace law](/know/concept/young-laplace-law/)
-to agree about the pressure $p$,
-where $p_0$ is the external air pressure:
+to agree about the pressure $$p$$,
+where $$p_0$$ is the external air pressure:
$$\begin{aligned}
p_0 - \rho g y
= p_0 - \frac{\alpha}{R}
\end{aligned}$$
-Rearranging this yields that $R = L_c^2 / y$.
+Rearranging this yields that $$R = L_c^2 / y$$.
Inserting this into the curvature equation gives us:
$$\begin{aligned}
@@ -106,8 +106,8 @@ $$\begin{aligned}
= \frac{y}{L_c^2}
\end{aligned}$$
-By differentiating this equation with respect to $s$
-and using $\idv{y}{s} = \sin\theta$, we arrive at:
+By differentiating this equation with respect to $$s$$
+and using $$\idv{y}{s} = \sin\theta$$, we arrive at:
$$\begin{aligned}
\boxed{
@@ -115,7 +115,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-To solve this equation, we multiply it by $\idv{\theta}{s}$,
+To solve this equation, we multiply it by $$\idv{\theta}{s}$$,
which is nonzero close to the wall:
$$\begin{aligned}
@@ -123,16 +123,16 @@ $$\begin{aligned}
= \dv{\theta}{s} \sin\theta
\end{aligned}$$
-We integrate both sides with respect to $s$
-and set the integration constant to $1$,
-such that we get zero when $\theta \to 0$ away from the wall:
+We integrate both sides with respect to $$s$$
+and set the integration constant to $$1$$,
+such that we get zero when $$\theta \to 0$$ away from the wall:
$$\begin{aligned}
\frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2
= 1 - \cos\theta
\end{aligned}$$
-Isolating this for $\idv{\theta}{s}$ and using a trigonometric identity then yields:
+Isolating this for $$\idv{\theta}{s}$$ and using a trigonometric identity then yields:
$$\begin{aligned}
\dv{\theta}{s}
@@ -142,7 +142,7 @@ $$\begin{aligned}
\end{aligned}$$
We use trigonometric relations on the equations
-for $\idv{x}{s}$ and $\idv{y}{s}$ to get $\theta$-derivatives:
+for $$\idv{x}{s}$$ and $$\idv{y}{s}$$ to get $$\theta$$-derivatives:
$$\begin{aligned}
\dv{x}{\theta}
@@ -156,7 +156,7 @@ $$\begin{aligned}
= - L_c \cos\!\Big( \frac{\theta}{2} \Big)
\end{aligned}$$
-Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall.
+Let $$\theta_0 = \phi - \pi/2$$ be the initial elevation angle $$\theta(0)$$ at the wall.
Then, by integrating the above equations, we get the following solutions:
$$\begin{gathered}
@@ -172,9 +172,9 @@ $$\begin{gathered}
}
\end{gathered}$$
-Where the integration constant has been chosen such that $y \to 0$ for $\theta \to 0$ away from the wall,
-and $x = 0$ for $\theta = \theta_0$.
-This result is consistent with our earlier expression for $d$:
+Where the integration constant has been chosen such that $$y \to 0$$ for $$\theta \to 0$$ away from the wall,
+and $$x = 0$$ for $$\theta = \theta_0$$.
+This result is consistent with our earlier expression for $$d$$:
$$\begin{aligned}
d
diff --git a/source/know/concept/metacentric-height/index.md b/source/know/concept/metacentric-height/index.md
index e4e6e35..3d0c9a5 100644
--- a/source/know/concept/metacentric-height/index.md
+++ b/source/know/concept/metacentric-height/index.md
@@ -8,8 +8,8 @@ categories:
layout: "concept"
---
-Consider an object with center of mass $G$,
-floating in a large body of liquid whose surface is flat at $z = 0$.
+Consider an object with center of mass $$G$$,
+floating in a large body of liquid whose surface is flat at $$z = 0$$.
For our purposes, it is easiest to use a coordinate system
whose origin is at the area centroid
of the object's cross-section through the liquid's surface, namely:
@@ -19,9 +19,9 @@ $$\begin{aligned}
\equiv \frac{1}{A_{wl}} \iint_{wl} (x, y) \dd{A}
\end{aligned}$$
-Where $A_{wl}$ is the cross-sectional area
+Where $$A_{wl}$$ is the cross-sectional area
enclosed by the "waterline" around the "boat".
-Note that the boat's center of mass $G$
+Note that the boat's center of mass $$G$$
does not coincide with the origin in general,
as is illustrated in the following sketch
of our choice of coordinate system:
@@ -30,13 +30,13 @@ of our choice of coordinate system:
-Here, $B$ is the **center of buoyancy**, equal to
+Here, $$B$$ is the **center of buoyancy**, equal to
the center of mass of the volume of water displaced by the boat
as per [Archimedes' principle](/know/concept/archimedes-principle/).
-At equilibrium, the forces of buoyancy $\vb{F}_B$ and gravity $\vb{F}_G$
+At equilibrium, the forces of buoyancy $$\vb{F}_B$$ and gravity $$\vb{F}_G$$
have equal magnitudes in opposite directions,
-and $B$ is directly above or below $G$,
-or in other words, $x_B = x_G$ and $y_B = y_G$,
+and $$B$$ is directly above or below $$G$$,
+or in other words, $$x_B = x_G$$ and $$y_B = y_G$$,
which are calculated as follows:
$$\begin{aligned}
@@ -47,12 +47,12 @@ $$\begin{aligned}
&\equiv \frac{1}{V_{disp}} \iiint_{disp} (x, y, z) \dd{V}
\end{aligned}$$
-Where $V_{boat}$ is the volume of the whole boat,
-and $V_{disp}$ is the volume of liquid it displaces.
+Where $$V_{boat}$$ is the volume of the whole boat,
+and $$V_{disp}$$ is the volume of liquid it displaces.
Whether a given equilibrium is *stable* is more complicated.
-Suppose the ship is tilted by a small angle $\theta$ around the $x$-axis,
-in which case the old waterline, previously in the $z = 0$ plane,
+Suppose the ship is tilted by a small angle $$\theta$$ around the $$x$$-axis,
+in which case the old waterline, previously in the $$z = 0$$ plane,
gets shifted to a new plane, namely:
$$\begin{aligned}
@@ -61,8 +61,8 @@ $$\begin{aligned}
\approx \theta y
\end{aligned}$$
-Then $V_{disp}$ changes by $\Delta V_{disp}$, which is estimated below.
-If a point of the old waterline is raised by $z$,
+Then $$V_{disp}$$ changes by $$\Delta V_{disp}$$, which is estimated below.
+If a point of the old waterline is raised by $$z$$,
then the displaced liquid underneath it is reduced proportionally,
hence the sign:
@@ -73,17 +73,17 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-So $V_{disp}$ is unchanged, at least to first order in $\theta$.
+So $$V_{disp}$$ is unchanged, at least to first order in $$\theta$$.
However, the *shape* of the displaced volume may have changed significantly.
-Therefore, the shift of the position of the buoyancy center from $B$ to $B'$
-involves a correction $\Delta y_B$ in addition to the rotation by $\theta$:
+Therefore, the shift of the position of the buoyancy center from $$B$$ to $$B'$$
+involves a correction $$\Delta y_B$$ in addition to the rotation by $$\theta$$:
$$\begin{aligned}
y_B'
= y_B - \theta z_B + \Delta y_B
\end{aligned}$$
-We find $\Delta y_B$ by calculating the virtual buoyancy center of the shape difference:
+We find $$\Delta y_B$$ by calculating the virtual buoyancy center of the shape difference:
on the side of the boat that has been lifted by the rotation,
the center of buoyancy is "pushed" away due to the reduced displacement there,
and vice versa on the other side. Consequently:
@@ -95,7 +95,7 @@ $$\begin{aligned}
= - \frac{\theta I}{V_{disp}}
\end{aligned}$$
-Where we have defined the so-called **area moment** $I$ of the waterline as follows:
+Where we have defined the so-called **area moment** $$I$$ of the waterline as follows:
$$\begin{aligned}
\boxed{
@@ -104,8 +104,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Now that we have an expression for $\Delta y_B$,
-the new center's position $y_B'$ is found to be:
+Now that we have an expression for $$\Delta y_B$$,
+the new center's position $$y_B'$$ is found to be:
$$\begin{aligned}
y_B'
@@ -113,8 +113,8 @@ $$\begin{aligned}
\approx y_B - \sin(\theta) \: \Big( z_B + \frac{I}{V_{disp}} \Big)
\end{aligned}$$
-This looks like a rotation by $\theta$ around a so-called **metacenter** $M$,
-with a height $z_M$ known as the **metacentric height**, defined as:
+This looks like a rotation by $$\theta$$ around a so-called **metacenter** $$M$$,
+with a height $$z_M$$ known as the **metacentric height**, defined as:
$$\begin{aligned}
\boxed{
@@ -123,22 +123,22 @@ $$\begin{aligned}
}
\end{aligned}$$
-Meanwhile, the position of $M$ is defined such that it lies
-on the line between the old centers $G$ and $B$.
-Our calculation of $y_B'$ has shown that the new $B'$ always lies below $M$.
+Meanwhile, the position of $$M$$ is defined such that it lies
+on the line between the old centers $$G$$ and $$B$$.
+Our calculation of $$y_B'$$ has shown that the new $$B'$$ always lies below $$M$$.
After the rotation, the boat is not in equilibrium anymore,
-because the new $G'$ is not directly above or below $B'$.
-The force of gravity then causes a torque $\vb{T}$ given by:
+because the new $$G'$$ is not directly above or below $$B'$$.
+The force of gravity then causes a torque $$\vb{T}$$ given by:
$$\begin{aligned}
\vb{T}
= (\vb{r}_G' - \vb{r}_B') \cross m \vb{g}
\end{aligned}$$
-Where $\vb{g}$ points downwards.
-Since the rotation was around the $x$-axis,
-we are only interested in the $x$-component $T_x$, which becomes:
+Where $$\vb{g}$$ points downwards.
+Since the rotation was around the $$x$$-axis,
+we are only interested in the $$x$$-component $$T_x$$, which becomes:
$$\begin{aligned}
T_x
@@ -146,17 +146,17 @@ $$\begin{aligned}
= - \big((y_G - \theta z_G) - (y_B - \theta z_M)\big) m \mathrm{g}
\end{aligned}$$
-With $y_G' = y_G - \theta z_G$ being a simple rotation of $G$.
-At the initial equilibrium $y_G = y_B$, so:
+With $$y_G' = y_G - \theta z_G$$ being a simple rotation of $$G$$.
+At the initial equilibrium $$y_G = y_B$$, so:
$$\begin{aligned}
T_x
= \theta (z_G - z_M) m \mathrm{g}
\end{aligned}$$
-If $z_M < z_G$, then $T_x$ has the same sign as $\theta$,
-so $\vb{T}$ further destabilizes the boat.
-But if $z_M > z_G$, then $\vb{T}$ counteracts the rotation,
+If $$z_M < z_G$$, then $$T_x$$ has the same sign as $$\theta$$,
+so $$\vb{T}$$ further destabilizes the boat.
+But if $$z_M > z_G$$, then $$\vb{T}$$ counteracts the rotation,
and the boat returns to the original equilibrium,
leading us to the following stability condition:
@@ -167,7 +167,7 @@ $$\begin{aligned}
\end{aligned}$$
In other words, for a given boat design (or general shape)
-$z_G$ and $z_M$ can be calculated,
+$$z_G$$ and $$z_M$$ can be calculated,
and as long as they satisfy the above inequality,
it will float stably in water (or any other fluid,
although the buoyancy depends significantly on the density).
diff --git a/source/know/concept/microcanonical-ensemble/index.md b/source/know/concept/microcanonical-ensemble/index.md
index 03d37de..a56a506 100644
--- a/source/know/concept/microcanonical-ensemble/index.md
+++ b/source/know/concept/microcanonical-ensemble/index.md
@@ -10,18 +10,18 @@ layout: "concept"
---
The **microcanonical** or **NVE ensemble** is a statistical model
-of a theoretical system with constant internal energy $U$,
-volume $V$, and particle count $N$.
+of a theoretical system with constant internal energy $$U$$,
+volume $$V$$, and particle count $$N$$.
Consider a box with those properties.
We now put an imaginary rigid wall inside the box,
-thus dividing it into two subsystems $A$ and $B$,
+thus dividing it into two subsystems $$A$$ and $$B$$,
which can exchange energy (i.e. heat), but no particles.
-At any time, $A$ has energy $U_A$, and $B$ has $U_B$,
-so that in total $U = U_A \!+\! U_B$.
+At any time, $$A$$ has energy $$U_A$$, and $$B$$ has $$U_B$$,
+so that in total $$U = U_A \!+\! U_B$$.
The particles in each subsystem are in a certain **microstate** (configuration).
-For a given $U$, there is a certain number $c$
+For a given $$U$$, there is a certain number $$c$$
of possible whole-box microstates with that energy, given by:
$$\begin{aligned}
@@ -29,43 +29,43 @@ $$\begin{aligned}
= \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A)
\end{aligned}$$
-Where $c_A$ and $c_B$ are the numbers of subsystem microstates
+Where $$c_A$$ and $$c_B$$ are the numbers of subsystem microstates
at the given energy levels.
The core assumption of the microcanonical ensemble
-is that each of these microstates has the same probability $1 / c$.
-Consequently, the probability of finding an energy $U_A$ in $A$ is:
+is that each of these microstates has the same probability $$1 / c$$.
+Consequently, the probability of finding an energy $$U_A$$ in $$A$$ is:
$$\begin{aligned}
p_A(U_A)
= \frac{c_A(U_A) \: c_B(U - U_A)}{c(U)}
\end{aligned}$$
-If a certain $U_A$ has a higher probability,
-then there are more $A$-microstates with that energy,
+If a certain $$U_A$$ has a higher probability,
+then there are more $$A$$-microstates with that energy,
so, statistically, for an *ensemble* of many boxes,
-we expect that $U_A$ is more common.
+we expect that $$U_A$$ is more common.
-The maximum of $p_A$ will be the most common in the ensemble.
+The maximum of $$p_A$$ will be the most common in the ensemble.
Assuming that we have given the boxes enough time to settle,
we go one step further,
and refer to this maximum as "equilibrium".
In other words, the subsystem microstates at equilibrium
-are maxima of $p_A$ and $p_B$.
+are maxima of $$p_A$$ and $$p_B$$.
-We only need to look at $p_A$.
-Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$:
+We only need to look at $$p_A$$.
+Clearly, a maximum of $$p_A$$ is also a maximum of $$\ln p_A$$:
$$\begin{aligned}
\ln p_A(U_A)
= \ln{c_A(U_A)} + \ln{c_B(U - U_A)} - \ln{c(U)}
\end{aligned}$$
-Here, in the quantity $\ln{c_A}$,
+Here, in the quantity $$\ln{c_A}$$,
we recognize the definition of
-the entropy $S_A \equiv k \ln{c_A}$,
-where $k$ is Boltzmann's constant.
-We thus multiply by $k$:
+the entropy $$S_A \equiv k \ln{c_A}$$,
+where $$k$$ is Boltzmann's constant.
+We thus multiply by $$k$$:
$$\begin{aligned}
k \ln p_A(U_A)
@@ -73,10 +73,10 @@ $$\begin{aligned}
\end{aligned}$$
Since entropy is additive over subsystems,
-the total is $S = S_A + S_B$.
+the total is $$S = S_A + S_B$$.
To reach equilibrium, we are thus
**maximizing the total entropy**,
-meaning that $S$ is the [thermodynamic potential](/know/concept/thermodynamic-potential/)
+meaning that $$S$$ is the [thermodynamic potential](/know/concept/thermodynamic-potential/)
that corresponds to the microcanonical ensemble.
For our example, maximizing gives the following,
@@ -90,7 +90,7 @@ $$\begin{aligned}
\end{aligned}$$
By definition, the energy-derivative of the entropy
-is the reciprocal temperature $1 / T$.
+is the reciprocal temperature $$1 / T$$.
In other words,
equilibrium is reached when both subsystems
are at the same temperature:
@@ -102,16 +102,16 @@ $$\begin{aligned}
= \frac{1}{T_B}
\end{aligned}$$
-Recall that our partitioning into $A$ and $B$ was arbitrary,
-meaning that, in fact, the temperature $T$ must be uniform in the whole box.
+Recall that our partitioning into $$A$$ and $$B$$ was arbitrary,
+meaning that, in fact, the temperature $$T$$ must be uniform in the whole box.
We get this specific result because
-heat was the only thing that $A$ and $B$ could exchange.
+heat was the only thing that $$A$$ and $$B$$ could exchange.
The point is that the most likely state of the box
-maximizes the total entropy $S$.
+maximizes the total entropy $$S$$.
We also would have reached that conclusion
if our imaginary wall was permeable and flexible,
-i.e if it allowed changes in volume $V_A$ and particle count $N_A$.
+i.e if it allowed changes in volume $$V_A$$ and particle count $$N_A$$.
diff --git a/source/know/concept/modulational-instability/index.md b/source/know/concept/modulational-instability/index.md
index 628cbb4..a01293c 100644
--- a/source/know/concept/modulational-instability/index.md
+++ b/source/know/concept/modulational-instability/index.md
@@ -15,17 +15,17 @@ In fiber optics, **modulational instability** (MI)
is a nonlinear effect that leads to the exponential amplification
of background noise in certain frequency regions.
It only occurs in the [anomalous dispersion regime](/know/concept/dispersive-broadening/)
-($\beta_2 < 0$), which we will prove shortly.
+($$\beta_2 < 0$$), which we will prove shortly.
Consider the following simple solution to the nonlinear Schrödinger equation:
-a time-invariant constant power $P_0$ at the carrier frequency $\omega_0$,
+a time-invariant constant power $$P_0$$ at the carrier frequency $$\omega_0$$,
which is experiencing [self-phase modulation](/know/concept/self-phase-modulation/):
$$\begin{aligned}
A(z,t) = \sqrt{P_0} \exp( i \gamma P_0 z)
\end{aligned}$$
-We add a small perturbation $\varepsilon(z,t)$ to this signal,
+We add a small perturbation $$\varepsilon(z,t)$$ to this signal,
representing background noise:
$$\begin{aligned}
@@ -33,8 +33,8 @@ $$\begin{aligned}
\end{aligned}$$
We insert this into the nonlinear Schrödinger equation to get a perturbation equation,
-which we linearize by assuming that $|\varepsilon|^2$ is negligible compared to $P_0$,
-such that all higher-order terms of $\varepsilon$ can be dropped, yielding:
+which we linearize by assuming that $$|\varepsilon|^2$$ is negligible compared to $$P_0$$,
+such that all higher-order terms of $$\varepsilon$$ can be dropped, yielding:
$$\begin{aligned}
0
@@ -51,9 +51,9 @@ $$\begin{aligned}
\end{aligned}$$
We split the perturbation into real and imaginary parts
-$\varepsilon(z,t) = \varepsilon_r(z,t) + i \varepsilon_i(z,t)$,
+$$\varepsilon(z,t) = \varepsilon_r(z,t) + i \varepsilon_i(z,t)$$,
which we fill in in this equation.
-The point is that $\varepsilon_r$ and $\varepsilon_i$ are real functions:
+The point is that $$\varepsilon_r$$ and $$\varepsilon_i$$ are real functions:
$$\begin{aligned}
0
@@ -63,7 +63,7 @@ $$\begin{aligned}
\end{aligned}$$
Splitting this into its real and imaginary parts gives two PDEs
-relating $\varepsilon_r$ and $\varepsilon_i$:
+relating $$\varepsilon_r$$ and $$\varepsilon_i$$:
$$\begin{aligned}
\pdv{\varepsilon_r}{z} = \frac{\beta_2}{2} \pdvn{2}{\varepsilon_i}{t}
@@ -72,8 +72,8 @@ $$\begin{aligned}
\end{aligned}$$
We [Fourier transform](/know/concept/fourier-transform/)
-these in $t$ to turn them into ODEs relating
-$\tilde{\varepsilon}_r(z,\omega)$ and $\tilde{\varepsilon}_i(z,\omega)$:
+these in $$t$$ to turn them into ODEs relating
+$$\tilde{\varepsilon}_r(z,\omega)$$ and $$\tilde{\varepsilon}_i(z,\omega)$$:
$$\begin{aligned}
\pdv{\tilde{\varepsilon}_r}{z} = - \frac{\beta_2}{2} \omega^2 \tilde{\varepsilon}_i
@@ -82,7 +82,7 @@ $$\begin{aligned}
\end{aligned}$$
We are interested in exponential growth, so let us make the following ansatz,
-where $k$ may be a function of $\omega$, as long as it is $z$-invariant:
+where $$k$$ may be a function of $$\omega$$, as long as it is $$z$$-invariant:
$$\begin{aligned}
\tilde{\varepsilon}_r(z, \omega) = \tilde{\varepsilon}_r(0, \omega) \exp(k z)
@@ -91,7 +91,7 @@ $$\begin{aligned}
\end{aligned}$$
With this, we can write the system of ODEs for
-$\tilde{\varepsilon}_r(z,\omega)$ and $\tilde{\varepsilon}_i(z,\omega)$
+$$\tilde{\varepsilon}_r(z,\omega)$$ and $$\tilde{\varepsilon}_i(z,\omega)$$
in matrix form:
$$\begin{aligned}
@@ -112,7 +112,7 @@ $$\begin{aligned}
k = \pm \sqrt{ - \frac{\beta_2}{2} \omega^2 \Big( \frac{\beta_2}{2} \omega^2 + 2 \gamma P_0 \Big) }
\end{aligned}$$
-To get exponential growth, it is essential that $\mathrm{Re}\{k\} > 0$,
+To get exponential growth, it is essential that $$\mathrm{Re}\{k\} > 0$$,
so we discard the negative sign,
and get the following condition for MI:
@@ -124,22 +124,22 @@ $$\begin{aligned}
}
\end{aligned}$$
-Since $\omega^2$ is positive, $\beta_2$ must be negative,
+Since $$\omega^2$$ is positive, $$\beta_2$$ must be negative,
so MI can only occur in the ADR.
-It is worth noting that $\beta_2 = \beta_2(\omega_0)$,
+It is worth noting that $$\beta_2 = \beta_2(\omega_0)$$,
meaning there can only be exponential
noise growth when the "parent pulse" is in the anomalous dispersion regime,
but that growth may appear in areas of normal dispersion,
as long as the above condition is satisfied by the parent.
This result has been derived using perturbation,
-so only holds as long as $|\varepsilon|^2 \ll P_0$.
+so only holds as long as $$|\varepsilon|^2 \ll P_0$$.
Over time, the noise gets amplified so greatly
that this approximation breaks down.
-Next, we define the **gain** $g(\omega)$,
+Next, we define the **gain** $$g(\omega)$$,
which expresses how quickly the
-perturbation grows as a function of the frequency offset $\omega$:
+perturbation grows as a function of the frequency offset $$\omega$$:
$$\begin{aligned}
\boxed{
@@ -149,7 +149,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The frequencies with maximum gain are then found as extrema of $g(\omega)$,
+The frequencies with maximum gain are then found as extrema of $$g(\omega)$$,
which satisfy:
$$\begin{aligned}
@@ -162,10 +162,10 @@ $$\begin{aligned}
A simulation of MI is illustrated below.
The pulse considered was a solition of the following form
-with settings $T_0 = 10\:\mathrm{ps}$, $P_0 = 10\:\mathrm{kW}$,
-$\beta = -10\:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$,
+with settings $$T_0 = 10\:\mathrm{ps}$$, $$P_0 = 10\:\mathrm{kW}$$,
+$$\beta = -10\:\mathrm{ps}^2/\mathrm{m}$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$,
whose peak is approximately flat, so our derivation is valid there,
-hence it "wrinkles" in the $t$-domain:
+hence it "wrinkles" in the $$t$$-domain:
$$\begin{aligned}
A(0, t)
@@ -176,19 +176,19 @@ $$\begin{aligned}
-Where $L_\mathrm{NL} = 1/(\gamma P_0)$ is the characteristic length of nonlinear effects.
+Where $$L_\mathrm{NL} = 1/(\gamma P_0)$$ is the characteristic length of nonlinear effects.
Note that no noise was added to the simulation;
what you are seeing are pure numerical errors getting amplified.
-If one of the gain peaks accumulates a lot of energy quickly ($L_\mathrm{NL}$ is small),
+If one of the gain peaks accumulates a lot of energy quickly ($$L_\mathrm{NL}$$ is small),
and that peak is in the anomalous dispersion regime,
then it can in turn also cause MI in its own surroundings,
leading to a cascade of secondary and tertiary gain areas.
-This is seen above for $z > 30 L_\mathrm{NL}$.
+This is seen above for $$z > 30 L_\mathrm{NL}$$.
What we described is "pure" MI, but there also exists
a different type caused by Raman scattering.
-In that case, amplification occurs at the strongest peak of the Raman gain $\tilde{g}_R(\omega)$,
+In that case, amplification occurs at the strongest peak of the Raman gain $$\tilde{g}_R(\omega)$$,
even when the parent pulse is in the NDR.
This is an example of stimulated Raman scattering (SRS).
diff --git a/source/know/concept/multi-photon-absorption/index.md b/source/know/concept/multi-photon-absorption/index.md
index af433ac..5dd9887 100644
--- a/source/know/concept/multi-photon-absorption/index.md
+++ b/source/know/concept/multi-photon-absorption/index.md
@@ -11,10 +11,10 @@ categories:
layout: "concept"
---
-Consider a quantum system where there are many eigenstates $\Ket{n}$,
+Consider a quantum system where there are many eigenstates $$\Ket{n}$$,
e.g. atomic orbitals, for an electron to occupy.
Suppose an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
-passes by, such that its Hamiltonian gets perturbed by $\hat{H}_1$, given in the
+passes by, such that its Hamiltonian gets perturbed by $$\hat{H}_1$$, given in the
[electric dipole approximation](/know/concept/electric-dipole-approximation/) by:
$$\begin{aligned}
@@ -23,19 +23,19 @@ $$\begin{aligned}
\approx -\vu{p} \cdot \vb{E} e^{-i \omega t}
\end{aligned}$$
-Where $\vb{E}$ is the [electric field](/know/concept/electric-field/) amplitude,
-and $\vu{p} \equiv q \vu{x}$ is the transition dipole moment operator.
+Where $$\vb{E}$$ is the [electric field](/know/concept/electric-field/) amplitude,
+and $$\vu{p} \equiv q \vu{x}$$ is the transition dipole moment operator.
Here, we have made the
[rotating wave approximation](/know/concept/rotating-wave-approximation/)
-to neglect the $e^{i \omega t}$ term,
+to neglect the $$e^{i \omega t}$$ term,
because it turns out to be irrelevant in this discussion.
-We call the ground state $\Ket{0}$,
+We call the ground state $$\Ket{0}$$,
but other than that, the other states need *not* be sorted by energy.
However, we demand that the following holds
-for all even-numbered states $\Ket{e}$ and $\Ket{e'}$,
-and for all odd-numbered ($u$neven) states $\Ket{u}$ and $\Ket{u'}$:
+for all even-numbered states $$\Ket{e}$$ and $$\Ket{e'}$$,
+and for all odd-numbered ($$u$$neven) states $$\Ket{u}$$ and $$\Ket{u'}$$:
$$\begin{aligned}
\matrixel{e}{\hat{H}_1}{e'} = \matrixel{u}{\hat{H}_1}{u'} = 0
@@ -46,7 +46,7 @@ $$\begin{aligned}
This is justified for atomic orbitals thanks to
[Laporte's selection rule](/know/concept/selection-rules/).
Therefore, [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/)
-says that the $N$th-order coefficient corrections are:
+says that the $$N$$th-order coefficient corrections are:
$$\begin{aligned}
c_e^{(N)}(t)
@@ -56,8 +56,8 @@ $$\begin{aligned}
&= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(N-1)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau}
\end{aligned}$$
-Where $\omega_{eu} = (E_e \!-\! E_u) / \hbar$.
-For simplicity, the electron starts in the lowest-energy state $\Ket{0}$:
+Where $$\omega_{eu} = (E_e \!-\! E_u) / \hbar$$.
+For simplicity, the electron starts in the lowest-energy state $$\Ket{0}$$:
$$\begin{aligned}
c_0^{(0)} = 1
@@ -65,8 +65,8 @@ $$\begin{aligned}
c_u^{(0)} = c_{e \neq 0}^{(0)} = 0
\end{aligned}$$
-Finally, we prove the following useful relation for large $t$,
-involving a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
+Finally, we prove the following useful relation for large $$t$$,
+involving a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$:
$$\begin{aligned}
\lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
@@ -86,7 +86,7 @@ $$\begin{aligned}
= i e^{i x t / 2} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
\end{aligned}$$
-By taking the limit $t \to \infty$,
+By taking the limit $$t \to \infty$$,
it can be turned into a nascent Dirac delta function:
$$\begin{aligned}
@@ -102,7 +102,7 @@ $$\begin{aligned}
= 4 \pi^2 \delta^2(x)
\end{aligned}$$
-However, a squared delta function $\delta^2$ is not ideal,
+However, a squared delta function $$\delta^2$$ is not ideal,
so we take a step back:
$$\begin{aligned}
@@ -111,7 +111,7 @@ $$\begin{aligned}
= \delta(x) \lim_{t \to \infty} \frac{t}{2 \pi}
\end{aligned}$$
-Where we have set $x = 0$ according to the first delta function.
+Where we have set $$x = 0$$ according to the first delta function.
This gives the target:
$$\begin{aligned}
@@ -119,6 +119,7 @@ $$\begin{aligned}
= 4 \pi^2 \delta^2(x)
= 2 \pi \: \delta(x) \: t
\end{aligned}$$
+
@@ -127,7 +128,7 @@ $$\begin{aligned}
To warm up, we start at first-order perturbation theory.
Thanks to our choice of initial condition,
-nothing at all happens to any of the even-numbered states $\Ket{e}$:
+nothing at all happens to any of the even-numbered states $$\Ket{e}$$:
$$\begin{aligned}
c_e^{(1)}(t)
@@ -135,8 +136,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-While the odd-numbered states $\Ket{u}$ have a nonzero correction $c_u^{(1)}$,
-where $\vb{p}_{u0} = \matrixel{u}{\vu{p}}{0}$:
+While the odd-numbered states $$\Ket{u}$$ have a nonzero correction $$c_u^{(1)}$$,
+where $$\vb{p}_{u0} = \matrixel{u}{\vu{p}}{0}$$:
$$\begin{aligned}
c_u^{(1)}(t)
@@ -157,10 +158,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Since $\big| c_u^{(1)}(t) \big|^2$ is the probability
-of finding the electron in $\Ket{u}$,
-its transition rate $R_u^{(1)}(t)$ is as follows,
-averaged since the beginning $t = 0$:
+Since $$\big| c_u^{(1)}(t) \big|^2$$ is the probability
+of finding the electron in $$\Ket{u}$$,
+its transition rate $$R_u^{(1)}(t)$$ is as follows,
+averaged since the beginning $$t = 0$$:
$$\begin{aligned}
R_u^{(1)}(t)
@@ -169,7 +170,7 @@ $$\begin{aligned}
\cdot \bigg| \frac{e^{i (\omega_{u0} - \omega) t} - 1}{\omega_{u0} - \omega} \bigg|^2
\end{aligned}$$
-For large $t \to \infty$, we can use the formula we proved earlier
+For large $$t \to \infty$$, we can use the formula we proved earlier
to get [Fermi's golden rule](/know/concept/fermis-golden-rule/):
$$\begin{aligned}
@@ -180,17 +181,17 @@ $$\begin{aligned}
\end{aligned}$$
This well-known formula represents **one-photon absorption**:
-it peaks at $\omega_{u0} = \omega$, i.e. when one photon $\hbar \omega$
-has the exact energy of the transition $\hbar \omega_{u0}$.
-Note that this transition is only possible when $\matrixel{u}{\vu{p}}{0} \neq 0$,
-i.e. for any odd-numbered final state $\Ket{u}$.
+it peaks at $$\omega_{u0} = \omega$$, i.e. when one photon $$\hbar \omega$$
+has the exact energy of the transition $$\hbar \omega_{u0}$$.
+Note that this transition is only possible when $$\matrixel{u}{\vu{p}}{0} \neq 0$$,
+i.e. for any odd-numbered final state $$\Ket{u}$$.
## Two-photon absorption
Next, we go to second-order perturbation theory.
Based on the previous result, this time
-all odd-numbered states $\Ket{u}$ are unaffected:
+all odd-numbered states $$\Ket{u}$$ are unaffected:
$$\begin{aligned}
c_u^{(2)}(t)
@@ -198,8 +199,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-While the even-numbered states $\Ket{e}$ have the following correction,
-using $\omega_{eu} \!+\! \omega_{u0} = \omega_{e0}$:
+While the even-numbered states $$\Ket{e}$$ have the following correction,
+using $$\omega_{eu} \!+\! \omega_{u0} = \omega_{e0}$$:
$$\begin{aligned}
c_e^{(2)}(t)
@@ -213,7 +214,7 @@ $$\begin{aligned}
- \frac{e^{i (\omega_{eu} - \omega) \tau}}{i (\omega_{eu} - \omega)} \bigg]_0^t
\end{aligned}$$
-The second term represents one-photon absorption between $\Ket{u}$ and $\Ket{e}$.
+The second term represents one-photon absorption between $$\Ket{u}$$ and $$\Ket{e}$$.
We do not care about that, so we drop it, leaving only the first term:
$$\begin{aligned}
@@ -224,7 +225,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-As before, we can define a rate $R_e^{(2)}(t)$
+As before, we can define a rate $$R_e^{(2)}(t)$$
for all transitions represented by this term:
$$\begin{aligned}
@@ -234,7 +235,7 @@ $$\begin{aligned}
\cdot \bigg| \frac{e^{i (\omega_{e0} - 2 \omega) t} - 1}{\omega_{e0} - 2 \omega} \bigg|^2
\end{aligned}$$
-Which for $t \to \infty$ takes a similar form to Fermi's golden rule,
+Which for $$t \to \infty$$ takes a similar form to Fermi's golden rule,
using the formula we proved:
$$\begin{aligned}
@@ -245,19 +246,19 @@ $$\begin{aligned}
}
\end{aligned}$$
-This represents **two-photon absorption**, since it peaks at $\omega_{e0} = 2 \omega$:
-two identical photons $\hbar \omega$ are absorbed simultaneously
-to bridge the energy gap $\hbar \omega_{e0}$.
-Surprisingly, such a transition can only occur when $\matrixel{e}{\vu{p}}{0} = 0$,
-i.e. for any even-numbered final state $\Ket{e}$.
-Notice that the rate is proportional to $|\vb{E}|^4$,
+This represents **two-photon absorption**, since it peaks at $$\omega_{e0} = 2 \omega$$:
+two identical photons $$\hbar \omega$$ are absorbed simultaneously
+to bridge the energy gap $$\hbar \omega_{e0}$$.
+Surprisingly, such a transition can only occur when $$\matrixel{e}{\vu{p}}{0} = 0$$,
+i.e. for any even-numbered final state $$\Ket{e}$$.
+Notice that the rate is proportional to $$|\vb{E}|^4$$,
so this effect is only noticeable at high light intensities.
## Three-photon absorption
For third-order perturbation theory,
-all even-numbered states $\Ket{e}$ are unchanged:
+all even-numbered states $$\Ket{e}$$ are unchanged:
$$\begin{aligned}
c_e^{(3)}(t)
@@ -265,7 +266,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-And the odd-numbered states $\Ket{u}$ get the following third-order corrections:
+And the odd-numbered states $$\Ket{u}$$ get the following third-order corrections:
$$\begin{aligned}
c_u^{(3)}(t)
@@ -294,7 +295,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The resulting transition rate $R_u^{(3)}(t)$
+The resulting transition rate $$R_u^{(3)}(t)$$
is found to have the following familiar form:
$$\begin{aligned}
@@ -317,31 +318,31 @@ $$\begin{aligned}
}
\end{aligned}$$
-This represents **three-photon absorption**, since it peaks at $\omega_{u0} = 3 \omega$:
-three identical photons $\hbar \omega$ are absorbed simultaneously
-to bridge the energy gap $\hbar \omega_{u0}$.
+This represents **three-photon absorption**, since it peaks at $$\omega_{u0} = 3 \omega$$:
+three identical photons $$\hbar \omega$$ are absorbed simultaneously
+to bridge the energy gap $$\hbar \omega_{u0}$$.
This process is similar to one-photon absorption,
-in the sense that it can only occur if $\matrixel{u}{\vu{p}}{0} \neq 0$.
-The rate is proportional to $|\vb{E}|^6$,
+in the sense that it can only occur if $$\matrixel{u}{\vu{p}}{0} \neq 0$$.
+The rate is proportional to $$|\vb{E}|^6$$,
so this effect only appears at extremely high light intensities.
## N-photon absorption
A pattern has appeared in these calculations:
-in $N$th-order perturbation theory,
-we get a term representing $N$-photon absorption,
-with a transition rate proportional to $|\vb{E}|^{2N}$.
+in $$N$$th-order perturbation theory,
+we get a term representing $$N$$-photon absorption,
+with a transition rate proportional to $$|\vb{E}|^{2N}$$.
Indeed, we can derive infinitely many formulas in this way,
although the results become increasingly unrealistic
-due to the dependence on $\vb{E}$.
+due to the dependence on $$\vb{E}$$.
-If $N$ is odd, only odd-numbered destinations $\Ket{u}$ are allowed
-(assuming the electron starts in the ground state $\Ket{0}$),
-and if $N$ is even, only even-numbered destinations $\Ket{e}$.
+If $$N$$ is odd, only odd-numbered destinations $$\Ket{u}$$ are allowed
+(assuming the electron starts in the ground state $$\Ket{0}$$),
+and if $$N$$ is even, only even-numbered destinations $$\Ket{e}$$.
Note that nothing has been said about the energies of these states
-(other than $\Ket{0}$ being the minimum);
-everything is determined by the matrix elements $\matrixel{f}{\vu{p}}{i}$.
+(other than $$\Ket{0}$$ being the minimum);
+everything is determined by the matrix elements $$\matrixel{f}{\vu{p}}{i}$$.
diff --git a/source/know/concept/navier-cauchy-equation/index.md b/source/know/concept/navier-cauchy-equation/index.md
index 13a1ebb..5071c5f 100644
--- a/source/know/concept/navier-cauchy-equation/index.md
+++ b/source/know/concept/navier-cauchy-equation/index.md
@@ -12,9 +12,9 @@ The **Navier-Cauchy equation** describes **elastodynamics**:
the movements inside an elastic solid
in response to external forces and/or internal stresses.
-For a particle of the solid, whose position is given by the displacement field $\va{u}$,
+For a particle of the solid, whose position is given by the displacement field $$\va{u}$$,
Newton's second law is as follows,
-where $\dd{m}$ and $\dd{V}$ are the particle's mass and volume, respectively:
+where $$\dd{m}$$ and $$\dd{V}$$ are the particle's mass and volume, respectively:
$$\begin{aligned}
\va{f^*} \dd{V}
@@ -22,10 +22,10 @@ $$\begin{aligned}
= \rho \pdvn{2}{\va{u}}{t} \dd{V}
\end{aligned}$$
-Where $\rho$ is the mass density,
-and $\va{f^*}$ is the effective force density,
-defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
-like so, with $\va{f}$ being an external body force, e.g. from gravity:
+Where $$\rho$$ is the mass density,
+and $$\va{f^*}$$ is the effective force density,
+defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$
+like so, with $$\va{f}$$ being an external body force, e.g. from gravity:
$$\begin{aligned}
\va{f^*}
@@ -34,17 +34,17 @@ $$\begin{aligned}
We can therefore write Newton's second law as follows,
while switching to index notation,
-where $\nabla_j = \ipdv{}{x_j}$ is the partial derivative
-with respect to the $j$th coordinate:
+where $$\nabla_j = \ipdv{}{x_j}$$ is the partial derivative
+with respect to the $$j$$th coordinate:
$$\begin{aligned}
f_i + \sum_{j} \nabla_j \sigma_{ij}
= \rho \pdvn{2}{u_i}{t}
\end{aligned}$$
-The components $\sigma_{ij}$ of the Cauchy stress tensor
+The components $$\sigma_{ij}$$ of the Cauchy stress tensor
are given by [Hooke's law](/know/concept/hookes-law/),
-where $\mu$ and $\lambda$ are the Lamé coefficients,
+where $$\mu$$ and $$\lambda$$ are the Lamé coefficients,
which describe the material:
$$\begin{aligned}
@@ -52,10 +52,10 @@ $$\begin{aligned}
= 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk}
\end{aligned}$$
-In turn, the components $u_{ij}$ of the
+In turn, the components $$u_{ij}$$ of the
[Cauchy strain tensor](/know/concept/cauchy-strain-tensor/)
are defined as follows,
-where $u_i$ are once again the components of the displacement vector $\va{u}$:
+where $$u_i$$ are once again the components of the displacement vector $$\va{u}$$:
$$\begin{aligned}
u_{ij}
@@ -72,7 +72,7 @@ $$\begin{aligned}
&= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \nabla_i \sum_{j} u_{jj}
\end{aligned}$$
-And then into this we insert the definition of the strain components $u_{ij}$, yielding:
+And then into this we insert the definition of the strain components $$u_{ij}$$, yielding:
$$\begin{aligned}
\rho \pdvn{2}{u_i}{t}
diff --git a/source/know/concept/navier-stokes-equations/index.md b/source/know/concept/navier-stokes-equations/index.md
index fd26860..964acda 100644
--- a/source/know/concept/navier-stokes-equations/index.md
+++ b/source/know/concept/navier-stokes-equations/index.md
@@ -33,10 +33,10 @@ $$\begin{aligned}
= \va{f^*}
\end{aligned}$$
-$\mathrm{D}/\mathrm{D}t$ is the [material derivative](/know/concept/material-derivative/),
-$\rho$ is the density, and $\va{f^*}$ is the effective force density,
-expressed in terms of an external body force $\va{f}$ (e.g. gravity)
-and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$:
+$$\mathrm{D}/\mathrm{D}t$$ is the [material derivative](/know/concept/material-derivative/),
+$$\rho$$ is the density, and $$\va{f^*}$$ is the effective force density,
+expressed in terms of an external body force $$\va{f}$$ (e.g. gravity)
+and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$:
$$\begin{aligned}
\va{f^*}
@@ -51,8 +51,8 @@ $$\begin{aligned}
= - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
\end{aligned}$$
-Where $\eta$ is the dynamic viscosity.
-Inserting this, we calculate $\nabla \cdot \hat{\sigma}^\top$ in index notation:
+Where $$\eta$$ is the dynamic viscosity.
+Inserting this, we calculate $$\nabla \cdot \hat{\sigma}^\top$$ in index notation:
$$\begin{aligned}
\big( \nabla \cdot \hat{\sigma}^\top \big)_i
@@ -62,7 +62,7 @@ $$\begin{aligned}
&= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i
\end{aligned}$$
-Thanks to incompressibility $\nabla \cdot \va{v} = 0$,
+Thanks to incompressibility $$\nabla \cdot \va{v} = 0$$,
the middle term vanishes, leaving us with:
$$\begin{aligned}
@@ -70,7 +70,7 @@ $$\begin{aligned}
= \va{f} - \nabla p + \eta \nabla^2 \va{v}
\end{aligned}$$
-We assume that the only body force is gravity $\va{f} = \rho \va{g}$.
+We assume that the only body force is gravity $$\va{f} = \rho \va{g}$$.
Newton's second law then becomes:
$$\begin{aligned}
@@ -78,8 +78,8 @@ $$\begin{aligned}
= \rho \va{g} - \nabla p + \eta \nabla^2 \va{v}
\end{aligned}$$
-Dividing by $\rho$, and replacing $\eta$
-with the kinematic viscosity $\nu = \eta/\rho$,
+Dividing by $$\rho$$, and replacing $$\eta$$
+with the kinematic viscosity $$\nu = \eta/\rho$$,
yields the main equation:
$$\begin{aligned}
@@ -90,7 +90,7 @@ $$\begin{aligned}
\end{aligned}$$
Finally, we can optionally allow incompressible fluids
-with an inhomogeneous "lumpy" density $\rho$,
+with an inhomogeneous "lumpy" density $$\rho$$,
by demanding conservation of mass,
just like for the Euler equations:
@@ -115,11 +115,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Due to the definition of viscosity $\nu$ as the molecular "stickiness",
-we have boundary conditions for the velocity field $\va{v}$:
-at any interface, $\va{v}$ must be continuous.
+Due to the definition of viscosity $$\nu$$ as the molecular "stickiness",
+we have boundary conditions for the velocity field $$\va{v}$$:
+at any interface, $$\va{v}$$ must be continuous.
Likewise, Newton's third law demands that the normal component
-of stress $\hat{\sigma} \cdot \vu{n}$ is continuous there.
+of stress $$\hat{\sigma} \cdot \vu{n}$$ is continuous there.
diff --git a/source/know/concept/newtons-bucket/index.md b/source/know/concept/newtons-bucket/index.md
index 757262d..21740b9 100644
--- a/source/know/concept/newtons-bucket/index.md
+++ b/source/know/concept/newtons-bucket/index.md
@@ -10,23 +10,23 @@ layout: "concept"
---
**Newton's bucket** is a cylindrical bucket
-that rotates at angular velocity $\omega$.
+that rotates at angular velocity $$\omega$$.
Due to [viscosity](/know/concept/viscosity/),
any liquid in the bucket is affected by the rotation,
-eventually achieving the exact same $\omega$.
+eventually achieving the exact same $$\omega$$.
However, once in equilibrium, the liquid's surface is not flat,
but curved upwards from the center.
-This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$:
+This is due to the centrifugal force $$\va{F}_\mathrm{f} = m \va{f}$$ on a molecule with mass $$m$$:
$$\begin{aligned}
\va{f}
= \omega^2 \va{r}
\end{aligned}$$
-Where $\va{r}$ is the molecule's position relative to the axis of rotation.
+Where $$\va{r}$$ is the molecule's position relative to the axis of rotation.
This (fictitious) force can be written as the gradient
-of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$:
+of a potential $$\Phi_\mathrm{f}$$, such that $$\va{f} = - \nabla \Phi_\mathrm{f}$$:
$$\begin{aligned}
\Phi_\mathrm{f}
@@ -34,8 +34,8 @@ $$\begin{aligned}
= - \frac{\omega^2}{2} (x^2 + y^2)
\end{aligned}$$
-In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$,
-where $\va{g} = - \nabla \Phi_\mathrm{g}$:
+In addition, each molecule feels a gravitational force $$\va{F}_\mathrm{g} = m \va{g}$$,
+where $$\va{g} = - \nabla \Phi_\mathrm{g}$$:
$$\begin{aligned}
\Phi_\mathrm{g}
@@ -43,7 +43,7 @@ $$\begin{aligned}
\end{aligned}$$
Overall, the molecule therefore feels an "effective" force
-with a potential $\Phi$ given by:
+with a potential $$\Phi$$ given by:
$$\begin{aligned}
\Phi
@@ -51,7 +51,7 @@ $$\begin{aligned}
= \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2)
\end{aligned}$$
-At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$
+At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $$p$$
in the liquid is the one that satisfies:
$$\begin{aligned}
@@ -59,7 +59,7 @@ $$\begin{aligned}
= - \nabla \Phi
\end{aligned}$$
-Removing the gradients gives integration constants $p_0$ and $\Phi_0$,
+Removing the gradients gives integration constants $$p_0$$ and $$\Phi_0$$,
so the equilibrium equation is:
$$\begin{aligned}
@@ -67,16 +67,16 @@ $$\begin{aligned}
= - \rho (\Phi - \Phi_0)
\end{aligned}$$
-We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$,
-where $z_0$ is the liquid height at the center:
+We isolate this for $$p$$ and rewrite $$\Phi_0 = \mathrm{g} z_0$$,
+where $$z_0$$ is the liquid height at the center:
$$\begin{aligned}
p
= p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2)
\end{aligned}$$
-At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure.
-The $z$-coordinate at which this is satisfied is as follows,
+At the surface, we demand that $$p = p_0$$, where $$p_0$$ is the air pressure.
+The $$z$$-coordinate at which this is satisfied is as follows,
telling us that the surface is parabolic:
$$\begin{aligned}
diff --git a/source/know/concept/no-cloning-theorem/index.md b/source/know/concept/no-cloning-theorem/index.md
index d4ca0d4..a91ae6f 100644
--- a/source/know/concept/no-cloning-theorem/index.md
+++ b/source/know/concept/no-cloning-theorem/index.md
@@ -10,13 +10,13 @@ layout: "concept"
---
In quantum mechanics, the **no-cloning theorem** states
-there is no general way to make copies of an arbitrary quantum state $\ket{\psi}$.
+there is no general way to make copies of an arbitrary quantum state $$\ket{\psi}$$.
This has profound implications for quantum information.
To prove this theorem, let us pretend that a machine exists
that can do just that: copy arbitrary quantum states.
-Given an input $\ket{\psi}$ and a blank $\ket{?}$,
-this machines turns $\ket{?}$ into $\ket{\psi}$:
+Given an input $$\ket{\psi}$$ and a blank $$\ket{?}$$,
+this machines turns $$\ket{?}$$ into $$\ket{\psi}$$:
$$\begin{aligned}
\ket{\psi} \ket{?}
@@ -24,7 +24,7 @@ $$\begin{aligned}
\ket{\psi} \ket{\psi}
\end{aligned}$$
-We can use this device to make copies of the basis vectors $\ket{0}$ and $\ket{1}$:
+We can use this device to make copies of the basis vectors $$\ket{0}$$ and $$\ket{1}$$:
$$\begin{aligned}
\ket{0} \ket{?}
@@ -36,7 +36,7 @@ $$\begin{aligned}
\ket{1} \ket{1}
\end{aligned}$$
-If we feed this machine a superposition $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$,
+If we feed this machine a superposition $$\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$$,
we *want* the following behaviour:
$$\begin{aligned}
@@ -47,7 +47,7 @@ $$\begin{aligned}
&= \Big( \alpha^2 \ket{0} \ket{0} + \alpha \beta \ket{0} \ket{1} + \alpha \beta \ket{1} \ket{0} + \beta^2 \ket{1} \ket{1} \Big)
\end{aligned}$$
-Note the appearance of the cross terms with a factor of $\alpha \beta$.
+Note the appearance of the cross terms with a factor of $$\alpha \beta$$.
The problem is that the fundamental linearity of quantum mechanics
dictates different behaviour:
diff --git a/source/know/concept/optical-wave-breaking/index.md b/source/know/concept/optical-wave-breaking/index.md
index aa43d25..42064ff 100644
--- a/source/know/concept/optical-wave-breaking/index.md
+++ b/source/know/concept/optical-wave-breaking/index.md
@@ -14,18 +14,18 @@ In fiber optics, **optical wave breaking** (OWB) is a nonlinear effect
caused by interaction between
[group velocity dispersion](/know/concept/dispersive-broadening/) (GVD) and
[self-phase modulation](/know/concept/self-phase-modulation/) (SPM).
-It only happens in the normal dispersion regime ($\beta_2 > 0$)
+It only happens in the normal dispersion regime ($$\beta_2 > 0$$)
for pulses meeting a certain criterium, as we will see.
SPM creates low frequencies at the front of the pulse, and high ones at the back,
-and if $\beta_2 > 0$, GVD lets low frequencies travel faster than high ones.
+and if $$\beta_2 > 0$$, GVD lets low frequencies travel faster than high ones.
When those effects interact, the pulse gets temporally stretched
in a surprisingly sophisticated way.
-To illustrate this, the instantaneous frequency $\omega_i(z, t) = -\ipdv{\phi}{t}$
+To illustrate this, the instantaneous frequency $$\omega_i(z, t) = -\ipdv{\phi}{t}$$
has been plotted below for a theoretical Gaussian input pulse experiencing OWB,
-with settings $T_0 = 100\:\mathrm{fs}$, $P_0 = 5\:\mathrm{kW}$,
-$\beta_2 = 2\:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$.
+with settings $$T_0 = 100\:\mathrm{fs}$$, $$P_0 = 5\:\mathrm{kW}$$,
+$$\beta_2 = 2\:\mathrm{ps}^2/\mathrm{m}$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$.
In the left panel, we see the typical S-shape caused by SPM,
and the arrows indicate the direction that GVD is pushing the curve in.
@@ -41,30 +41,30 @@ hence the name *wave breaking*:
Several interesting things happen around this moment.
To demonstrate this, spectrograms of the same simulation
have been plotted below, together with pulse profiles
-in both the $t$-domain and $\omega$-domain on an arbitrary linear scale
+in both the $$t$$-domain and $$\omega$$-domain on an arbitrary linear scale
(click the image to get a better look).
Initially, the spectrum broadens due to SPM in the usual way,
but shortly after OWB, this process is stopped by the appearance
-of so-called **sidelobes** in the $\omega$-domain on either side of the pulse.
+of so-called **sidelobes** in the $$\omega$$-domain on either side of the pulse.
In the meantime, in the time domain,
the pulse steepens at the edges, but flattens at the peak.
After OWB, a train of small waves falls off the edges,
-which eventually melt together, leading to a trapezoid shape in the $t$-domain.
+which eventually melt together, leading to a trapezoid shape in the $$t$$-domain.
Dispersive broadening then continues normally:
-We call the distance at which the wave breaks $L_\mathrm{WB}$,
+We call the distance at which the wave breaks $$L_\mathrm{WB}$$,
and would like to analytically predict it.
-We do this using the instantaneous frequency $\omega_i$,
+We do this using the instantaneous frequency $$\omega_i$$,
by estimating when the SPM fluctuations overtake their own base,
as was illustrated earlier.
-To get $\omega_i$ of a Gaussian pulse experiencing both GVD and SPM,
-it is a reasonable approximation, for small $z$, to simply add up
+To get $$\omega_i$$ of a Gaussian pulse experiencing both GVD and SPM,
+it is a reasonable approximation, for small $$z$$, to simply add up
the instantaneous frequencies for these separate effects:
$$\begin{aligned}
@@ -74,7 +74,7 @@ $$\begin{aligned}
+ 2\gamma P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
\end{aligned}$$
-Assuming that $z$ is small enough such that $z^2 \approx 0$, this
+Assuming that $$z$$ is small enough such that $$z^2 \approx 0$$, this
expression can be reduced to:
$$\begin{aligned}
@@ -83,8 +83,8 @@ $$\begin{aligned}
= \frac{\beta_2 t z}{T_0^4} \bigg( 1 + 2 N_\mathrm{sol}^2 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
\end{aligned}$$
-Where we have assumed $\beta_2 > 0$,
-and $N_\mathrm{sol}$ is the **soliton number**,
+Where we have assumed $$\beta_2 > 0$$,
+and $$N_\mathrm{sol}$$ is the **soliton number**,
which is defined as:
$$\begin{aligned}
@@ -93,11 +93,11 @@ $$\begin{aligned}
This quantity is very important in anomalous dispersion,
but even in normal dispesion, it is still a useful measure of the relative strengths of GVD and SPM.
-As was illustrated earlier, $\omega_i$ overtakes itself at the edges,
-so OWB occurs when $\omega_i$ oscillates there,
-which starts when its $t$-derivative,
-the **instantaneous chirpyness** $\xi_i$,
-has *two* real roots for $t^2$:
+As was illustrated earlier, $$\omega_i$$ overtakes itself at the edges,
+so OWB occurs when $$\omega_i$$ oscillates there,
+which starts when its $$t$$-derivative,
+the **instantaneous chirpyness** $$\xi_i$$,
+has *two* real roots for $$t^2$$:
$$\begin{aligned}
0
@@ -107,10 +107,10 @@ $$\begin{aligned}
= \frac{\beta_2 z}{T_0^4} \: f\Big(\frac{t^2}{T_0^2}\Big)
\end{aligned}$$
-Where the function $f(x)$ has been defined for convenience. As it turns
-out, this equation can be solved analytically using the Lambert $W$ function,
-leading to the following exact minimum value $N_\mathrm{min}^2$ for $N_\mathrm{sol}^2$,
-such that OWB can only occur when $N_\mathrm{sol}^2 > N_\mathrm{min}^2$:
+Where the function $$f(x)$$ has been defined for convenience. As it turns
+out, this equation can be solved analytically using the Lambert $$W$$ function,
+leading to the following exact minimum value $$N_\mathrm{min}^2$$ for $$N_\mathrm{sol}^2$$,
+such that OWB can only occur when $$N_\mathrm{sol}^2 > N_\mathrm{min}^2$$:
$$\begin{aligned}
\boxed{
@@ -118,15 +118,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-If this condition $N_\mathrm{sol}^2 > N_\mathrm{min}^2$ is not satisfied,
-$\xi_i$ cannot have two roots for $t^2$, meaning $\omega_i$ cannot overtake itself.
+If this condition $$N_\mathrm{sol}^2 > N_\mathrm{min}^2$$ is not satisfied,
+$$\xi_i$$ cannot have two roots for $$t^2$$, meaning $$\omega_i$$ cannot overtake itself.
GVD is unable to keep up with SPM, so OWB will not occur.
-Next, consider two points at $t_1$ and $t_2$ in the pulse,
-separated by a small initial interval $(t_2 - t_1)$.
-The frequency difference between these points due to $\omega_i$
+Next, consider two points at $$t_1$$ and $$t_2$$ in the pulse,
+separated by a small initial interval $$(t_2 - t_1)$$.
+The frequency difference between these points due to $$\omega_i$$
will cause them to displace relative to each other
-after a short distance $z$ by some amount $\Delta t$,
+after a short distance $$z$$ by some amount $$\Delta t$$,
estimated by:
$$\begin{aligned}
@@ -144,12 +144,12 @@ $$\begin{aligned}
&&\Delta\xi_i \equiv \xi_i(z,t_2) - \xi_i(z,t_1)
\end{aligned}$$
-Where $\beta_1(\omega)$ is the inverse of the group velocity.
-OWB takes place when $t_2$ and $t_1$ catch up to each other,
-which is when $-\Delta t = (t_2 - t_1)$.
-The distance where this happens first, $z = L_\mathrm{WB}$,
+Where $$\beta_1(\omega)$$ is the inverse of the group velocity.
+OWB takes place when $$t_2$$ and $$t_1$$ catch up to each other,
+which is when $$-\Delta t = (t_2 - t_1)$$.
+The distance where this happens first, $$z = L_\mathrm{WB}$$,
must therefore satisfy the following condition
-for a particular value of $t$:
+for a particular value of $$t$$:
$$\begin{aligned}
L_\mathrm{WB} \, \beta_2 \, \xi_i(L_\mathrm{WB}, t) = -1
@@ -157,9 +157,9 @@ $$\begin{aligned}
L_\mathrm{WB}^2 = - \frac{T_0^4}{\beta_2^2 \, f(t^2/T_0^2)}
\end{aligned}$$
-The time $t$ of OWB must be where $\omega_i(t)$ has its steepest slope,
-which is at the minimum value of $\xi_i(t)$, and by extension $f(x)$.
-This turns out to be $f(3/2)$:
+The time $$t$$ of OWB must be where $$\omega_i(t)$$ has its steepest slope,
+which is at the minimum value of $$\xi_i(t)$$, and by extension $$f(x)$$.
+This turns out to be $$f(3/2)$$:
$$\begin{aligned}
f_\mathrm{min} = f(3/2)
@@ -167,10 +167,10 @@ $$\begin{aligned}
= 1 - N_\mathrm{sol}^2 / N_\mathrm{min}^2
\end{aligned}$$
-Clearly, $f_\mathrm{min} \ge 0$ when $N_\mathrm{sol}^2 \le N_\mathrm{min}^2$,
-which, when inserted above, leads to an imaginary $L_\mathrm{WB}$,
+Clearly, $$f_\mathrm{min} \ge 0$$ when $$N_\mathrm{sol}^2 \le N_\mathrm{min}^2$$,
+which, when inserted above, leads to an imaginary $$L_\mathrm{WB}$$,
confirming that OWB cannot occur in that case.
-Otherwise, if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then:
+Otherwise, if $$N_\mathrm{sol}^2 > N_\mathrm{min}^2$$, then:
$$\begin{aligned}
\boxed{
@@ -180,24 +180,24 @@ $$\begin{aligned}
}
\end{aligned}$$
-This prediction for $L_\mathrm{WB}$ appears to agree well
+This prediction for $$L_\mathrm{WB}$$ appears to agree well
with the OWB observed in the simulation:
-Because all spectral broadening up to $L_\mathrm{WB}$ is caused by SPM,
+Because all spectral broadening up to $$L_\mathrm{WB}$$ is caused by SPM,
whose frequency behaviour is known, it is in fact possible to draw
some analytical conclusions about the achieved bandwidth when OWB sets in.
-Filling $L_\mathrm{WB}$ in into $\omega_\mathrm{SPM}$ gives:
+Filling $$L_\mathrm{WB}$$ in into $$\omega_\mathrm{SPM}$$ gives:
$$\begin{aligned}
\omega_{\mathrm{SPM}}(L_\mathrm{WB},t)
= \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
\end{aligned}$$
-Assuming that $N_\mathrm{sol}^2$ is large in the denominator, this can
+Assuming that $$N_\mathrm{sol}^2$$ is large in the denominator, this can
be approximately reduced to:
$$\begin{aligned}
@@ -206,18 +206,18 @@ $$\begin{aligned}
= 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
\end{aligned}$$
-The expression $x \exp(-x^2)$ has its global extrema
-$\pm 1 / \sqrt{2 e}$ at $x^2 = 1/2$. The maximum SPM frequency shift
-achieved at $L_\mathrm{WB}$ is therefore given by:
+The expression $$x \exp(-x^2)$$ has its global extrema
+$$\pm 1 / \sqrt{2 e}$$ at $$x^2 = 1/2$$. The maximum SPM frequency shift
+achieved at $$L_\mathrm{WB}$$ is therefore given by:
$$\begin{aligned}
\omega_\mathrm{max} = \sqrt{\frac{2 \gamma P_0}{e \beta_2}}
\end{aligned}$$
-Interestingly, this expression does not contain $T_0$ at all,
+Interestingly, this expression does not contain $$T_0$$ at all,
so the achieved spectrum when SPM is halted by OWB
is independent of the pulse width,
-for sufficiently large $N_\mathrm{sol}$.
+for sufficiently large $$N_\mathrm{sol}$$.
## References
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index e1d73a7..df90244 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -8,11 +8,11 @@ categories:
layout: "concept"
---
-**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$,
+**Parseval's theorem** is a relation between the inner product of two functions $$f(x)$$ and $$g(x)$$,
and the inner product of their [Fourier transforms](/know/concept/fourier-transform/)
-$\tilde{f}(k)$ and $\tilde{g}(k)$.
+$$\tilde{f}(k)$$ and $$\tilde{g}(k)$$.
There are two equivalent ways of stating it,
-where $A$, $B$, and $s$ are constants from the FT's definition:
+where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition:
$$\begin{aligned}
\boxed{
@@ -48,7 +48,7 @@ $$\begin{aligned}
= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
\end{aligned}$$
-Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Where $$\delta(k)$$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
Note that we can equally well do this proof in the opposite direction,
which yields an equivalent result:
@@ -68,11 +68,12 @@ $$\begin{aligned}
&= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
= \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
\end{aligned}$$
+
For this reason, physicists like to define the Fourier transform
-with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
+with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely
conserves the functions' normalization.
diff --git a/source/know/concept/partial-fraction-decomposition/index.md b/source/know/concept/partial-fraction-decomposition/index.md
index 03c1c76..bb7faa2 100644
--- a/source/know/concept/partial-fraction-decomposition/index.md
+++ b/source/know/concept/partial-fraction-decomposition/index.md
@@ -8,15 +8,15 @@ layout: "concept"
---
**Partial fraction decomposition** or **partial fraction expansion**
-is a method to rewrite quotients of two polynomials $g(x)$ and $h(x)$,
-where the numerator $g(x)$ is of lower order than $h(x)$,
-as sums of fractions with $x$ in the denominator:
+is a method to rewrite quotients of two polynomials $$g(x)$$ and $$h(x)$$,
+where the numerator $$g(x)$$ is of lower order than $$h(x)$$,
+as sums of fractions with $$x$$ in the denominator:
$$\begin{aligned}
f(x) = \frac{g(x)}{h(x)} = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ...
\end{aligned}$$
-Where $h_n$ etc. are the roots of the denominator $h(x)$. If all $N$ of
+Where $$h_n$$ etc. are the roots of the denominator $$h(x)$$. If all $$N$$ of
these roots are distinct, then it is sufficient to simply posit:
$$\begin{aligned}
@@ -25,8 +25,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-The constants $c_n$ can either be found the hard way,
-by multiplying the denominators around and solving a system of $N$
+The constants $$c_n$$ can either be found the hard way,
+by multiplying the denominators around and solving a system of $$N$$
equations, or the easy way by using this trick:
$$\begin{aligned}
@@ -35,7 +35,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $h_1$ is a root with multiplicity $m > 1$, then the sum takes the form of:
+If $$h_1$$ is a root with multiplicity $$m > 1$$, then the sum takes the form of:
$$\begin{aligned}
\boxed{
@@ -44,15 +44,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $c_{1,j}$ are found by putting the terms on a common denominator, e.g.
+Where $$c_{1,j}$$ are found by putting the terms on a common denominator, e.g.
$$\begin{aligned}
\frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2}
= \frac{c_{1,1} (x - h_1) + c_{1,2}}{(x - h_1)^2}
\end{aligned}$$
-And then, using the linear independence of $x^0, x^1, x^2, ...$, solving
-a system of $m$ equations to find all $c_{1,1}, ..., c_{1,m}$.
+And then, using the linear independence of $$x^0, x^1, x^2, ...$$, solving
+a system of $$m$$ equations to find all $$c_{1,1}, ..., c_{1,m}$$.
diff --git a/source/know/concept/path-integral-formulation/index.md b/source/know/concept/path-integral-formulation/index.md
index b92cf5f..a8dcc76 100644
--- a/source/know/concept/path-integral-formulation/index.md
+++ b/source/know/concept/path-integral-formulation/index.md
@@ -14,10 +14,10 @@ which is equivalent to the "traditional" Schrödinger equation.
Whereas the latter is based on [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
the former comes from [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
-It expresses the [propagator](/know/concept/propagator/) $K$
-using the following sum over all possible paths $x(t)$,
-which all go from the initial position $x_0$ at time $t_0$
-to the destination $x_N$ at time $t_N$:
+It expresses the [propagator](/know/concept/propagator/) $$K$$
+using the following sum over all possible paths $$x(t)$$,
+which all go from the initial position $$x_0$$ at time $$t_0$$
+to the destination $$x_N$$ at time $$t_N$$:
$$\begin{aligned}
\boxed{
@@ -26,32 +26,32 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $A$ normalizes.
-$S[x]$ is the classical action of the path $x$, whose minimization yields
+Where $$A$$ normalizes.
+$$S[x]$$ is the classical action of the path $$x$$, whose minimization yields
the Euler-Lagrange equation from Lagrangian mechanics.
Note that each path is given an equal weight,
even unrealistic paths that make big detours.
This apparent problem solves itself,
-thanks to the fact that paths close to the classical optimum $x_c(t)$
-have an action close to $S_c = S[x_c]$,
+thanks to the fact that paths close to the classical optimum $$x_c(t)$$
+have an action close to $$S_c = S[x_c]$$,
while the paths far away have very different actions.
-Since $S[x]$ is inside a complex exponential,
-this means that paths close to $x_c$ add contructively,
+Since $$S[x]$$ is inside a complex exponential,
+this means that paths close to $$x_c$$ add contructively,
and the others add destructively and cancel out.
-An interesting way too look at it is by varying $\hbar$:
+An interesting way too look at it is by varying $$\hbar$$:
as its value decreases, minor action differences yield big phase differences,
-which make the quantum wave function stay closer to $x_c$.
-In the limit $\hbar \to 0$, quantum mechanics thus turns into classical mechanics.
+which make the quantum wave function stay closer to $$x_c$$.
+In the limit $$\hbar \to 0$$, quantum mechanics thus turns into classical mechanics.
## Time-slicing derivation
The most popular way to derive the path integral formulation proceeds as follows:
-starting from the definition of the propagator $K$,
-we divide the time interval $t_N - t_0$ into $N$ "slices"
-of equal width $\Delta t = (t_N - t_0) / N$,
-where $N$ is large:
+starting from the definition of the propagator $$K$$,
+we divide the time interval $$t_N - t_0$$ into $$N$$ "slices"
+of equal width $$\Delta t = (t_N - t_0) / N$$,
+where $$N$$ is large:
$$\begin{aligned}
K(x_N, t_N; x_0, t_0)
@@ -59,9 +59,9 @@ $$\begin{aligned}
= \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar} \cdots e^{- i \hat{H} \Delta t / \hbar}}{x_0}
\end{aligned}$$
-Between the exponentials we insert $N\!-\!1$ identity operators
-$\hat{I} = \int \Ket{x} \Bra{x} \dd{x}$,
-and define $x_j = x(t_j)$ for an arbitrary path $x(t)$:
+Between the exponentials we insert $$N\!-\!1$$ identity operators
+$$\hat{I} = \int \Ket{x} \Bra{x} \dd{x}$$,
+and define $$x_j = x(t_j)$$ for an arbitrary path $$x(t)$$:
$$\begin{aligned}
K
@@ -69,10 +69,10 @@ $$\begin{aligned}
\dd{x_1} \cdots \dd{x_{N - 1}}
\end{aligned}$$
-For sufficiently small time steps $\Delta t$ (i.e. large $N$
+For sufficiently small time steps $$\Delta t$$ (i.e. large $$N$$
we make the following approximation
(which would be exact, were it not for the fact that
-$\hat{T}$ and $\hat{V}$ are operators):
+$$\hat{T}$$ and $$\hat{V}$$ are operators):
$$\begin{aligned}
e^{- i \hat{H} \Delta t / \hbar}
@@ -80,7 +80,7 @@ $$\begin{aligned}
\approx e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}
\end{aligned}$$
-Since $\hat{V} = V(x_j)$,
+Since $$\hat{V} = V(x_j)$$,
we can take it out of the inner product as a constant factor:
$$\begin{aligned}
@@ -89,15 +89,15 @@ $$\begin{aligned}
\end{aligned}$$
Here we insert the identity operator
-expanded in the momentum basis $\hat{I} = \int \Ket{p} \Bra{p} \dd{p}$,
-and commute it with the kinetic energy $\hat{T} = \hat{p}^2 / (2m)$ to get:
+expanded in the momentum basis $$\hat{I} = \int \Ket{p} \Bra{p} \dd{p}$$,
+and commute it with the kinetic energy $$\hat{T} = \hat{p}^2 / (2m)$$ to get:
$$\begin{aligned}
\matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
= \int_{-\infty}^\infty \Inprod{x_{j+1}}{p} \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \Inprod{p}{x_j} \dd{p}
\end{aligned}$$
-In the momentum basis $\Ket{p}$,
+In the momentum basis $$\Ket{p}$$,
the position basis vectors
are represented by plane waves:
@@ -119,7 +119,7 @@ $$\begin{aligned}
&= \frac{1}{2 \pi \hbar} \sqrt{\frac{2 \pi m \hbar}{i \Delta t}} \exp\!\Big( i \frac{m (x_{j+1} - x_j)^2}{2 \hbar \Delta t} \Big)
\end{aligned}$$
-Inserting this back into the definition of the propagator $K(x_N, t_N; x_0, t_0)$ yields:
+Inserting this back into the definition of the propagator $$K(x_N, t_N; x_0, t_0)$$ yields:
$$\begin{aligned}
K
@@ -129,7 +129,7 @@ $$\begin{aligned}
\dd{x_1} \cdots \dd{x_{N-1}}
\end{aligned}$$
-For large $N$ and small $\Delta t$, the sum in the exponent becomes an integral:
+For large $$N$$ and small $$\Delta t$$, the sum in the exponent becomes an integral:
$$\begin{aligned}
\frac{i}{\hbar} \sum_{j = 0}^{N - 1} \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \Delta t^2} - V(x_j) \Big) \Delta t
@@ -137,8 +137,8 @@ $$\begin{aligned}
\frac{i}{\hbar} \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
\end{aligned}$$
-Upon closer inspection, this integral turns out to be the classical action $S[x]$,
-with the integrand being the Lagrangian $L$:
+Upon closer inspection, this integral turns out to be the classical action $$S[x]$$,
+with the integrand being the Lagrangian $$L$$:
$$\begin{aligned}
S[x(t)]
@@ -146,7 +146,7 @@ $$\begin{aligned}
= \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
\end{aligned}$$
-The definition of the propagator $K$ is then further reduced to the following:
+The definition of the propagator $$K$$ is then further reduced to the following:
$$\begin{aligned}
K
@@ -155,8 +155,8 @@ $$\begin{aligned}
\end{aligned}$$
Finally, for the purpose of normalization,
-we define the integral over all paths $x(t)$ as follows,
-where we write $D[x]$ instead of $\dd{x}$:
+we define the integral over all paths $$x(t)$$ as follows,
+where we write $$D[x]$$ instead of $$\dd{x}$$:
$$\begin{aligned}
\int D[x]
@@ -164,7 +164,7 @@ $$\begin{aligned}
\end{aligned}$$
We thus arrive at **Feynman's path integral**,
-which sums over all possible paths $x(t)$:
+which sums over all possible paths $$x(t)$$:
$$\begin{aligned}
K
diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
index 58a3f69..9821718 100644
--- a/source/know/concept/pauli-exclusion-principle/index.md
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -12,9 +12,9 @@ In quantum mechanics, the **Pauli exclusion principle** is a theorem with
profound consequences for how the world works.
Suppose we have a composite state
-$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$, where the two
-identical particles $x_1$ and $x_2$ each can occupy the same two allowed
-states $a$ and $b$. We then define the permutation operator $\hat{P}$ as
+$$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$$, where the two
+identical particles $$x_1$$ and $$x_2$$ each can occupy the same two allowed
+states $$a$$ and $$b$$. We then define the permutation operator $$\hat{P}$$ as
follows:
$$\begin{aligned}
@@ -28,22 +28,22 @@ $$\begin{aligned}
\hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b}
\end{aligned}$$
-Therefore, $\Ket{a}\Ket{b}$ is an eigenvector of $\hat{P}^2$ with
-eigenvalue $1$. Since $[\hat{P}, \hat{P}^2] = 0$, $\Ket{a}\Ket{b}$
-must also be an eigenket of $\hat{P}$ with eigenvalue $\lambda$,
-satisfying $\lambda^2 = 1$, so we know that $\lambda = 1$ or $\lambda = -1$:
+Therefore, $$\Ket{a}\Ket{b}$$ is an eigenvector of $$\hat{P}^2$$ with
+eigenvalue $$1$$. Since $$[\hat{P}, \hat{P}^2] = 0$$, $$\Ket{a}\Ket{b}$$
+must also be an eigenket of $$\hat{P}$$ with eigenvalue $$\lambda$$,
+satisfying $$\lambda^2 = 1$$, so we know that $$\lambda = 1$$ or $$\lambda = -1$$:
$$\begin{aligned}
\hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
\end{aligned}$$
As it turns out, in nature, each class of particle has a single
-associated permutation eigenvalue $\lambda$, or in other words: whether
-$\lambda$ is $-1$ or $1$ depends on the type of particle that $x_1$
-and $x_2$ are. Particles with $\lambda = -1$ are called
-**fermions**, and those with $\lambda = 1$ are known as **bosons**. We
-define $\hat{P}_f$ with $\lambda = -1$ and $\hat{P}_b$ with
-$\lambda = 1$, such that:
+associated permutation eigenvalue $$\lambda$$, or in other words: whether
+$$\lambda$$ is $$-1$$ or $$1$$ depends on the type of particle that $$x_1$$
+and $$x_2$$ are. Particles with $$\lambda = -1$$ are called
+**fermions**, and those with $$\lambda = 1$$ are known as **bosons**. We
+define $$\hat{P}_f$$ with $$\lambda = -1$$ and $$\hat{P}_b$$ with
+$$\lambda = 1$$, such that:
$$\begin{aligned}
\hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
@@ -53,20 +53,20 @@ $$\begin{aligned}
Another fundamental fact of nature is that identical particles cannot be
distinguished by any observation. Therefore it is impossible to tell
-apart $\Ket{a}\Ket{b}$ and the permuted state $\Ket{b}\Ket{a}$,
-regardless of the eigenvalue $\lambda$. There is no physical difference!
+apart $$\Ket{a}\Ket{b}$$ and the permuted state $$\Ket{b}\Ket{a}$$,
+regardless of the eigenvalue $$\lambda$$. There is no physical difference!
-But this does not mean that $\hat{P}$ is useless: despite not having any
+But this does not mean that $$\hat{P}$$ is useless: despite not having any
observable effect, the resulting difference between fermions and bosons
is absolutely fundamental. Consider the following superposition state,
-where $\alpha$ and $\beta$ are unknown:
+where $$\alpha$$ and $$\beta$$ are unknown:
$$\begin{aligned}
\Ket{\Psi(a, b)}
= \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a}
\end{aligned}$$
-When we apply $\hat{P}$, we can "choose" between two "intepretations" of
+When we apply $$\hat{P}$$, we can "choose" between two "intepretations" of
its action, both shown below. Obviously, since the left-hand sides are
equal, the right-hand sides must be equal too:
@@ -78,28 +78,28 @@ $$\begin{aligned}
&= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b}
\end{aligned}$$
-This gives us the equations $\lambda \alpha = \beta$ and
-$\lambda \beta = \alpha$. In fact, just from this we could have deduced
-that $\lambda$ can be either $-1$ or $1$. In any case, for bosons
-($\lambda = 1$), we thus find that $\alpha = \beta$:
+This gives us the equations $$\lambda \alpha = \beta$$ and
+$$\lambda \beta = \alpha$$. In fact, just from this we could have deduced
+that $$\lambda$$ can be either $$-1$$ or $$1$$. In any case, for bosons
+($$\lambda = 1$$), we thus find that $$\alpha = \beta$$:
$$\begin{aligned}
\Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
\end{aligned}$$
-Where $C$ is a normalization constant. As expected, this state is
-**symmetric**: switching $a$ and $b$ gives the same result. Meanwhile, for
-fermions ($\lambda = -1$), we find that $\alpha = -\beta$:
+Where $$C$$ is a normalization constant. As expected, this state is
+**symmetric**: switching $$a$$ and $$b$$ gives the same result. Meanwhile, for
+fermions ($$\lambda = -1$$), we find that $$\alpha = -\beta$$:
$$\begin{aligned}
\Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
\end{aligned}$$
-This state is called **antisymmetric** under exchange: switching $a$ and $b$
+This state is called **antisymmetric** under exchange: switching $$a$$ and $$b$$
causes a sign change, as we would expect for fermions.
-Now, what if the particles $x_1$ and $x_2$ are in the same state $a$?
-For bosons, we just need to update the normalization constant $C$:
+Now, what if the particles $$x_1$$ and $$x_2$$ are in the same state $$a$$?
+For bosons, we just need to update the normalization constant $$C$$:
$$\begin{aligned}
\Ket{\Psi(a, a)}_b
diff --git a/source/know/concept/plancks-law/index.md b/source/know/concept/plancks-law/index.md
index 2db783d..6c2cd2e 100644
--- a/source/know/concept/plancks-law/index.md
+++ b/source/know/concept/plancks-law/index.md
@@ -18,14 +18,14 @@ and photons are bosons
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
this system must obey the
[Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
-with a chemical potential $\mu = 0$ (due to the freely varying population):
+with a chemical potential $$\mu = 0$$ (due to the freely varying population):
$$\begin{aligned}
f_B(E)
= \frac{1}{\exp(\beta E) - 1}
\end{aligned}$$
-Each photon has an energy $E = \hbar \omega = \hbar c k$,
+Each photon has an energy $$E = \hbar \omega = \hbar c k$$,
so the [density of states](/know/concept/density-of-states/)
is as follows in 3D:
@@ -37,9 +37,9 @@ $$\begin{aligned}
= \frac{8 \pi V E^2}{h^3 c^3}
\end{aligned}$$
-Where the factor of $2$ accounts for the photon's polarization degeneracy.
-We thus expect that the number of photons $N(E)$
-with an energy between $E$ and $E + \dd{E}$ is given by:
+Where the factor of $$2$$ accounts for the photon's polarization degeneracy.
+We thus expect that the number of photons $$N(E)$$
+with an energy between $$E$$ and $$E + \dd{E}$$ is given by:
$$\begin{aligned}
N(E) \dd{E}
@@ -47,16 +47,16 @@ $$\begin{aligned}
= \frac{8 \pi V}{h^3 c^3} \frac{E^2}{\exp(\beta E) - 1} \dd{E}
\end{aligned}$$
-By substituting $E = h \nu$, we find that the number of photons $N(\nu)$
-with a frequency between $\nu$ and $\nu + \dd{\nu}$ must be as follows:
+By substituting $$E = h \nu$$, we find that the number of photons $$N(\nu)$$
+with a frequency between $$\nu$$ and $$\nu + \dd{\nu}$$ must be as follows:
$$\begin{aligned}
N(\nu) \dd{\nu}
= \frac{8 \pi V}{c^3} \frac{\nu^2}{\exp(\beta h \nu) - 1} \dd{\nu}
\end{aligned}$$
-Multiplying by the energy $h \nu$ yields the distribution of the radiated energy,
-which we divide by the volume $V$ to get Planck's law,
+Multiplying by the energy $$h \nu$$ yields the distribution of the radiated energy,
+which we divide by the volume $$V$$ to get Planck's law,
also called the **Plank distribution**,
describing a black body's radiated spectral energy density per unit volume:
@@ -70,8 +70,8 @@ $$\begin{aligned}
## Wien's displacement law
-The Planck distribution peaks at a particular frequency $\nu_{\mathrm{max}}$,
-which can be found by solving the following equation for $\nu$:
+The Planck distribution peaks at a particular frequency $$\nu_{\mathrm{max}}$$,
+which can be found by solving the following equation for $$\nu$$:
$$\begin{aligned}
0
@@ -81,7 +81,7 @@ $$\begin{aligned}
= 3 \nu^2 (\exp(\beta h \nu) - 1) - \nu^3 \beta h \exp(\beta h \nu)
\end{aligned}$$
-By defining $x \equiv \beta h \nu_{\mathrm{max}}$,
+By defining $$x \equiv \beta h \nu_{\mathrm{max}}$$,
this turns into the following transcendental equation:
$$\begin{aligned}
@@ -98,14 +98,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Which states that the peak frequency $\nu_{\mathrm{max}}$
-is proportional to the temperature $T$.
+Which states that the peak frequency $$\nu_{\mathrm{max}}$$
+is proportional to the temperature $$T$$.
## Stefan-Boltzmann law
-Because $u(\nu)$ represents the radiated spectral energy density,
-we can find the total radiated energy $U$ per unit volume by integrating over $\nu$:
+Because $$u(\nu)$$ represents the radiated spectral energy density,
+we can find the total radiated energy $$U$$ per unit volume by integrating over $$\nu$$:
$$\begin{aligned}
U
@@ -116,9 +116,9 @@ $$\begin{aligned}
= \frac{8 \pi}{\beta^4 h^3 c^3} \int_0^\infty \frac{x^3}{\exp(x) - 1} \dd{x}
\end{aligned}$$
-This definite integral turns out to be $\pi^4/15$,
+This definite integral turns out to be $$\pi^4/15$$,
leading us to the **Stefan-Boltzmann law**,
-which states that the radiated energy is proportional to $T^4$:
+which states that the radiated energy is proportional to $$T^4$$:
$$\begin{aligned}
\boxed{
@@ -126,7 +126,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\sigma$ is the **Stefan-Boltzmann constant**, which is defined as follows:
+Where $$\sigma$$ is the **Stefan-Boltzmann constant**, which is defined as follows:
$$\begin{aligned}
\sigma
diff --git a/source/know/concept/prandtl-equations/index.md b/source/know/concept/prandtl-equations/index.md
index 3afe405..d82657c 100644
--- a/source/know/concept/prandtl-equations/index.md
+++ b/source/know/concept/prandtl-equations/index.md
@@ -11,7 +11,7 @@ layout: "concept"
In fluid dynamics, the **Prandtl equations** or **boundary layer equations**
describe the movement of a [viscous](/know/concept/viscosity/) fluid
-with a large [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \gg 1$
+with a large [Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re} \gg 1$$
close to a solid surface.
Fluids with a large Reynolds number
@@ -27,23 +27,23 @@ where viscosity plays an important role.
This is in contrast to the ideal flow far away from the surface.
We consider a simple theoretical case in 2D:
-a large flat surface located at $y = 0$ for all $x \in \mathbb{R}$,
-with a fluid *trying* to flow parallel to it at $U$.
-The 2D treatment can be justified by assuming that everything is constant in the $z$-direction.
+a large flat surface located at $$y = 0$$ for all $$x \in \mathbb{R}$$,
+with a fluid *trying* to flow parallel to it at $$U$$.
+The 2D treatment can be justified by assuming that everything is constant in the $$z$$-direction.
We will not solve this case,
but instead derive general equations
to describe the flow close to a flat surface.
-At the wall, there is a very thin boundary layer of thickness $\delta$,
-where the fluid is assumed to be completely stationary $\va{v} = 0$.
-We are mainly interested in the region $\delta < y \ll L$,
-where $L$ is the distance at which the fluid becomes practically ideal.
+At the wall, there is a very thin boundary layer of thickness $$\delta$$,
+where the fluid is assumed to be completely stationary $$\va{v} = 0$$.
+We are mainly interested in the region $$\delta < y \ll L$$,
+where $$L$$ is the distance at which the fluid becomes practically ideal.
This the so-called **slip-flow** region,
in which the fluid is not stationary,
but still viscosity-dominated.
In 2D, the steady Navier-Stokes equations are as follows,
-where the flow $\va{v} = (v_x, v_y)$:
+where the flow $$\va{v} = (v_x, v_y)$$:
$$\begin{aligned}
v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
@@ -58,10 +58,10 @@ $$\begin{aligned}
The latter represents the fluid's incompressibility.
We non-dimensionalize these equations,
-and assume that changes along the $y$-axis
-happen on a short scale (say, $\delta$),
-and along the $x$-axis on a longer scale (say, $L$).
-Let $\tilde{x}$ and $\tilde{y}$ be dimenionless variables of order $1$:
+and assume that changes along the $$y$$-axis
+happen on a short scale (say, $$\delta$$),
+and along the $$x$$-axis on a longer scale (say, $$L$$).
+Let $$\tilde{x}$$ and $$\tilde{y}$$ be dimenionless variables of order $$1$$:
$$\begin{aligned}
x
@@ -106,7 +106,7 @@ $$\begin{aligned}
\end{aligned}$$
For future convenience,
-we multiply the former equation by $L / U^2$, and the latter by $\delta / U^2$:
+we multiply the former equation by $$L / U^2$$, and the latter by $$\delta / U^2$$:
$$\begin{aligned}
\tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
@@ -118,12 +118,12 @@ $$\begin{aligned}
+ \nu \Big( \frac{\delta^2}{U L^3} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
\end{aligned}$$
-We would like to estimate $\delta$.
-Intuitively, we expect that higher viscosities $\nu$ give thicker layers,
-and that faster velocities $U$ give thinner layers.
+We would like to estimate $$\delta$$.
+Intuitively, we expect that higher viscosities $$\nu$$ give thicker layers,
+and that faster velocities $$U$$ give thinner layers.
Furthermore, we expect *downstream thickening*:
-with distance $x$, viscous stresses slow down the slip-flow,
-leading to a gradual increase of $\delta(x)$.
+with distance $$x$$, viscous stresses slow down the slip-flow,
+leading to a gradual increase of $$\delta(x)$$.
Some dimensional analysis thus yields the following estimate:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
\sim \sqrt{\frac{\nu L}{U}}
\end{aligned}$$
-We thus insert $\delta = \sqrt{\nu L / U}$ into the Navier-Stokes equations, giving us:
+We thus insert $$\delta = \sqrt{\nu L / U}$$ into the Navier-Stokes equations, giving us:
$$\begin{aligned}
\tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
@@ -144,7 +144,7 @@ $$\begin{aligned}
+ \nu \Big( \frac{\nu}{U^2 L^2} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
\end{aligned}$$
-Here, we recognize the definition of the Reynolds number $\mathrm{Re} = U L / \nu$:
+Here, we recognize the definition of the Reynolds number $$\mathrm{Re} = U L / \nu$$:
$$\begin{aligned}
\tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
@@ -156,8 +156,8 @@ $$\begin{aligned}
+ \frac{1}{\mathrm{Re}^2} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \pdvn{2}{\tilde{v}_y}{\tilde{y}}
\end{aligned}$$
-Recall that we are only considering large Reynolds numbers $\mathrm{Re} \gg 1$,
-in which case $\mathrm{Re}^{-1} \ll 1$,
+Recall that we are only considering large Reynolds numbers $$\mathrm{Re} \gg 1$$,
+in which case $$\mathrm{Re}^{-1} \ll 1$$,
so we can drop many terms, leaving us with these redimensionalized equations:
$$\begin{aligned}
@@ -168,11 +168,11 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-The second one tells us that for a given $x$-value,
+The second one tells us that for a given $$x$$-value,
the pressure is the same at the surface
-as in the main flow $y > L$, where the fluid is ideal.
-In the latter regime, we apply Bernoulli's theorem to rewrite $p$,
-using the *Bernoulli head* $H$ and the mainstream velocity $U(x)$:
+as in the main flow $$y > L$$, where the fluid is ideal.
+In the latter regime, we apply Bernoulli's theorem to rewrite $$p$$,
+using the *Bernoulli head* $$H$$ and the mainstream velocity $$U(x)$$:
$$\begin{aligned}
p
diff --git a/source/know/concept/probability-current/index.md b/source/know/concept/probability-current/index.md
index 43683a3..bd41dab 100644
--- a/source/know/concept/probability-current/index.md
+++ b/source/know/concept/probability-current/index.md
@@ -10,15 +10,15 @@ layout: "concept"
In quantum mechanics, the **probability current** describes the movement
of the probability of finding a particle at given point in space.
-In other words, it treats the particle as a heterogeneous fluid with density $|\psi|^2$.
-Now, the probability of finding the particle within a volume $V$ is:
+In other words, it treats the particle as a heterogeneous fluid with density $$|\psi|^2$$.
+Now, the probability of finding the particle within a volume $$V$$ is:
$$\begin{aligned}
P = \int_{V} | \psi |^2 \ddn{3}{\vb{r}}
\end{aligned}$$
As the system evolves in time, this probability may change, so we take
-its derivative with respect to time $t$, and when necessary substitute
+its derivative with respect to time $$t$$, and when necessary substitute
in the other side of the Schrödinger equation to get:
$$\begin{aligned}
@@ -33,8 +33,8 @@ $$\begin{aligned}
= - \int_{V} \nabla \cdot \vb{J} \ddn{3}{\vb{r}}
\end{aligned}$$
-Where we have defined the probability current $\vb{J}$ as follows in
-the $\vb{r}$-basis:
+Where we have defined the probability current $$\vb{J}$$ as follows in
+the $$\vb{r}$$-basis:
$$\begin{aligned}
\vb{J}
@@ -43,8 +43,8 @@ $$\begin{aligned}
\end{aligned}$$
Let us rewrite this using the momentum operator
-$\vu{p} = -i \hbar \nabla$ as follows, noting that $\vu{p} / m$ is
-simply the velocity operator $\vu{v}$:
+$$\vu{p} = -i \hbar \nabla$$ as follows, noting that $$\vu{p} / m$$ is
+simply the velocity operator $$\vu{v}$$:
$$\begin{aligned}
\boxed{
@@ -55,7 +55,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Returning to the derivation of $\vb{J}$, we now have the following
+Returning to the derivation of $$\vb{J}$$, we now have the following
equation:
$$\begin{aligned}
@@ -65,7 +65,7 @@ $$\begin{aligned}
\end{aligned}$$
By removing the integrals, we thus arrive at the **continuity equation**
-for $\vb{J}$:
+for $$\vb{J}$$:
$$\begin{aligned}
\boxed{
@@ -77,13 +77,13 @@ $$\begin{aligned}
This states that the total probability is conserved, and is reminiscent of charge
conservation in electromagnetism. In other words, the probability at a
point can only change by letting it "flow" towards or away from it. Thus
-$\vb{J}$ represents the flow of probability, which is analogous to the
+$$\vb{J}$$ represents the flow of probability, which is analogous to the
motion of a particle.
As a bonus, this still holds for a particle in an electromagnetic vector
-potential $\vb{A}$, thanks to the gauge invariance of the Schrödinger
+potential $$\vb{A}$$, thanks to the gauge invariance of the Schrödinger
equation. We can thus extend the definition to a particle with charge
-$q$ in an SI-unit field, neglecting spin:
+$$q$$ in an SI-unit field, neglecting spin:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/propagator/index.md b/source/know/concept/propagator/index.md
index 437c57e..54e9eb6 100644
--- a/source/know/concept/propagator/index.md
+++ b/source/know/concept/propagator/index.md
@@ -8,9 +8,9 @@ categories:
layout: "concept"
---
-In quantum mechanics, the **propagator** $K(x_f, t_f; x_i, t_i)$
+In quantum mechanics, the **propagator** $$K(x_f, t_f; x_i, t_i)$$
gives the probability amplitude that a particle
-starting at $x_i$ at $t_i$ ends up at position $x_f$ at $t_f$.
+starting at $$x_i$$ at $$t_i$$ ends up at position $$x_f$$ at $$t_f$$.
It is defined as follows:
$$\begin{aligned}
@@ -20,24 +20,24 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\hat{U} \equiv \exp(- i t \hat{H} / \hbar)$ is the time-evolution operator.
+Where $$\hat{U} \equiv \exp(- i t \hat{H} / \hbar)$$ is the time-evolution operator.
The probability that a particle travels
-from $(x_i, t_i)$ to $(x_f, t_f)$ is then given by:
+from $$(x_i, t_i)$$ to $$(x_f, t_f)$$ is then given by:
$$\begin{aligned}
P
&= \big| K(x_f, t_f; x_i, t_i) \big|^2
\end{aligned}$$
-Given a general (i.e. non-collapsed) initial state $\psi_i(x) \equiv \psi(x, t_i)$,
-we must integrate over $x_i$:
+Given a general (i.e. non-collapsed) initial state $$\psi_i(x) \equiv \psi(x, t_i)$$,
+we must integrate over $$x_i$$:
$$\begin{aligned}
P
&= \bigg| \int_{-\infty}^\infty K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \bigg|^2
\end{aligned}$$
-And if the final state $\psi_f(x) \equiv \psi(x, t_f)$
+And if the final state $$\psi_f(x) \equiv \psi(x, t_f)$$
is not a basis vector either, then we integrate twice:
$$\begin{aligned}
@@ -45,7 +45,7 @@ $$\begin{aligned}
&= \bigg| \iint_{-\infty}^\infty \psi_f^*(x_f) \: K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \dd{x_f} \bigg|^2
\end{aligned}$$
-Given a $\psi_i(x)$, the propagator can also be used
+Given a $$\psi_i(x)$$, the propagator can also be used
to find the full final wave function:
$$\begin{aligned}
@@ -56,9 +56,9 @@ $$\begin{aligned}
\end{aligned}$$
Sometimes the name "propagator" is also used to refer to
-the [fundamental solution](/know/concept/fundamental-solution/) $G$
+the [fundamental solution](/know/concept/fundamental-solution/) $$G$$
of the time-dependent Schrödinger equation,
-which is related to $K$ by:
+which is related to $$K$$ by:
$$\begin{aligned}
\boxed{
@@ -67,4 +67,4 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Theta(t)$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
+Where $$\Theta(t)$$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
diff --git a/source/know/concept/pulay-mixing/index.md b/source/know/concept/pulay-mixing/index.md
index 91beb03..6e809dd 100644
--- a/source/know/concept/pulay-mixing/index.md
+++ b/source/know/concept/pulay-mixing/index.md
@@ -8,15 +8,15 @@ layout: "concept"
---
Some numerical problems are most easily solved *iteratively*,
-by generating a series $\rho_1$, $\rho_2$, etc.
-converging towards the desired solution $\rho_*$.
+by generating a series $$\rho_1$$, $$\rho_2$$, etc.
+converging towards the desired solution $$\rho_*$$.
**Pulay mixing**, also often called
**direct inversion in the iterative subspace** (DIIS),
can speed up the convergence for some types of problems,
and also helps to avoid periodic divergences.
-The key concept it relies on is the **residual vector** $R_n$
-of the $n$th iteration, which in some way measures the error of the current $\rho_n$.
+The key concept it relies on is the **residual vector** $$R_n$$
+of the $$n$$th iteration, which in some way measures the error of the current $$\rho_n$$.
Its exact definition varies,
but is generally along the lines of the difference between
the input of the iteration and the raw resulting output:
@@ -27,16 +27,16 @@ $$\begin{aligned}
= \rho_n^\mathrm{new}[\rho_n] - \rho_n
\end{aligned}$$
-It is not always clear what to do with $\rho_n^\mathrm{new}$.
-Directly using it as the next input ($\rho_{n+1} = \rho_n^\mathrm{new}$)
+It is not always clear what to do with $$\rho_n^\mathrm{new}$$.
+Directly using it as the next input ($$\rho_{n+1} = \rho_n^\mathrm{new}$$)
often leads to oscillation,
-and linear mixing ($\rho_{n+1} = (1\!-\!f) \rho_n + f \rho_n^\mathrm{new}$)
+and linear mixing ($$\rho_{n+1} = (1\!-\!f) \rho_n + f \rho_n^\mathrm{new}$$)
can take a very long time to converge properly.
Pulay mixing offers an improvement.
-The idea is to construct the next iteration's input $\rho_{n+1}$
-as a linear combination of the previous inputs $\rho_1$, $\rho_2$, ..., $\rho_n$,
-such that it is as close as possible to the optimal $\rho_*$:
+The idea is to construct the next iteration's input $$\rho_{n+1}$$
+as a linear combination of the previous inputs $$\rho_1$$, $$\rho_2$$, ..., $$\rho_n$$,
+such that it is as close as possible to the optimal $$\rho_*$$:
$$\begin{aligned}
\boxed{
@@ -46,10 +46,10 @@ $$\begin{aligned}
\end{aligned}$$
To do so, we make two assumptions.
-Firstly, the current $\rho_n$ is already close to $\rho_*$,
+Firstly, the current $$\rho_n$$ is already close to $$\rho_*$$,
so that such a linear combination makes sense.
Secondly, the iteration is linear,
-such that the raw output $\rho_{n+1}^\mathrm{new}$
+such that the raw output $$\rho_{n+1}^\mathrm{new}$$
is also a linear combination with the *same coefficients*:
$$\begin{aligned}
@@ -58,7 +58,7 @@ $$\begin{aligned}
\end{aligned}$$
We will return to these assumptions later.
-The point is that $R_{n+1}$ is also a linear combination:
+The point is that $$R_{n+1}$$ is also a linear combination:
$$\begin{aligned}
R_{n+1}
@@ -67,23 +67,23 @@ $$\begin{aligned}
= \sum_{m = 1}^n \alpha_m R_m
\end{aligned}$$
-The goal is to choose the coefficients $\alpha_m$ such that
-the norm of the error $|R_{n+1}| \approx 0$,
-subject to the following constraint to preserve the normalization of $\rho_{n+1}$:
+The goal is to choose the coefficients $$\alpha_m$$ such that
+the norm of the error $$|R_{n+1}| \approx 0$$,
+subject to the following constraint to preserve the normalization of $$\rho_{n+1}$$:
$$\begin{aligned}
\sum_{m=1}^n \alpha_m = 1
\end{aligned}$$
We thus want to minimize the following quantity,
-where $\lambda$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+where $$\lambda$$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/):
$$\begin{aligned}
\Inprod{R_{n+1}}{R_{n+1}} + \lambda \sum_{m = 1}^n \alpha_m^*
= \sum_{m=1}^n \alpha_m^* \Big( \sum_{k=1}^n \alpha_k \Inprod{R_m}{R_k} + \lambda \Big)
\end{aligned}$$
-By differentiating the right-hand side with respect to $\alpha_m^*$
+By differentiating the right-hand side with respect to $$\alpha_m^*$$
and demanding that the result is zero,
we get a system of equations that we can write in matrix form,
which is cheap to solve:
@@ -106,38 +106,38 @@ $$\begin{aligned}
\end{aligned}$$
From this, we can also see that the Lagrange multiplier
-$\lambda = - \Inprod{R_{n+1}}{R_{n+1}}$,
-where $R_{n+1}$ is the *predicted* residual of the next iteration,
+$$\lambda = - \Inprod{R_{n+1}}{R_{n+1}}$$,
+where $$R_{n+1}$$ is the *predicted* residual of the next iteration,
subject to the two assumptions.
-However, in practice, the earlier inputs $\rho_1$, $\rho_2$, etc.
-are much further from $\rho_*$ than $\rho_n$,
-so usually only the most recent $N\!+\!1$ inputs $\rho_{n - N}$, ..., $\rho_n$ are used:
+However, in practice, the earlier inputs $$\rho_1$$, $$\rho_2$$, etc.
+are much further from $$\rho_*$$ than $$\rho_n$$,
+so usually only the most recent $$N\!+\!1$$ inputs $$\rho_{n - N}$$, ..., $$\rho_n$$ are used:
$$\begin{aligned}
\rho_{n+1}
= \sum_{m = n-N}^n \alpha_m \rho_m
\end{aligned}$$
-You might be confused by the absence of any $\rho_m^\mathrm{new}$
-in the creation of $\rho_{n+1}$, as if the iteration's outputs are being ignored.
+You might be confused by the absence of any $$\rho_m^\mathrm{new}$$
+in the creation of $$\rho_{n+1}$$, as if the iteration's outputs are being ignored.
This is due to the first assumption,
-which states that $\rho_n^\mathrm{new}$ and $\rho_n$ are already similar,
+which states that $$\rho_n^\mathrm{new}$$ and $$\rho_n$$ are already similar,
such that they are basically interchangeable.
Speaking of which, about those assumptions:
-while they will clearly become more accurate as $\rho_n$ approaches $\rho_*$,
+while they will clearly become more accurate as $$\rho_n$$ approaches $$\rho_*$$,
they might be very dubious in the beginning.
A consequence of this is that the early iterations might get "trapped"
-in a suboptimal subspace spanned by $\rho_1$, $\rho_2$, etc.
-To say it another way, we would be varying $n$ coefficients $\alpha_m$
-to try to optimize a $D$-dimensional $\rho_{n+1}$,
-where in general $D \gg n$, at least in the beginning.
+in a suboptimal subspace spanned by $$\rho_1$$, $$\rho_2$$, etc.
+To say it another way, we would be varying $$n$$ coefficients $$\alpha_m$$
+to try to optimize a $$D$$-dimensional $$\rho_{n+1}$$,
+where in general $$D \gg n$$, at least in the beginning.
There is an easy fix to this problem:
-add a small amount of the raw residual $R_m$
-to "nudge" $\rho_{n+1}$ towards the right subspace,
-where $\beta \in [0,1]$ is a tunable parameter:
+add a small amount of the raw residual $$R_m$$
+to "nudge" $$\rho_{n+1}$$ towards the right subspace,
+where $$\beta \in [0,1]$$ is a tunable parameter:
$$\begin{aligned}
\boxed{
@@ -146,7 +146,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-In other words, we end up introducing a small amount of the raw outputs $\rho_m^\mathrm{new}$,
+In other words, we end up introducing a small amount of the raw outputs $$\rho_m^\mathrm{new}$$,
while still giving more weight to iterations with smaller residuals.
Pulay mixing is very effective for certain types of problems,
diff --git a/source/know/concept/quantum-entanglement/index.md b/source/know/concept/quantum-entanglement/index.md
index 5ae5c92..cfb6721 100644
--- a/source/know/concept/quantum-entanglement/index.md
+++ b/source/know/concept/quantum-entanglement/index.md
@@ -9,19 +9,19 @@ categories:
layout: "concept"
---
-Consider a composite quantum system which consists of two subsystems $A$ and $B$,
-respectively with basis states $\Ket{a_n}$ and $\Ket{b_n}$.
-All accessible states of the sytem $\Ket{\Psi}$ lie in
+Consider a composite quantum system which consists of two subsystems $$A$$ and $$B$$,
+respectively with basis states $$\Ket{a_n}$$ and $$\Ket{b_n}$$.
+All accessible states of the sytem $$\Ket{\Psi}$$ lie in
the tensor product of the subsystems'
-[Hilbert spaces](/know/concept/hilbert-space/) $\mathbb{H}_A$ and $\mathbb{H}_B$:
+[Hilbert spaces](/know/concept/hilbert-space/) $$\mathbb{H}_A$$ and $$\mathbb{H}_B$$:
$$\begin{aligned}
\Ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B
\end{aligned}$$
A subset of these states can be written as the tensor product (i.e. Kronecker product in a basis)
-of a state $\Ket{\alpha}$ in $A$ and a state $\Ket{\beta}$ in $B$,
-often abbreviated as $\Ket{\alpha} \Ket{\beta}$:
+of a state $$\Ket{\alpha}$$ in $$A$$ and a state $$\Ket{\beta}$$ in $$B$$,
+often abbreviated as $$\Ket{\alpha} \Ket{\beta}$$:
$$\begin{aligned}
\Ket{\Psi}
@@ -32,20 +32,20 @@ $$\begin{aligned}
The states that can be written in this way are called **separable**,
and states that cannot are called **entangled**.
Therefore, we are dealing with **quantum entanglement**
-if the state of subsystem $A$ cannot be fully described
-independently of the state of subsystem $B$, and vice versa.
+if the state of subsystem $$A$$ cannot be fully described
+independently of the state of subsystem $$B$$, and vice versa.
To detect and quantify entanglement,
-we can use the [density operator](/know/concept/density-operator/) $\hat{\rho}$.
-For a pure ensemble in a given (possibly entangled) state $\Ket{\Psi}$,
-$\hat{\rho}$ is given by:
+we can use the [density operator](/know/concept/density-operator/) $$\hat{\rho}$$.
+For a pure ensemble in a given (possibly entangled) state $$\Ket{\Psi}$$,
+$$\hat{\rho}$$ is given by:
$$\begin{aligned}
\hat{\rho} = \Ket{\Psi} \Bra{\Psi}
\end{aligned}$$
-From this, we would like to extract the corresponding state of subsystem $A$.
-For that purpose, we define the **reduced density operator** $\hat{\rho}_A$ of subsystem $A$ as follows:
+From this, we would like to extract the corresponding state of subsystem $$A$$.
+For that purpose, we define the **reduced density operator** $$\hat{\rho}_A$$ of subsystem $$A$$ as follows:
$$\begin{aligned}
\boxed{
@@ -55,11 +55,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Tr_B(\hat{\rho})$ is called the **partial trace** of $\hat{\rho}$,
-which basically eliminates subsystem $B$ from $\hat{\rho}$.
-For a pure composite state $\Ket{\Psi}$,
-the resulting $\hat{\rho}_A$ describes a pure state in $A$ if $\Ket{\Psi}$ is separable,
-else, if $\Ket{\Psi}$ is entangled, it describes a mixed state in $A$.
+Where $$\Tr_B(\hat{\rho})$$ is called the **partial trace** of $$\hat{\rho}$$,
+which basically eliminates subsystem $$B$$ from $$\hat{\rho}$$.
+For a pure composite state $$\Ket{\Psi}$$,
+the resulting $$\hat{\rho}_A$$ describes a pure state in $$A$$ if $$\Ket{\Psi}$$ is separable,
+else, if $$\Ket{\Psi}$$ is entangled, it describes a mixed state in $$A$$.
In the former case we simply find:
$$\begin{aligned}
@@ -70,9 +70,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-We call $\Ket{\Psi}$ **maximally entangled**
+We call $$\Ket{\Psi}$$ **maximally entangled**
if its reduced density operators are **maximally mixed**,
-where $N$ is the dimension of $\mathbb{H}_A$ and $\hat{I}$ is the identity matrix:
+where $$N$$ is the dimension of $$\mathbb{H}_A$$ and $$\hat{I}$$ is the identity matrix:
$$\begin{aligned}
\hat{\rho}_A
@@ -80,10 +80,10 @@ $$\begin{aligned}
\end{aligned}$$
Suppose that we are given an entangled pure state
-$\Ket{\Psi} \neq \Ket{\alpha} \otimes \Ket{\beta}$.
-Then the partial traces $\hat{\rho}_A$ and $\hat{\rho}_B$
-of $\hat{\rho} = \Ket{\Psi} \Bra{\Psi}$ are mixed states with the same probabilities $p_n$
-(assuming $\mathbb{H}_A$ and $\mathbb{H}_B$ have the same dimensions,
+$$\Ket{\Psi} \neq \Ket{\alpha} \otimes \Ket{\beta}$$.
+Then the partial traces $$\hat{\rho}_A$$ and $$\hat{\rho}_B$$
+of $$\hat{\rho} = \Ket{\Psi} \Bra{\Psi}$$ are mixed states with the same probabilities $$p_n$$
+(assuming $$\mathbb{H}_A$$ and $$\mathbb{H}_B$$ have the same dimensions,
which is usually the case):
$$\begin{aligned}
@@ -97,9 +97,9 @@ $$\begin{aligned}
\end{aligned}$$
There exists an orthonormal choice
-of the subsystem basis states $\Ket{a_n}$ and $\Ket{b_n}$,
-such that $\Ket{\Psi}$ can be written as follows,
-where $p_n$ are the probabilities in the reduced density operators:
+of the subsystem basis states $$\Ket{a_n}$$ and $$\Ket{b_n}$$,
+such that $$\Ket{\Psi}$$ can be written as follows,
+where $$p_n$$ are the probabilities in the reduced density operators:
$$\begin{aligned}
\Ket{\Psi}
@@ -108,18 +108,18 @@ $$\begin{aligned}
This is the **Schmidt decomposition**,
and the **Schmidt number** is the number of nonzero terms in the summation,
-which can be used to determine if the state $\Ket{\Psi}$
+which can be used to determine if the state $$\Ket{\Psi}$$
is entangled (greater than one) or separable (equal to one).
By looking at the Schmidt decomposition, we can notice that,
-if $\hat{O}_A$ and $\hat{O}_B$ are the subsystem observables
-with basis eigenstates $\Ket{a_n}$ and $\Ket{b_n}$,
+if $$\hat{O}_A$$ and $$\hat{O}_B$$ are the subsystem observables
+with basis eigenstates $$\Ket{a_n}$$ and $$\Ket{b_n}$$,
then measurement results of these operators
-will be perfectly correlated across $A$ and $B$.
+will be perfectly correlated across $$A$$ and $$B$$.
This is a general property of entangled systems,
but beware: correlation does not imply entanglement!
-But what if the composite system is in a mixed state $\hat{\rho}$?
+But what if the composite system is in a mixed state $$\hat{\rho}$$?
The state is separable if and only if:
$$\begin{aligned}
@@ -129,13 +129,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $p_m$ are probabilities,
-and $\hat{\rho}_A$ and $\hat{\rho}_B$ can be any subsystem states.
+Where $$p_m$$ are probabilities,
+and $$\hat{\rho}_A$$ and $$\hat{\rho}_B$$ can be any subsystem states.
In reality, it is very hard to determine, using this criterium,
-whether an arbitrary given $\hat{\rho}$ is separable or not.
+whether an arbitrary given $$\hat{\rho}$$ is separable or not.
As a final side note, the expectation value
-of an obervable $\hat{O}_A$ acting only on $A$ is given by:
+of an obervable $$\hat{O}_A$$ acting only on $$A$$ is given by:
$$\begin{aligned}
\expval{\hat{O}_A}
diff --git a/source/know/concept/quantum-fourier-transform/index.md b/source/know/concept/quantum-fourier-transform/index.md
index 5595fa2..113367c 100644
--- a/source/know/concept/quantum-fourier-transform/index.md
+++ b/source/know/concept/quantum-fourier-transform/index.md
@@ -10,9 +10,9 @@ layout: "concept"
The **quantum Fourier transform (QFT)** is a quantum counterpart
of the classical discrete Fourier transform.
-It is defined like so, where $\Ket{x}$
-is an $n$-qubit computational basis state $\Ket{x_1} \cdots \Ket{x_n}$,
-and $\omega_N$ is an $N$th complex root of unity $\omega_N^N = 1$ with $N = 2^n$:
+It is defined like so, where $$\Ket{x}$$
+is an $$n$$-qubit computational basis state $$\Ket{x_1} \cdots \Ket{x_n}$$,
+and $$\omega_N$$ is an $$N$$th complex root of unity $$\omega_N^N = 1$$ with $$N = 2^n$$:
$$\begin{aligned}
\boxed{
@@ -22,9 +22,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that $\Ket{x}$ and $\Ket{k}$ refer to the same basis set;
+Note that $$\Ket{x}$$ and $$\Ket{k}$$ refer to the same basis set;
we use these names to clarify which "space" we are considering.
-Furthermore, note the sign of the exponent of $\omega_N$,
+Furthermore, note the sign of the exponent of $$\omega_N$$,
which is the opposite of the classical DFT convention.
In other words, the *forward* QFT corresponds to the *inverse* DFT.
@@ -39,7 +39,7 @@ $$\begin{aligned}
\end{aligned}$$
The above definitions of the QFT and iQFT describe
-the effect on a single basis vector $\Ket{x}$,
+the effect on a single basis vector $$\Ket{x}$$,
so the effect on an abitrary superposition follows from linearity,
e.g. for the QFT:
@@ -50,14 +50,14 @@ $$\begin{aligned}
\end{aligned}$$
Classically, such a double sum takes
-$\mathcal{O}(N^2) = \mathcal{O}(2^{2n})$ time to evaluate naively,
-with a potential improvement to $\mathcal{O}(N \log{N}) = \mathcal{O}(n 2^n)$
+$$\mathcal{O}(N^2) = \mathcal{O}(2^{2n})$$ time to evaluate naively,
+with a potential improvement to $$\mathcal{O}(N \log{N}) = \mathcal{O}(n 2^n)$$
for smarter algorithms.
-Quantum computers can do a QFT in $\mathcal{O}(\log^2(N)) = \mathcal{O}(n^2)$ time,
-and approximate it in $\mathcal{O}(n \log{n})$ time.
+Quantum computers can do a QFT in $$\mathcal{O}(\log^2(N)) = \mathcal{O}(n^2)$$ time,
+and approximate it in $$\mathcal{O}(n \log{n})$$ time.
To find out how, we look at the forward QFT,
-which maps a basis state $\Ket{x}$ to another state $\Ket{\tilde{x}}$:
+which maps a basis state $$\Ket{x}$$ to another state $$\Ket{\tilde{x}}$$:
$$\begin{aligned}
\Ket{x}
@@ -67,8 +67,8 @@ $$\begin{aligned}
= \frac{1}{\sqrt{N}} \sum_{k = 0}^{N - 1} \exp\!\bigg( \frac{i 2 \pi x k}{N} \bigg) \Ket{k}
\end{aligned}$$
-We can decompose $k$ into its binary representation
-$k_1 2^{n-1} + k_2 2^{n-2} + ... + k_n 2^{0}$:
+We can decompose $$k$$ into its binary representation
+$$k_1 2^{n-1} + k_2 2^{n-2} + ... + k_n 2^{0}$$:
$$\begin{aligned}
\Ket{\tilde{x}}
@@ -84,7 +84,7 @@ $$\begin{aligned}
&= \frac{1}{\sqrt{2^n}} \sum_{k = 0}^{2^n - 1} \prod_{j = 1}^{n} \exp\!\bigg( i 2 \pi x \frac{k_j}{2^j} \bigg) \Ket{k}
\end{aligned}$$
-If $k_j = 0$, the exponential is zero.
+If $$k_j = 0$$, the exponential is zero.
We can thus separate this state into its individual qubits:
$$\begin{aligned}
@@ -93,14 +93,14 @@ $$\begin{aligned}
\end{aligned}$$
Next, we use the trick from before:
-decompose $x$ as $x_1 2^{n-1} + x_2 2^{n-2} + ... + x_n 2^{0}$:
+decompose $$x$$ as $$x_1 2^{n-1} + x_2 2^{n-2} + ... + x_n 2^{0}$$:
$$\begin{aligned}
\Ket{\tilde{x}}
&= \bigotimes_{j = 1}^{n} \frac{1}{\sqrt{2}} \bigg( \Ket{0} + \exp\!\Big( i 2 \pi \sum_{r = 1}^{n} x_r 2^{n-r-j} \Big) \Ket{1} \bigg)
\end{aligned}$$
-The factor $2^{n-r-j}$ may be smaller or bigger than $1$,
+The factor $$2^{n-r-j}$$ may be smaller or bigger than $$1$$,
so it is convenient for us to use the following notation
for non-integer binary numbers.
Note the decimal point in the middle:
@@ -110,7 +110,7 @@ $$\begin{aligned}
= \sum_{r = 1}^{n} a_r 2^{d - r}
\end{aligned}$$
-In the above QFT state, the position of the decimal point is $d = n - j$,
+In the above QFT state, the position of the decimal point is $$d = n - j$$,
so we write it as:
$$\begin{aligned}
@@ -119,7 +119,7 @@ $$\begin{aligned}
+ \exp\!\Big( i 2 \pi \big[ x_1 \cdots x_{n-j} \:.\: x_{n-j+1} \cdots x_n \big] \Big) \Ket{1} \bigg)
\end{aligned}$$
-Because $\exp(i 2 \pi m) = 1$ for all integers $m$,
+Because $$\exp(i 2 \pi m) = 1$$ for all integers $$m$$,
we can discard bits before the decimal point:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, each exponential can be factorized,
-with every factor containing one bit of $x$:
+with every factor containing one bit of $$x$$:
$$\begin{aligned}
\exp\!\Big( i 2 \pi \big[ 0\:.\: x_{n-j+1} \cdots x_n \big] \Big)
@@ -141,42 +141,42 @@ $$\begin{aligned}
This suggests a way to implement the QFT using
[quantum gates](/know/concept/quantum-gate/) in a circuit.
-If the $j$th qubit is in $\Ket{+} = (\Ket{0} + \Ket{1})/\sqrt{2}$,
-then for each bit $x_{n-j+r}$ where $r \in \{1, ..., j\}$:
+If the $$j$$th qubit is in $$\Ket{+} = (\Ket{0} + \Ket{1})/\sqrt{2}$$,
+then for each bit $$x_{n-j+r}$$ where $$r \in \{1, ..., j\}$$:
-+ If $x_{n-j+r} = 0$, do nothing.
-+ If $x_{n-j+r} = 1$, add a relative phase $2 \pi / 2^{r}$.
++ If $$x_{n-j+r} = 0$$, do nothing.
++ If $$x_{n-j+r} = 1$$, add a relative phase $$2 \pi / 2^{r}$$.
The full QFT algorithm therefore proceeds as follows,
-for the $(n\!-\!j\!+\!1)$'th input $\Ket{x_{n-j+1}}$:
-
-1. Apply the Hadamard gate $H$.
- If $x_{n-j+1} = 0$, this puts the qubit in $\Ket{+}$.
- If $x_{n-j+1} = 1$, this puts the qubit in $\Ket{+}$,
- and then adds a phase $\pi$, yielding $\Ket{-}$.
-2. Apply the phase shift gate $R_{\phi}$ controlled by $x_{n-j+2}$,
- with angle $\phi = 2 \pi / 2^{2}$.
-3. Apply $R_{\phi}$ controlled by $x_{n-j+3}$,
- with angle $\phi = 2 \pi / 2^{3}$...
-4. And so on, until the $n$th bit $x_n$ is reached,
- and used to control $R_\phi$ with $\phi = 2 \pi / 2^{j}$.
-
-And so on, for each $j \in \{1, ..., n\}$,
-and we reach the above expression for $\Ket{\tilde{x}}$.
-We started from the $(n\!-\!j\!+\!1)$'th input qubit,
+for the $$(n\!-\!j\!+\!1)$$'th input $$\Ket{x_{n-j+1}}$$:
+
+1. Apply the Hadamard gate $$H$$.
+ If $$x_{n-j+1} = 0$$, this puts the qubit in $$\Ket{+}$$.
+ If $$x_{n-j+1} = 1$$, this puts the qubit in $$\Ket{+}$$,
+ and then adds a phase $$\pi$$, yielding $$\Ket{-}$$.
+2. Apply the phase shift gate $$R_{\phi}$$ controlled by $$x_{n-j+2}$$,
+ with angle $$\phi = 2 \pi / 2^{2}$$.
+3. Apply $$R_{\phi}$$ controlled by $$x_{n-j+3}$$,
+ with angle $$\phi = 2 \pi / 2^{3}$$...
+4. And so on, until the $$n$$th bit $$x_n$$ is reached,
+ and used to control $$R_\phi$$ with $$\phi = 2 \pi / 2^{j}$$.
+
+And so on, for each $$j \in \{1, ..., n\}$$,
+and we reach the above expression for $$\Ket{\tilde{x}}$$.
+We started from the $$(n\!-\!j\!+\!1)$$'th input qubit,
i.e. we read the input in reverse order.
Therefore all the qubits need to be swapped back to front,
either before or after the above algorithm is run.
The quantum circuit to execute the mentioned steps is illustrated below,
excluding the swapping part to get the right order.
-Here, $R_m$ means $R_\phi$ with $\phi = 2 \pi / 2^m$:
+Here, $$R_m$$ means $$R_\phi$$ with $$\phi = 2 \pi / 2^m$$:
-Again, note how the inputs $\Ket{x_j}$ and outputs $\Ket{k_j}$ are in the opposite order.
+Again, note how the inputs $$\Ket{x_j}$$ and outputs $$\Ket{k_j}$$ are in the opposite order.
The complete circuit, including the swapping at the end,
therefore looks like this:
@@ -184,8 +184,8 @@ therefore looks like this:
-For each of the $n$ qubits, $\mathcal{O}(n)$ gates are applied,
-so overall the QFT algorithm is $\mathcal{O}(n^2)$.
+For each of the $$n$$ qubits, $$\mathcal{O}(n)$$ gates are applied,
+so overall the QFT algorithm is $$\mathcal{O}(n^2)$$.
diff --git a/source/know/concept/quantum-gate/index.md b/source/know/concept/quantum-gate/index.md
index 38c39a1..8c251be 100644
--- a/source/know/concept/quantum-gate/index.md
+++ b/source/know/concept/quantum-gate/index.md
@@ -8,7 +8,7 @@ layout: "concept"
---
In quantum computing, **quantum gates** are the equivalent
-of classical binary logic gates such as $\mathrm{NOT}$, $\mathrm{AND}$, etc.
+of classical binary logic gates such as $$\mathrm{NOT}$$, $$\mathrm{AND}$$, etc.
Because of the continuous nature of qubits,
the number of possible quantum gates is uncountably infinite,
so we only consider the most important examples here.
@@ -16,7 +16,7 @@ so we only consider the most important examples here.
## One-qubit gates
-As an example, consider the following must general single-qubit state $\Ket{\psi}$:
+As an example, consider the following must general single-qubit state $$\Ket{\psi}$$:
$$\begin{aligned}
\Ket{\psi}
@@ -52,10 +52,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-They have the following effect on $\Ket{\psi}$.
-Note that $X$ is equivalent to the classical $\mathrm{NOT}$ gate
+They have the following effect on $$\Ket{\psi}$$.
+Note that $$X$$ is equivalent to the classical $$\mathrm{NOT}$$ gate
(and is often given that name),
-and $Z$ is sometimes called the **phase-flip gate**:
+and $$Z$$ is sometimes called the **phase-flip gate**:
$$\begin{aligned}
X \Ket{\psi}
@@ -68,9 +68,9 @@ $$\begin{aligned}
= \begin{bmatrix} \alpha \\ -\beta \end{bmatrix}
\end{aligned}$$
-In fact, $Z$ is a specific case of the **phase shift gate** $R_\phi$,
+In fact, $$Z$$ is a specific case of the **phase shift gate** $$R_\phi$$,
which modifies the qubit's phase without changing its amplitudes.
-For an angle $\phi$, it is given by:
+For an angle $$\phi$$, it is given by:
$$\begin{aligned}
\boxed{
@@ -82,17 +82,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-For $\phi = \pi$, we recover the Pauli-$Z$ gate.
-In general, the action of $R_\phi$ is as follows:
+For $$\phi = \pi$$, we recover the Pauli-$$Z$$ gate.
+In general, the action of $$R_\phi$$ is as follows:
$$\begin{aligned}
R_\phi \Ket{\psi}
= \begin{bmatrix} \alpha \\ e^{i \phi} \beta \end{bmatrix}
\end{aligned}$$
-Two common special cases of $R_\phi$
-are $\phi = \pi/2$ and $\phi = \pi/4$,
-respectively called $S$ and $T$:
+Two common special cases of $$R_\phi$$
+are $$\phi = \pi/2$$ and $$\phi = \pi/4$$,
+respectively called $$S$$ and $$T$$:
$$\begin{aligned}
\boxed{
@@ -113,7 +113,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, we have the **Hadamard gate** $H$,
+Finally, we have the **Hadamard gate** $$H$$,
which is defined as follows:
$$\begin{aligned}
@@ -127,14 +127,14 @@ $$\begin{aligned}
\end{aligned}$$
Its action consists of rotating the qubit
-by $\pi$ around the axis $(X + Z) / \sqrt{2}$ of the Bloch sphere:
+by $$\pi$$ around the axis $$(X + Z) / \sqrt{2}$$ of the Bloch sphere:
$$\begin{aligned}
H \Ket{\psi}
= \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix}
\end{aligned}$$
-Notably, it maps the eigenstates of $X$ and $Z$ to each other,
+Notably, it maps the eigenstates of $$X$$ and $$Z$$ to each other,
and is its own inverse (i.e. unitary):
$$\begin{aligned}
@@ -147,20 +147,20 @@ $$\begin{aligned}
H \Ket{-} = \Ket{1}
\end{aligned}$$
-The **Clifford gates** are a set including $X$, $Y$, $Z$, $H$ and $S$,
+The **Clifford gates** are a set including $$X$$, $$Y$$, $$Z$$, $$H$$ and $$S$$,
or more generally any gates that rotate
-by multiples of $\pi/2$ around the Bloch sphere.
-This set is **not universal**, meaning that if we start from $\Ket{0}$,
-we can only reach $\Ket{0}$, $\Ket{1}$, $\Ket{+}$, $\Ket{-}$, $\Ket{+i}$ $\Ket{-i}$ using these gates.
+by multiples of $$\pi/2$$ around the Bloch sphere.
+This set is **not universal**, meaning that if we start from $$\Ket{0}$$,
+we can only reach $$\Ket{0}$$, $$\Ket{1}$$, $$\Ket{+}$$, $$\Ket{-}$$, $$\Ket{+i}$$ $$\Ket{-i}$$ using these gates.
-If we add *any* non-Clifford gate, for example $T$,
+If we add *any* non-Clifford gate, for example $$T$$,
then we can reach any point on the Bloch sphere,
which means that the set is **universal**.
However, there is a problem: a qubit has an uncountable infinity of states,
but a quantum circuit consists of a countably infinite sequence of gates, at most.
Therefore, technically, we can never reach the whole Bloch sphere,
-but we *can* come up with circuits that approximate a target state to some degree $\varepsilon$.
+but we *can* come up with circuits that approximate a target state to some degree $$\varepsilon$$.
This is the definition of universality:
any state can be approximated.
@@ -168,7 +168,7 @@ any state can be approximated.
## Two-qubit gates
As an example, let us consider
-the following two pure one-qubit states $\Ket{\psi_1}$ and $\Ket{\psi_2}$:
+the following two pure one-qubit states $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$:
$$\begin{aligned}
\Ket{\psi_1}
@@ -181,7 +181,7 @@ $$\begin{aligned}
\end{aligned}$$
The composite state of both qubits, assuming they are pure,
-is then their tensor product $\otimes$:
+is then their tensor product $$\otimes$$:
$$\begin{aligned}
\Ket{\psi_1 \psi_2}
@@ -192,15 +192,15 @@ $$\begin{aligned}
\end{aligned}$$
Note that a two-qubit system may be [entangled](/know/concept/quantum-entanglement/),
-in which case the coefficients $c_{00}$ etc. cannot be written as products,
-i.e. $\Ket{\psi_2}$ cannot be expressed separately from $\Ket{\psi_1}$, and vice versa.
+in which case the coefficients $$c_{00}$$ etc. cannot be written as products,
+i.e. $$\Ket{\psi_2}$$ cannot be expressed separately from $$\Ket{\psi_1}$$, and vice versa.
In other words, the general action of a two-qubit quantum gate
-can be expressed in the basis of $\Ket{00}$, $\Ket{01}$, $\Ket{10}$ and $\Ket{11}$,
-but not always in the basis of $\Ket{0}_1$, $\Ket{1}_1$, $\Ket{0}_2$ and $\Ket{1}_2$.
+can be expressed in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$,
+but not always in the basis of $$\Ket{0}_1$$, $$\Ket{1}_1$$, $$\Ket{0}_2$$ and $$\Ket{1}_2$$.
-With that said, the first two-qubit gate is $\mathrm{SWAP}$,
-which simply swaps $\Ket{\psi_1}$ and $\Ket{\psi_2}$:
+With that said, the first two-qubit gate is $$\mathrm{SWAP}$$,
+which simply swaps $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$:
@@ -218,8 +218,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-This matrix is given in the basis of $\Ket{00}$, $\Ket{01}$, $\Ket{10}$ and $\Ket{11}$.
-Note that $\mathrm{SWAP}$ cannot generate entanglement,
+This matrix is given in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$.
+Note that $$\mathrm{SWAP}$$ cannot generate entanglement,
so if its input is separable, its output is too.
In any case, its effect is clear:
@@ -228,8 +228,8 @@ $$\begin{aligned}
&= c_{00} \Ket{00} + c_{10} \Ket{01} + c_{01} \Ket{10} + c_{11} \Ket{11}
\end{aligned}$$
-Next, there is the **controlled NOT gate** $\mathrm{CNOT}$,
-which "flips" (applies $X$ to) $\Ket{\psi_2}$ if $\Ket{\psi_1}$ is true:
+Next, there is the **controlled NOT gate** $$\mathrm{CNOT}$$,
+which "flips" (applies $$X$$ to) $$\Ket{\psi_2}$$ if $$\Ket{\psi_1}$$ is true:
@@ -247,16 +247,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-That is, it swaps the last two coefficients $c_{10}$ and $c_{11}$ in the composite state vector:
+That is, it swaps the last two coefficients $$c_{10}$$ and $$c_{11}$$ in the composite state vector:
$$\begin{aligned}
\mathrm{CNOT} \Ket{\psi_1 \psi_2}
&= c_{00} \Ket{00} + c_{01} \Ket{01} + c_{11} \Ket{10} + c_{10} \Ket{11}
\end{aligned}$$
-More generally, from every one-qubit gate $U$,
-we can define a two-qubit **controlled U gate** $\mathrm{CU}$,
-which applies $U$ to $\Ket{\psi_2}$ if $\Ket{\psi_1}$ is true:
+More generally, from every one-qubit gate $$U$$,
+we can define a two-qubit **controlled U gate** $$\mathrm{CU}$$,
+which applies $$U$$ to $$\Ket{\psi_2}$$ if $$\Ket{\psi_1}$$ is true:
@@ -274,7 +274,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the lower-right 2x2 block is simply $U$.
+Where the lower-right 2x2 block is simply $$U$$.
The general action of this gate is given by:
$$\begin{aligned}
@@ -283,8 +283,8 @@ $$\begin{aligned}
\end{aligned}$$
A set of gates is **universal** if all possible mappings
-from $n$ to $n$ qubits can be approximated using only these gates.
-A minimal universal set is $\{\mathrm{CNOT}, T, S\}$,
+from $$n$$ to $$n$$ qubits can be approximated using only these gates.
+A minimal universal set is $$\{\mathrm{CNOT}, T, S\}$$,
and there exist many others.
diff --git a/source/know/concept/quantum-teleportation/index.md b/source/know/concept/quantum-teleportation/index.md
index c7b9295..095c2c6 100644
--- a/source/know/concept/quantum-teleportation/index.md
+++ b/source/know/concept/quantum-teleportation/index.md
@@ -11,8 +11,8 @@ layout: "concept"
between systems without the use of a quantum channel.
It is based on [quantum entanglement](/know/concept/quantum-entanglement/).
-Suppose that Alice has a qubit $\Ket{q}_{A'}$ that she wants to send to Bob.
-Since she has not measured it yet, she does not know $\alpha$ or $\beta$;
+Suppose that Alice has a qubit $$\Ket{q}_{A'}$$ that she wants to send to Bob.
+Since she has not measured it yet, she does not know $$\alpha$$ or $$\beta$$;
she just wants Bob to get the same qubit:
$$\begin{aligned}
@@ -21,14 +21,14 @@ $$\begin{aligned}
\end{aligned}$$
She can only directly communicate with Bob over a classical channel.
-This is not enough: even if Alice did know $\alpha$ and $\beta$ exactly
+This is not enough: even if Alice did know $$\alpha$$ and $$\beta$$ exactly
(which would need her having infinitely many copies to measure),
sending an arbitrary real number requires an infinite amount of classical data.
However, between them, she and Bob also have an entangled [Bell state](/know/concept/bell-state/),
-e.g. $\ket{\Phi^+}_{AB}$ (it does not matter which Bell state it is)
+e.g. $$\ket{\Phi^+}_{AB}$$ (it does not matter which Bell state it is)
The state of the composite system is then as follows,
-with $A'$ being Alice' qubit, $A$ her side of the Bell state, and $B$ Bob's side:
+with $$A'$$ being Alice' qubit, $$A$$ her side of the Bell state, and $$B$$ Bob's side:
$$\begin{aligned}
\Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB}
@@ -38,7 +38,7 @@ $$\begin{aligned}
+ \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB}
\end{aligned}$$
-Now, observe that we can write any combination of $\Ket{0}$ and $\Ket{1}$
+Now, observe that we can write any combination of $$\Ket{0}$$ and $$\Ket{1}$$
in the Bell basis like so:
$$\begin{aligned}
@@ -79,7 +79,7 @@ $$\begin{aligned}
\end{aligned}$$
Thus, purely due to entanglement,
-Bob's qubit $B$ is in a superposition of the following states:
+Bob's qubit $$B$$ is in a superposition of the following states:
$$\begin{aligned}
\Ket{q}
@@ -106,38 +106,38 @@ $$\begin{aligned}
\end{aligned}$$
The point is that, thanks to the initial entanglement between Alice and Bob,
-adding $\Ket{q}_{A'}$ into the mix somehow "teleports" that information to Bob,
+adding $$\Ket{q}_{A'}$$ into the mix somehow "teleports" that information to Bob,
although it is not in a usable form yet.
-To finish the process, Alice measures her side $A'A$ in the Bell basis.
-Consequently, $A'A$ collapses into one of
-$\ket{\Phi^{+}}$, $\ket{\Phi^{-}}$, $\ket{\Psi^{+}}$, $\ket{\Psi^{-}}$
+To finish the process, Alice measures her side $$A'A$$ in the Bell basis.
+Consequently, $$A'A$$ collapses into one of
+$$\ket{\Phi^{+}}$$, $$\ket{\Phi^{-}}$$, $$\ket{\Psi^{+}}$$, $$\ket{\Psi^{-}}$$
with equal probability, and she knows which.
-This collapse leaves Bob's side $B$ in $\Ket{q}$, $\hat{\sigma}_z \Ket{q}$,
-$\hat{\sigma}_x \Ket{q}$, or $\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$, respectively.
-The entanglement between $A$ and $B$ is thus broken,
-and instead Alice has local entanglement between $A'$ and $A$.
+This collapse leaves Bob's side $$B$$ in $$\Ket{q}$$, $$\hat{\sigma}_z \Ket{q}$$,
+$$\hat{\sigma}_x \Ket{q}$$, or $$\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$$, respectively.
+The entanglement between $$A$$ and $$B$$ is thus broken,
+and instead Alice has local entanglement between $$A'$$ and $$A$$.
She then uses the classical channel to tell Bob her result,
-who then either does nothing (for $\Ket{q}$),
-applies $\hat{\sigma}_z$ (for $\hat{\sigma}_z \Ket{q}$),
-applies $\hat{\sigma}_x$ (for $\hat{\sigma}_x \Ket{q}$),
-or applies $\hat{\sigma}_z \hat{\sigma}_x$ (for $\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$).
-Then, due to the fact that $\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$,
-he recovers $\Ket{q}$ in his local qubit $B$.
+who then either does nothing (for $$\Ket{q}$$),
+applies $$\hat{\sigma}_z$$ (for $$\hat{\sigma}_z \Ket{q}$$),
+applies $$\hat{\sigma}_x$$ (for $$\hat{\sigma}_x \Ket{q}$$),
+or applies $$\hat{\sigma}_z \hat{\sigma}_x$$ (for $$\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$$).
+Then, due to the fact that $$\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$$,
+he recovers $$\Ket{q}$$ in his local qubit $$B$$.
This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem)
-because Alice does not require any knowledge of $\Ket{q}$,
-and after the measurement, her qubit $A'$ will no longer be in that state.
+because Alice does not require any knowledge of $$\Ket{q}$$,
+and after the measurement, her qubit $$A'$$ will no longer be in that state.
In other words, quantum teleportation *moves* states,
rather than copying them.
Nor does this conflict with Einstein's relativity,
since the information travels no faster than light:
-the entangled $\ket{\Phi^{+}}_{AB}$ state must be distributed in advance,
+the entangled $$\ket{\Phi^{+}}_{AB}$$ state must be distributed in advance,
and Alice' declaration of her result is sent classically.
Before receiving that, Bob only sees his side of the maximally entangled
-Bell state $\ket{\Phi^{+}}_{AB}$, which contains nothing of $\Ket{q}$.
+Bell state $$\ket{\Phi^{+}}_{AB}$$, which contains nothing of $$\Ket{q}$$.
## References
diff --git a/source/know/concept/rabi-oscillation/index.md b/source/know/concept/rabi-oscillation/index.md
index 9077cce..07f8b25 100644
--- a/source/know/concept/rabi-oscillation/index.md
+++ b/source/know/concept/rabi-oscillation/index.md
@@ -12,21 +12,21 @@ layout: "concept"
In quantum mechanics, from the derivation of
[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
-we know that a time-dependent term $\hat{H}_1$ in the Hamiltonian
+we know that a time-dependent term $$\hat{H}_1$$ in the Hamiltonian
affects the state as follows,
-where $c_n(t)$ are the coefficients of the linear combination
-of basis states $\Ket{n} \exp(-i E_n t / \hbar)$:
+where $$c_n(t)$$ are the coefficients of the linear combination
+of basis states $$\Ket{n} \exp(-i E_n t / \hbar)$$:
$$\begin{aligned}
i \hbar \dv{c_m}{t}
= \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} \exp(i \omega_{mn} t)
\end{aligned}$$
-Where $\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$
-for energies $E_m$ and $E_n$.
+Where $$\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$$
+for energies $$E_m$$ and $$E_n$$.
Note that this equation is exact,
despite being used for deriving perturbation theory.
-Consider a two-level system where $n \in \{a, b\}$,
+Consider a two-level system where $$n \in \{a, b\}$$,
in which case the above equation can be expanded to the following:
$$\begin{aligned}
@@ -37,8 +37,8 @@ $$\begin{aligned}
&= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} \: c_b
\end{aligned}$$
-Where $\omega_0 \equiv \omega_{ba}$ is positive.
-We assume that $\hat{H}_1$ has odd spatial parity,
+Where $$\omega_0 \equiv \omega_{ba}$$ is positive.
+We assume that $$\hat{H}_1$$ has odd spatial parity,
in which case [Laporte's selection rule](/know/concept/selection-rules/)
states that the diagonal matrix elements vanish, leaving:
@@ -50,8 +50,8 @@ $$\begin{aligned}
&= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a
\end{aligned}$$
-We now choose $\hat{H}_1$ to be as follows,
-sinusoidally oscillating with a spatially odd $V(\vec{r})$:
+We now choose $$\hat{H}_1$$ to be as follows,
+sinusoidally oscillating with a spatially odd $$V(\vec{r})$$:
$$\begin{aligned}
\hat{H}_1(t)
@@ -59,8 +59,8 @@ $$\begin{aligned}
= \frac{V}{2} \Big( \exp(i \omega t) + \exp(-i \omega t) \Big)
\end{aligned}$$
-We insert this into the equations for $c_a$ and $c_b$,
-and define $V_{ab} \equiv \matrixel{a}{V}{b}$, leading us to:
+We insert this into the equations for $$c_a$$ and $$c_b$$,
+and define $$V_{ab} \equiv \matrixel{a}{V}{b}$$, leading us to:
$$\begin{aligned}
\dv{c_a}{t}
@@ -72,8 +72,8 @@ $$\begin{aligned}
Here, we make the
[rotating wave approximation](/know/concept/rotating-wave-approximation/):
-assuming we are close to resonance $\omega \approx \omega_0$,
-we argue that $\exp(i (\omega \!+\! \omega_0) t)$
+assuming we are close to resonance $$\omega \approx \omega_0$$,
+we argue that $$\exp(i (\omega \!+\! \omega_0) t)$$
oscillates so fast that its effect is negligible
when the system is observed over a reasonable time interval.
Dropping those terms leaves us with:
@@ -91,8 +91,8 @@ $$\begin{aligned}
\end{aligned}$$
Now we can solve this system of coupled equations exactly.
-We differentiate the first equation with respect to $t$,
-and then substitute $\idv{c_b}{t}$ for the second equation:
+We differentiate the first equation with respect to $$t$$,
+and then substitute $$\idv{c_b}{t}$$ for the second equation:
$$\begin{aligned}
\dvn{2}{c_a}{t}
@@ -105,15 +105,15 @@ $$\begin{aligned}
&= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} c_a
\end{aligned}$$
-In the first term, we recognize $\idv{c_a}{t}$,
-which we insert to arrive at an equation for $c_a(t)$:
+In the first term, we recognize $$\idv{c_a}{t}$$,
+which we insert to arrive at an equation for $$c_a(t)$$:
$$\begin{aligned}
0
= \dvn{2}{c_a}{t} - i (\omega - \omega_0) \dv{c_a}{t} + \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a
\end{aligned}$$
-To solve this, we make the ansatz $c_a(t) = \exp(\lambda t)$,
+To solve this, we make the ansatz $$c_a(t) = \exp(\lambda t)$$,
which, upon insertion, gives us:
$$\begin{aligned}
@@ -121,7 +121,7 @@ $$\begin{aligned}
= \lambda^2 - i (\omega - \omega_0) \lambda + \frac{|V_{ab}|^2}{(2 \hbar)^2}
\end{aligned}$$
-This quadratic equation has two complex roots $\lambda_1$ and $\lambda_2$,
+This quadratic equation has two complex roots $$\lambda_1$$ and $$\lambda_2$$,
which are found to be:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
= i \frac{\omega - \omega_0 - \tilde{\Omega}}{2}
\end{aligned}$$
-Where we have defined the **generalized Rabi frequency** $\tilde{\Omega}$ to be given by:
+Where we have defined the **generalized Rabi frequency** $$\tilde{\Omega}$$ to be given by:
$$\begin{aligned}
\boxed{
@@ -141,8 +141,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-So that the general solution $c_a(t)$ is as follows,
-where $A$ and $B$ are arbitrary constants,
+So that the general solution $$c_a(t)$$ is as follows,
+where $$A$$ and $$B$$ are arbitrary constants,
to be determined from initial conditions (and normalization):
$$\begin{aligned}
@@ -152,11 +152,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-And then the corresponding $c_b(t)$ can be found
+And then the corresponding $$c_b(t)$$ can be found
from the coupled equation we started at,
-or, if we only care about the probability density $|c_a|^2$,
-we can use $|c_b|^2 = 1 - |c_a|^2$.
-For example, if $A = 0$ and $B = 1$,
+or, if we only care about the probability density $$|c_a|^2$$,
+we can use $$|c_b|^2 = 1 - |c_a|^2$$.
+For example, if $$A = 0$$ and $$B = 1$$,
we get the following probabilities
$$\begin{aligned}
@@ -170,15 +170,15 @@ $$\begin{aligned}
\end{aligned}$$
Note that the period was halved by squaring.
-This periodic "flopping" of the particle between $\Ket{a}$ and $\Ket{b}$
+This periodic "flopping" of the particle between $$\Ket{a}$$ and $$\Ket{b}$$
is known as **Rabi oscillation**, **Rabi flopping** or the **Rabi cycle**.
This is a more accurate treatment
of the flopping found from first-order perturbation theory.
The name **generalized Rabi frequency** suggests
that there is a non-general version.
-Indeed, the **Rabi frequency** $\Omega$ is based on
-the special case of exact resonance $\omega = \omega_0$:
+Indeed, the **Rabi frequency** $$\Omega$$ is based on
+the special case of exact resonance $$\omega = \omega_0$$:
$$\begin{aligned}
\Omega
@@ -187,7 +187,7 @@ $$\begin{aligned}
As an example, Rabi oscillation arises
in the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
-where $\hat{H}_1$ is:
+where $$\hat{H}_1$$ is:
$$\begin{aligned}
\hat{H}_1(t)
@@ -202,10 +202,10 @@ $$\begin{aligned}
= - \frac{\vec{d} \cdot \vec{E}_0}{\hbar}
\end{aligned}$$
-Where $\vec{E}_0$ is the [electric field](/know/concept/electric-field/) amplitude,
-and $\vec{d} \equiv q \matrixel{b}{\vec{r}}{a}$ is the transition dipole moment
-of the electron between orbitals $\Ket{a}$ and $\Ket{b}$.
-Apparently, some authors define $\vec{d}$ with the opposite sign,
+Where $$\vec{E}_0$$ is the [electric field](/know/concept/electric-field/) amplitude,
+and $$\vec{d} \equiv q \matrixel{b}{\vec{r}}{a}$$ is the transition dipole moment
+of the electron between orbitals $$\Ket{a}$$ and $$\Ket{b}$$.
+Apparently, some authors define $$\vec{d}$$ with the opposite sign,
thereby departing from its classical interpretation.
diff --git a/source/know/concept/random-phase-approximation/index.md b/source/know/concept/random-phase-approximation/index.md
index ac007eb..0f53136 100644
--- a/source/know/concept/random-phase-approximation/index.md
+++ b/source/know/concept/random-phase-approximation/index.md
@@ -8,19 +8,19 @@ categories:
layout: "concept"
---
-Recall that the [self-energy](/know/concept/self-energy/) $\Sigma$
+Recall that the [self-energy](/know/concept/self-energy/) $$\Sigma$$
is defined as a sum of [Feynman diagrams](/know/concept/feynman-diagram/),
-which each have an order $n$ equal to the number of interaction lines.
+which each have an order $$n$$ equal to the number of interaction lines.
We consider the self-energy in the context of [jellium](/know/concept/jellium/),
-so the interaction lines $W$ represent Coulomb repulsion,
+so the interaction lines $$W$$ represent Coulomb repulsion,
and we use [imaginary time](/know/concept/imaginary-time/).
Let us non-dimensionalize the Feynman diagrams in the self-energy,
-by measuring momenta in units of $\hbar k_F$,
-and energies in $\epsilon_F = \hbar^2 k_F^2 / (2 m)$.
-Each internal variable then gives a factor $k_F^5$,
-where $k_F^3$ comes from the 3D momentum integral,
-and $k_F^2$ from the energy $1 / \beta$:
+by measuring momenta in units of $$\hbar k_F$$,
+and energies in $$\epsilon_F = \hbar^2 k_F^2 / (2 m)$$.
+Each internal variable then gives a factor $$k_F^5$$,
+where $$k_F^3$$ comes from the 3D momentum integral,
+and $$k_F^2$$ from the energy $$1 / \beta$$:
$$\begin{aligned}
\frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty \cdots \:\dd{\vb{k}}
@@ -28,8 +28,8 @@ $$\begin{aligned}
k_F^5
\end{aligned}$$
-Meanwhile, every line gives a factor $1 / k_F^2$.
-The [Matsubara Green's function](/know/concept/matsubara-greens-function/) $G^0$
+Meanwhile, every line gives a factor $$1 / k_F^2$$.
+The [Matsubara Green's function](/know/concept/matsubara-greens-function/) $$G^0$$
for a system with continuous translational symmetry
is found from [equation-of-motion theory](/know/concept/equation-of-motion-theory/):
@@ -44,38 +44,38 @@ $$\begin{aligned}
\frac{1}{k_F^2}
\end{aligned}$$
-An $n$th-order diagram in $\Sigma$ contains $n$ interaction lines,
-$2n\!-\!1$ fermion lines, and $n$ integrals,
-so in total it evolves as $1 / k_F^{n-2}$.
-In jellium, we know that the electron density is proportional to $k_F^3$,
-so for high densities we can rest assured that higher-order terms in $\Sigma$
+An $$n$$th-order diagram in $$\Sigma$$ contains $$n$$ interaction lines,
+$$2n\!-\!1$$ fermion lines, and $$n$$ integrals,
+so in total it evolves as $$1 / k_F^{n-2}$$.
+In jellium, we know that the electron density is proportional to $$k_F^3$$,
+so for high densities we can rest assured that higher-order terms in $$\Sigma$$
converge to zero faster than lower-order terms.
-However, at a given order $n$, not all diagrams are equally important.
+However, at a given order $$n$$, not all diagrams are equally important.
In a given diagram, due to momentum conservation,
some interaction lines carry the same momentum variable.
-Because $W(\vb{k}) \propto 1 / |\vb{k}|^2$,
-small $\vb{k}$ make a large contribution,
-and the more interaction lines depend on the same $\vb{k}$,
+Because $$W(\vb{k}) \propto 1 / |\vb{k}|^2$$,
+small $$\vb{k}$$ make a large contribution,
+and the more interaction lines depend on the same $$\vb{k}$$,
the larger the contribution becomes.
In other words, each diagram is dominated by contributions
from the momentum carried by the largest number of interactions.
-At order $n$, there is one diagram
-where all $n$ interactions carry the same momentum,
+At order $$n$$, there is one diagram
+where all $$n$$ interactions carry the same momentum,
and this one dominates all others at this order.
The **random phase approximation** consists of removing most diagrams
-from the defintion of the full self-energy $\Sigma$,
-leaving only the single most divergent one at each order $n$,
-i.e. the ones where all $n$ interaction lines
+from the defintion of the full self-energy $$\Sigma$$,
+leaving only the single most divergent one at each order $$n$$,
+i.e. the ones where all $$n$$ interaction lines
carry the same momentum and energy:
-Where we have defined the **screened interaction** $W^\mathrm{RPA}$,
+Where we have defined the **screened interaction** $$W^\mathrm{RPA}$$,
denoted by a double wavy line:
@@ -99,19 +99,19 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined the **pair-bubble** $\Pi_0$ as follows,
-with an internal wavevector $\vb{q}$, fermionic frequency $i \omega_m^F$, and spin $s$.
-Abbreviating $\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n^B)$
-and $\tilde{\vb{q}} \equiv (\vb{q}, i \omega_n^F)$:
+Where we have defined the **pair-bubble** $$\Pi_0$$ as follows,
+with an internal wavevector $$\vb{q}$$, fermionic frequency $$i \omega_m^F$$, and spin $$s$$.
+Abbreviating $$\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n^B)$$
+and $$\tilde{\vb{q}} \equiv (\vb{q}, i \omega_n^F)$$:
-We isolate the Dyson equation for $W^\mathrm{RPA}$,
+We isolate the Dyson equation for $$W^\mathrm{RPA}$$,
which reveals its physical interpretation as a *screened* interaction:
-the "raw" interaction $W \!=\! e^2 / (\varepsilon_0 |\vb{k}|^2)$
-is weakened by a term containing $\Pi_0$:
+the "raw" interaction $$W \!=\! e^2 / (\varepsilon_0 |\vb{k}|^2)$$
+is weakened by a term containing $$\Pi_0$$:
$$\begin{aligned}
W^\mathrm{RPA}(\vb{k}, i \omega_n^B)
@@ -119,7 +119,7 @@ $$\begin{aligned}
= \frac{e^2}{\varepsilon_0 |\vb{k}|^2 - e^2 \Pi_0(\vb{k}, i \omega_n^B)}
\end{aligned}$$
-Let us evaluate the pair-bubble $\Pi_0$ more concretely.
+Let us evaluate the pair-bubble $$\Pi_0$$ more concretely.
The Feynman diagram translates to:
$$\begin{aligned}
@@ -133,9 +133,9 @@ $$\begin{aligned}
Here we recognize a [Matsubara sum](/know/concept/matsubara-sum/),
and rewrite accordingly.
-Note that the residues of $n_F$ are $1 / (\hbar \beta)$
+Note that the residues of $$n_F$$ are $$1 / (\hbar \beta)$$
when it is a function of frequency,
-and $1 / \beta$ when it is a function of energy, so:
+and $$1 / \beta$$ when it is a function of energy, so:
$$\begin{aligned}
\Pi_0(\vb{k}, i \omega_n^B)
@@ -147,12 +147,12 @@ $$\begin{aligned}
{i \hbar \omega_n^B + \varepsilon_{\vb{q}} - \varepsilon_{\vb{k}+\vb{q}}} \dd{\vb{q}}
\end{aligned}$$
-Where we have used that $n_F(\varepsilon \!+\! i \hbar \omega_n^B) = n_F(\varepsilon)$.
-Analogously to extracting the retarded Green's function $G^R(\omega)$
-from the Matsubara Green's function $G^0(i \omega_n^F)$,
-we replace $i \omega_n^F \to \omega \!+\! i \eta$,
-where $\eta \to 0^+$ is a positive infinitesimal,
-yielding the retarded pair-bubble $\Pi_0^R$:
+Where we have used that $$n_F(\varepsilon \!+\! i \hbar \omega_n^B) = n_F(\varepsilon)$$.
+Analogously to extracting the retarded Green's function $$G^R(\omega)$$
+from the Matsubara Green's function $$G^0(i \omega_n^F)$$,
+we replace $$i \omega_n^F \to \omega \!+\! i \eta$$,
+where $$\eta \to 0^+$$ is a positive infinitesimal,
+yielding the retarded pair-bubble $$\Pi_0^R$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/random-variable/index.md b/source/know/concept/random-variable/index.md
index 1781c8b..ecb8e96 100644
--- a/source/know/concept/random-variable/index.md
+++ b/source/know/concept/random-variable/index.md
@@ -19,50 +19,50 @@ of a random variable.
## Probability space
-A **probability space** or **probability triple** $(\Omega, \mathcal{F}, P)$
+A **probability space** or **probability triple** $$(\Omega, \mathcal{F}, P)$$
is the formal mathematical model of a given **stochastic experiment**,
i.e. a process with a random outcome.
-The **sample space** $\Omega$ is the set
-of all possible outcomes $\omega$ of the experimement.
-Those $\omega$ are selected randomly according to certain criteria.
-A subset $A \subset \Omega$ is called an **event**,
-and can be regarded as a true statement about all $\omega$ in that $A$.
+The **sample space** $$\Omega$$ is the set
+of all possible outcomes $$\omega$$ of the experimement.
+Those $$\omega$$ are selected randomly according to certain criteria.
+A subset $$A \subset \Omega$$ is called an **event**,
+and can be regarded as a true statement about all $$\omega$$ in that $$A$$.
-The **event space** $\mathcal{F}$ is a set of events $A$
+The **event space** $$\mathcal{F}$$ is a set of events $$A$$
that are interesting to us,
-i.e. we have subjectively chosen $\mathcal{F}$
+i.e. we have subjectively chosen $$\mathcal{F}$$
based on the problem at hand.
-Since events $A$ represent statements about outcomes $\omega$,
+Since events $$A$$ represent statements about outcomes $$\omega$$,
and we would like to use logic on those statemenets,
-we demand that $\mathcal{F}$ is a [$\sigma$-algebra](/know/concept/sigma-algebra/).
+we demand that $$\mathcal{F}$$ is a [$$\sigma$$-algebra](/know/concept/sigma-algebra/).
-Finally, the **probability measure** or **probability function** $P$
-is a function that maps $A$ events to probabilities $P(A)$.
-Formally, $P : \mathcal{F} \to \mathbb{R}$ is defined to satisfy:
+Finally, the **probability measure** or **probability function** $$P$$
+is a function that maps $$A$$ events to probabilities $$P(A)$$.
+Formally, $$P : \mathcal{F} \to \mathbb{R}$$ is defined to satisfy:
-1. If $A \in \mathcal{F}$, then $P(A) \in [0, 1]$.
-2. If $A, B \in \mathcal{F}$ do not overlap $A \cap B = \varnothing$,
- then $P(A \cup B) = P(A) + P(B)$.
-3. The total probability $P(\Omega) = 1$.
+1. If $$A \in \mathcal{F}$$, then $$P(A) \in [0, 1]$$.
+2. If $$A, B \in \mathcal{F}$$ do not overlap $$A \cap B = \varnothing$$,
+ then $$P(A \cup B) = P(A) + P(B)$$.
+3. The total probability $$P(\Omega) = 1$$.
-The reason we only assign probability to events $A$
-rather than individual outcomes $\omega$ is that
-if $\Omega$ is continuous, all $\omega$ have zero probability,
-while intervals $A$ can have nonzero probability.
+The reason we only assign probability to events $$A$$
+rather than individual outcomes $$\omega$$ is that
+if $$\Omega$$ is continuous, all $$\omega$$ have zero probability,
+while intervals $$A$$ can have nonzero probability.
## Random variable
-Once we have a probability space $(\Omega, \mathcal{F}, P)$,
-we can define a **random variable** $X$
-as a function that maps outcomes $\omega$
+Once we have a probability space $$(\Omega, \mathcal{F}, P)$$,
+we can define a **random variable** $$X$$
+as a function that maps outcomes $$\omega$$
to another set, usually the real numbers.
To be a valid real-valued random variable,
-a function $X : \Omega \to \mathbb{R}^n$ must satisfy the following condition,
-in which case $X$ is said to be **measurable**
-from $(\Omega, \mathcal{F})$ to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$:
+a function $$X : \Omega \to \mathbb{R}^n$$ must satisfy the following condition,
+in which case $$X$$ is said to be **measurable**
+from $$(\Omega, \mathcal{F})$$ to $$(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$$:
$$\begin{aligned}
\{ \omega \in \Omega : X(\omega) \in B \} \in \mathcal{F}
@@ -70,16 +70,16 @@ $$\begin{aligned}
\end{aligned}$$
In other words, for a given Borel set
-(see [$\sigma$-algebra](/know/concept/sigma-algebra/)) $B \in \mathcal{B}(\mathbb{R}^n)$,
-the set of all outcomes $\omega \in \Omega$ that satisfy $X(\omega) \in B$
-must form a valid event; this set must be in $\mathcal{F}$.
+(see [$$\sigma$$-algebra](/know/concept/sigma-algebra/)) $$B \in \mathcal{B}(\mathbb{R}^n)$$,
+the set of all outcomes $$\omega \in \Omega$$ that satisfy $$X(\omega) \in B$$
+must form a valid event; this set must be in $$\mathcal{F}$$.
The point is that we need to be able to assign probabilities
-to statements of the form $X \in [a, b]$ for all $a < b$,
-which is only possible if that statement corresponds to an event in $\mathcal{F}$,
-since $P$'s domain is $\mathcal{F}$.
+to statements of the form $$X \in [a, b]$$ for all $$a < b$$,
+which is only possible if that statement corresponds to an event in $$\mathcal{F}$$,
+since $$P$$'s domain is $$\mathcal{F}$$.
-Given such an $X$, and a set $B \subseteq \mathbb{R}$,
-the **preimage** or **inverse image** $X^{-1}$ is defined as:
+Given such an $$X$$, and a set $$B \subseteq \mathbb{R}$$,
+the **preimage** or **inverse image** $$X^{-1}$$ is defined as:
$$\begin{aligned}
X^{-1}(B)
@@ -87,16 +87,16 @@ $$\begin{aligned}
\end{aligned}$$
As suggested by the notation,
-$X^{-1}$ can be regarded as the inverse of $X$:
-it maps $B$ to the event for which $X \in B$.
-With this, our earlier requirement that $X$ be measurable
-can be written as: $X^{-1}(B) \in \mathcal{F}$ for any $B \in \mathcal{B}(\mathbb{R}^n)$.
-This is also often stated as "$X$ is *$\mathcal{F}$-measurable"*.
-
-Related to $\mathcal{F}$ is the **information**
-obtained by observing a random variable $X$.
-Let $\sigma(X)$ be the information generated by observing $X$,
-i.e. the events whose occurrence can be deduced from the value of $X$,
+$$X^{-1}$$ can be regarded as the inverse of $$X$$:
+it maps $$B$$ to the event for which $$X \in B$$.
+With this, our earlier requirement that $$X$$ be measurable
+can be written as: $$X^{-1}(B) \in \mathcal{F}$$ for any $$B \in \mathcal{B}(\mathbb{R}^n)$$.
+This is also often stated as "$$X$$ is *$$\mathcal{F}$$-measurable"*.
+
+Related to $$\mathcal{F}$$ is the **information**
+obtained by observing a random variable $$X$$.
+Let $$\sigma(X)$$ be the information generated by observing $$X$$,
+i.e. the events whose occurrence can be deduced from the value of $$X$$,
or, more formally:
$$\begin{aligned}
@@ -105,29 +105,29 @@ $$\begin{aligned}
= \{ A \in \mathcal{F} : A = X^{-1}(B) \mathrm{\:for\:some\:} B \in \mathcal{B}(\mathbb{R}^n) \}
\end{aligned}$$
-In other words, if the realized value of $X$ is
-found to be in a certain Borel set $B \in \mathcal{B}(\mathbb{R}^n)$,
-then the preimage $X^{-1}(B)$ (i.e. the event yielding this $B$)
+In other words, if the realized value of $$X$$ is
+found to be in a certain Borel set $$B \in \mathcal{B}(\mathbb{R}^n)$$,
+then the preimage $$X^{-1}(B)$$ (i.e. the event yielding this $$B$$)
is known to have occurred.
-In general, given any $\sigma$-algebra $\mathcal{H}$,
-a variable $Y$ is said to be *"$\mathcal{H}$-measurable"*
-if $\sigma(Y) \subseteq \mathcal{H}$,
-so that $\mathcal{H}$ contains at least
-all information extractable from $Y$.
+In general, given any $$\sigma$$-algebra $$\mathcal{H}$$,
+a variable $$Y$$ is said to be *"$$\mathcal{H}$$-measurable"*
+if $$\sigma(Y) \subseteq \mathcal{H}$$,
+so that $$\mathcal{H}$$ contains at least
+all information extractable from $$Y$$.
-Note that $\mathcal{H}$ can be generated by another random variable $X$,
-i.e. $\mathcal{H} = \sigma(X)$.
+Note that $$\mathcal{H}$$ can be generated by another random variable $$X$$,
+i.e. $$\mathcal{H} = \sigma(X)$$.
In that case, the **Doob-Dynkin lemma** states
-that $Y$ is only $\sigma(X)$-measurable
-if $Y$ can always be computed from $X$,
-i.e. there exists a function $f$ such that
-$Y(\omega) = f(X(\omega))$ for all $\omega \in \Omega$.
+that $$Y$$ is only $$\sigma(X)$$-measurable
+if $$Y$$ can always be computed from $$X$$,
+i.e. there exists a function $$f$$ such that
+$$Y(\omega) = f(X(\omega))$$ for all $$\omega \in \Omega$$.
Now, we are ready to define some familiar concepts from probability theory.
-The **cumulative distribution function** $F_X(x)$ is
-the probability of the event where the realized value of $X$
-is smaller than some given $x \in \mathbb{R}$:
+The **cumulative distribution function** $$F_X(x)$$ is
+the probability of the event where the realized value of $$X$$
+is smaller than some given $$x \in \mathbb{R}$$:
$$\begin{aligned}
F_X(x)
@@ -136,8 +136,8 @@ $$\begin{aligned}
= P(X^{-1}(]\!-\!\infty, x]))
\end{aligned}$$
-If $F_X(x)$ is differentiable,
-then the **probability density function** $f_X(x)$ is defined as:
+If $$F_X(x)$$ is differentiable,
+then the **probability density function** $$f_X(x)$$ is defined as:
$$\begin{aligned}
f_X(x)
@@ -147,10 +147,10 @@ $$\begin{aligned}
## Expectation value
-The **expectation value** $\mathbf{E}[X]$ of a random variable $X$
+The **expectation value** $$\mathbf{E}[X]$$ of a random variable $$X$$
can be defined in the familiar way, as the sum/integral
-of every possible value of $X$ mutliplied by the corresponding probability (density).
-For continuous and discrete sample spaces $\Omega$, respectively:
+of every possible value of $$X$$ mutliplied by the corresponding probability (density).
+For continuous and discrete sample spaces $$\Omega$$, respectively:
$$\begin{aligned}
\mathbf{E}[X]
@@ -160,18 +160,18 @@ $$\begin{aligned}
= \sum_{i = 1}^N x_i \: P(X \!=\! x_i)
\end{aligned}$$
-However, $f_X(x)$ is not guaranteed to exist,
+However, $$f_X(x)$$ is not guaranteed to exist,
and the distinction between continuous and discrete is cumbersome.
-A more general definition of $\mathbf{E}[X]$
+A more general definition of $$\mathbf{E}[X]$$
is the following Lebesgue-Stieltjes integral,
-since $F_X(x)$ always exists:
+since $$F_X(x)$$ always exists:
$$\begin{aligned}
\mathbf{E}[X]
= \int_{-\infty}^\infty x \dd{F_X(x)}
\end{aligned}$$
-This is valid for any sample space $\Omega$.
+This is valid for any sample space $$\Omega$$.
Or, equivalently, a Lebesgue integral can be used:
$$\begin{aligned}
@@ -182,8 +182,8 @@ $$\begin{aligned}
An expectation value defined in this way has many useful properties,
most notably linearity.
-We can also define the familiar **variance** $\mathbf{V}[X]$
-of a random variable $X$ as follows:
+We can also define the familiar **variance** $$\mathbf{V}[X]$$
+of a random variable $$X$$ as follows:
$$\begin{aligned}
\mathbf{V}[X]
diff --git a/source/know/concept/rayleigh-plateau-instability/index.md b/source/know/concept/rayleigh-plateau-instability/index.md
index 2ba060c..5dcf77f 100644
--- a/source/know/concept/rayleigh-plateau-instability/index.md
+++ b/source/know/concept/rayleigh-plateau-instability/index.md
@@ -16,11 +16,11 @@ It is the reason why a smooth stream of water (e.g. from a tap)
eventually breaks into droplets as it falls.
Consider an infinitely long cylinder of liquid
-with radius $R_0$ and surface tension $\alpha$.
+with radius $$R_0$$ and surface tension $$\alpha$$.
In this case, the [Young-Laplace equation](/know/concept/young-laplace-law/)
states that its internal pressure
-is a constant $p_i$ expressed as follows,
-where $p_o$ is the exterior air pressure:
+is a constant $$p_i$$ expressed as follows,
+where $$p_o$$ is the exterior air pressure:
$$\begin{aligned}
p_i
@@ -28,12 +28,12 @@ $$\begin{aligned}
\end{aligned}$$
We assume that the liquid is at rest.
-Alternatively, if it is moving in the $z$-direction,
+Alternatively, if it is moving in the $$z$$-direction,
we can also let our coordinate system travel at the same speed.
Anyway, for convenience,
we neglect any motion or acceleration of the liquid column.
-Next, we add a perturbation $p_\epsilon$, assumed to be small,
+Next, we add a perturbation $$p_\epsilon$$, assumed to be small,
to the internal pressure, which we allow to vary with time and space.
We use cylindrical coordinates:
@@ -42,7 +42,7 @@ $$\begin{aligned}
\end{aligned}$$
This internal pressure difference will cause the liquid to start to flow.
-We express the flow velocity as a vector $\vec{u} = (u_r, u_\phi, u_z)$,
+We express the flow velocity as a vector $$\vec{u} = (u_r, u_\phi, u_z)$$,
which obeys the following Euler equations:
$$\begin{aligned}
@@ -53,7 +53,7 @@ $$\begin{aligned}
\end{aligned}$$
The latter equation states that the fluid is incompressible.
-We assume that $\vec{u}$ is so small that we can ignore
+We assume that $$\vec{u}$$ is so small that we can ignore
the quadratic term in the former equation, leaving:
$$\begin{aligned}
@@ -62,7 +62,7 @@ $$\begin{aligned}
\end{aligned}$$
Taking the divergence and using incompressibility
-yields the Laplace equation for $p_\epsilon$:
+yields the Laplace equation for $$p_\epsilon$$:
$$\begin{aligned}
- \frac{1}{\rho} \nabla^2 p_\epsilon
@@ -81,7 +81,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Finally, we add a perturbation $R_\epsilon \ll R_0$
+Finally, we add a perturbation $$R_\epsilon \ll R_0$$
to the radius of the surface of the liquid column:
$$\begin{aligned}
@@ -89,13 +89,13 @@ $$\begin{aligned}
= R_0 + R_\epsilon(z, t)
\end{aligned}$$
-Note that there is no dependence on the angle $\phi$;
+Note that there is no dependence on the angle $$\phi$$;
the deformation is assumed to be symmetric.
Imagine the cross-section of the cylinder,
and convince yourself that all asymmetric deformations
will be removed by surface tension, which prefers a circular shape.
-We thus assume that $R_\epsilon$, $p_\epsilon$ and $\vec{u}$
-do not depend on $\phi$.
+We thus assume that $$R_\epsilon$$, $$p_\epsilon$$ and $$\vec{u}$$
+do not depend on $$\phi$$.
The Laplace equation then reduces to:
$$\begin{aligned}
@@ -105,9 +105,9 @@ $$\begin{aligned}
\end{aligned}$$
Before solving this, we need boundary conditions.
-The radial fluid velocity $u_r$ (the $r$-component of $\vec{u}$)
-at the column surface $r\!=\!R$ is the
-[material derivative](/know/concept/material-derivative/) of $R_\epsilon$:
+The radial fluid velocity $$u_r$$ (the $$r$$-component of $$\vec{u}$$)
+at the column surface $$r\!=\!R$$ is the
+[material derivative](/know/concept/material-derivative/) of $$R_\epsilon$$:
$$\begin{aligned}
u_r(r\!=\!R)
@@ -115,17 +115,17 @@ $$\begin{aligned}
= \pdv{R_\epsilon}{t} + u_z(r\!=\!R) \pdv{R_\epsilon}{z}
\end{aligned}$$
-We linearize this by assuming that the deformation $R_\epsilon$
-varies slowly with respect to $z$:
+We linearize this by assuming that the deformation $$R_\epsilon$$
+varies slowly with respect to $$z$$:
$$\begin{aligned}
u_r(r\!=\!R)
\approx \pdv{R_\epsilon}{t}
\end{aligned}$$
-Meanwhile, we can write the boundary condition of the pressure $p$
+Meanwhile, we can write the boundary condition of the pressure $$p$$
in two ways, respectively from the Young-Laplace equation
-and the definition of the perturbation $p_\epsilon$:
+and the definition of the perturbation $$p_\epsilon$$:
$$\begin{aligned}
p(r\!=\!R)
@@ -135,32 +135,32 @@ $$\begin{aligned}
= p_i + p_\epsilon(r\!=\!R)
\end{aligned}$$
-Where $R_1$ and $R_2$ are the principal curvature radii of the column surface.
+Where $$R_1$$ and $$R_2$$ are the principal curvature radii of the column surface.
These two expressions must be equivalent,
-so, by inserting the definition of $p_i = p_o + \alpha / R_0$:
+so, by inserting the definition of $$p_i = p_o + \alpha / R_0$$:
$$\begin{aligned}
p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
= p_o + \frac{\alpha}{R_0} + p_\epsilon(r\!=\!R)
\end{aligned}$$
-Isolating this equation for $p_\epsilon$ yields the desired boundary condition:
+Isolating this equation for $$p_\epsilon$$ yields the desired boundary condition:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
= \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big) - \frac{\alpha}{R_0}
\end{aligned}$$
-The principal radius around the circumference is $R_0 + R_\epsilon$,
+The principal radius around the circumference is $$R_0 + R_\epsilon$$,
while the curvature along the length can be approximated
-using the second $z$-derivative of $R_\epsilon$:
+using the second $$z$$-derivative of $$R_\epsilon$$:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
\approx \alpha \Big( \frac{1}{R_0 + R_\epsilon} - \pdvn{2}{R_\epsilon}{z} \Big) - \frac{\alpha}{R_0}
\end{aligned}$$
-This can be simplified a bit by using the assumption that $R_\epsilon$ is small:
+This can be simplified a bit by using the assumption that $$R_\epsilon$$ is small:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
@@ -170,8 +170,8 @@ $$\begin{aligned}
At last, we have all the necessary boundary condition.
We now make the following ansatz,
-where $k$ is the wavenumber
-and $\sigma$ describes exponential growth or decay:
+where $$k$$ is the wavenumber
+and $$\sigma$$ describes exponential growth or decay:
$$\begin{aligned}
\vec{u}(r, z, t)
@@ -187,7 +187,7 @@ $$\begin{aligned}
This is justified by the fact that we can Fourier-expand any perturbation;
this ansatz is simply the dominant term of the resulting series.
-Inserting this into the Laplace equation for $p_\epsilon$ yields
+Inserting this into the Laplace equation for $$p_\epsilon$$ yields
Bessel's modified equation of order zero:
$$\begin{aligned}
@@ -195,8 +195,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-This has well-known solutions: the modified Bessel functions $I_0$ and $K_0$.
-However, because $K_0$ diverges at $r = 0$, we must set the constant $B = 0$:
+This has well-known solutions: the modified Bessel functions $$I_0$$ and $$K_0$$.
+However, because $$K_0$$ diverges at $$r = 0$$, we must set the constant $$B = 0$$:
$$\begin{aligned}
p_\epsilon(r)
@@ -204,7 +204,7 @@ $$\begin{aligned}
= A I_0(kr)
\end{aligned}$$
-Inserting the ansatz into the boundary condition for $p_\epsilon$
+Inserting the ansatz into the boundary condition for $$p_\epsilon$$
gives us the following relation:
$$\begin{aligned}
@@ -213,8 +213,8 @@ $$\begin{aligned}
= A I_0(k R)
\end{aligned}$$
-Meanwhile, the linearized Euler equation governing $\vec{u}$
-states that $u_r$ is given by:
+Meanwhile, the linearized Euler equation governing $$\vec{u}$$
+states that $$u_r$$ is given by:
$$\begin{aligned}
\sigma u_r
@@ -222,7 +222,7 @@ $$\begin{aligned}
= - \frac{A k}{\rho} I_0'(kr)
\end{aligned}$$
-Now that we have an expression for $u_r$,
+Now that we have an expression for $$u_r$$,
we can revisit its boundary condition:
$$\begin{aligned}
@@ -231,8 +231,8 @@ $$\begin{aligned}
= \sigma R_\epsilon
\end{aligned}$$
-Isolating this for $R_\epsilon$ and inserting it
-into the boundary condition for $p_\epsilon$ yields:
+Isolating this for $$R_\epsilon$$ and inserting it
+into the boundary condition for $$p_\epsilon$$ yields:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
@@ -240,18 +240,18 @@ $$\begin{aligned}
= \alpha \Big( \frac{1}{R_0^2} + k^2 \Big) \Big( \frac{A k}{\rho \sigma^2} I_0'(k R) \Big)
\end{aligned}$$
-Isolating this for the exponential growth/decay parameter $\sigma$
+Isolating this for the exponential growth/decay parameter $$\sigma$$
gives us the desired result,
-where we have also used the fact that $R \approx R_0$:
+where we have also used the fact that $$R \approx R_0$$:
$$\begin{aligned}
\sigma^2
= \frac{\alpha k}{\rho R_0^2} (1 - k^2 R_0^2) \frac{I_0'(kR_0)}{I_0(kR_0)}
\end{aligned}$$
-To get exponential growth (i.e. instability), we need $\sigma^2 > 0$.
-Since $(1 - k^2 R_0^2)$ is the only factor that can be negative,
-we need $k R_0 < 1$, leading us to the **critical wavelength** $\lambda_c$:
+To get exponential growth (i.e. instability), we need $$\sigma^2 > 0$$.
+Since $$(1 - k^2 R_0^2)$$ is the only factor that can be negative,
+we need $$k R_0 < 1$$, leading us to the **critical wavelength** $$\lambda_c$$:
$$\begin{aligned}
\boxed{
@@ -261,12 +261,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-If the perturbation wavelength $\lambda$ is larger than $\lambda_c$,
+If the perturbation wavelength $$\lambda$$ is larger than $$\lambda_c$$,
surface tension creates a higher pressure in the narrower sections
compared to the wider ones, thereby pumping the liquid into the bulges,
further increasing their size until they become droplets.
-Else, if $\lambda < \lambda_c$, the tighter curvatures
+Else, if $$\lambda < \lambda_c$$, the tighter curvatures
dominate the action of surface tension,
which will then try to smoothen the surface by shrinking the bulges
and widening the constrictions.
diff --git a/source/know/concept/rayleigh-plesset-equation/index.md b/source/know/concept/rayleigh-plesset-equation/index.md
index 4718683..b3ec8f5 100644
--- a/source/know/concept/rayleigh-plesset-equation/index.md
+++ b/source/know/concept/rayleigh-plesset-equation/index.md
@@ -16,7 +16,7 @@ Notably, it leads to [cavitation](/know/concept/cavitation/).
Consider the main
[Navier-Stokes equation](/know/concept/navier-stokes-equations/)
-for the velocity field $\va{v}$:
+for the velocity field $$\va{v}$$:
$$\begin{aligned}
\frac{\mathrm{D} \va{v}}{\mathrm{D} t}
@@ -24,9 +24,9 @@ $$\begin{aligned}
= - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
\end{aligned}$$
-We make the ansatz $\va{v} = v(r, t) \vu{e}_r$,
-where $\vu{e}_r$ is the basis vector;
-in other words, we demand that the only spatial variation of the flow is in $r$.
+We make the ansatz $$\va{v} = v(r, t) \vu{e}_r$$,
+where $$\vu{e}_r$$ is the basis vector;
+in other words, we demand that the only spatial variation of the flow is in $$r$$.
The above equation then becomes:
$$\begin{aligned}
@@ -44,17 +44,17 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-This is only satisfied if $r^2 v$ is constant with respect to $r$,
-leading us to a solution $v(r)$ given by:
+This is only satisfied if $$r^2 v$$ is constant with respect to $$r$$,
+leading us to a solution $$v(r)$$ given by:
$$\begin{aligned}
v(r)
= \frac{C(t)}{r^2}
\end{aligned}$$
-Where $C(t)$ is an unknown function that does not depend on $r$.
+Where $$C(t)$$ is an unknown function that does not depend on $$r$$.
We then insert this result in the main Navier-Stokes equation,
-and isolate it for $\ipdv{p}{r}$, yielding:
+and isolate it for $$\ipdv{p}{r}$$, yielding:
$$\begin{aligned}
\pdv{p}{r}
@@ -63,8 +63,8 @@ $$\begin{aligned}
= - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg)
\end{aligned}$$
-Integrating this with respect to $r$ yields the following expression for $p$,
-where $p_\infty(t)$ is the (possibly time-dependent) pressure at $r = \infty$:
+Integrating this with respect to $$r$$ yields the following expression for $$p$$,
+where $$p_\infty(t)$$ is the (possibly time-dependent) pressure at $$r = \infty$$:
$$\begin{aligned}
p(r)
@@ -73,7 +73,7 @@ $$\begin{aligned}
From the definition of [viscosity](/know/concept/viscosity/),
we know that the normal [stress](/know/concept/cauchy-stress-tensor/)
-$\sigma_{rr}$ in the liquid is given by:
+$$\sigma_{rr}$$ in the liquid is given by:
$$\begin{aligned}
\sigma_{rr}(r)
@@ -81,11 +81,11 @@ $$\begin{aligned}
\end{aligned}$$
We now consider a spherical bubble
-with radius $R(t)$ and interior pressure $P(t)$ along its surface.
-Since we know the liquid pressure $p(r)$,
-we can find $P$ from $\sigma_{rr}(r)$.
+with radius $$R(t)$$ and interior pressure $$P(t)$$ along its surface.
+Since we know the liquid pressure $$p(r)$$,
+we can find $$P$$ from $$\sigma_{rr}(r)$$.
Furthermore, to include the effects of surface tension, we simply add
-the [Young-Laplace law](/know/concept/young-laplace-law/) to $P$:
+the [Young-Laplace law](/know/concept/young-laplace-law/) to $$P$$:
$$\begin{aligned}
P
@@ -93,18 +93,18 @@ $$\begin{aligned}
= p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R}
\end{aligned}$$
-We isolate this for $p(R)$, and equate it to
-our expression for $p(r)$
-at the surface $r\!=\!R$:
+We isolate this for $$p(R)$$, and equate it to
+our expression for $$p(r)$$
+at the surface $$r\!=\!R$$:
$$\begin{aligned}
P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R}
= p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg)
\end{aligned}$$
-Isolating for $P$,
-and inserting the fact that $R'(t) = v(t)$,
-such that $C = r^2 v = R^2 R'$,
+Isolating for $$P$$,
+and inserting the fact that $$R'(t) = v(t)$$,
+such that $$C = r^2 v = R^2 R'$$,
yields:
$$\begin{aligned}
@@ -115,7 +115,7 @@ $$\begin{aligned}
&= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R}
\end{aligned}$$
-Rearranging this and defining $\Delta p \equiv P - p_\infty$
+Rearranging this and defining $$\Delta p \equiv P - p_\infty$$
leads to the Rayleigh-Plesset equation:
$$\begin{aligned}
diff --git a/source/know/concept/reduced-mass/index.md b/source/know/concept/reduced-mass/index.md
index 6718895..1a05b5c 100644
--- a/source/know/concept/reduced-mass/index.md
+++ b/source/know/concept/reduced-mass/index.md
@@ -8,9 +8,9 @@ layout: "concept"
---
Problems with two interacting objects can be simplified
-by combining them into a pseudo-object with **reduced mass** $\mu$,
+by combining them into a pseudo-object with **reduced mass** $$\mu$$,
whose position equals the relative position of the objects.
-For bodies 1 and 2 with respective masses $m_1$ and $m_2$:
+For bodies 1 and 2 with respective masses $$m_1$$ and $$m_2$$:
$$\begin{aligned}
\boxed{
@@ -18,11 +18,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\va{x}_1$ and $\va{x}_2$ are the objects' respective positions,
+If $$\va{x}_1$$ and $$\va{x}_2$$ are the objects' respective positions,
then we define
-the relative position $\va{x}_r$,
-the relative velocity $\va{v}_r$,
-and the relative acceleration $\va{a}_r$:
+the relative position $$\va{x}_r$$,
+the relative velocity $$\va{v}_r$$,
+and the relative acceleration $$\va{a}_r$$:
$$\begin{aligned}
\va{x}_r
@@ -69,9 +69,9 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, Newton's third law states that
-if object 1 experiences a force $\va{F}_1 = m_1 \va{a}_1$ caused by object 2,
-then object 2 experiences an opposite and equal force $\va{F}_2 = - \va{F}_1$.
-In fact, our earlier relation between $\va{a}_1$ and $\va{a}_1$
+if object 1 experiences a force $$\va{F}_1 = m_1 \va{a}_1$$ caused by object 2,
+then object 2 experiences an opposite and equal force $$\va{F}_2 = - \va{F}_1$$.
+In fact, our earlier relation between $$\va{a}_1$$ and $$\va{a}_1$$
boils down to Newton's third law:
$$\begin{aligned}
@@ -80,7 +80,7 @@ $$\begin{aligned}
\va{a}_2 = - \frac{m_1}{m_2} \va{a}_1
\end{aligned}$$
-With all that in mind, let us take a closer look at the relative acceleration $\va{a}_r$:
+With all that in mind, let us take a closer look at the relative acceleration $$\va{a}_r$$:
$$\begin{aligned}
\va{a}_r
@@ -90,15 +90,15 @@ $$\begin{aligned}
= - \frac{\va{F}_2}{\mu}
\end{aligned}$$
-Where $\mu$ is the reduced mass, as defined above.
-In other words, the relative acceleration $\va{a}_r$
-is just $\va{a}_1 = \va{F}_1 / m_1$ multiplied by $m_1 / \mu$.
+Where $$\mu$$ is the reduced mass, as defined above.
+In other words, the relative acceleration $$\va{a}_r$$
+is just $$\va{a}_1 = \va{F}_1 / m_1$$ multiplied by $$m_1 / \mu$$.
This can be regarded as focusing on the dynamics of body 1,
while correcting for the effects of body 2.
This also suggests the following way
-to recover the original positions $\va{x}_1$ and $\va{x}_2$
-from $\va{x}_r$, which you can easily verify for yourself:
+to recover the original positions $$\va{x}_1$$ and $$\va{x}_2$$
+from $$\va{x}_r$$, which you can easily verify for yourself:
$$\begin{aligned}
\va{x}_1
@@ -110,7 +110,7 @@ $$\begin{aligned}
= - \frac{m_1}{m_1 + m_2} \va{x}_r
\end{aligned}$$
-With this, we can rewrite the total kinetic energy $T$ in an elegant way:
+With this, we can rewrite the total kinetic energy $$T$$ in an elegant way:
$$\begin{aligned}
T
@@ -125,11 +125,11 @@ $$\begin{aligned}
= \frac{1}{2} \mu \va{v}_r^2
\end{aligned}$$
-Then, assuming that the system's potential energy $V$
+Then, assuming that the system's potential energy $$V$$
only depends on the distance between the two objects,
-i.e. $V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$,
-we just showed that we can rewrite both $T$ and $V$
-to contain only $\mu$ and relative quantities.
+i.e. $$V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$$,
+we just showed that we can rewrite both $$T$$ and $$V$$
+to contain only $$\mu$$ and relative quantities.
This is relevant for both [Lagrangian mechanics](/know/concept/lagrangian-mechanics/)
and [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
-where $L = T - V$ and $H = T + V$ respectively.
+where $$L = T - V$$ and $$H = T + V$$ respectively.
diff --git a/source/know/concept/renyi-entropy/index.md b/source/know/concept/renyi-entropy/index.md
index d53ccc0..55f233c 100644
--- a/source/know/concept/renyi-entropy/index.md
+++ b/source/know/concept/renyi-entropy/index.md
@@ -9,7 +9,7 @@ layout: "concept"
In information theory, the **Rényi entropy** is a measure
(or family of measures) of the "suprise" or "information"
-contained in a random variable $X$.
+contained in a random variable $$X$$.
It is defined as follows:
$$\begin{aligned}
@@ -19,12 +19,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\alpha \ge 0$ is a free parameter.
+Where $$\alpha \ge 0$$ is a free parameter.
The logarithm is usually base-2, but variations exist.
-The case $\alpha = 0$ is known as the **Hartley entropy** or **max-entropy**,
-and quantifies the "surprise" of an event from $X$,
-if $X$ is uniformly distributed:
+The case $$\alpha = 0$$ is known as the **Hartley entropy** or **max-entropy**,
+and quantifies the "surprise" of an event from $$X$$,
+if $$X$$ is uniformly distributed:
$$\begin{aligned}
\boxed{
@@ -33,9 +33,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $N$ is the cardinality of $X$; the number of different possible events.
-The most famous case, however, is $\alpha = 1$.
-Since $H_\alpha$ is problematic for $\alpha \to 1$, we must take the limit:
+Where $$N$$ is the cardinality of $$X$$; the number of different possible events.
+The most famous case, however, is $$\alpha = 1$$.
+Since $$H_\alpha$$ is problematic for $$\alpha \to 1$$, we must take the limit:
$$\begin{aligned}
H_1(X)
@@ -44,7 +44,7 @@ $$\begin{aligned}
\end{aligned}$$
We then apply L'Hôpital's rule to evaluate this limit,
-and use the fact that all $p_i$ sum to $1$:
+and use the fact that all $$p_i$$ sum to $$1$$:
$$\begin{aligned}
H_1(X)
@@ -64,9 +64,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Next, for $\alpha = 2$, we get the **collision entropy**, which describes
+Next, for $$\alpha = 2$$, we get the **collision entropy**, which describes
the surprise of two independent and identically distributed variables
-$X$ and $Y$ yielding the same event:
+$$X$$ and $$Y$$ yielding the same event:
$$\begin{aligned}
\boxed{
@@ -76,9 +76,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, in the limit $\alpha \to \infty$,
+Finally, in the limit $$\alpha \to \infty$$,
the largest probability dominates the sum,
-leading to the definition of the **min-entropy** $H_\infty$,
+leading to the definition of the **min-entropy** $$H_\infty$$,
describing the surprise of the most likely event:
$$\begin{aligned}
diff --git a/source/know/concept/repetition-code/index.md b/source/know/concept/repetition-code/index.md
index 31181bb..99ac630 100644
--- a/source/know/concept/repetition-code/index.md
+++ b/source/know/concept/repetition-code/index.md
@@ -8,7 +8,7 @@ layout: "concept"
---
A **repetition code** is a simple approach to error correction:
-to protect a bit $x$, make two copies:
+to protect a bit $$x$$, make two copies:
$$\begin{aligned}
0 \to 000
@@ -16,7 +16,7 @@ $$\begin{aligned}
1 \to 111
\end{aligned}$$
-If a single-bit error occurs, e.g. $000 \to 100$,
+If a single-bit error occurs, e.g. $$000 \to 100$$,
a majority vote resets the minority bit.
Clearly, this does not protect against multi-bit errors,
but that is usually not necessary.
@@ -30,7 +30,7 @@ as discussed below.
## Bit flip code
Suppose that we want to detect errors in
-the following arbitrary qubit state $\Ket{\psi}$:
+the following arbitrary qubit state $$\Ket{\psi}$$:
$$\begin{aligned}
\Ket{\psi}
@@ -38,7 +38,7 @@ $$\begin{aligned}
\end{aligned}$$
For now, let us limit ourselves to detecting **bit flips**,
-where $\alpha$ and $\beta$ get switched:
+where $$\alpha$$ and $$\beta$$ get switched:
$$\begin{aligned}
\alpha \Ket{0} + \beta \Ket{1}
@@ -56,7 +56,7 @@ $$\begin{aligned}
= \alpha \Ket{000} + \beta \Ket{111}
\end{aligned}$$
-In other words, a *logical* $\Ket{0}$ (written $\ket{\overline{0}}$)
+In other words, a *logical* $$\Ket{0}$$ (written $$\ket{\overline{0}}$$)
is represented by 3 *physical* qubits, and vice versa:
$$\begin{aligned}
@@ -80,7 +80,7 @@ of [quantum gates](/know/concept/quantum-gate/):
-So, a little while after encoding the state $\Ket{\psi}$ like that,
+So, a little while after encoding the state $$\Ket{\psi}$$ like that,
a bit flip occurs on the 2nd qubit:
$$\begin{aligned}
@@ -94,8 +94,8 @@ We could measure the state, but that would make it collapse,
which is probably not what we want.
The trick is to use operators called **stabilizers**,
-in this case for example $ZZI = Z_1 \otimes Z_2 \otimes I_3$,
-where $I$ is identity and $Z$ is the Pauli-$Z$ gate.
+in this case for example $$ZZI = Z_1 \otimes Z_2 \otimes I_3$$,
+where $$I$$ is identity and $$Z$$ is the Pauli-$$Z$$ gate.
The 3-qubit basis states are its eigenvectors:
$$\begin{alignedat}{2}
@@ -124,13 +124,13 @@ $$\begin{alignedat}{2}
&&= + \Ket{111}
\end{alignedat}$$
-We could measure $ZZI$ for $\ket{\overline{\psi}}$,
-and if the eigenvalue is $-1$,
+We could measure $$ZZI$$ for $$\ket{\overline{\psi}}$$,
+and if the eigenvalue is $$-1$$,
we know that a bit flip has occurred,
-whereas if the eigenvalue is $+1$,
-there is *maybe* no error ($\Ket{001}$ and $\Ket{110}$ are false negatives).
+whereas if the eigenvalue is $$+1$$,
+there is *maybe* no error ($$\Ket{001}$$ and $$\Ket{110}$$ are false negatives).
-These false negatives are fixed by including another stabilizer $IZZ$,
+These false negatives are fixed by including another stabilizer $$IZZ$$,
with these eigenvectors:
$$\begin{alignedat}{2}
@@ -159,56 +159,56 @@ $$\begin{alignedat}{2}
&&= + \Ket{111}
\end{alignedat}$$
-In which case $\Ket{100}$ and $\Ket{011}$ are false negatives.
-In other words, $IZZ$ cannot detect if the 1st qubit was flipped,
-while $ZZI$ cannot protect the 3rd qubit.
+In which case $$\Ket{100}$$ and $$\Ket{011}$$ are false negatives.
+In other words, $$IZZ$$ cannot detect if the 1st qubit was flipped,
+while $$ZZI$$ cannot protect the 3rd qubit.
But by using both, we know exactly which qubit was flipped
thanks to the eigenvalues:
-We know that the angular momentum $z$-component operator $\hat{L}_z$ satisfies:
+We know that the angular momentum $$z$$-component operator $$\hat{L}_z$$ satisfies:
$$\begin{aligned}
\comm{\hat{L}_z}{\hat{x}} = i \hbar \hat{y}
@@ -103,7 +103,7 @@ $$\begin{aligned}
\end{aligned}$$
We take the first relation,
-and wrap it in $\Bra{\ell_f m_f}$ and $\Ket{\ell_i m_i}$, giving:
+and wrap it in $$\Bra{\ell_f m_f}$$ and $$\Ket{\ell_i m_i}$$, giving:
$$\begin{aligned}
i \hbar \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
@@ -114,7 +114,7 @@ $$\begin{aligned}
&= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
\end{aligned}$$
-Next, we do the same thing with the second relation, for $[\hat{L}_z, \hat{y}]$, giving:
+Next, we do the same thing with the second relation, for $$[\hat{L}_z, \hat{y}]$$, giving:
$$\begin{aligned}
- i \hbar \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
@@ -125,7 +125,7 @@ $$\begin{aligned}
&= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
\end{aligned}$$
-Respectively isolating the two above results for $\hat{x}$ and $\hat{y}$,
+Respectively isolating the two above results for $$\hat{x}$$ and $$\hat{y}$$,
we arrive at these equations:
$$\begin{aligned}
@@ -144,11 +144,11 @@ $$\begin{aligned}
&= (m_f - m_i)^2 \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
\end{aligned}$$
-This can only be true if $\Delta m = \pm 1$,
-unless the inner products of $\hat{x}$ and $\hat{y}$ are zero,
-in which case we cannot say anything about $\Delta m$ yet.
+This can only be true if $$\Delta m = \pm 1$$,
+unless the inner products of $$\hat{x}$$ and $$\hat{y}$$ are zero,
+in which case we cannot say anything about $$\Delta m$$ yet.
Assuming the latter, we take the inner product of
-the commutator $\comm{\hat{L}_z}{\hat{z}} = 0$, and find:
+the commutator $$\comm{\hat{L}_z}{\hat{z}} = 0$$, and find:
$$\begin{aligned}
0
@@ -159,17 +159,17 @@ $$\begin{aligned}
&= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i}
\end{aligned}$$
-If $\matrixel{f}{\hat{z}}{i} \neq 0$, we require $\Delta m = 0$.
-The previous requirement was $\Delta m = \pm 1$,
-implying that $\matrixel{f}{\hat{x}}{i} = \matrixel{f}{\hat{y}}{i} = 0$
-whenever $\matrixel{f}{\hat{z}}{i} \neq 0$.
-Only if $\matrixel{f}{\hat{z}}{i} = 0$
-does the previous rule $\Delta m = \pm 1$ hold,
-in which case the inner products of $\hat{x}$ and $\hat{y}$ are nonzero.
+If $$\matrixel{f}{\hat{z}}{i} \neq 0$$, we require $$\Delta m = 0$$.
+The previous requirement was $$\Delta m = \pm 1$$,
+implying that $$\matrixel{f}{\hat{x}}{i} = \matrixel{f}{\hat{y}}{i} = 0$$
+whenever $$\matrixel{f}{\hat{z}}{i} \neq 0$$.
+Only if $$\matrixel{f}{\hat{z}}{i} = 0$$
+does the previous rule $$\Delta m = \pm 1$$ hold,
+in which case the inner products of $$\hat{x}$$ and $$\hat{y}$$ are nonzero.
-Meanwhile, for the total angular momentum $\ell$ we have the following:
+Meanwhile, for the total angular momentum $$\ell$$ we have the following:
$$\begin{aligned}
\boxed{
@@ -195,7 +195,7 @@ $$\begin{aligned}
-To begin with, we want to find the commutator of $\hat{L}^2$ and $\hat{x}$:
+To begin with, we want to find the commutator of $$\hat{L}^2$$ and $$\hat{x}$$:
$$\begin{aligned}
\comm{\hat{L}^2}{\hat{x}}
@@ -263,7 +263,7 @@ $$\begin{aligned}
&= 2 i \hbar (\hat{y} \hat{L}_z - \hat{z} \hat{L}_y - i \hbar \hat{x})
\end{aligned}$$
-Repeating this process for $\comm{\hat{L}^2}{\hat{y}}$ and $\comm{\hat{L}^2}{\hat{z}}$,
+Repeating this process for $$\comm{\hat{L}^2}{\hat{y}}$$ and $$\comm{\hat{L}^2}{\hat{z}}$$,
we find analogous expressions:
$$\begin{aligned}
@@ -274,7 +274,7 @@ $$\begin{aligned}
&= 2 i \hbar (\hat{x} \hat{L}_y - \hat{y} \hat{L}_x - i \hbar \hat{z})
\end{aligned}$$
-Next, we take the commutator with $\hat{L}^2$ of the commutator we just found:
+Next, we take the commutator with $$\hat{L}^2$$ of the commutator we just found:
$$\begin{aligned}
\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
@@ -285,7 +285,7 @@ $$\begin{aligned}
- i \hbar \comm{\hat{L}^2}{\hat{x}} \big)
\end{aligned}$$
-Where we used that $\comm{\hat{L}^2}{\hat{L}_y} = \comm{\hat{L}^2}{\hat{L}_z} = 0$.
+Where we used that $$\comm{\hat{L}^2}{\hat{L}_y} = \comm{\hat{L}^2}{\hat{L}_z} = 0$$.
The other commutators look familiar:
$$\begin{aligned}
@@ -305,8 +305,8 @@ $$\begin{aligned}
\end{aligned}$$
Substituting the well-known commutators
-$i \hbar \hat{L}_y = \comm{\hat{L}_z}{\hat{L}_x}$ and
-$i \hbar \hat{L}_z = \comm{\hat{L}_x}{\hat{L}_y}$:
+$$i \hbar \hat{L}_y = \comm{\hat{L}_z}{\hat{L}_x}$$ and
+$$i \hbar \hat{L}_z = \comm{\hat{L}_x}{\hat{L}_y}$$:
$$\begin{aligned}
\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
@@ -320,7 +320,7 @@ $$\begin{aligned}
+ 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
\end{aligned}$$
-By definition, $\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2$,
+By definition, $$\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2$$,
which we use to arrive at:
$$\begin{aligned}
@@ -343,12 +343,12 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Where $\vu{L} = \vu{r} \cross \vu{p}$ by definition,
+Where $$\vu{L} = \vu{r} \cross \vu{p}$$ by definition,
and the cross product of a vector with itself is zero.
This process can be repeated for
-$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}}$ and
-$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}$,
+$$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}}$$ and
+$$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}$$,
leading us to:
$$\begin{aligned}
@@ -362,13 +362,13 @@ $$\begin{aligned}
&= 2 \hbar^2 (\hat{z} \hat{L}^2 + \hat{L}^2 \hat{z})
\end{aligned}$$
-At last, this brings us to the desired equation for $\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}$,
-with $\vu{r} = (\hat{x}, \hat{y}, \hat{z})$.
+At last, this brings us to the desired equation for $$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}$$,
+with $$\vu{r} = (\hat{x}, \hat{y}, \hat{z})$$.
-We then multiply this relation by $\Bra{f} = \Bra{\ell_f m_f}$ on the left
-and $\Ket{i} = \Ket{\ell_i m_i}$ on the right,
+We then multiply this relation by $$\Bra{f} = \Bra{\ell_f m_f}$$ on the left
+and $$\Ket{i} = \Ket{\ell_i m_i}$$ on the right,
so the right-hand side becomes:
$$\begin{aligned}
@@ -450,21 +450,22 @@ $$\begin{aligned}
&= \big( (\ell_f + \ell_i + 1)^2 - 1 \big) \big( (\ell_f - \ell_i)^2 - 1 \big)
\end{aligned}$$
-The first factor is zero if $\ell_f = \ell_i = 0$,
-in which case the matrix element $\matrixel{f}{\vu{r}}{i} = 0$ anyway.
+The first factor is zero if $$\ell_f = \ell_i = 0$$,
+in which case the matrix element $$\matrixel{f}{\vu{r}}{i} = 0$$ anyway.
The other, non-trivial option is therefore:
$$\begin{aligned}
(\ell_f - \ell_i)^2
= 1
\end{aligned}$$
+
## Rotational rules
-Given a general (pseudo)scalar operator $\hat{s}$,
+Given a general (pseudo)scalar operator $$\hat{s}$$,
which, by nature, must satisfy the
following relations with the angular momentum operators:
@@ -476,8 +477,8 @@ $$\begin{aligned}
\comm{\hat{L}_{\pm}}{\hat{s}} = 0
\end{aligned}$$
-Where $\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$.
-The inner product of any such $\hat{s}$ must obey these selection rules:
+Where $$\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$$.
+The inner product of any such $$\hat{s}$$ must obey these selection rules:
$$\begin{aligned}
\boxed{
@@ -490,9 +491,9 @@ $$\begin{aligned}
\end{aligned}$$
It is common to write this in the following more complete way, where
-$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$ is the **reduced matrix element**,
-which is identical to $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$, but
-with a different notation to say that it does not depend on $m_f$ or $m_i$:
+$$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$$ is the **reduced matrix element**,
+which is identical to $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$$, but
+with a different notation to say that it does not depend on $$m_f$$ or $$m_i$$:
$$\begin{aligned}
\boxed{
@@ -506,7 +507,7 @@ $$\begin{aligned}
-Firstly, we look at the commutator of $\hat{s}$ with the $z$-component $\hat{L}_z$:
+Firstly, we look at the commutator of $$\hat{s}$$ with the $$z$$-component $$\hat{L}_z$$:
$$\begin{aligned}
0
= \matrixel{\ell_f m_f}{\comm{\hat{L}_z}{\hat{s}}}{\ell_i m_i}
@@ -515,10 +516,10 @@ $$\begin{aligned}
&= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
\end{aligned}$$
-Which can only be true if $m_f \!-\! m_i = 0$, unless,
-of course, $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$ by itself.
+Which can only be true if $$m_f \!-\! m_i = 0$$, unless,
+of course, $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} = 0$$ by itself.
-Secondly, we look at the commutator of $\hat{s}$ with the total angular momentum $\hat{L}^2$:
+Secondly, we look at the commutator of $$\hat{s}$$ with the total angular momentum $$\hat{L}^2$$:
$$\begin{aligned}
0
@@ -528,7 +529,7 @@ $$\begin{aligned}
&= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
\end{aligned}$$
-Assuming $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$,
+Assuming $$\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$$,
this can only be satisfied if the following holds:
$$\begin{aligned}
@@ -537,11 +538,11 @@ $$\begin{aligned}
= (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i)
\end{aligned}$$
-If $\ell_f = \ell_i = 0$ this equation is trivially satisfied.
-Otherwise, the only option is $\ell_f \!-\! \ell_i = 0$,
+If $$\ell_f = \ell_i = 0$$ this equation is trivially satisfied.
+Otherwise, the only option is $$\ell_f \!-\! \ell_i = 0$$,
which is another part of the selection rule.
-Thirdly, we look at the commutator of $\hat{s}$ with the ladder operators $\hat{L}_\pm$:
+Thirdly, we look at the commutator of $$\hat{s}$$ with the ladder operators $$\hat{L}_\pm$$:
$$\begin{aligned}
0
@@ -551,9 +552,9 @@ $$\begin{aligned}
&= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
\end{aligned}$$
-Where $C_f$ and $C_i$ are constants given below.
-We already know that $\Delta \ell = 0$ and $\Delta m = 0$,
-so the above matrix elements are only nonzero if $m_f = m_i \pm 1$.
+Where $$C_f$$ and $$C_i$$ are constants given below.
+We already know that $$\Delta \ell = 0$$ and $$\Delta m = 0$$,
+so the above matrix elements are only nonzero if $$m_f = m_i \pm 1$$.
Therefore:
$$\begin{aligned}
@@ -568,7 +569,7 @@ $$\begin{aligned}
&= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)}
\end{aligned}$$
-In other words, $C_f = C_i$. The above equation therefore reduces to:
+In other words, $$C_f = C_i$$. The above equation therefore reduces to:
$$\begin{aligned}
\matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i}
@@ -576,13 +577,13 @@ $$\begin{aligned}
\end{aligned}$$
Which means that the value of the matrix element
-does not depend on $m_i$ (or $m_f$) at all.
+does not depend on $$m_i$$ (or $$m_f$$) at all.
-Similarly, given a general (pseudo)vector operator $\vu{V}$,
+Similarly, given a general (pseudo)vector operator $$\vu{V}$$,
which, by nature, must satisfy the following commutation relations,
-where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$:
+where $$\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$$:
$$\begin{gathered}
\comm{\hat{L}_z}{\hat{V}_z} = 0
@@ -596,7 +597,7 @@ $$\begin{gathered}
\comm{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z
\end{gathered}$$
-The inner product of any such $\vu{V}$ must obey the following selection rules:
+The inner product of any such $$\vu{V}$$ must obey the following selection rules:
$$\begin{aligned}
\boxed{
@@ -637,17 +638,17 @@ Selection rules are not always about atomic electron transitions, or angular mom
According to the **principle of indistinguishability**,
permuting identical particles never leads to an observable difference.
In other words, the particles are fundamentally indistinguishable,
-so for any observable $\hat{O}$ and multi-particle state $\Ket{\Psi}$, we can say:
+so for any observable $$\hat{O}$$ and multi-particle state $$\Ket{\Psi}$$, we can say:
$$\begin{aligned}
\matrixel{\Psi}{\hat{O}}{\Psi}
= \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi}
\end{aligned}$$
-Where $\hat{P}$ is an arbitrary permutation operator.
-Indistinguishability implies that $\comm{\hat{P}}{\hat{O}} = 0$
-for all $\hat{O}$ and $\hat{P}$,
-which lets us prove the above equation, using that $\hat{P}$ is unitary:
+Where $$\hat{P}$$ is an arbitrary permutation operator.
+Indistinguishability implies that $$\comm{\hat{P}}{\hat{O}} = 0$$
+for all $$\hat{O}$$ and $$\hat{P}$$,
+which lets us prove the above equation, using that $$\hat{P}$$ is unitary:
$$\begin{aligned}
\matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi}
@@ -656,9 +657,9 @@ $$\begin{aligned}
= \matrixel{\Psi}{\hat{O}}{\Psi}
\end{aligned}$$
-Consider a symmetric state $\Ket{s}$ and an antisymmetric state $\Ket{a}$
+Consider a symmetric state $$\Ket{s}$$ and an antisymmetric state $$\Ket{a}$$
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
-which obey the following for a permutation $\hat{P}$:
+which obey the following for a permutation $$\hat{P}$$:
$$\begin{aligned}
\hat{P} \Ket{s}
@@ -668,8 +669,8 @@ $$\begin{aligned}
= - \Ket{a}
\end{aligned}$$
-Any obervable $\hat{O}$ then satisfies the equation below,
-again thanks to the fact that $\hat{P} = \hat{P}^{-1}$:
+Any obervable $$\hat{O}$$ then satisfies the equation below,
+again thanks to the fact that $$\hat{P} = \hat{P}^{-1}$$:
$$\begin{aligned}
\matrixel{s}{\hat{O}}{a}
diff --git a/source/know/concept/self-energy/index.md b/source/know/concept/self-energy/index.md
index 25ba3fb..005f135 100644
--- a/source/know/concept/self-energy/index.md
+++ b/source/know/concept/self-energy/index.md
@@ -8,17 +8,17 @@ categories:
layout: "concept"
---
-Suppose we have a time-independent Hamiltonian $\hat{H} = \hat{H}_0 + \hat{W}$,
-consisting of a simple $\hat{H}_0$ and a difficult interaction $\hat{W}$,
+Suppose we have a time-independent Hamiltonian $$\hat{H} = \hat{H}_0 + \hat{W}$$,
+consisting of a simple $$\hat{H}_0$$ and a difficult interaction $$\hat{W}$$,
for example describing Coulomb repulsion between electrons.
The concept of [imaginary time](/know/concept/imaginary-time/)
exists to handle such difficult time-independent Hamiltonians
at nonzero temperatures. Therefore, we know that the
[Matsubara Green's function](/know/concept/matsubara-greens-function/)
-$G$ can be written as follows, where $\mathcal{T}$ is the
+$$G$$ can be written as follows, where $$\mathcal{T}$$ is the
[time-ordered product](/know/concept/time-ordered-product/),
-and $\beta = 1 / (k_B T)$:
+and $$\beta = 1 / (k_B T)$$:
$$\begin{aligned}
G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)
@@ -26,7 +26,7 @@ $$\begin{aligned}
{\hbar \Expval{\hat{K}(\hbar \beta, 0)}}
\end{aligned}$$
-Where we know that the time evolution operator $\hat{K}$
+Where we know that the time evolution operator $$\hat{K}$$
is as follows in the [interaction picture](/know/concept/interaction-picture/):
$$\begin{aligned}
@@ -37,11 +37,11 @@ $$\begin{aligned}
\mathcal{T}\bigg\{ \bigg( \int_{\tau_1}^{\tau_2} \hat{W}(\tau) \dd{\tau} \bigg)^n \bigg\}
\end{aligned}$$
-Where $\hat{W}$ is the two-body operator in the interaction picture.
+Where $$\hat{W}$$ is the two-body operator in the interaction picture.
We insert this into the full Green's function above,
and abbreviate
-$G_{ba} \equiv G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)$
-and $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
+$$G_{ba} \equiv G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)$$
+and $$\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$$:
$$\begin{aligned}
G_{ba}
@@ -51,11 +51,11 @@ $$\begin{aligned}
\Expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
\end{aligned}$$
-Next, we write out the interaction operator $\hat{W}$
+Next, we write out the interaction operator $$\hat{W}$$
in the [second quantization](/know/concept/second-quantization/),
assuming there is no spin-flipping,
-and that $W(\vb{r}_1, \vb{r}_2) = W(\vb{r}_2, \vb{r}_1)$
-(hence $1/2$ to avoid double-counting):
+and that $$W(\vb{r}_1, \vb{r}_2) = W(\vb{r}_2, \vb{r}_1)$$
+(hence $$1/2$$ to avoid double-counting):
$$\begin{aligned}
\hat{W}(\tau_1)
@@ -63,8 +63,8 @@ $$\begin{aligned}
W(\vb{r}_1, \vb{r}_2) \hat{\Psi}_{s_2}(\vb{r}_2, \tau_1) \hat{\Psi}_{s_1}(\vb{r}_1, \tau_1) \dd{\vb{r}_1} \dd{\vb{r}_2}
\end{aligned}$$
-We integrate this over $\tau_1$ and over a dummy $\tau_2$.
-Defining $W_{j'j} \equiv W(\vb{r}_j', \vb{r}_j) \: \delta(\tau_1 \!-\! \tau_2)$ we get:
+We integrate this over $$\tau_1$$ and over a dummy $$\tau_2$$.
+Defining $$W_{j'j} \equiv W(\vb{r}_j', \vb{r}_j) \: \delta(\tau_1 \!-\! \tau_2)$$ we get:
$$\begin{aligned}
\int_0^{\hbar \beta} \hat{W}(\tau_1) \dd{\tau_1}
@@ -74,8 +74,8 @@ $$\begin{aligned}
&= \frac{1}{2} \iint \hat{\Psi}_1^\dagger \hat{\Psi}_2^\dagger W_{1,2} \hat{\Psi}_2 \hat{\Psi}_1 \dd{1} \dd{2}
\end{aligned}$$
-Where we have further abbreviated $\int \dd{j} \equiv \sum_{s_j} \int \dd{\vb{r}_j} \int \dd{\tau_j}$.
-The full $G_{ba}$ thus becomes:
+Where we have further abbreviated $$\int \dd{j} \equiv \sum_{s_j} \int \dd{\vb{r}_j} \int \dd{\tau_j}$$.
+The full $$G_{ba}$$ thus becomes:
$$\begin{aligned}
G_{ba}
@@ -102,7 +102,7 @@ $$\begin{aligned}
By applying [Wick's theorem](/know/concept/wicks-theorem/),
we can rewrite these as a sum of products of single-particle Green's functions,
-so for instance $G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)$ becomes:
+so for instance $$G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)$$ becomes:
$$\begin{aligned}
G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)
@@ -115,14 +115,14 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-And analogously for $G^0_\mathrm{den}$.
+And analogously for $$G^0_\mathrm{den}$$.
If we are studying bosons instead of fermions,
the above determinant would need to be replaced by a *permanent*.
We assume fermions from now on.
-We thus have sums over all permutations $p$
+We thus have sums over all permutations $$p$$
of products of single-particle Green's function,
-times $(-1)^p$ to account for swaps of fermionic operators:
+times $$(-1)^p$$ to account for swaps of fermionic operators:
$$\begin{aligned}
G_{ba}
@@ -134,30 +134,30 @@ $$\begin{aligned}
These integrals over products of interactions and Green's functions
are the perfect place to apply [Feynman diagrams](/know/concept/feynman-diagram/).
-Conveniently, it turns out that the factor $(-1)^p$
-is equivalent to the rule that each diagram must be multiplied by $(-1)^F$,
-with $F$ the number of fermion loops.
-Keep in mind that fermion lines absorb a factor $-\hbar$ each (see above),
-and interactions $-1/\hbar$.
+Conveniently, it turns out that the factor $$(-1)^p$$
+is equivalent to the rule that each diagram must be multiplied by $$(-1)^F$$,
+with $$F$$ the number of fermion loops.
+Keep in mind that fermion lines absorb a factor $$-\hbar$$ each (see above),
+and interactions $$-1/\hbar$$.
The denominator turns into a sum of all possible diagrams
-(including equivalent ones) for each total order $n$
+(including equivalent ones) for each total order $$n$$
(the order is the number of interaction lines).
-The endpoints $a$ and $b$ do not appear here,
+The endpoints $$a$$ and $$b$$ do not appear here,
so we conclude that all those diagrams only have internal vertices;
we will therefore refer to them as **internal diagrams**.
-And in the numerator, we sum over all diagrams of total order $n$
-containing the external vertices $a$ and $b$.
+And in the numerator, we sum over all diagrams of total order $$n$$
+containing the external vertices $$a$$ and $$b$$.
Some of them are **connected**,
-so all vertices (including $a$ and $b$) are in the same graph,
+so all vertices (including $$a$$ and $$b$$) are in the same graph,
but most are **disconnected**.
Because disconnected diagrams have no shared lines or vertices to integrate over,
they can simply be factored into separate diagrams.
-If it contains $a$ and $b$, we call it an **external diagram**,
+If it contains $$a$$ and $$b$$, we call it an **external diagram**,
and then clearly all disconnected parts must be internal diagrams
-($a$ and $b$ are always connected,
+($$a$$ and $$b$$ are always connected,
since they are the only vertices with just one fermion line;
all internal vertices must have two).
We thus find:
@@ -173,7 +173,7 @@ $$\begin{aligned}
Where the total order is the sum of the orders of all considered diagrams,
and the new factor is needed for all the possible choices
of vertices to put in the external part.
-Note that the external diagram does not directly depend on $n$,
+Note that the external diagram does not directly depend on $$n$$,
so we reorganize:
$$\begin{aligned}
@@ -184,9 +184,9 @@ $$\begin{aligned}
{\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
\end{aligned}$$
-Since both $n$ and $m$ start at zero,
+Since both $$n$$ and $$m$$ start at zero,
and the sums include all possible diagrams,
-we see that the second sum in the numerator does not actually depend on $m$:
+we see that the second sum in the numerator does not actually depend on $$m$$:
$$\begin{aligned}
\hbar G_{ba}
@@ -199,10 +199,10 @@ $$\begin{aligned}
In other words, all the disconnected diagrams simply cancel out,
and we are left with a sum over all possible fully connected diagrams
-that contain $a$ and $b$. Furthermore, it can be shown using combinatorics
-that exactly $2^m m!$ diagrams at each order are topologically equivalent,
+that contain $$a$$ and $$b$$. Furthermore, it can be shown using combinatorics
+that exactly $$2^m m!$$ diagrams at each order are topologically equivalent,
so we are left with non-equivalent diagrams only.
-Let $G(b,a) = G_{ba}$:
+Let $$G(b,a) = G_{ba}$$:
@@ -213,9 +213,9 @@ that can be cut in two valid diagrams
by removing just one fermion line,
while an **irreducible diagram** cannot be split like that.
-At last, we define the **self-energy** $\Sigma(y,x)$
-as the sum of all irreducible terms in $G(b,a)$,
-after removing the two external lines from/to $a$ and $b$:
+At last, we define the **self-energy** $$\Sigma(y,x)$$
+as the sum of all irreducible terms in $$G(b,a)$$,
+after removing the two external lines from/to $$a$$ and $$b$$:
@@ -224,27 +224,27 @@ after removing the two external lines from/to $a$ and $b$:
Despite its appearance, the self-energy has the semantics of a line,
so it has two endpoints over which to integrate if necessary.
-By construction, by reattaching $G^0(x,a)$ and $G^0(b,y)$ to the self-energy,
+By construction, by reattaching $$G^0(x,a)$$ and $$G^0(b,y)$$ to the self-energy,
we get all irreducible diagrams,
and by connecting multiple irreducible diagrams with single fermion lines,
-we get all fully connected diagrams containing the endpoints $a$ and $b$.
+we get all fully connected diagrams containing the endpoints $$a$$ and $$b$$.
-In other words, the full $G(b,a)$ is constructed
-by taking the unperturbed $G^0(b,a)$
-and inserting one or more irreducible diagrams between $a$ and $b$.
+In other words, the full $$G(b,a)$$ is constructed
+by taking the unperturbed $$G^0(b,a)$$
+and inserting one or more irreducible diagrams between $$a$$ and $$b$$.
We can equally well insert a single irreducible diagram
as a sequence of connected irreducible diagrams.
Thanks to this recursive structure,
-you can convince youself that $G(b,a)$ obeys
-a [Dyson equation](/know/concept/dyson-equation/) involving $\Sigma(y, x)$:
+you can convince youself that $$G(b,a)$$ obeys
+a [Dyson equation](/know/concept/dyson-equation/) involving $$\Sigma(y, x)$$:
This makes sense: in the "normal" Dyson equation
-we have a one-body perturbation instead of $\Sigma$,
-while $\Sigma$ represents a two-body effect
+we have a one-body perturbation instead of $$\Sigma$$,
+while $$\Sigma$$ represents a two-body effect
as an infinite sum of one-body diagrams.
Interpreting this diagrammatic Dyson equation yields:
@@ -255,7 +255,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Keep in mind that $\int \dd{x} \equiv \sum_{s_x} \int \dd{\vb{r}_x} \int \dd{\tau_x}$.
+Keep in mind that $$\int \dd{x} \equiv \sum_{s_x} \int \dd{\vb{r}_x} \int \dd{\tau_x}$$.
In the special case of a system with continuous translational symmetry
and no spin dependence, this simplifies to:
@@ -266,12 +266,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n)$,
-with $\omega_n$ being a fermionic Matsubara frequency.
-Note that conservation of spin, $\vb{k}$ and $\omega_n$,
+Where $$\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n)$$,
+with $$\omega_n$$ being a fermionic Matsubara frequency.
+Note that conservation of spin, $$\vb{k}$$ and $$\omega_n$$,
together with the linear structure of the Dyson equation,
-makes $\Sigma$ diagonal in all of those quantities.
-Isolating for $G$:
+makes $$\Sigma$$ diagonal in all of those quantities.
+Isolating for $$G$$:
$$\begin{aligned}
G_{s}(\tilde{\vb{k}})
@@ -280,7 +280,7 @@ $$\begin{aligned}
\end{aligned}$$
From [equation-of-motion theory](/know/concept/equation-of-motion-theory/),
-we already know an expression for $G$ in diagonal $\vb{k}$-space:
+we already know an expression for $$G$$ in diagonal $$\vb{k}$$-space:
$$\begin{aligned}
G_s^0(\vb{k}, i \omega_n)
@@ -294,10 +294,10 @@ The self-energy thus corrects the non-interacting energies for interactions.
It can therefore be regarded as the energy
a particle has due to changes it has caused in its environment.
-Unfortunately, in practice, $\Sigma$ is rarely as simple as
+Unfortunately, in practice, $$\Sigma$$ is rarely as simple as
in the translationally-invariant example above;
in fact, it does not even need to be Hermitian,
-i.e. $\Sigma(y,x) \neq \Sigma^*(x,y)$,
+i.e. $$\Sigma(y,x) \neq \Sigma^*(x,y)$$,
in which case it resists the standard techniques for analysis.
diff --git a/source/know/concept/self-phase-modulation/index.md b/source/know/concept/self-phase-modulation/index.md
index 3b46c4e..f13ad2f 100644
--- a/source/know/concept/self-phase-modulation/index.md
+++ b/source/know/concept/self-phase-modulation/index.md
@@ -12,13 +12,13 @@ layout: "concept"
In fiber optics, **self-phase modulation** (SPM) is a nonlinear effect
that gradually broadens pulses' spectra.
-Unlike dispersion, SPM does create new frequencies: in the $\omega$-domain,
+Unlike dispersion, SPM does create new frequencies: in the $$\omega$$-domain,
the pulse steadily spreads out with a distinctive "accordion" peak.
Lower frequencies are created at the front of the
pulse and higher ones at the back, giving S-shaped spectrograms.
-A pulse envelope $A(z, t)$ inside a fiber must obey the nonlinear Schrödinger equation,
-where the parameters $\beta_2$ and $\gamma$ respectively
+A pulse envelope $$A(z, t)$$ inside a fiber must obey the nonlinear Schrödinger equation,
+where the parameters $$\beta_2$$ and $$\gamma$$ respectively
control dispersion and nonlinearity:
$$\begin{aligned}
@@ -26,20 +26,20 @@ $$\begin{aligned}
= i \pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma |A|^2 A
\end{aligned}$$
-By setting $\beta_2 = 0$ to neglect dispersion,
+By setting $$\beta_2 = 0$$ to neglect dispersion,
solving this equation becomes trivial.
-For any arbitrary input pulse $A_0(t) = A(0, t)$,
+For any arbitrary input pulse $$A_0(t) = A(0, t)$$,
we arrive at the following analytical solution:
$$\begin{aligned}
A(z,t) = A_0 \exp\!\big( i \gamma |A_0|^2 z\big)
\end{aligned}$$
-The intensity $|A|^2$ in the time domain is thus unchanged,
+The intensity $$|A|^2$$ in the time domain is thus unchanged,
and only its phase is modified.
It is also clear that the largest phase increase occurs at the peak of the pulse,
-where the intensity is $P_0$.
-To quantify this, it is useful to define the **nonlinear length** $L_N$,
+where the intensity is $$P_0$$.
+To quantify this, it is useful to define the **nonlinear length** $$L_N$$,
which gives the distance after which the phase of the
peak has increased by exactly 1 radian:
@@ -52,15 +52,15 @@ $$\begin{aligned}
\end{aligned}$$
SPM is illustrated below for the following Gaussian initial pulse envelope,
-with parameter values $T_0 = 6\:\mathrm{ps}$, $P_0 = 1\:\mathrm{kW}$,
-$\beta_2 = 0$, and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$:
+with parameter values $$T_0 = 6\:\mathrm{ps}$$, $$P_0 = 1\:\mathrm{kW}$$,
+$$\beta_2 = 0$$, and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$:
$$\begin{aligned}
A(0, t)
= \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big)
\end{aligned}$$
-From earlier, we then know the analytical solution for the $z$-evolution:
+From earlier, we then know the analytical solution for the $$z$$-evolution:
$$\begin{aligned}
A(z, t) = \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big) \exp\!\bigg( i \gamma z P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
@@ -70,10 +70,10 @@ $$\begin{aligned}
-The **instantaneous frequency** $\omega_\mathrm{SPM}(z, t)$,
+The **instantaneous frequency** $$\omega_\mathrm{SPM}(z, t)$$,
which describes the dominant angular frequency at a given point in the time domain,
is found to be as follows for the Gaussian pulse,
-where $\phi(z, t)$ is the phase of $A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$:
+where $$\phi(z, t)$$ is the phase of $$A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$$:
$$\begin{aligned}
\omega_{\mathrm{SPM}}(z,t)
@@ -82,8 +82,8 @@ $$\begin{aligned}
\end{aligned}$$
This result gives the S-shaped spectrograms seen in the illustration.
-The frequency shift thus not only depends on $L_N$,
-but also on $T_0$: the spectra of narrow pulses broaden much faster.
+The frequency shift thus not only depends on $$L_N$$,
+but also on $$T_0$$: the spectra of narrow pulses broaden much faster.
The interaction between self-phase modulation
and [dispersion](/know/concept/dispersive-broadening/)
diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
index 4cc6201..a8bc3c2 100644
--- a/source/know/concept/self-steepening/index.md
+++ b/source/know/concept/self-steepening/index.md
@@ -22,17 +22,17 @@ $$\begin{aligned}
= i\pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{}{t} \Big) \big(|A|^2 A\big)
\end{aligned}$$
-Where $\omega_0$ is the angular frequency of the pump.
+Where $$\omega_0$$ is the angular frequency of the pump.
We will use the following ansatz,
-consisting of an arbitrary power profile $P$ with a phase $\phi$:
+consisting of an arbitrary power profile $$P$$ with a phase $$\phi$$:
$$\begin{aligned}
A(z,t) = \sqrt{P(z,t)} \, \exp\!\big(i \phi(z,t)\big)
\end{aligned}$$
For a long pulse travelling over a short distance, it is reasonable to
-neglect dispersion ($\beta_2 = 0$).
-Inserting the ansatz then gives the following, where $\varepsilon = \gamma / \omega_0$:
+neglect dispersion ($$\beta_2 = 0$$).
+Inserting the ansatz then gives the following, where $$\varepsilon = \gamma / \omega_0$$:
$$\begin{aligned}
0 &= i \frac{1}{2} \frac{P_z}{\sqrt{P}} - \sqrt{P} \phi_z + \gamma P \sqrt{P} + i \varepsilon \frac{3}{2} P_t \sqrt{P} - \varepsilon P \sqrt{P} \phi_t
@@ -47,7 +47,7 @@ $$\begin{aligned}
0 &= P_z + \varepsilon 3 P_t P
\end{aligned}$$
-The phase $\phi$ is not so interesting, so we focus on the latter equation for $P$.
+The phase $$\phi$$ is not so interesting, so we focus on the latter equation for $$P$$.
As it turns out, it has a general solution of the form below, which shows that
more intense parts of the pulse will tend to lag behind compared to the rest:
@@ -55,8 +55,8 @@ $$\begin{aligned}
P(z,t) = f(t - 3 \varepsilon z P)
\end{aligned}$$
-Where $f$ is the initial power profile: $f(t) = P(0,t)$.
-The derivatives $P_t$ and $P_z$ are then given by:
+Where $$f$$ is the initial power profile: $$f(t) = P(0,t)$$.
+The derivatives $$P_t$$ and $$P_z$$ are then given by:
$$\begin{aligned}
P_t
@@ -73,10 +73,10 @@ $$\begin{aligned}
\end{aligned}$$
These derivatives both go to infinity when their denominator is zero,
-which, since $\varepsilon$ is positive, will happen earliest where $f'$
-has its most negative value, called $f_\mathrm{min}'$,
+which, since $$\varepsilon$$ is positive, will happen earliest where $$f'$$
+has its most negative value, called $$f_\mathrm{min}'$$,
which is located on the trailing edge of the pulse.
-At the propagation distance where this occurs, $L_\mathrm{shock}$,
+At the propagation distance where this occurs, $$L_\mathrm{shock}$$,
the pulse will "tip over", creating a discontinuous shock:
$$\begin{aligned}
@@ -86,21 +86,21 @@ $$\begin{aligned}
\end{aligned}$$
In practice, however, this will never actually happen, because by the time
-$L_\mathrm{shock}$ is reached, the pulse spectrum will have become so
+$$L_\mathrm{shock}$$ is reached, the pulse spectrum will have become so
broad that dispersion can no longer be neglected.
A simulation of self-steepening without dispersion is illustrated below
for the following Gaussian initial power distribution,
-with $T_0 = 25\:\mathrm{fs}$, $P_0 = 3\:\mathrm{kW}$,
-$\beta_2 = 0$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$:
+with $$T_0 = 25\:\mathrm{fs}$$, $$P_0 = 3\:\mathrm{kW}$$,
+$$\beta_2 = 0$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$:
$$\begin{aligned}
f(t) = P(0,t) = P_0 \exp\!\Big(\! -\!\frac{t^2}{T_0^2} \Big)
\end{aligned}$$
-Its steepest points are found to be at $2 t^2 = T_0^2$, so
-$f_\mathrm{min}'$ and $L_\mathrm{shock}$ are given by:
+Its steepest points are found to be at $$2 t^2 = T_0^2$$, so
+$$f_\mathrm{min}'$$ and $$L_\mathrm{shock}$$ are given by:
$$\begin{aligned}
f_\mathrm{min}' = - \frac{\sqrt{2} P_0}{T_0} \exp\!\Big(\!-\!\frac{1}{2}\Big)
@@ -109,7 +109,7 @@ $$\begin{aligned}
\end{aligned}$$
This example Gaussian pulse therefore has a theoretical
-$L_\mathrm{shock} = 0.847\,\mathrm{m}$,
+$$L_\mathrm{shock} = 0.847\,\mathrm{m}$$,
which turns out to be accurate,
although the simulation breaks down due to insufficient resolution:
@@ -118,7 +118,7 @@ although the simulation breaks down due to insufficient resolution:
Unfortunately, self-steepening cannot be simulated perfectly: as the
-pulse approaches $L_\mathrm{shock}$, its spectrum broadens to infinite
+pulse approaches $$L_\mathrm{shock}$$, its spectrum broadens to infinite
frequencies to represent the singularity in its slope.
The simulation thus collapses into chaos when the edge of the frequency window is reached.
Nevertheless, the general trends are nicely visible:
@@ -126,9 +126,9 @@ the trailing slope becomes extremely steep, and the spectrum
broadens so much that dispersion cannot be neglected anymore.
When self-steepening is added to the nonlinear Schrödinger equation,
-it no longer conserves the total pulse energy $\int |A|^2 \dd{t}$.
-Fortunately, the photon number $N_\mathrm{ph}$ is still
-conserved, which for the physical envelope $A(z,t)$ is defined as:
+it no longer conserves the total pulse energy $$\int |A|^2 \dd{t}$$.
+Fortunately, the photon number $$N_\mathrm{ph}$$ is still
+conserved, which for the physical envelope $$A(z,t)$$ is defined as:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/shors-algorithm/index.md b/source/know/concept/shors-algorithm/index.md
index 0ff35c5..0241f9f 100644
--- a/source/know/concept/shors-algorithm/index.md
+++ b/source/know/concept/shors-algorithm/index.md
@@ -17,8 +17,8 @@ It weakens widely-used cryptographic schemes,
such as RSA and [Diffie-Hellman](/know/concept/diffie-hellman-key-exchange/).
In essence, Shor's algorithm's revolutionary achievement
-is that it can efficiently find the periods $s_1, ..., s_A$
-of a function $f(x_1, ..., x_A)$ on a discrete finite field, where:
+is that it can efficiently find the periods $$s_1, ..., s_A$$
+of a function $$f(x_1, ..., x_A)$$ on a discrete finite field, where:
$$\begin{aligned}
f(x_1, ..., x_A)
@@ -32,27 +32,27 @@ Shor's algorithm can solve practically every such problem.
## Integer factorization
-Originally, Shor's algorithm was designed to factorize an integer $N$,
-in which case the goal is to find the period $s$ of
-the modular exponentiation function $f$ (for reasons explained later):
+Originally, Shor's algorithm was designed to factorize an integer $$N$$,
+in which case the goal is to find the period $$s$$ of
+the modular exponentiation function $$f$$ (for reasons explained later):
$$\begin{aligned}
f(x)
= a^x \bmod N
\end{aligned}$$
-For a given $a$ and $N$.
-The period $s$ is the smallest integer satisfying $f(x) = f(x+s)$.
-To do this, the following $2q$-qubit quantum circuit is used,
-with $q$ chosen so that $N^2 \le 2^q < 2 N^2$:
+For a given $$a$$ and $$N$$.
+The period $$s$$ is the smallest integer satisfying $$f(x) = f(x+s)$$.
+To do this, the following $$2q$$-qubit quantum circuit is used,
+with $$q$$ chosen so that $$N^2 \le 2^q < 2 N^2$$:
-Here, $\mathrm{QFT}_q$ refers to the $q$-qubit
+Here, $$\mathrm{QFT}_q$$ refers to the $$q$$-qubit
[quantum Fourier transform](/know/concept/quantum-fourier-transform/),
-and the oracle $U_f$ calculates $f(x)$ for predetermined values of $a$ and $N$.
+and the oracle $$U_f$$ calculates $$f(x)$$ for predetermined values of $$a$$ and $$N$$.
It is an XOR oracle, working as follows:
$$\begin{aligned}
@@ -61,8 +61,8 @@ $$\begin{aligned}
\Ket{x} \Ket{y \oplus f(x)}
\end{aligned}$$
-Execution starts by applying the [Hadamard gate](/know/concept/quantum-gate/) $H$
-to the first $q$ qubits, yielding:
+Execution starts by applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
+to the first $$q$$ qubits, yielding:
$$\begin{aligned}
\Ket{0}^{\otimes q} \Ket{0}^{\otimes q}
@@ -71,8 +71,8 @@ $$\begin{aligned}
= \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
\end{aligned}$$
-Where $Q = 2^q$, and $\Ket{x}$ is the computational basis state $\Ket{x_1} \cdots \Ket{x_q}$.
-Moving on to $U_f$:
+Where $$Q = 2^q$$, and $$\Ket{x}$$ is the computational basis state $$\Ket{x_1} \cdots \Ket{x_q}$$.
+Moving on to $$U_f$$:
$$\begin{aligned}
\frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
@@ -80,15 +80,15 @@ $$\begin{aligned}
\frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{f(x)}
\end{aligned}$$
-Then we measure $f(x)$, causing it collapse as follows,
-for an unknown arbitrary value of $x_0$:
+Then we measure $$f(x)$$, causing it collapse as follows,
+for an unknown arbitrary value of $$x_0$$:
$$\begin{aligned}
f(x_0) = f(x_0 + s) = f(x_0 + 2s) = \cdots = f(x_0 + (L-1) s)
\end{aligned}$$
Due to [entanglement](/know/concept/quantum-entanglement/),
-the unmeasured (top $q$) qubits change state into a superposition:
+the unmeasured (top $$q$$) qubits change state into a superposition:
$$\begin{aligned}
\frac{1}{\sqrt{L}} \sum_{\ell = 0}^{L - 1} \Ket{x_0 + \ell s}
@@ -96,7 +96,7 @@ $$\begin{aligned}
Clearly, there is a periodic structure here,
but we cannot measure it directly,
-because we do not know the value of $x_0$,
+because we do not know the value of $$x_0$$,
which, to make matters worse, changes every time we run the algorithm.
This is where the QFT comes in, which outputs the following state:
@@ -104,8 +104,8 @@ $$\begin{aligned}
\frac{1}{\sqrt{QL}} \sum_{k = 0}^{Q - 1} \bigg( \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg) \Ket{k}
\end{aligned}$$
-Where $\omega_Q$ is a $Q$th root of unity.
-Measuring this state yields a $\Ket{k}$, with a probability $P(k)$:
+Where $$\omega_Q$$ is a $$Q$$th root of unity.
+Measuring this state yields a $$\Ket{k}$$, with a probability $$P(k)$$:
$$\begin{aligned}
P(k)
@@ -114,13 +114,13 @@ $$\begin{aligned}
= \frac{1}{QL} \bigg| \sum_{\ell = 0}^{L - 1} \omega_Q^{\ell s k} \bigg|^2
\end{aligned}$$
-The last step holds because $|\omega_Q| = 1$.
+The last step holds because $$|\omega_Q| = 1$$.
Surprisingly, this implies that we did not need
-to perform the measurement of $f(x)$ earlier!
-This makes sense: the period $s$ does not depend on $x_0$,
-so why would we need an implicit $x_0$ to determine $s$?
+to perform the measurement of $$f(x)$$ earlier!
+This makes sense: the period $$s$$ does not depend on $$x_0$$,
+so why would we need an implicit $$x_0$$ to determine $$s$$?
-So, what does the above probability $P(k)$ work out to?
+So, what does the above probability $$P(k)$$ work out to?
There are two cases:
$$\begin{alignedat}{2}
@@ -132,22 +132,22 @@ $$\begin{alignedat}{2}
\end{alignedat}$$
Where the latter case was evaluated as a geometric series.
-The condition $\omega_Q^{sk}\!=\!1$ is equivalent to asking
-if $sk$ is a multiple of $Q$, i.e. if $sk = cQ$, for an integer $c$.
+The condition $$\omega_Q^{sk}\!=\!1$$ is equivalent to asking
+if $$sk$$ is a multiple of $$Q$$, i.e. if $$sk = cQ$$, for an integer $$c$$.
-Recall that $L$ is the number of times that $s$ fits in $Q$,
-so $L\!=\!\lfloor Q / s \rfloor$.
-Assuming $Q/s$ is an integer, then $L\!=\!Q/s$ and $Q\!=\!s L$,
+Recall that $$L$$ is the number of times that $$s$$ fits in $$Q$$,
+so $$L\!=\!\lfloor Q / s \rfloor$$.
+Assuming $$Q/s$$ is an integer, then $$L\!=\!Q/s$$ and $$Q\!=\!s L$$,
which tells us that
-$\omega_Q^{sk}\!=\!\omega_{s L}^{s k}\!=\!\omega_L^k$.
-This implies that if $k$ is a multiple of $L$ (i.e. $k\!=\!c L$),
-then $\omega_L^k\!=\!1$, so $P(k) = L / Q$,
+$$\omega_Q^{sk}\!=\!\omega_{s L}^{s k}\!=\!\omega_L^k$$.
+This implies that if $$k$$ is a multiple of $$L$$ (i.e. $$k\!=\!c L$$),
+then $$\omega_L^k\!=\!1$$, so $$P(k) = L / Q$$,
which is exactly what we got earlier!
-In other words, the condition $\omega_Q^{sk}\!=\!1$
-is equivalent to $Q/s$ being an integer.
-In that case, we have that $Q\!=\!sL$,
-which we substitute into $P(k)$ from earlier:
+In other words, the condition $$\omega_Q^{sk}\!=\!1$$
+is equivalent to $$Q/s$$ being an integer.
+In that case, we have that $$Q\!=\!sL$$,
+which we substitute into $$P(k)$$ from earlier:
$$\begin{aligned}
\mathrm{if} \: (Q/s) \in \mathbb{N}: \qquad
@@ -156,25 +156,25 @@ $$\begin{aligned}
= \frac{1}{s}
\end{aligned}$$
-And because $k$ is a multiple of $L$,
-and $L$ fits $s$ times in $Q$,
-there must be exactly $s$ values of $k$ that satisfy $P(k) = 1/s$.
-Therefore the probability of all other $k$-values is zero!
-This becomes clearer when you look at the sum used to calculate $P(k)$:
-if $Q\!=\!sL$, then it sums $\omega_L^{\ell k}$ over $\ell$,
-leading to perfect destructive interference for the "bad" $k$-values,
+And because $$k$$ is a multiple of $$L$$,
+and $$L$$ fits $$s$$ times in $$Q$$,
+there must be exactly $$s$$ values of $$k$$ that satisfy $$P(k) = 1/s$$.
+Therefore the probability of all other $$k$$-values is zero!
+This becomes clearer when you look at the sum used to calculate $$P(k)$$:
+if $$Q\!=\!sL$$, then it sums $$\omega_L^{\ell k}$$ over $$\ell$$,
+leading to perfect destructive interference for the "bad" $$k$$-values,
leaving only the "good" ones.
-**So, to summarize: if** $Q/s$ **is an integer**,
-then measuring only yields $k$-values that are multiples of $L\!=\!Q/s$.
+**So, to summarize: if** $$Q/s$$ **is an integer**,
+then measuring only yields $$k$$-values that are multiples of $$L\!=\!Q/s$$.
Running Shor's algorithm several times then gives
-several $k$-values separated by $L$.
-That tells us what $L$ is, and we already know $Q$,
-so we *finally* find the period $s = Q/L$.
+several $$k$$-values separated by $$L$$.
+That tells us what $$L$$ is, and we already know $$Q$$,
+so we *finally* find the period $$s = Q/L$$.
-That begs the question: what if $Q/s$ is not an integer?
-We cannot *check* this, since $s$ is unknown!
-Instead, we rewrite the probability $P(k)$ as follows:
+That begs the question: what if $$Q/s$$ is not an integer?
+We cannot *check* this, since $$s$$ is unknown!
+Instead, we rewrite the probability $$P(k)$$ as follows:
$$\begin{aligned}
\mathrm{if} \: (Q/s) \not\in \mathbb{N}: \qquad
@@ -183,7 +183,7 @@ $$\begin{aligned}
= \frac{1}{QL} \Bigg| \frac{\sin(\pi s k L / Q)}{\sin(\pi s k / Q)} \Bigg|^2
\end{aligned}$$
-This function peaks if $s k$ is close to a multiple of $Q$, i.e. $s k \approx c Q$,
+This function peaks if $$s k$$ is close to a multiple of $$Q$$, i.e. $$s k \approx c Q$$,
which we rearrange:
$$\begin{aligned}
@@ -191,62 +191,62 @@ $$\begin{aligned}
\end{aligned}$$
We know the left-hand side,
-and, from the definition of $f(x)$,
-clearly $s \le N$.
-We chose $Q \sim N^2$,
-so $s$ is quite small,
-and consequently $c$ is too, since $k < Q$.
+and, from the definition of $$f(x)$$,
+clearly $$s \le N$$.
+We chose $$Q \sim N^2$$,
+so $$s$$ is quite small,
+and consequently $$c$$ is too, since $$k < Q$$.
-In other words, $c/s$ is a "simple" fraction,
+In other words, $$c/s$$ is a "simple" fraction,
so our goal is to find a "simple" fraction
-that is close to the "complicated" fraction $k/Q$.
-For example, if $k/Q\!=\!0.332$,
-then probably $c/s\!=\!1/3$.
+that is close to the "complicated" fraction $$k/Q$$.
+For example, if $$k/Q\!=\!0.332$$,
+then probably $$c/s\!=\!1/3$$.
This can be done rigorously using the **continued fractions algorithm**:
-write $k/Q$ as a continued fraction,
+write $$k/Q$$ as a continued fraction,
until the non-integer part of the denominator becomes small enough.
This part is then neglected,
-and we calculate whatever is left, to get an estimate of $c/s$.
+and we calculate whatever is left, to get an estimate of $$c/s$$.
-Of course, $P(k)$ is a probability distribution,
+Of course, $$P(k)$$ is a probability distribution,
so even though the odds are in our favour,
-we might occasionally measure a misleading $k$-value.
+we might occasionally measure a misleading $$k$$-value.
Running Shor's algorithm several times "fixes" this.
-**So, to summarize: if** $Q/s$ **is not an integer**,
-the measured $k$-values are generally close to $c Q / s$ for an integer $c$.
-By approximating $k/Q$ using the continued fraction algorithm,
-we estimate $c/s$.
-Repeating this procedure gives several values of $c/s$,
-such that $s$ is easy to deduce
+**So, to summarize: if** $$Q/s$$ **is not an integer**,
+the measured $$k$$-values are generally close to $$c Q / s$$ for an integer $$c$$.
+By approximating $$k/Q$$ using the continued fraction algorithm,
+we estimate $$c/s$$.
+Repeating this procedure gives several values of $$c/s$$,
+such that $$s$$ is easy to deduce
by taking the least common multiple of the denominators.
-In any case, once we think we have $s$,
-we can easily verify that $f(x)\!=\!f(x\!+\!s)$.
-Whether $s$ is the *smallest* such integer depends on how lucky we are,
+In any case, once we think we have $$s$$,
+we can easily verify that $$f(x)\!=\!f(x\!+\!s)$$.
+Whether $$s$$ is the *smallest* such integer depends on how lucky we are,
but fortunately, for most applications of this algorithm,
that does not actually matter,
-and usually we find the smallest $s$ anyway.
+and usually we find the smallest $$s$$ anyway.
-You typically need to repeat the algorithm $\mathcal{O}(\log{q})$ times,
-and the QFT is $\mathcal{O}(q^2)$.
-The bottleneck is modular exponentiation $f$,
-which is $\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$
+You typically need to repeat the algorithm $$\mathcal{O}(\log{q})$$ times,
+and the QFT is $$\mathcal{O}(q^2)$$.
+The bottleneck is modular exponentiation $$f$$,
+which is $$\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$$
and therefore worse than the QFT,
-yielding a total complexity of $\mathcal{O}(q^2 (\log{q})^2 \log{\log{q}})$.
+yielding a total complexity of $$\mathcal{O}(q^2 (\log{q})^2 \log{\log{q}})$$.
-OK, but what does $s$ have to do with factorizing integers?
-Well, recall that $f$ is given by:
+OK, but what does $$s$$ have to do with factorizing integers?
+Well, recall that $$f$$ is given by:
$$\begin{aligned}
f(x)
= a^x \bmod N
\end{aligned}$$
-$N$ is the number to factorize, and $a$ is a random integer *coprime* to $N$,
-meaning $\gcd(a, N) = 1$.
-The fact that $s$ is the period of $f$ for a certain $a$-value, implies that:
+$$N$$ is the number to factorize, and $$a$$ is a random integer *coprime* to $$N$$,
+meaning $$\gcd(a, N) = 1$$.
+The fact that $$s$$ is the period of $$f$$ for a certain $$a$$-value, implies that:
$$\begin{aligned}
a^x
@@ -256,7 +256,7 @@ $$\begin{aligned}
= a^s \bmod N
\end{aligned}$$
-Suppose that $s$ is even. In that case,
+Suppose that $$s$$ is even. In that case,
we can rewrite the above equation as follows:
$$\begin{aligned}
@@ -264,16 +264,16 @@ $$\begin{aligned}
= 0 \bmod N
\end{aligned}$$
-In other words, $(a^{s/2})^2 \!-\! 1$ is a multiple of $N$.
-We then use that $(a\!-\!b) (a\!+\!b) = a^2\!-\!b^2$:
+In other words, $$(a^{s/2})^2 \!-\! 1$$ is a multiple of $$N$$.
+We then use that $$(a\!-\!b) (a\!+\!b) = a^2\!-\!b^2$$:
$$\begin{aligned}
\big( a^{s/2} - 1 \big) \big( a^{s/2} + 1 \big)
= 0 \bmod N
\end{aligned}$$
-Because $s$ is even by assumption, the two factors on the left are integers,
-and as just mentioned, their product is a multiple of $N$.
+Because $$s$$ is even by assumption, the two factors on the left are integers,
+and as just mentioned, their product is a multiple of $$N$$.
Then we only need to calculate:
$$\begin{aligned}
@@ -282,13 +282,13 @@ $$\begin{aligned}
\gcd\!\big( a^{s/2}\!+\!1, N \big) > 1
\end{aligned}$$
-And there we have the factors of $N$!
-The $\gcd$ can be calculated efficiently in $\mathcal{O}(q^2)$ time.
+And there we have the factors of $$N$$!
+The $$\gcd$$ can be calculated efficiently in $$\mathcal{O}(q^2)$$ time.
-But what if $s$ is odd?
-No problem, then we just choose a new $a$ coprime to $N$,
-and keep repeating Shor's algorithm until we do find an even $s$.
-We do the same if $a^{s/2}\!\pm\!1$ is itself a multiple of $N$.
+But what if $$s$$ is odd?
+No problem, then we just choose a new $$a$$ coprime to $$N$$,
+and keep repeating Shor's algorithm until we do find an even $$s$$.
+We do the same if $$a^{s/2}\!\pm\!1$$ is itself a multiple of $$N$$.
diff --git a/source/know/concept/sigma-algebra/index.md b/source/know/concept/sigma-algebra/index.md
index 5ba7b14..ac607a7 100644
--- a/source/know/concept/sigma-algebra/index.md
+++ b/source/know/concept/sigma-algebra/index.md
@@ -8,15 +8,15 @@ categories:
layout: "concept"
---
-In set theory, given a set $\Omega$, a $\sigma$**-algebra**
-is a family $\mathcal{F}$ of subsets of $\Omega$
+In set theory, given a set $$\Omega$$, a $$\sigma$$**-algebra**
+is a family $$\mathcal{F}$$ of subsets of $$\Omega$$
with these properties:
-1. The full set is included $\Omega \in \mathcal{F}$.
-2. For all subsets $A$, if $A \in \mathcal{F}$,
- then its complement $\Omega \!-\! A \in \mathcal{F}$ too.
-3. If two events $A, B \in \mathcal{F}$,
- then their union $A \cup B \in \mathcal{F}$ too.
+1. The full set is included $$\Omega \in \mathcal{F}$$.
+2. For all subsets $$A$$, if $$A \in \mathcal{F}$$,
+ then its complement $$\Omega \!-\! A \in \mathcal{F}$$ too.
+3. If two events $$A, B \in \mathcal{F}$$,
+ then their union $$A \cup B \in \mathcal{F}$$ too.
This forms a Boolean algebra:
property (1) represents TRUE,
@@ -24,28 +24,28 @@ property (1) represents TRUE,
and that is all we need to define all logic.
For example, FALSE and OR follow from the above points:
-4. The empty set is included $\varnothing \in \mathcal{F}$.
-5. If two events $A, B \in \mathcal{F}$,
- then their intersection $A \cap B \in \mathcal{F}$ too.
+4. The empty set is included $$\varnothing \in \mathcal{F}$$.
+5. If two events $$A, B \in \mathcal{F}$$,
+ then their intersection $$A \cap B \in \mathcal{F}$$ too.
-For a given $\Omega$, there are typically multiple valid $\mathcal{F}$,
+For a given $$\Omega$$, there are typically multiple valid $$\mathcal{F}$$,
in which case you need to specify your choice.
-Usually this would be the smallest $\mathcal{F}$
+Usually this would be the smallest $$\mathcal{F}$$
(i.e. smallest family of subsets)
that contains all subsets of special interest
for the topic at hand.
-Likewise, a **sub-$\sigma$-algebra**
-is a sub-family of a certain $\mathcal{F}$,
-which is a valid $\sigma$-algebra in its own right.
-
-A notable $\sigma$-algebra is the **Borel algebra** $\mathcal{B}(\Omega)$,
-which is defined when $\Omega$ is a metric space,
-such as the real numbers $\mathbb{R}$.
-Using that as an example, the Borel algebra $\mathcal{B}(\mathbb{R})$
+Likewise, a **sub-$$\sigma$$-algebra**
+is a sub-family of a certain $$\mathcal{F}$$,
+which is a valid $$\sigma$$-algebra in its own right.
+
+A notable $$\sigma$$-algebra is the **Borel algebra** $$\mathcal{B}(\Omega)$$,
+which is defined when $$\Omega$$ is a metric space,
+such as the real numbers $$\mathbb{R}$$.
+Using that as an example, the Borel algebra $$\mathcal{B}(\mathbb{R})$$
is defined as the family of all open intervals of the real line,
-and all the subsets of $\mathbb{R}$ obtained by countable sequences
+and all the subsets of $$\mathbb{R}$$ obtained by countable sequences
of unions and intersections of those intervals.
-The elements of $\mathcal{B}$ are **Borel sets**.
+The elements of $$\mathcal{B}$$ are **Borel sets**.
diff --git a/source/know/concept/simons-algorithm/index.md b/source/know/concept/simons-algorithm/index.md
index fef10b9..5502837 100644
--- a/source/know/concept/simons-algorithm/index.md
+++ b/source/know/concept/simons-algorithm/index.md
@@ -19,10 +19,10 @@ is of no practical use,
but nevertheless Simon's algorithm is an important landmark.
Simon's problem is this:
-we are given a "black box" function $f(x)$
-that takes an $n$-bit input $x$
-and returns an $n$-bit output.
-We are promised that there exists an $s$ such that for all $x_1$ and $x_2$:
+we are given a "black box" function $$f(x)$$
+that takes an $$n$$-bit input $$x$$
+and returns an $$n$$-bit output.
+We are promised that there exists an $$s$$ such that for all $$x_1$$ and $$x_2$$:
$$\begin{aligned}
f(x_1)
@@ -31,24 +31,24 @@ $$\begin{aligned}
x_2 = s \oplus x_1
\end{aligned}$$
-In other words, regardless of what $f(x)$ does behind the scenes,
-its output is the same for inputs $x_1$ and $x_2$
-if and only if $x_2 = s \oplus x_1$,
-or, equivalently, $x_1 = s \oplus x_2$.
+In other words, regardless of what $$f(x)$$ does behind the scenes,
+its output is the same for inputs $$x_1$$ and $$x_2$$
+if and only if $$x_2 = s \oplus x_1$$,
+or, equivalently, $$x_1 = s \oplus x_2$$.
-The goal is to find the $n$-bit number $s$, using as few calls to $f$ as possible.
+The goal is to find the $$n$$-bit number $$s$$, using as few calls to $$f$$ as possible.
There are two cases:
-if $s = 0$, then $f$ is one-to-one, since $x_2 = 0 \oplus x_1 = x_1$.
-Otherwise, if $s \neq 0$, then $f$ is two-to-one by definition:
-for every $x_1$ there exists exactly one $x_2$ such that $x_2 = s \oplus x_1$.
+if $$s = 0$$, then $$f$$ is one-to-one, since $$x_2 = 0 \oplus x_1 = x_1$$.
+Otherwise, if $$s \neq 0$$, then $$f$$ is two-to-one by definition:
+for every $$x_1$$ there exists exactly one $$x_2$$ such that $$x_2 = s \oplus x_1$$.
A classical computer solves this by randomly guessing inputs,
until it finds two that give the same output,
-and then $s = x_1 \oplus x_2$.
-For $n$-bit numbers, this takes $\mathcal{O}(\sqrt{2^n})$ guesses
+and then $$s = x_1 \oplus x_2$$.
+For $$n$$-bit numbers, this takes $$\mathcal{O}(\sqrt{2^n})$$ guesses
(the square root is due to the birthday paradox).
-A quantum computer needs to query $f$ only $\mathcal{O}(n)$ times,
+A quantum computer needs to query $$f$$ only $$\mathcal{O}(n)$$ times,
although the exact number varies due to the algorithm's probabilistic nature.
It uses the following circuit:
@@ -56,8 +56,8 @@ It uses the following circuit:
-The XOR oracle $U_f$ implements $f$,
-and has the following action for $n$-bit $a$ and $b$:
+The XOR oracle $$U_f$$ implements $$f$$,
+and has the following action for $$n$$-bit $$a$$ and $$b$$:
$$\begin{aligned}
\Ket{a} \Ket{b}
@@ -65,9 +65,9 @@ $$\begin{aligned}
\Ket{a} \Ket{b \oplus f(a)}
\end{aligned}$$
-Starting from the state $\Ket{0}^{\otimes 2 n}$,
-we apply the [Hadamard gate](/know/concept/quantum-gate/) $H$
-to each of the first $n$ qubits:
+Starting from the state $$\Ket{0}^{\otimes 2 n}$$,
+we apply the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
+to each of the first $$n$$ qubits:
$$\begin{aligned}
\Ket{0}^{\otimes n} \Ket{0}^{\otimes n}
@@ -76,10 +76,10 @@ $$\begin{aligned}
= \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{0}^{\otimes n}
\end{aligned}$$
-Where $\Ket{x}$ is shorthand for $\Ket{x}_1 \cdots \Ket{x}_n$.
-In other words, we now have an equal superposition of all possible inputs $x$,
-with a constant $\Ket{0}^{\otimes n}$ beside it.
-We give this to the oracle $U_f$:
+Where $$\Ket{x}$$ is shorthand for $$\Ket{x}_1 \cdots \Ket{x}_n$$.
+In other words, we now have an equal superposition of all possible inputs $$x$$,
+with a constant $$\Ket{0}^{\otimes n}$$ beside it.
+We give this to the oracle $$U_f$$:
$$\begin{aligned}
\frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{0}^{\otimes n}
@@ -87,10 +87,10 @@ $$\begin{aligned}
\frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{f(x)}
\end{aligned}$$
-Then we apply $H^{\otimes n}$ to the first $n$ qubits again,
+Then we apply $$H^{\otimes n}$$ to the first $$n$$ qubits again,
which, thanks to the definition of the Hadamard transform,
yields the following,
-where $x \cdot y$ is the bitwise dot product:
+where $$x \cdot y$$ is the bitwise dot product:
$$\begin{aligned}
\frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{f(x)}
@@ -101,10 +101,10 @@ $$\begin{aligned}
Next, we measure all qubits.
The order in which we do this does not matter,
-but, for clarity, let us measure the last $n$ qubits first,
-yielding $\Ket{f(x_1)}$ for some $x_1$.
-Doing this leaves the $2n$ qubits in the following state,
-where $f(x_1) = f(x_2)$ and $x_2 = s \oplus x_1$:
+but, for clarity, let us measure the last $$n$$ qubits first,
+yielding $$\Ket{f(x_1)}$$ for some $$x_1$$.
+Doing this leaves the $$2n$$ qubits in the following state,
+where $$f(x_1) = f(x_2)$$ and $$x_2 = s \oplus x_1$$:
$$\begin{alignedat}{2}
&\mathrm{if} \: s = 0: \qquad
@@ -114,18 +114,18 @@ $$\begin{alignedat}{2}
&&\frac{1}{\sqrt{2^{n+1}}} \sum_{y = 0}^{2^n - 1} \Big( (-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \Big) \Ket{y} \Ket{f(x_1)}
\end{alignedat}$$
-If $s = 0$, we get an equiprobable superposition of all $y$.
-So, when we measure the first $n$ qubits, the result is a uniformly random number,
-regardless of the phase $(-1)^{x_1 \cdot y}$.
+If $$s = 0$$, we get an equiprobable superposition of all $$y$$.
+So, when we measure the first $$n$$ qubits, the result is a uniformly random number,
+regardless of the phase $$(-1)^{x_1 \cdot y}$$.
-If $s \neq 0$, the situation is more interesting,
-because we can only measure $y$-values where:
+If $$s \neq 0$$, the situation is more interesting,
+because we can only measure $$y$$-values where:
$$\begin{aligned}
(-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \neq 0
\end{aligned}$$
-Since $x_2 = s \oplus x_1$ by definition,
+Since $$x_2 = s \oplus x_1$$ by definition,
we can rewrite this as follows:
$$\begin{aligned}
@@ -134,21 +134,21 @@ $$\begin{aligned}
\neq 0
\end{aligned}$$
-Clearly, the expression can only be nonzero if $s \cdot y$ is even.
-In other words, when we measure the first $n$ qubits,
-we get a random $y$-value,
-for which $s \cdot y$ is guaranteed to be even.
+Clearly, the expression can only be nonzero if $$s \cdot y$$ is even.
+In other words, when we measure the first $$n$$ qubits,
+we get a random $$y$$-value,
+for which $$s \cdot y$$ is guaranteed to be even.
-In both cases $s = 0$ and $s \neq 0$,
-we measure a $y$-value that satisfies the equation:
+In both cases $$s = 0$$ and $$s \neq 0$$,
+we measure a $$y$$-value that satisfies the equation:
$$\begin{aligned}
s \cdot y = 0 \:\:(\bmod 2)
\end{aligned}$$
-This tells us something about $s$, albeit not much.
-But if we run Simon's algorithm $N$ times,
-we get various $y$-values $y_1, ..., y_N$,
+This tells us something about $$s$$, albeit not much.
+But if we run Simon's algorithm $$N$$ times,
+we get various $$y$$-values $$y_1, ..., y_N$$,
from which we can build a system of linear equations:
$$\begin{aligned}
@@ -162,10 +162,10 @@ $$\begin{aligned}
\end{aligned}$$
This can be solved efficiently by a classical computer.
-In the best-case scenario, all those $y$-values would be linearly independent
+In the best-case scenario, all those $$y$$-values would be linearly independent
(when regarded as vectors of bits),
-in which case only $N = n - 1$ equations would be necessary.
-Simon's algorithm is therefore $\mathcal{O}(n)$.
+in which case only $$N = n - 1$$ equations would be necessary.
+Simon's algorithm is therefore $$\mathcal{O}(n)$$.
It may feel like "cheating" to use a classical computer at the end.
Remember that the point of this algorithm is to limit the number of oracle queries,
diff --git a/source/know/concept/slater-determinant/index.md b/source/know/concept/slater-determinant/index.md
index 03ec153..72c8cf2 100644
--- a/source/know/concept/slater-determinant/index.md
+++ b/source/know/concept/slater-determinant/index.md
@@ -9,13 +9,13 @@ layout: "concept"
---
In quantum mechanics, the **Slater determinant** is a trick
-to create a many-particle wave function for a system of $N$ fermions,
+to create a many-particle wave function for a system of $$N$$ fermions,
with the necessary antisymmetry.
-Given an orthogonal set of individual states $\psi_n(x)$, we write
-$\psi_n(x_n)$ to say that particle $x_n$ is in state $\psi_n$. Now the
+Given an orthogonal set of individual states $$\psi_n(x)$$, we write
+$$\psi_n(x_n)$$ to say that particle $$x_n$$ is in state $$\psi_n$$. Now the
goal is to find an expression for an overall many-particle wave
-function $\Psi(x_1, ..., x_N)$ that satisfies the
+function $$\Psi(x_1, ..., x_N)$$ that satisfies the
[Pauli exclusion principle](/know/concept/pauli-exclusion-principle/).
Enter the Slater determinant:
@@ -34,13 +34,13 @@ Swapping the state of two particles corresponds to exchanging two rows,
which flips the sign of the determinant.
Similarly, switching two columns means swapping two states,
which also results in a sign change.
-Finally, putting two particles into the same state makes $\Psi$ vanish.
+Finally, putting two particles into the same state makes $$\Psi$$ vanish.
Not all valid many-fermion wave functions can be
written as a single Slater determinant; a linear combination of multiple
may be needed. Nevertheless, an appropriate choice of the input set
-$\psi_n(x)$ can optimize how well a single determinant approximates a
-given $\Psi$.
+$$\psi_n(x)$$ can optimize how well a single determinant approximates a
+given $$\Psi$$.
In fact, there exists a similar trick for bosons, where the goal is to
create a symmetric wave function which allows multiple particles to
diff --git a/source/know/concept/sokhotski-plemelj-theorem/index.md b/source/know/concept/sokhotski-plemelj-theorem/index.md
index 984a558..66e89bc 100644
--- a/source/know/concept/sokhotski-plemelj-theorem/index.md
+++ b/source/know/concept/sokhotski-plemelj-theorem/index.md
@@ -9,8 +9,8 @@ categories:
layout: "concept"
---
-The goal is to evaluate integrals of the following form, where $a < 0 < b$,
-and $f(x)$ is assumed to be continuous in the integration interval $[a, b]$:
+The goal is to evaluate integrals of the following form, where $$a < 0 < b$$,
+and $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$:
$$\begin{aligned}
\lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
@@ -26,7 +26,7 @@ $$\begin{aligned}
\end{aligned}$$
To evaluate the real part,
-we notice that for $\eta \to 0^+$ the integrand diverges for $x \to 0$,
+we notice that for $$\eta \to 0^+$$ the integrand diverges for $$x \to 0$$,
and thus split the integral as follows:
$$\begin{aligned}
@@ -35,7 +35,7 @@ $$\begin{aligned}
\end{aligned}$$
This is simply the definition of the
-[Cauchy principal value](/know/concept/cauchy-principal-value/) $\mathcal{P}$,
+[Cauchy principal value](/know/concept/cauchy-principal-value/) $$\mathcal{P}$$,
so the real part is given by:
$$\begin{aligned}
@@ -45,7 +45,7 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, in the imaginary part,
-we substitute $\eta$ for $1 / m$, and introduce $\pi$:
+we substitute $$\eta$$ for $$1 / m$$, and introduce $$\pi$$:
$$\begin{aligned}
\lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x}
@@ -54,8 +54,8 @@ $$\begin{aligned}
&= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{m}{1 + m^2 x^2} f(x) \dd{x}
\end{aligned}$$
-The expression $m / \pi (1 + m^2 x^2)$ is a so-called *nascent delta function*,
-meaning that in the limit $m \to +\infty$ it converges to
+The expression $$m / \pi (1 + m^2 x^2)$$ is a so-called *nascent delta function*,
+meaning that in the limit $$m \to +\infty$$ it converges to
the [Dirac delta function](/know/concept/dirac-delta-function/):
$$\begin{aligned}
@@ -76,9 +76,9 @@ $$\begin{aligned}
\end{aligned}$$
However, this theorem is often written in the following sloppy way,
-where $\eta$ is defined up front to be small,
-the integral is hidden, and $f(x)$ is set to $1$.
-This awkwardly leaves $\mathcal{P}$ behind:
+where $$\eta$$ is defined up front to be small,
+the integral is hidden, and $$f(x)$$ is set to $$1$$.
+This awkwardly leaves $$\mathcal{P}$$ behind:
$$\begin{aligned}
\frac{1}{x + i \eta}
@@ -87,13 +87,13 @@ $$\begin{aligned}
The full, complex version of the Sokhotski-Plemelj theorem
evaluates integrals of the following form
-over a contour $C$ in the complex plane:
+over a contour $$C$$ in the complex plane:
$$\begin{aligned}
\phi(z) = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta}
\end{aligned}$$
-Where $f(z)$ must be [holomorphic](/know/concept/holomorphic-function/).
+Where $$f(z)$$ must be [holomorphic](/know/concept/holomorphic-function/).
The Sokhotski-Plemelj theorem then states:
$$\begin{aligned}
@@ -103,8 +103,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the sign is positive if $z$ is inside $C$, and negative if it is outside.
-The real version follows by letting $C$ follow the whole real axis,
-making $C$ an infinitely large semicircle,
+Where the sign is positive if $$z$$ is inside $$C$$, and negative if it is outside.
+The real version follows by letting $$C$$ follow the whole real axis,
+making $$C$$ an infinitely large semicircle,
so that the integrand vanishes away from the real axis,
-because $1 / (\zeta \!-\! z) \to 0$ for $|\zeta| \to \infty$.
+because $$1 / (\zeta \!-\! z) \to 0$$ for $$|\zeta| \to \infty$$.
diff --git a/source/know/concept/spherical-coordinates/index.md b/source/know/concept/spherical-coordinates/index.md
index bc01c95..9df5d51 100644
--- a/source/know/concept/spherical-coordinates/index.md
+++ b/source/know/concept/spherical-coordinates/index.md
@@ -10,16 +10,16 @@ layout: "concept"
**Spherical coordinates** are an extension of polar coordinates to 3D.
The position of a given point in space is described by
-three coordinates $(r, \theta, \varphi)$, defined as:
+three coordinates $$(r, \theta, \varphi)$$, defined as:
-* $r$: the **radius** or **radial distance**: distance to the origin.
-* $\theta$: the **elevation**, **polar angle** or **colatitude**:
- angle to the positive $z$-axis, or **zenith**, i.e. the "north pole".
-* $\varphi$: the **azimuth**, **azimuthal angle** or **longitude**:
- angle from the positive $x$-axis, typically in the counter-clockwise sense.
+* $$r$$: the **radius** or **radial distance**: distance to the origin.
+* $$\theta$$: the **elevation**, **polar angle** or **colatitude**:
+ angle to the positive $$z$$-axis, or **zenith**, i.e. the "north pole".
+* $$\varphi$$: the **azimuth**, **azimuthal angle** or **longitude**:
+ angle from the positive $$x$$-axis, typically in the counter-clockwise sense.
-Cartesian coordinates $(x, y, z)$ and the spherical system
-$(r, \theta, \varphi)$ are related by:
+Cartesian coordinates $$(x, y, z)$$ and the spherical system
+$$(r, \theta, \varphi)$$ are related by:
$$\begin{aligned}
\boxed{
@@ -31,8 +31,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Conversely, a point given in $(x, y, z)$
-can be converted to $(r, \theta, \varphi)$
+Conversely, a point given in $$(x, y, z)$$
+can be converted to $$(r, \theta, \varphi)$$
using these formulae:
$$\begin{aligned}
@@ -47,7 +47,7 @@ $$\begin{aligned}
The spherical coordinate system is an orthogonal
[curvilinear system](/know/concept/curvilinear-coordinates/),
-whose scale factors $h_r$, $h_\theta$ and $h_\varphi$ we want to find.
+whose scale factors $$h_r$$, $$h_\theta$$ and $$h_\varphi$$ we want to find.
To do so, we calculate the differentials of the Cartesian coordinates:
$$\begin{aligned}
@@ -58,7 +58,7 @@ $$\begin{aligned}
\dd{z} &= \dd{r} \cos\theta - \dd{\theta} r \sin\theta
\end{aligned}$$
-And then we calculate the line element $\dd{\ell}^2$,
+And then we calculate the line element $$\dd{\ell}^2$$,
skipping many terms thanks to orthogonality:
$$\begin{aligned}
@@ -74,7 +74,7 @@ $$\begin{aligned}
Finally, we can simply read off
the squares of the desired scale factors
-$h_r^2$, $h_\theta^2$ and $h_\varphi^2$:
+$$h_r^2$$, $$h_\theta^2$$ and $$h_\varphi^2$$:
$$\begin{aligned}
\boxed{
@@ -146,7 +146,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The differential element of volume $\dd{V}$
+The differential element of volume $$\dd{V}$$
takes the following form:
$$\begin{aligned}
@@ -163,7 +163,7 @@ $$\begin{aligned}
= \int_0^{2\pi} \int_0^\pi \int_0^\infty f(r, \theta, \varphi) \: r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
\end{aligned}$$
-The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+The isosurface elements are as follows, where $$S_r$$ is a surface at constant $$r$$, etc.:
$$\begin{aligned}
\boxed{
@@ -177,7 +177,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+Similarly, the normal vector element $$\dd{\vu{S}}$$ for an arbitrary surface is given by:
$$\begin{aligned}
\boxed{
@@ -188,7 +188,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+And finally, the tangent vector element $$\dd{\vu{\ell}}$$ of a given curve is as follows:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/spitzer-resistivity/index.md b/source/know/concept/spitzer-resistivity/index.md
index f9fd969..283b7fd 100644
--- a/source/know/concept/spitzer-resistivity/index.md
+++ b/source/know/concept/spitzer-resistivity/index.md
@@ -9,17 +9,17 @@ layout: "concept"
---
If an [electric field](/know/concept/electric-field/)
-with magnitude $E$ is applied to the plasma, the electrons experience
-a [Lorentz force](/know/concept/lorentz-force/) $q_e E$
+with magnitude $$E$$ is applied to the plasma, the electrons experience
+a [Lorentz force](/know/concept/lorentz-force/) $$q_e E$$
(we neglect the ions due to their mass),
-where $q_e$ is the electron charge.
+where $$q_e$$ is the electron charge.
However, collisions slow them down while they travel through the plasma.,
-This can be modelled as a drag force $f_{ei} m_e v_e$,
-where $f_{ei}$ is the electron-ion collision frequency
-(we neglect $f_{ee}$ since all electrons are moving together),
-$m_e$ is their mass,
-and $v_e$ their typical velocity relative to the ions in the background.
+This can be modelled as a drag force $$f_{ei} m_e v_e$$,
+where $$f_{ei}$$ is the electron-ion collision frequency
+(we neglect $$f_{ee}$$ since all electrons are moving together),
+$$m_e$$ is their mass,
+and $$v_e$$ their typical velocity relative to the ions in the background.
Balancing the two forces yields the following relation:
$$\begin{aligned}
@@ -27,7 +27,7 @@ $$\begin{aligned}
= f_{ei} m_e v_e
\end{aligned}$$
-Using that the current density $J = q_e n_e v_e$,
+Using that the current density $$J = q_e n_e v_e$$,
we can rearrange this like so:
$$\begin{aligned}
@@ -37,12 +37,12 @@ $$\begin{aligned}
= \eta J
\end{aligned}$$
-This is Ohm's law, where $\eta$ is the resistivity.
-From our derivation of the [Coulomb logarithm](/know/concept/coulomb-logarithm/) $\ln(\Lambda)$,
-we estimate $f_{ei}$ to be as follows,
-where $n_i$ is the ion density,
-$\sigma$ is the collision cross-section,
-and $\mu$ is the [reduced mass](/know/concept/reduced-mass/)
+This is Ohm's law, where $$\eta$$ is the resistivity.
+From our derivation of the [Coulomb logarithm](/know/concept/coulomb-logarithm/) $$\ln(\Lambda)$$,
+we estimate $$f_{ei}$$ to be as follows,
+where $$n_i$$ is the ion density,
+$$\sigma$$ is the collision cross-section,
+and $$\mu$$ is the [reduced mass](/know/concept/reduced-mass/)
of the electron-ion system:
$$\begin{aligned}
@@ -52,15 +52,15 @@ $$\begin{aligned}
\approx \frac{1}{2 \pi} \frac{Z q_e^4}{\varepsilon_0^2 m_e^2} \frac{n_e}{v_e^3} \ln(\Lambda)
\end{aligned}$$
-Where we used that $\mu \approx m_e$,
-and $q_i = -Z q_e$ for some ionization $Z$,
-and as a result $n_e \approx Z n_i$ due to the plasma's quasi-neutrality.
-Beware: authors disagree about the constant factors in $f_{ei}$;
+Where we used that $$\mu \approx m_e$$,
+and $$q_i = -Z q_e$$ for some ionization $$Z$$,
+and as a result $$n_e \approx Z n_i$$ due to the plasma's quasi-neutrality.
+Beware: authors disagree about the constant factors in $$f_{ei}$$;
recall that it was derived from fairly rough estimates.
This article follows Bellan.
-Inserting this expression for $f_{ei}$ into
-the so-called **Spitzer resistivity** $\eta$ then yields:
+Inserting this expression for $$f_{ei}$$ into
+the so-called **Spitzer resistivity** $$\eta$$ then yields:
$$\begin{aligned}
\boxed{
@@ -70,10 +70,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-A reasonable estimate for the typical velocity $v_e$
+A reasonable estimate for the typical velocity $$v_e$$
at thermal equilibrium is as follows,
-where $k_B$ is Boltzmann's constant,
-and $T_e$ is the electron temperature:
+where $$k_B$$ is Boltzmann's constant,
+and $$T_e$$ is the electron temperature:
$$\begin{aligned}
\frac{1}{2} m_e v_e^2
@@ -85,8 +85,8 @@ $$\begin{aligned}
Other choices exist,
see e.g. the [Maxwell-Boltzmann distribution](/know/concept/maxwell-boltzmann-distribution/),
-but always $v_e \propto \sqrt{T_e/m_e}$.
-Inserting this $v_e$ into $\eta$ then gives:
+but always $$v_e \propto \sqrt{T_e/m_e}$$.
+Inserting this $$v_e$$ into $$\eta$$ then gives:
$$\begin{aligned}
\eta
diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
index e0b26af..dd83334 100644
--- a/source/know/concept/step-index-fiber/index.md
+++ b/source/know/concept/step-index-fiber/index.md
@@ -9,31 +9,31 @@ categories:
layout: "concept"
---
-As light propagates in the $z$-direction through an optical fiber,
-the transverse profile $F(x,y)$ of the [electric field](/know/concept/electric-field/)
+As light propagates in the $$z$$-direction through an optical fiber,
+the transverse profile $$F(x,y)$$ of the [electric field](/know/concept/electric-field/)
can be shown to obey the *Helmholtz equation* in 2D:
$$\begin{aligned}
\nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F = 0
\end{aligned}$$
-With $n$ being the position-dependent refractive index,
-$k$ the vacuum wavenumber $\omega / c$,
-and $\beta$ the mode's propagation constant, to be determined later.
+With $$n$$ being the position-dependent refractive index,
+$$k$$ the vacuum wavenumber $$\omega / c$$,
+and $$\beta$$ the mode's propagation constant, to be determined later.
In [polar coordinates](/know/concept/cylindrical-polar-coordinates/)
-$(r,\phi)$ this equation can be rewritten as follows:
+$$(r,\phi)$$ this equation can be rewritten as follows:
$$\begin{aligned}
\pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F = 0
\end{aligned}$$
-Where we have defined $\mu \equiv n^2 k^2 \!-\! \beta^2$ for brevity.
-From now on, we only consider choices of $\mu$ that do not depend on $\phi$ or $z$,
-but may vary with $r$.
+Where we have defined $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ for brevity.
+From now on, we only consider choices of $$\mu$$ that do not depend on $$\phi$$ or $$z$$,
+but may vary with $$r$$.
This Helmholtz equation can be solved by *separation of variables*:
-we assume that there exist two functions $R(r)$ and $\Phi(\phi)$
-such that $F(r,\phi) = R(r) \, \Phi(\phi)$.
+we assume that there exist two functions $$R(r)$$ and $$\Phi(\phi)$$
+such that $$F(r,\phi) = R(r) \, \Phi(\phi)$$.
Inserting this ansatz:
$$\begin{aligned}
@@ -41,10 +41,10 @@ $$\begin{aligned}
\end{aligned}$$
We rearrange this such that each side only depends on one variable,
-by dividing by $R\Phi$ (ignoring the fact that it may be zero),
-and multiplying by $r^2$.
-Since this equation should hold for *all* values of $r$ and $\phi$,
-this means that both sides must equal a constant $\ell^2$:
+by dividing by $$R\Phi$$ (ignoring the fact that it may be zero),
+and multiplying by $$r^2$$.
+Since this equation should hold for *all* values of $$r$$ and $$\phi$$,
+this means that both sides must equal a constant $$\ell^2$$:
$$\begin{aligned}
r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2
@@ -52,8 +52,8 @@ $$\begin{aligned}
= \ell^2
\end{aligned}$$
-This gives an eigenvalue problem for $\Phi$,
-and the well-known *Bessel equation* for $R$:
+This gives an eigenvalue problem for $$\Phi$$,
+and the well-known *Bessel equation* for $$R$$:
$$\begin{aligned}
\boxed{
@@ -65,9 +65,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-We will return to $R$ later; we start with $\Phi$, because it has the
-simplest equation. Since the angle $\phi$ is limited to $[0,2\pi]$,
-$\Phi$ must be $2 \pi$-periodic, so:
+We will return to $$R$$ later; we start with $$\Phi$$, because it has the
+simplest equation. Since the angle $$\phi$$ is limited to $$[0,2\pi]$$,
+$$\Phi$$ must be $$2 \pi$$-periodic, so:
$$\begin{aligned}
\Phi(0) = \Phi(2\pi)
@@ -75,18 +75,18 @@ $$\begin{aligned}
\Phi'(0) = \Phi'(2\pi)
\end{aligned}$$
-The above equation for $\Phi$ with these periodic boundary conditions
+The above equation for $$\Phi$$ with these periodic boundary conditions
is a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
-Consequently, there are infinitely many allowed values of $\ell^2$,
+Consequently, there are infinitely many allowed values of $$\ell^2$$,
all real, and one of them is lowest, known as the *ground state*.
-To find the eigenvalues $\ell^2$ and their corresponding $\Phi$,
-we in turn assume that $\ell^2 < 0$, $\ell^2 = 0$, or $\ell^2 > 0$,
-and check if we can then arrive at a non-trivial $\Phi$ for each case.
+To find the eigenvalues $$\ell^2$$ and their corresponding $$\Phi$$,
+we in turn assume that $$\ell^2 < 0$$, $$\ell^2 = 0$$, or $$\ell^2 > 0$$,
+and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
-* For $\ell^2 < 0$, solutions have the form $\Phi(\phi) = A \sinh(\phi \ell) + B \cosh(\phi \ell)$,
- where $A$ and $B$ are unknown linearity constants.
- At least one of these constants must be nonzero for $\Phi$ to be non-trivial,
+* For $$\ell^2 < 0$$, solutions have the form $$\Phi(\phi) = A \sinh(\phi \ell) + B \cosh(\phi \ell)$$,
+ where $$A$$ and $$B$$ are unknown linearity constants.
+ At least one of these constants must be nonzero for $$\Phi$$ to be non-trivial,
but the challenge is to satisfy the boundary conditions:
$$\begin{alignedat}{3}
@@ -112,12 +112,12 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
= 2 \big( \cosh(2 \pi \ell) - 1 \big)
\end{aligned}$$
- This can only be zero if $\ell = 0$,
- which contradicts the premise that $\ell^2 < 0$,
- so we conclude that $\ell^2$ cannot be negative,
+ This can only be zero if $$\ell = 0$$,
+ which contradicts the premise that $$\ell^2 < 0$$,
+ so we conclude that $$\ell^2$$ cannot be negative,
because no non-trivial solutions exist here.
-* For $\ell^2 = 0$, the solution is $\Phi(\phi) = A \phi + B$.
+* For $$\ell^2 = 0$$, the solution is $$\Phi(\phi) = A \phi + B$$.
Putting this in the boundary conditions:
$$\begin{alignedat}{3}
@@ -130,11 +130,11 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
B &&= B
\end{alignedat}$$
- $B$ can be nonzero, so this a valid solution.
- We conclude that $\ell^2 = 0$ is the ground state.
+ $$B$$ can be nonzero, so this a valid solution.
+ We conclude that $$\ell^2 = 0$$ is the ground state.
-* For $\ell^2 > 0$, all solutions have the form
- $\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$, therefore:
+* For $$\ell^2 > 0$$, all solutions have the form
+ $$\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$$, therefore:
$$\begin{alignedat}{3}
\Phi(0) &= \Phi(2 \pi)
@@ -159,7 +159,7 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
= 2 \big(\cos(2 \pi \ell) - 1\big)
\end{aligned}$$
- Meaning that $\ell$ must be an integer.
+ Meaning that $$\ell$$ must be an integer.
We revisit the boundary conditions and indeed see:
$$\begin{alignedat}{3}
@@ -172,16 +172,16 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
0 &&= 0
\end{alignedat}$$
- So $A$ and $B$ are *both* unconstrained,
- and each integer $\ell$ is a doubly-degenerate eigenvalue.
+ So $$A$$ and $$B$$ are *both* unconstrained,
+ and each integer $$\ell$$ is a doubly-degenerate eigenvalue.
The two linearly independent solutions,
- $\sin(\phi \ell)$ and $\cos(\phi \ell)$,
+ $$\sin(\phi \ell)$$ and $$\cos(\phi \ell)$$,
represent the polarization of light in the mode.
For simplicity, we assume that all light is in a single polarization,
- so only $\cos(\phi \ell)$ will be considered from now on.
+ so only $$\cos(\phi \ell)$$ will be considered from now on.
-By combining our result for $\ell^2 = 0$ and $\ell^2 > 0$,
-we get the following for $\ell = 0, 1, 2, ...$:
+By combining our result for $$\ell^2 = 0$$ and $$\ell^2 > 0$$,
+we get the following for $$\ell = 0, 1, 2, ...$$:
$$\begin{aligned}
\boxed{
@@ -189,28 +189,28 @@ $$\begin{aligned}
}
\end{aligned}$$
-Here, $\ell$ is called the **primary mode index**.
-We exclude $\ell < 0$ because $\cos(x) \propto \cos(-x)$
-and $\sin(x) \propto \sin(-x)$,
-and because $A$ is free to choose thanks to linearity.
+Here, $$\ell$$ is called the **primary mode index**.
+We exclude $$\ell < 0$$ because $$\cos(x) \propto \cos(-x)$$
+and $$\sin(x) \propto \sin(-x)$$,
+and because $$A$$ is free to choose thanks to linearity.
-Let us now revisit the Bessel equation for the radial function $R(r)$,
+Let us now revisit the Bessel equation for the radial function $$R(r)$$,
which should be continuous and differentiable throughout the fiber:
$$\begin{aligned}
r^2 R'' + r R' + \mu r^2 R - \ell^2 R = 0
\end{aligned}$$
-To continue, we need to specify the refractive index $n(r)$, contained in $\mu(r)$.
+To continue, we need to specify the refractive index $$n(r)$$, contained in $$\mu(r)$$.
We choose a **step-index fiber**,
-whose cross-section consists of a **core** with radius $a$,
-surrounded by a **cladding** that extends to infinity $r \to \infty$.
-In the core $r < a$, the index $n$ is a constant $n_i$,
-while in the cladding $r > a$ it is another constant $n_o$.
+whose cross-section consists of a **core** with radius $$a$$,
+surrounded by a **cladding** that extends to infinity $$r \to \infty$$.
+In the core $$r < a$$, the index $$n$$ is a constant $$n_i$$,
+while in the cladding $$r > a$$ it is another constant $$n_o$$.
-Since $\mu$ is different in the core and cladding,
-we will get different solutions $R_i$ and $R_o$ there,
-so we must demand that the field is continuous at the boundary $r = a$:
+Since $$\mu$$ is different in the core and cladding,
+we will get different solutions $$R_i$$ and $$R_o$$ there,
+so we must demand that the field is continuous at the boundary $$r = a$$:
$$\begin{aligned}
R_i(a) = R_o(a)
@@ -219,14 +219,14 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, for a physically plausible solution,
-we require that $R_i$ is finite
-and that $R_o$ decays monotonically to zero when $r \to \infty$.
-These constraints will turn out to restrict $\mu$.
+we require that $$R_i$$ is finite
+and that $$R_o$$ decays monotonically to zero when $$r \to \infty$$.
+These constraints will turn out to restrict $$\mu$$.
-Introducing a new coordinate $\rho \equiv r \sqrt{|\mu|}$
+Introducing a new coordinate $$\rho \equiv r \sqrt{|\mu|}$$
gives the Bessel equation's standard form,
which has well-known solutions called *Bessel functions*, shown below.
-Let $\pm$ be the sign of $\mu$:
+Let $$\pm$$ be the sign of $$\mu$$:
$$\begin{aligned}
\begin{cases}
@@ -244,11 +244,11 @@ $$\begin{aligned}
-Looking at these solutions with our constraints for $R_o$ in mind,
-we see that for $\mu > 0$ none of the solutions decay
-*monotonically* to zero, so we must have $\mu \le 0$ in the cladding.
-Of the remaining candidates, $\ln\!(r)$, $r^\ell$ and $I_\ell(\rho)$ do not decay at all,
-leading to the following $R_o$:
+Looking at these solutions with our constraints for $$R_o$$ in mind,
+we see that for $$\mu > 0$$ none of the solutions decay
+*monotonically* to zero, so we must have $$\mu \le 0$$ in the cladding.
+Of the remaining candidates, $$\ln\!(r)$$, $$r^\ell$$ and $$I_\ell(\rho)$$ do not decay at all,
+leading to the following $$R_o$$:
$$\begin{aligned}
R_{o,\ell}(r) =
@@ -261,12 +261,12 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-Next, for $R_i$, we see that when $\mu < 0$ all solutions are invalid
-since they diverge at $r = 0$,
-and so do $\ln\!(r)$, $r^{-\ell}$ and $Y_\ell(\rho)$.
-Of the remaining candidates, $r^0$ and $r^\ell$ have a non-negative slope
-at the boundary $r = a$, so they can never be continuous with $R_o'$.
-This leaves $J_\ell(\rho)$ for $\mu > 0$:
+Next, for $$R_i$$, we see that when $$\mu < 0$$ all solutions are invalid
+since they diverge at $$r = 0$$,
+and so do $$\ln\!(r)$$, $$r^{-\ell}$$ and $$Y_\ell(\rho)$$.
+Of the remaining candidates, $$r^0$$ and $$r^\ell$$ have a non-negative slope
+at the boundary $$r = a$$, so they can never be continuous with $$R_o'$$.
+This leaves $$J_\ell(\rho)$$ for $$\mu > 0$$:
$$\begin{aligned}
R_{i,\ell}(r) =
@@ -274,7 +274,7 @@ $$\begin{aligned}
\qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,...
\end{aligned}$$
-Putting this all together, we now know what the full solution for $F$ should look like:
+Putting this all together, we now know what the full solution for $$F$$ should look like:
$$\begin{aligned}
F_\ell(r, \phi)
@@ -289,25 +289,25 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-Where $A_\ell$ and $B_\ell$ are constants to be chosen
-based on the light's intensity, and to satisfy the continuity condition at $r = a$.
+Where $$A_\ell$$ and $$B_\ell$$ are constants to be chosen
+based on the light's intensity, and to satisfy the continuity condition at $$r = a$$.
-We found that $\mu \le 0$ in the cladding and $\mu > 0$ in the core.
-Since $\mu \equiv n^2 k^2 \!-\! \beta^2$ by definition,
-this discovery places a constraint on the propagation constant $\beta$:
+We found that $$\mu \le 0$$ in the cladding and $$\mu > 0$$ in the core.
+Since $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ by definition,
+this discovery places a constraint on the propagation constant $$\beta$$:
$$\begin{aligned}
n_i^2 k^2 > \beta^2 \ge n_o^2 k^2
\end{aligned}$$
-Therefore, $n_i > n_o$ in a step-index fiber,
-and there is only a limited range of allowed $\beta$-values;
+Therefore, $$n_i > n_o$$ in a step-index fiber,
+and there is only a limited range of allowed $$\beta$$-values;
the fiber is not able to guide the light outside this range.
-However, not all $\beta$ in this range are created equal for all $k$.
+However, not all $$\beta$$ in this range are created equal for all $$k$$.
To investigate further, let us define the quantities
-$\xi_\mathrm{core}$ and $\xi_\mathrm{clad}$ like so,
-assuming $n_i$ and $n_o$ do not depend on $k$:
+$$\xi_\mathrm{core}$$ and $$\xi_\mathrm{clad}$$ like so,
+assuming $$n_i$$ and $$n_o$$ do not depend on $$k$$:
$$\begin{aligned}
\xi_i(k)
@@ -317,13 +317,13 @@ $$\begin{aligned}
\equiv \sqrt{ \beta^2(k) - n_o^2 k^2 }
\end{aligned}$$
-It is important to note that the sum of their squares is constant with respect to $\beta$:
+It is important to note that the sum of their squares is constant with respect to $$\beta$$:
$$\begin{aligned}
\xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2
\end{aligned}$$
-Where $\mathrm{NA}$ is the so-called **numerical aperture**,
+Where $$\mathrm{NA}$$ is the so-called **numerical aperture**,
often mentioned in papers and datasheets as one of a fiber's key parameters.
It is defined as:
@@ -334,7 +334,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-From this, we define a new fiber parameter: the $V$-**number**,
+From this, we define a new fiber parameter: the $$V$$-**number**,
which is extremely useful:
$$\begin{aligned}
@@ -345,8 +345,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Now, the allowed values of $\beta$ are found
-by fulfilling the boundary conditions (for $\mu \neq 0$):
+Now, the allowed values of $$\beta$$ are found
+by fulfilling the boundary conditions (for $$\mu \neq 0$$):
$$\begin{aligned}
A_\ell J_\ell(a \xi_i)
@@ -356,10 +356,10 @@ $$\begin{aligned}
&= B_\ell \xi_o K_\ell'(a \xi_o)
\end{aligned}$$
-To remove $A_\ell$ and $B_\ell$,
+To remove $$A_\ell$$ and $$B_\ell$$,
we divide the latter equation by the former,
-meanwhile defining $X \equiv a \xi_i$ and $Y \equiv a \xi_o$
-for convenience, such that $X^2 + Y^2 = V^2$:
+meanwhile defining $$X \equiv a \xi_i$$ and $$Y \equiv a \xi_o$$
+for convenience, such that $$X^2 + Y^2 = V^2$$:
$$\begin{aligned}
X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)}
@@ -374,7 +374,7 @@ $$\begin{aligned}
K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x}
\end{aligned}$$
-With this, the transcendental equation for $\beta$
+With this, the transcendental equation for $$\beta$$
takes this convenient form:
$$\begin{aligned}
@@ -383,27 +383,27 @@ $$\begin{aligned}
}
\end{aligned}$$
-All $\beta$ that satisfy this indicate the existence
+All $$\beta$$ that satisfy this indicate the existence
of a **linearly polarized** mode.
-These modes are called $\mathrm{LP}_{\ell m}$,
-where $\ell$ is the primary (azimuthal) mode index,
-and $m$ the secondary (radial) mode index,
-which is needed because multiple $\beta$ may exist for a single $\ell$.
+These modes are called $$\mathrm{LP}_{\ell m}$$,
+where $$\ell$$ is the primary (azimuthal) mode index,
+and $$m$$ the secondary (radial) mode index,
+which is needed because multiple $$\beta$$ may exist for a single $$\ell$$.
An example graphical solution of the transcendental equation
-is illustrated below for a fiber with $V = 5$,
+is illustrated below for a fiber with $$V = 5$$,
where red and blue respectively denote the left and right-hand side:
-This shows that each $\mathrm{LP}_{\ell m}$ has an associated cut-off $V_{\ell m}$,
-so that if $V > V_{\ell m}$ then $\mathrm{LP}_{lm}$ exists,
-as long as $\beta$ stays in the allowed range.
-The cut-offs of the secondary modes for a given $\ell$
-are found as the $m$th roots of $J_{\ell-1}(V_{\ell m}) = 0$.
-In the above figure, they are $V_{01} = 0$, $V_{11} = 2.405$, and $V_{02} = V_{21} = 3.832$.
+This shows that each $$\mathrm{LP}_{\ell m}$$ has an associated cut-off $$V_{\ell m}$$,
+so that if $$V > V_{\ell m}$$ then $$\mathrm{LP}_{lm}$$ exists,
+as long as $$\beta$$ stays in the allowed range.
+The cut-offs of the secondary modes for a given $$\ell$$
+are found as the $$m$$th roots of $$J_{\ell-1}(V_{\ell m}) = 0$$.
+In the above figure, they are $$V_{01} = 0$$, $$V_{11} = 2.405$$, and $$V_{02} = V_{21} = 3.832$$.
All differential equations have been linear,
so a linear combination of these solutions is also valid.
diff --git a/source/know/concept/stochastic-process/index.md b/source/know/concept/stochastic-process/index.md
index dc3d30e..68f028d 100644
--- a/source/know/concept/stochastic-process/index.md
+++ b/source/know/concept/stochastic-process/index.md
@@ -9,24 +9,24 @@ categories:
layout: "concept"
---
-A **stochastic process** $X_t$ is a time-indexed
+A **stochastic process** $$X_t$$ is a time-indexed
[random variable](/know/concept/random-variable/),
-$\{ X_t : t > 0 \}$, i.e. a set of (usually correlated)
-random variables, each labelled with a unique timestamp $t$.
+$$\{ X_t : t > 0 \}$$, i.e. a set of (usually correlated)
+random variables, each labelled with a unique timestamp $$t$$.
Whereas "ordinary" random variables are defined on
-a probability space $(\Omega, \mathcal{F}, P)$,
+a probability space $$(\Omega, \mathcal{F}, P)$$,
stochastic process are defined on
-a **filtered probability space** $(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$.
-As before, $\Omega$ is the sample space,
-$\mathcal{F}$ is the event space,
-and $P$ is the probability measure.
+a **filtered probability space** $$(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$$.
+As before, $$\Omega$$ is the sample space,
+$$\mathcal{F}$$ is the event space,
+and $$P$$ is the probability measure.
-The **filtration** $\{ \mathcal{F}_t : t \ge 0 \}$
-is a time-indexed set of [$\sigma$-algebras](/know/concept/sigma-algebra/) on $\Omega$,
+The **filtration** $$\{ \mathcal{F}_t : t \ge 0 \}$$
+is a time-indexed set of [$$\sigma$$-algebras](/know/concept/sigma-algebra/) on $$\Omega$$,
which contains at least all the information generated
-by $X_t$ up to the current time $t$,
-and is a subset of $\mathcal{F}_t$:
+by $$X_t$$ up to the current time $$t$$,
+and is a subset of $$\mathcal{F}_t$$:
$$\begin{aligned}
\mathcal{F}
@@ -34,18 +34,18 @@ $$\begin{aligned}
\supseteq \sigma(X_s : 0 \le s \le t)
\end{aligned}$$
-In other words, $\mathcal{F}_t$ is the "accumulated" $\sigma$-algebra
-of all information extractable from $X_t$,
-and hence grows with time: $\mathcal{F}_s \subseteq \mathcal{F}_t$ for $s < t$.
-Given $\mathcal{F}_t$, all values $X_s$ for $s \le t$ can be computed,
-i.e. if you know $\mathcal{F}_t$, then the present and past of $X_t$ can be reconstructed.
-
-Given any filtration $\mathcal{H}_t$, a stochastic process $X_t$
-is said to be *"$\mathcal{H}_t$-adapted"*
-if $X_t$'s own filtration $\sigma(X_s : 0 \le s \le t) \subseteq \mathcal{H}_t$,
-meaning $\mathcal{H}_t$ contains enough information
-to determine the current and past values of $X_t$.
-Clearly, $X_t$ is always adapted to its own filtration.
+In other words, $$\mathcal{F}_t$$ is the "accumulated" $$\sigma$$-algebra
+of all information extractable from $$X_t$$,
+and hence grows with time: $$\mathcal{F}_s \subseteq \mathcal{F}_t$$ for $$s < t$$.
+Given $$\mathcal{F}_t$$, all values $$X_s$$ for $$s \le t$$ can be computed,
+i.e. if you know $$\mathcal{F}_t$$, then the present and past of $$X_t$$ can be reconstructed.
+
+Given any filtration $$\mathcal{H}_t$$, a stochastic process $$X_t$$
+is said to be *"$$\mathcal{H}_t$$-adapted"*
+if $$X_t$$'s own filtration $$\sigma(X_s : 0 \le s \le t) \subseteq \mathcal{H}_t$$,
+meaning $$\mathcal{H}_t$$ contains enough information
+to determine the current and past values of $$X_t$$.
+Clearly, $$X_t$$ is always adapted to its own filtration.
Filtration and their adaptations are very useful
for working with stochastic processes,
diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md
index e3b4526..3a02a83 100644
--- a/source/know/concept/stokes-law/index.md
+++ b/source/know/concept/stokes-law/index.md
@@ -9,17 +9,17 @@ categories:
layout: "concept"
---
-**Stokes' law** describes the size of the drag force $D$
-at low [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \ll 1$
-experienced by a spherical object in a steady, uniform flow at velocity $U$.
+**Stokes' law** describes the size of the drag force $$D$$
+at low [Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re} \ll 1$$
+experienced by a spherical object in a steady, uniform flow at velocity $$U$$.
## Flow field
-Imagine a sphere with radius $a$ sinking in a viscous liquid.
+Imagine a sphere with radius $$a$$ sinking in a viscous liquid.
To model this situation, let us pretend that the sphere is fixed instead,
-and the fluid comes from infinity at velocity $U$ along the $z$-axis,
-flows past the sphere, and continues to infinity at the same $U$.
+and the fluid comes from infinity at velocity $$U$$ along the $$z$$-axis,
+flows past the sphere, and continues to infinity at the same $$U$$.
The Reynolds number is:
$$\begin{aligned}
@@ -27,7 +27,7 @@ $$\begin{aligned}
= \frac{2 a U}{\nu}
\end{aligned}$$
-We assume that $\mathrm{Re} \ll 1$, in which case
+We assume that $$\mathrm{Re} \ll 1$$, in which case
the incompressible [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
are reduced to the **steady Stokes equations**:
@@ -39,10 +39,10 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-The goal is to solve for $p$ and $\va{v}$.
+The goal is to solve for $$p$$ and $$\va{v}$$.
We make the following ansatz in
-[spherical coordinates](/know/concept/spherical-coordinates/) $(r, \theta, \phi)$,
-where $q(r)$, $f(r)$ and $g(r)$ are unknown functions:
+[spherical coordinates](/know/concept/spherical-coordinates/) $$(r, \theta, \phi)$$,
+where $$q(r)$$, $$f(r)$$ and $$g(r)$$ are unknown functions:
$$\begin{gathered}
p
@@ -59,12 +59,12 @@ $$\begin{gathered}
\end{gathered}$$
The fluid hits the sphere head on,
-so the solution is taken to be $\phi$-independent due to symmetry.
-Note that $\theta$ is the angle to the positive $z$-axis,
-which is the direction of $\va{U} = U \vu{e}_z$.
-Moreover, note that $\va{U} \cdot \vu{e}_r = U \cos\theta$
-and $\va{U} \cdot \vu{e}_\theta = - U \sin\theta$,
-where $\vu{e}_r$ and $\vu{e}_\theta$ are basis vectors.
+so the solution is taken to be $$\phi$$-independent due to symmetry.
+Note that $$\theta$$ is the angle to the positive $$z$$-axis,
+which is the direction of $$\va{U} = U \vu{e}_z$$.
+Moreover, note that $$\va{U} \cdot \vu{e}_r = U \cos\theta$$
+and $$\va{U} \cdot \vu{e}_\theta = - U \sin\theta$$,
+where $$\vu{e}_r$$ and $$\vu{e}_\theta$$ are basis vectors.
To begin with, we insert this ansatz into the incompressibility condition,
yielding:
@@ -79,7 +79,7 @@ $$\begin{aligned}
&= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big)
\end{aligned}$$
-The parenthesized expression must be zero for all $r$,
+The parenthesized expression must be zero for all $$r$$,
leading us to the following relation:
$$\begin{aligned}
@@ -98,7 +98,7 @@ $$\begin{aligned}
\end{aligned}$$
This is simply the Laplace equation,
-which is as follows for our ansatz $p(r, \theta)$:
+which is as follows for our ansatz $$p(r, \theta)$$:
$$\begin{aligned}
0
@@ -115,8 +115,8 @@ $$\begin{aligned}
&= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big)
\end{aligned}$$
-Again, the parenthesized expression must be zero for all $r$,
-meaning it is an ODE for $q(r)$,
+Again, the parenthesized expression must be zero for all $$r$$,
+meaning it is an ODE for $$q(r)$$,
whose solution is straightforwardly found to be:
$$\begin{aligned}
@@ -124,7 +124,7 @@ $$\begin{aligned}
= \frac{C_3}{r^2} + C_4 r
\end{aligned}$$
-Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later).
+Where $$C_3$$ and $$C_4$$ are linearity constants ($$C_1$$ and $$C_2$$ appear later).
The pressure is therefore:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
= \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big)
\end{aligned}$$
-Consequently, its gradient $\nabla p$ in spherical coordinates is as follows:
+Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows:
$$\begin{aligned}
\nabla p
@@ -140,8 +140,8 @@ $$\begin{aligned}
= \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big)
\end{aligned}$$
-According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$.
-Let us look at the $r$-component of $\nabla^2 \va{v}$:
+According to the Stokes equation, this equals $$\eta \nabla^2 \va{v}$$.
+Let us look at the $$r$$-component of $$\nabla^2 \va{v}$$:
$$\begin{aligned}
(\nabla^2 \va{v})_r
@@ -154,22 +154,22 @@ $$\begin{aligned}
&= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big)
\end{aligned}$$
-Substituting $g$ for the expression we found from incompressibility lets us simplify this:
+Substituting $$g$$ for the expression we found from incompressibility lets us simplify this:
$$\begin{aligned}
\eta (\nabla^2 \va{v})_r
&= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
\end{aligned}$$
-The Stokes equation says that this must be equal to the $r$-component of $\nabla p$:
+The Stokes equation says that this must be equal to the $$r$$-component of $$\nabla p$$:
$$\begin{aligned}
\eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
= \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big)
\end{aligned}$$
-Where we have inserted $\idv{q}{r}$.
-Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$,
+Where we have inserted $$\idv{q}{r}$$.
+Dividing out $$\eta U \cos\theta$$ leaves an ODE for $$f(r)$$,
satisfied by:
$$\begin{aligned}
@@ -178,17 +178,17 @@ $$\begin{aligned}
\end{aligned}$$
Then, thanks to our earlier relation again,
-we know that $g(r)$ is as follows:
+we know that $$g(r)$$ is as follows:
$$\begin{aligned}
g(r)
= C_1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} + \frac{C_4 r^2}{5}
\end{aligned}$$
-So what about $C_1$, $C_2$, $C_3$ and $C_4$?
-For $r\!\to\!\infty$, we expect that $\va{v}\!\to\!\va{U}$,
-meaning that $f(r)\!\to\!1$ and $g(r)\!\to\!1$.
-This implies that $C_4 = 0$ and $C_1 = 1$, leaving:
+So what about $$C_1$$, $$C_2$$, $$C_3$$ and $$C_4$$?
+For $$r\!\to\!\infty$$, we expect that $$\va{v}\!\to\!\va{U}$$,
+meaning that $$f(r)\!\to\!1$$ and $$g(r)\!\to\!1$$.
+This implies that $$C_4 = 0$$ and $$C_1 = 1$$, leaving:
$$\begin{aligned}
f(r)
@@ -199,10 +199,10 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, the viscous *no-slip* condition demands
-that $\va{v} = 0$ at the sphere's surface $r = a$, so $f(a) = g(a) = 0$ there.
-Inserting $a$ into $f$ and $g$, setting them to zero,
+that $$\va{v} = 0$$ at the sphere's surface $$r = a$$, so $$f(a) = g(a) = 0$$ there.
+Inserting $$a$$ into $$f$$ and $$g$$, setting them to zero,
and solving the resulting system of equations
-yields $C_2 = a^3 / 2$ and $C_3 = -3 a / 2$.
+yields $$C_2 = a^3 / 2$$ and $$C_3 = -3 a / 2$$.
Therefore the full solution is:
$$\begin{gathered}
@@ -225,8 +225,8 @@ $$\begin{gathered}
From the definition of [viscosity](/know/concept/viscosity/),
we know that there must be shear stresses at the sphere surface,
-described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$.
-The drag force $\va{D}$ on the surface is:
+described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$.
+The drag force $$\va{D}$$ on the surface is:
$$\begin{aligned}
\va{D}
@@ -234,7 +234,7 @@ $$\begin{aligned}
= \int_0^{2\pi} \!\!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:a^2 \sin\theta \dd{\theta} \dd{\phi}
\end{aligned}$$
-Where $\vu{e}_r$ is the sphere's surface normal vector.
+Where $$\vu{e}_r$$ is the sphere's surface normal vector.
The integrand can be expanded as follows:
$$\begin{aligned}
@@ -242,7 +242,7 @@ $$\begin{aligned}
= \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r}
\end{aligned}$$
-To calculate this, we start by taking the gradient of the velocity field $\va{v}$:
+To calculate this, we start by taking the gradient of the velocity field $$\va{v}$$:
$$\begin{aligned}
\nabla\va{v}
@@ -253,7 +253,7 @@ $$\begin{aligned}
+ \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big)
\end{aligned}$$
-Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$:
+Some of these terms are necessary to calculate the stress elements $$\sigma_{rr}$$ and $$\sigma_{\theta r}$$:
$$\begin{aligned}
\sigma_{rr}
@@ -264,6 +264,7 @@ $$\begin{aligned}
\\
&= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big)
\end{aligned}$$
+
$$\begin{aligned}
\sigma_{\theta r}
&= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big)
@@ -276,7 +277,7 @@ $$\begin{aligned}
&= - \frac{3 \eta U a^3}{2 r^4} \sin\theta
\end{aligned}$$
-At the sphere's surface we set $r = a$, so these expressions reduce to the following:
+At the sphere's surface we set $$r = a$$, so these expressions reduce to the following:
$$\begin{aligned}
\sigma_{rr}
@@ -287,7 +288,7 @@ $$\begin{aligned}
\end{aligned}$$
Now we can finally calculate the effective stress on the surface,
-by converting the basis vectors $\vu{e}_r$ and $\vu{e}_\theta$ to Cartesian coordinates:
+by converting the basis vectors $$\vu{e}_r$$ and $$\vu{e}_\theta$$ to Cartesian coordinates:
$$\begin{aligned}
\hat{\sigma} \cdot \vu{e}_r
@@ -301,11 +302,11 @@ $$\begin{aligned}
= \vu{e}_z \frac{3 \eta U}{2 a}
\end{aligned}$$
-Remarkably, the stress at every point on the sphere is purely in the $z$-direction!
+Remarkably, the stress at every point on the sphere is purely in the $$z$$-direction!
This is not entirely unexpected though: symmetry cancels out all other components.
-With this, we can do the integrals for $\va{D}$,
-which reduce to a surface area factor $4 \pi a^2$:
+With this, we can do the integrals for $$\va{D}$$,
+which reduce to a surface area factor $$4 \pi a^2$$:
$$\begin{aligned}
\va{D}
@@ -315,7 +316,7 @@ $$\begin{aligned}
\end{aligned}$$
At last, we arrive at Stokes' law,
-which simply expresses the magnitude of $\va{D}$:
+which simply expresses the magnitude of $$\va{D}$$:
$$\begin{aligned}
\boxed{
@@ -327,8 +328,8 @@ $$\begin{aligned}
To arrive at this result,
we assumed that the sphere was fixed, and the fluid was flowing past it.
We can equally well let the fluid be at rest,
-with the sphere falling through it at $U$.
-The force of gravity then exerts the following force $G$ on it,
+with the sphere falling through it at $$U$$.
+The force of gravity then exerts the following force $$G$$ on it,
subtracting [buoyancy](/know/concept/archimedes-principle/):
$$\begin{aligned}
@@ -336,9 +337,9 @@ $$\begin{aligned}
= \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
\end{aligned}$$
-Where $\rho_s$ and $\rho_f$ are the sphere's and fluid's densities,
-and $g_0$ is the gravitational acceleration.
-Since $D$ acts in the opposite sense of $G$,
+Where $$\rho_s$$ and $$\rho_f$$ are the sphere's and fluid's densities,
+and $$g_0$$ is the gravitational acceleration.
+Since $$D$$ acts in the opposite sense of $$G$$,
after some time, they cancel out:
$$\begin{aligned}
@@ -346,7 +347,7 @@ $$\begin{aligned}
= \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
\end{aligned}$$
-This is an equation for the **terminal velocity** $U_t$,
+This is an equation for the **terminal velocity** $$U_t$$,
which we find to be as follows:
$$\begin{aligned}
@@ -356,7 +357,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The falling sphere will accelerate until $U_t$,
+The falling sphere will accelerate until $$U_t$$,
and then continue falling at constant speed.
diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md
index 23922d5..75daae3 100644
--- a/source/know/concept/sturm-liouville-theory/index.md
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -21,23 +21,23 @@ of eigenfunctions.
## General operator
Consider the most general form of a second-order linear
-differential operator $\hat{L}$, where $p_0(x)$, $p_1(x)$, and $p_2(x)$
-are real functions of $x \in [a,b]$ which are non-zero for all $x \in ]a, b[$:
+differential operator $$\hat{L}$$, where $$p_0(x)$$, $$p_1(x)$$, and $$p_2(x)$$
+are real functions of $$x \in [a,b]$$ which are non-zero for all $$x \in ]a, b[$$:
$$\begin{aligned}
\hat{L} \{u(x)\} = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x)
\end{aligned}$$
We now define the **adjoint** or **Hermitian** operator
-$\hat{L}^\dagger$ analogously to matrices:
+$$\hat{L}^\dagger$$ analogously to matrices:
$$\begin{aligned}
\inprod{f}{\hat{L} g}
= \inprod{\hat{L}^\dagger f}{g}
\end{aligned}$$
-What is $\hat{L}^\dagger$, given the above definition of $\hat{L}$?
-We start from the inner product $\inprod{f}{\hat{L} g}$:
+What is $$\hat{L}^\dagger$$, given the above definition of $$\hat{L}$$?
+We start from the inner product $$\inprod{f}{\hat{L} g}$$:
$$\begin{aligned}
\inprod{f}{\hat{L} g}
@@ -51,7 +51,7 @@ $$\begin{aligned}
&= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x}
\end{aligned}$$
-We now have an expression for $\hat{L}^\dagger$, but are left with an
+We now have an expression for $$\hat{L}^\dagger$$, but are left with an
annoying boundary term:
$$\begin{aligned}
@@ -60,8 +60,8 @@ $$\begin{aligned}
\end{aligned}$$
To fix this,
-let us demand that $p_1(x) = p_0'(x)$ and that
-$[p_0(f^* g' - (f^*)' g)]_a^b = 0$, leaving:
+let us demand that $$p_1(x) = p_0'(x)$$ and that
+$$[p_0(f^* g' - (f^*)' g)]_a^b = 0$$, leaving:
$$\begin{aligned}
\inprod{f}{\hat{L} g}
@@ -69,8 +69,8 @@ $$\begin{aligned}
= \inprod{\hat{L}^\dagger f}{g}
\end{aligned}$$
-Using the aforementioned restriction $p_1(x) = p_0'(x)$,
-we then take a look at the definition of $\hat{L}^\dagger$:
+Using the aforementioned restriction $$p_1(x) = p_0'(x)$$,
+we then take a look at the definition of $$\hat{L}^\dagger$$:
$$\begin{aligned}
\hat{L}^\dagger \{f\}
@@ -83,7 +83,7 @@ $$\begin{aligned}
&= (p_0 f')' + p_2 f
\end{aligned}$$
-The original operator $\hat{L}$ reduces to the same form,
+The original operator $$\hat{L}$$ reduces to the same form,
so it is **self-adjoint**:
$$\begin{aligned}
@@ -93,19 +93,19 @@ $$\begin{aligned}
= \hat{L}^\dagger \{f\}
\end{aligned}$$
-Consequently, every such second-order linear operator $\hat{L}$ is self-adjoint,
-as long as it satisfies the constraints $p_1(x) = p_0'(x)$ and $[p_0 (f^* g' - (f^*)' g)]_a^b = 0$.
+Consequently, every such second-order linear operator $$\hat{L}$$ is self-adjoint,
+as long as it satisfies the constraints $$p_1(x) = p_0'(x)$$ and $$[p_0 (f^* g' - (f^*)' g)]_a^b = 0$$.
Let us ignore the latter constraint for now (it will return later),
-and focus on the former: what if $\hat{L}$ does not satisfy $p_0' \neq p_1$?
-We multiply it by an unknown $p(x) \neq 0$, and divide by $p_0(x) \neq 0$:
+and focus on the former: what if $$\hat{L}$$ does not satisfy $$p_0' \neq p_1$$?
+We multiply it by an unknown $$p(x) \neq 0$$, and divide by $$p_0(x) \neq 0$$:
$$\begin{aligned}
\frac{p(x)}{p_0(x)} \hat{L} \{u\} = p(x) u'' + p(x) \frac{p_1(x)}{p_0(x)} u' + p(x) \frac{p_2(x)}{p_0(x)} u
\end{aligned}$$
-We now define $q(x)$,
-and demand that the derivative $p'(x)$ of the unknown $p(x)$ satisfies:
+We now define $$q(x)$$,
+and demand that the derivative $$p'(x)$$ of the unknown $$p(x)$$ satisfies:
$$\begin{aligned}
q(x) = p(x) \frac{p_2(x)}{p_0(x)}
@@ -113,7 +113,7 @@ $$\begin{aligned}
p'(x) = p(x) \frac{p_1(x)}{p_0(x)}
\end{aligned}$$
-The latter is a differential equation for $p(x)$, which we solve by integration:
+The latter is a differential equation for $$p(x)$$, which we solve by integration:
$$\begin{gathered}
\frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x}
@@ -128,7 +128,7 @@ $$\begin{gathered}
p(x) = p(a) \exp\!\Big( \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} \Big)
\end{gathered}$$
-Now that we have $p(x)$ and $q(x)$, we can define a new operator $\hat{L}_p$ as follows:
+Now that we have $$p(x)$$ and $$q(x)$$, we can define a new operator $$\hat{L}_p$$ as follows:
$$\begin{aligned}
\hat{L}_p \{u\}
@@ -138,11 +138,11 @@ $$\begin{aligned}
\end{aligned}$$
This is the self-adjoint form from earlier!
-So even if $p_0' \neq p_1$, any second-order linear operator with $p_0(x) \neq 0$
+So even if $$p_0' \neq p_1$$, any second-order linear operator with $$p_0(x) \neq 0$$
can easily be put in self-adjoint form.
-This general form is known as the **Sturm-Liouville operator** $\hat{L}_{SL}$,
-where $p(x)$ and $q(x)$ are non-zero real functions of the variable $x \in [a,b]$:
+This general form is known as the **Sturm-Liouville operator** $$\hat{L}_{SL}$$,
+where $$p(x)$$ and $$q(x)$$ are non-zero real functions of the variable $$x \in [a,b]$$:
$$\begin{aligned}
\boxed{
@@ -156,8 +156,8 @@ $$\begin{aligned}
## Eigenvalue problem
A **Sturm-Liouville problem** (SLP) is analogous to a matrix eigenvalue problem,
-where $w(x)$ is a real weight function, $\lambda$ is the **eigenvalue**,
-and $u(x)$ is the corresponding **eigenfunction**:
+where $$w(x)$$ is a real weight function, $$\lambda$$ is the **eigenvalue**,
+and $$u(x)$$ is the corresponding **eigenfunction**:
$$\begin{aligned}
\boxed{
@@ -165,14 +165,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Necessarily, $w(x) > 0$ except in isolated points, where $w(x) = 0$ is allowed;
-the point is that any inner product $\Inprod{f}{w g}$ may never be zero due to $w$'s fault.
-Furthermore, the convention is that $u(x)$ cannot be trivially zero.
+Necessarily, $$w(x) > 0$$ except in isolated points, where $$w(x) = 0$$ is allowed;
+the point is that any inner product $$\Inprod{f}{w g}$$ may never be zero due to $$w$$'s fault.
+Furthermore, the convention is that $$u(x)$$ cannot be trivially zero.
-In our derivation of $\hat{L}_{SL}$,
+In our derivation of $$\hat{L}_{SL}$$,
we removed a boundary term to get self-adjointness.
Consequently, to have a valid SLP, the boundary conditions for
-$u(x)$ must be as follows, otherwise the operator cannot be self-adjoint:
+$$u(x)$$ must be as follows, otherwise the operator cannot be self-adjoint:
$$\begin{aligned}
\Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0
@@ -181,16 +181,16 @@ $$\begin{aligned}
There are many boundary conditions (BCs) which satisfy this requirement.
Some notable ones are listed here non-exhaustively:
-+ **Dirichlet BCs**: $u(a) = u(b) = 0$
-+ **Neumann BCs**: $u'(a) = u'(b) = 0$
-+ **Robin BCs**: $\alpha_1 u(a) + \beta_1 u'(a) = \alpha_2 u(b) + \beta_2 u'(b) = 0$ with $\alpha_{1,2}, \beta_{1,2} \in \mathbb{R}$
-+ **Periodic BCs**: $p(a) = p(b)$, $u(a) = u(b)$, and $u'(a) = u'(b)$
-+ **Legendre "BCs"**: $p(a) = p(b) = 0$
++ **Dirichlet BCs**: $$u(a) = u(b) = 0$$
++ **Neumann BCs**: $$u'(a) = u'(b) = 0$$
++ **Robin BCs**: $$\alpha_1 u(a) + \beta_1 u'(a) = \alpha_2 u(b) + \beta_2 u'(b) = 0$$ with $$\alpha_{1,2}, \beta_{1,2} \in \mathbb{R}$$
++ **Periodic BCs**: $$p(a) = p(b)$$, $$u(a) = u(b)$$, and $$u'(a) = u'(b)$$
++ **Legendre "BCs"**: $$p(a) = p(b) = 0$$
Once this requirement is satisfied, Sturm-Liouville theory gives us
-some very useful information about $\lambda$ and $u(x)$.
+some very useful information about $$\lambda$$ and $$u(x)$$.
From the definition of an SLP, we know that, given two arbitrary (and possibly identical)
-eigenfunctions $u_n$ and $u_m$, the following must be satisfied:
+eigenfunctions $$u_n$$ and $$u_m$$, the following must be satisfied:
$$\begin{aligned}
0 = \hat{L}_{SL}\{u_n\} + \lambda_n w u_n = \hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*
@@ -215,7 +215,7 @@ $$\begin{aligned}
&= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
\end{aligned}$$
-The operator $\hat{L}_{SL}$ is self-adjoint by definition,
+The operator $$\hat{L}_{SL}$$ is self-adjoint by definition,
so the left-hand side vanishes, leaving us with:
$$\begin{aligned}
@@ -223,20 +223,20 @@ $$\begin{aligned}
&= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
\end{aligned}$$
-When $m = n$, the inner product $\Inprod{u_n}{w u_n}$ is real and positive
-(assuming $u_n$ is not trivially zero, in which case it would be disqualified anyway).
-In this case we thus know that $\lambda_n^* = \lambda_n$,
-i.e. the eigenvalue $\lambda_n$ is real for any $n$.
+When $$m = n$$, the inner product $$\Inprod{u_n}{w u_n}$$ is real and positive
+(assuming $$u_n$$ is not trivially zero, in which case it would be disqualified anyway).
+In this case we thus know that $$\lambda_n^* = \lambda_n$$,
+i.e. the eigenvalue $$\lambda_n$$ is real for any $$n$$.
-When $m \neq n$, then $\lambda_m^* - \lambda_n$ may or may not be zero,
+When $$m \neq n$$, then $$\lambda_m^* - \lambda_n$$ may or may not be zero,
depending on the degeneracy. If there is no degeneracy, we
-see that $\Inprod{u_m}{w u_n} = 0$, i.e. the eigenfunctions are orthogonal.
+see that $$\Inprod{u_m}{w u_n} = 0$$, i.e. the eigenfunctions are orthogonal.
In case of degeneracy, manual orthogonalization is needed, but as it turns out,
this is guaranteed to be doable, using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method/).
-In conclusion, **a Sturm-Liouville problem has real eigenvalues $\lambda$,
-and all the corresponding eigenfunctions $u(x)$ are mutually orthogonal**:
+In conclusion, **a Sturm-Liouville problem has real eigenvalues $$\lambda$$,
+and all the corresponding eigenfunctions $$u(x)$$ are mutually orthogonal**:
$$\begin{aligned}
\boxed{
@@ -252,16 +252,16 @@ so it is always worth checking whether you are dealing with an SLP.
Another useful fact of SLPs is that they always
have an infinite number of discrete eigenvalues.
-Furthermore, the eigenvalues always ascend to $+\infty$;
-in other words, there always exists a *lowest* eigenvalue $\lambda_0 > -\infty$,
+Furthermore, the eigenvalues always ascend to $$+\infty$$;
+in other words, there always exists a *lowest* eigenvalue $$\lambda_0 > -\infty$$,
known as the **ground state**.
## Completeness
-Not only are the eigenfunctions $u_n(x)$ of an SLP orthogonal, they
-also form a **complete basis**, meaning that any well-behaved function $f(x)$ can be
-expanded as a **generalized Fourier series** with coefficients $a_n$:
+Not only are the eigenfunctions $$u_n(x)$$ of an SLP orthogonal, they
+also form a **complete basis**, meaning that any well-behaved function $$f(x)$$ can be
+expanded as a **generalized Fourier series** with coefficients $$a_n$$:
$$\begin{aligned}
\boxed{
@@ -271,12 +271,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-This series will converge significantly faster if $f(x)$
-satisfies the same BCs as $u_n(x)$. In that case the
-expansion will even be valid for the inclusive interval $x \in [a, b]$.
+This series will converge significantly faster if $$f(x)$$
+satisfies the same BCs as $$u_n(x)$$. In that case the
+expansion will even be valid for the inclusive interval $$x \in [a, b]$$.
-To find an expression for the coefficients $a_n$,
-we multiply the above generalized Fourier series by $w(x) u_m^*(x)$ for an arbitrary $m$:
+To find an expression for the coefficients $$a_n$$,
+we multiply the above generalized Fourier series by $$w(x) u_m^*(x)$$ for an arbitrary $$m$$:
$$\begin{aligned}
f(x) w(x) u_m^*(x)
@@ -303,9 +303,9 @@ $$\begin{aligned}
= a_m A_m
\end{aligned}$$
-After isolating this for $a_n$, we see that
+After isolating this for $$a_n$$, we see that
the coefficients are given by the projection of the target
-function $f(x)$ onto the normalized eigenfunctions $u_n(x) / A_n$:
+function $$f(x)$$ onto the normalized eigenfunctions $$u_n(x) / A_n$$:
$$\begin{aligned}
\boxed{
@@ -317,7 +317,7 @@ $$\begin{aligned}
As a final remark, we can see something interesting
by rearranging the generalized Fourier series
-after inserting the expression for $a_n$:
+after inserting the expression for $$a_n$$:
$$\begin{aligned}
f(x)
@@ -328,7 +328,7 @@ $$\begin{aligned}
\end{aligned}$$
Upon closer inspection, the parenthesized summation
-must be the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$
+must be the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$
for the integral to work out.
This is in fact the underlying requirement for completeness:
diff --git a/source/know/concept/superdense-coding/index.md b/source/know/concept/superdense-coding/index.md
index 5c1e4ca..ba6e898 100644
--- a/source/know/concept/superdense-coding/index.md
+++ b/source/know/concept/superdense-coding/index.md
@@ -20,48 +20,78 @@ She could send a qubit, which has a larger state space than a classical bit,
but it can only be measured once, thereby yielding only one bit of data.
However, they are already sharing an entangled pair of qubits
-in the [Bell state](/know/concept/bell-state/) $\ket{\Phi^{+}}_{AB}$,
-where $A$ and $B$ are qubits belonging to Alice and Bob, respectively.
+in the [Bell state](/know/concept/bell-state/) $$\ket{\Phi^{+}}_{AB}$$,
+where $$A$$ and $$B$$ are qubits belonging to Alice and Bob, respectively.
-Based on the values of the two classical bits $(a_1, a_2)$,
-Alice performs the following operations on her side $A$
+Based on the values of the two classical bits $$(a_1, a_2)$$,
+Alice performs the following operations on her side $$A$$
of the Bell state:
-Where $U_f$ is a phase oracle,
+Where $$U_f$$ is a phase oracle,
whose action is defined as follows,
-where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$:
+where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$:
$$\begin{aligned}
\Ket{x}
@@ -51,14 +51,14 @@ $$\begin{aligned}
= (-1)^{s \cdot x} \Ket{x}
\end{aligned}$$
-That is, it introduces a phase flip based on the value of $f(x)$.
+That is, it introduces a phase flip based on the value of $$f(x)$$.
For an example implementation of such an oracle,
see the Deutsch-Jozsa algorithm:
its circuit is identical to this one,
-but describes $U_f$ in a different (but equivalent) way.
+but describes $$U_f$$ in a different (but equivalent) way.
-Starting from the state $\Ket{0}^{\otimes N}$,
-applying the [Hadamard gate](/know/concept/quantum-gate/) $H$
+Starting from the state $$\Ket{0}^{\otimes N}$$,
+applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
to all qubits yields:
$$\begin{aligned}
@@ -68,7 +68,7 @@ $$\begin{aligned}
= \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x}
\end{aligned}$$
-This is an equal superposition of all candidates $\Ket{x}$,
+This is an equal superposition of all candidates $$\Ket{x}$$,
which we feed to the oracle:
$$\begin{aligned}
@@ -78,7 +78,7 @@ $$\begin{aligned}
\end{aligned}$$
Then, thanks to the definition of the Hadamard transform,
-a final set of $H$-gates leads us to:
+a final set of $$H$$-gates leads us to:
$$\begin{aligned}
\frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \Ket{x}
@@ -87,9 +87,9 @@ $$\begin{aligned}
= \Ket{s_1} \cdots \Ket{s_N}
\end{aligned}$$
-Which, upon measurement, gives us the desired binary representation of $s$.
-For comparison, the Deutsch-Jozsa algorithm only cares whether $s = 0$ or $s \neq 0$,
-whereas this algorithm is interested in the exact value of $s$.
+Which, upon measurement, gives us the desired binary representation of $$s$$.
+For comparison, the Deutsch-Jozsa algorithm only cares whether $$s = 0$$ or $$s \neq 0$$,
+whereas this algorithm is interested in the exact value of $$s$$.
diff --git a/source/know/concept/berry-phase/index.md b/source/know/concept/berry-phase/index.md
index eedc548..d237ea5 100644
--- a/source/know/concept/berry-phase/index.md
+++ b/source/know/concept/berry-phase/index.md
@@ -8,8 +8,8 @@ categories:
layout: "concept"
---
-Consider a Hamiltonian $\hat{H}$ that does not explicitly depend on time,
-but does depend on a given parameter $\vb{R}$.
+Consider a Hamiltonian $$\hat{H}$$ that does not explicitly depend on time,
+but does depend on a given parameter $$\vb{R}$$.
The Schrödinger equations then read:
$$\begin{aligned}
@@ -20,9 +20,9 @@ $$\begin{aligned}
&= E_n(\vb{R}) \Ket{\psi_n(\vb{R})}
\end{aligned}$$
-The general full solution $\Ket{\Psi_n}$ has the following form,
-where we allow $\vb{R}$ to evolve in time,
-and we have abbreviated the traditional phase of the "wiggle factor" as $L_n$:
+The general full solution $$\Ket{\Psi_n}$$ has the following form,
+where we allow $$\vb{R}$$ to evolve in time,
+and we have abbreviated the traditional phase of the "wiggle factor" as $$L_n$$:
$$\begin{aligned}
\Ket{\Psi_n(t)}
@@ -31,11 +31,11 @@ $$\begin{aligned}
L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'}
\end{aligned}$$
-The **geometric phase** $\gamma_n(t)$ is more interesting.
-It is not included in $\Ket{\psi_n}$,
-because it depends on the path $\vb{R}(t)$
-rather than only the present $\vb{R}$ and $t$.
-Its dynamics can be found by inserting the above $\Ket{\Psi_n}$
+The **geometric phase** $$\gamma_n(t)$$ is more interesting.
+It is not included in $$\Ket{\psi_n}$$,
+because it depends on the path $$\vb{R}(t)$$
+rather than only the present $$\vb{R}$$ and $$t$$.
+Its dynamics can be found by inserting the above $$\Ket{\Psi_n}$$
into the time-dependent Schrödinger equation:
$$\begin{aligned}
@@ -58,8 +58,8 @@ $$\begin{aligned}
&= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
\end{aligned}$$
-Front-multiplying by $i \Bra{\Psi_n}$ gives us
-the equation of motion of the geometric phase $\gamma_n$:
+Front-multiplying by $$i \Bra{\Psi_n}$$ gives us
+the equation of motion of the geometric phase $$\gamma_n$$:
$$\begin{aligned}
\boxed{
@@ -68,7 +68,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows:
+Where we have defined the so-called **Berry connection** $$\vb{A}_n$$ as follows:
$$\begin{aligned}
\boxed{
@@ -77,9 +77,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Importantly, note that $\vb{A}_n$ is real,
-provided that $\Ket{\psi_n}$ is always normalized for all $\vb{R}$.
-To prove this, we start from the fact that $\nabla_\vb{R} 1 = 0$:
+Importantly, note that $$\vb{A}_n$$ is real,
+provided that $$\Ket{\psi_n}$$ is always normalized for all $$\vb{R}$$.
+To prove this, we start from the fact that $$\nabla_\vb{R} 1 = 0$$:
$$\begin{aligned}
0
@@ -91,14 +91,14 @@ $$\begin{aligned}
= 2 \Imag\{ \vb{A}_n \}
\end{aligned}$$
-Consequently, $\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is always real,
-because $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary.
+Consequently, $$\vb{A}_n = \Imag \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is always real,
+because $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary.
-Suppose now that the parameter $\vb{R}(t)$ is changed adiabatically
+Suppose now that the parameter $$\vb{R}(t)$$ is changed adiabatically
(i.e. so slow that the system stays in the same eigenstate)
-for $t \in [0, T]$, along a circuit $C$ with $\vb{R}(0) \!=\! \vb{R}(T)$.
-Integrating the phase $\gamma_n(t)$ over this contour $C$ then yields
-the **Berry phase** $\gamma_n(C)$:
+for $$t \in [0, T]$$, along a circuit $$C$$ with $$\vb{R}(0) \!=\! \vb{R}(T)$$.
+Integrating the phase $$\gamma_n(t)$$ over this contour $$C$$ then yields
+the **Berry phase** $$\gamma_n(C)$$:
$$\begin{aligned}
\boxed{
@@ -107,9 +107,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-But we have a problem: $\vb{A}_n$ is not unique!
+But we have a problem: $$\vb{A}_n$$ is not unique!
Due to the Schrödinger equation's gauge invariance,
-any function $f(\vb{R}(t))$ can be added to $\gamma_n(t)$
+any function $$f(\vb{R}(t))$$ can be added to $$\gamma_n(t)$$
without making an immediate physical difference to the state.
Consider the following general gauge transformation:
@@ -118,9 +118,9 @@ $$\begin{aligned}
\equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})}
\end{aligned}$$
-To find $\vb{A}_n$ for a particular choice of $f$,
+To find $$\vb{A}_n$$ for a particular choice of $$f$$,
we need to evaluate the inner product
-$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:
+$$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$$:
$$\begin{aligned}
\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}
@@ -131,16 +131,16 @@ $$\begin{aligned}
&= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
\end{aligned}$$
-Unfortunately, $f$ does not vanish as we would have liked,
-so $\vb{A}_n$ depends on our choice of $f$.
+Unfortunately, $$f$$ does not vanish as we would have liked,
+so $$\vb{A}_n$$ depends on our choice of $$f$$.
However, the curl of a gradient is always zero,
-so although $\vb{A}_n$ is not unique,
-its curl $\nabla_\vb{R} \cross \vb{A}_n$ is guaranteed to be.
-Conveniently, we can introduce a curl in the definition of $\gamma_n(C)$
+so although $$\vb{A}_n$$ is not unique,
+its curl $$\nabla_\vb{R} \cross \vb{A}_n$$ is guaranteed to be.
+Conveniently, we can introduce a curl in the definition of $$\gamma_n(C)$$
by applying Stokes' theorem, under the assumption
-that $\vb{A}_n$ has no singularities in the area enclosed by $C$
-(fortunately, $\vb{A}_n$ can always be chosen to satisfy this):
+that $$\vb{A}_n$$ has no singularities in the area enclosed by $$C$$
+(fortunately, $$\vb{A}_n$$ can always be chosen to satisfy this):
$$\begin{aligned}
\boxed{
@@ -149,10 +149,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we defined $\vb{B}_n$ as the curl of $\vb{A}_n$.
-Now $\gamma_n(C)$ is guaranteed to be unique.
-Note that $\vb{B}_n$ is analogous to a magnetic field,
-and $\vb{A}_n$ to a magnetic vector potential:
+Where we defined $$\vb{B}_n$$ as the curl of $$\vb{A}_n$$.
+Now $$\gamma_n(C)$$ is guaranteed to be unique.
+Note that $$\vb{B}_n$$ is analogous to a magnetic field,
+and $$\vb{A}_n$$ to a magnetic vector potential:
$$\begin{aligned}
\vb{B}_n(\vb{R})
@@ -160,9 +160,9 @@ $$\begin{aligned}
= \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\}
\end{aligned}$$
-Unfortunately, $\nabla_\vb{R} \psi_n$ is difficult to evaluate explicitly,
-so we would like to rewrite $\vb{B}_n$ such that it does not enter.
-We do this as follows, inserting $1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$ along the way:
+Unfortunately, $$\nabla_\vb{R} \psi_n$$ is difficult to evaluate explicitly,
+so we would like to rewrite $$\vb{B}_n$$ such that it does not enter.
+We do this as follows, inserting $$1 = \sum_{m} \Ket{\psi_m} \Bra{\psi_m}$$ along the way:
$$\begin{aligned}
i \vb{B}_n
@@ -172,9 +172,9 @@ $$\begin{aligned}
&= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
\end{aligned}$$
-The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary
+The fact that $$\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$$ is imaginary
means it is parallel to its complex conjugate,
-and thus the cross product vanishes, so we exclude $n$ from the sum:
+and thus the cross product vanishes, so we exclude $$n$$ from the sum:
$$\begin{aligned}
\vb{B}_n
@@ -200,8 +200,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Which only involves $\nabla_\vb{R} \hat{H}$,
-and is therefore easier to evaluate than any $\Ket{\nabla_\vb{R} \psi_n}$.
+Which only involves $$\nabla_\vb{R} \hat{H}$$,
+and is therefore easier to evaluate than any $$\Ket{\nabla_\vb{R} \psi_n}$$.
diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md
index 14ba4cb..1193a93 100644
--- a/source/know/concept/binomial-distribution/index.md
+++ b/source/know/concept/binomial-distribution/index.md
@@ -9,11 +9,11 @@ layout: "concept"
---
The **binomial distribution** is a discrete probability distribution
-describing a **Bernoulli process**: a set of independent $N$ trials where
+describing a **Bernoulli process**: a set of independent $$N$$ trials where
each has only two possible outcomes, "success" and "failure",
-the former with probability $p$ and the latter with $q = 1 - p$.
+the former with probability $$p$$ and the latter with $$q = 1 - p$$.
The binomial distribution then gives the probability
-that $n$ out of the $N$ trials succeed:
+that $$n$$ out of the $$N$$ trials succeed:
$$\begin{aligned}
\boxed{
@@ -22,8 +22,8 @@ $$\begin{aligned}
\end{aligned}$$
The first factor is known as the **binomial coefficient**, which describes the
-number of microstates (i.e. permutations) that have $n$ successes out of $N$ trials.
-These happen to be the coefficients in the polynomial $(a + b)^N$,
+number of microstates (i.e. permutations) that have $$n$$ successes out of $$N$$ trials.
+These happen to be the coefficients in the polynomial $$(a + b)^N$$,
and can be read off of Pascal's triangle.
It is defined as follows:
@@ -33,10 +33,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-The remaining factor $p^n (1 - p)^{N - n}$ is then just the
+The remaining factor $$p^n (1 - p)^{N - n}$$ is then just the
probability of attaining each microstate.
-The expected or mean number of successes $\mu$ after $N$ trials is as follows:
+The expected or mean number of successes $$\mu$$ after $$N$$ trials is as follows:
$$\begin{aligned}
\boxed{
@@ -49,7 +49,7 @@ $$\begin{aligned}
-The $x$, $y$ and $z$-axes represent the components of a spin-1/2-alike system,
+The $$x$$, $$y$$ and $$z$$-axes represent the components of a spin-1/2-alike system,
and their extremes are the eigenstates of the Pauli matrices:
$$\begin{aligned}
@@ -31,7 +31,7 @@ $$\begin{aligned}
\to \{\Ket{+i}, \Ket{-i}\}
\end{aligned}$$
-Where the latter two states are expressed as follows in the conventional $z$-basis:
+Where the latter two states are expressed as follows in the conventional $$z$$-basis:
$$\begin{aligned}
\Ket{\pm}
@@ -42,15 +42,15 @@ $$\begin{aligned}
\end{aligned}$$
More generally, every point on the surface of the sphere
-describes a pure qubit state in terms of the angles $\theta$ and $\varphi$,
+describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
respectively the elevation and azimuth:
$$\begin{aligned}
\Ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \Ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \Ket{1}
\end{aligned}$$
-We can generalize this further by describing points using the **Bloch vector** $\vec{r}$,
-with radius $r \le 1$:
+We can generalize this further by describing points using the **Bloch vector** $$\vec{r}$$,
+with radius $$r \le 1$$:
$$\begin{aligned}
\boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that $\vec{r}$ is not actually a qubit state,
+Note that $$\vec{r}$$ is not actually a qubit state,
but rather an implicit description of one,
meaning that it does not need to be normalized.
The main point of the Bloch vector is that it allows us
@@ -73,9 +73,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$ is the Pauli "vector".
-Now, we know that $\hat{\rho}$ represents a pure ensemble
-if and only if it is idempotent, i.e. $\hat{\rho}^2 = \hat{\rho}$:
+Where $$\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$$ is the Pauli "vector".
+Now, we know that $$\hat{\rho}$$ represents a pure ensemble
+if and only if it is idempotent, i.e. $$\hat{\rho}^2 = \hat{\rho}$$:
$$\begin{aligned}
\hat{\rho}^2
@@ -83,8 +83,8 @@ $$\begin{aligned}
= \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
\end{aligned}$$
-You can easily convince yourself that if $(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$,
-then we get $\hat{\rho}$ again, and the state is pure:
+You can easily convince yourself that if $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
+then we get $$\hat{\rho}$$ again, and the state is pure:
$$\begin{aligned}
(\vec{r} \cdot \vec{\sigma})^2
@@ -105,13 +105,13 @@ $$\begin{aligned}
= r^2 \hat{I}
\end{aligned}$$
-Therefore, if the radius $r = 1$, the ensemble is pure,
-else if $r < 1$ it is mixed.
+Therefore, if the radius $$r = 1$$, the ensemble is pure,
+else if $$r < 1$$ it is mixed.
Another useful property of the Bloch vector
is that the expectation value of the Pauli matrices
-are given by the corresponding component of $\vec{r}$,
-for example for $\hat{\sigma}_z$:
+are given by the corresponding component of $$\vec{r}$$,
+for example for $$\hat{\sigma}_z$$:
$$\begin{aligned}
\Expval{\hat{\sigma}_z}
diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
index e049a71..6f445f1 100644
--- a/source/know/concept/blochs-theorem/index.md
+++ b/source/know/concept/blochs-theorem/index.md
@@ -8,14 +8,14 @@ layout: "concept"
---
In quantum mechanics, **Bloch's theorem** states that,
-given a potential $V(\vb{r})$ which is periodic on a lattice,
-i.e. $V(\vb{r}) = V(\vb{r} + \vb{a})$
-for a primitive lattice vector $\vb{a}$,
-then it follows that the solutions $\psi(\vb{r})$
+given a potential $$V(\vb{r})$$ which is periodic on a lattice,
+i.e. $$V(\vb{r}) = V(\vb{r} + \vb{a})$$
+for a primitive lattice vector $$\vb{a}$$,
+then it follows that the solutions $$\psi(\vb{r})$$
to the time-independent Schrödinger equation
take the following form,
-where the function $u(\vb{r})$ is periodic on the same lattice,
-i.e. $u(\vb{r}) = u(\vb{r} + \vb{a})$:
+where the function $$u(\vb{r})$$ is periodic on the same lattice,
+i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$:
$$
\begin{aligned}
@@ -30,8 +30,8 @@ the solutions are simply plane waves with a periodic modulation,
known as **Bloch functions** or **Bloch states**.
This is suprisingly easy to prove:
-if the Hamiltonian $\hat{H}$ is lattice-periodic,
-then both $\psi(\vb{r})$ and $\psi(\vb{r} + \vb{a})$
+if the Hamiltonian $$\hat{H}$$ is lattice-periodic,
+then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$
are eigenstates with the same energy:
$$
@@ -42,8 +42,8 @@ $$
\end{aligned}
$$
-Now define the unitary translation operator $\hat{T}(\vb{a})$ such that
-$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$.
+Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that
+$$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$$.
From the previous equation, we then know that:
$$
@@ -55,10 +55,10 @@ $$
\end{aligned}
$$
-In other words, if $\hat{H}$ is lattice-periodic,
-then it will commute with $\hat{T}(\vb{a})$,
-i.e. $[\hat{H}, \hat{T}(\vb{a})] = 0$.
-Consequently, $\hat{H}$ and $\hat{T}(\vb{a})$ must share eigenstates $\psi(\vb{r})$:
+In other words, if $$\hat{H}$$ is lattice-periodic,
+then it will commute with $$\hat{T}(\vb{a})$$,
+i.e. $$[\hat{H}, \hat{T}(\vb{a})] = 0$$.
+Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ must share eigenstates $$\psi(\vb{r})$$:
$$
\begin{aligned}
@@ -68,10 +68,10 @@ $$
\end{aligned}
$$
-Since $\hat{T}$ is unitary,
-its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real.
-Therefore a translation by $\vb{a}$ causes a phase shift,
-for some vector $\vb{k}$:
+Since $$\hat{T}$$ is unitary,
+its eigenvalues $$\tau$$ must have the form $$e^{i \theta}$$, with $$\theta$$ real.
+Therefore a translation by $$\vb{a}$$ causes a phase shift,
+for some vector $$\vb{k}$$:
$$
\begin{aligned}
@@ -83,7 +83,7 @@ $$
$$
Let us now define the following function,
-keeping our arbitrary choice of $\vb{k}$:
+keeping our arbitrary choice of $$\vb{k}$$:
$$
\begin{aligned}
@@ -92,7 +92,7 @@ $$
\end{aligned}
$$
-As it turns out, this function is guaranteed to be lattice-periodic for any $\vb{k}$:
+As it turns out, this function is guaranteed to be lattice-periodic for any $$\vb{k}$$:
$$
\begin{aligned}
@@ -108,4 +108,4 @@ $$
$$
Then Bloch's theorem follows from
-isolating the definition of $u(\vb{r})$ for $\psi(\vb{r})$.
+isolating the definition of $$u(\vb{r})$$ for $$\psi(\vb{r})$$.
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 20399df..9ed2fd2 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -10,17 +10,17 @@ layout: "concept"
---
Consider a collection of particles,
-each with its own position $\vb{r}$ and velocity $\vb{v}$.
-We can thus define a probability density function $f(\vb{r}, \vb{v}, t)$
-describing the expected number of particles at $(\vb{r}, \vb{v})$ at time $t$.
-Let the total number of particles $N$ be conserved, then clearly:
+each with its own position $$\vb{r}$$ and velocity $$\vb{v}$$.
+We can thus define a probability density function $$f(\vb{r}, \vb{v}, t)$$
+describing the expected number of particles at $$(\vb{r}, \vb{v})$$ at time $$t$$.
+Let the total number of particles $$N$$ be conserved, then clearly:
$$\begin{aligned}
N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}}
\end{aligned}$$
At equilibrium, all processes affecting the particles
-no longer have a net effect, so $f$ is fixed:
+no longer have a net effect, so $$f$$ is fixed:
$$\begin{aligned}
\dv{f}{t}
@@ -35,7 +35,7 @@ $$\begin{aligned}
= \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
\end{aligned}$$
-Where the right-hand side simply means "all changes in $f$ due to collisions".
+Where the right-hand side simply means "all changes in $$f$$ due to collisions".
Applying the chain rule to the left-hand side then yields:
$$\begin{aligned}
@@ -49,8 +49,8 @@ $$\begin{aligned}
&= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}}
\end{aligned}$$
-Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$.
-Inserting Newton's second law $\vb{F} = m \vb{a}$
+Where we have introduced the shorthand $$\ipdv{f}{\vb{v}}$$.
+Inserting Newton's second law $$\vb{F} = m \vb{a}$$
leads us to the **Boltzmann equation** or
**Boltzmann transport equation** (BTE):
@@ -64,41 +64,41 @@ $$\begin{aligned}
But what about the collision term?
Expressions for it exist, which are almost exact in many cases,
but unfortunately also quite difficult to work with.
-In addition, $f$ is a 7-dimensional function,
+In addition, $$f$$ is a 7-dimensional function,
so the BTE is already hard to solve without collisions.
We only present the simplest case,
known as the **Bhatnagar-Gross-Krook approximation**:
-if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known,
-then each collision brings the system closer to $f_0$:
+if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known,
+then each collision brings the system closer to $$f_0$$:
$$\begin{aligned}
\pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}}
= \frac{f_0 - f}{\tau}
\end{aligned}$$
-Where $\tau$ is the average collision period.
+Where $$\tau$$ is the average collision period.
The right-hand side is called the **Krook term**.
## Moment equations
-From the definition of $f$,
-we see that integrating over all $\vb{v}$ yields the particle density $n$:
+From the definition of $$f$$,
+we see that integrating over all $$\vb{v}$$ yields the particle density $$n$$:
$$\begin{aligned}
n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
\end{aligned}$$
-Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so:
+Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so:
$$\begin{aligned}
\Expval{Q}
= \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
\end{aligned}$$
-With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate,
-assuming that $\vb{F}$ does not depend on $\vb{v}$:
+With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate,
+assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$:
$$\begin{aligned}
0
@@ -110,10 +110,10 @@ $$\begin{aligned}
+ \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{}{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}}
\end{aligned}$$
-The first integral is simply $n \Expval{Q}$.
-In the second integral, note that $\vb{v}$ is a coordinate
-and hence not dependent on $\vb{r}$, so $\nabla \cdot \vb{v} = 0$.
-Since $f$ is a probability density, $f \to 0$ for $\vb{v} \to \pm\infty$,
+The first integral is simply $$n \Expval{Q}$$.
+In the second integral, note that $$\vb{v}$$ is a coordinate
+and hence not dependent on $$\vb{r}$$, so $$\nabla \cdot \vb{v} = 0$$.
+Since $$f$$ is a probability density, $$f \to 0$$ for $$\vb{v} \to \pm\infty$$,
so the first term in the third integral vanishes after it is integrated:
$$\begin{aligned}
@@ -134,9 +134,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-If we set $Q = m$, then the mass density $\rho = n \Expval{Q}$,
+If we set $$Q = m$$, then the mass density $$\rho = n \Expval{Q}$$,
and we find that the **zeroth moment** of the BTE describes conservation of mass,
-where $\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$ is the fluid velocity:
+where $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity:
$$\begin{aligned}
\boxed{
@@ -150,8 +150,8 @@ $$\begin{aligned}
-There are $N$ qubits in initial state $\Ket{0}$, and one in $\Ket{1}$.
-For clarity, the oracle $U_f$ works like so:
+There are $$N$$ qubits in initial state $$\Ket{0}$$, and one in $$\Ket{1}$$.
+For clarity, the oracle $$U_f$$ works like so:
$$\begin{aligned}
\Ket{x_1} \Ket{x_2} \cdots \Ket{x_N} \Ket{y}
@@ -154,7 +154,7 @@ $$\begin{aligned}
\Ket{x_1} \cdots \Ket{x_N} \Ket{y \oplus f(x_1, ..., x_N)}
\end{aligned}$$
-Applying the $N + 1$ Hadamard gates to the initial state
+Applying the $$N + 1$$ Hadamard gates to the initial state
yields the following superposition:
$$\begin{aligned}
@@ -164,9 +164,9 @@ $$\begin{aligned}
= \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x} \Ket{-}
\end{aligned}$$
-Where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$ denotes a classical binary state.
-For example, if $x = 5 = 2^0 + 2^2$ in the summation,
-then $\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$
+Where $$\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$$ denotes a classical binary state.
+For example, if $$x = 5 = 2^0 + 2^2$$ in the summation,
+then $$\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$$
(from least to most significant).
We give this state to the oracle,
@@ -178,8 +178,8 @@ $$\begin{aligned}
\frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \Ket{x} \Ket{-}
\end{aligned}$$
-The last qubit $\Ket{-}$ is garbage.
-Next, applying the Hadamard transform to the other $N$ gives:
+The last qubit $$\Ket{-}$$ is garbage.
+Next, applying the Hadamard transform to the other $$N$$ gives:
$$\begin{aligned}
\to \boxed{H^{\otimes N}} \to \quad
@@ -187,8 +187,8 @@ $$\begin{aligned}
\bigg( \frac{1}{\sqrt{2^N}} \sum_{y = 0}^{2^N - 1} (-1)^{x \cdot y} \Ket{y} \bigg)
\end{aligned}$$
-Where $x \cdot y$ is the bitwise dot product of the binary representations of $x$ and $y$,
-so, for example, if $N = 2$, then $x \cdot y = x_1 y_1 + x_2 y_2$.
+Where $$x \cdot y$$ is the bitwise dot product of the binary representations of $$x$$ and $$y$$,
+so, for example, if $$N = 2$$, then $$x \cdot y = x_1 y_1 + x_2 y_2$$.
Note that the above expression has not been reduced at all;
it follows from the definition of the Hadamard transform.
We can rewrite it like so:
@@ -200,24 +200,24 @@ $$\begin{aligned}
\end{aligned}$$
The parenthesized expression can be interpreted as the coefficients
-of a superposition of several $y$-values.
-Therefore, the probability that a measurement yields $y = 0$,
-i.e. $\Ket{y} = \Ket{0}^{\otimes N}$, is:
+of a superposition of several $$y$$-values.
+Therefore, the probability that a measurement yields $$y = 0$$,
+i.e. $$\Ket{y} = \Ket{0}^{\otimes N}$$, is:
$$\begin{aligned}
|c_0|^2
= \bigg| \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \bigg|^2
\end{aligned}$$
-The summation always contains an even number of terms, for all values of $N$.
-Consequently, if $f$ is constant, then $|c_0|^2 = |\!\pm\! 2^N / 2^N|^2 = 1$.
-Otherwise, if $f$ is balanced, all the terms cancel out, so we are left with $|c_0|^2 = 0$.
+The summation always contains an even number of terms, for all values of $$N$$.
+Consequently, if $$f$$ is constant, then $$|c_0|^2 = |\!\pm\! 2^N / 2^N|^2 = 1$$.
+Otherwise, if $$f$$ is balanced, all the terms cancel out, so we are left with $$|c_0|^2 = 0$$.
In other words, we reach the same result as the Deutsch algorithm:
-we only need to measure the $N$ qubits once;
-$f$ is constant if and only if all are zero.
+we only need to measure the $$N$$ qubits once;
+$$f$$ is constant if and only if all are zero.
The Deutsch-Jozsa algorithm needs only one oracle query to give an error-free result,
-whereas a classical computer needs $2^{N-1} + 1$ queries in the worst case;
+whereas a classical computer needs $$2^{N-1} + 1$$ queries in the worst case;
a revolutionary discovery.
diff --git a/source/know/concept/dielectric-function/index.md b/source/know/concept/dielectric-function/index.md
index 30622f4..529ce2a 100644
--- a/source/know/concept/dielectric-function/index.md
+++ b/source/know/concept/dielectric-function/index.md
@@ -9,11 +9,11 @@ categories:
layout: "concept"
---
-The **dielectric function** or **relative permittivity** $\varepsilon_r$
+The **dielectric function** or **relative permittivity** $$\varepsilon_r$$
is a measure of how strongly a given medium counteracts
[electric fields](/know/concept/electric-field/) compared to a vacuum.
-Let $\vb{D}$ be the applied external field,
-and $\vb{E}$ the effective field inside the material:
+Let $$\vb{D}$$ be the applied external field,
+and $$\vb{E}$$ the effective field inside the material:
$$\begin{aligned}
\boxed{
@@ -21,15 +21,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\varepsilon_r$ is large, then $\vb{D}$ is strongly suppressed,
+If $$\varepsilon_r$$ is large, then $$\vb{D}$$ is strongly suppressed,
because the material's electrons and nuclei move to create an opposing field.
-In order for $\varepsilon_r$ to be well defined, we only consider linear media,
-where the induced polarization $\vb{P}$ is proportional to $\vb{E}$.
+In order for $$\varepsilon_r$$ to be well defined, we only consider linear media,
+where the induced polarization $$\vb{P}$$ is proportional to $$\vb{E}$$.
-We would like to find an alternative definition of $\varepsilon_r$.
-Consider that the usual electric fields $\vb{E}$, $\vb{D}$, and $\vb{P}$
+We would like to find an alternative definition of $$\varepsilon_r$$.
+Consider that the usual electric fields $$\vb{E}$$, $$\vb{D}$$, and $$\vb{P}$$
can each be written as the gradient of an electrostatic potential like so,
-where $\Phi_\mathrm{tot}$, $\Phi_\mathrm{ext}$ and $\Phi_\mathrm{ind}$
+where $$\Phi_\mathrm{tot}$$, $$\Phi_\mathrm{ext}$$ and $$\Phi_\mathrm{ind}$$
are the total, external and induced potentials, respectively:
$$\begin{aligned}
@@ -43,8 +43,8 @@ $$\begin{aligned}
= \varepsilon_0 \nabla \Phi_\mathrm{ind}
\end{aligned}$$
-Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$.
-Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$
+Such that $$\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$$.
+Inserting this into $$\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$$
then suggests defining:
$$\begin{aligned}
@@ -57,10 +57,10 @@ $$\begin{aligned}
## From induced charge density
-A common way to calculate $\varepsilon_r$ is from
-the induced charge density $\rho_\mathrm{ind}$,
+A common way to calculate $$\varepsilon_r$$ is from
+the induced charge density $$\rho_\mathrm{ind}$$,
i.e. the offset caused by the material's particles responding to the field.
-We start from [Gauss' law](/know/concept/maxwells-equations/) for $\vb{P}$:
+We start from [Gauss' law](/know/concept/maxwells-equations/) for $$\vb{P}$$:
$$\begin{aligned}
\nabla \cdot \vb{P}
@@ -77,8 +77,8 @@ $$\begin{aligned}
= V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
\end{aligned}$$
-Where $V(\vb{q})$ represents Coulomb interactions,
-and $V(0) = 0$ to ensure overall neutrality:
+Where $$V(\vb{q})$$ represents Coulomb interactions,
+and $$V(0) = 0$$ to ensure overall neutrality:
$$\begin{aligned}
V(\vb{q})
@@ -89,7 +89,7 @@ $$\begin{aligned}
\end{aligned}$$
The [convolution theorem](/know/concept/convolution-theorem/)
-then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain:
+then gives us the solution $$\Phi_\mathrm{ind}$$ in the $$\vb{r}$$-domain:
$$\begin{aligned}
\Phi_\mathrm{ind}(\vb{r})
@@ -97,13 +97,13 @@ $$\begin{aligned}
= \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'}
\end{aligned}$$
-To proceed, we need to find an expression for $\rho_\mathrm{ind}$
-that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$,
+To proceed, we need to find an expression for $$\rho_\mathrm{ind}$$
+that is proportional to $$\Phi_\mathrm{tot}$$ or $$\Phi_\mathrm{ext}$$,
or some linear combination thereof.
Such an expression must exist for a linear material.
-Suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$,
-for some $C_\mathrm{ext}$, which may depend on $\vb{q}$. Then:
+Suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{ext} \Phi_\mathrm{ext}$$,
+for some $$C_\mathrm{ext}$$, which may depend on $$\vb{q}$$. Then:
$$\begin{aligned}
\Phi_\mathrm{tot}
@@ -115,8 +115,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$,
-for some quantity $C_\mathrm{tot}$, then:
+Similarly, suppose we can show that $$\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$$,
+for some quantity $$C_\mathrm{tot}$$, then:
$$\begin{aligned}
\Phi_\mathrm{ext}
diff --git a/source/know/concept/diffie-hellman-key-exchange/index.md b/source/know/concept/diffie-hellman-key-exchange/index.md
index 951d5f5..4735209 100644
--- a/source/know/concept/diffie-hellman-key-exchange/index.md
+++ b/source/know/concept/diffie-hellman-key-exchange/index.md
@@ -13,30 +13,30 @@ when they can only communicate over an insecure channel.
The fundamental assumption of the Diffie-Hellman scheme,
upon which its security rests,
-is that the following function $f(n)$ is a **trapdoor function**,
-which means that calculating $f$ is easy,
-but its inverse $f^{-1}$ is extremely hard to find:
+is that the following function $$f(n)$$ is a **trapdoor function**,
+which means that calculating $$f$$ is easy,
+but its inverse $$f^{-1}$$ is extremely hard to find:
$$\begin{aligned}
f(n) = g^n \bmod p
\end{aligned}$$
-Where $n$ is a natural number, and $p$ is a prime.
-The natural number $g$ is a so-called *primitive root modulo* $p$.
-Importantly, $g$ and $p$ have been specifically chosen
-such that $f(n)$ can take any value in $\{1, ..., p \!-\! 1\}$
-for $n$ in $\{0, ..., p \!-\! 2\}$.
-The trapdoor assumption is that, given $g$, $p$ and $f(n)$,
-there is no efficient algorithm to recover $n$.
+Where $$n$$ is a natural number, and $$p$$ is a prime.
+The natural number $$g$$ is a so-called *primitive root modulo* $$p$$.
+Importantly, $$g$$ and $$p$$ have been specifically chosen
+such that $$f(n)$$ can take any value in $$\{1, ..., p \!-\! 1\}$$
+for $$n$$ in $$\{0, ..., p \!-\! 2\}$$.
+The trapdoor assumption is that, given $$g$$, $$p$$ and $$f(n)$$,
+there is no efficient algorithm to recover $$n$$.
Suppose that Alice and Bob want to exchange encrypted data in the future,
so they need to agree on an encryption key to use.
However, they can only exchange messages with each other over
an insecure channel, which is being eavesdropped.
-After they publicly agree on the values of $g$ and $p$,
-Alice and Bob each choose a secret number from $\{0, ..., p \!-\! 2\}$, respectively $a$ and $b$,
-and then privately calculate $A$ and $B$ as follows:
+After they publicly agree on the values of $$g$$ and $$p$$,
+Alice and Bob each choose a secret number from $$\{0, ..., p \!-\! 2\}$$, respectively $$a$$ and $$b$$,
+and then privately calculate $$A$$ and $$B$$ as follows:
$$\begin{aligned}
A = g^a \bmod p
@@ -44,9 +44,9 @@ $$\begin{aligned}
B = g^b \bmod p
\end{aligned}$$
-Finally, they transmit these numbers $A$ and $B$
+Finally, they transmit these numbers $$A$$ and $$B$$
to each other over the insecure connection,
-and then each side calculates $k$, which is the desired secret key:
+and then each side calculates $$k$$, which is the desired secret key:
$$\begin{aligned}
\boxed{
@@ -54,11 +54,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-The point is that $k$ includes both $a$ *and* $b$,
-but each side only needs to know *either* $a$ *or* $b$.
+The point is that $$k$$ includes both $$a$$ *and* $$b$$,
+but each side only needs to know *either* $$a$$ *or* $$b$$.
And, due to the trapdoor assumption,
-the eavesdropper knows $A$ and $B$,
-but cannot recover $a$ or $b$.
+the eavesdropper knows $$A$$ and $$B$$,
+but cannot recover $$a$$ or $$b$$.
This assumption is just that: an assumption.
So far, nobody has been able to prove or disprove it
diff --git a/source/know/concept/dirac-delta-function/index.md b/source/know/concept/dirac-delta-function/index.md
index 88a08cb..518eba1 100644
--- a/source/know/concept/dirac-delta-function/index.md
+++ b/source/know/concept/dirac-delta-function/index.md
@@ -8,10 +8,10 @@ categories:
layout: "concept"
---
-The **Dirac delta function** $\delta(x)$, often just the **delta function**,
+The **Dirac delta function** $$\delta(x)$$, often just the **delta function**,
is a function (or, more accurately, a [Schwartz distribution](/know/concept/schwartz-distribution/))
that is commonly used in physics.
-It is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is
+It is an infinitely narrow discontinuous "spike" at $$x = 0$$ whose area is
defined to be 1:
$$\begin{aligned}
@@ -35,7 +35,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-$\delta(x)$ is thus quite an effective weapon against integrals. This may not seem very
+$$\delta(x)$$ is thus quite an effective weapon against integrals. This may not seem very
useful due to its "unnatural" definition, but in fact it appears as the
limit of several reasonable functions:
@@ -57,7 +57,7 @@ $$\begin{aligned}
\:\:\propto\:\: \hat{\mathcal{F}}\{1\}
\end{aligned}$$
-When the argument of $\delta(x)$ is scaled, the delta function is itself scaled:
+When the argument of $$\delta(x)$$ is scaled, the delta function is itself scaled:
$$\begin{aligned}
\boxed{
@@ -70,17 +70,18 @@ $$\begin{aligned}
-The **instantaneous frequency** $\omega_\mathrm{GVD}(z, t)$,
+The **instantaneous frequency** $$\omega_\mathrm{GVD}(z, t)$$,
which describes the dominant angular frequency at a given point in the time domain,
is found to be as follows for the Gaussian pulse,
-where $\phi(z, t)$ is the phase of $A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$:
+where $$\phi(z, t)$$ is the phase of $$A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$$:
$$\begin{aligned}
\omega_{\mathrm{GVD}}(z,t)
@@ -74,20 +74,20 @@ $$\begin{aligned}
= \frac{\beta_2 z / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} \frac{t}{T_0^2}
\end{aligned}$$
-This expression is linear in time, and depending on the sign of $\beta_2$,
+This expression is linear in time, and depending on the sign of $$\beta_2$$,
frequencies on one side of the pulse arrive first,
and those on the other side arrive last.
-The effect is stronger for smaller $T_0$:
+The effect is stronger for smaller $$T_0$$:
this makes sense, since short pulses are spectrally wider.
The interaction between dispersion and [self-phase modulation](/know/concept/self-phase-modulation/)
leads to many interesting effects,
such as [modulational instability](/know/concept/modulational-instability/)
and [optical wave breaking](/know/concept/optical-wave-breaking/).
-Of great importance is the sign of $\beta_2$:
-in the **anomalous dispersion regime** ($\beta_2 < 0$),
+Of great importance is the sign of $$\beta_2$$:
+in the **anomalous dispersion regime** ($$\beta_2 < 0$$),
lower frequencies travel more slowly than higher ones,
-and vice versa in the **normal dispersion regime** ($\beta_2 > 0$).
+and vice versa in the **normal dispersion regime** ($$\beta_2 > 0$$).
diff --git a/source/know/concept/drude-model/index.md b/source/know/concept/drude-model/index.md
index 28e6dc2..b175e64 100644
--- a/source/know/concept/drude-model/index.md
+++ b/source/know/concept/drude-model/index.md
@@ -18,19 +18,19 @@ as found in metals and doped semiconductors.
An [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
has an oscillating [electric field](/know/concept/electric-field/)
-$E(t) = E_0 \exp(- i \omega t)$
+$$E(t) = E_0 \exp(- i \omega t)$$
that exerts a force on the charge carriers,
-which have mass $m$ and charge $q$.
+which have mass $$m$$ and charge $$q$$.
They thus obey the following equation of motion,
-where $\gamma$ is a frictional damping coefficient:
+where $$\gamma$$ is a frictional damping coefficient:
$$\begin{aligned}
m \dvn{2}{x}{t} + m \gamma \dv{x}{t}
= q E_0 \exp(- i \omega t)
\end{aligned}$$
-Inserting the ansatz $x(t) = x_0 \exp(- i \omega t)$
-and isolating for the displacement $x_0$ yields:
+Inserting the ansatz $$x(t) = x_0 \exp(- i \omega t)$$
+and isolating for the displacement $$x_0$$ yields:
$$\begin{aligned}
- x_0 m \omega^2 - i x_0 m \gamma \omega
@@ -40,10 +40,10 @@ $$\begin{aligned}
= - \frac{q E_0}{m (\omega^2 + i \gamma \omega)}
\end{aligned}$$
-The polarization density $P(t)$ is therefore as shown below.
-Note that the dipole moment $p$ goes from negative to positive,
-and the electric field $E$ from positive to negative.
-Let $N$ be the density of carriers in the gas, then:
+The polarization density $$P(t)$$ is therefore as shown below.
+Note that the dipole moment $$p$$ goes from negative to positive,
+and the electric field $$E$$ from positive to negative.
+Let $$N$$ be the density of carriers in the gas, then:
$$\begin{aligned}
P(t)
@@ -52,8 +52,8 @@ $$\begin{aligned}
= - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} E(t)
\end{aligned}$$
-The electric displacement field $D$ is thus as follows,
-where $\varepsilon_r$ is the unknown relative permittivity of the gas,
+The electric displacement field $$D$$ is thus as follows,
+where $$\varepsilon_r$$ is the unknown relative permittivity of the gas,
which we will find shortly:
$$\begin{aligned}
@@ -63,8 +63,8 @@ $$\begin{aligned}
= \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) E
\end{aligned}$$
-The parenthesized expression is the desired dielectric function $\varepsilon_r$,
-which depends on $\omega$:
+The parenthesized expression is the desired dielectric function $$\varepsilon_r$$,
+which depends on $$\omega$$:
$$\begin{aligned}
\boxed{
@@ -82,26 +82,26 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\gamma = 0$, then $\varepsilon_r$ is
-negative $\omega < \omega_p$,
-positive for $\omega > \omega_p$,
-and zero for $\omega = \omega_p$.
+If $$\gamma = 0$$, then $$\varepsilon_r$$ is
+negative $$\omega < \omega_p$$,
+positive for $$\omega > \omega_p$$,
+and zero for $$\omega = \omega_p$$.
Respectively, this leads to
-an imaginary index $\sqrt{\varepsilon_r}$ (high absorption),
-a real index tending to $1$ (transparency),
+an imaginary index $$\sqrt{\varepsilon_r}$$ (high absorption),
+a real index tending to $$1$$ (transparency),
and the possibility of self-sustained plasma oscillations.
-For metals, $\omega_p$ lies in the UV.
+For metals, $$\omega_p$$ lies in the UV.
-We can refine this result for $\varepsilon_r$,
-by recognizing the (mean) velocity $v = \idv{x}{t}$,
+We can refine this result for $$\varepsilon_r$$,
+by recognizing the (mean) velocity $$v = \idv{x}{t}$$,
and rewriting the equation of motion accordingly:
$$\begin{aligned}
m \dv{v}{t} + m \gamma v = q E(t)
\end{aligned}$$
-Note that $m v$ is simply the momentum $p$.
-We define the **momentum scattering time** $\tau \equiv 1 / \gamma$,
+Note that $$m v$$ is simply the momentum $$p$$.
+We define the **momentum scattering time** $$\tau \equiv 1 / \gamma$$,
which represents the average time between collisions,
where each collision resets the involved particles' momentums to zero.
Or, more formally:
@@ -111,9 +111,9 @@ $$\begin{aligned}
= - \frac{p}{\tau} + q E
\end{aligned}$$
-Returning to the equation for the mean velocity $v$,
-we insert the ansatz $v(t) = v_0 \exp(- i \omega t)$,
-for the same electric field $E(t) = E_0 \exp(-i \omega t)$ as before:
+Returning to the equation for the mean velocity $$v$$,
+we insert the ansatz $$v(t) = v_0 \exp(- i \omega t)$$,
+for the same electric field $$E(t) = E_0 \exp(-i \omega t)$$ as before:
$$\begin{aligned}
- i m \omega v_0 + \frac{m}{\tau} v_0 = q E_0
@@ -121,7 +121,7 @@ $$\begin{aligned}
v_0 = \frac{q \tau}{m (1 - i \omega \tau)} E_0
\end{aligned}$$
-From $v(t)$, we find the resulting average current density $J(t)$ to be as follows:
+From $$v(t)$$, we find the resulting average current density $$J(t)$$ to be as follows:
$$\begin{aligned}
J(t)
@@ -129,8 +129,8 @@ $$\begin{aligned}
= \sigma E(t)
\end{aligned}$$
-Where $\sigma(\omega)$ is the **AC conductivity**,
-which depends on the **DC conductivity** $\sigma_0$:
+Where $$\sigma(\omega)$$ is the **AC conductivity**,
+which depends on the **DC conductivity** $$\sigma_0$$:
$$\begin{aligned}
\boxed{
@@ -145,7 +145,7 @@ $$\begin{aligned}
\end{aligned}$$
We can use these quantities to rewrite
-the dielectric function $\varepsilon_r$ from earlier:
+the dielectric function $$\varepsilon_r$$ from earlier:
$$\begin{aligned}
\boxed{
@@ -164,12 +164,12 @@ which can be treated as free particles
moving in the bands of the material.
The Drude model can also be used in this case,
-by replacing the actual carrier mass $m$
-by the effective mass $m^*$.
+by replacing the actual carrier mass $$m$$
+by the effective mass $$m^*$$.
Furthermore, semiconductors already have
-a high intrinsic permittivity $\varepsilon_{\mathrm{int}}$
+a high intrinsic permittivity $$\varepsilon_{\mathrm{int}}$$
before the dopant is added,
-so the diplacement field $D$ is:
+so the diplacement field $$D$$ is:
$$\begin{aligned}
D
@@ -177,9 +177,9 @@ $$\begin{aligned}
= \varepsilon_{\mathrm{int}} \varepsilon_0 E - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} E
\end{aligned}$$
-Where $P_{\mathrm{int}}$ is the intrinsic undoped polarization,
-and $P_{\mathrm{free}}$ is the contribution of the free carriers.
-The dielectric function $\varepsilon_r(\omega)$ is therefore given by:
+Where $$P_{\mathrm{int}}$$ is the intrinsic undoped polarization,
+and $$P_{\mathrm{free}}$$ is the contribution of the free carriers.
+The dielectric function $$\varepsilon_r(\omega)$$ is therefore given by:
$$\begin{aligned}
\boxed{
@@ -188,8 +188,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the plasma frequency $\omega_p$ has been redefined as follows
-to include $\varepsilon_\mathrm{int}$:
+Where the plasma frequency $$\omega_p$$ has been redefined as follows
+to include $$\varepsilon_\mathrm{int}$$:
$$\begin{aligned}
\boxed{
@@ -198,14 +198,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-The meaning of $\omega_p$ is the same as for metals,
-with high absorption for $\omega < \omega_p$.
-However, due to the lower carrier density $N$ in a semiconductor,
-$\omega_p$ lies in the IR rather than UV.
+The meaning of $$\omega_p$$ is the same as for metals,
+with high absorption for $$\omega < \omega_p$$.
+However, due to the lower carrier density $$N$$ in a semiconductor,
+$$\omega_p$$ lies in the IR rather than UV.
-However, instead of asymptotically going to $1$ for $\omega > \omega_p$ like a metal,
-$\varepsilon_r$ tends to $\varepsilon_\mathrm{int}$ instead,
-and crosses $1$ along the way,
+However, instead of asymptotically going to $$1$$ for $$\omega > \omega_p$$ like a metal,
+$$\varepsilon_r$$ tends to $$\varepsilon_\mathrm{int}$$ instead,
+and crosses $$1$$ along the way,
at which point the reflectivity is zero.
This occurs at:
@@ -214,9 +214,9 @@ $$\begin{aligned}
= \frac{\varepsilon_{\mathrm{int}}}{\varepsilon_{\mathrm{int}} - 1} \omega_p^2
\end{aligned}$$
-This is used to experimentally determine the effective mass $m^*$
+This is used to experimentally determine the effective mass $$m^*$$
of the doped semiconductor,
-by finding which value of $m^*$ gives the measured $\omega$.
+by finding which value of $$m^*$$ gives the measured $$\omega$$.
diff --git a/source/know/concept/dynkins-formula/index.md b/source/know/concept/dynkins-formula/index.md
index 53da86f..c0d20c5 100644
--- a/source/know/concept/dynkins-formula/index.md
+++ b/source/know/concept/dynkins-formula/index.md
@@ -8,12 +8,12 @@ categories:
layout: "concept"
---
-Given an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
-with a time-independent drift $f$ and intensity $g$
-such that the diffusion uniquely exists on the $t$-axis.
-We define the **infinitesimal generator** $\hat{A}$
-as an operator with the following action on a given function $h(x)$,
-where $\mathbf{E}$ is a
+Given an [Itō diffusion](/know/concept/ito-calculus/) $$X_t$$
+with a time-independent drift $$f$$ and intensity $$g$$
+such that the diffusion uniquely exists on the $$t$$-axis.
+We define the **infinitesimal generator** $$\hat{A}$$
+as an operator with the following action on a given function $$h(x)$$,
+where $$\mathbf{E}$$ is a
[conditional expectation](/know/concept/conditional-expectation/):
$$\begin{aligned}
@@ -23,13 +23,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Which only makes sense for $h$ where this limit exists.
-The assumption that $X_t$ does not have any explicit time-dependence
-means that $X_0$ need not be the true initial condition;
-it can also be the state $X_s$ at any $s$ infinitesimally smaller than $t$.
+Which only makes sense for $$h$$ where this limit exists.
+The assumption that $$X_t$$ does not have any explicit time-dependence
+means that $$X_0$$ need not be the true initial condition;
+it can also be the state $$X_s$$ at any $$s$$ infinitesimally smaller than $$t$$.
-Conveniently, for a sufficiently well-behaved $h$,
-the generator $\hat{A}$ is identical to the Kolmogorov operator $\hat{L}$
+Conveniently, for a sufficiently well-behaved $$h$$,
+the generator $$\hat{A}$$ is identical to the Kolmogorov operator $$\hat{L}$$
found in the [backward Kolmogorov equation](/know/concept/kolmogorov-equations/):
$$\begin{aligned}
@@ -44,7 +44,7 @@ $$\begin{aligned}
@@ -28,8 +28,8 @@ Let $n_L$, $n_C$ and $n_R$ be the respective domains' refractive indices:
In its simplest "microscopic" form, the reflection at the boundaries
is simply caused by the index differences there.
-Consider this ansatz for the [electric field](/know/concept/electric-field/) $E_m(x)$,
-where $m$ is the mode:
+Consider this ansatz for the [electric field](/know/concept/electric-field/) $$E_m(x)$$,
+where $$m$$ is the mode:
$$\begin{aligned}
E_m(x)
@@ -40,9 +40,9 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-The goal is to find the modes' wavenumbers $k_m$.
-First, we demand that $E_m$ and its derivative $\idv{E_m}{x}$
-are continuous at the boundaries $x = \pm \ell/2$:
+The goal is to find the modes' wavenumbers $$k_m$$.
+First, we demand that $$E_m$$ and its derivative $$\idv{E_m}{x}$$
+are continuous at the boundaries $$x = \pm \ell/2$$:
$$\begin{aligned}
A_1 e^{i k_m n_L \ell/2}
@@ -95,8 +95,8 @@ $$\begin{aligned}
\end{aligned}$$
We want non-trivial solutions, where we
-cannot simply satisfy the system by setting $A_1$, $A_2$, $A_3$ and
-$A_4$; this constraint will give us an equation for $k_m$. Therefore, we
+cannot simply satisfy the system by setting $$A_1$$, $$A_2$$, $$A_3$$ and
+$$A_4$$; this constraint will give us an equation for $$k_m$$. Therefore, we
demand that the system matrix is singular, i.e. its determinant is zero:
$$\begin{aligned}
@@ -106,7 +106,7 @@ $$\begin{aligned}
&+ (n_C^2 + n_L n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} - e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
\end{aligned}$$
-We multiply by $e^{i k_m (n_L + n_R) \ell / 2}$ and
+We multiply by $$e^{i k_m (n_L + n_R) \ell / 2}$$ and
decompose the exponentials into sines and cosines:
$$\begin{aligned}
@@ -125,8 +125,8 @@ $$\begin{aligned}
\end{aligned}$$
Thanks to linearity, we can choose one of the amplitudes
-$A_1$, $A_2$, $A_3$ or $A_4$ freely,
-and then the others are determined by $k_m$ and the field's continuity.
+$$A_1$$, $$A_2$$, $$A_3$$ or $$A_4$$ freely,
+and then the others are determined by $$k_m$$ and the field's continuity.
## Macroscopic cavity
@@ -134,7 +134,7 @@ and then the others are determined by $k_m$ and the field's continuity.
Next, consider a "macroscopic" Fabry-Pérot cavity
with complex mirror structures at boundaries, e.g. Bragg reflectors.
If the cavity is large enough, we can neglect the mirrors' thicknesses,
-and just use their reflection coefficients $r_L$ and $r_R$.
+and just use their reflection coefficients $$r_L$$ and $$r_R$$.
We use the same ansatz:
$$\begin{aligned}
@@ -147,10 +147,10 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-On the left, $A_3$ is the reflection of $A_2$,
-and on the right, $A_2$ is the reflection of $A_3$,
+On the left, $$A_3$$ is the reflection of $$A_2$$,
+and on the right, $$A_2$$ is the reflection of $$A_3$$,
where the reflected amplitudes are determined
-by the coefficients $r_L$ and $r_R$, respectively:
+by the coefficients $$r_L$$ and $$r_R$$, respectively:
$$\begin{aligned}
A_3 e^{- i k_m n_C \ell/2}
@@ -185,8 +185,8 @@ $$\begin{aligned}
&= 1 - r_L r_R e^{i 2 k_m n_C \ell}
\end{aligned}$$
-Isolating this for $k_m$ yields the following modes,
-where $m$ is an arbitrary integer:
+Isolating this for $$k_m$$ yields the following modes,
+where $$m$$ is an arbitrary integer:
$$\begin{aligned}
\boxed{
@@ -195,12 +195,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-These $k_m$ satisfy the matrix equation above.
-Thanks to linearity, we can choose one of $A_2$ or $A_3$,
+These $$k_m$$ satisfy the matrix equation above.
+Thanks to linearity, we can choose one of $$A_2$$ or $$A_3$$,
and then the other is determined by the corresponding reflection equation.
Finally, we look at the light transmitted through the mirrors,
-according to $1 \!-\! r_L$ and $1 \!-\! r_R$:
+according to $$1 \!-\! r_L$$ and $$1 \!-\! r_R$$:
$$\begin{aligned}
A_1 e^{i k_m n_L \ell/2}
@@ -210,7 +210,7 @@ $$\begin{aligned}
&= (1 - r_R) A_3 e^{i k_m n_C \ell/2}
\end{aligned}$$
-We simply isolate for $A_1$ and $A_4$ respectively,
+We simply isolate for $$A_1$$ and $$A_4$$ respectively,
yielding the following amplitudes:
$$\begin{aligned}
diff --git a/source/know/concept/fermi-dirac-distribution/index.md b/source/know/concept/fermi-dirac-distribution/index.md
index ea1f8b8..09a3e76 100644
--- a/source/know/concept/fermi-dirac-distribution/index.md
+++ b/source/know/concept/fermi-dirac-distribution/index.md
@@ -13,12 +13,12 @@ layout: "concept"
which obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
will distribute themselves across the available states in a system at equilibrium.
-Consider one single-particle state $s$,
-which can contain $0$ or $1$ fermions.
-Because the occupation number $N$ is variable,
+Consider one single-particle state $$s$$,
+which can contain $$0$$ or $$1$$ fermions.
+Because the occupation number $$N$$ is variable,
we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
-whose grand partition function $\mathcal{Z}$ is as follows,
-where we sum over all microstates of $s$:
+whose grand partition function $$\mathcal{Z}$$ is as follows,
+where we sum over all microstates of $$s$$:
$$\begin{aligned}
\mathcal{Z}
@@ -26,12 +26,12 @@ $$\begin{aligned}
= 1 + \exp(- \beta (\varepsilon - \mu))
\end{aligned}$$
-Where $\mu$ is the chemical potential,
-and $\varepsilon$ is the energy contribution per particle in $s$,
-i.e. the total energy of all particles $E = \varepsilon N$.
+Where $$\mu$$ is the chemical potential,
+and $$\varepsilon$$ is the energy contribution per particle in $$s$$,
+i.e. the total energy of all particles $$E = \varepsilon N$$.
The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
-is the Landau potential $\Omega$, given by:
+is the Landau potential $$\Omega$$, given by:
$$\begin{aligned}
\Omega
@@ -39,8 +39,8 @@ $$\begin{aligned}
= - k T \ln\!\Big( 1 + \exp(- \beta (\varepsilon - \mu)) \Big)
\end{aligned}$$
-The average number of particles $\Expval{N}$
-in state $s$ is then found to be as follows:
+The average number of particles $$\Expval{N}$$
+in state $$s$$ is then found to be as follows:
$$\begin{aligned}
\Expval{N}
@@ -49,9 +49,9 @@ $$\begin{aligned}
= \frac{\exp(- \beta (\varepsilon - \mu))}{1 + \exp(- \beta (\varepsilon - \mu))}
\end{aligned}$$
-By multiplying both the numerator and the denominator by $\exp(\beta (\varepsilon \!-\! \mu))$,
+By multiplying both the numerator and the denominator by $$\exp(\beta (\varepsilon \!-\! \mu))$$,
we arrive at the standard form of
-the **Fermi-Dirac distribution** or **Fermi function** $f_F$:
+the **Fermi-Dirac distribution** or **Fermi function** $$f_F$$:
$$\begin{aligned}
\boxed{
@@ -61,9 +61,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-This tells the expected occupation number $\Expval{N}$ of state $s$,
-given a temperature $T$ and chemical potential $\mu$.
-The corresponding variance $\sigma^2$ of $N$ is found to be:
+This tells the expected occupation number $$\Expval{N}$$ of state $$s$$,
+given a temperature $$T$$ and chemical potential $$\mu$$.
+The corresponding variance $$\sigma^2$$ of $$N$$ is found to be:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md
index 18fcfd8..021c8e4 100644
--- a/source/know/concept/fermis-golden-rule/index.md
+++ b/source/know/concept/fermis-golden-rule/index.md
@@ -13,29 +13,29 @@ layout: "concept"
In quantum mechanics, **Fermi's golden rule** expresses
the transition rate between two states of a system,
when a sinusoidal perturbation is applied
-at the resonance frequency $\omega = E_g / \hbar$ of the
-energy gap $E_g$. The main conclusion is that the rate is independent of
+at the resonance frequency $$\omega = E_g / \hbar$$ of the
+energy gap $$E_g$$. The main conclusion is that the rate is independent of
time.
From [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
we know that the transition probability
-for a particle in state $\Ket{a}$ to go to $\Ket{b}$
-is as follows for a periodic perturbation at frequency $\omega$:
+for a particle in state $$\Ket{a}$$ to go to $$\Ket{b}$$
+is as follows for a periodic perturbation at frequency $$\omega$$:
$$\begin{aligned}
P_{ab}
= \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2}
\end{aligned}$$
-Where $\omega_{ba} \equiv (E_b - E_a) / \hbar$.
-If we assume that $\Ket{b}$ irreversibly absorbs an unlimited number of particles,
-then we can interpret $P_{ab}$ as the "amount" of the current particle
-that has transitioned since the last period $2 \pi n / (\omega_{ba} \!-\! \omega)$.
+Where $$\omega_{ba} \equiv (E_b - E_a) / \hbar$$.
+If we assume that $$\Ket{b}$$ irreversibly absorbs an unlimited number of particles,
+then we can interpret $$P_{ab}$$ as the "amount" of the current particle
+that has transitioned since the last period $$2 \pi n / (\omega_{ba} \!-\! \omega)$$.
-For generality, let $E_b$ be the center
-of a state continuum with width $\Delta E$.
-In that case, $P_{ab}$ must be modified as follows,
-where $\rho(E_x)$ is the destination's
+For generality, let $$E_b$$ be the center
+of a state continuum with width $$\Delta E$$.
+In that case, $$P_{ab}$$ must be modified as follows,
+where $$\rho(E_x)$$ is the destination's
[density of states](/know/concept/density-of-states/):
$$\begin{aligned}
@@ -44,26 +44,26 @@ $$\begin{aligned}
\frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
\end{aligned}$$
-If $E_b$ is not in a continuum, then $\rho(E_x) = \delta(E_x - E_b)$.
-The integrand is a sharp sinc-function around $E_x$.
-For large $t$, it is so sharp that we can take out $\rho(E_x)$.
+If $$E_b$$ is not in a continuum, then $$\rho(E_x) = \delta(E_x - E_b)$$.
+The integrand is a sharp sinc-function around $$E_x$$.
+For large $$t$$, it is so sharp that we can take out $$\rho(E_x)$$.
In that case, we also simplify the integration limits.
-Then we substitute $x \equiv (\omega_{xa}\!-\!\omega) / 2$ to get:
+Then we substitute $$x \equiv (\omega_{xa}\!-\!\omega) / 2$$ to get:
$$\begin{aligned}
P_{ab}
&\approx \frac{2}{\hbar} |V_{ba}|^2 \rho(E_b) \int_{-\infty}^\infty \frac{\sin^2(x t)}{x^2} \:dx
\end{aligned}$$
-This definite integral turns out to be $\pi |t|$,
-so we find, because clearly $t > 0$:
+This definite integral turns out to be $$\pi |t|$$,
+so we find, because clearly $$t > 0$$:
$$\begin{aligned}
P_{ab}
&= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t
\end{aligned}$$
-The transition rate $R_{ab}$,
+The transition rate $$R_{ab}$$,
i.e. the number of particles per unit time,
then takes this form:
@@ -75,9 +75,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that the $t$-dependence has disappeared,
-and all that remains is a constant factor involving $E_b = E_a \!+\! \hbar \omega$,
-where $\omega$ is the resonance frequency.
+Note that the $$t$$-dependence has disappeared,
+and all that remains is a constant factor involving $$E_b = E_a \!+\! \hbar \omega$$,
+where $$\omega$$ is the resonance frequency.
diff --git a/source/know/concept/feynman-diagram/index.md b/source/know/concept/feynman-diagram/index.md
index d117562..c36e7df 100644
--- a/source/know/concept/feynman-diagram/index.md
+++ b/source/know/concept/feynman-diagram/index.md
@@ -13,13 +13,13 @@ When doing calculations in the context of condensed matter physics and quantum f
that would be tedious or error-prone to work with directly.
This article is about condensed matter physics.
-Suppose we have a many-particle Hamiltonian $\hat{H} = \hat{H}_0 + \hat{H}_1$,
-consisting of an "easy" term $\hat{H}_0$,
-and then a "difficult" term $\hat{H}_1$
+Suppose we have a many-particle Hamiltonian $$\hat{H} = \hat{H}_0 + \hat{H}_1$$,
+consisting of an "easy" term $$\hat{H}_0$$,
+and then a "difficult" term $$\hat{H}_1$$
with time-dependent and/or interacting parts.
-Let $\Ket{\Phi_0}$ be a known eigenstate (or superposition thereof)
-of the easily solvable part $\hat{H}_0$,
-with respect to which we will take expectation values $\Expval{}$.
+Let $$\Ket{\Phi_0}$$ be a known eigenstate (or superposition thereof)
+of the easily solvable part $$\hat{H}_0$$,
+with respect to which we will take expectation values $$\Expval{}$$.
Below, we go through the most notable components of Feynman diagrams
and how to translate them into a mathematical expression.
@@ -28,26 +28,27 @@ and how to translate them into a mathematical expression.
## Real space
The most common component is a **fermion line**, which represents
-a [Green's function](/know/concept/greens-functions/) $G^0$
-for the simple Hamiltonian $\hat{H}_0$.
+a [Green's function](/know/concept/greens-functions/) $$G^0$$
+for the simple Hamiltonian $$\hat{H}_0$$.
Any type of Green's function is possible in theory (e.g. a retarded),
but usually the *causal* function is used.
-Let the subscript $I$ refer to the
+Let the subscript $$I$$ refer to the
[interaction picture](/know/concept/interaction-picture/),
-and $\mathcal{T}\{\}$ denote the
+and $$\mathcal{T}\{\}$$ denote the
[time-ordered product](/know/concept/time-ordered-product/):
+
$$\begin{aligned}
= i \hbar G_{s_2 s_1}^0(\vb{r}_2, t_2; \vb{r}_1, t_1)
= \Expval{\mathcal{T} \Big\{ \hat{\Psi}_{s_2 I}(\vb{r}_2, t_2) \hat{\Psi}_{s_1 I}^\dagger(\vb{r}_1, t_1) \Big\}}
\end{aligned}$$
The arrow points in the direction of time, or more generally,
-from the point of creation $\hat{\Psi}{}^\dagger$
-to the point of annihilation $\hat{\Psi}$.
+from the point of creation $$\hat{\Psi}{}^\dagger$$
+to the point of annihilation $$\hat{\Psi}$$.
The dots at the ends are called **vertices**,
which represent points in space and time with a spin.
Vertices can be
@@ -55,19 +56,20 @@ Vertices can be
or **external** (either one Green's function entering OR one leaving).
Less common is a **heavy fermion line**, representing
-a causal Green's function $G$ for the entire Hamiltonian $\hat{H}$,
-where the subscript $H$ refers to the [Heisenberg picture](/know/concept/heisenberg-picture/):
+a causal Green's function $$G$$ for the entire Hamiltonian $$\hat{H}$$,
+where the subscript $$H$$ refers to the [Heisenberg picture](/know/concept/heisenberg-picture/):
+
$$\begin{aligned}
= i \hbar G_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
= \Expval{\mathcal{T} \Big\{ \hat{\Psi}_{s_2 H}(\vb{r}_2, t_2) \hat{\Psi}_{s_1 H}^\dagger(\vb{r}_1, t_1) \Big\}}
\end{aligned}$$
Next, an **interaction line** or **boson line** represents
-a two-body interaction operator $\hat{W}$ (in $\hat{H}_1$),
+a two-body interaction operator $$\hat{W}$$ (in $$\hat{H}_1$$),
which we assume to be instantaneous, i.e. time-independent
(in quantum field theory this is *not* assumed),
hence it starts and ends at the same time,
@@ -76,16 +78,17 @@ and no arrow is drawn:
+
$$\begin{aligned}
= \frac{1}{i \hbar} W_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
= \frac{1}{i \hbar} W(\vb{r}_2, \vb{r}_1; t_1) \: \delta(t_2 - t_1)
\end{aligned}$$
We have chosen to disallow spin flipping,
-so $W$ does not depend on $s_1$ or $s_2$.
-For reference, this function $W$
+so $$W$$ does not depend on $$s_1$$ or $$s_2$$.
+For reference, this function $$W$$
has a time-dependence coming only from the interaction picture,
-and is to be used as follows to get the full two-body operator $\hat{W}$:
+and is to be used as follows to get the full two-body operator $$\hat{W}$$:
$$\begin{aligned}
\hat{W}
@@ -93,12 +96,13 @@ $$\begin{aligned}
W(\vb{r}_1, \vb{r}_2) \hat{\Psi}_{s_2}(\vb{r}_2) \hat{\Psi}_{s_1}(\vb{r}_1) \dd{\vb{r}_1} \dd{\vb{r}_2}
\end{aligned}$$
-One-body (time-dependent) operators $\hat{V}$ in $\hat{H}_1$
+One-body (time-dependent) operators $$\hat{V}$$ in $$\hat{H}_1$$
are instead represented by a special vertex:
+
$$\begin{aligned}
= \frac{1}{i \hbar} V_s(\vb{r}, t)
\end{aligned}$$
@@ -114,27 +118,27 @@ a. Each vertex must be connected to one or two fermion lines,
at most one of which leaves,
and at most one of which enters.
b. Each internal vertex contains at most one "event";
- which could be $V$ or $W$.
+ which could be $$V$$ or $$W$$.
Finally, we need some additional rules to convert
diagrams into mathematical expressions:
1. Disallow spin flipping by multiplying
- each internal vertex by $\delta_{s_\mathrm{in} s_\mathrm{out}}$.
-2. If both ends of a line are at the same time (always the case for $W$),
- an infinitesimal $\eta \to 0^+$ must be added
+ each internal vertex by $$\delta_{s_\mathrm{in} s_\mathrm{out}}$$.
+2. If both ends of a line are at the same time (always the case for $$W$$),
+ an infinitesimal $$\eta \to 0^+$$ must be added
to the time of all creation operators,
- so e.g. $G(t, t) \to G(t, t\!+\!\eta)$.
-3. Integrate over spacetime coordinates $(\vb{r}, t)$
- and sum over the spin $s$ of all internal vertices,
+ so e.g. $$G(t, t) \to G(t, t\!+\!\eta)$$.
+3. Integrate over spacetime coordinates $$(\vb{r}, t)$$
+ and sum over the spin $$s$$ of all internal vertices,
but not external ones.
-4. Multiply the result by $(-1)^F$,
- where $F$ is the number of closed fermion loops.
+4. Multiply the result by $$(-1)^F$$,
+ where $$F$$ is the number of closed fermion loops.
5. Depending on the context, additional constant factors may be required;
sometimes they are changed on-the-fly during a calculation.
Note that rules 4 and 5 are convention,
-just like the factors $i \hbar$ in $G^0$, $G$, $V$ and $W$;
+just like the factors $$i \hbar$$ in $$G^0$$, $$G$$, $$V$$ and $$W$$;
it simply turns out to be nicer to do it this way
when using Feynman diagrams in the wild.
@@ -171,11 +175,12 @@ in the Fourier transform definitions for space and time.
Working in Fourier space allows us to simplify calculations.
Consider the following diagram and the resulting expression,
-where $\tilde{\vb{r}} = (\vb{r}, t)$, and $\tilde{\vb{k}} = (\vb{k}, \omega)$:
+where $$\tilde{\vb{r}} = (\vb{r}, t)$$, and $$\tilde{\vb{k}} = (\vb{k}, \omega)$$:
+
$$\begin{aligned}
&= (i \hbar)^3 \sum_{s s'} \!\!\iint \dd{\tilde{\vb{r}}} \dd{\tilde{\vb{r}}'}
G_{s_1's}^0(\tilde{\vb{r}}_1', \tilde{\vb{r}}) G_{s s_1}^0(\tilde{\vb{r}}, \tilde{\vb{r}}_1) \delta_{s_1 s_1'}
@@ -230,8 +235,8 @@ $$\begin{aligned}
\end{aligned}$$
Conveniently, the Dirac delta functions that appear from the integrals
-represent conservation of wavevector $\vb{k}$ (momentum $\hbar \vb{k}$)
-and angular frequency $\omega$ (energy $\hbar \omega$).
+represent conservation of wavevector $$\vb{k}$$ (momentum $$\hbar \vb{k}$$)
+and angular frequency $$\omega$$ (energy $$\hbar \omega$$).
In Fourier space, it makes more sense
to regard the incoming energies and momenta and spins as given,
@@ -246,21 +251,21 @@ $$\begin{aligned}
\: G_{s_2}^0(\tilde{\vb{q}}_1 \!+\! \tilde{\vb{p}}) \: G_{s_2}^0(\tilde{\vb{q}}_1)
\end{aligned}$$
-Therefore, we say that fermion lines represent $i \hbar G_{s}^0(\vb{k}, \omega)$,
-interaction lines $W(\vb{k}) / i \hbar$, etc.,
+Therefore, we say that fermion lines represent $$i \hbar G_{s}^0(\vb{k}, \omega)$$,
+interaction lines $$W(\vb{k}) / i \hbar$$, etc.,
and the other interpretation rules are modified to the following:
-1. Each line has a momentum $\vb{k}$ and energy $\omega$,
- and each fermion line has a spin $s$;
+1. Each line has a momentum $$\vb{k}$$ and energy $$\omega$$,
+ and each fermion line has a spin $$s$$;
these must all be conserved at each vertex.
2. If both ends of a *fermion* line would be at the same time,
- multiply it by $e^{i \omega \eta}$,
- where $\eta \to 0^+$ is a positive infinitesimal,
- so e.g. $G(\tau, \tau) \to e^{i \omega \eta} G(\tau, \tau)$.
-3. Integrate over all internal $(\vb{k}, \omega)$,
- and sum over all internal spins $s$.
- Let each $(\vb{k}, \omega)$ integral contribute a factor $1 / (2 \pi)^4$.
-4. Multiply the end result by $(-1)^F$, where $F$ is the number of closed fermion loops.
+ multiply it by $$e^{i \omega \eta}$$,
+ where $$\eta \to 0^+$$ is a positive infinitesimal,
+ so e.g. $$G(\tau, \tau) \to e^{i \omega \eta} G(\tau, \tau)$$.
+3. Integrate over all internal $$(\vb{k}, \omega)$$,
+ and sum over all internal spins $$s$$.
+ Let each $$(\vb{k}, \omega)$$ integral contribute a factor $$1 / (2 \pi)^4$$.
+4. Multiply the end result by $$(-1)^F$$, where $$F$$ is the number of closed fermion loops.
5. Depending on the context, additional constant factors may be required;
sometimes they are changed on-the-fly during a calculation.
@@ -288,7 +293,7 @@ $$\begin{aligned}
= \Expval{\mathcal{T} \Big\{ \hat{\Psi}_H(\vb{r}_2, \tau_2) \hat{\Psi}_H^\dagger(\vb{r}_1, \tau_1) \Big\}}
\end{aligned}$$
-Where the time-ordering is with respect to $\tau$.
+Where the time-ordering is with respect to $$\tau$$.
Interaction lines are modified like so:
$$\begin{aligned}
@@ -298,14 +303,14 @@ $$\begin{aligned}
= -\frac{1}{\hbar} W(\vb{r}_2, \vb{r}_1; \tau_1) \delta(\tau_2 \!-\! \tau_1)
\end{aligned}$$
-One-body $V$-vertices are usually not used,
+One-body $$V$$-vertices are usually not used,
because they are intended for real-time-dependent operators,
-but in theory they would get a factor $-1/\hbar$ too.
+but in theory they would get a factor $$-1/\hbar$$ too.
For imaginary time, the Fourier transform is defined differently,
and a distinction must be made between
-fermionic Matsubara frequencies $i \omega_n^f$ (for $G$ and $G^0$)
-and bosonic Matsubara ones $i \omega_n^b$ (for $W$).
+fermionic Matsubara frequencies $$i \omega_n^f$$ (for $$G$$ and $$G^0$$)
+and bosonic Matsubara ones $$i \omega_n^b$$ (for $$W$$).
This distinction is compatible with frequency conservation,
since a sum of two fermionic frequencies is always bosonic.
We have:
@@ -322,8 +327,8 @@ $$\begin{aligned}
The interpretation in Fourier space is the same,
except that each internal integral/sum
-instead gives a constant $1 / \big(\hbar \beta (2 \pi)^3\big)$,
-and same-time fermion lines need a factor of $e^{i \omega_n^f \eta}$.
+instead gives a constant $$1 / \big(\hbar \beta (2 \pi)^3\big)$$,
+and same-time fermion lines need a factor of $$e^{i \omega_n^f \eta}$$.
diff --git a/source/know/concept/ficks-laws/index.md b/source/know/concept/ficks-laws/index.md
index f978ef5..b205af9 100644
--- a/source/know/concept/ficks-laws/index.md
+++ b/source/know/concept/ficks-laws/index.md
@@ -20,10 +20,10 @@ as opposed to **non-Fickian** or **anomalous diffusion**.
moves from regions of high concentration to regions of lower concentration,
at a rate proportional to the difference in concentration.
-Let $\vec{J}$ be the **diffusion flux** (with unit $\mathrm{m}^{-2} \mathrm{s}^{-1}$),
+Let $$\vec{J}$$ be the **diffusion flux** (with unit $$\mathrm{m}^{-2} \mathrm{s}^{-1}$$),
whose magnitude and direction describe the "flow" of diffusing matter.
Formally, Fick's first law predicts that the flux
-is proportional to the gradient of the concentration $C$ (with unit $\mathrm{m}^{-3}$):
+is proportional to the gradient of the concentration $$C$$ (with unit $$\mathrm{m}^{-3}$$):
$$\begin{aligned}
\boxed{
@@ -31,7 +31,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $D$ (with unit $\mathrm{m}^{2}/\mathrm{s}$)
+Where $$D$$ (with unit $$\mathrm{m}^{2}/\mathrm{s}$$)
is known as the **diffusion coefficient** or **diffusivity**,
and depends on both the medium and the diffusing substance.
@@ -47,17 +47,17 @@ than about diffusion in general.
To derive **Fick's second law**, we demand that matter is conserved,
i.e. the diffusing species is not created or destroyed anywhere.
-Suppose that an arbitrary volume $V$ contains an amount $M$ of diffusing matter,
-distributed in space according to $C(\vec{r})$, such that:
+Suppose that an arbitrary volume $$V$$ contains an amount $$M$$ of diffusing matter,
+distributed in space according to $$C(\vec{r})$$, such that:
$$\begin{aligned}
M
\equiv \int_V C \dd{V}
\end{aligned}$$
-Over time $t$, matter enters/leaves $V$.
-Let $S$ be the surface of $V$, and $\vec{J}$ the diffusion flux,
-then $M$ changes as follows, to which we apply the divergence theorem:
+Over time $$t$$, matter enters/leaves $$V$$.
+Let $$S$$ be the surface of $$V$$, and $$\vec{J}$$ the diffusion flux,
+then $$M$$ changes as follows, to which we apply the divergence theorem:
$$\begin{aligned}
\dv{M}{t}
@@ -65,8 +65,8 @@ $$\begin{aligned}
= - \int_V \nabla \cdot \vec{J} \dd{V}
\end{aligned}$$
-For comparison, we differentiate the definition of $M$,
-and exploit that the integral ignores $t$:
+For comparison, we differentiate the definition of $$M$$,
+and exploit that the integral ignores $$t$$:
$$\begin{aligned}
\dv{M}{t}
@@ -74,14 +74,14 @@ $$\begin{aligned}
= \int_V \pdv{C}{t} \dd{V}
\end{aligned}$$
-Both $\idv{M}{t}$ are equal, so stripping the integrals leads to this **continuity equation**:
+Both $$\idv{M}{t}$$ are equal, so stripping the integrals leads to this **continuity equation**:
$$\begin{aligned}
\pdv{C}{t}
= - \nabla \cdot \vec{J}
\end{aligned}$$
-From Fick's first law, we already have an expression for $\vec{J}$.
+From Fick's first law, we already have an expression for $$\vec{J}$$.
Substituting this into the continuity equation yields
the general form of Fick's second law:
@@ -92,8 +92,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Usually, it is assumed that $D$ is constant
-with respect to space $\vec{r}$ and concentration $C$,
+Usually, it is assumed that $$D$$ is constant
+with respect to space $$\vec{r}$$ and concentration $$C$$,
in which case Fick's second law reduces to:
$$\begin{aligned}
@@ -105,8 +105,8 @@ $$\begin{aligned}
Fick's second law has exact solutions for many situations,
but the most important one is arguably the **fundamental solution**.
-Consider a 1D system (for simplicity) with constant diffusivity $D$,
-where the initial concentration $C(x, 0)$ is
+Consider a 1D system (for simplicity) with constant diffusivity $$D$$,
+where the initial concentration $$C(x, 0)$$ is
a [Dirac delta function](/know/concept/dirac-delta-function/):
$$\begin{aligned}
@@ -114,7 +114,7 @@ $$\begin{aligned}
\end{aligned}$$
According to Fick's second law,
-the concentration's time evolution of $C$ turns out to be:
+the concentration's time evolution of $$C$$ turns out to be:
$$\begin{aligned}
H(x - x_0, t)
@@ -127,11 +127,11 @@ as a consequence of
the [central limit theorem](/know/concept/central-limit-theorem/):
the diffusion behaviour is a sum of many independent steps
(i.e. molecular collisions).
-The standard deviation is $\sqrt{2 D t}$,
-meaning that the distance of a diffusion is proportional to $\sqrt{t}$.
+The standard deviation is $$\sqrt{2 D t}$$,
+meaning that the distance of a diffusion is proportional to $$\sqrt{t}$$.
-This solution $H$ is extremely useful,
-because any initial concentration $C(x, 0)$ can be written as
+This solution $$H$$ is extremely useful,
+because any initial concentration $$C(x, 0)$$ can be written as
a convolution of itself with a delta function:
$$\begin{aligned}
@@ -142,7 +142,7 @@ $$\begin{aligned}
In other words, any function is a linear combination of delta functions.
Fick's second law is linear,
-so the overall solution $C(x, t)$ is the same combination of fundamental solutions $H$:
+so the overall solution $$C(x, t)$$ is the same combination of fundamental solutions $$H$$:
$$\begin{aligned}
C(x, t)
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md
index f8b5b2a..0bc849b 100644
--- a/source/know/concept/fourier-transform/index.md
+++ b/source/know/concept/fourier-transform/index.md
@@ -10,11 +10,11 @@ layout: "concept"
---
The **Fourier transform** (FT) is an integral transform which converts a
-function $f(x)$ into its frequency representation $\tilde{f}(k)$.
+function $$f(x)$$ into its frequency representation $$\tilde{f}(k)$$.
Great volumes have already been written about this subject,
so let us focus on the aspects that are useful to physicists.
-The **forward** FT is defined as follows, where $A$, $B$, and $s$ are unspecified constants
+The **forward** FT is defined as follows, where $$A$$, $$B$$, and $$s$$ are unspecified constants
(for now):
$$\begin{aligned}
@@ -35,9 +35,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Clearly, the inverse FT of the forward FT of $f(x)$ must equal $f(x)$
+Clearly, the inverse FT of the forward FT of $$f(x)$$ must equal $$f(x)$$
again. Let us verify this, by rearranging the integrals to get the
-[Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$:
+[Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$:
$$\begin{aligned}
\hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\}
@@ -49,29 +49,29 @@ $$\begin{aligned}
= \frac{2 \pi A B}{|s|} f(x)
\end{aligned}$$
-Therefore, the constants $A$, $B$, and $s$ are subject to the following
+Therefore, the constants $$A$$, $$B$$, and $$s$$ are subject to the following
constraint:
$$\begin{aligned}
\boxed{\frac{2\pi A B}{|s|} = 1}
\end{aligned}$$
-But that still gives a lot of freedom. The exact choices of $A$ and $B$
+But that still gives a lot of freedom. The exact choices of $$A$$ and $$B$$
are generally motivated by the [convolution theorem](/know/concept/convolution-theorem/)
and [Parseval's theorem](/know/concept/parsevals-theorem/).
-The choice of $|s|$ depends on whether the frequency variable $k$
-represents the angular ($|s| = 1$) or the physical ($|s| = 2\pi$)
-frequency. The sign of $s$ is not so important, but is generally based
-on whether the analysis is for forward ($s > 0$) or backward-propagating
-($s < 0$) waves.
+The choice of $$|s|$$ depends on whether the frequency variable $$k$$
+represents the angular ($$|s| = 1$$) or the physical ($$|s| = 2\pi$$)
+frequency. The sign of $$s$$ is not so important, but is generally based
+on whether the analysis is for forward ($$s > 0$$) or backward-propagating
+($$s < 0$$) waves.
## Derivatives
The FT of a derivative has a very useful property.
Below, after integrating by parts, we remove the boundary term by
-assuming that $f(x)$ is localized, i.e. $f(x) \to 0$ for $x \to \pm \infty$:
+assuming that $$f(x)$$ is localized, i.e. $$f(x) \to 0$$ for $$x \to \pm \infty$$:
$$\begin{aligned}
\hat{\mathcal{F}}\{f'(x)\}
@@ -82,7 +82,7 @@ $$\begin{aligned}
&= (- i s k) \tilde{f}(k)
\end{aligned}$$
-Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives
+Therefore, as long as $$f(x)$$ is localized, the FT eliminates derivatives
of the transformed variable, which makes it useful against PDEs:
$$\begin{aligned}
@@ -92,8 +92,8 @@ $$\begin{aligned}
\end{aligned}$$
This generalizes to higher-order derivatives, as long as these
-derivatives are also localized in the $x$-domain, which is practically
-guaranteed if $f(x)$ itself is localized:
+derivatives are also localized in the $$x$$-domain, which is practically
+guaranteed if $$f(x)$$ itself is localized:
$$\begin{aligned}
\boxed{
@@ -115,9 +115,9 @@ $$\begin{aligned}
## Multiple dimensions
-The Fourier transform is straightforward to generalize to $N$ dimensions.
-Given a scalar field $f(\vb{x})$ with $\vb{x} = (x_1, ..., x_N)$,
-its FT $\tilde{f}(\vb{k})$ is defined as follows:
+The Fourier transform is straightforward to generalize to $$N$$ dimensions.
+Given a scalar field $$f(\vb{x})$$ with $$\vb{x} = (x_1, ..., x_N)$$,
+its FT $$\tilde{f}(\vb{k})$$ is defined as follows:
$$\begin{aligned}
\boxed{
@@ -127,7 +127,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the wavevector $\vb{k} = (k_1, ..., k_N)$.
+Where the wavevector $$\vb{k} = (k_1, ..., k_N)$$.
Likewise, the inverse FT is given by:
$$\begin{aligned}
@@ -138,9 +138,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-In practice, in $N$D, there is not as much disagreement about
-the constants $A$, $B$ and $s$ as in 1D:
-typically $A = 1$ and $B = 1 / (2 \pi)^N$, with $s = \pm 1$.
+In practice, in $$N$$D, there is not as much disagreement about
+the constants $$A$$, $$B$$ and $$s$$ as in 1D:
+typically $$A = 1$$ and $$B = 1 / (2 \pi)^N$$, with $$s = \pm 1$$.
Any choice will do, as long as:
$$\begin{aligned}
@@ -155,7 +155,7 @@ $$\begin{aligned}
-Here, $B$ is the **center of buoyancy**, equal to
+Here, $$B$$ is the **center of buoyancy**, equal to
the center of mass of the volume of water displaced by the boat
as per [Archimedes' principle](/know/concept/archimedes-principle/).
-At equilibrium, the forces of buoyancy $\vb{F}_B$ and gravity $\vb{F}_G$
+At equilibrium, the forces of buoyancy $$\vb{F}_B$$ and gravity $$\vb{F}_G$$
have equal magnitudes in opposite directions,
-and $B$ is directly above or below $G$,
-or in other words, $x_B = x_G$ and $y_B = y_G$,
+and $$B$$ is directly above or below $$G$$,
+or in other words, $$x_B = x_G$$ and $$y_B = y_G$$,
which are calculated as follows:
$$\begin{aligned}
@@ -47,12 +47,12 @@ $$\begin{aligned}
&\equiv \frac{1}{V_{disp}} \iiint_{disp} (x, y, z) \dd{V}
\end{aligned}$$
-Where $V_{boat}$ is the volume of the whole boat,
-and $V_{disp}$ is the volume of liquid it displaces.
+Where $$V_{boat}$$ is the volume of the whole boat,
+and $$V_{disp}$$ is the volume of liquid it displaces.
Whether a given equilibrium is *stable* is more complicated.
-Suppose the ship is tilted by a small angle $\theta$ around the $x$-axis,
-in which case the old waterline, previously in the $z = 0$ plane,
+Suppose the ship is tilted by a small angle $$\theta$$ around the $$x$$-axis,
+in which case the old waterline, previously in the $$z = 0$$ plane,
gets shifted to a new plane, namely:
$$\begin{aligned}
@@ -61,8 +61,8 @@ $$\begin{aligned}
\approx \theta y
\end{aligned}$$
-Then $V_{disp}$ changes by $\Delta V_{disp}$, which is estimated below.
-If a point of the old waterline is raised by $z$,
+Then $$V_{disp}$$ changes by $$\Delta V_{disp}$$, which is estimated below.
+If a point of the old waterline is raised by $$z$$,
then the displaced liquid underneath it is reduced proportionally,
hence the sign:
@@ -73,17 +73,17 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-So $V_{disp}$ is unchanged, at least to first order in $\theta$.
+So $$V_{disp}$$ is unchanged, at least to first order in $$\theta$$.
However, the *shape* of the displaced volume may have changed significantly.
-Therefore, the shift of the position of the buoyancy center from $B$ to $B'$
-involves a correction $\Delta y_B$ in addition to the rotation by $\theta$:
+Therefore, the shift of the position of the buoyancy center from $$B$$ to $$B'$$
+involves a correction $$\Delta y_B$$ in addition to the rotation by $$\theta$$:
$$\begin{aligned}
y_B'
= y_B - \theta z_B + \Delta y_B
\end{aligned}$$
-We find $\Delta y_B$ by calculating the virtual buoyancy center of the shape difference:
+We find $$\Delta y_B$$ by calculating the virtual buoyancy center of the shape difference:
on the side of the boat that has been lifted by the rotation,
the center of buoyancy is "pushed" away due to the reduced displacement there,
and vice versa on the other side. Consequently:
@@ -95,7 +95,7 @@ $$\begin{aligned}
= - \frac{\theta I}{V_{disp}}
\end{aligned}$$
-Where we have defined the so-called **area moment** $I$ of the waterline as follows:
+Where we have defined the so-called **area moment** $$I$$ of the waterline as follows:
$$\begin{aligned}
\boxed{
@@ -104,8 +104,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Now that we have an expression for $\Delta y_B$,
-the new center's position $y_B'$ is found to be:
+Now that we have an expression for $$\Delta y_B$$,
+the new center's position $$y_B'$$ is found to be:
$$\begin{aligned}
y_B'
@@ -113,8 +113,8 @@ $$\begin{aligned}
\approx y_B - \sin(\theta) \: \Big( z_B + \frac{I}{V_{disp}} \Big)
\end{aligned}$$
-This looks like a rotation by $\theta$ around a so-called **metacenter** $M$,
-with a height $z_M$ known as the **metacentric height**, defined as:
+This looks like a rotation by $$\theta$$ around a so-called **metacenter** $$M$$,
+with a height $$z_M$$ known as the **metacentric height**, defined as:
$$\begin{aligned}
\boxed{
@@ -123,22 +123,22 @@ $$\begin{aligned}
}
\end{aligned}$$
-Meanwhile, the position of $M$ is defined such that it lies
-on the line between the old centers $G$ and $B$.
-Our calculation of $y_B'$ has shown that the new $B'$ always lies below $M$.
+Meanwhile, the position of $$M$$ is defined such that it lies
+on the line between the old centers $$G$$ and $$B$$.
+Our calculation of $$y_B'$$ has shown that the new $$B'$$ always lies below $$M$$.
After the rotation, the boat is not in equilibrium anymore,
-because the new $G'$ is not directly above or below $B'$.
-The force of gravity then causes a torque $\vb{T}$ given by:
+because the new $$G'$$ is not directly above or below $$B'$$.
+The force of gravity then causes a torque $$\vb{T}$$ given by:
$$\begin{aligned}
\vb{T}
= (\vb{r}_G' - \vb{r}_B') \cross m \vb{g}
\end{aligned}$$
-Where $\vb{g}$ points downwards.
-Since the rotation was around the $x$-axis,
-we are only interested in the $x$-component $T_x$, which becomes:
+Where $$\vb{g}$$ points downwards.
+Since the rotation was around the $$x$$-axis,
+we are only interested in the $$x$$-component $$T_x$$, which becomes:
$$\begin{aligned}
T_x
@@ -146,17 +146,17 @@ $$\begin{aligned}
= - \big((y_G - \theta z_G) - (y_B - \theta z_M)\big) m \mathrm{g}
\end{aligned}$$
-With $y_G' = y_G - \theta z_G$ being a simple rotation of $G$.
-At the initial equilibrium $y_G = y_B$, so:
+With $$y_G' = y_G - \theta z_G$$ being a simple rotation of $$G$$.
+At the initial equilibrium $$y_G = y_B$$, so:
$$\begin{aligned}
T_x
= \theta (z_G - z_M) m \mathrm{g}
\end{aligned}$$
-If $z_M < z_G$, then $T_x$ has the same sign as $\theta$,
-so $\vb{T}$ further destabilizes the boat.
-But if $z_M > z_G$, then $\vb{T}$ counteracts the rotation,
+If $$z_M < z_G$$, then $$T_x$$ has the same sign as $$\theta$$,
+so $$\vb{T}$$ further destabilizes the boat.
+But if $$z_M > z_G$$, then $$\vb{T}$$ counteracts the rotation,
and the boat returns to the original equilibrium,
leading us to the following stability condition:
@@ -167,7 +167,7 @@ $$\begin{aligned}
\end{aligned}$$
In other words, for a given boat design (or general shape)
-$z_G$ and $z_M$ can be calculated,
+$$z_G$$ and $$z_M$$ can be calculated,
and as long as they satisfy the above inequality,
it will float stably in water (or any other fluid,
although the buoyancy depends significantly on the density).
diff --git a/source/know/concept/microcanonical-ensemble/index.md b/source/know/concept/microcanonical-ensemble/index.md
index 03d37de..a56a506 100644
--- a/source/know/concept/microcanonical-ensemble/index.md
+++ b/source/know/concept/microcanonical-ensemble/index.md
@@ -10,18 +10,18 @@ layout: "concept"
---
The **microcanonical** or **NVE ensemble** is a statistical model
-of a theoretical system with constant internal energy $U$,
-volume $V$, and particle count $N$.
+of a theoretical system with constant internal energy $$U$$,
+volume $$V$$, and particle count $$N$$.
Consider a box with those properties.
We now put an imaginary rigid wall inside the box,
-thus dividing it into two subsystems $A$ and $B$,
+thus dividing it into two subsystems $$A$$ and $$B$$,
which can exchange energy (i.e. heat), but no particles.
-At any time, $A$ has energy $U_A$, and $B$ has $U_B$,
-so that in total $U = U_A \!+\! U_B$.
+At any time, $$A$$ has energy $$U_A$$, and $$B$$ has $$U_B$$,
+so that in total $$U = U_A \!+\! U_B$$.
The particles in each subsystem are in a certain **microstate** (configuration).
-For a given $U$, there is a certain number $c$
+For a given $$U$$, there is a certain number $$c$$
of possible whole-box microstates with that energy, given by:
$$\begin{aligned}
@@ -29,43 +29,43 @@ $$\begin{aligned}
= \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A)
\end{aligned}$$
-Where $c_A$ and $c_B$ are the numbers of subsystem microstates
+Where $$c_A$$ and $$c_B$$ are the numbers of subsystem microstates
at the given energy levels.
The core assumption of the microcanonical ensemble
-is that each of these microstates has the same probability $1 / c$.
-Consequently, the probability of finding an energy $U_A$ in $A$ is:
+is that each of these microstates has the same probability $$1 / c$$.
+Consequently, the probability of finding an energy $$U_A$$ in $$A$$ is:
$$\begin{aligned}
p_A(U_A)
= \frac{c_A(U_A) \: c_B(U - U_A)}{c(U)}
\end{aligned}$$
-If a certain $U_A$ has a higher probability,
-then there are more $A$-microstates with that energy,
+If a certain $$U_A$$ has a higher probability,
+then there are more $$A$$-microstates with that energy,
so, statistically, for an *ensemble* of many boxes,
-we expect that $U_A$ is more common.
+we expect that $$U_A$$ is more common.
-The maximum of $p_A$ will be the most common in the ensemble.
+The maximum of $$p_A$$ will be the most common in the ensemble.
Assuming that we have given the boxes enough time to settle,
we go one step further,
and refer to this maximum as "equilibrium".
In other words, the subsystem microstates at equilibrium
-are maxima of $p_A$ and $p_B$.
+are maxima of $$p_A$$ and $$p_B$$.
-We only need to look at $p_A$.
-Clearly, a maximum of $p_A$ is also a maximum of $\ln p_A$:
+We only need to look at $$p_A$$.
+Clearly, a maximum of $$p_A$$ is also a maximum of $$\ln p_A$$:
$$\begin{aligned}
\ln p_A(U_A)
= \ln{c_A(U_A)} + \ln{c_B(U - U_A)} - \ln{c(U)}
\end{aligned}$$
-Here, in the quantity $\ln{c_A}$,
+Here, in the quantity $$\ln{c_A}$$,
we recognize the definition of
-the entropy $S_A \equiv k \ln{c_A}$,
-where $k$ is Boltzmann's constant.
-We thus multiply by $k$:
+the entropy $$S_A \equiv k \ln{c_A}$$,
+where $$k$$ is Boltzmann's constant.
+We thus multiply by $$k$$:
$$\begin{aligned}
k \ln p_A(U_A)
@@ -73,10 +73,10 @@ $$\begin{aligned}
\end{aligned}$$
Since entropy is additive over subsystems,
-the total is $S = S_A + S_B$.
+the total is $$S = S_A + S_B$$.
To reach equilibrium, we are thus
**maximizing the total entropy**,
-meaning that $S$ is the [thermodynamic potential](/know/concept/thermodynamic-potential/)
+meaning that $$S$$ is the [thermodynamic potential](/know/concept/thermodynamic-potential/)
that corresponds to the microcanonical ensemble.
For our example, maximizing gives the following,
@@ -90,7 +90,7 @@ $$\begin{aligned}
\end{aligned}$$
By definition, the energy-derivative of the entropy
-is the reciprocal temperature $1 / T$.
+is the reciprocal temperature $$1 / T$$.
In other words,
equilibrium is reached when both subsystems
are at the same temperature:
@@ -102,16 +102,16 @@ $$\begin{aligned}
= \frac{1}{T_B}
\end{aligned}$$
-Recall that our partitioning into $A$ and $B$ was arbitrary,
-meaning that, in fact, the temperature $T$ must be uniform in the whole box.
+Recall that our partitioning into $$A$$ and $$B$$ was arbitrary,
+meaning that, in fact, the temperature $$T$$ must be uniform in the whole box.
We get this specific result because
-heat was the only thing that $A$ and $B$ could exchange.
+heat was the only thing that $$A$$ and $$B$$ could exchange.
The point is that the most likely state of the box
-maximizes the total entropy $S$.
+maximizes the total entropy $$S$$.
We also would have reached that conclusion
if our imaginary wall was permeable and flexible,
-i.e if it allowed changes in volume $V_A$ and particle count $N_A$.
+i.e if it allowed changes in volume $$V_A$$ and particle count $$N_A$$.
diff --git a/source/know/concept/modulational-instability/index.md b/source/know/concept/modulational-instability/index.md
index 628cbb4..a01293c 100644
--- a/source/know/concept/modulational-instability/index.md
+++ b/source/know/concept/modulational-instability/index.md
@@ -15,17 +15,17 @@ In fiber optics, **modulational instability** (MI)
is a nonlinear effect that leads to the exponential amplification
of background noise in certain frequency regions.
It only occurs in the [anomalous dispersion regime](/know/concept/dispersive-broadening/)
-($\beta_2 < 0$), which we will prove shortly.
+($$\beta_2 < 0$$), which we will prove shortly.
Consider the following simple solution to the nonlinear Schrödinger equation:
-a time-invariant constant power $P_0$ at the carrier frequency $\omega_0$,
+a time-invariant constant power $$P_0$$ at the carrier frequency $$\omega_0$$,
which is experiencing [self-phase modulation](/know/concept/self-phase-modulation/):
$$\begin{aligned}
A(z,t) = \sqrt{P_0} \exp( i \gamma P_0 z)
\end{aligned}$$
-We add a small perturbation $\varepsilon(z,t)$ to this signal,
+We add a small perturbation $$\varepsilon(z,t)$$ to this signal,
representing background noise:
$$\begin{aligned}
@@ -33,8 +33,8 @@ $$\begin{aligned}
\end{aligned}$$
We insert this into the nonlinear Schrödinger equation to get a perturbation equation,
-which we linearize by assuming that $|\varepsilon|^2$ is negligible compared to $P_0$,
-such that all higher-order terms of $\varepsilon$ can be dropped, yielding:
+which we linearize by assuming that $$|\varepsilon|^2$$ is negligible compared to $$P_0$$,
+such that all higher-order terms of $$\varepsilon$$ can be dropped, yielding:
$$\begin{aligned}
0
@@ -51,9 +51,9 @@ $$\begin{aligned}
\end{aligned}$$
We split the perturbation into real and imaginary parts
-$\varepsilon(z,t) = \varepsilon_r(z,t) + i \varepsilon_i(z,t)$,
+$$\varepsilon(z,t) = \varepsilon_r(z,t) + i \varepsilon_i(z,t)$$,
which we fill in in this equation.
-The point is that $\varepsilon_r$ and $\varepsilon_i$ are real functions:
+The point is that $$\varepsilon_r$$ and $$\varepsilon_i$$ are real functions:
$$\begin{aligned}
0
@@ -63,7 +63,7 @@ $$\begin{aligned}
\end{aligned}$$
Splitting this into its real and imaginary parts gives two PDEs
-relating $\varepsilon_r$ and $\varepsilon_i$:
+relating $$\varepsilon_r$$ and $$\varepsilon_i$$:
$$\begin{aligned}
\pdv{\varepsilon_r}{z} = \frac{\beta_2}{2} \pdvn{2}{\varepsilon_i}{t}
@@ -72,8 +72,8 @@ $$\begin{aligned}
\end{aligned}$$
We [Fourier transform](/know/concept/fourier-transform/)
-these in $t$ to turn them into ODEs relating
-$\tilde{\varepsilon}_r(z,\omega)$ and $\tilde{\varepsilon}_i(z,\omega)$:
+these in $$t$$ to turn them into ODEs relating
+$$\tilde{\varepsilon}_r(z,\omega)$$ and $$\tilde{\varepsilon}_i(z,\omega)$$:
$$\begin{aligned}
\pdv{\tilde{\varepsilon}_r}{z} = - \frac{\beta_2}{2} \omega^2 \tilde{\varepsilon}_i
@@ -82,7 +82,7 @@ $$\begin{aligned}
\end{aligned}$$
We are interested in exponential growth, so let us make the following ansatz,
-where $k$ may be a function of $\omega$, as long as it is $z$-invariant:
+where $$k$$ may be a function of $$\omega$$, as long as it is $$z$$-invariant:
$$\begin{aligned}
\tilde{\varepsilon}_r(z, \omega) = \tilde{\varepsilon}_r(0, \omega) \exp(k z)
@@ -91,7 +91,7 @@ $$\begin{aligned}
\end{aligned}$$
With this, we can write the system of ODEs for
-$\tilde{\varepsilon}_r(z,\omega)$ and $\tilde{\varepsilon}_i(z,\omega)$
+$$\tilde{\varepsilon}_r(z,\omega)$$ and $$\tilde{\varepsilon}_i(z,\omega)$$
in matrix form:
$$\begin{aligned}
@@ -112,7 +112,7 @@ $$\begin{aligned}
k = \pm \sqrt{ - \frac{\beta_2}{2} \omega^2 \Big( \frac{\beta_2}{2} \omega^2 + 2 \gamma P_0 \Big) }
\end{aligned}$$
-To get exponential growth, it is essential that $\mathrm{Re}\{k\} > 0$,
+To get exponential growth, it is essential that $$\mathrm{Re}\{k\} > 0$$,
so we discard the negative sign,
and get the following condition for MI:
@@ -124,22 +124,22 @@ $$\begin{aligned}
}
\end{aligned}$$
-Since $\omega^2$ is positive, $\beta_2$ must be negative,
+Since $$\omega^2$$ is positive, $$\beta_2$$ must be negative,
so MI can only occur in the ADR.
-It is worth noting that $\beta_2 = \beta_2(\omega_0)$,
+It is worth noting that $$\beta_2 = \beta_2(\omega_0)$$,
meaning there can only be exponential
noise growth when the "parent pulse" is in the anomalous dispersion regime,
but that growth may appear in areas of normal dispersion,
as long as the above condition is satisfied by the parent.
This result has been derived using perturbation,
-so only holds as long as $|\varepsilon|^2 \ll P_0$.
+so only holds as long as $$|\varepsilon|^2 \ll P_0$$.
Over time, the noise gets amplified so greatly
that this approximation breaks down.
-Next, we define the **gain** $g(\omega)$,
+Next, we define the **gain** $$g(\omega)$$,
which expresses how quickly the
-perturbation grows as a function of the frequency offset $\omega$:
+perturbation grows as a function of the frequency offset $$\omega$$:
$$\begin{aligned}
\boxed{
@@ -149,7 +149,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The frequencies with maximum gain are then found as extrema of $g(\omega)$,
+The frequencies with maximum gain are then found as extrema of $$g(\omega)$$,
which satisfy:
$$\begin{aligned}
@@ -162,10 +162,10 @@ $$\begin{aligned}
A simulation of MI is illustrated below.
The pulse considered was a solition of the following form
-with settings $T_0 = 10\:\mathrm{ps}$, $P_0 = 10\:\mathrm{kW}$,
-$\beta = -10\:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$,
+with settings $$T_0 = 10\:\mathrm{ps}$$, $$P_0 = 10\:\mathrm{kW}$$,
+$$\beta = -10\:\mathrm{ps}^2/\mathrm{m}$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$,
whose peak is approximately flat, so our derivation is valid there,
-hence it "wrinkles" in the $t$-domain:
+hence it "wrinkles" in the $$t$$-domain:
$$\begin{aligned}
A(0, t)
@@ -176,19 +176,19 @@ $$\begin{aligned}
-Where $L_\mathrm{NL} = 1/(\gamma P_0)$ is the characteristic length of nonlinear effects.
+Where $$L_\mathrm{NL} = 1/(\gamma P_0)$$ is the characteristic length of nonlinear effects.
Note that no noise was added to the simulation;
what you are seeing are pure numerical errors getting amplified.
-If one of the gain peaks accumulates a lot of energy quickly ($L_\mathrm{NL}$ is small),
+If one of the gain peaks accumulates a lot of energy quickly ($$L_\mathrm{NL}$$ is small),
and that peak is in the anomalous dispersion regime,
then it can in turn also cause MI in its own surroundings,
leading to a cascade of secondary and tertiary gain areas.
-This is seen above for $z > 30 L_\mathrm{NL}$.
+This is seen above for $$z > 30 L_\mathrm{NL}$$.
What we described is "pure" MI, but there also exists
a different type caused by Raman scattering.
-In that case, amplification occurs at the strongest peak of the Raman gain $\tilde{g}_R(\omega)$,
+In that case, amplification occurs at the strongest peak of the Raman gain $$\tilde{g}_R(\omega)$$,
even when the parent pulse is in the NDR.
This is an example of stimulated Raman scattering (SRS).
diff --git a/source/know/concept/multi-photon-absorption/index.md b/source/know/concept/multi-photon-absorption/index.md
index af433ac..5dd9887 100644
--- a/source/know/concept/multi-photon-absorption/index.md
+++ b/source/know/concept/multi-photon-absorption/index.md
@@ -11,10 +11,10 @@ categories:
layout: "concept"
---
-Consider a quantum system where there are many eigenstates $\Ket{n}$,
+Consider a quantum system where there are many eigenstates $$\Ket{n}$$,
e.g. atomic orbitals, for an electron to occupy.
Suppose an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
-passes by, such that its Hamiltonian gets perturbed by $\hat{H}_1$, given in the
+passes by, such that its Hamiltonian gets perturbed by $$\hat{H}_1$$, given in the
[electric dipole approximation](/know/concept/electric-dipole-approximation/) by:
$$\begin{aligned}
@@ -23,19 +23,19 @@ $$\begin{aligned}
\approx -\vu{p} \cdot \vb{E} e^{-i \omega t}
\end{aligned}$$
-Where $\vb{E}$ is the [electric field](/know/concept/electric-field/) amplitude,
-and $\vu{p} \equiv q \vu{x}$ is the transition dipole moment operator.
+Where $$\vb{E}$$ is the [electric field](/know/concept/electric-field/) amplitude,
+and $$\vu{p} \equiv q \vu{x}$$ is the transition dipole moment operator.
Here, we have made the
[rotating wave approximation](/know/concept/rotating-wave-approximation/)
-to neglect the $e^{i \omega t}$ term,
+to neglect the $$e^{i \omega t}$$ term,
because it turns out to be irrelevant in this discussion.
-We call the ground state $\Ket{0}$,
+We call the ground state $$\Ket{0}$$,
but other than that, the other states need *not* be sorted by energy.
However, we demand that the following holds
-for all even-numbered states $\Ket{e}$ and $\Ket{e'}$,
-and for all odd-numbered ($u$neven) states $\Ket{u}$ and $\Ket{u'}$:
+for all even-numbered states $$\Ket{e}$$ and $$\Ket{e'}$$,
+and for all odd-numbered ($$u$$neven) states $$\Ket{u}$$ and $$\Ket{u'}$$:
$$\begin{aligned}
\matrixel{e}{\hat{H}_1}{e'} = \matrixel{u}{\hat{H}_1}{u'} = 0
@@ -46,7 +46,7 @@ $$\begin{aligned}
This is justified for atomic orbitals thanks to
[Laporte's selection rule](/know/concept/selection-rules/).
Therefore, [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/)
-says that the $N$th-order coefficient corrections are:
+says that the $$N$$th-order coefficient corrections are:
$$\begin{aligned}
c_e^{(N)}(t)
@@ -56,8 +56,8 @@ $$\begin{aligned}
&= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(N-1)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau}
\end{aligned}$$
-Where $\omega_{eu} = (E_e \!-\! E_u) / \hbar$.
-For simplicity, the electron starts in the lowest-energy state $\Ket{0}$:
+Where $$\omega_{eu} = (E_e \!-\! E_u) / \hbar$$.
+For simplicity, the electron starts in the lowest-energy state $$\Ket{0}$$:
$$\begin{aligned}
c_0^{(0)} = 1
@@ -65,8 +65,8 @@ $$\begin{aligned}
c_u^{(0)} = c_{e \neq 0}^{(0)} = 0
\end{aligned}$$
-Finally, we prove the following useful relation for large $t$,
-involving a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
+Finally, we prove the following useful relation for large $$t$$,
+involving a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta$$:
$$\begin{aligned}
\lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
@@ -86,7 +86,7 @@ $$\begin{aligned}
= i e^{i x t / 2} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
\end{aligned}$$
-By taking the limit $t \to \infty$,
+By taking the limit $$t \to \infty$$,
it can be turned into a nascent Dirac delta function:
$$\begin{aligned}
@@ -102,7 +102,7 @@ $$\begin{aligned}
= 4 \pi^2 \delta^2(x)
\end{aligned}$$
-However, a squared delta function $\delta^2$ is not ideal,
+However, a squared delta function $$\delta^2$$ is not ideal,
so we take a step back:
$$\begin{aligned}
@@ -111,7 +111,7 @@ $$\begin{aligned}
= \delta(x) \lim_{t \to \infty} \frac{t}{2 \pi}
\end{aligned}$$
-Where we have set $x = 0$ according to the first delta function.
+Where we have set $$x = 0$$ according to the first delta function.
This gives the target:
$$\begin{aligned}
@@ -119,6 +119,7 @@ $$\begin{aligned}
= 4 \pi^2 \delta^2(x)
= 2 \pi \: \delta(x) \: t
\end{aligned}$$
+
@@ -127,7 +128,7 @@ $$\begin{aligned}
To warm up, we start at first-order perturbation theory.
Thanks to our choice of initial condition,
-nothing at all happens to any of the even-numbered states $\Ket{e}$:
+nothing at all happens to any of the even-numbered states $$\Ket{e}$$:
$$\begin{aligned}
c_e^{(1)}(t)
@@ -135,8 +136,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-While the odd-numbered states $\Ket{u}$ have a nonzero correction $c_u^{(1)}$,
-where $\vb{p}_{u0} = \matrixel{u}{\vu{p}}{0}$:
+While the odd-numbered states $$\Ket{u}$$ have a nonzero correction $$c_u^{(1)}$$,
+where $$\vb{p}_{u0} = \matrixel{u}{\vu{p}}{0}$$:
$$\begin{aligned}
c_u^{(1)}(t)
@@ -157,10 +158,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Since $\big| c_u^{(1)}(t) \big|^2$ is the probability
-of finding the electron in $\Ket{u}$,
-its transition rate $R_u^{(1)}(t)$ is as follows,
-averaged since the beginning $t = 0$:
+Since $$\big| c_u^{(1)}(t) \big|^2$$ is the probability
+of finding the electron in $$\Ket{u}$$,
+its transition rate $$R_u^{(1)}(t)$$ is as follows,
+averaged since the beginning $$t = 0$$:
$$\begin{aligned}
R_u^{(1)}(t)
@@ -169,7 +170,7 @@ $$\begin{aligned}
\cdot \bigg| \frac{e^{i (\omega_{u0} - \omega) t} - 1}{\omega_{u0} - \omega} \bigg|^2
\end{aligned}$$
-For large $t \to \infty$, we can use the formula we proved earlier
+For large $$t \to \infty$$, we can use the formula we proved earlier
to get [Fermi's golden rule](/know/concept/fermis-golden-rule/):
$$\begin{aligned}
@@ -180,17 +181,17 @@ $$\begin{aligned}
\end{aligned}$$
This well-known formula represents **one-photon absorption**:
-it peaks at $\omega_{u0} = \omega$, i.e. when one photon $\hbar \omega$
-has the exact energy of the transition $\hbar \omega_{u0}$.
-Note that this transition is only possible when $\matrixel{u}{\vu{p}}{0} \neq 0$,
-i.e. for any odd-numbered final state $\Ket{u}$.
+it peaks at $$\omega_{u0} = \omega$$, i.e. when one photon $$\hbar \omega$$
+has the exact energy of the transition $$\hbar \omega_{u0}$$.
+Note that this transition is only possible when $$\matrixel{u}{\vu{p}}{0} \neq 0$$,
+i.e. for any odd-numbered final state $$\Ket{u}$$.
## Two-photon absorption
Next, we go to second-order perturbation theory.
Based on the previous result, this time
-all odd-numbered states $\Ket{u}$ are unaffected:
+all odd-numbered states $$\Ket{u}$$ are unaffected:
$$\begin{aligned}
c_u^{(2)}(t)
@@ -198,8 +199,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-While the even-numbered states $\Ket{e}$ have the following correction,
-using $\omega_{eu} \!+\! \omega_{u0} = \omega_{e0}$:
+While the even-numbered states $$\Ket{e}$$ have the following correction,
+using $$\omega_{eu} \!+\! \omega_{u0} = \omega_{e0}$$:
$$\begin{aligned}
c_e^{(2)}(t)
@@ -213,7 +214,7 @@ $$\begin{aligned}
- \frac{e^{i (\omega_{eu} - \omega) \tau}}{i (\omega_{eu} - \omega)} \bigg]_0^t
\end{aligned}$$
-The second term represents one-photon absorption between $\Ket{u}$ and $\Ket{e}$.
+The second term represents one-photon absorption between $$\Ket{u}$$ and $$\Ket{e}$$.
We do not care about that, so we drop it, leaving only the first term:
$$\begin{aligned}
@@ -224,7 +225,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-As before, we can define a rate $R_e^{(2)}(t)$
+As before, we can define a rate $$R_e^{(2)}(t)$$
for all transitions represented by this term:
$$\begin{aligned}
@@ -234,7 +235,7 @@ $$\begin{aligned}
\cdot \bigg| \frac{e^{i (\omega_{e0} - 2 \omega) t} - 1}{\omega_{e0} - 2 \omega} \bigg|^2
\end{aligned}$$
-Which for $t \to \infty$ takes a similar form to Fermi's golden rule,
+Which for $$t \to \infty$$ takes a similar form to Fermi's golden rule,
using the formula we proved:
$$\begin{aligned}
@@ -245,19 +246,19 @@ $$\begin{aligned}
}
\end{aligned}$$
-This represents **two-photon absorption**, since it peaks at $\omega_{e0} = 2 \omega$:
-two identical photons $\hbar \omega$ are absorbed simultaneously
-to bridge the energy gap $\hbar \omega_{e0}$.
-Surprisingly, such a transition can only occur when $\matrixel{e}{\vu{p}}{0} = 0$,
-i.e. for any even-numbered final state $\Ket{e}$.
-Notice that the rate is proportional to $|\vb{E}|^4$,
+This represents **two-photon absorption**, since it peaks at $$\omega_{e0} = 2 \omega$$:
+two identical photons $$\hbar \omega$$ are absorbed simultaneously
+to bridge the energy gap $$\hbar \omega_{e0}$$.
+Surprisingly, such a transition can only occur when $$\matrixel{e}{\vu{p}}{0} = 0$$,
+i.e. for any even-numbered final state $$\Ket{e}$$.
+Notice that the rate is proportional to $$|\vb{E}|^4$$,
so this effect is only noticeable at high light intensities.
## Three-photon absorption
For third-order perturbation theory,
-all even-numbered states $\Ket{e}$ are unchanged:
+all even-numbered states $$\Ket{e}$$ are unchanged:
$$\begin{aligned}
c_e^{(3)}(t)
@@ -265,7 +266,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-And the odd-numbered states $\Ket{u}$ get the following third-order corrections:
+And the odd-numbered states $$\Ket{u}$$ get the following third-order corrections:
$$\begin{aligned}
c_u^{(3)}(t)
@@ -294,7 +295,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The resulting transition rate $R_u^{(3)}(t)$
+The resulting transition rate $$R_u^{(3)}(t)$$
is found to have the following familiar form:
$$\begin{aligned}
@@ -317,31 +318,31 @@ $$\begin{aligned}
}
\end{aligned}$$
-This represents **three-photon absorption**, since it peaks at $\omega_{u0} = 3 \omega$:
-three identical photons $\hbar \omega$ are absorbed simultaneously
-to bridge the energy gap $\hbar \omega_{u0}$.
+This represents **three-photon absorption**, since it peaks at $$\omega_{u0} = 3 \omega$$:
+three identical photons $$\hbar \omega$$ are absorbed simultaneously
+to bridge the energy gap $$\hbar \omega_{u0}$$.
This process is similar to one-photon absorption,
-in the sense that it can only occur if $\matrixel{u}{\vu{p}}{0} \neq 0$.
-The rate is proportional to $|\vb{E}|^6$,
+in the sense that it can only occur if $$\matrixel{u}{\vu{p}}{0} \neq 0$$.
+The rate is proportional to $$|\vb{E}|^6$$,
so this effect only appears at extremely high light intensities.
## N-photon absorption
A pattern has appeared in these calculations:
-in $N$th-order perturbation theory,
-we get a term representing $N$-photon absorption,
-with a transition rate proportional to $|\vb{E}|^{2N}$.
+in $$N$$th-order perturbation theory,
+we get a term representing $$N$$-photon absorption,
+with a transition rate proportional to $$|\vb{E}|^{2N}$$.
Indeed, we can derive infinitely many formulas in this way,
although the results become increasingly unrealistic
-due to the dependence on $\vb{E}$.
+due to the dependence on $$\vb{E}$$.
-If $N$ is odd, only odd-numbered destinations $\Ket{u}$ are allowed
-(assuming the electron starts in the ground state $\Ket{0}$),
-and if $N$ is even, only even-numbered destinations $\Ket{e}$.
+If $$N$$ is odd, only odd-numbered destinations $$\Ket{u}$$ are allowed
+(assuming the electron starts in the ground state $$\Ket{0}$$),
+and if $$N$$ is even, only even-numbered destinations $$\Ket{e}$$.
Note that nothing has been said about the energies of these states
-(other than $\Ket{0}$ being the minimum);
-everything is determined by the matrix elements $\matrixel{f}{\vu{p}}{i}$.
+(other than $$\Ket{0}$$ being the minimum);
+everything is determined by the matrix elements $$\matrixel{f}{\vu{p}}{i}$$.
diff --git a/source/know/concept/navier-cauchy-equation/index.md b/source/know/concept/navier-cauchy-equation/index.md
index 13a1ebb..5071c5f 100644
--- a/source/know/concept/navier-cauchy-equation/index.md
+++ b/source/know/concept/navier-cauchy-equation/index.md
@@ -12,9 +12,9 @@ The **Navier-Cauchy equation** describes **elastodynamics**:
the movements inside an elastic solid
in response to external forces and/or internal stresses.
-For a particle of the solid, whose position is given by the displacement field $\va{u}$,
+For a particle of the solid, whose position is given by the displacement field $$\va{u}$$,
Newton's second law is as follows,
-where $\dd{m}$ and $\dd{V}$ are the particle's mass and volume, respectively:
+where $$\dd{m}$$ and $$\dd{V}$$ are the particle's mass and volume, respectively:
$$\begin{aligned}
\va{f^*} \dd{V}
@@ -22,10 +22,10 @@ $$\begin{aligned}
= \rho \pdvn{2}{\va{u}}{t} \dd{V}
\end{aligned}$$
-Where $\rho$ is the mass density,
-and $\va{f^*}$ is the effective force density,
-defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
-like so, with $\va{f}$ being an external body force, e.g. from gravity:
+Where $$\rho$$ is the mass density,
+and $$\va{f^*}$$ is the effective force density,
+defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$
+like so, with $$\va{f}$$ being an external body force, e.g. from gravity:
$$\begin{aligned}
\va{f^*}
@@ -34,17 +34,17 @@ $$\begin{aligned}
We can therefore write Newton's second law as follows,
while switching to index notation,
-where $\nabla_j = \ipdv{}{x_j}$ is the partial derivative
-with respect to the $j$th coordinate:
+where $$\nabla_j = \ipdv{}{x_j}$$ is the partial derivative
+with respect to the $$j$$th coordinate:
$$\begin{aligned}
f_i + \sum_{j} \nabla_j \sigma_{ij}
= \rho \pdvn{2}{u_i}{t}
\end{aligned}$$
-The components $\sigma_{ij}$ of the Cauchy stress tensor
+The components $$\sigma_{ij}$$ of the Cauchy stress tensor
are given by [Hooke's law](/know/concept/hookes-law/),
-where $\mu$ and $\lambda$ are the Lamé coefficients,
+where $$\mu$$ and $$\lambda$$ are the Lamé coefficients,
which describe the material:
$$\begin{aligned}
@@ -52,10 +52,10 @@ $$\begin{aligned}
= 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk}
\end{aligned}$$
-In turn, the components $u_{ij}$ of the
+In turn, the components $$u_{ij}$$ of the
[Cauchy strain tensor](/know/concept/cauchy-strain-tensor/)
are defined as follows,
-where $u_i$ are once again the components of the displacement vector $\va{u}$:
+where $$u_i$$ are once again the components of the displacement vector $$\va{u}$$:
$$\begin{aligned}
u_{ij}
@@ -72,7 +72,7 @@ $$\begin{aligned}
&= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \nabla_i \sum_{j} u_{jj}
\end{aligned}$$
-And then into this we insert the definition of the strain components $u_{ij}$, yielding:
+And then into this we insert the definition of the strain components $$u_{ij}$$, yielding:
$$\begin{aligned}
\rho \pdvn{2}{u_i}{t}
diff --git a/source/know/concept/navier-stokes-equations/index.md b/source/know/concept/navier-stokes-equations/index.md
index fd26860..964acda 100644
--- a/source/know/concept/navier-stokes-equations/index.md
+++ b/source/know/concept/navier-stokes-equations/index.md
@@ -33,10 +33,10 @@ $$\begin{aligned}
= \va{f^*}
\end{aligned}$$
-$\mathrm{D}/\mathrm{D}t$ is the [material derivative](/know/concept/material-derivative/),
-$\rho$ is the density, and $\va{f^*}$ is the effective force density,
-expressed in terms of an external body force $\va{f}$ (e.g. gravity)
-and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$:
+$$\mathrm{D}/\mathrm{D}t$$ is the [material derivative](/know/concept/material-derivative/),
+$$\rho$$ is the density, and $$\va{f^*}$$ is the effective force density,
+expressed in terms of an external body force $$\va{f}$$ (e.g. gravity)
+and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$:
$$\begin{aligned}
\va{f^*}
@@ -51,8 +51,8 @@ $$\begin{aligned}
= - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
\end{aligned}$$
-Where $\eta$ is the dynamic viscosity.
-Inserting this, we calculate $\nabla \cdot \hat{\sigma}^\top$ in index notation:
+Where $$\eta$$ is the dynamic viscosity.
+Inserting this, we calculate $$\nabla \cdot \hat{\sigma}^\top$$ in index notation:
$$\begin{aligned}
\big( \nabla \cdot \hat{\sigma}^\top \big)_i
@@ -62,7 +62,7 @@ $$\begin{aligned}
&= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i
\end{aligned}$$
-Thanks to incompressibility $\nabla \cdot \va{v} = 0$,
+Thanks to incompressibility $$\nabla \cdot \va{v} = 0$$,
the middle term vanishes, leaving us with:
$$\begin{aligned}
@@ -70,7 +70,7 @@ $$\begin{aligned}
= \va{f} - \nabla p + \eta \nabla^2 \va{v}
\end{aligned}$$
-We assume that the only body force is gravity $\va{f} = \rho \va{g}$.
+We assume that the only body force is gravity $$\va{f} = \rho \va{g}$$.
Newton's second law then becomes:
$$\begin{aligned}
@@ -78,8 +78,8 @@ $$\begin{aligned}
= \rho \va{g} - \nabla p + \eta \nabla^2 \va{v}
\end{aligned}$$
-Dividing by $\rho$, and replacing $\eta$
-with the kinematic viscosity $\nu = \eta/\rho$,
+Dividing by $$\rho$$, and replacing $$\eta$$
+with the kinematic viscosity $$\nu = \eta/\rho$$,
yields the main equation:
$$\begin{aligned}
@@ -90,7 +90,7 @@ $$\begin{aligned}
\end{aligned}$$
Finally, we can optionally allow incompressible fluids
-with an inhomogeneous "lumpy" density $\rho$,
+with an inhomogeneous "lumpy" density $$\rho$$,
by demanding conservation of mass,
just like for the Euler equations:
@@ -115,11 +115,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Due to the definition of viscosity $\nu$ as the molecular "stickiness",
-we have boundary conditions for the velocity field $\va{v}$:
-at any interface, $\va{v}$ must be continuous.
+Due to the definition of viscosity $$\nu$$ as the molecular "stickiness",
+we have boundary conditions for the velocity field $$\va{v}$$:
+at any interface, $$\va{v}$$ must be continuous.
Likewise, Newton's third law demands that the normal component
-of stress $\hat{\sigma} \cdot \vu{n}$ is continuous there.
+of stress $$\hat{\sigma} \cdot \vu{n}$$ is continuous there.
diff --git a/source/know/concept/newtons-bucket/index.md b/source/know/concept/newtons-bucket/index.md
index 757262d..21740b9 100644
--- a/source/know/concept/newtons-bucket/index.md
+++ b/source/know/concept/newtons-bucket/index.md
@@ -10,23 +10,23 @@ layout: "concept"
---
**Newton's bucket** is a cylindrical bucket
-that rotates at angular velocity $\omega$.
+that rotates at angular velocity $$\omega$$.
Due to [viscosity](/know/concept/viscosity/),
any liquid in the bucket is affected by the rotation,
-eventually achieving the exact same $\omega$.
+eventually achieving the exact same $$\omega$$.
However, once in equilibrium, the liquid's surface is not flat,
but curved upwards from the center.
-This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$:
+This is due to the centrifugal force $$\va{F}_\mathrm{f} = m \va{f}$$ on a molecule with mass $$m$$:
$$\begin{aligned}
\va{f}
= \omega^2 \va{r}
\end{aligned}$$
-Where $\va{r}$ is the molecule's position relative to the axis of rotation.
+Where $$\va{r}$$ is the molecule's position relative to the axis of rotation.
This (fictitious) force can be written as the gradient
-of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$:
+of a potential $$\Phi_\mathrm{f}$$, such that $$\va{f} = - \nabla \Phi_\mathrm{f}$$:
$$\begin{aligned}
\Phi_\mathrm{f}
@@ -34,8 +34,8 @@ $$\begin{aligned}
= - \frac{\omega^2}{2} (x^2 + y^2)
\end{aligned}$$
-In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$,
-where $\va{g} = - \nabla \Phi_\mathrm{g}$:
+In addition, each molecule feels a gravitational force $$\va{F}_\mathrm{g} = m \va{g}$$,
+where $$\va{g} = - \nabla \Phi_\mathrm{g}$$:
$$\begin{aligned}
\Phi_\mathrm{g}
@@ -43,7 +43,7 @@ $$\begin{aligned}
\end{aligned}$$
Overall, the molecule therefore feels an "effective" force
-with a potential $\Phi$ given by:
+with a potential $$\Phi$$ given by:
$$\begin{aligned}
\Phi
@@ -51,7 +51,7 @@ $$\begin{aligned}
= \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2)
\end{aligned}$$
-At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$
+At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $$p$$
in the liquid is the one that satisfies:
$$\begin{aligned}
@@ -59,7 +59,7 @@ $$\begin{aligned}
= - \nabla \Phi
\end{aligned}$$
-Removing the gradients gives integration constants $p_0$ and $\Phi_0$,
+Removing the gradients gives integration constants $$p_0$$ and $$\Phi_0$$,
so the equilibrium equation is:
$$\begin{aligned}
@@ -67,16 +67,16 @@ $$\begin{aligned}
= - \rho (\Phi - \Phi_0)
\end{aligned}$$
-We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$,
-where $z_0$ is the liquid height at the center:
+We isolate this for $$p$$ and rewrite $$\Phi_0 = \mathrm{g} z_0$$,
+where $$z_0$$ is the liquid height at the center:
$$\begin{aligned}
p
= p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2)
\end{aligned}$$
-At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure.
-The $z$-coordinate at which this is satisfied is as follows,
+At the surface, we demand that $$p = p_0$$, where $$p_0$$ is the air pressure.
+The $$z$$-coordinate at which this is satisfied is as follows,
telling us that the surface is parabolic:
$$\begin{aligned}
diff --git a/source/know/concept/no-cloning-theorem/index.md b/source/know/concept/no-cloning-theorem/index.md
index d4ca0d4..a91ae6f 100644
--- a/source/know/concept/no-cloning-theorem/index.md
+++ b/source/know/concept/no-cloning-theorem/index.md
@@ -10,13 +10,13 @@ layout: "concept"
---
In quantum mechanics, the **no-cloning theorem** states
-there is no general way to make copies of an arbitrary quantum state $\ket{\psi}$.
+there is no general way to make copies of an arbitrary quantum state $$\ket{\psi}$$.
This has profound implications for quantum information.
To prove this theorem, let us pretend that a machine exists
that can do just that: copy arbitrary quantum states.
-Given an input $\ket{\psi}$ and a blank $\ket{?}$,
-this machines turns $\ket{?}$ into $\ket{\psi}$:
+Given an input $$\ket{\psi}$$ and a blank $$\ket{?}$$,
+this machines turns $$\ket{?}$$ into $$\ket{\psi}$$:
$$\begin{aligned}
\ket{\psi} \ket{?}
@@ -24,7 +24,7 @@ $$\begin{aligned}
\ket{\psi} \ket{\psi}
\end{aligned}$$
-We can use this device to make copies of the basis vectors $\ket{0}$ and $\ket{1}$:
+We can use this device to make copies of the basis vectors $$\ket{0}$$ and $$\ket{1}$$:
$$\begin{aligned}
\ket{0} \ket{?}
@@ -36,7 +36,7 @@ $$\begin{aligned}
\ket{1} \ket{1}
\end{aligned}$$
-If we feed this machine a superposition $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$,
+If we feed this machine a superposition $$\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$$,
we *want* the following behaviour:
$$\begin{aligned}
@@ -47,7 +47,7 @@ $$\begin{aligned}
&= \Big( \alpha^2 \ket{0} \ket{0} + \alpha \beta \ket{0} \ket{1} + \alpha \beta \ket{1} \ket{0} + \beta^2 \ket{1} \ket{1} \Big)
\end{aligned}$$
-Note the appearance of the cross terms with a factor of $\alpha \beta$.
+Note the appearance of the cross terms with a factor of $$\alpha \beta$$.
The problem is that the fundamental linearity of quantum mechanics
dictates different behaviour:
diff --git a/source/know/concept/optical-wave-breaking/index.md b/source/know/concept/optical-wave-breaking/index.md
index aa43d25..42064ff 100644
--- a/source/know/concept/optical-wave-breaking/index.md
+++ b/source/know/concept/optical-wave-breaking/index.md
@@ -14,18 +14,18 @@ In fiber optics, **optical wave breaking** (OWB) is a nonlinear effect
caused by interaction between
[group velocity dispersion](/know/concept/dispersive-broadening/) (GVD) and
[self-phase modulation](/know/concept/self-phase-modulation/) (SPM).
-It only happens in the normal dispersion regime ($\beta_2 > 0$)
+It only happens in the normal dispersion regime ($$\beta_2 > 0$$)
for pulses meeting a certain criterium, as we will see.
SPM creates low frequencies at the front of the pulse, and high ones at the back,
-and if $\beta_2 > 0$, GVD lets low frequencies travel faster than high ones.
+and if $$\beta_2 > 0$$, GVD lets low frequencies travel faster than high ones.
When those effects interact, the pulse gets temporally stretched
in a surprisingly sophisticated way.
-To illustrate this, the instantaneous frequency $\omega_i(z, t) = -\ipdv{\phi}{t}$
+To illustrate this, the instantaneous frequency $$\omega_i(z, t) = -\ipdv{\phi}{t}$$
has been plotted below for a theoretical Gaussian input pulse experiencing OWB,
-with settings $T_0 = 100\:\mathrm{fs}$, $P_0 = 5\:\mathrm{kW}$,
-$\beta_2 = 2\:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$.
+with settings $$T_0 = 100\:\mathrm{fs}$$, $$P_0 = 5\:\mathrm{kW}$$,
+$$\beta_2 = 2\:\mathrm{ps}^2/\mathrm{m}$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$.
In the left panel, we see the typical S-shape caused by SPM,
and the arrows indicate the direction that GVD is pushing the curve in.
@@ -41,30 +41,30 @@ hence the name *wave breaking*:
Several interesting things happen around this moment.
To demonstrate this, spectrograms of the same simulation
have been plotted below, together with pulse profiles
-in both the $t$-domain and $\omega$-domain on an arbitrary linear scale
+in both the $$t$$-domain and $$\omega$$-domain on an arbitrary linear scale
(click the image to get a better look).
Initially, the spectrum broadens due to SPM in the usual way,
but shortly after OWB, this process is stopped by the appearance
-of so-called **sidelobes** in the $\omega$-domain on either side of the pulse.
+of so-called **sidelobes** in the $$\omega$$-domain on either side of the pulse.
In the meantime, in the time domain,
the pulse steepens at the edges, but flattens at the peak.
After OWB, a train of small waves falls off the edges,
-which eventually melt together, leading to a trapezoid shape in the $t$-domain.
+which eventually melt together, leading to a trapezoid shape in the $$t$$-domain.
Dispersive broadening then continues normally:
-We call the distance at which the wave breaks $L_\mathrm{WB}$,
+We call the distance at which the wave breaks $$L_\mathrm{WB}$$,
and would like to analytically predict it.
-We do this using the instantaneous frequency $\omega_i$,
+We do this using the instantaneous frequency $$\omega_i$$,
by estimating when the SPM fluctuations overtake their own base,
as was illustrated earlier.
-To get $\omega_i$ of a Gaussian pulse experiencing both GVD and SPM,
-it is a reasonable approximation, for small $z$, to simply add up
+To get $$\omega_i$$ of a Gaussian pulse experiencing both GVD and SPM,
+it is a reasonable approximation, for small $$z$$, to simply add up
the instantaneous frequencies for these separate effects:
$$\begin{aligned}
@@ -74,7 +74,7 @@ $$\begin{aligned}
+ 2\gamma P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
\end{aligned}$$
-Assuming that $z$ is small enough such that $z^2 \approx 0$, this
+Assuming that $$z$$ is small enough such that $$z^2 \approx 0$$, this
expression can be reduced to:
$$\begin{aligned}
@@ -83,8 +83,8 @@ $$\begin{aligned}
= \frac{\beta_2 t z}{T_0^4} \bigg( 1 + 2 N_\mathrm{sol}^2 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
\end{aligned}$$
-Where we have assumed $\beta_2 > 0$,
-and $N_\mathrm{sol}$ is the **soliton number**,
+Where we have assumed $$\beta_2 > 0$$,
+and $$N_\mathrm{sol}$$ is the **soliton number**,
which is defined as:
$$\begin{aligned}
@@ -93,11 +93,11 @@ $$\begin{aligned}
This quantity is very important in anomalous dispersion,
but even in normal dispesion, it is still a useful measure of the relative strengths of GVD and SPM.
-As was illustrated earlier, $\omega_i$ overtakes itself at the edges,
-so OWB occurs when $\omega_i$ oscillates there,
-which starts when its $t$-derivative,
-the **instantaneous chirpyness** $\xi_i$,
-has *two* real roots for $t^2$:
+As was illustrated earlier, $$\omega_i$$ overtakes itself at the edges,
+so OWB occurs when $$\omega_i$$ oscillates there,
+which starts when its $$t$$-derivative,
+the **instantaneous chirpyness** $$\xi_i$$,
+has *two* real roots for $$t^2$$:
$$\begin{aligned}
0
@@ -107,10 +107,10 @@ $$\begin{aligned}
= \frac{\beta_2 z}{T_0^4} \: f\Big(\frac{t^2}{T_0^2}\Big)
\end{aligned}$$
-Where the function $f(x)$ has been defined for convenience. As it turns
-out, this equation can be solved analytically using the Lambert $W$ function,
-leading to the following exact minimum value $N_\mathrm{min}^2$ for $N_\mathrm{sol}^2$,
-such that OWB can only occur when $N_\mathrm{sol}^2 > N_\mathrm{min}^2$:
+Where the function $$f(x)$$ has been defined for convenience. As it turns
+out, this equation can be solved analytically using the Lambert $$W$$ function,
+leading to the following exact minimum value $$N_\mathrm{min}^2$$ for $$N_\mathrm{sol}^2$$,
+such that OWB can only occur when $$N_\mathrm{sol}^2 > N_\mathrm{min}^2$$:
$$\begin{aligned}
\boxed{
@@ -118,15 +118,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-If this condition $N_\mathrm{sol}^2 > N_\mathrm{min}^2$ is not satisfied,
-$\xi_i$ cannot have two roots for $t^2$, meaning $\omega_i$ cannot overtake itself.
+If this condition $$N_\mathrm{sol}^2 > N_\mathrm{min}^2$$ is not satisfied,
+$$\xi_i$$ cannot have two roots for $$t^2$$, meaning $$\omega_i$$ cannot overtake itself.
GVD is unable to keep up with SPM, so OWB will not occur.
-Next, consider two points at $t_1$ and $t_2$ in the pulse,
-separated by a small initial interval $(t_2 - t_1)$.
-The frequency difference between these points due to $\omega_i$
+Next, consider two points at $$t_1$$ and $$t_2$$ in the pulse,
+separated by a small initial interval $$(t_2 - t_1)$$.
+The frequency difference between these points due to $$\omega_i$$
will cause them to displace relative to each other
-after a short distance $z$ by some amount $\Delta t$,
+after a short distance $$z$$ by some amount $$\Delta t$$,
estimated by:
$$\begin{aligned}
@@ -144,12 +144,12 @@ $$\begin{aligned}
&&\Delta\xi_i \equiv \xi_i(z,t_2) - \xi_i(z,t_1)
\end{aligned}$$
-Where $\beta_1(\omega)$ is the inverse of the group velocity.
-OWB takes place when $t_2$ and $t_1$ catch up to each other,
-which is when $-\Delta t = (t_2 - t_1)$.
-The distance where this happens first, $z = L_\mathrm{WB}$,
+Where $$\beta_1(\omega)$$ is the inverse of the group velocity.
+OWB takes place when $$t_2$$ and $$t_1$$ catch up to each other,
+which is when $$-\Delta t = (t_2 - t_1)$$.
+The distance where this happens first, $$z = L_\mathrm{WB}$$,
must therefore satisfy the following condition
-for a particular value of $t$:
+for a particular value of $$t$$:
$$\begin{aligned}
L_\mathrm{WB} \, \beta_2 \, \xi_i(L_\mathrm{WB}, t) = -1
@@ -157,9 +157,9 @@ $$\begin{aligned}
L_\mathrm{WB}^2 = - \frac{T_0^4}{\beta_2^2 \, f(t^2/T_0^2)}
\end{aligned}$$
-The time $t$ of OWB must be where $\omega_i(t)$ has its steepest slope,
-which is at the minimum value of $\xi_i(t)$, and by extension $f(x)$.
-This turns out to be $f(3/2)$:
+The time $$t$$ of OWB must be where $$\omega_i(t)$$ has its steepest slope,
+which is at the minimum value of $$\xi_i(t)$$, and by extension $$f(x)$$.
+This turns out to be $$f(3/2)$$:
$$\begin{aligned}
f_\mathrm{min} = f(3/2)
@@ -167,10 +167,10 @@ $$\begin{aligned}
= 1 - N_\mathrm{sol}^2 / N_\mathrm{min}^2
\end{aligned}$$
-Clearly, $f_\mathrm{min} \ge 0$ when $N_\mathrm{sol}^2 \le N_\mathrm{min}^2$,
-which, when inserted above, leads to an imaginary $L_\mathrm{WB}$,
+Clearly, $$f_\mathrm{min} \ge 0$$ when $$N_\mathrm{sol}^2 \le N_\mathrm{min}^2$$,
+which, when inserted above, leads to an imaginary $$L_\mathrm{WB}$$,
confirming that OWB cannot occur in that case.
-Otherwise, if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then:
+Otherwise, if $$N_\mathrm{sol}^2 > N_\mathrm{min}^2$$, then:
$$\begin{aligned}
\boxed{
@@ -180,24 +180,24 @@ $$\begin{aligned}
}
\end{aligned}$$
-This prediction for $L_\mathrm{WB}$ appears to agree well
+This prediction for $$L_\mathrm{WB}$$ appears to agree well
with the OWB observed in the simulation:
-Because all spectral broadening up to $L_\mathrm{WB}$ is caused by SPM,
+Because all spectral broadening up to $$L_\mathrm{WB}$$ is caused by SPM,
whose frequency behaviour is known, it is in fact possible to draw
some analytical conclusions about the achieved bandwidth when OWB sets in.
-Filling $L_\mathrm{WB}$ in into $\omega_\mathrm{SPM}$ gives:
+Filling $$L_\mathrm{WB}$$ in into $$\omega_\mathrm{SPM}$$ gives:
$$\begin{aligned}
\omega_{\mathrm{SPM}}(L_\mathrm{WB},t)
= \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
\end{aligned}$$
-Assuming that $N_\mathrm{sol}^2$ is large in the denominator, this can
+Assuming that $$N_\mathrm{sol}^2$$ is large in the denominator, this can
be approximately reduced to:
$$\begin{aligned}
@@ -206,18 +206,18 @@ $$\begin{aligned}
= 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
\end{aligned}$$
-The expression $x \exp(-x^2)$ has its global extrema
-$\pm 1 / \sqrt{2 e}$ at $x^2 = 1/2$. The maximum SPM frequency shift
-achieved at $L_\mathrm{WB}$ is therefore given by:
+The expression $$x \exp(-x^2)$$ has its global extrema
+$$\pm 1 / \sqrt{2 e}$$ at $$x^2 = 1/2$$. The maximum SPM frequency shift
+achieved at $$L_\mathrm{WB}$$ is therefore given by:
$$\begin{aligned}
\omega_\mathrm{max} = \sqrt{\frac{2 \gamma P_0}{e \beta_2}}
\end{aligned}$$
-Interestingly, this expression does not contain $T_0$ at all,
+Interestingly, this expression does not contain $$T_0$$ at all,
so the achieved spectrum when SPM is halted by OWB
is independent of the pulse width,
-for sufficiently large $N_\mathrm{sol}$.
+for sufficiently large $$N_\mathrm{sol}$$.
## References
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index e1d73a7..df90244 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -8,11 +8,11 @@ categories:
layout: "concept"
---
-**Parseval's theorem** is a relation between the inner product of two functions $f(x)$ and $g(x)$,
+**Parseval's theorem** is a relation between the inner product of two functions $$f(x)$$ and $$g(x)$$,
and the inner product of their [Fourier transforms](/know/concept/fourier-transform/)
-$\tilde{f}(k)$ and $\tilde{g}(k)$.
+$$\tilde{f}(k)$$ and $$\tilde{g}(k)$$.
There are two equivalent ways of stating it,
-where $A$, $B$, and $s$ are constants from the FT's definition:
+where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition:
$$\begin{aligned}
\boxed{
@@ -48,7 +48,7 @@ $$\begin{aligned}
= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
\end{aligned}$$
-Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Where $$\delta(k)$$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
Note that we can equally well do this proof in the opposite direction,
which yields an equivalent result:
@@ -68,11 +68,12 @@ $$\begin{aligned}
&= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
= \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
\end{aligned}$$
+
For this reason, physicists like to define the Fourier transform
-with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
+with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely
conserves the functions' normalization.
diff --git a/source/know/concept/partial-fraction-decomposition/index.md b/source/know/concept/partial-fraction-decomposition/index.md
index 03c1c76..bb7faa2 100644
--- a/source/know/concept/partial-fraction-decomposition/index.md
+++ b/source/know/concept/partial-fraction-decomposition/index.md
@@ -8,15 +8,15 @@ layout: "concept"
---
**Partial fraction decomposition** or **partial fraction expansion**
-is a method to rewrite quotients of two polynomials $g(x)$ and $h(x)$,
-where the numerator $g(x)$ is of lower order than $h(x)$,
-as sums of fractions with $x$ in the denominator:
+is a method to rewrite quotients of two polynomials $$g(x)$$ and $$h(x)$$,
+where the numerator $$g(x)$$ is of lower order than $$h(x)$$,
+as sums of fractions with $$x$$ in the denominator:
$$\begin{aligned}
f(x) = \frac{g(x)}{h(x)} = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ...
\end{aligned}$$
-Where $h_n$ etc. are the roots of the denominator $h(x)$. If all $N$ of
+Where $$h_n$$ etc. are the roots of the denominator $$h(x)$$. If all $$N$$ of
these roots are distinct, then it is sufficient to simply posit:
$$\begin{aligned}
@@ -25,8 +25,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-The constants $c_n$ can either be found the hard way,
-by multiplying the denominators around and solving a system of $N$
+The constants $$c_n$$ can either be found the hard way,
+by multiplying the denominators around and solving a system of $$N$$
equations, or the easy way by using this trick:
$$\begin{aligned}
@@ -35,7 +35,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $h_1$ is a root with multiplicity $m > 1$, then the sum takes the form of:
+If $$h_1$$ is a root with multiplicity $$m > 1$$, then the sum takes the form of:
$$\begin{aligned}
\boxed{
@@ -44,15 +44,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $c_{1,j}$ are found by putting the terms on a common denominator, e.g.
+Where $$c_{1,j}$$ are found by putting the terms on a common denominator, e.g.
$$\begin{aligned}
\frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2}
= \frac{c_{1,1} (x - h_1) + c_{1,2}}{(x - h_1)^2}
\end{aligned}$$
-And then, using the linear independence of $x^0, x^1, x^2, ...$, solving
-a system of $m$ equations to find all $c_{1,1}, ..., c_{1,m}$.
+And then, using the linear independence of $$x^0, x^1, x^2, ...$$, solving
+a system of $$m$$ equations to find all $$c_{1,1}, ..., c_{1,m}$$.
diff --git a/source/know/concept/path-integral-formulation/index.md b/source/know/concept/path-integral-formulation/index.md
index b92cf5f..a8dcc76 100644
--- a/source/know/concept/path-integral-formulation/index.md
+++ b/source/know/concept/path-integral-formulation/index.md
@@ -14,10 +14,10 @@ which is equivalent to the "traditional" Schrödinger equation.
Whereas the latter is based on [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
the former comes from [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
-It expresses the [propagator](/know/concept/propagator/) $K$
-using the following sum over all possible paths $x(t)$,
-which all go from the initial position $x_0$ at time $t_0$
-to the destination $x_N$ at time $t_N$:
+It expresses the [propagator](/know/concept/propagator/) $$K$$
+using the following sum over all possible paths $$x(t)$$,
+which all go from the initial position $$x_0$$ at time $$t_0$$
+to the destination $$x_N$$ at time $$t_N$$:
$$\begin{aligned}
\boxed{
@@ -26,32 +26,32 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $A$ normalizes.
-$S[x]$ is the classical action of the path $x$, whose minimization yields
+Where $$A$$ normalizes.
+$$S[x]$$ is the classical action of the path $$x$$, whose minimization yields
the Euler-Lagrange equation from Lagrangian mechanics.
Note that each path is given an equal weight,
even unrealistic paths that make big detours.
This apparent problem solves itself,
-thanks to the fact that paths close to the classical optimum $x_c(t)$
-have an action close to $S_c = S[x_c]$,
+thanks to the fact that paths close to the classical optimum $$x_c(t)$$
+have an action close to $$S_c = S[x_c]$$,
while the paths far away have very different actions.
-Since $S[x]$ is inside a complex exponential,
-this means that paths close to $x_c$ add contructively,
+Since $$S[x]$$ is inside a complex exponential,
+this means that paths close to $$x_c$$ add contructively,
and the others add destructively and cancel out.
-An interesting way too look at it is by varying $\hbar$:
+An interesting way too look at it is by varying $$\hbar$$:
as its value decreases, minor action differences yield big phase differences,
-which make the quantum wave function stay closer to $x_c$.
-In the limit $\hbar \to 0$, quantum mechanics thus turns into classical mechanics.
+which make the quantum wave function stay closer to $$x_c$$.
+In the limit $$\hbar \to 0$$, quantum mechanics thus turns into classical mechanics.
## Time-slicing derivation
The most popular way to derive the path integral formulation proceeds as follows:
-starting from the definition of the propagator $K$,
-we divide the time interval $t_N - t_0$ into $N$ "slices"
-of equal width $\Delta t = (t_N - t_0) / N$,
-where $N$ is large:
+starting from the definition of the propagator $$K$$,
+we divide the time interval $$t_N - t_0$$ into $$N$$ "slices"
+of equal width $$\Delta t = (t_N - t_0) / N$$,
+where $$N$$ is large:
$$\begin{aligned}
K(x_N, t_N; x_0, t_0)
@@ -59,9 +59,9 @@ $$\begin{aligned}
= \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar} \cdots e^{- i \hat{H} \Delta t / \hbar}}{x_0}
\end{aligned}$$
-Between the exponentials we insert $N\!-\!1$ identity operators
-$\hat{I} = \int \Ket{x} \Bra{x} \dd{x}$,
-and define $x_j = x(t_j)$ for an arbitrary path $x(t)$:
+Between the exponentials we insert $$N\!-\!1$$ identity operators
+$$\hat{I} = \int \Ket{x} \Bra{x} \dd{x}$$,
+and define $$x_j = x(t_j)$$ for an arbitrary path $$x(t)$$:
$$\begin{aligned}
K
@@ -69,10 +69,10 @@ $$\begin{aligned}
\dd{x_1} \cdots \dd{x_{N - 1}}
\end{aligned}$$
-For sufficiently small time steps $\Delta t$ (i.e. large $N$
+For sufficiently small time steps $$\Delta t$$ (i.e. large $$N$$
we make the following approximation
(which would be exact, were it not for the fact that
-$\hat{T}$ and $\hat{V}$ are operators):
+$$\hat{T}$$ and $$\hat{V}$$ are operators):
$$\begin{aligned}
e^{- i \hat{H} \Delta t / \hbar}
@@ -80,7 +80,7 @@ $$\begin{aligned}
\approx e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}
\end{aligned}$$
-Since $\hat{V} = V(x_j)$,
+Since $$\hat{V} = V(x_j)$$,
we can take it out of the inner product as a constant factor:
$$\begin{aligned}
@@ -89,15 +89,15 @@ $$\begin{aligned}
\end{aligned}$$
Here we insert the identity operator
-expanded in the momentum basis $\hat{I} = \int \Ket{p} \Bra{p} \dd{p}$,
-and commute it with the kinetic energy $\hat{T} = \hat{p}^2 / (2m)$ to get:
+expanded in the momentum basis $$\hat{I} = \int \Ket{p} \Bra{p} \dd{p}$$,
+and commute it with the kinetic energy $$\hat{T} = \hat{p}^2 / (2m)$$ to get:
$$\begin{aligned}
\matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
= \int_{-\infty}^\infty \Inprod{x_{j+1}}{p} \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \Inprod{p}{x_j} \dd{p}
\end{aligned}$$
-In the momentum basis $\Ket{p}$,
+In the momentum basis $$\Ket{p}$$,
the position basis vectors
are represented by plane waves:
@@ -119,7 +119,7 @@ $$\begin{aligned}
&= \frac{1}{2 \pi \hbar} \sqrt{\frac{2 \pi m \hbar}{i \Delta t}} \exp\!\Big( i \frac{m (x_{j+1} - x_j)^2}{2 \hbar \Delta t} \Big)
\end{aligned}$$
-Inserting this back into the definition of the propagator $K(x_N, t_N; x_0, t_0)$ yields:
+Inserting this back into the definition of the propagator $$K(x_N, t_N; x_0, t_0)$$ yields:
$$\begin{aligned}
K
@@ -129,7 +129,7 @@ $$\begin{aligned}
\dd{x_1} \cdots \dd{x_{N-1}}
\end{aligned}$$
-For large $N$ and small $\Delta t$, the sum in the exponent becomes an integral:
+For large $$N$$ and small $$\Delta t$$, the sum in the exponent becomes an integral:
$$\begin{aligned}
\frac{i}{\hbar} \sum_{j = 0}^{N - 1} \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \Delta t^2} - V(x_j) \Big) \Delta t
@@ -137,8 +137,8 @@ $$\begin{aligned}
\frac{i}{\hbar} \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
\end{aligned}$$
-Upon closer inspection, this integral turns out to be the classical action $S[x]$,
-with the integrand being the Lagrangian $L$:
+Upon closer inspection, this integral turns out to be the classical action $$S[x]$$,
+with the integrand being the Lagrangian $$L$$:
$$\begin{aligned}
S[x(t)]
@@ -146,7 +146,7 @@ $$\begin{aligned}
= \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
\end{aligned}$$
-The definition of the propagator $K$ is then further reduced to the following:
+The definition of the propagator $$K$$ is then further reduced to the following:
$$\begin{aligned}
K
@@ -155,8 +155,8 @@ $$\begin{aligned}
\end{aligned}$$
Finally, for the purpose of normalization,
-we define the integral over all paths $x(t)$ as follows,
-where we write $D[x]$ instead of $\dd{x}$:
+we define the integral over all paths $$x(t)$$ as follows,
+where we write $$D[x]$$ instead of $$\dd{x}$$:
$$\begin{aligned}
\int D[x]
@@ -164,7 +164,7 @@ $$\begin{aligned}
\end{aligned}$$
We thus arrive at **Feynman's path integral**,
-which sums over all possible paths $x(t)$:
+which sums over all possible paths $$x(t)$$:
$$\begin{aligned}
K
diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
index 58a3f69..9821718 100644
--- a/source/know/concept/pauli-exclusion-principle/index.md
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -12,9 +12,9 @@ In quantum mechanics, the **Pauli exclusion principle** is a theorem with
profound consequences for how the world works.
Suppose we have a composite state
-$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$, where the two
-identical particles $x_1$ and $x_2$ each can occupy the same two allowed
-states $a$ and $b$. We then define the permutation operator $\hat{P}$ as
+$$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$$, where the two
+identical particles $$x_1$$ and $$x_2$$ each can occupy the same two allowed
+states $$a$$ and $$b$$. We then define the permutation operator $$\hat{P}$$ as
follows:
$$\begin{aligned}
@@ -28,22 +28,22 @@ $$\begin{aligned}
\hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b}
\end{aligned}$$
-Therefore, $\Ket{a}\Ket{b}$ is an eigenvector of $\hat{P}^2$ with
-eigenvalue $1$. Since $[\hat{P}, \hat{P}^2] = 0$, $\Ket{a}\Ket{b}$
-must also be an eigenket of $\hat{P}$ with eigenvalue $\lambda$,
-satisfying $\lambda^2 = 1$, so we know that $\lambda = 1$ or $\lambda = -1$:
+Therefore, $$\Ket{a}\Ket{b}$$ is an eigenvector of $$\hat{P}^2$$ with
+eigenvalue $$1$$. Since $$[\hat{P}, \hat{P}^2] = 0$$, $$\Ket{a}\Ket{b}$$
+must also be an eigenket of $$\hat{P}$$ with eigenvalue $$\lambda$$,
+satisfying $$\lambda^2 = 1$$, so we know that $$\lambda = 1$$ or $$\lambda = -1$$:
$$\begin{aligned}
\hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
\end{aligned}$$
As it turns out, in nature, each class of particle has a single
-associated permutation eigenvalue $\lambda$, or in other words: whether
-$\lambda$ is $-1$ or $1$ depends on the type of particle that $x_1$
-and $x_2$ are. Particles with $\lambda = -1$ are called
-**fermions**, and those with $\lambda = 1$ are known as **bosons**. We
-define $\hat{P}_f$ with $\lambda = -1$ and $\hat{P}_b$ with
-$\lambda = 1$, such that:
+associated permutation eigenvalue $$\lambda$$, or in other words: whether
+$$\lambda$$ is $$-1$$ or $$1$$ depends on the type of particle that $$x_1$$
+and $$x_2$$ are. Particles with $$\lambda = -1$$ are called
+**fermions**, and those with $$\lambda = 1$$ are known as **bosons**. We
+define $$\hat{P}_f$$ with $$\lambda = -1$$ and $$\hat{P}_b$$ with
+$$\lambda = 1$$, such that:
$$\begin{aligned}
\hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
@@ -53,20 +53,20 @@ $$\begin{aligned}
Another fundamental fact of nature is that identical particles cannot be
distinguished by any observation. Therefore it is impossible to tell
-apart $\Ket{a}\Ket{b}$ and the permuted state $\Ket{b}\Ket{a}$,
-regardless of the eigenvalue $\lambda$. There is no physical difference!
+apart $$\Ket{a}\Ket{b}$$ and the permuted state $$\Ket{b}\Ket{a}$$,
+regardless of the eigenvalue $$\lambda$$. There is no physical difference!
-But this does not mean that $\hat{P}$ is useless: despite not having any
+But this does not mean that $$\hat{P}$$ is useless: despite not having any
observable effect, the resulting difference between fermions and bosons
is absolutely fundamental. Consider the following superposition state,
-where $\alpha$ and $\beta$ are unknown:
+where $$\alpha$$ and $$\beta$$ are unknown:
$$\begin{aligned}
\Ket{\Psi(a, b)}
= \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a}
\end{aligned}$$
-When we apply $\hat{P}$, we can "choose" between two "intepretations" of
+When we apply $$\hat{P}$$, we can "choose" between two "intepretations" of
its action, both shown below. Obviously, since the left-hand sides are
equal, the right-hand sides must be equal too:
@@ -78,28 +78,28 @@ $$\begin{aligned}
&= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b}
\end{aligned}$$
-This gives us the equations $\lambda \alpha = \beta$ and
-$\lambda \beta = \alpha$. In fact, just from this we could have deduced
-that $\lambda$ can be either $-1$ or $1$. In any case, for bosons
-($\lambda = 1$), we thus find that $\alpha = \beta$:
+This gives us the equations $$\lambda \alpha = \beta$$ and
+$$\lambda \beta = \alpha$$. In fact, just from this we could have deduced
+that $$\lambda$$ can be either $$-1$$ or $$1$$. In any case, for bosons
+($$\lambda = 1$$), we thus find that $$\alpha = \beta$$:
$$\begin{aligned}
\Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
\end{aligned}$$
-Where $C$ is a normalization constant. As expected, this state is
-**symmetric**: switching $a$ and $b$ gives the same result. Meanwhile, for
-fermions ($\lambda = -1$), we find that $\alpha = -\beta$:
+Where $$C$$ is a normalization constant. As expected, this state is
+**symmetric**: switching $$a$$ and $$b$$ gives the same result. Meanwhile, for
+fermions ($$\lambda = -1$$), we find that $$\alpha = -\beta$$:
$$\begin{aligned}
\Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
\end{aligned}$$
-This state is called **antisymmetric** under exchange: switching $a$ and $b$
+This state is called **antisymmetric** under exchange: switching $$a$$ and $$b$$
causes a sign change, as we would expect for fermions.
-Now, what if the particles $x_1$ and $x_2$ are in the same state $a$?
-For bosons, we just need to update the normalization constant $C$:
+Now, what if the particles $$x_1$$ and $$x_2$$ are in the same state $$a$$?
+For bosons, we just need to update the normalization constant $$C$$:
$$\begin{aligned}
\Ket{\Psi(a, a)}_b
diff --git a/source/know/concept/plancks-law/index.md b/source/know/concept/plancks-law/index.md
index 2db783d..6c2cd2e 100644
--- a/source/know/concept/plancks-law/index.md
+++ b/source/know/concept/plancks-law/index.md
@@ -18,14 +18,14 @@ and photons are bosons
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
this system must obey the
[Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
-with a chemical potential $\mu = 0$ (due to the freely varying population):
+with a chemical potential $$\mu = 0$$ (due to the freely varying population):
$$\begin{aligned}
f_B(E)
= \frac{1}{\exp(\beta E) - 1}
\end{aligned}$$
-Each photon has an energy $E = \hbar \omega = \hbar c k$,
+Each photon has an energy $$E = \hbar \omega = \hbar c k$$,
so the [density of states](/know/concept/density-of-states/)
is as follows in 3D:
@@ -37,9 +37,9 @@ $$\begin{aligned}
= \frac{8 \pi V E^2}{h^3 c^3}
\end{aligned}$$
-Where the factor of $2$ accounts for the photon's polarization degeneracy.
-We thus expect that the number of photons $N(E)$
-with an energy between $E$ and $E + \dd{E}$ is given by:
+Where the factor of $$2$$ accounts for the photon's polarization degeneracy.
+We thus expect that the number of photons $$N(E)$$
+with an energy between $$E$$ and $$E + \dd{E}$$ is given by:
$$\begin{aligned}
N(E) \dd{E}
@@ -47,16 +47,16 @@ $$\begin{aligned}
= \frac{8 \pi V}{h^3 c^3} \frac{E^2}{\exp(\beta E) - 1} \dd{E}
\end{aligned}$$
-By substituting $E = h \nu$, we find that the number of photons $N(\nu)$
-with a frequency between $\nu$ and $\nu + \dd{\nu}$ must be as follows:
+By substituting $$E = h \nu$$, we find that the number of photons $$N(\nu)$$
+with a frequency between $$\nu$$ and $$\nu + \dd{\nu}$$ must be as follows:
$$\begin{aligned}
N(\nu) \dd{\nu}
= \frac{8 \pi V}{c^3} \frac{\nu^2}{\exp(\beta h \nu) - 1} \dd{\nu}
\end{aligned}$$
-Multiplying by the energy $h \nu$ yields the distribution of the radiated energy,
-which we divide by the volume $V$ to get Planck's law,
+Multiplying by the energy $$h \nu$$ yields the distribution of the radiated energy,
+which we divide by the volume $$V$$ to get Planck's law,
also called the **Plank distribution**,
describing a black body's radiated spectral energy density per unit volume:
@@ -70,8 +70,8 @@ $$\begin{aligned}
## Wien's displacement law
-The Planck distribution peaks at a particular frequency $\nu_{\mathrm{max}}$,
-which can be found by solving the following equation for $\nu$:
+The Planck distribution peaks at a particular frequency $$\nu_{\mathrm{max}}$$,
+which can be found by solving the following equation for $$\nu$$:
$$\begin{aligned}
0
@@ -81,7 +81,7 @@ $$\begin{aligned}
= 3 \nu^2 (\exp(\beta h \nu) - 1) - \nu^3 \beta h \exp(\beta h \nu)
\end{aligned}$$
-By defining $x \equiv \beta h \nu_{\mathrm{max}}$,
+By defining $$x \equiv \beta h \nu_{\mathrm{max}}$$,
this turns into the following transcendental equation:
$$\begin{aligned}
@@ -98,14 +98,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Which states that the peak frequency $\nu_{\mathrm{max}}$
-is proportional to the temperature $T$.
+Which states that the peak frequency $$\nu_{\mathrm{max}}$$
+is proportional to the temperature $$T$$.
## Stefan-Boltzmann law
-Because $u(\nu)$ represents the radiated spectral energy density,
-we can find the total radiated energy $U$ per unit volume by integrating over $\nu$:
+Because $$u(\nu)$$ represents the radiated spectral energy density,
+we can find the total radiated energy $$U$$ per unit volume by integrating over $$\nu$$:
$$\begin{aligned}
U
@@ -116,9 +116,9 @@ $$\begin{aligned}
= \frac{8 \pi}{\beta^4 h^3 c^3} \int_0^\infty \frac{x^3}{\exp(x) - 1} \dd{x}
\end{aligned}$$
-This definite integral turns out to be $\pi^4/15$,
+This definite integral turns out to be $$\pi^4/15$$,
leading us to the **Stefan-Boltzmann law**,
-which states that the radiated energy is proportional to $T^4$:
+which states that the radiated energy is proportional to $$T^4$$:
$$\begin{aligned}
\boxed{
@@ -126,7 +126,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\sigma$ is the **Stefan-Boltzmann constant**, which is defined as follows:
+Where $$\sigma$$ is the **Stefan-Boltzmann constant**, which is defined as follows:
$$\begin{aligned}
\sigma
diff --git a/source/know/concept/prandtl-equations/index.md b/source/know/concept/prandtl-equations/index.md
index 3afe405..d82657c 100644
--- a/source/know/concept/prandtl-equations/index.md
+++ b/source/know/concept/prandtl-equations/index.md
@@ -11,7 +11,7 @@ layout: "concept"
In fluid dynamics, the **Prandtl equations** or **boundary layer equations**
describe the movement of a [viscous](/know/concept/viscosity/) fluid
-with a large [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \gg 1$
+with a large [Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re} \gg 1$$
close to a solid surface.
Fluids with a large Reynolds number
@@ -27,23 +27,23 @@ where viscosity plays an important role.
This is in contrast to the ideal flow far away from the surface.
We consider a simple theoretical case in 2D:
-a large flat surface located at $y = 0$ for all $x \in \mathbb{R}$,
-with a fluid *trying* to flow parallel to it at $U$.
-The 2D treatment can be justified by assuming that everything is constant in the $z$-direction.
+a large flat surface located at $$y = 0$$ for all $$x \in \mathbb{R}$$,
+with a fluid *trying* to flow parallel to it at $$U$$.
+The 2D treatment can be justified by assuming that everything is constant in the $$z$$-direction.
We will not solve this case,
but instead derive general equations
to describe the flow close to a flat surface.
-At the wall, there is a very thin boundary layer of thickness $\delta$,
-where the fluid is assumed to be completely stationary $\va{v} = 0$.
-We are mainly interested in the region $\delta < y \ll L$,
-where $L$ is the distance at which the fluid becomes practically ideal.
+At the wall, there is a very thin boundary layer of thickness $$\delta$$,
+where the fluid is assumed to be completely stationary $$\va{v} = 0$$.
+We are mainly interested in the region $$\delta < y \ll L$$,
+where $$L$$ is the distance at which the fluid becomes practically ideal.
This the so-called **slip-flow** region,
in which the fluid is not stationary,
but still viscosity-dominated.
In 2D, the steady Navier-Stokes equations are as follows,
-where the flow $\va{v} = (v_x, v_y)$:
+where the flow $$\va{v} = (v_x, v_y)$$:
$$\begin{aligned}
v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
@@ -58,10 +58,10 @@ $$\begin{aligned}
The latter represents the fluid's incompressibility.
We non-dimensionalize these equations,
-and assume that changes along the $y$-axis
-happen on a short scale (say, $\delta$),
-and along the $x$-axis on a longer scale (say, $L$).
-Let $\tilde{x}$ and $\tilde{y}$ be dimenionless variables of order $1$:
+and assume that changes along the $$y$$-axis
+happen on a short scale (say, $$\delta$$),
+and along the $$x$$-axis on a longer scale (say, $$L$$).
+Let $$\tilde{x}$$ and $$\tilde{y}$$ be dimenionless variables of order $$1$$:
$$\begin{aligned}
x
@@ -106,7 +106,7 @@ $$\begin{aligned}
\end{aligned}$$
For future convenience,
-we multiply the former equation by $L / U^2$, and the latter by $\delta / U^2$:
+we multiply the former equation by $$L / U^2$$, and the latter by $$\delta / U^2$$:
$$\begin{aligned}
\tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
@@ -118,12 +118,12 @@ $$\begin{aligned}
+ \nu \Big( \frac{\delta^2}{U L^3} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
\end{aligned}$$
-We would like to estimate $\delta$.
-Intuitively, we expect that higher viscosities $\nu$ give thicker layers,
-and that faster velocities $U$ give thinner layers.
+We would like to estimate $$\delta$$.
+Intuitively, we expect that higher viscosities $$\nu$$ give thicker layers,
+and that faster velocities $$U$$ give thinner layers.
Furthermore, we expect *downstream thickening*:
-with distance $x$, viscous stresses slow down the slip-flow,
-leading to a gradual increase of $\delta(x)$.
+with distance $$x$$, viscous stresses slow down the slip-flow,
+leading to a gradual increase of $$\delta(x)$$.
Some dimensional analysis thus yields the following estimate:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
\sim \sqrt{\frac{\nu L}{U}}
\end{aligned}$$
-We thus insert $\delta = \sqrt{\nu L / U}$ into the Navier-Stokes equations, giving us:
+We thus insert $$\delta = \sqrt{\nu L / U}$$ into the Navier-Stokes equations, giving us:
$$\begin{aligned}
\tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
@@ -144,7 +144,7 @@ $$\begin{aligned}
+ \nu \Big( \frac{\nu}{U^2 L^2} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
\end{aligned}$$
-Here, we recognize the definition of the Reynolds number $\mathrm{Re} = U L / \nu$:
+Here, we recognize the definition of the Reynolds number $$\mathrm{Re} = U L / \nu$$:
$$\begin{aligned}
\tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
@@ -156,8 +156,8 @@ $$\begin{aligned}
+ \frac{1}{\mathrm{Re}^2} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \pdvn{2}{\tilde{v}_y}{\tilde{y}}
\end{aligned}$$
-Recall that we are only considering large Reynolds numbers $\mathrm{Re} \gg 1$,
-in which case $\mathrm{Re}^{-1} \ll 1$,
+Recall that we are only considering large Reynolds numbers $$\mathrm{Re} \gg 1$$,
+in which case $$\mathrm{Re}^{-1} \ll 1$$,
so we can drop many terms, leaving us with these redimensionalized equations:
$$\begin{aligned}
@@ -168,11 +168,11 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-The second one tells us that for a given $x$-value,
+The second one tells us that for a given $$x$$-value,
the pressure is the same at the surface
-as in the main flow $y > L$, where the fluid is ideal.
-In the latter regime, we apply Bernoulli's theorem to rewrite $p$,
-using the *Bernoulli head* $H$ and the mainstream velocity $U(x)$:
+as in the main flow $$y > L$$, where the fluid is ideal.
+In the latter regime, we apply Bernoulli's theorem to rewrite $$p$$,
+using the *Bernoulli head* $$H$$ and the mainstream velocity $$U(x)$$:
$$\begin{aligned}
p
diff --git a/source/know/concept/probability-current/index.md b/source/know/concept/probability-current/index.md
index 43683a3..bd41dab 100644
--- a/source/know/concept/probability-current/index.md
+++ b/source/know/concept/probability-current/index.md
@@ -10,15 +10,15 @@ layout: "concept"
In quantum mechanics, the **probability current** describes the movement
of the probability of finding a particle at given point in space.
-In other words, it treats the particle as a heterogeneous fluid with density $|\psi|^2$.
-Now, the probability of finding the particle within a volume $V$ is:
+In other words, it treats the particle as a heterogeneous fluid with density $$|\psi|^2$$.
+Now, the probability of finding the particle within a volume $$V$$ is:
$$\begin{aligned}
P = \int_{V} | \psi |^2 \ddn{3}{\vb{r}}
\end{aligned}$$
As the system evolves in time, this probability may change, so we take
-its derivative with respect to time $t$, and when necessary substitute
+its derivative with respect to time $$t$$, and when necessary substitute
in the other side of the Schrödinger equation to get:
$$\begin{aligned}
@@ -33,8 +33,8 @@ $$\begin{aligned}
= - \int_{V} \nabla \cdot \vb{J} \ddn{3}{\vb{r}}
\end{aligned}$$
-Where we have defined the probability current $\vb{J}$ as follows in
-the $\vb{r}$-basis:
+Where we have defined the probability current $$\vb{J}$$ as follows in
+the $$\vb{r}$$-basis:
$$\begin{aligned}
\vb{J}
@@ -43,8 +43,8 @@ $$\begin{aligned}
\end{aligned}$$
Let us rewrite this using the momentum operator
-$\vu{p} = -i \hbar \nabla$ as follows, noting that $\vu{p} / m$ is
-simply the velocity operator $\vu{v}$:
+$$\vu{p} = -i \hbar \nabla$$ as follows, noting that $$\vu{p} / m$$ is
+simply the velocity operator $$\vu{v}$$:
$$\begin{aligned}
\boxed{
@@ -55,7 +55,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Returning to the derivation of $\vb{J}$, we now have the following
+Returning to the derivation of $$\vb{J}$$, we now have the following
equation:
$$\begin{aligned}
@@ -65,7 +65,7 @@ $$\begin{aligned}
\end{aligned}$$
By removing the integrals, we thus arrive at the **continuity equation**
-for $\vb{J}$:
+for $$\vb{J}$$:
$$\begin{aligned}
\boxed{
@@ -77,13 +77,13 @@ $$\begin{aligned}
This states that the total probability is conserved, and is reminiscent of charge
conservation in electromagnetism. In other words, the probability at a
point can only change by letting it "flow" towards or away from it. Thus
-$\vb{J}$ represents the flow of probability, which is analogous to the
+$$\vb{J}$$ represents the flow of probability, which is analogous to the
motion of a particle.
As a bonus, this still holds for a particle in an electromagnetic vector
-potential $\vb{A}$, thanks to the gauge invariance of the Schrödinger
+potential $$\vb{A}$$, thanks to the gauge invariance of the Schrödinger
equation. We can thus extend the definition to a particle with charge
-$q$ in an SI-unit field, neglecting spin:
+$$q$$ in an SI-unit field, neglecting spin:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/propagator/index.md b/source/know/concept/propagator/index.md
index 437c57e..54e9eb6 100644
--- a/source/know/concept/propagator/index.md
+++ b/source/know/concept/propagator/index.md
@@ -8,9 +8,9 @@ categories:
layout: "concept"
---
-In quantum mechanics, the **propagator** $K(x_f, t_f; x_i, t_i)$
+In quantum mechanics, the **propagator** $$K(x_f, t_f; x_i, t_i)$$
gives the probability amplitude that a particle
-starting at $x_i$ at $t_i$ ends up at position $x_f$ at $t_f$.
+starting at $$x_i$$ at $$t_i$$ ends up at position $$x_f$$ at $$t_f$$.
It is defined as follows:
$$\begin{aligned}
@@ -20,24 +20,24 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\hat{U} \equiv \exp(- i t \hat{H} / \hbar)$ is the time-evolution operator.
+Where $$\hat{U} \equiv \exp(- i t \hat{H} / \hbar)$$ is the time-evolution operator.
The probability that a particle travels
-from $(x_i, t_i)$ to $(x_f, t_f)$ is then given by:
+from $$(x_i, t_i)$$ to $$(x_f, t_f)$$ is then given by:
$$\begin{aligned}
P
&= \big| K(x_f, t_f; x_i, t_i) \big|^2
\end{aligned}$$
-Given a general (i.e. non-collapsed) initial state $\psi_i(x) \equiv \psi(x, t_i)$,
-we must integrate over $x_i$:
+Given a general (i.e. non-collapsed) initial state $$\psi_i(x) \equiv \psi(x, t_i)$$,
+we must integrate over $$x_i$$:
$$\begin{aligned}
P
&= \bigg| \int_{-\infty}^\infty K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \bigg|^2
\end{aligned}$$
-And if the final state $\psi_f(x) \equiv \psi(x, t_f)$
+And if the final state $$\psi_f(x) \equiv \psi(x, t_f)$$
is not a basis vector either, then we integrate twice:
$$\begin{aligned}
@@ -45,7 +45,7 @@ $$\begin{aligned}
&= \bigg| \iint_{-\infty}^\infty \psi_f^*(x_f) \: K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \dd{x_f} \bigg|^2
\end{aligned}$$
-Given a $\psi_i(x)$, the propagator can also be used
+Given a $$\psi_i(x)$$, the propagator can also be used
to find the full final wave function:
$$\begin{aligned}
@@ -56,9 +56,9 @@ $$\begin{aligned}
\end{aligned}$$
Sometimes the name "propagator" is also used to refer to
-the [fundamental solution](/know/concept/fundamental-solution/) $G$
+the [fundamental solution](/know/concept/fundamental-solution/) $$G$$
of the time-dependent Schrödinger equation,
-which is related to $K$ by:
+which is related to $$K$$ by:
$$\begin{aligned}
\boxed{
@@ -67,4 +67,4 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Theta(t)$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
+Where $$\Theta(t)$$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
diff --git a/source/know/concept/pulay-mixing/index.md b/source/know/concept/pulay-mixing/index.md
index 91beb03..6e809dd 100644
--- a/source/know/concept/pulay-mixing/index.md
+++ b/source/know/concept/pulay-mixing/index.md
@@ -8,15 +8,15 @@ layout: "concept"
---
Some numerical problems are most easily solved *iteratively*,
-by generating a series $\rho_1$, $\rho_2$, etc.
-converging towards the desired solution $\rho_*$.
+by generating a series $$\rho_1$$, $$\rho_2$$, etc.
+converging towards the desired solution $$\rho_*$$.
**Pulay mixing**, also often called
**direct inversion in the iterative subspace** (DIIS),
can speed up the convergence for some types of problems,
and also helps to avoid periodic divergences.
-The key concept it relies on is the **residual vector** $R_n$
-of the $n$th iteration, which in some way measures the error of the current $\rho_n$.
+The key concept it relies on is the **residual vector** $$R_n$$
+of the $$n$$th iteration, which in some way measures the error of the current $$\rho_n$$.
Its exact definition varies,
but is generally along the lines of the difference between
the input of the iteration and the raw resulting output:
@@ -27,16 +27,16 @@ $$\begin{aligned}
= \rho_n^\mathrm{new}[\rho_n] - \rho_n
\end{aligned}$$
-It is not always clear what to do with $\rho_n^\mathrm{new}$.
-Directly using it as the next input ($\rho_{n+1} = \rho_n^\mathrm{new}$)
+It is not always clear what to do with $$\rho_n^\mathrm{new}$$.
+Directly using it as the next input ($$\rho_{n+1} = \rho_n^\mathrm{new}$$)
often leads to oscillation,
-and linear mixing ($\rho_{n+1} = (1\!-\!f) \rho_n + f \rho_n^\mathrm{new}$)
+and linear mixing ($$\rho_{n+1} = (1\!-\!f) \rho_n + f \rho_n^\mathrm{new}$$)
can take a very long time to converge properly.
Pulay mixing offers an improvement.
-The idea is to construct the next iteration's input $\rho_{n+1}$
-as a linear combination of the previous inputs $\rho_1$, $\rho_2$, ..., $\rho_n$,
-such that it is as close as possible to the optimal $\rho_*$:
+The idea is to construct the next iteration's input $$\rho_{n+1}$$
+as a linear combination of the previous inputs $$\rho_1$$, $$\rho_2$$, ..., $$\rho_n$$,
+such that it is as close as possible to the optimal $$\rho_*$$:
$$\begin{aligned}
\boxed{
@@ -46,10 +46,10 @@ $$\begin{aligned}
\end{aligned}$$
To do so, we make two assumptions.
-Firstly, the current $\rho_n$ is already close to $\rho_*$,
+Firstly, the current $$\rho_n$$ is already close to $$\rho_*$$,
so that such a linear combination makes sense.
Secondly, the iteration is linear,
-such that the raw output $\rho_{n+1}^\mathrm{new}$
+such that the raw output $$\rho_{n+1}^\mathrm{new}$$
is also a linear combination with the *same coefficients*:
$$\begin{aligned}
@@ -58,7 +58,7 @@ $$\begin{aligned}
\end{aligned}$$
We will return to these assumptions later.
-The point is that $R_{n+1}$ is also a linear combination:
+The point is that $$R_{n+1}$$ is also a linear combination:
$$\begin{aligned}
R_{n+1}
@@ -67,23 +67,23 @@ $$\begin{aligned}
= \sum_{m = 1}^n \alpha_m R_m
\end{aligned}$$
-The goal is to choose the coefficients $\alpha_m$ such that
-the norm of the error $|R_{n+1}| \approx 0$,
-subject to the following constraint to preserve the normalization of $\rho_{n+1}$:
+The goal is to choose the coefficients $$\alpha_m$$ such that
+the norm of the error $$|R_{n+1}| \approx 0$$,
+subject to the following constraint to preserve the normalization of $$\rho_{n+1}$$:
$$\begin{aligned}
\sum_{m=1}^n \alpha_m = 1
\end{aligned}$$
We thus want to minimize the following quantity,
-where $\lambda$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+where $$\lambda$$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/):
$$\begin{aligned}
\Inprod{R_{n+1}}{R_{n+1}} + \lambda \sum_{m = 1}^n \alpha_m^*
= \sum_{m=1}^n \alpha_m^* \Big( \sum_{k=1}^n \alpha_k \Inprod{R_m}{R_k} + \lambda \Big)
\end{aligned}$$
-By differentiating the right-hand side with respect to $\alpha_m^*$
+By differentiating the right-hand side with respect to $$\alpha_m^*$$
and demanding that the result is zero,
we get a system of equations that we can write in matrix form,
which is cheap to solve:
@@ -106,38 +106,38 @@ $$\begin{aligned}
\end{aligned}$$
From this, we can also see that the Lagrange multiplier
-$\lambda = - \Inprod{R_{n+1}}{R_{n+1}}$,
-where $R_{n+1}$ is the *predicted* residual of the next iteration,
+$$\lambda = - \Inprod{R_{n+1}}{R_{n+1}}$$,
+where $$R_{n+1}$$ is the *predicted* residual of the next iteration,
subject to the two assumptions.
-However, in practice, the earlier inputs $\rho_1$, $\rho_2$, etc.
-are much further from $\rho_*$ than $\rho_n$,
-so usually only the most recent $N\!+\!1$ inputs $\rho_{n - N}$, ..., $\rho_n$ are used:
+However, in practice, the earlier inputs $$\rho_1$$, $$\rho_2$$, etc.
+are much further from $$\rho_*$$ than $$\rho_n$$,
+so usually only the most recent $$N\!+\!1$$ inputs $$\rho_{n - N}$$, ..., $$\rho_n$$ are used:
$$\begin{aligned}
\rho_{n+1}
= \sum_{m = n-N}^n \alpha_m \rho_m
\end{aligned}$$
-You might be confused by the absence of any $\rho_m^\mathrm{new}$
-in the creation of $\rho_{n+1}$, as if the iteration's outputs are being ignored.
+You might be confused by the absence of any $$\rho_m^\mathrm{new}$$
+in the creation of $$\rho_{n+1}$$, as if the iteration's outputs are being ignored.
This is due to the first assumption,
-which states that $\rho_n^\mathrm{new}$ and $\rho_n$ are already similar,
+which states that $$\rho_n^\mathrm{new}$$ and $$\rho_n$$ are already similar,
such that they are basically interchangeable.
Speaking of which, about those assumptions:
-while they will clearly become more accurate as $\rho_n$ approaches $\rho_*$,
+while they will clearly become more accurate as $$\rho_n$$ approaches $$\rho_*$$,
they might be very dubious in the beginning.
A consequence of this is that the early iterations might get "trapped"
-in a suboptimal subspace spanned by $\rho_1$, $\rho_2$, etc.
-To say it another way, we would be varying $n$ coefficients $\alpha_m$
-to try to optimize a $D$-dimensional $\rho_{n+1}$,
-where in general $D \gg n$, at least in the beginning.
+in a suboptimal subspace spanned by $$\rho_1$$, $$\rho_2$$, etc.
+To say it another way, we would be varying $$n$$ coefficients $$\alpha_m$$
+to try to optimize a $$D$$-dimensional $$\rho_{n+1}$$,
+where in general $$D \gg n$$, at least in the beginning.
There is an easy fix to this problem:
-add a small amount of the raw residual $R_m$
-to "nudge" $\rho_{n+1}$ towards the right subspace,
-where $\beta \in [0,1]$ is a tunable parameter:
+add a small amount of the raw residual $$R_m$$
+to "nudge" $$\rho_{n+1}$$ towards the right subspace,
+where $$\beta \in [0,1]$$ is a tunable parameter:
$$\begin{aligned}
\boxed{
@@ -146,7 +146,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-In other words, we end up introducing a small amount of the raw outputs $\rho_m^\mathrm{new}$,
+In other words, we end up introducing a small amount of the raw outputs $$\rho_m^\mathrm{new}$$,
while still giving more weight to iterations with smaller residuals.
Pulay mixing is very effective for certain types of problems,
diff --git a/source/know/concept/quantum-entanglement/index.md b/source/know/concept/quantum-entanglement/index.md
index 5ae5c92..cfb6721 100644
--- a/source/know/concept/quantum-entanglement/index.md
+++ b/source/know/concept/quantum-entanglement/index.md
@@ -9,19 +9,19 @@ categories:
layout: "concept"
---
-Consider a composite quantum system which consists of two subsystems $A$ and $B$,
-respectively with basis states $\Ket{a_n}$ and $\Ket{b_n}$.
-All accessible states of the sytem $\Ket{\Psi}$ lie in
+Consider a composite quantum system which consists of two subsystems $$A$$ and $$B$$,
+respectively with basis states $$\Ket{a_n}$$ and $$\Ket{b_n}$$.
+All accessible states of the sytem $$\Ket{\Psi}$$ lie in
the tensor product of the subsystems'
-[Hilbert spaces](/know/concept/hilbert-space/) $\mathbb{H}_A$ and $\mathbb{H}_B$:
+[Hilbert spaces](/know/concept/hilbert-space/) $$\mathbb{H}_A$$ and $$\mathbb{H}_B$$:
$$\begin{aligned}
\Ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B
\end{aligned}$$
A subset of these states can be written as the tensor product (i.e. Kronecker product in a basis)
-of a state $\Ket{\alpha}$ in $A$ and a state $\Ket{\beta}$ in $B$,
-often abbreviated as $\Ket{\alpha} \Ket{\beta}$:
+of a state $$\Ket{\alpha}$$ in $$A$$ and a state $$\Ket{\beta}$$ in $$B$$,
+often abbreviated as $$\Ket{\alpha} \Ket{\beta}$$:
$$\begin{aligned}
\Ket{\Psi}
@@ -32,20 +32,20 @@ $$\begin{aligned}
The states that can be written in this way are called **separable**,
and states that cannot are called **entangled**.
Therefore, we are dealing with **quantum entanglement**
-if the state of subsystem $A$ cannot be fully described
-independently of the state of subsystem $B$, and vice versa.
+if the state of subsystem $$A$$ cannot be fully described
+independently of the state of subsystem $$B$$, and vice versa.
To detect and quantify entanglement,
-we can use the [density operator](/know/concept/density-operator/) $\hat{\rho}$.
-For a pure ensemble in a given (possibly entangled) state $\Ket{\Psi}$,
-$\hat{\rho}$ is given by:
+we can use the [density operator](/know/concept/density-operator/) $$\hat{\rho}$$.
+For a pure ensemble in a given (possibly entangled) state $$\Ket{\Psi}$$,
+$$\hat{\rho}$$ is given by:
$$\begin{aligned}
\hat{\rho} = \Ket{\Psi} \Bra{\Psi}
\end{aligned}$$
-From this, we would like to extract the corresponding state of subsystem $A$.
-For that purpose, we define the **reduced density operator** $\hat{\rho}_A$ of subsystem $A$ as follows:
+From this, we would like to extract the corresponding state of subsystem $$A$$.
+For that purpose, we define the **reduced density operator** $$\hat{\rho}_A$$ of subsystem $$A$$ as follows:
$$\begin{aligned}
\boxed{
@@ -55,11 +55,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\Tr_B(\hat{\rho})$ is called the **partial trace** of $\hat{\rho}$,
-which basically eliminates subsystem $B$ from $\hat{\rho}$.
-For a pure composite state $\Ket{\Psi}$,
-the resulting $\hat{\rho}_A$ describes a pure state in $A$ if $\Ket{\Psi}$ is separable,
-else, if $\Ket{\Psi}$ is entangled, it describes a mixed state in $A$.
+Where $$\Tr_B(\hat{\rho})$$ is called the **partial trace** of $$\hat{\rho}$$,
+which basically eliminates subsystem $$B$$ from $$\hat{\rho}$$.
+For a pure composite state $$\Ket{\Psi}$$,
+the resulting $$\hat{\rho}_A$$ describes a pure state in $$A$$ if $$\Ket{\Psi}$$ is separable,
+else, if $$\Ket{\Psi}$$ is entangled, it describes a mixed state in $$A$$.
In the former case we simply find:
$$\begin{aligned}
@@ -70,9 +70,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-We call $\Ket{\Psi}$ **maximally entangled**
+We call $$\Ket{\Psi}$$ **maximally entangled**
if its reduced density operators are **maximally mixed**,
-where $N$ is the dimension of $\mathbb{H}_A$ and $\hat{I}$ is the identity matrix:
+where $$N$$ is the dimension of $$\mathbb{H}_A$$ and $$\hat{I}$$ is the identity matrix:
$$\begin{aligned}
\hat{\rho}_A
@@ -80,10 +80,10 @@ $$\begin{aligned}
\end{aligned}$$
Suppose that we are given an entangled pure state
-$\Ket{\Psi} \neq \Ket{\alpha} \otimes \Ket{\beta}$.
-Then the partial traces $\hat{\rho}_A$ and $\hat{\rho}_B$
-of $\hat{\rho} = \Ket{\Psi} \Bra{\Psi}$ are mixed states with the same probabilities $p_n$
-(assuming $\mathbb{H}_A$ and $\mathbb{H}_B$ have the same dimensions,
+$$\Ket{\Psi} \neq \Ket{\alpha} \otimes \Ket{\beta}$$.
+Then the partial traces $$\hat{\rho}_A$$ and $$\hat{\rho}_B$$
+of $$\hat{\rho} = \Ket{\Psi} \Bra{\Psi}$$ are mixed states with the same probabilities $$p_n$$
+(assuming $$\mathbb{H}_A$$ and $$\mathbb{H}_B$$ have the same dimensions,
which is usually the case):
$$\begin{aligned}
@@ -97,9 +97,9 @@ $$\begin{aligned}
\end{aligned}$$
There exists an orthonormal choice
-of the subsystem basis states $\Ket{a_n}$ and $\Ket{b_n}$,
-such that $\Ket{\Psi}$ can be written as follows,
-where $p_n$ are the probabilities in the reduced density operators:
+of the subsystem basis states $$\Ket{a_n}$$ and $$\Ket{b_n}$$,
+such that $$\Ket{\Psi}$$ can be written as follows,
+where $$p_n$$ are the probabilities in the reduced density operators:
$$\begin{aligned}
\Ket{\Psi}
@@ -108,18 +108,18 @@ $$\begin{aligned}
This is the **Schmidt decomposition**,
and the **Schmidt number** is the number of nonzero terms in the summation,
-which can be used to determine if the state $\Ket{\Psi}$
+which can be used to determine if the state $$\Ket{\Psi}$$
is entangled (greater than one) or separable (equal to one).
By looking at the Schmidt decomposition, we can notice that,
-if $\hat{O}_A$ and $\hat{O}_B$ are the subsystem observables
-with basis eigenstates $\Ket{a_n}$ and $\Ket{b_n}$,
+if $$\hat{O}_A$$ and $$\hat{O}_B$$ are the subsystem observables
+with basis eigenstates $$\Ket{a_n}$$ and $$\Ket{b_n}$$,
then measurement results of these operators
-will be perfectly correlated across $A$ and $B$.
+will be perfectly correlated across $$A$$ and $$B$$.
This is a general property of entangled systems,
but beware: correlation does not imply entanglement!
-But what if the composite system is in a mixed state $\hat{\rho}$?
+But what if the composite system is in a mixed state $$\hat{\rho}$$?
The state is separable if and only if:
$$\begin{aligned}
@@ -129,13 +129,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $p_m$ are probabilities,
-and $\hat{\rho}_A$ and $\hat{\rho}_B$ can be any subsystem states.
+Where $$p_m$$ are probabilities,
+and $$\hat{\rho}_A$$ and $$\hat{\rho}_B$$ can be any subsystem states.
In reality, it is very hard to determine, using this criterium,
-whether an arbitrary given $\hat{\rho}$ is separable or not.
+whether an arbitrary given $$\hat{\rho}$$ is separable or not.
As a final side note, the expectation value
-of an obervable $\hat{O}_A$ acting only on $A$ is given by:
+of an obervable $$\hat{O}_A$$ acting only on $$A$$ is given by:
$$\begin{aligned}
\expval{\hat{O}_A}
diff --git a/source/know/concept/quantum-fourier-transform/index.md b/source/know/concept/quantum-fourier-transform/index.md
index 5595fa2..113367c 100644
--- a/source/know/concept/quantum-fourier-transform/index.md
+++ b/source/know/concept/quantum-fourier-transform/index.md
@@ -10,9 +10,9 @@ layout: "concept"
The **quantum Fourier transform (QFT)** is a quantum counterpart
of the classical discrete Fourier transform.
-It is defined like so, where $\Ket{x}$
-is an $n$-qubit computational basis state $\Ket{x_1} \cdots \Ket{x_n}$,
-and $\omega_N$ is an $N$th complex root of unity $\omega_N^N = 1$ with $N = 2^n$:
+It is defined like so, where $$\Ket{x}$$
+is an $$n$$-qubit computational basis state $$\Ket{x_1} \cdots \Ket{x_n}$$,
+and $$\omega_N$$ is an $$N$$th complex root of unity $$\omega_N^N = 1$$ with $$N = 2^n$$:
$$\begin{aligned}
\boxed{
@@ -22,9 +22,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that $\Ket{x}$ and $\Ket{k}$ refer to the same basis set;
+Note that $$\Ket{x}$$ and $$\Ket{k}$$ refer to the same basis set;
we use these names to clarify which "space" we are considering.
-Furthermore, note the sign of the exponent of $\omega_N$,
+Furthermore, note the sign of the exponent of $$\omega_N$$,
which is the opposite of the classical DFT convention.
In other words, the *forward* QFT corresponds to the *inverse* DFT.
@@ -39,7 +39,7 @@ $$\begin{aligned}
\end{aligned}$$
The above definitions of the QFT and iQFT describe
-the effect on a single basis vector $\Ket{x}$,
+the effect on a single basis vector $$\Ket{x}$$,
so the effect on an abitrary superposition follows from linearity,
e.g. for the QFT:
@@ -50,14 +50,14 @@ $$\begin{aligned}
\end{aligned}$$
Classically, such a double sum takes
-$\mathcal{O}(N^2) = \mathcal{O}(2^{2n})$ time to evaluate naively,
-with a potential improvement to $\mathcal{O}(N \log{N}) = \mathcal{O}(n 2^n)$
+$$\mathcal{O}(N^2) = \mathcal{O}(2^{2n})$$ time to evaluate naively,
+with a potential improvement to $$\mathcal{O}(N \log{N}) = \mathcal{O}(n 2^n)$$
for smarter algorithms.
-Quantum computers can do a QFT in $\mathcal{O}(\log^2(N)) = \mathcal{O}(n^2)$ time,
-and approximate it in $\mathcal{O}(n \log{n})$ time.
+Quantum computers can do a QFT in $$\mathcal{O}(\log^2(N)) = \mathcal{O}(n^2)$$ time,
+and approximate it in $$\mathcal{O}(n \log{n})$$ time.
To find out how, we look at the forward QFT,
-which maps a basis state $\Ket{x}$ to another state $\Ket{\tilde{x}}$:
+which maps a basis state $$\Ket{x}$$ to another state $$\Ket{\tilde{x}}$$:
$$\begin{aligned}
\Ket{x}
@@ -67,8 +67,8 @@ $$\begin{aligned}
= \frac{1}{\sqrt{N}} \sum_{k = 0}^{N - 1} \exp\!\bigg( \frac{i 2 \pi x k}{N} \bigg) \Ket{k}
\end{aligned}$$
-We can decompose $k$ into its binary representation
-$k_1 2^{n-1} + k_2 2^{n-2} + ... + k_n 2^{0}$:
+We can decompose $$k$$ into its binary representation
+$$k_1 2^{n-1} + k_2 2^{n-2} + ... + k_n 2^{0}$$:
$$\begin{aligned}
\Ket{\tilde{x}}
@@ -84,7 +84,7 @@ $$\begin{aligned}
&= \frac{1}{\sqrt{2^n}} \sum_{k = 0}^{2^n - 1} \prod_{j = 1}^{n} \exp\!\bigg( i 2 \pi x \frac{k_j}{2^j} \bigg) \Ket{k}
\end{aligned}$$
-If $k_j = 0$, the exponential is zero.
+If $$k_j = 0$$, the exponential is zero.
We can thus separate this state into its individual qubits:
$$\begin{aligned}
@@ -93,14 +93,14 @@ $$\begin{aligned}
\end{aligned}$$
Next, we use the trick from before:
-decompose $x$ as $x_1 2^{n-1} + x_2 2^{n-2} + ... + x_n 2^{0}$:
+decompose $$x$$ as $$x_1 2^{n-1} + x_2 2^{n-2} + ... + x_n 2^{0}$$:
$$\begin{aligned}
\Ket{\tilde{x}}
&= \bigotimes_{j = 1}^{n} \frac{1}{\sqrt{2}} \bigg( \Ket{0} + \exp\!\Big( i 2 \pi \sum_{r = 1}^{n} x_r 2^{n-r-j} \Big) \Ket{1} \bigg)
\end{aligned}$$
-The factor $2^{n-r-j}$ may be smaller or bigger than $1$,
+The factor $$2^{n-r-j}$$ may be smaller or bigger than $$1$$,
so it is convenient for us to use the following notation
for non-integer binary numbers.
Note the decimal point in the middle:
@@ -110,7 +110,7 @@ $$\begin{aligned}
= \sum_{r = 1}^{n} a_r 2^{d - r}
\end{aligned}$$
-In the above QFT state, the position of the decimal point is $d = n - j$,
+In the above QFT state, the position of the decimal point is $$d = n - j$$,
so we write it as:
$$\begin{aligned}
@@ -119,7 +119,7 @@ $$\begin{aligned}
+ \exp\!\Big( i 2 \pi \big[ x_1 \cdots x_{n-j} \:.\: x_{n-j+1} \cdots x_n \big] \Big) \Ket{1} \bigg)
\end{aligned}$$
-Because $\exp(i 2 \pi m) = 1$ for all integers $m$,
+Because $$\exp(i 2 \pi m) = 1$$ for all integers $$m$$,
we can discard bits before the decimal point:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, each exponential can be factorized,
-with every factor containing one bit of $x$:
+with every factor containing one bit of $$x$$:
$$\begin{aligned}
\exp\!\Big( i 2 \pi \big[ 0\:.\: x_{n-j+1} \cdots x_n \big] \Big)
@@ -141,42 +141,42 @@ $$\begin{aligned}
This suggests a way to implement the QFT using
[quantum gates](/know/concept/quantum-gate/) in a circuit.
-If the $j$th qubit is in $\Ket{+} = (\Ket{0} + \Ket{1})/\sqrt{2}$,
-then for each bit $x_{n-j+r}$ where $r \in \{1, ..., j\}$:
+If the $$j$$th qubit is in $$\Ket{+} = (\Ket{0} + \Ket{1})/\sqrt{2}$$,
+then for each bit $$x_{n-j+r}$$ where $$r \in \{1, ..., j\}$$:
-+ If $x_{n-j+r} = 0$, do nothing.
-+ If $x_{n-j+r} = 1$, add a relative phase $2 \pi / 2^{r}$.
++ If $$x_{n-j+r} = 0$$, do nothing.
++ If $$x_{n-j+r} = 1$$, add a relative phase $$2 \pi / 2^{r}$$.
The full QFT algorithm therefore proceeds as follows,
-for the $(n\!-\!j\!+\!1)$'th input $\Ket{x_{n-j+1}}$:
-
-1. Apply the Hadamard gate $H$.
- If $x_{n-j+1} = 0$, this puts the qubit in $\Ket{+}$.
- If $x_{n-j+1} = 1$, this puts the qubit in $\Ket{+}$,
- and then adds a phase $\pi$, yielding $\Ket{-}$.
-2. Apply the phase shift gate $R_{\phi}$ controlled by $x_{n-j+2}$,
- with angle $\phi = 2 \pi / 2^{2}$.
-3. Apply $R_{\phi}$ controlled by $x_{n-j+3}$,
- with angle $\phi = 2 \pi / 2^{3}$...
-4. And so on, until the $n$th bit $x_n$ is reached,
- and used to control $R_\phi$ with $\phi = 2 \pi / 2^{j}$.
-
-And so on, for each $j \in \{1, ..., n\}$,
-and we reach the above expression for $\Ket{\tilde{x}}$.
-We started from the $(n\!-\!j\!+\!1)$'th input qubit,
+for the $$(n\!-\!j\!+\!1)$$'th input $$\Ket{x_{n-j+1}}$$:
+
+1. Apply the Hadamard gate $$H$$.
+ If $$x_{n-j+1} = 0$$, this puts the qubit in $$\Ket{+}$$.
+ If $$x_{n-j+1} = 1$$, this puts the qubit in $$\Ket{+}$$,
+ and then adds a phase $$\pi$$, yielding $$\Ket{-}$$.
+2. Apply the phase shift gate $$R_{\phi}$$ controlled by $$x_{n-j+2}$$,
+ with angle $$\phi = 2 \pi / 2^{2}$$.
+3. Apply $$R_{\phi}$$ controlled by $$x_{n-j+3}$$,
+ with angle $$\phi = 2 \pi / 2^{3}$$...
+4. And so on, until the $$n$$th bit $$x_n$$ is reached,
+ and used to control $$R_\phi$$ with $$\phi = 2 \pi / 2^{j}$$.
+
+And so on, for each $$j \in \{1, ..., n\}$$,
+and we reach the above expression for $$\Ket{\tilde{x}}$$.
+We started from the $$(n\!-\!j\!+\!1)$$'th input qubit,
i.e. we read the input in reverse order.
Therefore all the qubits need to be swapped back to front,
either before or after the above algorithm is run.
The quantum circuit to execute the mentioned steps is illustrated below,
excluding the swapping part to get the right order.
-Here, $R_m$ means $R_\phi$ with $\phi = 2 \pi / 2^m$:
+Here, $$R_m$$ means $$R_\phi$$ with $$\phi = 2 \pi / 2^m$$:
-Again, note how the inputs $\Ket{x_j}$ and outputs $\Ket{k_j}$ are in the opposite order.
+Again, note how the inputs $$\Ket{x_j}$$ and outputs $$\Ket{k_j}$$ are in the opposite order.
The complete circuit, including the swapping at the end,
therefore looks like this:
@@ -184,8 +184,8 @@ therefore looks like this:
-For each of the $n$ qubits, $\mathcal{O}(n)$ gates are applied,
-so overall the QFT algorithm is $\mathcal{O}(n^2)$.
+For each of the $$n$$ qubits, $$\mathcal{O}(n)$$ gates are applied,
+so overall the QFT algorithm is $$\mathcal{O}(n^2)$$.
diff --git a/source/know/concept/quantum-gate/index.md b/source/know/concept/quantum-gate/index.md
index 38c39a1..8c251be 100644
--- a/source/know/concept/quantum-gate/index.md
+++ b/source/know/concept/quantum-gate/index.md
@@ -8,7 +8,7 @@ layout: "concept"
---
In quantum computing, **quantum gates** are the equivalent
-of classical binary logic gates such as $\mathrm{NOT}$, $\mathrm{AND}$, etc.
+of classical binary logic gates such as $$\mathrm{NOT}$$, $$\mathrm{AND}$$, etc.
Because of the continuous nature of qubits,
the number of possible quantum gates is uncountably infinite,
so we only consider the most important examples here.
@@ -16,7 +16,7 @@ so we only consider the most important examples here.
## One-qubit gates
-As an example, consider the following must general single-qubit state $\Ket{\psi}$:
+As an example, consider the following must general single-qubit state $$\Ket{\psi}$$:
$$\begin{aligned}
\Ket{\psi}
@@ -52,10 +52,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-They have the following effect on $\Ket{\psi}$.
-Note that $X$ is equivalent to the classical $\mathrm{NOT}$ gate
+They have the following effect on $$\Ket{\psi}$$.
+Note that $$X$$ is equivalent to the classical $$\mathrm{NOT}$$ gate
(and is often given that name),
-and $Z$ is sometimes called the **phase-flip gate**:
+and $$Z$$ is sometimes called the **phase-flip gate**:
$$\begin{aligned}
X \Ket{\psi}
@@ -68,9 +68,9 @@ $$\begin{aligned}
= \begin{bmatrix} \alpha \\ -\beta \end{bmatrix}
\end{aligned}$$
-In fact, $Z$ is a specific case of the **phase shift gate** $R_\phi$,
+In fact, $$Z$$ is a specific case of the **phase shift gate** $$R_\phi$$,
which modifies the qubit's phase without changing its amplitudes.
-For an angle $\phi$, it is given by:
+For an angle $$\phi$$, it is given by:
$$\begin{aligned}
\boxed{
@@ -82,17 +82,17 @@ $$\begin{aligned}
}
\end{aligned}$$
-For $\phi = \pi$, we recover the Pauli-$Z$ gate.
-In general, the action of $R_\phi$ is as follows:
+For $$\phi = \pi$$, we recover the Pauli-$$Z$$ gate.
+In general, the action of $$R_\phi$$ is as follows:
$$\begin{aligned}
R_\phi \Ket{\psi}
= \begin{bmatrix} \alpha \\ e^{i \phi} \beta \end{bmatrix}
\end{aligned}$$
-Two common special cases of $R_\phi$
-are $\phi = \pi/2$ and $\phi = \pi/4$,
-respectively called $S$ and $T$:
+Two common special cases of $$R_\phi$$
+are $$\phi = \pi/2$$ and $$\phi = \pi/4$$,
+respectively called $$S$$ and $$T$$:
$$\begin{aligned}
\boxed{
@@ -113,7 +113,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, we have the **Hadamard gate** $H$,
+Finally, we have the **Hadamard gate** $$H$$,
which is defined as follows:
$$\begin{aligned}
@@ -127,14 +127,14 @@ $$\begin{aligned}
\end{aligned}$$
Its action consists of rotating the qubit
-by $\pi$ around the axis $(X + Z) / \sqrt{2}$ of the Bloch sphere:
+by $$\pi$$ around the axis $$(X + Z) / \sqrt{2}$$ of the Bloch sphere:
$$\begin{aligned}
H \Ket{\psi}
= \frac{1}{\sqrt{2}} \begin{bmatrix} \alpha + \beta \\ \alpha - \beta \end{bmatrix}
\end{aligned}$$
-Notably, it maps the eigenstates of $X$ and $Z$ to each other,
+Notably, it maps the eigenstates of $$X$$ and $$Z$$ to each other,
and is its own inverse (i.e. unitary):
$$\begin{aligned}
@@ -147,20 +147,20 @@ $$\begin{aligned}
H \Ket{-} = \Ket{1}
\end{aligned}$$
-The **Clifford gates** are a set including $X$, $Y$, $Z$, $H$ and $S$,
+The **Clifford gates** are a set including $$X$$, $$Y$$, $$Z$$, $$H$$ and $$S$$,
or more generally any gates that rotate
-by multiples of $\pi/2$ around the Bloch sphere.
-This set is **not universal**, meaning that if we start from $\Ket{0}$,
-we can only reach $\Ket{0}$, $\Ket{1}$, $\Ket{+}$, $\Ket{-}$, $\Ket{+i}$ $\Ket{-i}$ using these gates.
+by multiples of $$\pi/2$$ around the Bloch sphere.
+This set is **not universal**, meaning that if we start from $$\Ket{0}$$,
+we can only reach $$\Ket{0}$$, $$\Ket{1}$$, $$\Ket{+}$$, $$\Ket{-}$$, $$\Ket{+i}$$ $$\Ket{-i}$$ using these gates.
-If we add *any* non-Clifford gate, for example $T$,
+If we add *any* non-Clifford gate, for example $$T$$,
then we can reach any point on the Bloch sphere,
which means that the set is **universal**.
However, there is a problem: a qubit has an uncountable infinity of states,
but a quantum circuit consists of a countably infinite sequence of gates, at most.
Therefore, technically, we can never reach the whole Bloch sphere,
-but we *can* come up with circuits that approximate a target state to some degree $\varepsilon$.
+but we *can* come up with circuits that approximate a target state to some degree $$\varepsilon$$.
This is the definition of universality:
any state can be approximated.
@@ -168,7 +168,7 @@ any state can be approximated.
## Two-qubit gates
As an example, let us consider
-the following two pure one-qubit states $\Ket{\psi_1}$ and $\Ket{\psi_2}$:
+the following two pure one-qubit states $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$:
$$\begin{aligned}
\Ket{\psi_1}
@@ -181,7 +181,7 @@ $$\begin{aligned}
\end{aligned}$$
The composite state of both qubits, assuming they are pure,
-is then their tensor product $\otimes$:
+is then their tensor product $$\otimes$$:
$$\begin{aligned}
\Ket{\psi_1 \psi_2}
@@ -192,15 +192,15 @@ $$\begin{aligned}
\end{aligned}$$
Note that a two-qubit system may be [entangled](/know/concept/quantum-entanglement/),
-in which case the coefficients $c_{00}$ etc. cannot be written as products,
-i.e. $\Ket{\psi_2}$ cannot be expressed separately from $\Ket{\psi_1}$, and vice versa.
+in which case the coefficients $$c_{00}$$ etc. cannot be written as products,
+i.e. $$\Ket{\psi_2}$$ cannot be expressed separately from $$\Ket{\psi_1}$$, and vice versa.
In other words, the general action of a two-qubit quantum gate
-can be expressed in the basis of $\Ket{00}$, $\Ket{01}$, $\Ket{10}$ and $\Ket{11}$,
-but not always in the basis of $\Ket{0}_1$, $\Ket{1}_1$, $\Ket{0}_2$ and $\Ket{1}_2$.
+can be expressed in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$,
+but not always in the basis of $$\Ket{0}_1$$, $$\Ket{1}_1$$, $$\Ket{0}_2$$ and $$\Ket{1}_2$$.
-With that said, the first two-qubit gate is $\mathrm{SWAP}$,
-which simply swaps $\Ket{\psi_1}$ and $\Ket{\psi_2}$:
+With that said, the first two-qubit gate is $$\mathrm{SWAP}$$,
+which simply swaps $$\Ket{\psi_1}$$ and $$\Ket{\psi_2}$$:
@@ -218,8 +218,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-This matrix is given in the basis of $\Ket{00}$, $\Ket{01}$, $\Ket{10}$ and $\Ket{11}$.
-Note that $\mathrm{SWAP}$ cannot generate entanglement,
+This matrix is given in the basis of $$\Ket{00}$$, $$\Ket{01}$$, $$\Ket{10}$$ and $$\Ket{11}$$.
+Note that $$\mathrm{SWAP}$$ cannot generate entanglement,
so if its input is separable, its output is too.
In any case, its effect is clear:
@@ -228,8 +228,8 @@ $$\begin{aligned}
&= c_{00} \Ket{00} + c_{10} \Ket{01} + c_{01} \Ket{10} + c_{11} \Ket{11}
\end{aligned}$$
-Next, there is the **controlled NOT gate** $\mathrm{CNOT}$,
-which "flips" (applies $X$ to) $\Ket{\psi_2}$ if $\Ket{\psi_1}$ is true:
+Next, there is the **controlled NOT gate** $$\mathrm{CNOT}$$,
+which "flips" (applies $$X$$ to) $$\Ket{\psi_2}$$ if $$\Ket{\psi_1}$$ is true:
@@ -247,16 +247,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-That is, it swaps the last two coefficients $c_{10}$ and $c_{11}$ in the composite state vector:
+That is, it swaps the last two coefficients $$c_{10}$$ and $$c_{11}$$ in the composite state vector:
$$\begin{aligned}
\mathrm{CNOT} \Ket{\psi_1 \psi_2}
&= c_{00} \Ket{00} + c_{01} \Ket{01} + c_{11} \Ket{10} + c_{10} \Ket{11}
\end{aligned}$$
-More generally, from every one-qubit gate $U$,
-we can define a two-qubit **controlled U gate** $\mathrm{CU}$,
-which applies $U$ to $\Ket{\psi_2}$ if $\Ket{\psi_1}$ is true:
+More generally, from every one-qubit gate $$U$$,
+we can define a two-qubit **controlled U gate** $$\mathrm{CU}$$,
+which applies $$U$$ to $$\Ket{\psi_2}$$ if $$\Ket{\psi_1}$$ is true:
@@ -274,7 +274,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the lower-right 2x2 block is simply $U$.
+Where the lower-right 2x2 block is simply $$U$$.
The general action of this gate is given by:
$$\begin{aligned}
@@ -283,8 +283,8 @@ $$\begin{aligned}
\end{aligned}$$
A set of gates is **universal** if all possible mappings
-from $n$ to $n$ qubits can be approximated using only these gates.
-A minimal universal set is $\{\mathrm{CNOT}, T, S\}$,
+from $$n$$ to $$n$$ qubits can be approximated using only these gates.
+A minimal universal set is $$\{\mathrm{CNOT}, T, S\}$$,
and there exist many others.
diff --git a/source/know/concept/quantum-teleportation/index.md b/source/know/concept/quantum-teleportation/index.md
index c7b9295..095c2c6 100644
--- a/source/know/concept/quantum-teleportation/index.md
+++ b/source/know/concept/quantum-teleportation/index.md
@@ -11,8 +11,8 @@ layout: "concept"
between systems without the use of a quantum channel.
It is based on [quantum entanglement](/know/concept/quantum-entanglement/).
-Suppose that Alice has a qubit $\Ket{q}_{A'}$ that she wants to send to Bob.
-Since she has not measured it yet, she does not know $\alpha$ or $\beta$;
+Suppose that Alice has a qubit $$\Ket{q}_{A'}$$ that she wants to send to Bob.
+Since she has not measured it yet, she does not know $$\alpha$$ or $$\beta$$;
she just wants Bob to get the same qubit:
$$\begin{aligned}
@@ -21,14 +21,14 @@ $$\begin{aligned}
\end{aligned}$$
She can only directly communicate with Bob over a classical channel.
-This is not enough: even if Alice did know $\alpha$ and $\beta$ exactly
+This is not enough: even if Alice did know $$\alpha$$ and $$\beta$$ exactly
(which would need her having infinitely many copies to measure),
sending an arbitrary real number requires an infinite amount of classical data.
However, between them, she and Bob also have an entangled [Bell state](/know/concept/bell-state/),
-e.g. $\ket{\Phi^+}_{AB}$ (it does not matter which Bell state it is)
+e.g. $$\ket{\Phi^+}_{AB}$$ (it does not matter which Bell state it is)
The state of the composite system is then as follows,
-with $A'$ being Alice' qubit, $A$ her side of the Bell state, and $B$ Bob's side:
+with $$A'$$ being Alice' qubit, $$A$$ her side of the Bell state, and $$B$$ Bob's side:
$$\begin{aligned}
\Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB}
@@ -38,7 +38,7 @@ $$\begin{aligned}
+ \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB}
\end{aligned}$$
-Now, observe that we can write any combination of $\Ket{0}$ and $\Ket{1}$
+Now, observe that we can write any combination of $$\Ket{0}$$ and $$\Ket{1}$$
in the Bell basis like so:
$$\begin{aligned}
@@ -79,7 +79,7 @@ $$\begin{aligned}
\end{aligned}$$
Thus, purely due to entanglement,
-Bob's qubit $B$ is in a superposition of the following states:
+Bob's qubit $$B$$ is in a superposition of the following states:
$$\begin{aligned}
\Ket{q}
@@ -106,38 +106,38 @@ $$\begin{aligned}
\end{aligned}$$
The point is that, thanks to the initial entanglement between Alice and Bob,
-adding $\Ket{q}_{A'}$ into the mix somehow "teleports" that information to Bob,
+adding $$\Ket{q}_{A'}$$ into the mix somehow "teleports" that information to Bob,
although it is not in a usable form yet.
-To finish the process, Alice measures her side $A'A$ in the Bell basis.
-Consequently, $A'A$ collapses into one of
-$\ket{\Phi^{+}}$, $\ket{\Phi^{-}}$, $\ket{\Psi^{+}}$, $\ket{\Psi^{-}}$
+To finish the process, Alice measures her side $$A'A$$ in the Bell basis.
+Consequently, $$A'A$$ collapses into one of
+$$\ket{\Phi^{+}}$$, $$\ket{\Phi^{-}}$$, $$\ket{\Psi^{+}}$$, $$\ket{\Psi^{-}}$$
with equal probability, and she knows which.
-This collapse leaves Bob's side $B$ in $\Ket{q}$, $\hat{\sigma}_z \Ket{q}$,
-$\hat{\sigma}_x \Ket{q}$, or $\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$, respectively.
-The entanglement between $A$ and $B$ is thus broken,
-and instead Alice has local entanglement between $A'$ and $A$.
+This collapse leaves Bob's side $$B$$ in $$\Ket{q}$$, $$\hat{\sigma}_z \Ket{q}$$,
+$$\hat{\sigma}_x \Ket{q}$$, or $$\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$$, respectively.
+The entanglement between $$A$$ and $$B$$ is thus broken,
+and instead Alice has local entanglement between $$A'$$ and $$A$$.
She then uses the classical channel to tell Bob her result,
-who then either does nothing (for $\Ket{q}$),
-applies $\hat{\sigma}_z$ (for $\hat{\sigma}_z \Ket{q}$),
-applies $\hat{\sigma}_x$ (for $\hat{\sigma}_x \Ket{q}$),
-or applies $\hat{\sigma}_z \hat{\sigma}_x$ (for $\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$).
-Then, due to the fact that $\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$,
-he recovers $\Ket{q}$ in his local qubit $B$.
+who then either does nothing (for $$\Ket{q}$$),
+applies $$\hat{\sigma}_z$$ (for $$\hat{\sigma}_z \Ket{q}$$),
+applies $$\hat{\sigma}_x$$ (for $$\hat{\sigma}_x \Ket{q}$$),
+or applies $$\hat{\sigma}_z \hat{\sigma}_x$$ (for $$\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$$).
+Then, due to the fact that $$\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$$,
+he recovers $$\Ket{q}$$ in his local qubit $$B$$.
This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem)
-because Alice does not require any knowledge of $\Ket{q}$,
-and after the measurement, her qubit $A'$ will no longer be in that state.
+because Alice does not require any knowledge of $$\Ket{q}$$,
+and after the measurement, her qubit $$A'$$ will no longer be in that state.
In other words, quantum teleportation *moves* states,
rather than copying them.
Nor does this conflict with Einstein's relativity,
since the information travels no faster than light:
-the entangled $\ket{\Phi^{+}}_{AB}$ state must be distributed in advance,
+the entangled $$\ket{\Phi^{+}}_{AB}$$ state must be distributed in advance,
and Alice' declaration of her result is sent classically.
Before receiving that, Bob only sees his side of the maximally entangled
-Bell state $\ket{\Phi^{+}}_{AB}$, which contains nothing of $\Ket{q}$.
+Bell state $$\ket{\Phi^{+}}_{AB}$$, which contains nothing of $$\Ket{q}$$.
## References
diff --git a/source/know/concept/rabi-oscillation/index.md b/source/know/concept/rabi-oscillation/index.md
index 9077cce..07f8b25 100644
--- a/source/know/concept/rabi-oscillation/index.md
+++ b/source/know/concept/rabi-oscillation/index.md
@@ -12,21 +12,21 @@ layout: "concept"
In quantum mechanics, from the derivation of
[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
-we know that a time-dependent term $\hat{H}_1$ in the Hamiltonian
+we know that a time-dependent term $$\hat{H}_1$$ in the Hamiltonian
affects the state as follows,
-where $c_n(t)$ are the coefficients of the linear combination
-of basis states $\Ket{n} \exp(-i E_n t / \hbar)$:
+where $$c_n(t)$$ are the coefficients of the linear combination
+of basis states $$\Ket{n} \exp(-i E_n t / \hbar)$$:
$$\begin{aligned}
i \hbar \dv{c_m}{t}
= \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} \exp(i \omega_{mn} t)
\end{aligned}$$
-Where $\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$
-for energies $E_m$ and $E_n$.
+Where $$\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$$
+for energies $$E_m$$ and $$E_n$$.
Note that this equation is exact,
despite being used for deriving perturbation theory.
-Consider a two-level system where $n \in \{a, b\}$,
+Consider a two-level system where $$n \in \{a, b\}$$,
in which case the above equation can be expanded to the following:
$$\begin{aligned}
@@ -37,8 +37,8 @@ $$\begin{aligned}
&= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} \: c_b
\end{aligned}$$
-Where $\omega_0 \equiv \omega_{ba}$ is positive.
-We assume that $\hat{H}_1$ has odd spatial parity,
+Where $$\omega_0 \equiv \omega_{ba}$$ is positive.
+We assume that $$\hat{H}_1$$ has odd spatial parity,
in which case [Laporte's selection rule](/know/concept/selection-rules/)
states that the diagonal matrix elements vanish, leaving:
@@ -50,8 +50,8 @@ $$\begin{aligned}
&= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a
\end{aligned}$$
-We now choose $\hat{H}_1$ to be as follows,
-sinusoidally oscillating with a spatially odd $V(\vec{r})$:
+We now choose $$\hat{H}_1$$ to be as follows,
+sinusoidally oscillating with a spatially odd $$V(\vec{r})$$:
$$\begin{aligned}
\hat{H}_1(t)
@@ -59,8 +59,8 @@ $$\begin{aligned}
= \frac{V}{2} \Big( \exp(i \omega t) + \exp(-i \omega t) \Big)
\end{aligned}$$
-We insert this into the equations for $c_a$ and $c_b$,
-and define $V_{ab} \equiv \matrixel{a}{V}{b}$, leading us to:
+We insert this into the equations for $$c_a$$ and $$c_b$$,
+and define $$V_{ab} \equiv \matrixel{a}{V}{b}$$, leading us to:
$$\begin{aligned}
\dv{c_a}{t}
@@ -72,8 +72,8 @@ $$\begin{aligned}
Here, we make the
[rotating wave approximation](/know/concept/rotating-wave-approximation/):
-assuming we are close to resonance $\omega \approx \omega_0$,
-we argue that $\exp(i (\omega \!+\! \omega_0) t)$
+assuming we are close to resonance $$\omega \approx \omega_0$$,
+we argue that $$\exp(i (\omega \!+\! \omega_0) t)$$
oscillates so fast that its effect is negligible
when the system is observed over a reasonable time interval.
Dropping those terms leaves us with:
@@ -91,8 +91,8 @@ $$\begin{aligned}
\end{aligned}$$
Now we can solve this system of coupled equations exactly.
-We differentiate the first equation with respect to $t$,
-and then substitute $\idv{c_b}{t}$ for the second equation:
+We differentiate the first equation with respect to $$t$$,
+and then substitute $$\idv{c_b}{t}$$ for the second equation:
$$\begin{aligned}
\dvn{2}{c_a}{t}
@@ -105,15 +105,15 @@ $$\begin{aligned}
&= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} c_a
\end{aligned}$$
-In the first term, we recognize $\idv{c_a}{t}$,
-which we insert to arrive at an equation for $c_a(t)$:
+In the first term, we recognize $$\idv{c_a}{t}$$,
+which we insert to arrive at an equation for $$c_a(t)$$:
$$\begin{aligned}
0
= \dvn{2}{c_a}{t} - i (\omega - \omega_0) \dv{c_a}{t} + \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a
\end{aligned}$$
-To solve this, we make the ansatz $c_a(t) = \exp(\lambda t)$,
+To solve this, we make the ansatz $$c_a(t) = \exp(\lambda t)$$,
which, upon insertion, gives us:
$$\begin{aligned}
@@ -121,7 +121,7 @@ $$\begin{aligned}
= \lambda^2 - i (\omega - \omega_0) \lambda + \frac{|V_{ab}|^2}{(2 \hbar)^2}
\end{aligned}$$
-This quadratic equation has two complex roots $\lambda_1$ and $\lambda_2$,
+This quadratic equation has two complex roots $$\lambda_1$$ and $$\lambda_2$$,
which are found to be:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
= i \frac{\omega - \omega_0 - \tilde{\Omega}}{2}
\end{aligned}$$
-Where we have defined the **generalized Rabi frequency** $\tilde{\Omega}$ to be given by:
+Where we have defined the **generalized Rabi frequency** $$\tilde{\Omega}$$ to be given by:
$$\begin{aligned}
\boxed{
@@ -141,8 +141,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-So that the general solution $c_a(t)$ is as follows,
-where $A$ and $B$ are arbitrary constants,
+So that the general solution $$c_a(t)$$ is as follows,
+where $$A$$ and $$B$$ are arbitrary constants,
to be determined from initial conditions (and normalization):
$$\begin{aligned}
@@ -152,11 +152,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-And then the corresponding $c_b(t)$ can be found
+And then the corresponding $$c_b(t)$$ can be found
from the coupled equation we started at,
-or, if we only care about the probability density $|c_a|^2$,
-we can use $|c_b|^2 = 1 - |c_a|^2$.
-For example, if $A = 0$ and $B = 1$,
+or, if we only care about the probability density $$|c_a|^2$$,
+we can use $$|c_b|^2 = 1 - |c_a|^2$$.
+For example, if $$A = 0$$ and $$B = 1$$,
we get the following probabilities
$$\begin{aligned}
@@ -170,15 +170,15 @@ $$\begin{aligned}
\end{aligned}$$
Note that the period was halved by squaring.
-This periodic "flopping" of the particle between $\Ket{a}$ and $\Ket{b}$
+This periodic "flopping" of the particle between $$\Ket{a}$$ and $$\Ket{b}$$
is known as **Rabi oscillation**, **Rabi flopping** or the **Rabi cycle**.
This is a more accurate treatment
of the flopping found from first-order perturbation theory.
The name **generalized Rabi frequency** suggests
that there is a non-general version.
-Indeed, the **Rabi frequency** $\Omega$ is based on
-the special case of exact resonance $\omega = \omega_0$:
+Indeed, the **Rabi frequency** $$\Omega$$ is based on
+the special case of exact resonance $$\omega = \omega_0$$:
$$\begin{aligned}
\Omega
@@ -187,7 +187,7 @@ $$\begin{aligned}
As an example, Rabi oscillation arises
in the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
-where $\hat{H}_1$ is:
+where $$\hat{H}_1$$ is:
$$\begin{aligned}
\hat{H}_1(t)
@@ -202,10 +202,10 @@ $$\begin{aligned}
= - \frac{\vec{d} \cdot \vec{E}_0}{\hbar}
\end{aligned}$$
-Where $\vec{E}_0$ is the [electric field](/know/concept/electric-field/) amplitude,
-and $\vec{d} \equiv q \matrixel{b}{\vec{r}}{a}$ is the transition dipole moment
-of the electron between orbitals $\Ket{a}$ and $\Ket{b}$.
-Apparently, some authors define $\vec{d}$ with the opposite sign,
+Where $$\vec{E}_0$$ is the [electric field](/know/concept/electric-field/) amplitude,
+and $$\vec{d} \equiv q \matrixel{b}{\vec{r}}{a}$$ is the transition dipole moment
+of the electron between orbitals $$\Ket{a}$$ and $$\Ket{b}$$.
+Apparently, some authors define $$\vec{d}$$ with the opposite sign,
thereby departing from its classical interpretation.
diff --git a/source/know/concept/random-phase-approximation/index.md b/source/know/concept/random-phase-approximation/index.md
index ac007eb..0f53136 100644
--- a/source/know/concept/random-phase-approximation/index.md
+++ b/source/know/concept/random-phase-approximation/index.md
@@ -8,19 +8,19 @@ categories:
layout: "concept"
---
-Recall that the [self-energy](/know/concept/self-energy/) $\Sigma$
+Recall that the [self-energy](/know/concept/self-energy/) $$\Sigma$$
is defined as a sum of [Feynman diagrams](/know/concept/feynman-diagram/),
-which each have an order $n$ equal to the number of interaction lines.
+which each have an order $$n$$ equal to the number of interaction lines.
We consider the self-energy in the context of [jellium](/know/concept/jellium/),
-so the interaction lines $W$ represent Coulomb repulsion,
+so the interaction lines $$W$$ represent Coulomb repulsion,
and we use [imaginary time](/know/concept/imaginary-time/).
Let us non-dimensionalize the Feynman diagrams in the self-energy,
-by measuring momenta in units of $\hbar k_F$,
-and energies in $\epsilon_F = \hbar^2 k_F^2 / (2 m)$.
-Each internal variable then gives a factor $k_F^5$,
-where $k_F^3$ comes from the 3D momentum integral,
-and $k_F^2$ from the energy $1 / \beta$:
+by measuring momenta in units of $$\hbar k_F$$,
+and energies in $$\epsilon_F = \hbar^2 k_F^2 / (2 m)$$.
+Each internal variable then gives a factor $$k_F^5$$,
+where $$k_F^3$$ comes from the 3D momentum integral,
+and $$k_F^2$$ from the energy $$1 / \beta$$:
$$\begin{aligned}
\frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty \cdots \:\dd{\vb{k}}
@@ -28,8 +28,8 @@ $$\begin{aligned}
k_F^5
\end{aligned}$$
-Meanwhile, every line gives a factor $1 / k_F^2$.
-The [Matsubara Green's function](/know/concept/matsubara-greens-function/) $G^0$
+Meanwhile, every line gives a factor $$1 / k_F^2$$.
+The [Matsubara Green's function](/know/concept/matsubara-greens-function/) $$G^0$$
for a system with continuous translational symmetry
is found from [equation-of-motion theory](/know/concept/equation-of-motion-theory/):
@@ -44,38 +44,38 @@ $$\begin{aligned}
\frac{1}{k_F^2}
\end{aligned}$$
-An $n$th-order diagram in $\Sigma$ contains $n$ interaction lines,
-$2n\!-\!1$ fermion lines, and $n$ integrals,
-so in total it evolves as $1 / k_F^{n-2}$.
-In jellium, we know that the electron density is proportional to $k_F^3$,
-so for high densities we can rest assured that higher-order terms in $\Sigma$
+An $$n$$th-order diagram in $$\Sigma$$ contains $$n$$ interaction lines,
+$$2n\!-\!1$$ fermion lines, and $$n$$ integrals,
+so in total it evolves as $$1 / k_F^{n-2}$$.
+In jellium, we know that the electron density is proportional to $$k_F^3$$,
+so for high densities we can rest assured that higher-order terms in $$\Sigma$$
converge to zero faster than lower-order terms.
-However, at a given order $n$, not all diagrams are equally important.
+However, at a given order $$n$$, not all diagrams are equally important.
In a given diagram, due to momentum conservation,
some interaction lines carry the same momentum variable.
-Because $W(\vb{k}) \propto 1 / |\vb{k}|^2$,
-small $\vb{k}$ make a large contribution,
-and the more interaction lines depend on the same $\vb{k}$,
+Because $$W(\vb{k}) \propto 1 / |\vb{k}|^2$$,
+small $$\vb{k}$$ make a large contribution,
+and the more interaction lines depend on the same $$\vb{k}$$,
the larger the contribution becomes.
In other words, each diagram is dominated by contributions
from the momentum carried by the largest number of interactions.
-At order $n$, there is one diagram
-where all $n$ interactions carry the same momentum,
+At order $$n$$, there is one diagram
+where all $$n$$ interactions carry the same momentum,
and this one dominates all others at this order.
The **random phase approximation** consists of removing most diagrams
-from the defintion of the full self-energy $\Sigma$,
-leaving only the single most divergent one at each order $n$,
-i.e. the ones where all $n$ interaction lines
+from the defintion of the full self-energy $$\Sigma$$,
+leaving only the single most divergent one at each order $$n$$,
+i.e. the ones where all $$n$$ interaction lines
carry the same momentum and energy:
-Where we have defined the **screened interaction** $W^\mathrm{RPA}$,
+Where we have defined the **screened interaction** $$W^\mathrm{RPA}$$,
denoted by a double wavy line:
@@ -99,19 +99,19 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined the **pair-bubble** $\Pi_0$ as follows,
-with an internal wavevector $\vb{q}$, fermionic frequency $i \omega_m^F$, and spin $s$.
-Abbreviating $\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n^B)$
-and $\tilde{\vb{q}} \equiv (\vb{q}, i \omega_n^F)$:
+Where we have defined the **pair-bubble** $$\Pi_0$$ as follows,
+with an internal wavevector $$\vb{q}$$, fermionic frequency $$i \omega_m^F$$, and spin $$s$$.
+Abbreviating $$\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n^B)$$
+and $$\tilde{\vb{q}} \equiv (\vb{q}, i \omega_n^F)$$:
-We isolate the Dyson equation for $W^\mathrm{RPA}$,
+We isolate the Dyson equation for $$W^\mathrm{RPA}$$,
which reveals its physical interpretation as a *screened* interaction:
-the "raw" interaction $W \!=\! e^2 / (\varepsilon_0 |\vb{k}|^2)$
-is weakened by a term containing $\Pi_0$:
+the "raw" interaction $$W \!=\! e^2 / (\varepsilon_0 |\vb{k}|^2)$$
+is weakened by a term containing $$\Pi_0$$:
$$\begin{aligned}
W^\mathrm{RPA}(\vb{k}, i \omega_n^B)
@@ -119,7 +119,7 @@ $$\begin{aligned}
= \frac{e^2}{\varepsilon_0 |\vb{k}|^2 - e^2 \Pi_0(\vb{k}, i \omega_n^B)}
\end{aligned}$$
-Let us evaluate the pair-bubble $\Pi_0$ more concretely.
+Let us evaluate the pair-bubble $$\Pi_0$$ more concretely.
The Feynman diagram translates to:
$$\begin{aligned}
@@ -133,9 +133,9 @@ $$\begin{aligned}
Here we recognize a [Matsubara sum](/know/concept/matsubara-sum/),
and rewrite accordingly.
-Note that the residues of $n_F$ are $1 / (\hbar \beta)$
+Note that the residues of $$n_F$$ are $$1 / (\hbar \beta)$$
when it is a function of frequency,
-and $1 / \beta$ when it is a function of energy, so:
+and $$1 / \beta$$ when it is a function of energy, so:
$$\begin{aligned}
\Pi_0(\vb{k}, i \omega_n^B)
@@ -147,12 +147,12 @@ $$\begin{aligned}
{i \hbar \omega_n^B + \varepsilon_{\vb{q}} - \varepsilon_{\vb{k}+\vb{q}}} \dd{\vb{q}}
\end{aligned}$$
-Where we have used that $n_F(\varepsilon \!+\! i \hbar \omega_n^B) = n_F(\varepsilon)$.
-Analogously to extracting the retarded Green's function $G^R(\omega)$
-from the Matsubara Green's function $G^0(i \omega_n^F)$,
-we replace $i \omega_n^F \to \omega \!+\! i \eta$,
-where $\eta \to 0^+$ is a positive infinitesimal,
-yielding the retarded pair-bubble $\Pi_0^R$:
+Where we have used that $$n_F(\varepsilon \!+\! i \hbar \omega_n^B) = n_F(\varepsilon)$$.
+Analogously to extracting the retarded Green's function $$G^R(\omega)$$
+from the Matsubara Green's function $$G^0(i \omega_n^F)$$,
+we replace $$i \omega_n^F \to \omega \!+\! i \eta$$,
+where $$\eta \to 0^+$$ is a positive infinitesimal,
+yielding the retarded pair-bubble $$\Pi_0^R$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/random-variable/index.md b/source/know/concept/random-variable/index.md
index 1781c8b..ecb8e96 100644
--- a/source/know/concept/random-variable/index.md
+++ b/source/know/concept/random-variable/index.md
@@ -19,50 +19,50 @@ of a random variable.
## Probability space
-A **probability space** or **probability triple** $(\Omega, \mathcal{F}, P)$
+A **probability space** or **probability triple** $$(\Omega, \mathcal{F}, P)$$
is the formal mathematical model of a given **stochastic experiment**,
i.e. a process with a random outcome.
-The **sample space** $\Omega$ is the set
-of all possible outcomes $\omega$ of the experimement.
-Those $\omega$ are selected randomly according to certain criteria.
-A subset $A \subset \Omega$ is called an **event**,
-and can be regarded as a true statement about all $\omega$ in that $A$.
+The **sample space** $$\Omega$$ is the set
+of all possible outcomes $$\omega$$ of the experimement.
+Those $$\omega$$ are selected randomly according to certain criteria.
+A subset $$A \subset \Omega$$ is called an **event**,
+and can be regarded as a true statement about all $$\omega$$ in that $$A$$.
-The **event space** $\mathcal{F}$ is a set of events $A$
+The **event space** $$\mathcal{F}$$ is a set of events $$A$$
that are interesting to us,
-i.e. we have subjectively chosen $\mathcal{F}$
+i.e. we have subjectively chosen $$\mathcal{F}$$
based on the problem at hand.
-Since events $A$ represent statements about outcomes $\omega$,
+Since events $$A$$ represent statements about outcomes $$\omega$$,
and we would like to use logic on those statemenets,
-we demand that $\mathcal{F}$ is a [$\sigma$-algebra](/know/concept/sigma-algebra/).
+we demand that $$\mathcal{F}$$ is a [$$\sigma$$-algebra](/know/concept/sigma-algebra/).
-Finally, the **probability measure** or **probability function** $P$
-is a function that maps $A$ events to probabilities $P(A)$.
-Formally, $P : \mathcal{F} \to \mathbb{R}$ is defined to satisfy:
+Finally, the **probability measure** or **probability function** $$P$$
+is a function that maps $$A$$ events to probabilities $$P(A)$$.
+Formally, $$P : \mathcal{F} \to \mathbb{R}$$ is defined to satisfy:
-1. If $A \in \mathcal{F}$, then $P(A) \in [0, 1]$.
-2. If $A, B \in \mathcal{F}$ do not overlap $A \cap B = \varnothing$,
- then $P(A \cup B) = P(A) + P(B)$.
-3. The total probability $P(\Omega) = 1$.
+1. If $$A \in \mathcal{F}$$, then $$P(A) \in [0, 1]$$.
+2. If $$A, B \in \mathcal{F}$$ do not overlap $$A \cap B = \varnothing$$,
+ then $$P(A \cup B) = P(A) + P(B)$$.
+3. The total probability $$P(\Omega) = 1$$.
-The reason we only assign probability to events $A$
-rather than individual outcomes $\omega$ is that
-if $\Omega$ is continuous, all $\omega$ have zero probability,
-while intervals $A$ can have nonzero probability.
+The reason we only assign probability to events $$A$$
+rather than individual outcomes $$\omega$$ is that
+if $$\Omega$$ is continuous, all $$\omega$$ have zero probability,
+while intervals $$A$$ can have nonzero probability.
## Random variable
-Once we have a probability space $(\Omega, \mathcal{F}, P)$,
-we can define a **random variable** $X$
-as a function that maps outcomes $\omega$
+Once we have a probability space $$(\Omega, \mathcal{F}, P)$$,
+we can define a **random variable** $$X$$
+as a function that maps outcomes $$\omega$$
to another set, usually the real numbers.
To be a valid real-valued random variable,
-a function $X : \Omega \to \mathbb{R}^n$ must satisfy the following condition,
-in which case $X$ is said to be **measurable**
-from $(\Omega, \mathcal{F})$ to $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$:
+a function $$X : \Omega \to \mathbb{R}^n$$ must satisfy the following condition,
+in which case $$X$$ is said to be **measurable**
+from $$(\Omega, \mathcal{F})$$ to $$(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$$:
$$\begin{aligned}
\{ \omega \in \Omega : X(\omega) \in B \} \in \mathcal{F}
@@ -70,16 +70,16 @@ $$\begin{aligned}
\end{aligned}$$
In other words, for a given Borel set
-(see [$\sigma$-algebra](/know/concept/sigma-algebra/)) $B \in \mathcal{B}(\mathbb{R}^n)$,
-the set of all outcomes $\omega \in \Omega$ that satisfy $X(\omega) \in B$
-must form a valid event; this set must be in $\mathcal{F}$.
+(see [$$\sigma$$-algebra](/know/concept/sigma-algebra/)) $$B \in \mathcal{B}(\mathbb{R}^n)$$,
+the set of all outcomes $$\omega \in \Omega$$ that satisfy $$X(\omega) \in B$$
+must form a valid event; this set must be in $$\mathcal{F}$$.
The point is that we need to be able to assign probabilities
-to statements of the form $X \in [a, b]$ for all $a < b$,
-which is only possible if that statement corresponds to an event in $\mathcal{F}$,
-since $P$'s domain is $\mathcal{F}$.
+to statements of the form $$X \in [a, b]$$ for all $$a < b$$,
+which is only possible if that statement corresponds to an event in $$\mathcal{F}$$,
+since $$P$$'s domain is $$\mathcal{F}$$.
-Given such an $X$, and a set $B \subseteq \mathbb{R}$,
-the **preimage** or **inverse image** $X^{-1}$ is defined as:
+Given such an $$X$$, and a set $$B \subseteq \mathbb{R}$$,
+the **preimage** or **inverse image** $$X^{-1}$$ is defined as:
$$\begin{aligned}
X^{-1}(B)
@@ -87,16 +87,16 @@ $$\begin{aligned}
\end{aligned}$$
As suggested by the notation,
-$X^{-1}$ can be regarded as the inverse of $X$:
-it maps $B$ to the event for which $X \in B$.
-With this, our earlier requirement that $X$ be measurable
-can be written as: $X^{-1}(B) \in \mathcal{F}$ for any $B \in \mathcal{B}(\mathbb{R}^n)$.
-This is also often stated as "$X$ is *$\mathcal{F}$-measurable"*.
-
-Related to $\mathcal{F}$ is the **information**
-obtained by observing a random variable $X$.
-Let $\sigma(X)$ be the information generated by observing $X$,
-i.e. the events whose occurrence can be deduced from the value of $X$,
+$$X^{-1}$$ can be regarded as the inverse of $$X$$:
+it maps $$B$$ to the event for which $$X \in B$$.
+With this, our earlier requirement that $$X$$ be measurable
+can be written as: $$X^{-1}(B) \in \mathcal{F}$$ for any $$B \in \mathcal{B}(\mathbb{R}^n)$$.
+This is also often stated as "$$X$$ is *$$\mathcal{F}$$-measurable"*.
+
+Related to $$\mathcal{F}$$ is the **information**
+obtained by observing a random variable $$X$$.
+Let $$\sigma(X)$$ be the information generated by observing $$X$$,
+i.e. the events whose occurrence can be deduced from the value of $$X$$,
or, more formally:
$$\begin{aligned}
@@ -105,29 +105,29 @@ $$\begin{aligned}
= \{ A \in \mathcal{F} : A = X^{-1}(B) \mathrm{\:for\:some\:} B \in \mathcal{B}(\mathbb{R}^n) \}
\end{aligned}$$
-In other words, if the realized value of $X$ is
-found to be in a certain Borel set $B \in \mathcal{B}(\mathbb{R}^n)$,
-then the preimage $X^{-1}(B)$ (i.e. the event yielding this $B$)
+In other words, if the realized value of $$X$$ is
+found to be in a certain Borel set $$B \in \mathcal{B}(\mathbb{R}^n)$$,
+then the preimage $$X^{-1}(B)$$ (i.e. the event yielding this $$B$$)
is known to have occurred.
-In general, given any $\sigma$-algebra $\mathcal{H}$,
-a variable $Y$ is said to be *"$\mathcal{H}$-measurable"*
-if $\sigma(Y) \subseteq \mathcal{H}$,
-so that $\mathcal{H}$ contains at least
-all information extractable from $Y$.
+In general, given any $$\sigma$$-algebra $$\mathcal{H}$$,
+a variable $$Y$$ is said to be *"$$\mathcal{H}$$-measurable"*
+if $$\sigma(Y) \subseteq \mathcal{H}$$,
+so that $$\mathcal{H}$$ contains at least
+all information extractable from $$Y$$.
-Note that $\mathcal{H}$ can be generated by another random variable $X$,
-i.e. $\mathcal{H} = \sigma(X)$.
+Note that $$\mathcal{H}$$ can be generated by another random variable $$X$$,
+i.e. $$\mathcal{H} = \sigma(X)$$.
In that case, the **Doob-Dynkin lemma** states
-that $Y$ is only $\sigma(X)$-measurable
-if $Y$ can always be computed from $X$,
-i.e. there exists a function $f$ such that
-$Y(\omega) = f(X(\omega))$ for all $\omega \in \Omega$.
+that $$Y$$ is only $$\sigma(X)$$-measurable
+if $$Y$$ can always be computed from $$X$$,
+i.e. there exists a function $$f$$ such that
+$$Y(\omega) = f(X(\omega))$$ for all $$\omega \in \Omega$$.
Now, we are ready to define some familiar concepts from probability theory.
-The **cumulative distribution function** $F_X(x)$ is
-the probability of the event where the realized value of $X$
-is smaller than some given $x \in \mathbb{R}$:
+The **cumulative distribution function** $$F_X(x)$$ is
+the probability of the event where the realized value of $$X$$
+is smaller than some given $$x \in \mathbb{R}$$:
$$\begin{aligned}
F_X(x)
@@ -136,8 +136,8 @@ $$\begin{aligned}
= P(X^{-1}(]\!-\!\infty, x]))
\end{aligned}$$
-If $F_X(x)$ is differentiable,
-then the **probability density function** $f_X(x)$ is defined as:
+If $$F_X(x)$$ is differentiable,
+then the **probability density function** $$f_X(x)$$ is defined as:
$$\begin{aligned}
f_X(x)
@@ -147,10 +147,10 @@ $$\begin{aligned}
## Expectation value
-The **expectation value** $\mathbf{E}[X]$ of a random variable $X$
+The **expectation value** $$\mathbf{E}[X]$$ of a random variable $$X$$
can be defined in the familiar way, as the sum/integral
-of every possible value of $X$ mutliplied by the corresponding probability (density).
-For continuous and discrete sample spaces $\Omega$, respectively:
+of every possible value of $$X$$ mutliplied by the corresponding probability (density).
+For continuous and discrete sample spaces $$\Omega$$, respectively:
$$\begin{aligned}
\mathbf{E}[X]
@@ -160,18 +160,18 @@ $$\begin{aligned}
= \sum_{i = 1}^N x_i \: P(X \!=\! x_i)
\end{aligned}$$
-However, $f_X(x)$ is not guaranteed to exist,
+However, $$f_X(x)$$ is not guaranteed to exist,
and the distinction between continuous and discrete is cumbersome.
-A more general definition of $\mathbf{E}[X]$
+A more general definition of $$\mathbf{E}[X]$$
is the following Lebesgue-Stieltjes integral,
-since $F_X(x)$ always exists:
+since $$F_X(x)$$ always exists:
$$\begin{aligned}
\mathbf{E}[X]
= \int_{-\infty}^\infty x \dd{F_X(x)}
\end{aligned}$$
-This is valid for any sample space $\Omega$.
+This is valid for any sample space $$\Omega$$.
Or, equivalently, a Lebesgue integral can be used:
$$\begin{aligned}
@@ -182,8 +182,8 @@ $$\begin{aligned}
An expectation value defined in this way has many useful properties,
most notably linearity.
-We can also define the familiar **variance** $\mathbf{V}[X]$
-of a random variable $X$ as follows:
+We can also define the familiar **variance** $$\mathbf{V}[X]$$
+of a random variable $$X$$ as follows:
$$\begin{aligned}
\mathbf{V}[X]
diff --git a/source/know/concept/rayleigh-plateau-instability/index.md b/source/know/concept/rayleigh-plateau-instability/index.md
index 2ba060c..5dcf77f 100644
--- a/source/know/concept/rayleigh-plateau-instability/index.md
+++ b/source/know/concept/rayleigh-plateau-instability/index.md
@@ -16,11 +16,11 @@ It is the reason why a smooth stream of water (e.g. from a tap)
eventually breaks into droplets as it falls.
Consider an infinitely long cylinder of liquid
-with radius $R_0$ and surface tension $\alpha$.
+with radius $$R_0$$ and surface tension $$\alpha$$.
In this case, the [Young-Laplace equation](/know/concept/young-laplace-law/)
states that its internal pressure
-is a constant $p_i$ expressed as follows,
-where $p_o$ is the exterior air pressure:
+is a constant $$p_i$$ expressed as follows,
+where $$p_o$$ is the exterior air pressure:
$$\begin{aligned}
p_i
@@ -28,12 +28,12 @@ $$\begin{aligned}
\end{aligned}$$
We assume that the liquid is at rest.
-Alternatively, if it is moving in the $z$-direction,
+Alternatively, if it is moving in the $$z$$-direction,
we can also let our coordinate system travel at the same speed.
Anyway, for convenience,
we neglect any motion or acceleration of the liquid column.
-Next, we add a perturbation $p_\epsilon$, assumed to be small,
+Next, we add a perturbation $$p_\epsilon$$, assumed to be small,
to the internal pressure, which we allow to vary with time and space.
We use cylindrical coordinates:
@@ -42,7 +42,7 @@ $$\begin{aligned}
\end{aligned}$$
This internal pressure difference will cause the liquid to start to flow.
-We express the flow velocity as a vector $\vec{u} = (u_r, u_\phi, u_z)$,
+We express the flow velocity as a vector $$\vec{u} = (u_r, u_\phi, u_z)$$,
which obeys the following Euler equations:
$$\begin{aligned}
@@ -53,7 +53,7 @@ $$\begin{aligned}
\end{aligned}$$
The latter equation states that the fluid is incompressible.
-We assume that $\vec{u}$ is so small that we can ignore
+We assume that $$\vec{u}$$ is so small that we can ignore
the quadratic term in the former equation, leaving:
$$\begin{aligned}
@@ -62,7 +62,7 @@ $$\begin{aligned}
\end{aligned}$$
Taking the divergence and using incompressibility
-yields the Laplace equation for $p_\epsilon$:
+yields the Laplace equation for $$p_\epsilon$$:
$$\begin{aligned}
- \frac{1}{\rho} \nabla^2 p_\epsilon
@@ -81,7 +81,7 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Finally, we add a perturbation $R_\epsilon \ll R_0$
+Finally, we add a perturbation $$R_\epsilon \ll R_0$$
to the radius of the surface of the liquid column:
$$\begin{aligned}
@@ -89,13 +89,13 @@ $$\begin{aligned}
= R_0 + R_\epsilon(z, t)
\end{aligned}$$
-Note that there is no dependence on the angle $\phi$;
+Note that there is no dependence on the angle $$\phi$$;
the deformation is assumed to be symmetric.
Imagine the cross-section of the cylinder,
and convince yourself that all asymmetric deformations
will be removed by surface tension, which prefers a circular shape.
-We thus assume that $R_\epsilon$, $p_\epsilon$ and $\vec{u}$
-do not depend on $\phi$.
+We thus assume that $$R_\epsilon$$, $$p_\epsilon$$ and $$\vec{u}$$
+do not depend on $$\phi$$.
The Laplace equation then reduces to:
$$\begin{aligned}
@@ -105,9 +105,9 @@ $$\begin{aligned}
\end{aligned}$$
Before solving this, we need boundary conditions.
-The radial fluid velocity $u_r$ (the $r$-component of $\vec{u}$)
-at the column surface $r\!=\!R$ is the
-[material derivative](/know/concept/material-derivative/) of $R_\epsilon$:
+The radial fluid velocity $$u_r$$ (the $$r$$-component of $$\vec{u}$$)
+at the column surface $$r\!=\!R$$ is the
+[material derivative](/know/concept/material-derivative/) of $$R_\epsilon$$:
$$\begin{aligned}
u_r(r\!=\!R)
@@ -115,17 +115,17 @@ $$\begin{aligned}
= \pdv{R_\epsilon}{t} + u_z(r\!=\!R) \pdv{R_\epsilon}{z}
\end{aligned}$$
-We linearize this by assuming that the deformation $R_\epsilon$
-varies slowly with respect to $z$:
+We linearize this by assuming that the deformation $$R_\epsilon$$
+varies slowly with respect to $$z$$:
$$\begin{aligned}
u_r(r\!=\!R)
\approx \pdv{R_\epsilon}{t}
\end{aligned}$$
-Meanwhile, we can write the boundary condition of the pressure $p$
+Meanwhile, we can write the boundary condition of the pressure $$p$$
in two ways, respectively from the Young-Laplace equation
-and the definition of the perturbation $p_\epsilon$:
+and the definition of the perturbation $$p_\epsilon$$:
$$\begin{aligned}
p(r\!=\!R)
@@ -135,32 +135,32 @@ $$\begin{aligned}
= p_i + p_\epsilon(r\!=\!R)
\end{aligned}$$
-Where $R_1$ and $R_2$ are the principal curvature radii of the column surface.
+Where $$R_1$$ and $$R_2$$ are the principal curvature radii of the column surface.
These two expressions must be equivalent,
-so, by inserting the definition of $p_i = p_o + \alpha / R_0$:
+so, by inserting the definition of $$p_i = p_o + \alpha / R_0$$:
$$\begin{aligned}
p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
= p_o + \frac{\alpha}{R_0} + p_\epsilon(r\!=\!R)
\end{aligned}$$
-Isolating this equation for $p_\epsilon$ yields the desired boundary condition:
+Isolating this equation for $$p_\epsilon$$ yields the desired boundary condition:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
= \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big) - \frac{\alpha}{R_0}
\end{aligned}$$
-The principal radius around the circumference is $R_0 + R_\epsilon$,
+The principal radius around the circumference is $$R_0 + R_\epsilon$$,
while the curvature along the length can be approximated
-using the second $z$-derivative of $R_\epsilon$:
+using the second $$z$$-derivative of $$R_\epsilon$$:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
\approx \alpha \Big( \frac{1}{R_0 + R_\epsilon} - \pdvn{2}{R_\epsilon}{z} \Big) - \frac{\alpha}{R_0}
\end{aligned}$$
-This can be simplified a bit by using the assumption that $R_\epsilon$ is small:
+This can be simplified a bit by using the assumption that $$R_\epsilon$$ is small:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
@@ -170,8 +170,8 @@ $$\begin{aligned}
At last, we have all the necessary boundary condition.
We now make the following ansatz,
-where $k$ is the wavenumber
-and $\sigma$ describes exponential growth or decay:
+where $$k$$ is the wavenumber
+and $$\sigma$$ describes exponential growth or decay:
$$\begin{aligned}
\vec{u}(r, z, t)
@@ -187,7 +187,7 @@ $$\begin{aligned}
This is justified by the fact that we can Fourier-expand any perturbation;
this ansatz is simply the dominant term of the resulting series.
-Inserting this into the Laplace equation for $p_\epsilon$ yields
+Inserting this into the Laplace equation for $$p_\epsilon$$ yields
Bessel's modified equation of order zero:
$$\begin{aligned}
@@ -195,8 +195,8 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-This has well-known solutions: the modified Bessel functions $I_0$ and $K_0$.
-However, because $K_0$ diverges at $r = 0$, we must set the constant $B = 0$:
+This has well-known solutions: the modified Bessel functions $$I_0$$ and $$K_0$$.
+However, because $$K_0$$ diverges at $$r = 0$$, we must set the constant $$B = 0$$:
$$\begin{aligned}
p_\epsilon(r)
@@ -204,7 +204,7 @@ $$\begin{aligned}
= A I_0(kr)
\end{aligned}$$
-Inserting the ansatz into the boundary condition for $p_\epsilon$
+Inserting the ansatz into the boundary condition for $$p_\epsilon$$
gives us the following relation:
$$\begin{aligned}
@@ -213,8 +213,8 @@ $$\begin{aligned}
= A I_0(k R)
\end{aligned}$$
-Meanwhile, the linearized Euler equation governing $\vec{u}$
-states that $u_r$ is given by:
+Meanwhile, the linearized Euler equation governing $$\vec{u}$$
+states that $$u_r$$ is given by:
$$\begin{aligned}
\sigma u_r
@@ -222,7 +222,7 @@ $$\begin{aligned}
= - \frac{A k}{\rho} I_0'(kr)
\end{aligned}$$
-Now that we have an expression for $u_r$,
+Now that we have an expression for $$u_r$$,
we can revisit its boundary condition:
$$\begin{aligned}
@@ -231,8 +231,8 @@ $$\begin{aligned}
= \sigma R_\epsilon
\end{aligned}$$
-Isolating this for $R_\epsilon$ and inserting it
-into the boundary condition for $p_\epsilon$ yields:
+Isolating this for $$R_\epsilon$$ and inserting it
+into the boundary condition for $$p_\epsilon$$ yields:
$$\begin{aligned}
p_\epsilon(r\!=\!R)
@@ -240,18 +240,18 @@ $$\begin{aligned}
= \alpha \Big( \frac{1}{R_0^2} + k^2 \Big) \Big( \frac{A k}{\rho \sigma^2} I_0'(k R) \Big)
\end{aligned}$$
-Isolating this for the exponential growth/decay parameter $\sigma$
+Isolating this for the exponential growth/decay parameter $$\sigma$$
gives us the desired result,
-where we have also used the fact that $R \approx R_0$:
+where we have also used the fact that $$R \approx R_0$$:
$$\begin{aligned}
\sigma^2
= \frac{\alpha k}{\rho R_0^2} (1 - k^2 R_0^2) \frac{I_0'(kR_0)}{I_0(kR_0)}
\end{aligned}$$
-To get exponential growth (i.e. instability), we need $\sigma^2 > 0$.
-Since $(1 - k^2 R_0^2)$ is the only factor that can be negative,
-we need $k R_0 < 1$, leading us to the **critical wavelength** $\lambda_c$:
+To get exponential growth (i.e. instability), we need $$\sigma^2 > 0$$.
+Since $$(1 - k^2 R_0^2)$$ is the only factor that can be negative,
+we need $$k R_0 < 1$$, leading us to the **critical wavelength** $$\lambda_c$$:
$$\begin{aligned}
\boxed{
@@ -261,12 +261,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-If the perturbation wavelength $\lambda$ is larger than $\lambda_c$,
+If the perturbation wavelength $$\lambda$$ is larger than $$\lambda_c$$,
surface tension creates a higher pressure in the narrower sections
compared to the wider ones, thereby pumping the liquid into the bulges,
further increasing their size until they become droplets.
-Else, if $\lambda < \lambda_c$, the tighter curvatures
+Else, if $$\lambda < \lambda_c$$, the tighter curvatures
dominate the action of surface tension,
which will then try to smoothen the surface by shrinking the bulges
and widening the constrictions.
diff --git a/source/know/concept/rayleigh-plesset-equation/index.md b/source/know/concept/rayleigh-plesset-equation/index.md
index 4718683..b3ec8f5 100644
--- a/source/know/concept/rayleigh-plesset-equation/index.md
+++ b/source/know/concept/rayleigh-plesset-equation/index.md
@@ -16,7 +16,7 @@ Notably, it leads to [cavitation](/know/concept/cavitation/).
Consider the main
[Navier-Stokes equation](/know/concept/navier-stokes-equations/)
-for the velocity field $\va{v}$:
+for the velocity field $$\va{v}$$:
$$\begin{aligned}
\frac{\mathrm{D} \va{v}}{\mathrm{D} t}
@@ -24,9 +24,9 @@ $$\begin{aligned}
= - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
\end{aligned}$$
-We make the ansatz $\va{v} = v(r, t) \vu{e}_r$,
-where $\vu{e}_r$ is the basis vector;
-in other words, we demand that the only spatial variation of the flow is in $r$.
+We make the ansatz $$\va{v} = v(r, t) \vu{e}_r$$,
+where $$\vu{e}_r$$ is the basis vector;
+in other words, we demand that the only spatial variation of the flow is in $$r$$.
The above equation then becomes:
$$\begin{aligned}
@@ -44,17 +44,17 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-This is only satisfied if $r^2 v$ is constant with respect to $r$,
-leading us to a solution $v(r)$ given by:
+This is only satisfied if $$r^2 v$$ is constant with respect to $$r$$,
+leading us to a solution $$v(r)$$ given by:
$$\begin{aligned}
v(r)
= \frac{C(t)}{r^2}
\end{aligned}$$
-Where $C(t)$ is an unknown function that does not depend on $r$.
+Where $$C(t)$$ is an unknown function that does not depend on $$r$$.
We then insert this result in the main Navier-Stokes equation,
-and isolate it for $\ipdv{p}{r}$, yielding:
+and isolate it for $$\ipdv{p}{r}$$, yielding:
$$\begin{aligned}
\pdv{p}{r}
@@ -63,8 +63,8 @@ $$\begin{aligned}
= - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg)
\end{aligned}$$
-Integrating this with respect to $r$ yields the following expression for $p$,
-where $p_\infty(t)$ is the (possibly time-dependent) pressure at $r = \infty$:
+Integrating this with respect to $$r$$ yields the following expression for $$p$$,
+where $$p_\infty(t)$$ is the (possibly time-dependent) pressure at $$r = \infty$$:
$$\begin{aligned}
p(r)
@@ -73,7 +73,7 @@ $$\begin{aligned}
From the definition of [viscosity](/know/concept/viscosity/),
we know that the normal [stress](/know/concept/cauchy-stress-tensor/)
-$\sigma_{rr}$ in the liquid is given by:
+$$\sigma_{rr}$$ in the liquid is given by:
$$\begin{aligned}
\sigma_{rr}(r)
@@ -81,11 +81,11 @@ $$\begin{aligned}
\end{aligned}$$
We now consider a spherical bubble
-with radius $R(t)$ and interior pressure $P(t)$ along its surface.
-Since we know the liquid pressure $p(r)$,
-we can find $P$ from $\sigma_{rr}(r)$.
+with radius $$R(t)$$ and interior pressure $$P(t)$$ along its surface.
+Since we know the liquid pressure $$p(r)$$,
+we can find $$P$$ from $$\sigma_{rr}(r)$$.
Furthermore, to include the effects of surface tension, we simply add
-the [Young-Laplace law](/know/concept/young-laplace-law/) to $P$:
+the [Young-Laplace law](/know/concept/young-laplace-law/) to $$P$$:
$$\begin{aligned}
P
@@ -93,18 +93,18 @@ $$\begin{aligned}
= p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R}
\end{aligned}$$
-We isolate this for $p(R)$, and equate it to
-our expression for $p(r)$
-at the surface $r\!=\!R$:
+We isolate this for $$p(R)$$, and equate it to
+our expression for $$p(r)$$
+at the surface $$r\!=\!R$$:
$$\begin{aligned}
P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R}
= p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg)
\end{aligned}$$
-Isolating for $P$,
-and inserting the fact that $R'(t) = v(t)$,
-such that $C = r^2 v = R^2 R'$,
+Isolating for $$P$$,
+and inserting the fact that $$R'(t) = v(t)$$,
+such that $$C = r^2 v = R^2 R'$$,
yields:
$$\begin{aligned}
@@ -115,7 +115,7 @@ $$\begin{aligned}
&= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R}
\end{aligned}$$
-Rearranging this and defining $\Delta p \equiv P - p_\infty$
+Rearranging this and defining $$\Delta p \equiv P - p_\infty$$
leads to the Rayleigh-Plesset equation:
$$\begin{aligned}
diff --git a/source/know/concept/reduced-mass/index.md b/source/know/concept/reduced-mass/index.md
index 6718895..1a05b5c 100644
--- a/source/know/concept/reduced-mass/index.md
+++ b/source/know/concept/reduced-mass/index.md
@@ -8,9 +8,9 @@ layout: "concept"
---
Problems with two interacting objects can be simplified
-by combining them into a pseudo-object with **reduced mass** $\mu$,
+by combining them into a pseudo-object with **reduced mass** $$\mu$$,
whose position equals the relative position of the objects.
-For bodies 1 and 2 with respective masses $m_1$ and $m_2$:
+For bodies 1 and 2 with respective masses $$m_1$$ and $$m_2$$:
$$\begin{aligned}
\boxed{
@@ -18,11 +18,11 @@ $$\begin{aligned}
}
\end{aligned}$$
-If $\va{x}_1$ and $\va{x}_2$ are the objects' respective positions,
+If $$\va{x}_1$$ and $$\va{x}_2$$ are the objects' respective positions,
then we define
-the relative position $\va{x}_r$,
-the relative velocity $\va{v}_r$,
-and the relative acceleration $\va{a}_r$:
+the relative position $$\va{x}_r$$,
+the relative velocity $$\va{v}_r$$,
+and the relative acceleration $$\va{a}_r$$:
$$\begin{aligned}
\va{x}_r
@@ -69,9 +69,9 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, Newton's third law states that
-if object 1 experiences a force $\va{F}_1 = m_1 \va{a}_1$ caused by object 2,
-then object 2 experiences an opposite and equal force $\va{F}_2 = - \va{F}_1$.
-In fact, our earlier relation between $\va{a}_1$ and $\va{a}_1$
+if object 1 experiences a force $$\va{F}_1 = m_1 \va{a}_1$$ caused by object 2,
+then object 2 experiences an opposite and equal force $$\va{F}_2 = - \va{F}_1$$.
+In fact, our earlier relation between $$\va{a}_1$$ and $$\va{a}_1$$
boils down to Newton's third law:
$$\begin{aligned}
@@ -80,7 +80,7 @@ $$\begin{aligned}
\va{a}_2 = - \frac{m_1}{m_2} \va{a}_1
\end{aligned}$$
-With all that in mind, let us take a closer look at the relative acceleration $\va{a}_r$:
+With all that in mind, let us take a closer look at the relative acceleration $$\va{a}_r$$:
$$\begin{aligned}
\va{a}_r
@@ -90,15 +90,15 @@ $$\begin{aligned}
= - \frac{\va{F}_2}{\mu}
\end{aligned}$$
-Where $\mu$ is the reduced mass, as defined above.
-In other words, the relative acceleration $\va{a}_r$
-is just $\va{a}_1 = \va{F}_1 / m_1$ multiplied by $m_1 / \mu$.
+Where $$\mu$$ is the reduced mass, as defined above.
+In other words, the relative acceleration $$\va{a}_r$$
+is just $$\va{a}_1 = \va{F}_1 / m_1$$ multiplied by $$m_1 / \mu$$.
This can be regarded as focusing on the dynamics of body 1,
while correcting for the effects of body 2.
This also suggests the following way
-to recover the original positions $\va{x}_1$ and $\va{x}_2$
-from $\va{x}_r$, which you can easily verify for yourself:
+to recover the original positions $$\va{x}_1$$ and $$\va{x}_2$$
+from $$\va{x}_r$$, which you can easily verify for yourself:
$$\begin{aligned}
\va{x}_1
@@ -110,7 +110,7 @@ $$\begin{aligned}
= - \frac{m_1}{m_1 + m_2} \va{x}_r
\end{aligned}$$
-With this, we can rewrite the total kinetic energy $T$ in an elegant way:
+With this, we can rewrite the total kinetic energy $$T$$ in an elegant way:
$$\begin{aligned}
T
@@ -125,11 +125,11 @@ $$\begin{aligned}
= \frac{1}{2} \mu \va{v}_r^2
\end{aligned}$$
-Then, assuming that the system's potential energy $V$
+Then, assuming that the system's potential energy $$V$$
only depends on the distance between the two objects,
-i.e. $V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$,
-we just showed that we can rewrite both $T$ and $V$
-to contain only $\mu$ and relative quantities.
+i.e. $$V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$$,
+we just showed that we can rewrite both $$T$$ and $$V$$
+to contain only $$\mu$$ and relative quantities.
This is relevant for both [Lagrangian mechanics](/know/concept/lagrangian-mechanics/)
and [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
-where $L = T - V$ and $H = T + V$ respectively.
+where $$L = T - V$$ and $$H = T + V$$ respectively.
diff --git a/source/know/concept/renyi-entropy/index.md b/source/know/concept/renyi-entropy/index.md
index d53ccc0..55f233c 100644
--- a/source/know/concept/renyi-entropy/index.md
+++ b/source/know/concept/renyi-entropy/index.md
@@ -9,7 +9,7 @@ layout: "concept"
In information theory, the **Rényi entropy** is a measure
(or family of measures) of the "suprise" or "information"
-contained in a random variable $X$.
+contained in a random variable $$X$$.
It is defined as follows:
$$\begin{aligned}
@@ -19,12 +19,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\alpha \ge 0$ is a free parameter.
+Where $$\alpha \ge 0$$ is a free parameter.
The logarithm is usually base-2, but variations exist.
-The case $\alpha = 0$ is known as the **Hartley entropy** or **max-entropy**,
-and quantifies the "surprise" of an event from $X$,
-if $X$ is uniformly distributed:
+The case $$\alpha = 0$$ is known as the **Hartley entropy** or **max-entropy**,
+and quantifies the "surprise" of an event from $$X$$,
+if $$X$$ is uniformly distributed:
$$\begin{aligned}
\boxed{
@@ -33,9 +33,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $N$ is the cardinality of $X$; the number of different possible events.
-The most famous case, however, is $\alpha = 1$.
-Since $H_\alpha$ is problematic for $\alpha \to 1$, we must take the limit:
+Where $$N$$ is the cardinality of $$X$$; the number of different possible events.
+The most famous case, however, is $$\alpha = 1$$.
+Since $$H_\alpha$$ is problematic for $$\alpha \to 1$$, we must take the limit:
$$\begin{aligned}
H_1(X)
@@ -44,7 +44,7 @@ $$\begin{aligned}
\end{aligned}$$
We then apply L'Hôpital's rule to evaluate this limit,
-and use the fact that all $p_i$ sum to $1$:
+and use the fact that all $$p_i$$ sum to $$1$$:
$$\begin{aligned}
H_1(X)
@@ -64,9 +64,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Next, for $\alpha = 2$, we get the **collision entropy**, which describes
+Next, for $$\alpha = 2$$, we get the **collision entropy**, which describes
the surprise of two independent and identically distributed variables
-$X$ and $Y$ yielding the same event:
+$$X$$ and $$Y$$ yielding the same event:
$$\begin{aligned}
\boxed{
@@ -76,9 +76,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-Finally, in the limit $\alpha \to \infty$,
+Finally, in the limit $$\alpha \to \infty$$,
the largest probability dominates the sum,
-leading to the definition of the **min-entropy** $H_\infty$,
+leading to the definition of the **min-entropy** $$H_\infty$$,
describing the surprise of the most likely event:
$$\begin{aligned}
diff --git a/source/know/concept/repetition-code/index.md b/source/know/concept/repetition-code/index.md
index 31181bb..99ac630 100644
--- a/source/know/concept/repetition-code/index.md
+++ b/source/know/concept/repetition-code/index.md
@@ -8,7 +8,7 @@ layout: "concept"
---
A **repetition code** is a simple approach to error correction:
-to protect a bit $x$, make two copies:
+to protect a bit $$x$$, make two copies:
$$\begin{aligned}
0 \to 000
@@ -16,7 +16,7 @@ $$\begin{aligned}
1 \to 111
\end{aligned}$$
-If a single-bit error occurs, e.g. $000 \to 100$,
+If a single-bit error occurs, e.g. $$000 \to 100$$,
a majority vote resets the minority bit.
Clearly, this does not protect against multi-bit errors,
but that is usually not necessary.
@@ -30,7 +30,7 @@ as discussed below.
## Bit flip code
Suppose that we want to detect errors in
-the following arbitrary qubit state $\Ket{\psi}$:
+the following arbitrary qubit state $$\Ket{\psi}$$:
$$\begin{aligned}
\Ket{\psi}
@@ -38,7 +38,7 @@ $$\begin{aligned}
\end{aligned}$$
For now, let us limit ourselves to detecting **bit flips**,
-where $\alpha$ and $\beta$ get switched:
+where $$\alpha$$ and $$\beta$$ get switched:
$$\begin{aligned}
\alpha \Ket{0} + \beta \Ket{1}
@@ -56,7 +56,7 @@ $$\begin{aligned}
= \alpha \Ket{000} + \beta \Ket{111}
\end{aligned}$$
-In other words, a *logical* $\Ket{0}$ (written $\ket{\overline{0}}$)
+In other words, a *logical* $$\Ket{0}$$ (written $$\ket{\overline{0}}$$)
is represented by 3 *physical* qubits, and vice versa:
$$\begin{aligned}
@@ -80,7 +80,7 @@ of [quantum gates](/know/concept/quantum-gate/):
-So, a little while after encoding the state $\Ket{\psi}$ like that,
+So, a little while after encoding the state $$\Ket{\psi}$$ like that,
a bit flip occurs on the 2nd qubit:
$$\begin{aligned}
@@ -94,8 +94,8 @@ We could measure the state, but that would make it collapse,
which is probably not what we want.
The trick is to use operators called **stabilizers**,
-in this case for example $ZZI = Z_1 \otimes Z_2 \otimes I_3$,
-where $I$ is identity and $Z$ is the Pauli-$Z$ gate.
+in this case for example $$ZZI = Z_1 \otimes Z_2 \otimes I_3$$,
+where $$I$$ is identity and $$Z$$ is the Pauli-$$Z$$ gate.
The 3-qubit basis states are its eigenvectors:
$$\begin{alignedat}{2}
@@ -124,13 +124,13 @@ $$\begin{alignedat}{2}
&&= + \Ket{111}
\end{alignedat}$$
-We could measure $ZZI$ for $\ket{\overline{\psi}}$,
-and if the eigenvalue is $-1$,
+We could measure $$ZZI$$ for $$\ket{\overline{\psi}}$$,
+and if the eigenvalue is $$-1$$,
we know that a bit flip has occurred,
-whereas if the eigenvalue is $+1$,
-there is *maybe* no error ($\Ket{001}$ and $\Ket{110}$ are false negatives).
+whereas if the eigenvalue is $$+1$$,
+there is *maybe* no error ($$\Ket{001}$$ and $$\Ket{110}$$ are false negatives).
-These false negatives are fixed by including another stabilizer $IZZ$,
+These false negatives are fixed by including another stabilizer $$IZZ$$,
with these eigenvectors:
$$\begin{alignedat}{2}
@@ -159,56 +159,56 @@ $$\begin{alignedat}{2}
&&= + \Ket{111}
\end{alignedat}$$
-In which case $\Ket{100}$ and $\Ket{011}$ are false negatives.
-In other words, $IZZ$ cannot detect if the 1st qubit was flipped,
-while $ZZI$ cannot protect the 3rd qubit.
+In which case $$\Ket{100}$$ and $$\Ket{011}$$ are false negatives.
+In other words, $$IZZ$$ cannot detect if the 1st qubit was flipped,
+while $$ZZI$$ cannot protect the 3rd qubit.
But by using both, we know exactly which qubit was flipped
thanks to the eigenvalues:
-The two measurements, respectively representing $ZZI$ and $IZZ$,
-yield $\Ket{1}$ if a bit flip definitely occurred,
-and $\Ket{0}$ otherwise.
+The two measurements, respectively representing $$ZZI$$ and $$IZZ$$,
+yield $$\Ket{1}$$ if a bit flip definitely occurred,
+and $$\Ket{0}$$ otherwise.
There is no entanglement,
so the input is untouched.
@@ -228,7 +228,7 @@ $$\begin{aligned}
How to detect that?
If we want to protect against phase flips *instead of* bit flips,
we can simply do the same as before,
-but along the $X$-axis intead of the $Z$-axis:
+but along the $$X$$-axis intead of the $$Z$$-axis:
$$\begin{aligned}
\boxed{
@@ -244,7 +244,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Such that an arbitrary state $\Ket{\psi}$ is encoded as follows,
+Such that an arbitrary state $$\Ket{\psi}$$ is encoded as follows,
by the circuit shown below:
$$\begin{aligned}
@@ -258,9 +258,9 @@ $$\begin{aligned}
-A phase flip along the $Z$-axis
-corresponds to a bit flip along the $X$-axis $\Ket{+} \to \Ket{-}$.
-In this case, the stabilizers are $XXI$ and $IXX$,
+A phase flip along the $$Z$$-axis
+corresponds to a bit flip along the $$X$$-axis $$\Ket{+} \to \Ket{-}$$.
+In this case, the stabilizers are $$XXI$$ and $$IXX$$,
and the error detection circuit is as follows:
@@ -314,7 +314,7 @@ followed by 3 copies of the bit flip encoder:
We thus use 9 physical qubits to store 1 logical qubit.
Fortunately, more efficient schemes exist.
-The bit flip stabilizers $ZZI$ and $IZZ$
+The bit flip stabilizers $$ZZI$$ and $$IZZ$$
are applied on a per-block basis, like so:
$$\begin{aligned}
@@ -323,7 +323,7 @@ $$\begin{aligned}
IZZ \: III \: III \qquad\quad III \: IZZ \: III \qquad\quad III \: III \: IZZ
\end{aligned}$$
-Whereas the phase flip stabilizers $XXI$ and $IXX$
+Whereas the phase flip stabilizers $$XXI$$ and $$IXX$$
are applied to entire blocks at once:
$$\begin{aligned}
diff --git a/source/know/concept/residue-theorem/index.md b/source/know/concept/residue-theorem/index.md
index bcbd6bd..b58e3c2 100644
--- a/source/know/concept/residue-theorem/index.md
+++ b/source/know/concept/residue-theorem/index.md
@@ -8,21 +8,21 @@ categories:
layout: "concept"
---
-A function $f(z)$ is **meromorphic** if it is
+A function $$f(z)$$ is **meromorphic** if it is
[holomorphic](/know/concept/holomorphic-function/)
except in a finite number of **simple poles**,
-which are points $z_p$ where $f(z_p)$ diverges,
-but where the product $(z - z_p) f(z)$ is non-zero
-and still holomorphic close to $z_p$.
-In other words, $f(z)$ can be approximated close to $z_p$:
+which are points $$z_p$$ where $$f(z_p)$$ diverges,
+but where the product $$(z - z_p) f(z)$$ is non-zero
+and still holomorphic close to $$z_p$$.
+In other words, $$f(z)$$ can be approximated close to $$z_p$$:
$$\begin{aligned}
f(z)
\approx \frac{R_p}{z - z_p}
\end{aligned}$$
-Where the **residue** $R_p$ of a simple pole $z_p$ is defined as follows, and
-represents the rate at which $f(z)$ diverges close to $z_p$:
+Where the **residue** $$R_p$$ of a simple pole $$z_p$$ is defined as follows, and
+represents the rate at which $$f(z)$$ diverges close to $$z_p$$:
$$\begin{aligned}
\boxed{
@@ -32,8 +32,8 @@ $$\begin{aligned}
**Cauchy's residue theorem** for meromorphic functions
is a generalization of Cauchy's integral theorem for holomorphic functions,
-and states that the integral on a contour $C$
-purely depends on the simple poles $z_p$ enclosed by $C$:
+and states that the integral on a contour $$C$$
+purely depends on the simple poles $$z_p$$ enclosed by $$C$$:
$$\begin{aligned}
\boxed{
@@ -47,14 +47,14 @@ $$\begin{aligned}
-Here, $b$ is called the **impact parameter**.
-Intuitively, we expect $\theta$ to be larger for smaller $b$.
+Here, $$b$$ is called the **impact parameter**.
+Intuitively, we expect $$\theta$$ to be larger for smaller $$b$$.
By combining Coulomb's law with Newton's laws,
these particles' equations of motion are found to be as follows,
-where $r = |\vb{r}_1 - \vb{r}_2|$ is the distance between 1 and 2:
+where $$r = |\vb{r}_1 - \vb{r}_2|$$ is the distance between 1 and 2:
$$\begin{aligned}
m_1 \dv{\vb{v}_1}{t}
@@ -41,7 +41,7 @@ $$\begin{aligned}
\end{aligned}$$
Using the [reduced mass](/know/concept/reduced-mass/)
-$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$,
+$$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$$,
we turn this into a one-body problem:
$$\begin{aligned}
@@ -49,11 +49,11 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3}
\end{aligned}$$
-Where $\vb{v} \equiv \vb{v}_1 \!-\! \vb{v}_2$ is the relative velocity,
-and $\vb{r} \equiv \vb{r}_1 \!-\! \vb{r}_2$ is the relative position.
+Where $$\vb{v} \equiv \vb{v}_1 \!-\! \vb{v}_2$$ is the relative velocity,
+and $$\vb{r} \equiv \vb{r}_1 \!-\! \vb{r}_2$$ is the relative position.
The latter is as follows in
[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/)
-$(r, \varphi, z)$:
+$$(r, \varphi, z)$$:
$$\begin{aligned}
\vb{r}
@@ -62,19 +62,19 @@ $$\begin{aligned}
\end{aligned}$$
These new coordinates are sketched below,
-where the origin represents $\vb{r}_1 = \vb{r}_2$.
+where the origin represents $$\vb{r}_1 = \vb{r}_2$$.
Crucially, note the symmetry:
-if the "collision" occurs at $t = 0$,
-then by comparing $t > 0$ and $t < 0$
-we can see that $v_x$ is unchanged for any given $\pm t$,
-while $v_y$ simply changes sign:
+if the "collision" occurs at $$t = 0$$,
+then by comparing $$t > 0$$ and $$t < 0$$
+we can see that $$v_x$$ is unchanged for any given $$\pm t$$,
+while $$v_y$$ simply changes sign:
-From our expression for $\vb{r}$,
-we can find $\vb{v}$ by differentiating with respect to time:
+From our expression for $$\vb{r}$$,
+we can find $$\vb{v}$$ by differentiating with respect to time:
$$\begin{aligned}
\vb{v}
@@ -87,9 +87,9 @@ $$\begin{aligned}
&= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi + z' \:\vu{e}_z
\end{aligned}$$
-Where we have recognized the basis vectors $\vu{e}_r$ and $\vu{e}_\varphi$.
-If we choose the coordinate system such that all dynamics are in the $(x,y)$-plane,
-i.e. $z(t) = 0$, we have:
+Where we have recognized the basis vectors $$\vu{e}_r$$ and $$\vu{e}_\varphi$$.
+If we choose the coordinate system such that all dynamics are in the $$(x,y)$$-plane,
+i.e. $$z(t) = 0$$, we have:
$$\begin{aligned}
\vb{r}
@@ -99,8 +99,8 @@ $$\begin{aligned}
= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi
\end{aligned}$$
-Consequently, the angular momentum $\vb{L}$ is as follows,
-pointing purely in the $z$-direction:
+Consequently, the angular momentum $$\vb{L}$$ is as follows,
+pointing purely in the $$z$$-direction:
$$\begin{aligned}
\vb{L}(t)
@@ -110,8 +110,8 @@ $$\begin{aligned}
\end{aligned}$$
Now, from the figure above,
-we can argue geometrically that at infinity $t = \pm \infty$,
-the ratio $b/r$ is related to the angle $\chi$ between $\vb{v}$ and $\vb{r}$ like so:
+we can argue geometrically that at infinity $$t = \pm \infty$$,
+the ratio $$b/r$$ is related to the angle $$\chi$$ between $$\vb{v}$$ and $$\vb{r}$$ like so:
$$\begin{aligned}
\frac{b}{r(\pm \infty)}
@@ -122,8 +122,8 @@ $$\begin{aligned}
\end{aligned}$$
With this, we can rewrite
-the magnitude of the angular momentum $\vb{L}$ as follows,
-where the total velocity $|\vb{v}|$ is a constant,
+the magnitude of the angular momentum $$\vb{L}$$ as follows,
+where the total velocity $$|\vb{v}|$$ is a constant,
thanks to conservation of energy:
$$\begin{aligned}
@@ -134,7 +134,7 @@ $$\begin{aligned}
\end{aligned}$$
However, conveniently,
-angular momentum is also conserved, i.e. $\vb{L}$ is constant in time:
+angular momentum is also conserved, i.e. $$\vb{L}$$ is constant in time:
$$\begin{aligned}
\vb{L}'(t)
@@ -144,12 +144,12 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-Where we have replaced $\mu \vb{v}'$ with the equation of motion.
-Thanks to this, we can equate the two preceding expressions for $\vb{L}$,
+Where we have replaced $$\mu \vb{v}'$$ with the equation of motion.
+Thanks to this, we can equate the two preceding expressions for $$\vb{L}$$,
leading to the relation below.
Note the appearance of a new minus,
-because the sketch shows that $\varphi' < 0$,
-i.e. $\varphi$ decreases with increasing $t$:
+because the sketch shows that $$\varphi' < 0$$,
+i.e. $$\varphi$$ decreases with increasing $$t$$:
$$\begin{aligned}
- \mu r^2 \dv{\varphi}{t}
@@ -160,7 +160,7 @@ $$\begin{aligned}
\end{aligned}$$
Now, at last, we turn to the main equation of motion.
-Its $y$-component is given by:
+Its $$y$$-component is given by:
$$\begin{aligned}
\mu \dv{v_y}{t}
@@ -170,8 +170,8 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \dd{t}
\end{aligned}$$
-We replace $\dd{t}$ with our earlier relation,
-and recognize geometrically that $y/r = \sin{\varphi}$:
+We replace $$\dd{t}$$ with our earlier relation,
+and recognize geometrically that $$y/r = \sin{\varphi}$$:
$$\begin{aligned}
\mu \dd{v_y}
@@ -180,8 +180,8 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})}
\end{aligned}$$
-Integrating this from the initial state $i$ at $t = -\infty$
-to the final state $f$ at $t = \infty$ yields:
+Integrating this from the initial state $$i$$ at $$t = -\infty$$
+to the final state $$f$$ at $$t = \infty$$ yields:
$$\begin{aligned}
\Delta v_y
@@ -189,8 +189,8 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big)
\end{aligned}$$
-From symmetry, we see that $\varphi_i = \pi \!-\! \varphi_f$,
-and that $\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$, such that:
+From symmetry, we see that $$\varphi_i = \pi \!-\! \varphi_f$$,
+and that $$\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$$, such that:
$$\begin{aligned}
2 v_{y,f}
@@ -198,8 +198,8 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big)
\end{aligned}$$
-Furthermore, geometrically, at $t = \infty$
-we notice that $v_{y,f} = |\vb{v}| \sin{\varphi_f}$,
+Furthermore, geometrically, at $$t = \infty$$
+we notice that $$v_{y,f} = |\vb{v}| \sin{\varphi_f}$$,
leading to:
$$\begin{aligned}
@@ -208,7 +208,7 @@ $$\begin{aligned}
\end{aligned}$$
Rearranging this yields the following equation
-for the final polar angle $\varphi_f \equiv \varphi(\infty)$:
+for the final polar angle $$\varphi_f \equiv \varphi(\infty)$$:
$$\begin{aligned}
\tan{\varphi_f}
@@ -216,9 +216,9 @@ $$\begin{aligned}
= \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
\end{aligned}$$
-However, we want $\theta$, not $\varphi_f$.
+However, we want $$\theta$$, not $$\varphi_f$$.
One last use of symmetry and geometry
-tells us that $\theta = 2 \varphi_f$,
+tells us that $$\theta = 2 \varphi_f$$,
and we thus arrive at the celebrated **Rutherford scattering formula**:
$$\begin{aligned}
@@ -228,7 +228,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-In fact, this formula is also valid if $q_1$ and $q_2$ have opposite signs;
+In fact, this formula is also valid if $$q_1$$ and $$q_2$$ have opposite signs;
in that case particle 2 is simply located on the other side
of particle 1's trajectory.
diff --git a/source/know/concept/salt-equation/index.md b/source/know/concept/salt-equation/index.md
index bfd69b9..f5f085d 100644
--- a/source/know/concept/salt-equation/index.md
+++ b/source/know/concept/salt-equation/index.md
@@ -15,11 +15,11 @@ makes it especially appropriate for microscopically small lasers.
Consider the [Maxwell-Bloch equations](/know/concept/maxwell-bloch-equations/),
governing the complex polarization
-vector $\vb{P}^{+}$ and the scalar population inversion $D$ of a set of
+vector $$\vb{P}^{+}$$ and the scalar population inversion $$D$$ of a set of
active atoms (or quantum dots) embedded in a passive linear background
-material with refractive index $c / v$.
+material with refractive index $$c / v$$.
The system is affected by a driving [electric field](/know/concept/electric-field/)
-$\vb{E}^{+}(t) = \vb{E}_0^{+} e^{-i \omega t}$,
+$$\vb{E}^{+}(t) = \vb{E}_0^{+} e^{-i \omega t}$$,
such that the set of equations is:
$$\begin{aligned}
@@ -34,13 +34,13 @@ $$\begin{aligned}
&= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
\end{aligned}$$
-Where $\hbar \omega_0$ is the band gap of the active atoms,
-and $\gamma_\perp$ and $\gamma_\parallel$ are relaxation rates
+Where $$\hbar \omega_0$$ is the band gap of the active atoms,
+and $$\gamma_\perp$$ and $$\gamma_\parallel$$ are relaxation rates
of the atoms' polarization and population inversion, respectively.
-$D_0$ is the equilibrium inversion, i.e. the value of $D$ if there is no lasing.
-Note that $D_0$ also represents the pump,
-and both $D_0$ and $v$ depend on position $\vb{x}$.
-Finally, the transition dipole matrix elements $\vb{p}_0^{-}$ and $\vb{p}_0^{+}$ are given by:
+$$D_0$$ is the equilibrium inversion, i.e. the value of $$D$$ if there is no lasing.
+Note that $$D_0$$ also represents the pump,
+and both $$D_0$$ and $$v$$ depend on position $$\vb{x}$$.
+Finally, the transition dipole matrix elements $$\vb{p}_0^{-}$$ and $$\vb{p}_0^{+}$$ are given by:
$$\begin{aligned}
\vb{p}_0^{-}
@@ -51,12 +51,12 @@ $$\begin{aligned}
= (\vb{p}_0^{-})^*
\end{aligned}$$
-With $q < 0$ the electron charge, $\vu{x}$ the quantum position operator,
-and $\Ket{g}$ and $\Ket{e}$ respectively
+With $$q < 0$$ the electron charge, $$\vu{x}$$ the quantum position operator,
+and $$\Ket{g}$$ and $$\Ket{e}$$ respectively
the ground state and first excitation of the active atoms.
-We start by assuming that the cavity has $N$ quasinormal modes $\Psi_n$,
-each with a corresponding polarization $\vb{p}_n$ of the active matter.
+We start by assuming that the cavity has $$N$$ quasinormal modes $$\Psi_n$$,
+each with a corresponding polarization $$\vb{p}_n$$ of the active matter.
Note that this ansatz already suggests
that the interactions between the modes are limited:
@@ -80,8 +80,8 @@ $$\begin{aligned}
+ \frac{i}{\hbar} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n \: D
\end{aligned}$$
-With being $\vb{p}_0^{+} \vb{p}_0^{-}$ a dyadic product.
-Isolating the latter equation for $\vb{p}_n$ gives us:
+With being $$\vb{p}_0^{+} \vb{p}_0^{-}$$ a dyadic product.
+Isolating the latter equation for $$\vb{p}_n$$ gives us:
$$\begin{aligned}
\vb{p}_n
@@ -89,7 +89,7 @@ $$\begin{aligned}
= \frac{\gamma(\omega_n) D}{\hbar \gamma_\perp} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n
\end{aligned}$$
-Where we have defined the Lorentzian gain curve $\gamma(\omega_n)$ as follows,
+Where we have defined the Lorentzian gain curve $$\gamma(\omega_n)$$ as follows,
which represents the laser's preferred frequencies for amplification:
$$\begin{aligned}
@@ -97,10 +97,10 @@ $$\begin{aligned}
\equiv \frac{\gamma_\perp}{(\omega_n - \omega_0) + i \gamma_\perp}
\end{aligned}$$
-Inserting this expression for $\vb{p}_n$
+Inserting this expression for $$\vb{p}_n$$
into the first Maxwell-Bloch equation yields
the prototypical form of the SALT equation,
-where we still need to replace $D$ with known quantities:
+where we still need to replace $$D$$ with known quantities:
$$\begin{aligned}
0
@@ -108,8 +108,8 @@ $$\begin{aligned}
- \omega_n^2 \frac{\mu_0 \gamma(\omega_n) D}{\hbar \gamma_\perp} (\vb{p}_0^{+} \vb{p}_0^{-}) \cdot \bigg) \Psi_n
\end{aligned}$$
-To rewrite $D$, we turn to its (Maxwell-Bloch) equation of motion,
-making the crucial **stationary inversion approximation** $\ipdv{D}{t} = 0$:
+To rewrite $$D$$, we turn to its (Maxwell-Bloch) equation of motion,
+making the crucial **stationary inversion approximation** $$\ipdv{D}{t} = 0$$:
$$\begin{aligned}
D
@@ -119,8 +119,8 @@ $$\begin{aligned}
This is the most aggressive approximation we will make:
it removes all definite phase relations between modes,
and effectively eliminates time as a variable.
-We insert our ansatz for $\vb{E}^{+}$ and $\vb{P}^{+}$,
-and point out that only excited lasing modes contribute to $D$:
+We insert our ansatz for $$\vb{E}^{+}$$ and $$\vb{P}^{+}$$,
+and point out that only excited lasing modes contribute to $$D$$:
$$\begin{aligned}
D
@@ -130,9 +130,9 @@ $$\begin{aligned}
\end{aligned}$$
Here, we make the [rotating wave approximation](/know/concept/rotating-wave-approximation/)
-to neglect all terms where $\nu \neq \mu$
+to neglect all terms where $$\nu \neq \mu$$
on the basis that they oscillate too quickly,
-leaving only $\nu = \mu$:
+leaving only $$\nu = \mu$$:
$$\begin{aligned}
D
@@ -140,8 +140,8 @@ $$\begin{aligned}
\bigg( \vb{p}_\nu^* \cdot \Psi_\nu - \vb{p}_\nu \cdot \Psi_\nu^* \bigg)
\end{aligned}$$
-Inserting our earlier equation for $\vb{p}_n$
-and using the fact that $\vb{p}_0^{+} = (\vb{p}_0^{-})^*$ leads us to:
+Inserting our earlier equation for $$\vb{p}_n$$
+and using the fact that $$\vb{p}_0^{+} = (\vb{p}_0^{-})^*$$ leads us to:
$$\begin{aligned}
D
@@ -168,7 +168,7 @@ $$\begin{aligned}
= i 2 \big|\gamma(\omega_\nu)\big|^2
\end{aligned}$$
-Inserting this into our equation for $D$ gives the following expression:
+Inserting this into our equation for $$D$$ gives the following expression:
$$\begin{aligned}
D
@@ -176,7 +176,7 @@ $$\begin{aligned}
\Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu\Big|^2
\end{aligned}$$
-We then properly isolate this for $D$ to get its final form, namely:
+We then properly isolate this for $$D$$ to get its final form, namely:
$$\begin{aligned}
D
@@ -197,7 +197,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where we have defined **spatial hole burning** function $h(\vb{x})$ like so,
+Where we have defined **spatial hole burning** function $$h(\vb{x})$$ like so,
representing the depletion of the supply of charge
carriers as they are consumed by the active lasing modes:
@@ -209,10 +209,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-Many authors assume that $\vb{p}_0^- \parallel \Psi_n$,
-so that only its amplitude $|g|^2 \equiv \vb{p}_0^{+} \cdot \vb{p}_0^{-}$ matters.
-In that case, they often non-dimensionalize $D$ and $\Psi_n$
-by dividing out the units $d_c$ and $e_c$:
+Many authors assume that $$\vb{p}_0^- \parallel \Psi_n$$,
+so that only its amplitude $$|g|^2 \equiv \vb{p}_0^{+} \cdot \vb{p}_0^{-}$$ matters.
+In that case, they often non-dimensionalize $$D$$ and $$\Psi_n$$
+by dividing out the units $$d_c$$ and $$e_c$$:
$$\begin{aligned}
\tilde{\Psi}_n
@@ -228,8 +228,8 @@ $$\begin{aligned}
\equiv \frac{\varepsilon_0 \hbar \gamma_\perp}{|g|^2}
\end{aligned}$$
-And then the SALT equation and hole burning function $h$ are reduced to the following,
-where the vacuum wavenumber $k_n = \omega_n / c$:
+And then the SALT equation and hole burning function $$h$$ are reduced to the following,
+where the vacuum wavenumber $$k_n = \omega_n / c$$:
$$\begin{aligned}
0
@@ -249,7 +249,7 @@ $$\begin{aligned}
= \nabla (\nabla \cdot \Psi) - \nabla^2 \Psi
\end{aligned}$$
-Where $\nabla \cdot \Psi = 0$ thanks to [Gauss' law](/know/concept/maxwells-equations/),
+Where $$\nabla \cdot \Psi = 0$$ thanks to [Gauss' law](/know/concept/maxwells-equations/),
so we get an even further simplified SALT equation:
$$\begin{aligned}
@@ -258,21 +258,21 @@ $$\begin{aligned}
+ \gamma(c k_n) \frac{\tilde{D}_0}{1 + h} \bigg] \bigg) \tilde{\Psi}_n
\end{aligned}$$
-The challenge is to solve this equation for a given $\varepsilon_r(\vb{x})$ and $D_0(\vb{x})$,
-with the boundary condition that $\Psi_n$ is a plane wave at infinity,
+The challenge is to solve this equation for a given $$\varepsilon_r(\vb{x})$$ and $$D_0(\vb{x})$$,
+with the boundary condition that $$\Psi_n$$ is a plane wave at infinity,
i.e. that there is light leaving the cavity.
-If $k_n$ has a negative imaginary part, then that mode is behaving as an LED.
-Gradually increasing the pump $D_0$ in a chosen region
-causes the $k_n$'s imaginary parts become less negative,
+If $$k_n$$ has a negative imaginary part, then that mode is behaving as an LED.
+Gradually increasing the pump $$D_0$$ in a chosen region
+causes the $$k_n$$'s imaginary parts become less negative,
until one of them hits the real axis, at which point that mode starts lasing.
-After that, $D_0$ can be increased even further until some other $k_n$ become real.
+After that, $$D_0$$ can be increased even further until some other $$k_n$$ become real.
-Below threshold (i.e. before any mode is lasing), the problem is linear in $\Psi_n$,
-but above threshold it is nonlinear, and the amplitude of $\Psi_n$ is adjusted
-such that the corresponding $k_n$ never leaves the real axis.
+Below threshold (i.e. before any mode is lasing), the problem is linear in $$\Psi_n$$,
+but above threshold it is nonlinear, and the amplitude of $$\Psi_n$$ is adjusted
+such that the corresponding $$k_n$$ never leaves the real axis.
When any mode is lasing, hole burning makes it harder for other modes to activate,
-since it effectively reduces the pump $D_0$.
+since it effectively reduces the pump $$D_0$$.
## References
diff --git a/source/know/concept/schwartz-distribution/index.md b/source/know/concept/schwartz-distribution/index.md
index e74dc4a..2492d91 100644
--- a/source/know/concept/schwartz-distribution/index.md
+++ b/source/know/concept/schwartz-distribution/index.md
@@ -16,10 +16,10 @@ the [Dirac delta function](/know/concept/dirac-delta-function/)
and the [Heaviside step function](/know/concept/heaviside-step-function/),
whose unusual properties are justified by this generalization.
-We define the **Schwartz space** $\mathcal{S}$ of functions,
+We define the **Schwartz space** $$\mathcal{S}$$ of functions,
whose members are often called **test functions**.
-Every such $\phi(x) \in \mathcal{S}$ must satisfy
-the following constraint for any $p, q \in \mathbb{N}$:
+Every such $$\phi(x) \in \mathcal{S}$$ must satisfy
+the following constraint for any $$p, q \in \mathbb{N}$$:
$$\begin{aligned}
\mathrm{max} \big| x^p \phi^{(q)}(x) \big| < \infty
@@ -30,15 +30,15 @@ decay faster than any polynomial.
Furthermore, all test functions must be infinitely differentiable.
These are quite strict requirements.
-The **space of distributions** $\mathcal{S}'$ (note the prime)
-is then said to consist of *functionals* $f[\phi]$
-which map a test function $\phi$ from $\mathcal{S}$,
-to a number from $\mathbb{C}$;
-this is often written as $\Inprod{f}{\phi}$.
+The **space of distributions** $$\mathcal{S}'$$ (note the prime)
+is then said to consist of *functionals* $$f[\phi]$$
+which map a test function $$\phi$$ from $$\mathcal{S}$$,
+to a number from $$\mathbb{C}$$;
+this is often written as $$\Inprod{f}{\phi}$$.
This notation looks like the inner product of
a [Hilbert space](/know/concept/hilbert-space/),
-for good reason: any well-behaved function $f(x)$ can be embedded
-into $\mathcal{S}'$ by defining the corresponding functional $f[\phi]$ as follows:
+for good reason: any well-behaved function $$f(x)$$ can be embedded
+into $$\mathcal{S}'$$ by defining the corresponding functional $$f[\phi]$$ as follows:
$$\begin{aligned}
f[\phi]
@@ -46,20 +46,20 @@ $$\begin{aligned}
= \int_{-\infty}^\infty f(x) \: \phi(x) \dd{x}
\end{aligned}$$
-Not all functionals qualify for $\mathcal{S}'$:
-they also need to be linear in $\phi$, and **continuous**,
-which in this context means: if a series $\phi_n$
-converges to $\phi$, then $\Inprod{f}{\phi_n}$
-converges to $\Inprod{f}{\phi}$ for all $f$.
+Not all functionals qualify for $$\mathcal{S}'$$:
+they also need to be linear in $$\phi$$, and **continuous**,
+which in this context means: if a series $$\phi_n$$
+converges to $$\phi$$, then $$\Inprod{f}{\phi_n}$$
+converges to $$\Inprod{f}{\phi}$$ for all $$f$$.
-The power of this generalization is that $f(x)$ does not need to be well-behaved:
+The power of this generalization is that $$f(x)$$ does not need to be well-behaved:
for example, the Dirac delta function can also be used,
whose definition is nonsensical *outside* of an integral,
but perfectly reasonable *inside* one.
By treating it as a distribution,
we gain the ability to sanely define e.g. its derivatives.
-Using the example of embedding a well-behaved function $f(x)$ into $\mathcal{S}$,
+Using the example of embedding a well-behaved function $$f(x)$$ into $$\mathcal{S}$$,
we can work out what the derivative of a distribution is:
$$\begin{aligned}
@@ -69,7 +69,7 @@ $$\begin{aligned}
\end{aligned}$$
The test function removes the boundary term, yielding the result
-$- \Inprod{f}{\phi'}$. Although this was an example for a specific $f(x)$,
+$$- \Inprod{f}{\phi'}$$. Although this was an example for a specific $$f(x)$$,
we use it to define the derivative of any distribution:
$$\begin{aligned}
@@ -82,14 +82,14 @@ Using the same trick, we can find the
[Fourier transform](/know/concept/fourier-transform/) (FT)
of a generalized function.
We define the FT as follows,
-but be prepared for some switching of the names $k$ and $x$:
+but be prepared for some switching of the names $$k$$ and $$x$$:
$$\begin{aligned}
\tilde{\phi}(x)
= \int_{-\infty}^\infty \phi(k) \exp(- i k x) \dd{k}
\end{aligned}$$
-The FT of a Schwartz distribution $f$ then turns out to be as follows:
+The FT of a Schwartz distribution $$f$$ then turns out to be as follows:
$$\begin{aligned}
\inprod{\tilde{f}}{\phi}
@@ -100,10 +100,10 @@ $$\begin{aligned}
= \inprod{f}{\tilde{\phi}}
\end{aligned}$$
-Note that the ordinary FT $\tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\}$ is
-already a 1:1 mapping of test functions $\phi \leftrightarrow \tilde{\phi}$.
+Note that the ordinary FT $$\tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\}$$ is
+already a 1:1 mapping of test functions $$\phi \leftrightarrow \tilde{\phi}$$.
As it turns out,
-in this generalization it is also a 1:1 mapping of distributions in $\mathcal{S}'$,
+in this generalization it is also a 1:1 mapping of distributions in $$\mathcal{S}'$$,
defined as:
$$\begin{aligned}
diff --git a/source/know/concept/screw-pinch/index.md b/source/know/concept/screw-pinch/index.md
index c7477b3..0a787dc 100644
--- a/source/know/concept/screw-pinch/index.md
+++ b/source/know/concept/screw-pinch/index.md
@@ -19,13 +19,13 @@ The general way of doing this is called a **screw pinch**.
For simplicity, let the cylinder be infinitely long,
so that it is natural to work in
[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/)
-$(r, \theta, z)$.
+$$(r, \theta, z)$$.
Using the framework of ideal [magnetohydrodynamics](/know/concept/magnetohydrodynamics/) (MHD),
let us start by assuming that the fluid is stationary,
-and that the confining field $\vb{B}$ is fixed.
+and that the confining field $$\vb{B}$$ is fixed.
From the (ideal) generalized Ohm's law, it then follows
-that the [electric field](/know/concept/electric-field/) $\vb{E} = 0$:
+that the [electric field](/know/concept/electric-field/) $$\vb{E} = 0$$:
$$\begin{aligned}
\vb{u}
@@ -41,10 +41,10 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-To get the plasma's equilibrium state for a given $\vb{B}$,
+To get the plasma's equilibrium state for a given $$\vb{B}$$,
we first solve [Ampère's law](/know/concept/maxwells-equations/)
-for the current density $\vb{J}$,
-and then the MHD momentum equation for the pressure $p$.
+for the current density $$\vb{J}$$,
+and then the MHD momentum equation for the pressure $$p$$.
Symmetries should be used whenever possible to reduce these equations:
$$\begin{aligned}
@@ -55,15 +55,15 @@ $$\begin{aligned}
= \nabla p
\end{aligned}$$
-Note that the latter implies that $\nabla p$ is always orthogonal to $\vb{J}$ and $\vb{B}$,
+Note that the latter implies that $$\nabla p$$ is always orthogonal to $$\vb{J}$$ and $$\vb{B}$$,
meaning that the current density and magnetic field must follow
surfaces of constant pressure.
## ϴ-pinch
-In a so-called **ϴ-pinch**, the confining field $\vb{B}$
-is parallel to the $z$-axis, and its magntiude $B_z$ may only depend on $r$.
+In a so-called **ϴ-pinch**, the confining field $$\vb{B}$$
+is parallel to the $$z$$-axis, and its magntiude $$B_z$$ may only depend on $$r$$.
Concretely, we have:
$$\begin{aligned}
@@ -71,13 +71,13 @@ $$\begin{aligned}
= B_z(r) \: \vu{e}_z
\end{aligned}$$
-Where $\vu{e}_z$ is the basis vector of the $z$-axis.
-This $\vb{B}$ confines the plasma thanks to
+Where $$\vu{e}_z$$ is the basis vector of the $$z$$-axis.
+This $$\vb{B}$$ confines the plasma thanks to
the [Lorentz force](/know/concept/lorentz-force/),
which makes charged particles gyrate around magnetic field lines.
-Using Ampère's law, we find that the resulting current density $\vb{J}$,
-expressed in $(r, \theta, z)$:
+Using Ampère's law, we find that the resulting current density $$\vb{J}$$,
+expressed in $$(r, \theta, z)$$:
$$\begin{aligned}
\vb{J}
@@ -91,15 +91,15 @@ $$\begin{aligned}
= -\frac{1}{\mu_0} \pdv{B_z}{r} \: \vu{e}_\theta
\end{aligned}$$
-Where we have used that only $B_z$ is nonzero,
-and that it only depends on $r$.
-This yields a circular current parallel to $\vu{e}_\theta$,
+Where we have used that only $$B_z$$ is nonzero,
+and that it only depends on $$r$$.
+This yields a circular current parallel to $$\vu{e}_\theta$$,
hence the name *ϴ-pinch*.
-Next, we use the MHD momentum equation to find the pressure gradient $\nabla p$.
+Next, we use the MHD momentum equation to find the pressure gradient $$\nabla p$$.
The cross product is easy to evaluate,
-since $\vb{B}$ is parallel to $\vu{e}_z$,
-and $\vb{J}$ is parallel to $\vu{e}_\theta$:
+since $$\vb{B}$$ is parallel to $$\vu{e}_z$$,
+and $$\vb{J}$$ is parallel to $$\vu{e}_\theta$$:
$$\begin{aligned}
\nabla p
@@ -109,9 +109,9 @@ $$\begin{aligned}
= - \frac{1}{\mu_0} \pdv{B_z}{r} B_z \: \vu{e}_r
\end{aligned}$$
-Consequently, $\nabla p$ is parallel to $\vu{e}_r$,
-and only depends on $r$ through $B_z$.
-Along the $r$-direction, the above equation can be rewritten
+Consequently, $$\nabla p$$ is parallel to $$\vu{e}_r$$,
+and only depends on $$r$$ through $$B_z$$.
+Along the $$r$$-direction, the above equation can be rewritten
into the following equilibrium condition:
$$\begin{aligned}
@@ -121,22 +121,22 @@ $$\begin{aligned}
}
\end{aligned}$$
-In other words, the parenthesized expression does not depend on $r$.
+In other words, the parenthesized expression does not depend on $$r$$.
## Z-pinch
Meanwhile, in a so-called **Z-pinch**,
-we create an $r$-dependent current $\vb{J}$ parallel to the $z$-axis:
+we create an $$r$$-dependent current $$\vb{J}$$ parallel to the $$z$$-axis:
$$\begin{aligned}
\vb{J}
= J_z(r) \: \vu{e}_z
\end{aligned}$$
-We can then deduce $\vb{B}$ from Ampère's law,
-using that only $J_z$ is nonzero,
-and that $\ipdv{B_r}{\theta} = 0$ due to circular symmetry:
+We can then deduce $$\vb{B}$$ from Ampère's law,
+using that only $$J_z$$ is nonzero,
+and that $$\ipdv{B_r}{\theta} = 0$$ due to circular symmetry:
$$\begin{aligned}
\vb{J}
@@ -150,11 +150,11 @@ $$\begin{aligned}
= \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} \: \vu{e}_z
\end{aligned}$$
-Therefore, $\vb{J}$ induces a circular $\vb{B} = B_\theta(r) \: \vu{e}_\theta$,
+Therefore, $$\vb{J}$$ induces a circular $$\vb{B} = B_\theta(r) \: \vu{e}_\theta$$,
which confines the plasma for the same reason as in the ϴ-pinch:
the Lorentz force makes particles gyrate around magnetic field lines.
-Next, the resulting pressure gradient $\nabla p$ is found from the MHD momentum equation:
+Next, the resulting pressure gradient $$\nabla p$$ is found from the MHD momentum equation:
$$\begin{aligned}
\nabla p
@@ -164,8 +164,8 @@ $$\begin{aligned}
= - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta \: \vu{e}_r
\end{aligned}$$
-Once again, $\nabla p$ is parallel to $\vu{e}_r$ and only depends on $r$.
-After rearranging, we thus arrive at the following equilibrium condition in the $r$-direction:
+Once again, $$\nabla p$$ is parallel to $$\vu{e}_r$$ and only depends on $$r$$.
+After rearranging, we thus arrive at the following equilibrium condition in the $$r$$-direction:
$$\begin{aligned}
\boxed{
@@ -179,7 +179,7 @@ $$\begin{aligned}
Thanks to the linearity of electromagnetism,
a ϴ-pinch and Z-pinch can be combined to create a **screw pinch**,
-where $\vb{J}$ and $\vb{B}$ both have nonzero $\theta$ and $z$-components.
+where $$\vb{J}$$ and $$\vb{B}$$ both have nonzero $$\theta$$ and $$z$$-components.
By performing the above procedure again,
the following equilibrium condition is obtained:
diff --git a/source/know/concept/second-quantization/index.md b/source/know/concept/second-quantization/index.md
index 975921c..e446557 100644
--- a/source/know/concept/second-quantization/index.md
+++ b/source/know/concept/second-quantization/index.md
@@ -20,13 +20,13 @@ known as the **Fock states**, which are specific members of a **Fock space**,
a special kind of [Hilbert space](/know/concept/hilbert-space/),
with a well-defined number of particles.
-For a set of $N$ single-particle energy eigenstates
-$\psi_n(x)$ and $N$ identical particles $x_n$, the Fock states are
-all the wave functions which contain $n$ particles, for $n$ going from $0$ to $N$.
+For a set of $$N$$ single-particle energy eigenstates
+$$\psi_n(x)$$ and $$N$$ identical particles $$x_n$$, the Fock states are
+all the wave functions which contain $$n$$ particles, for $$n$$ going from $$0$$ to $$N$$.
-So for $n = 0$, there is one basis vector with $0$ particles,
-for $n = 1$, there are $N$ basis vectors with $1$ particle each,
-for $n = 2$, there are $N (N \!-\! 1)$ basis vectors with $2$ particles,
+So for $$n = 0$$, there is one basis vector with $$0$$ particles,
+for $$n = 1$$, there are $$N$$ basis vectors with $$1$$ particle each,
+for $$n = 2$$, there are $$N (N \!-\! 1)$$ basis vectors with $$2$$ particles,
etc.
In this basis, we define the **particle creation operators**
@@ -60,10 +60,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-The notation $\Ket{N_\alpha, N_\beta, ...}$ is shorthand for
+The notation $$\Ket{N_\alpha, N_\beta, ...}$$ is shorthand for
the appropriate [Slater determinants](/know/concept/slater-determinant/).
-As an example, take $\Ket{0, 1, 0, 1, 1}$,
-which contains three particles $a$, $b$ and $c$
+As an example, take $$\Ket{0, 1, 0, 1, 1}$$,
+which contains three particles $$a$$, $$b$$ and $$c$$
in states 2, 4 and 5:
$$\begin{aligned}
@@ -77,9 +77,9 @@ $$\begin{aligned}
\end{bmatrix}
\end{aligned}$$
-The creation operator $\hat{c}_\alpha^\dagger$ and annihilation
-operator $\hat{c}_\alpha$ are defined to live up to their name:
-they create or destroy a particle in the state $\psi_\alpha$:
+The creation operator $$\hat{c}_\alpha^\dagger$$ and annihilation
+operator $$\hat{c}_\alpha$$ are defined to live up to their name:
+they create or destroy a particle in the state $$\psi_\alpha$$:
$$\begin{aligned}
\boxed{
@@ -93,16 +93,16 @@ $$\begin{aligned}
}
\end{aligned}$$
-The factor $J_\alpha$ is sometimes known as the **Jordan-Wigner string**,
+The factor $$J_\alpha$$ is sometimes known as the **Jordan-Wigner string**,
and is necessary here to enforce the fermionic antisymmetry,
-when creating or destroying a particle in the $\alpha$th state:
+when creating or destroying a particle in the $$\alpha$$th state:
$$\begin{aligned}
J_\alpha = (-1)^{\sum_{j < \alpha} N_j}
\end{aligned}$$
So, for example, when creating a particle in state 4
-of $\Ket{0, 1, 1, 0, 1}$, we get the following:
+of $$\Ket{0, 1, 1, 0, 1}$$, we get the following:
$$\begin{aligned}
\hat{c}_4^\dagger \Ket{0, 1, 1, 0, 1}
@@ -122,8 +122,8 @@ $$\begin{aligned}
= - \Ket{1, 0}
\end{aligned}$$
-In other words, $\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$,
-meaning that the anticommutator $\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$.
+In other words, $$\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$$,
+meaning that the anticommutator $$\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$$.
You can verify for youself that
the general anticommutators of these operators are given by:
@@ -154,7 +154,7 @@ $$\begin{aligned}
= \matrixel{...(N_\alpha\!=\!0) ...}{\hat{c}_\alpha}{... (N_\alpha\!=\!1) ...}
\end{aligned}$$
-Let us now use these operators to define the **number operator** $\hat{N}_\alpha$ as follows:
+Let us now use these operators to define the **number operator** $$\hat{N}_\alpha$$ as follows:
$$\begin{aligned}
\boxed{
@@ -162,7 +162,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Its eigenvalue is the number of particles residing in state $\psi_\alpha$
+Its eigenvalue is the number of particles residing in state $$\psi_\alpha$$
(look at the hats):
$$\begin{aligned}
@@ -198,8 +198,8 @@ They must be symmetric under the exchange of two bosons.
To achieve this, the Fock states are represented by Slater *permanents*
rather than determinants.
-The boson creation and annihilation operators $\hat{c}_\alpha^\dagger$ and
-$\hat{c}_\alpha$ are straightforward:
+The boson creation and annihilation operators $$\hat{c}_\alpha^\dagger$$ and
+$$\hat{c}_\alpha$$ are straightforward:
$$\begin{gathered}
\boxed{
@@ -212,8 +212,8 @@ $$\begin{gathered}
\end{aligned}
}\end{gathered}$$
-Applying the annihilation operator $\hat{c}_\alpha$ when there are zero
-particles in $\alpha$ will quench the state:
+Applying the annihilation operator $$\hat{c}_\alpha$$ when there are zero
+particles in $$\alpha$$ will quench the state:
$$\begin{aligned}
\boxed{
@@ -232,8 +232,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-The constant factors applied by $\hat{c}_\alpha^\dagger$ and $\hat{c}_\alpha$
-ensure that $\hat{N}_\alpha$ keeps the same nice form:
+The constant factors applied by $$\hat{c}_\alpha^\dagger$$ and $$\hat{c}_\alpha$$
+ensure that $$\hat{N}_\alpha$$ keeps the same nice form:
$$\begin{aligned}
\boxed{
@@ -244,8 +244,8 @@ $$\begin{aligned}
## Operators
-Traditionally, an operator $\hat{V}$ simultaneously acting on $N$ indentical particles
-is the sum of the individual single-particle operators $\hat{V}_1$ acting on the $n$th particle:
+Traditionally, an operator $$\hat{V}$$ simultaneously acting on $$N$$ indentical particles
+is the sum of the individual single-particle operators $$\hat{V}_1$$ acting on the $$n$$th particle:
$$\begin{aligned}
\hat{V}
@@ -261,7 +261,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the matrix element $\matrixel{\alpha}{\hat{V}_1}{\beta}$ is to be
+Where the matrix element $$\matrixel{\alpha}{\hat{V}_1}{\beta}$$ is to be
evaluated in the normal way:
$$\begin{aligned}
@@ -269,7 +269,7 @@ $$\begin{aligned}
= \int \psi_\alpha^*(\vec{r}) \: \hat{V}_1(\vec{r}) \: \psi_\beta(\vec{r}) \dd{\vec{r}}
\end{aligned}$$
-Similarly, given some two-particle operator $\hat{V}$ in first-quantized form:
+Similarly, given some two-particle operator $$\hat{V}$$ in first-quantized form:
$$\begin{aligned}
\hat{V}
@@ -287,7 +287,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the constant $v_{\alpha \beta \gamma \delta}$ is defined from the
+Where the constant $$v_{\alpha \beta \gamma \delta}$$ is defined from the
single-particle wave functions:
$$\begin{aligned}
@@ -306,9 +306,9 @@ $$\begin{aligned}
= \sum_{\alpha} \Inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0}
\end{aligned}$$
-Where $\alpha$ and $b$ need not be in the same basis.
+Where $$\alpha$$ and $$b$$ need not be in the same basis.
With this, we can define the **field operators**,
-which create or destroy a particle at a given position $\vec{r}$:
+which create or destroy a particle at a given position $$\vec{r}$$:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/selection-rules/index.md b/source/know/concept/selection-rules/index.md
index 84227ba..373486e 100644
--- a/source/know/concept/selection-rules/index.md
+++ b/source/know/concept/selection-rules/index.md
@@ -10,33 +10,33 @@ layout: "concept"
In quantum mechanics, it is often necessary to evaluate
matrix elements of the following form,
-where $\ell$ and $m$ respectively represent
-the total angular momentum and its $z$-component:
+where $$\ell$$ and $$m$$ respectively represent
+the total angular momentum and its $$z$$-component:
$$\begin{aligned}
\matrixel{f}{\hat{O}}{i}
= \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
\end{aligned}$$
-Where $\hat{O}$ is an operator, $\Ket{i}$ is an initial state, and
-$\Ket{f}$ is a final state (usually at least; $\Ket{i}$ and $\Ket{f}$
+Where $$\hat{O}$$ is an operator, $$\Ket{i}$$ is an initial state, and
+$$\Ket{f}$$ is a final state (usually at least; $$\Ket{i}$$ and $$\Ket{f}$$
can be any states). **Selection rules** are requirements on the relations
-between $\ell_i$, $\ell_f$, $m_i$ and $m_f$, which, if not met,
+between $$\ell_i$$, $$\ell_f$$, $$m_i$$ and $$m_f$$, which, if not met,
guarantee that the above matrix element is zero.
## Parity rules
-Let $\hat{O}$ denote any operator which is odd under spatial inversion
+Let $$\hat{O}$$ denote any operator which is odd under spatial inversion
(parity):
$$\begin{aligned}
\hat{\Pi}^\dagger \hat{O} \hat{\Pi} = - \hat{O}
\end{aligned}$$
-Where $\hat{\Pi}$ is the parity operator.
-We wrap this property of $\hat{O}$
-in the states $\Ket{\ell_f m_f}$ and $\Ket{\ell_i m_i}$:
+Where $$\hat{\Pi}$$ is the parity operator.
+We wrap this property of $$\hat{O}$$
+in the states $$\Ket{\ell_f m_f}$$ and $$\Ket{\ell_i m_i}$$:
$$\begin{aligned}
\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
@@ -48,7 +48,7 @@ $$\begin{aligned}
\end{aligned}$$
Which clearly can only be true if the exponent is even,
-so $\Delta \ell \equiv \ell_f - \ell_i$ must be odd.
+so $$\Delta \ell \equiv \ell_f - \ell_i$$ must be odd.
This leads to the following selection rule,
often referred to as **Laporte's rule**:
@@ -60,9 +60,9 @@ $$\begin{aligned}
If this is not the case,
then the only possible way that the above equation can be satisfied
-is if the matrix element vanishes $\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$.
+is if the matrix element vanishes $$\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$$.
We can derive an analogous rule for
-any operator $\hat{E}$ which is even under parity:
+any operator $$\hat{E}$$ which is even under parity:
$$\begin{aligned}
\hat{\Pi}^\dagger \hat{E} \hat{\Pi} = \hat{E}
@@ -76,10 +76,10 @@ $$\begin{aligned}
## Dipole rules
Arguably the most common operator found in such matrix elements
-is a position vector operator, like $\vu{r}$ or $\hat{x}$,
+is a position vector operator, like $$\vu{r}$$ or $$\hat{x}$$,
and the associated selection rules are known as **dipole rules**.
-For the $z$-component of angular momentum $m$ we have the following:
+For the $$z$$-component of angular momentum $$m$$ we have the following:
$$\begin{aligned}
\boxed{
@@ -92,7 +92,7 @@ $$\begin{aligned}
@@ -224,27 +224,27 @@ after removing the two external lines from/to $a$ and $b$:
Despite its appearance, the self-energy has the semantics of a line,
so it has two endpoints over which to integrate if necessary.
-By construction, by reattaching $G^0(x,a)$ and $G^0(b,y)$ to the self-energy,
+By construction, by reattaching $$G^0(x,a)$$ and $$G^0(b,y)$$ to the self-energy,
we get all irreducible diagrams,
and by connecting multiple irreducible diagrams with single fermion lines,
-we get all fully connected diagrams containing the endpoints $a$ and $b$.
+we get all fully connected diagrams containing the endpoints $$a$$ and $$b$$.
-In other words, the full $G(b,a)$ is constructed
-by taking the unperturbed $G^0(b,a)$
-and inserting one or more irreducible diagrams between $a$ and $b$.
+In other words, the full $$G(b,a)$$ is constructed
+by taking the unperturbed $$G^0(b,a)$$
+and inserting one or more irreducible diagrams between $$a$$ and $$b$$.
We can equally well insert a single irreducible diagram
as a sequence of connected irreducible diagrams.
Thanks to this recursive structure,
-you can convince youself that $G(b,a)$ obeys
-a [Dyson equation](/know/concept/dyson-equation/) involving $\Sigma(y, x)$:
+you can convince youself that $$G(b,a)$$ obeys
+a [Dyson equation](/know/concept/dyson-equation/) involving $$\Sigma(y, x)$$:
This makes sense: in the "normal" Dyson equation
-we have a one-body perturbation instead of $\Sigma$,
-while $\Sigma$ represents a two-body effect
+we have a one-body perturbation instead of $$\Sigma$$,
+while $$\Sigma$$ represents a two-body effect
as an infinite sum of one-body diagrams.
Interpreting this diagrammatic Dyson equation yields:
@@ -255,7 +255,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Keep in mind that $\int \dd{x} \equiv \sum_{s_x} \int \dd{\vb{r}_x} \int \dd{\tau_x}$.
+Keep in mind that $$\int \dd{x} \equiv \sum_{s_x} \int \dd{\vb{r}_x} \int \dd{\tau_x}$$.
In the special case of a system with continuous translational symmetry
and no spin dependence, this simplifies to:
@@ -266,12 +266,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n)$,
-with $\omega_n$ being a fermionic Matsubara frequency.
-Note that conservation of spin, $\vb{k}$ and $\omega_n$,
+Where $$\tilde{\vb{k}} \equiv (\vb{k}, i \omega_n)$$,
+with $$\omega_n$$ being a fermionic Matsubara frequency.
+Note that conservation of spin, $$\vb{k}$$ and $$\omega_n$$,
together with the linear structure of the Dyson equation,
-makes $\Sigma$ diagonal in all of those quantities.
-Isolating for $G$:
+makes $$\Sigma$$ diagonal in all of those quantities.
+Isolating for $$G$$:
$$\begin{aligned}
G_{s}(\tilde{\vb{k}})
@@ -280,7 +280,7 @@ $$\begin{aligned}
\end{aligned}$$
From [equation-of-motion theory](/know/concept/equation-of-motion-theory/),
-we already know an expression for $G$ in diagonal $\vb{k}$-space:
+we already know an expression for $$G$$ in diagonal $$\vb{k}$$-space:
$$\begin{aligned}
G_s^0(\vb{k}, i \omega_n)
@@ -294,10 +294,10 @@ The self-energy thus corrects the non-interacting energies for interactions.
It can therefore be regarded as the energy
a particle has due to changes it has caused in its environment.
-Unfortunately, in practice, $\Sigma$ is rarely as simple as
+Unfortunately, in practice, $$\Sigma$$ is rarely as simple as
in the translationally-invariant example above;
in fact, it does not even need to be Hermitian,
-i.e. $\Sigma(y,x) \neq \Sigma^*(x,y)$,
+i.e. $$\Sigma(y,x) \neq \Sigma^*(x,y)$$,
in which case it resists the standard techniques for analysis.
diff --git a/source/know/concept/self-phase-modulation/index.md b/source/know/concept/self-phase-modulation/index.md
index 3b46c4e..f13ad2f 100644
--- a/source/know/concept/self-phase-modulation/index.md
+++ b/source/know/concept/self-phase-modulation/index.md
@@ -12,13 +12,13 @@ layout: "concept"
In fiber optics, **self-phase modulation** (SPM) is a nonlinear effect
that gradually broadens pulses' spectra.
-Unlike dispersion, SPM does create new frequencies: in the $\omega$-domain,
+Unlike dispersion, SPM does create new frequencies: in the $$\omega$$-domain,
the pulse steadily spreads out with a distinctive "accordion" peak.
Lower frequencies are created at the front of the
pulse and higher ones at the back, giving S-shaped spectrograms.
-A pulse envelope $A(z, t)$ inside a fiber must obey the nonlinear Schrödinger equation,
-where the parameters $\beta_2$ and $\gamma$ respectively
+A pulse envelope $$A(z, t)$$ inside a fiber must obey the nonlinear Schrödinger equation,
+where the parameters $$\beta_2$$ and $$\gamma$$ respectively
control dispersion and nonlinearity:
$$\begin{aligned}
@@ -26,20 +26,20 @@ $$\begin{aligned}
= i \pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma |A|^2 A
\end{aligned}$$
-By setting $\beta_2 = 0$ to neglect dispersion,
+By setting $$\beta_2 = 0$$ to neglect dispersion,
solving this equation becomes trivial.
-For any arbitrary input pulse $A_0(t) = A(0, t)$,
+For any arbitrary input pulse $$A_0(t) = A(0, t)$$,
we arrive at the following analytical solution:
$$\begin{aligned}
A(z,t) = A_0 \exp\!\big( i \gamma |A_0|^2 z\big)
\end{aligned}$$
-The intensity $|A|^2$ in the time domain is thus unchanged,
+The intensity $$|A|^2$$ in the time domain is thus unchanged,
and only its phase is modified.
It is also clear that the largest phase increase occurs at the peak of the pulse,
-where the intensity is $P_0$.
-To quantify this, it is useful to define the **nonlinear length** $L_N$,
+where the intensity is $$P_0$$.
+To quantify this, it is useful to define the **nonlinear length** $$L_N$$,
which gives the distance after which the phase of the
peak has increased by exactly 1 radian:
@@ -52,15 +52,15 @@ $$\begin{aligned}
\end{aligned}$$
SPM is illustrated below for the following Gaussian initial pulse envelope,
-with parameter values $T_0 = 6\:\mathrm{ps}$, $P_0 = 1\:\mathrm{kW}$,
-$\beta_2 = 0$, and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$:
+with parameter values $$T_0 = 6\:\mathrm{ps}$$, $$P_0 = 1\:\mathrm{kW}$$,
+$$\beta_2 = 0$$, and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$:
$$\begin{aligned}
A(0, t)
= \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big)
\end{aligned}$$
-From earlier, we then know the analytical solution for the $z$-evolution:
+From earlier, we then know the analytical solution for the $$z$$-evolution:
$$\begin{aligned}
A(z, t) = \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big) \exp\!\bigg( i \gamma z P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
@@ -70,10 +70,10 @@ $$\begin{aligned}
-The **instantaneous frequency** $\omega_\mathrm{SPM}(z, t)$,
+The **instantaneous frequency** $$\omega_\mathrm{SPM}(z, t)$$,
which describes the dominant angular frequency at a given point in the time domain,
is found to be as follows for the Gaussian pulse,
-where $\phi(z, t)$ is the phase of $A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$:
+where $$\phi(z, t)$$ is the phase of $$A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$$:
$$\begin{aligned}
\omega_{\mathrm{SPM}}(z,t)
@@ -82,8 +82,8 @@ $$\begin{aligned}
\end{aligned}$$
This result gives the S-shaped spectrograms seen in the illustration.
-The frequency shift thus not only depends on $L_N$,
-but also on $T_0$: the spectra of narrow pulses broaden much faster.
+The frequency shift thus not only depends on $$L_N$$,
+but also on $$T_0$$: the spectra of narrow pulses broaden much faster.
The interaction between self-phase modulation
and [dispersion](/know/concept/dispersive-broadening/)
diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
index 4cc6201..a8bc3c2 100644
--- a/source/know/concept/self-steepening/index.md
+++ b/source/know/concept/self-steepening/index.md
@@ -22,17 +22,17 @@ $$\begin{aligned}
= i\pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{}{t} \Big) \big(|A|^2 A\big)
\end{aligned}$$
-Where $\omega_0$ is the angular frequency of the pump.
+Where $$\omega_0$$ is the angular frequency of the pump.
We will use the following ansatz,
-consisting of an arbitrary power profile $P$ with a phase $\phi$:
+consisting of an arbitrary power profile $$P$$ with a phase $$\phi$$:
$$\begin{aligned}
A(z,t) = \sqrt{P(z,t)} \, \exp\!\big(i \phi(z,t)\big)
\end{aligned}$$
For a long pulse travelling over a short distance, it is reasonable to
-neglect dispersion ($\beta_2 = 0$).
-Inserting the ansatz then gives the following, where $\varepsilon = \gamma / \omega_0$:
+neglect dispersion ($$\beta_2 = 0$$).
+Inserting the ansatz then gives the following, where $$\varepsilon = \gamma / \omega_0$$:
$$\begin{aligned}
0 &= i \frac{1}{2} \frac{P_z}{\sqrt{P}} - \sqrt{P} \phi_z + \gamma P \sqrt{P} + i \varepsilon \frac{3}{2} P_t \sqrt{P} - \varepsilon P \sqrt{P} \phi_t
@@ -47,7 +47,7 @@ $$\begin{aligned}
0 &= P_z + \varepsilon 3 P_t P
\end{aligned}$$
-The phase $\phi$ is not so interesting, so we focus on the latter equation for $P$.
+The phase $$\phi$$ is not so interesting, so we focus on the latter equation for $$P$$.
As it turns out, it has a general solution of the form below, which shows that
more intense parts of the pulse will tend to lag behind compared to the rest:
@@ -55,8 +55,8 @@ $$\begin{aligned}
P(z,t) = f(t - 3 \varepsilon z P)
\end{aligned}$$
-Where $f$ is the initial power profile: $f(t) = P(0,t)$.
-The derivatives $P_t$ and $P_z$ are then given by:
+Where $$f$$ is the initial power profile: $$f(t) = P(0,t)$$.
+The derivatives $$P_t$$ and $$P_z$$ are then given by:
$$\begin{aligned}
P_t
@@ -73,10 +73,10 @@ $$\begin{aligned}
\end{aligned}$$
These derivatives both go to infinity when their denominator is zero,
-which, since $\varepsilon$ is positive, will happen earliest where $f'$
-has its most negative value, called $f_\mathrm{min}'$,
+which, since $$\varepsilon$$ is positive, will happen earliest where $$f'$$
+has its most negative value, called $$f_\mathrm{min}'$$,
which is located on the trailing edge of the pulse.
-At the propagation distance where this occurs, $L_\mathrm{shock}$,
+At the propagation distance where this occurs, $$L_\mathrm{shock}$$,
the pulse will "tip over", creating a discontinuous shock:
$$\begin{aligned}
@@ -86,21 +86,21 @@ $$\begin{aligned}
\end{aligned}$$
In practice, however, this will never actually happen, because by the time
-$L_\mathrm{shock}$ is reached, the pulse spectrum will have become so
+$$L_\mathrm{shock}$$ is reached, the pulse spectrum will have become so
broad that dispersion can no longer be neglected.
A simulation of self-steepening without dispersion is illustrated below
for the following Gaussian initial power distribution,
-with $T_0 = 25\:\mathrm{fs}$, $P_0 = 3\:\mathrm{kW}$,
-$\beta_2 = 0$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$:
+with $$T_0 = 25\:\mathrm{fs}$$, $$P_0 = 3\:\mathrm{kW}$$,
+$$\beta_2 = 0$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$:
$$\begin{aligned}
f(t) = P(0,t) = P_0 \exp\!\Big(\! -\!\frac{t^2}{T_0^2} \Big)
\end{aligned}$$
-Its steepest points are found to be at $2 t^2 = T_0^2$, so
-$f_\mathrm{min}'$ and $L_\mathrm{shock}$ are given by:
+Its steepest points are found to be at $$2 t^2 = T_0^2$$, so
+$$f_\mathrm{min}'$$ and $$L_\mathrm{shock}$$ are given by:
$$\begin{aligned}
f_\mathrm{min}' = - \frac{\sqrt{2} P_0}{T_0} \exp\!\Big(\!-\!\frac{1}{2}\Big)
@@ -109,7 +109,7 @@ $$\begin{aligned}
\end{aligned}$$
This example Gaussian pulse therefore has a theoretical
-$L_\mathrm{shock} = 0.847\,\mathrm{m}$,
+$$L_\mathrm{shock} = 0.847\,\mathrm{m}$$,
which turns out to be accurate,
although the simulation breaks down due to insufficient resolution:
@@ -118,7 +118,7 @@ although the simulation breaks down due to insufficient resolution:
Unfortunately, self-steepening cannot be simulated perfectly: as the
-pulse approaches $L_\mathrm{shock}$, its spectrum broadens to infinite
+pulse approaches $$L_\mathrm{shock}$$, its spectrum broadens to infinite
frequencies to represent the singularity in its slope.
The simulation thus collapses into chaos when the edge of the frequency window is reached.
Nevertheless, the general trends are nicely visible:
@@ -126,9 +126,9 @@ the trailing slope becomes extremely steep, and the spectrum
broadens so much that dispersion cannot be neglected anymore.
When self-steepening is added to the nonlinear Schrödinger equation,
-it no longer conserves the total pulse energy $\int |A|^2 \dd{t}$.
-Fortunately, the photon number $N_\mathrm{ph}$ is still
-conserved, which for the physical envelope $A(z,t)$ is defined as:
+it no longer conserves the total pulse energy $$\int |A|^2 \dd{t}$$.
+Fortunately, the photon number $$N_\mathrm{ph}$$ is still
+conserved, which for the physical envelope $$A(z,t)$$ is defined as:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/shors-algorithm/index.md b/source/know/concept/shors-algorithm/index.md
index 0ff35c5..0241f9f 100644
--- a/source/know/concept/shors-algorithm/index.md
+++ b/source/know/concept/shors-algorithm/index.md
@@ -17,8 +17,8 @@ It weakens widely-used cryptographic schemes,
such as RSA and [Diffie-Hellman](/know/concept/diffie-hellman-key-exchange/).
In essence, Shor's algorithm's revolutionary achievement
-is that it can efficiently find the periods $s_1, ..., s_A$
-of a function $f(x_1, ..., x_A)$ on a discrete finite field, where:
+is that it can efficiently find the periods $$s_1, ..., s_A$$
+of a function $$f(x_1, ..., x_A)$$ on a discrete finite field, where:
$$\begin{aligned}
f(x_1, ..., x_A)
@@ -32,27 +32,27 @@ Shor's algorithm can solve practically every such problem.
## Integer factorization
-Originally, Shor's algorithm was designed to factorize an integer $N$,
-in which case the goal is to find the period $s$ of
-the modular exponentiation function $f$ (for reasons explained later):
+Originally, Shor's algorithm was designed to factorize an integer $$N$$,
+in which case the goal is to find the period $$s$$ of
+the modular exponentiation function $$f$$ (for reasons explained later):
$$\begin{aligned}
f(x)
= a^x \bmod N
\end{aligned}$$
-For a given $a$ and $N$.
-The period $s$ is the smallest integer satisfying $f(x) = f(x+s)$.
-To do this, the following $2q$-qubit quantum circuit is used,
-with $q$ chosen so that $N^2 \le 2^q < 2 N^2$:
+For a given $$a$$ and $$N$$.
+The period $$s$$ is the smallest integer satisfying $$f(x) = f(x+s)$$.
+To do this, the following $$2q$$-qubit quantum circuit is used,
+with $$q$$ chosen so that $$N^2 \le 2^q < 2 N^2$$:
-Here, $\mathrm{QFT}_q$ refers to the $q$-qubit
+Here, $$\mathrm{QFT}_q$$ refers to the $$q$$-qubit
[quantum Fourier transform](/know/concept/quantum-fourier-transform/),
-and the oracle $U_f$ calculates $f(x)$ for predetermined values of $a$ and $N$.
+and the oracle $$U_f$$ calculates $$f(x)$$ for predetermined values of $$a$$ and $$N$$.
It is an XOR oracle, working as follows:
$$\begin{aligned}
@@ -61,8 +61,8 @@ $$\begin{aligned}
\Ket{x} \Ket{y \oplus f(x)}
\end{aligned}$$
-Execution starts by applying the [Hadamard gate](/know/concept/quantum-gate/) $H$
-to the first $q$ qubits, yielding:
+Execution starts by applying the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
+to the first $$q$$ qubits, yielding:
$$\begin{aligned}
\Ket{0}^{\otimes q} \Ket{0}^{\otimes q}
@@ -71,8 +71,8 @@ $$\begin{aligned}
= \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
\end{aligned}$$
-Where $Q = 2^q$, and $\Ket{x}$ is the computational basis state $\Ket{x_1} \cdots \Ket{x_q}$.
-Moving on to $U_f$:
+Where $$Q = 2^q$$, and $$\Ket{x}$$ is the computational basis state $$\Ket{x_1} \cdots \Ket{x_q}$$.
+Moving on to $$U_f$$:
$$\begin{aligned}
\frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
@@ -80,15 +80,15 @@ $$\begin{aligned}
\frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{f(x)}
\end{aligned}$$
-Then we measure $f(x)$, causing it collapse as follows,
-for an unknown arbitrary value of $x_0$:
+Then we measure $$f(x)$$, causing it collapse as follows,
+for an unknown arbitrary value of $$x_0$$:
$$\begin{aligned}
f(x_0) = f(x_0 + s) = f(x_0 + 2s) = \cdots = f(x_0 + (L-1) s)
\end{aligned}$$
Due to [entanglement](/know/concept/quantum-entanglement/),
-the unmeasured (top $q$) qubits change state into a superposition:
+the unmeasured (top $$q$$) qubits change state into a superposition:
$$\begin{aligned}
\frac{1}{\sqrt{L}} \sum_{\ell = 0}^{L - 1} \Ket{x_0 + \ell s}
@@ -96,7 +96,7 @@ $$\begin{aligned}
Clearly, there is a periodic structure here,
but we cannot measure it directly,
-because we do not know the value of $x_0$,
+because we do not know the value of $$x_0$$,
which, to make matters worse, changes every time we run the algorithm.
This is where the QFT comes in, which outputs the following state:
@@ -104,8 +104,8 @@ $$\begin{aligned}
\frac{1}{\sqrt{QL}} \sum_{k = 0}^{Q - 1} \bigg( \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg) \Ket{k}
\end{aligned}$$
-Where $\omega_Q$ is a $Q$th root of unity.
-Measuring this state yields a $\Ket{k}$, with a probability $P(k)$:
+Where $$\omega_Q$$ is a $$Q$$th root of unity.
+Measuring this state yields a $$\Ket{k}$$, with a probability $$P(k)$$:
$$\begin{aligned}
P(k)
@@ -114,13 +114,13 @@ $$\begin{aligned}
= \frac{1}{QL} \bigg| \sum_{\ell = 0}^{L - 1} \omega_Q^{\ell s k} \bigg|^2
\end{aligned}$$
-The last step holds because $|\omega_Q| = 1$.
+The last step holds because $$|\omega_Q| = 1$$.
Surprisingly, this implies that we did not need
-to perform the measurement of $f(x)$ earlier!
-This makes sense: the period $s$ does not depend on $x_0$,
-so why would we need an implicit $x_0$ to determine $s$?
+to perform the measurement of $$f(x)$$ earlier!
+This makes sense: the period $$s$$ does not depend on $$x_0$$,
+so why would we need an implicit $$x_0$$ to determine $$s$$?
-So, what does the above probability $P(k)$ work out to?
+So, what does the above probability $$P(k)$$ work out to?
There are two cases:
$$\begin{alignedat}{2}
@@ -132,22 +132,22 @@ $$\begin{alignedat}{2}
\end{alignedat}$$
Where the latter case was evaluated as a geometric series.
-The condition $\omega_Q^{sk}\!=\!1$ is equivalent to asking
-if $sk$ is a multiple of $Q$, i.e. if $sk = cQ$, for an integer $c$.
+The condition $$\omega_Q^{sk}\!=\!1$$ is equivalent to asking
+if $$sk$$ is a multiple of $$Q$$, i.e. if $$sk = cQ$$, for an integer $$c$$.
-Recall that $L$ is the number of times that $s$ fits in $Q$,
-so $L\!=\!\lfloor Q / s \rfloor$.
-Assuming $Q/s$ is an integer, then $L\!=\!Q/s$ and $Q\!=\!s L$,
+Recall that $$L$$ is the number of times that $$s$$ fits in $$Q$$,
+so $$L\!=\!\lfloor Q / s \rfloor$$.
+Assuming $$Q/s$$ is an integer, then $$L\!=\!Q/s$$ and $$Q\!=\!s L$$,
which tells us that
-$\omega_Q^{sk}\!=\!\omega_{s L}^{s k}\!=\!\omega_L^k$.
-This implies that if $k$ is a multiple of $L$ (i.e. $k\!=\!c L$),
-then $\omega_L^k\!=\!1$, so $P(k) = L / Q$,
+$$\omega_Q^{sk}\!=\!\omega_{s L}^{s k}\!=\!\omega_L^k$$.
+This implies that if $$k$$ is a multiple of $$L$$ (i.e. $$k\!=\!c L$$),
+then $$\omega_L^k\!=\!1$$, so $$P(k) = L / Q$$,
which is exactly what we got earlier!
-In other words, the condition $\omega_Q^{sk}\!=\!1$
-is equivalent to $Q/s$ being an integer.
-In that case, we have that $Q\!=\!sL$,
-which we substitute into $P(k)$ from earlier:
+In other words, the condition $$\omega_Q^{sk}\!=\!1$$
+is equivalent to $$Q/s$$ being an integer.
+In that case, we have that $$Q\!=\!sL$$,
+which we substitute into $$P(k)$$ from earlier:
$$\begin{aligned}
\mathrm{if} \: (Q/s) \in \mathbb{N}: \qquad
@@ -156,25 +156,25 @@ $$\begin{aligned}
= \frac{1}{s}
\end{aligned}$$
-And because $k$ is a multiple of $L$,
-and $L$ fits $s$ times in $Q$,
-there must be exactly $s$ values of $k$ that satisfy $P(k) = 1/s$.
-Therefore the probability of all other $k$-values is zero!
-This becomes clearer when you look at the sum used to calculate $P(k)$:
-if $Q\!=\!sL$, then it sums $\omega_L^{\ell k}$ over $\ell$,
-leading to perfect destructive interference for the "bad" $k$-values,
+And because $$k$$ is a multiple of $$L$$,
+and $$L$$ fits $$s$$ times in $$Q$$,
+there must be exactly $$s$$ values of $$k$$ that satisfy $$P(k) = 1/s$$.
+Therefore the probability of all other $$k$$-values is zero!
+This becomes clearer when you look at the sum used to calculate $$P(k)$$:
+if $$Q\!=\!sL$$, then it sums $$\omega_L^{\ell k}$$ over $$\ell$$,
+leading to perfect destructive interference for the "bad" $$k$$-values,
leaving only the "good" ones.
-**So, to summarize: if** $Q/s$ **is an integer**,
-then measuring only yields $k$-values that are multiples of $L\!=\!Q/s$.
+**So, to summarize: if** $$Q/s$$ **is an integer**,
+then measuring only yields $$k$$-values that are multiples of $$L\!=\!Q/s$$.
Running Shor's algorithm several times then gives
-several $k$-values separated by $L$.
-That tells us what $L$ is, and we already know $Q$,
-so we *finally* find the period $s = Q/L$.
+several $$k$$-values separated by $$L$$.
+That tells us what $$L$$ is, and we already know $$Q$$,
+so we *finally* find the period $$s = Q/L$$.
-That begs the question: what if $Q/s$ is not an integer?
-We cannot *check* this, since $s$ is unknown!
-Instead, we rewrite the probability $P(k)$ as follows:
+That begs the question: what if $$Q/s$$ is not an integer?
+We cannot *check* this, since $$s$$ is unknown!
+Instead, we rewrite the probability $$P(k)$$ as follows:
$$\begin{aligned}
\mathrm{if} \: (Q/s) \not\in \mathbb{N}: \qquad
@@ -183,7 +183,7 @@ $$\begin{aligned}
= \frac{1}{QL} \Bigg| \frac{\sin(\pi s k L / Q)}{\sin(\pi s k / Q)} \Bigg|^2
\end{aligned}$$
-This function peaks if $s k$ is close to a multiple of $Q$, i.e. $s k \approx c Q$,
+This function peaks if $$s k$$ is close to a multiple of $$Q$$, i.e. $$s k \approx c Q$$,
which we rearrange:
$$\begin{aligned}
@@ -191,62 +191,62 @@ $$\begin{aligned}
\end{aligned}$$
We know the left-hand side,
-and, from the definition of $f(x)$,
-clearly $s \le N$.
-We chose $Q \sim N^2$,
-so $s$ is quite small,
-and consequently $c$ is too, since $k < Q$.
+and, from the definition of $$f(x)$$,
+clearly $$s \le N$$.
+We chose $$Q \sim N^2$$,
+so $$s$$ is quite small,
+and consequently $$c$$ is too, since $$k < Q$$.
-In other words, $c/s$ is a "simple" fraction,
+In other words, $$c/s$$ is a "simple" fraction,
so our goal is to find a "simple" fraction
-that is close to the "complicated" fraction $k/Q$.
-For example, if $k/Q\!=\!0.332$,
-then probably $c/s\!=\!1/3$.
+that is close to the "complicated" fraction $$k/Q$$.
+For example, if $$k/Q\!=\!0.332$$,
+then probably $$c/s\!=\!1/3$$.
This can be done rigorously using the **continued fractions algorithm**:
-write $k/Q$ as a continued fraction,
+write $$k/Q$$ as a continued fraction,
until the non-integer part of the denominator becomes small enough.
This part is then neglected,
-and we calculate whatever is left, to get an estimate of $c/s$.
+and we calculate whatever is left, to get an estimate of $$c/s$$.
-Of course, $P(k)$ is a probability distribution,
+Of course, $$P(k)$$ is a probability distribution,
so even though the odds are in our favour,
-we might occasionally measure a misleading $k$-value.
+we might occasionally measure a misleading $$k$$-value.
Running Shor's algorithm several times "fixes" this.
-**So, to summarize: if** $Q/s$ **is not an integer**,
-the measured $k$-values are generally close to $c Q / s$ for an integer $c$.
-By approximating $k/Q$ using the continued fraction algorithm,
-we estimate $c/s$.
-Repeating this procedure gives several values of $c/s$,
-such that $s$ is easy to deduce
+**So, to summarize: if** $$Q/s$$ **is not an integer**,
+the measured $$k$$-values are generally close to $$c Q / s$$ for an integer $$c$$.
+By approximating $$k/Q$$ using the continued fraction algorithm,
+we estimate $$c/s$$.
+Repeating this procedure gives several values of $$c/s$$,
+such that $$s$$ is easy to deduce
by taking the least common multiple of the denominators.
-In any case, once we think we have $s$,
-we can easily verify that $f(x)\!=\!f(x\!+\!s)$.
-Whether $s$ is the *smallest* such integer depends on how lucky we are,
+In any case, once we think we have $$s$$,
+we can easily verify that $$f(x)\!=\!f(x\!+\!s)$$.
+Whether $$s$$ is the *smallest* such integer depends on how lucky we are,
but fortunately, for most applications of this algorithm,
that does not actually matter,
-and usually we find the smallest $s$ anyway.
+and usually we find the smallest $$s$$ anyway.
-You typically need to repeat the algorithm $\mathcal{O}(\log{q})$ times,
-and the QFT is $\mathcal{O}(q^2)$.
-The bottleneck is modular exponentiation $f$,
-which is $\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$
+You typically need to repeat the algorithm $$\mathcal{O}(\log{q})$$ times,
+and the QFT is $$\mathcal{O}(q^2)$$.
+The bottleneck is modular exponentiation $$f$$,
+which is $$\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$$
and therefore worse than the QFT,
-yielding a total complexity of $\mathcal{O}(q^2 (\log{q})^2 \log{\log{q}})$.
+yielding a total complexity of $$\mathcal{O}(q^2 (\log{q})^2 \log{\log{q}})$$.
-OK, but what does $s$ have to do with factorizing integers?
-Well, recall that $f$ is given by:
+OK, but what does $$s$$ have to do with factorizing integers?
+Well, recall that $$f$$ is given by:
$$\begin{aligned}
f(x)
= a^x \bmod N
\end{aligned}$$
-$N$ is the number to factorize, and $a$ is a random integer *coprime* to $N$,
-meaning $\gcd(a, N) = 1$.
-The fact that $s$ is the period of $f$ for a certain $a$-value, implies that:
+$$N$$ is the number to factorize, and $$a$$ is a random integer *coprime* to $$N$$,
+meaning $$\gcd(a, N) = 1$$.
+The fact that $$s$$ is the period of $$f$$ for a certain $$a$$-value, implies that:
$$\begin{aligned}
a^x
@@ -256,7 +256,7 @@ $$\begin{aligned}
= a^s \bmod N
\end{aligned}$$
-Suppose that $s$ is even. In that case,
+Suppose that $$s$$ is even. In that case,
we can rewrite the above equation as follows:
$$\begin{aligned}
@@ -264,16 +264,16 @@ $$\begin{aligned}
= 0 \bmod N
\end{aligned}$$
-In other words, $(a^{s/2})^2 \!-\! 1$ is a multiple of $N$.
-We then use that $(a\!-\!b) (a\!+\!b) = a^2\!-\!b^2$:
+In other words, $$(a^{s/2})^2 \!-\! 1$$ is a multiple of $$N$$.
+We then use that $$(a\!-\!b) (a\!+\!b) = a^2\!-\!b^2$$:
$$\begin{aligned}
\big( a^{s/2} - 1 \big) \big( a^{s/2} + 1 \big)
= 0 \bmod N
\end{aligned}$$
-Because $s$ is even by assumption, the two factors on the left are integers,
-and as just mentioned, their product is a multiple of $N$.
+Because $$s$$ is even by assumption, the two factors on the left are integers,
+and as just mentioned, their product is a multiple of $$N$$.
Then we only need to calculate:
$$\begin{aligned}
@@ -282,13 +282,13 @@ $$\begin{aligned}
\gcd\!\big( a^{s/2}\!+\!1, N \big) > 1
\end{aligned}$$
-And there we have the factors of $N$!
-The $\gcd$ can be calculated efficiently in $\mathcal{O}(q^2)$ time.
+And there we have the factors of $$N$$!
+The $$\gcd$$ can be calculated efficiently in $$\mathcal{O}(q^2)$$ time.
-But what if $s$ is odd?
-No problem, then we just choose a new $a$ coprime to $N$,
-and keep repeating Shor's algorithm until we do find an even $s$.
-We do the same if $a^{s/2}\!\pm\!1$ is itself a multiple of $N$.
+But what if $$s$$ is odd?
+No problem, then we just choose a new $$a$$ coprime to $$N$$,
+and keep repeating Shor's algorithm until we do find an even $$s$$.
+We do the same if $$a^{s/2}\!\pm\!1$$ is itself a multiple of $$N$$.
diff --git a/source/know/concept/sigma-algebra/index.md b/source/know/concept/sigma-algebra/index.md
index 5ba7b14..ac607a7 100644
--- a/source/know/concept/sigma-algebra/index.md
+++ b/source/know/concept/sigma-algebra/index.md
@@ -8,15 +8,15 @@ categories:
layout: "concept"
---
-In set theory, given a set $\Omega$, a $\sigma$**-algebra**
-is a family $\mathcal{F}$ of subsets of $\Omega$
+In set theory, given a set $$\Omega$$, a $$\sigma$$**-algebra**
+is a family $$\mathcal{F}$$ of subsets of $$\Omega$$
with these properties:
-1. The full set is included $\Omega \in \mathcal{F}$.
-2. For all subsets $A$, if $A \in \mathcal{F}$,
- then its complement $\Omega \!-\! A \in \mathcal{F}$ too.
-3. If two events $A, B \in \mathcal{F}$,
- then their union $A \cup B \in \mathcal{F}$ too.
+1. The full set is included $$\Omega \in \mathcal{F}$$.
+2. For all subsets $$A$$, if $$A \in \mathcal{F}$$,
+ then its complement $$\Omega \!-\! A \in \mathcal{F}$$ too.
+3. If two events $$A, B \in \mathcal{F}$$,
+ then their union $$A \cup B \in \mathcal{F}$$ too.
This forms a Boolean algebra:
property (1) represents TRUE,
@@ -24,28 +24,28 @@ property (1) represents TRUE,
and that is all we need to define all logic.
For example, FALSE and OR follow from the above points:
-4. The empty set is included $\varnothing \in \mathcal{F}$.
-5. If two events $A, B \in \mathcal{F}$,
- then their intersection $A \cap B \in \mathcal{F}$ too.
+4. The empty set is included $$\varnothing \in \mathcal{F}$$.
+5. If two events $$A, B \in \mathcal{F}$$,
+ then their intersection $$A \cap B \in \mathcal{F}$$ too.
-For a given $\Omega$, there are typically multiple valid $\mathcal{F}$,
+For a given $$\Omega$$, there are typically multiple valid $$\mathcal{F}$$,
in which case you need to specify your choice.
-Usually this would be the smallest $\mathcal{F}$
+Usually this would be the smallest $$\mathcal{F}$$
(i.e. smallest family of subsets)
that contains all subsets of special interest
for the topic at hand.
-Likewise, a **sub-$\sigma$-algebra**
-is a sub-family of a certain $\mathcal{F}$,
-which is a valid $\sigma$-algebra in its own right.
-
-A notable $\sigma$-algebra is the **Borel algebra** $\mathcal{B}(\Omega)$,
-which is defined when $\Omega$ is a metric space,
-such as the real numbers $\mathbb{R}$.
-Using that as an example, the Borel algebra $\mathcal{B}(\mathbb{R})$
+Likewise, a **sub-$$\sigma$$-algebra**
+is a sub-family of a certain $$\mathcal{F}$$,
+which is a valid $$\sigma$$-algebra in its own right.
+
+A notable $$\sigma$$-algebra is the **Borel algebra** $$\mathcal{B}(\Omega)$$,
+which is defined when $$\Omega$$ is a metric space,
+such as the real numbers $$\mathbb{R}$$.
+Using that as an example, the Borel algebra $$\mathcal{B}(\mathbb{R})$$
is defined as the family of all open intervals of the real line,
-and all the subsets of $\mathbb{R}$ obtained by countable sequences
+and all the subsets of $$\mathbb{R}$$ obtained by countable sequences
of unions and intersections of those intervals.
-The elements of $\mathcal{B}$ are **Borel sets**.
+The elements of $$\mathcal{B}$$ are **Borel sets**.
diff --git a/source/know/concept/simons-algorithm/index.md b/source/know/concept/simons-algorithm/index.md
index fef10b9..5502837 100644
--- a/source/know/concept/simons-algorithm/index.md
+++ b/source/know/concept/simons-algorithm/index.md
@@ -19,10 +19,10 @@ is of no practical use,
but nevertheless Simon's algorithm is an important landmark.
Simon's problem is this:
-we are given a "black box" function $f(x)$
-that takes an $n$-bit input $x$
-and returns an $n$-bit output.
-We are promised that there exists an $s$ such that for all $x_1$ and $x_2$:
+we are given a "black box" function $$f(x)$$
+that takes an $$n$$-bit input $$x$$
+and returns an $$n$$-bit output.
+We are promised that there exists an $$s$$ such that for all $$x_1$$ and $$x_2$$:
$$\begin{aligned}
f(x_1)
@@ -31,24 +31,24 @@ $$\begin{aligned}
x_2 = s \oplus x_1
\end{aligned}$$
-In other words, regardless of what $f(x)$ does behind the scenes,
-its output is the same for inputs $x_1$ and $x_2$
-if and only if $x_2 = s \oplus x_1$,
-or, equivalently, $x_1 = s \oplus x_2$.
+In other words, regardless of what $$f(x)$$ does behind the scenes,
+its output is the same for inputs $$x_1$$ and $$x_2$$
+if and only if $$x_2 = s \oplus x_1$$,
+or, equivalently, $$x_1 = s \oplus x_2$$.
-The goal is to find the $n$-bit number $s$, using as few calls to $f$ as possible.
+The goal is to find the $$n$$-bit number $$s$$, using as few calls to $$f$$ as possible.
There are two cases:
-if $s = 0$, then $f$ is one-to-one, since $x_2 = 0 \oplus x_1 = x_1$.
-Otherwise, if $s \neq 0$, then $f$ is two-to-one by definition:
-for every $x_1$ there exists exactly one $x_2$ such that $x_2 = s \oplus x_1$.
+if $$s = 0$$, then $$f$$ is one-to-one, since $$x_2 = 0 \oplus x_1 = x_1$$.
+Otherwise, if $$s \neq 0$$, then $$f$$ is two-to-one by definition:
+for every $$x_1$$ there exists exactly one $$x_2$$ such that $$x_2 = s \oplus x_1$$.
A classical computer solves this by randomly guessing inputs,
until it finds two that give the same output,
-and then $s = x_1 \oplus x_2$.
-For $n$-bit numbers, this takes $\mathcal{O}(\sqrt{2^n})$ guesses
+and then $$s = x_1 \oplus x_2$$.
+For $$n$$-bit numbers, this takes $$\mathcal{O}(\sqrt{2^n})$$ guesses
(the square root is due to the birthday paradox).
-A quantum computer needs to query $f$ only $\mathcal{O}(n)$ times,
+A quantum computer needs to query $$f$$ only $$\mathcal{O}(n)$$ times,
although the exact number varies due to the algorithm's probabilistic nature.
It uses the following circuit:
@@ -56,8 +56,8 @@ It uses the following circuit:
-The XOR oracle $U_f$ implements $f$,
-and has the following action for $n$-bit $a$ and $b$:
+The XOR oracle $$U_f$$ implements $$f$$,
+and has the following action for $$n$$-bit $$a$$ and $$b$$:
$$\begin{aligned}
\Ket{a} \Ket{b}
@@ -65,9 +65,9 @@ $$\begin{aligned}
\Ket{a} \Ket{b \oplus f(a)}
\end{aligned}$$
-Starting from the state $\Ket{0}^{\otimes 2 n}$,
-we apply the [Hadamard gate](/know/concept/quantum-gate/) $H$
-to each of the first $n$ qubits:
+Starting from the state $$\Ket{0}^{\otimes 2 n}$$,
+we apply the [Hadamard gate](/know/concept/quantum-gate/) $$H$$
+to each of the first $$n$$ qubits:
$$\begin{aligned}
\Ket{0}^{\otimes n} \Ket{0}^{\otimes n}
@@ -76,10 +76,10 @@ $$\begin{aligned}
= \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{0}^{\otimes n}
\end{aligned}$$
-Where $\Ket{x}$ is shorthand for $\Ket{x}_1 \cdots \Ket{x}_n$.
-In other words, we now have an equal superposition of all possible inputs $x$,
-with a constant $\Ket{0}^{\otimes n}$ beside it.
-We give this to the oracle $U_f$:
+Where $$\Ket{x}$$ is shorthand for $$\Ket{x}_1 \cdots \Ket{x}_n$$.
+In other words, we now have an equal superposition of all possible inputs $$x$$,
+with a constant $$\Ket{0}^{\otimes n}$$ beside it.
+We give this to the oracle $$U_f$$:
$$\begin{aligned}
\frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{0}^{\otimes n}
@@ -87,10 +87,10 @@ $$\begin{aligned}
\frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{f(x)}
\end{aligned}$$
-Then we apply $H^{\otimes n}$ to the first $n$ qubits again,
+Then we apply $$H^{\otimes n}$$ to the first $$n$$ qubits again,
which, thanks to the definition of the Hadamard transform,
yields the following,
-where $x \cdot y$ is the bitwise dot product:
+where $$x \cdot y$$ is the bitwise dot product:
$$\begin{aligned}
\frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{f(x)}
@@ -101,10 +101,10 @@ $$\begin{aligned}
Next, we measure all qubits.
The order in which we do this does not matter,
-but, for clarity, let us measure the last $n$ qubits first,
-yielding $\Ket{f(x_1)}$ for some $x_1$.
-Doing this leaves the $2n$ qubits in the following state,
-where $f(x_1) = f(x_2)$ and $x_2 = s \oplus x_1$:
+but, for clarity, let us measure the last $$n$$ qubits first,
+yielding $$\Ket{f(x_1)}$$ for some $$x_1$$.
+Doing this leaves the $$2n$$ qubits in the following state,
+where $$f(x_1) = f(x_2)$$ and $$x_2 = s \oplus x_1$$:
$$\begin{alignedat}{2}
&\mathrm{if} \: s = 0: \qquad
@@ -114,18 +114,18 @@ $$\begin{alignedat}{2}
&&\frac{1}{\sqrt{2^{n+1}}} \sum_{y = 0}^{2^n - 1} \Big( (-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \Big) \Ket{y} \Ket{f(x_1)}
\end{alignedat}$$
-If $s = 0$, we get an equiprobable superposition of all $y$.
-So, when we measure the first $n$ qubits, the result is a uniformly random number,
-regardless of the phase $(-1)^{x_1 \cdot y}$.
+If $$s = 0$$, we get an equiprobable superposition of all $$y$$.
+So, when we measure the first $$n$$ qubits, the result is a uniformly random number,
+regardless of the phase $$(-1)^{x_1 \cdot y}$$.
-If $s \neq 0$, the situation is more interesting,
-because we can only measure $y$-values where:
+If $$s \neq 0$$, the situation is more interesting,
+because we can only measure $$y$$-values where:
$$\begin{aligned}
(-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \neq 0
\end{aligned}$$
-Since $x_2 = s \oplus x_1$ by definition,
+Since $$x_2 = s \oplus x_1$$ by definition,
we can rewrite this as follows:
$$\begin{aligned}
@@ -134,21 +134,21 @@ $$\begin{aligned}
\neq 0
\end{aligned}$$
-Clearly, the expression can only be nonzero if $s \cdot y$ is even.
-In other words, when we measure the first $n$ qubits,
-we get a random $y$-value,
-for which $s \cdot y$ is guaranteed to be even.
+Clearly, the expression can only be nonzero if $$s \cdot y$$ is even.
+In other words, when we measure the first $$n$$ qubits,
+we get a random $$y$$-value,
+for which $$s \cdot y$$ is guaranteed to be even.
-In both cases $s = 0$ and $s \neq 0$,
-we measure a $y$-value that satisfies the equation:
+In both cases $$s = 0$$ and $$s \neq 0$$,
+we measure a $$y$$-value that satisfies the equation:
$$\begin{aligned}
s \cdot y = 0 \:\:(\bmod 2)
\end{aligned}$$
-This tells us something about $s$, albeit not much.
-But if we run Simon's algorithm $N$ times,
-we get various $y$-values $y_1, ..., y_N$,
+This tells us something about $$s$$, albeit not much.
+But if we run Simon's algorithm $$N$$ times,
+we get various $$y$$-values $$y_1, ..., y_N$$,
from which we can build a system of linear equations:
$$\begin{aligned}
@@ -162,10 +162,10 @@ $$\begin{aligned}
\end{aligned}$$
This can be solved efficiently by a classical computer.
-In the best-case scenario, all those $y$-values would be linearly independent
+In the best-case scenario, all those $$y$$-values would be linearly independent
(when regarded as vectors of bits),
-in which case only $N = n - 1$ equations would be necessary.
-Simon's algorithm is therefore $\mathcal{O}(n)$.
+in which case only $$N = n - 1$$ equations would be necessary.
+Simon's algorithm is therefore $$\mathcal{O}(n)$$.
It may feel like "cheating" to use a classical computer at the end.
Remember that the point of this algorithm is to limit the number of oracle queries,
diff --git a/source/know/concept/slater-determinant/index.md b/source/know/concept/slater-determinant/index.md
index 03ec153..72c8cf2 100644
--- a/source/know/concept/slater-determinant/index.md
+++ b/source/know/concept/slater-determinant/index.md
@@ -9,13 +9,13 @@ layout: "concept"
---
In quantum mechanics, the **Slater determinant** is a trick
-to create a many-particle wave function for a system of $N$ fermions,
+to create a many-particle wave function for a system of $$N$$ fermions,
with the necessary antisymmetry.
-Given an orthogonal set of individual states $\psi_n(x)$, we write
-$\psi_n(x_n)$ to say that particle $x_n$ is in state $\psi_n$. Now the
+Given an orthogonal set of individual states $$\psi_n(x)$$, we write
+$$\psi_n(x_n)$$ to say that particle $$x_n$$ is in state $$\psi_n$$. Now the
goal is to find an expression for an overall many-particle wave
-function $\Psi(x_1, ..., x_N)$ that satisfies the
+function $$\Psi(x_1, ..., x_N)$$ that satisfies the
[Pauli exclusion principle](/know/concept/pauli-exclusion-principle/).
Enter the Slater determinant:
@@ -34,13 +34,13 @@ Swapping the state of two particles corresponds to exchanging two rows,
which flips the sign of the determinant.
Similarly, switching two columns means swapping two states,
which also results in a sign change.
-Finally, putting two particles into the same state makes $\Psi$ vanish.
+Finally, putting two particles into the same state makes $$\Psi$$ vanish.
Not all valid many-fermion wave functions can be
written as a single Slater determinant; a linear combination of multiple
may be needed. Nevertheless, an appropriate choice of the input set
-$\psi_n(x)$ can optimize how well a single determinant approximates a
-given $\Psi$.
+$$\psi_n(x)$$ can optimize how well a single determinant approximates a
+given $$\Psi$$.
In fact, there exists a similar trick for bosons, where the goal is to
create a symmetric wave function which allows multiple particles to
diff --git a/source/know/concept/sokhotski-plemelj-theorem/index.md b/source/know/concept/sokhotski-plemelj-theorem/index.md
index 984a558..66e89bc 100644
--- a/source/know/concept/sokhotski-plemelj-theorem/index.md
+++ b/source/know/concept/sokhotski-plemelj-theorem/index.md
@@ -9,8 +9,8 @@ categories:
layout: "concept"
---
-The goal is to evaluate integrals of the following form, where $a < 0 < b$,
-and $f(x)$ is assumed to be continuous in the integration interval $[a, b]$:
+The goal is to evaluate integrals of the following form, where $$a < 0 < b$$,
+and $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$:
$$\begin{aligned}
\lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
@@ -26,7 +26,7 @@ $$\begin{aligned}
\end{aligned}$$
To evaluate the real part,
-we notice that for $\eta \to 0^+$ the integrand diverges for $x \to 0$,
+we notice that for $$\eta \to 0^+$$ the integrand diverges for $$x \to 0$$,
and thus split the integral as follows:
$$\begin{aligned}
@@ -35,7 +35,7 @@ $$\begin{aligned}
\end{aligned}$$
This is simply the definition of the
-[Cauchy principal value](/know/concept/cauchy-principal-value/) $\mathcal{P}$,
+[Cauchy principal value](/know/concept/cauchy-principal-value/) $$\mathcal{P}$$,
so the real part is given by:
$$\begin{aligned}
@@ -45,7 +45,7 @@ $$\begin{aligned}
\end{aligned}$$
Meanwhile, in the imaginary part,
-we substitute $\eta$ for $1 / m$, and introduce $\pi$:
+we substitute $$\eta$$ for $$1 / m$$, and introduce $$\pi$$:
$$\begin{aligned}
\lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x}
@@ -54,8 +54,8 @@ $$\begin{aligned}
&= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{m}{1 + m^2 x^2} f(x) \dd{x}
\end{aligned}$$
-The expression $m / \pi (1 + m^2 x^2)$ is a so-called *nascent delta function*,
-meaning that in the limit $m \to +\infty$ it converges to
+The expression $$m / \pi (1 + m^2 x^2)$$ is a so-called *nascent delta function*,
+meaning that in the limit $$m \to +\infty$$ it converges to
the [Dirac delta function](/know/concept/dirac-delta-function/):
$$\begin{aligned}
@@ -76,9 +76,9 @@ $$\begin{aligned}
\end{aligned}$$
However, this theorem is often written in the following sloppy way,
-where $\eta$ is defined up front to be small,
-the integral is hidden, and $f(x)$ is set to $1$.
-This awkwardly leaves $\mathcal{P}$ behind:
+where $$\eta$$ is defined up front to be small,
+the integral is hidden, and $$f(x)$$ is set to $$1$$.
+This awkwardly leaves $$\mathcal{P}$$ behind:
$$\begin{aligned}
\frac{1}{x + i \eta}
@@ -87,13 +87,13 @@ $$\begin{aligned}
The full, complex version of the Sokhotski-Plemelj theorem
evaluates integrals of the following form
-over a contour $C$ in the complex plane:
+over a contour $$C$$ in the complex plane:
$$\begin{aligned}
\phi(z) = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta}
\end{aligned}$$
-Where $f(z)$ must be [holomorphic](/know/concept/holomorphic-function/).
+Where $$f(z)$$ must be [holomorphic](/know/concept/holomorphic-function/).
The Sokhotski-Plemelj theorem then states:
$$\begin{aligned}
@@ -103,8 +103,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where the sign is positive if $z$ is inside $C$, and negative if it is outside.
-The real version follows by letting $C$ follow the whole real axis,
-making $C$ an infinitely large semicircle,
+Where the sign is positive if $$z$$ is inside $$C$$, and negative if it is outside.
+The real version follows by letting $$C$$ follow the whole real axis,
+making $$C$$ an infinitely large semicircle,
so that the integrand vanishes away from the real axis,
-because $1 / (\zeta \!-\! z) \to 0$ for $|\zeta| \to \infty$.
+because $$1 / (\zeta \!-\! z) \to 0$$ for $$|\zeta| \to \infty$$.
diff --git a/source/know/concept/spherical-coordinates/index.md b/source/know/concept/spherical-coordinates/index.md
index bc01c95..9df5d51 100644
--- a/source/know/concept/spherical-coordinates/index.md
+++ b/source/know/concept/spherical-coordinates/index.md
@@ -10,16 +10,16 @@ layout: "concept"
**Spherical coordinates** are an extension of polar coordinates to 3D.
The position of a given point in space is described by
-three coordinates $(r, \theta, \varphi)$, defined as:
+three coordinates $$(r, \theta, \varphi)$$, defined as:
-* $r$: the **radius** or **radial distance**: distance to the origin.
-* $\theta$: the **elevation**, **polar angle** or **colatitude**:
- angle to the positive $z$-axis, or **zenith**, i.e. the "north pole".
-* $\varphi$: the **azimuth**, **azimuthal angle** or **longitude**:
- angle from the positive $x$-axis, typically in the counter-clockwise sense.
+* $$r$$: the **radius** or **radial distance**: distance to the origin.
+* $$\theta$$: the **elevation**, **polar angle** or **colatitude**:
+ angle to the positive $$z$$-axis, or **zenith**, i.e. the "north pole".
+* $$\varphi$$: the **azimuth**, **azimuthal angle** or **longitude**:
+ angle from the positive $$x$$-axis, typically in the counter-clockwise sense.
-Cartesian coordinates $(x, y, z)$ and the spherical system
-$(r, \theta, \varphi)$ are related by:
+Cartesian coordinates $$(x, y, z)$$ and the spherical system
+$$(r, \theta, \varphi)$$ are related by:
$$\begin{aligned}
\boxed{
@@ -31,8 +31,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Conversely, a point given in $(x, y, z)$
-can be converted to $(r, \theta, \varphi)$
+Conversely, a point given in $$(x, y, z)$$
+can be converted to $$(r, \theta, \varphi)$$
using these formulae:
$$\begin{aligned}
@@ -47,7 +47,7 @@ $$\begin{aligned}
The spherical coordinate system is an orthogonal
[curvilinear system](/know/concept/curvilinear-coordinates/),
-whose scale factors $h_r$, $h_\theta$ and $h_\varphi$ we want to find.
+whose scale factors $$h_r$$, $$h_\theta$$ and $$h_\varphi$$ we want to find.
To do so, we calculate the differentials of the Cartesian coordinates:
$$\begin{aligned}
@@ -58,7 +58,7 @@ $$\begin{aligned}
\dd{z} &= \dd{r} \cos\theta - \dd{\theta} r \sin\theta
\end{aligned}$$
-And then we calculate the line element $\dd{\ell}^2$,
+And then we calculate the line element $$\dd{\ell}^2$$,
skipping many terms thanks to orthogonality:
$$\begin{aligned}
@@ -74,7 +74,7 @@ $$\begin{aligned}
Finally, we can simply read off
the squares of the desired scale factors
-$h_r^2$, $h_\theta^2$ and $h_\varphi^2$:
+$$h_r^2$$, $$h_\theta^2$$ and $$h_\varphi^2$$:
$$\begin{aligned}
\boxed{
@@ -146,7 +146,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The differential element of volume $\dd{V}$
+The differential element of volume $$\dd{V}$$
takes the following form:
$$\begin{aligned}
@@ -163,7 +163,7 @@ $$\begin{aligned}
= \int_0^{2\pi} \int_0^\pi \int_0^\infty f(r, \theta, \varphi) \: r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
\end{aligned}$$
-The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+The isosurface elements are as follows, where $$S_r$$ is a surface at constant $$r$$, etc.:
$$\begin{aligned}
\boxed{
@@ -177,7 +177,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+Similarly, the normal vector element $$\dd{\vu{S}}$$ for an arbitrary surface is given by:
$$\begin{aligned}
\boxed{
@@ -188,7 +188,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+And finally, the tangent vector element $$\dd{\vu{\ell}}$$ of a given curve is as follows:
$$\begin{aligned}
\boxed{
diff --git a/source/know/concept/spitzer-resistivity/index.md b/source/know/concept/spitzer-resistivity/index.md
index f9fd969..283b7fd 100644
--- a/source/know/concept/spitzer-resistivity/index.md
+++ b/source/know/concept/spitzer-resistivity/index.md
@@ -9,17 +9,17 @@ layout: "concept"
---
If an [electric field](/know/concept/electric-field/)
-with magnitude $E$ is applied to the plasma, the electrons experience
-a [Lorentz force](/know/concept/lorentz-force/) $q_e E$
+with magnitude $$E$$ is applied to the plasma, the electrons experience
+a [Lorentz force](/know/concept/lorentz-force/) $$q_e E$$
(we neglect the ions due to their mass),
-where $q_e$ is the electron charge.
+where $$q_e$$ is the electron charge.
However, collisions slow them down while they travel through the plasma.,
-This can be modelled as a drag force $f_{ei} m_e v_e$,
-where $f_{ei}$ is the electron-ion collision frequency
-(we neglect $f_{ee}$ since all electrons are moving together),
-$m_e$ is their mass,
-and $v_e$ their typical velocity relative to the ions in the background.
+This can be modelled as a drag force $$f_{ei} m_e v_e$$,
+where $$f_{ei}$$ is the electron-ion collision frequency
+(we neglect $$f_{ee}$$ since all electrons are moving together),
+$$m_e$$ is their mass,
+and $$v_e$$ their typical velocity relative to the ions in the background.
Balancing the two forces yields the following relation:
$$\begin{aligned}
@@ -27,7 +27,7 @@ $$\begin{aligned}
= f_{ei} m_e v_e
\end{aligned}$$
-Using that the current density $J = q_e n_e v_e$,
+Using that the current density $$J = q_e n_e v_e$$,
we can rearrange this like so:
$$\begin{aligned}
@@ -37,12 +37,12 @@ $$\begin{aligned}
= \eta J
\end{aligned}$$
-This is Ohm's law, where $\eta$ is the resistivity.
-From our derivation of the [Coulomb logarithm](/know/concept/coulomb-logarithm/) $\ln(\Lambda)$,
-we estimate $f_{ei}$ to be as follows,
-where $n_i$ is the ion density,
-$\sigma$ is the collision cross-section,
-and $\mu$ is the [reduced mass](/know/concept/reduced-mass/)
+This is Ohm's law, where $$\eta$$ is the resistivity.
+From our derivation of the [Coulomb logarithm](/know/concept/coulomb-logarithm/) $$\ln(\Lambda)$$,
+we estimate $$f_{ei}$$ to be as follows,
+where $$n_i$$ is the ion density,
+$$\sigma$$ is the collision cross-section,
+and $$\mu$$ is the [reduced mass](/know/concept/reduced-mass/)
of the electron-ion system:
$$\begin{aligned}
@@ -52,15 +52,15 @@ $$\begin{aligned}
\approx \frac{1}{2 \pi} \frac{Z q_e^4}{\varepsilon_0^2 m_e^2} \frac{n_e}{v_e^3} \ln(\Lambda)
\end{aligned}$$
-Where we used that $\mu \approx m_e$,
-and $q_i = -Z q_e$ for some ionization $Z$,
-and as a result $n_e \approx Z n_i$ due to the plasma's quasi-neutrality.
-Beware: authors disagree about the constant factors in $f_{ei}$;
+Where we used that $$\mu \approx m_e$$,
+and $$q_i = -Z q_e$$ for some ionization $$Z$$,
+and as a result $$n_e \approx Z n_i$$ due to the plasma's quasi-neutrality.
+Beware: authors disagree about the constant factors in $$f_{ei}$$;
recall that it was derived from fairly rough estimates.
This article follows Bellan.
-Inserting this expression for $f_{ei}$ into
-the so-called **Spitzer resistivity** $\eta$ then yields:
+Inserting this expression for $$f_{ei}$$ into
+the so-called **Spitzer resistivity** $$\eta$$ then yields:
$$\begin{aligned}
\boxed{
@@ -70,10 +70,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-A reasonable estimate for the typical velocity $v_e$
+A reasonable estimate for the typical velocity $$v_e$$
at thermal equilibrium is as follows,
-where $k_B$ is Boltzmann's constant,
-and $T_e$ is the electron temperature:
+where $$k_B$$ is Boltzmann's constant,
+and $$T_e$$ is the electron temperature:
$$\begin{aligned}
\frac{1}{2} m_e v_e^2
@@ -85,8 +85,8 @@ $$\begin{aligned}
Other choices exist,
see e.g. the [Maxwell-Boltzmann distribution](/know/concept/maxwell-boltzmann-distribution/),
-but always $v_e \propto \sqrt{T_e/m_e}$.
-Inserting this $v_e$ into $\eta$ then gives:
+but always $$v_e \propto \sqrt{T_e/m_e}$$.
+Inserting this $$v_e$$ into $$\eta$$ then gives:
$$\begin{aligned}
\eta
diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
index e0b26af..dd83334 100644
--- a/source/know/concept/step-index-fiber/index.md
+++ b/source/know/concept/step-index-fiber/index.md
@@ -9,31 +9,31 @@ categories:
layout: "concept"
---
-As light propagates in the $z$-direction through an optical fiber,
-the transverse profile $F(x,y)$ of the [electric field](/know/concept/electric-field/)
+As light propagates in the $$z$$-direction through an optical fiber,
+the transverse profile $$F(x,y)$$ of the [electric field](/know/concept/electric-field/)
can be shown to obey the *Helmholtz equation* in 2D:
$$\begin{aligned}
\nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F = 0
\end{aligned}$$
-With $n$ being the position-dependent refractive index,
-$k$ the vacuum wavenumber $\omega / c$,
-and $\beta$ the mode's propagation constant, to be determined later.
+With $$n$$ being the position-dependent refractive index,
+$$k$$ the vacuum wavenumber $$\omega / c$$,
+and $$\beta$$ the mode's propagation constant, to be determined later.
In [polar coordinates](/know/concept/cylindrical-polar-coordinates/)
-$(r,\phi)$ this equation can be rewritten as follows:
+$$(r,\phi)$$ this equation can be rewritten as follows:
$$\begin{aligned}
\pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F = 0
\end{aligned}$$
-Where we have defined $\mu \equiv n^2 k^2 \!-\! \beta^2$ for brevity.
-From now on, we only consider choices of $\mu$ that do not depend on $\phi$ or $z$,
-but may vary with $r$.
+Where we have defined $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ for brevity.
+From now on, we only consider choices of $$\mu$$ that do not depend on $$\phi$$ or $$z$$,
+but may vary with $$r$$.
This Helmholtz equation can be solved by *separation of variables*:
-we assume that there exist two functions $R(r)$ and $\Phi(\phi)$
-such that $F(r,\phi) = R(r) \, \Phi(\phi)$.
+we assume that there exist two functions $$R(r)$$ and $$\Phi(\phi)$$
+such that $$F(r,\phi) = R(r) \, \Phi(\phi)$$.
Inserting this ansatz:
$$\begin{aligned}
@@ -41,10 +41,10 @@ $$\begin{aligned}
\end{aligned}$$
We rearrange this such that each side only depends on one variable,
-by dividing by $R\Phi$ (ignoring the fact that it may be zero),
-and multiplying by $r^2$.
-Since this equation should hold for *all* values of $r$ and $\phi$,
-this means that both sides must equal a constant $\ell^2$:
+by dividing by $$R\Phi$$ (ignoring the fact that it may be zero),
+and multiplying by $$r^2$$.
+Since this equation should hold for *all* values of $$r$$ and $$\phi$$,
+this means that both sides must equal a constant $$\ell^2$$:
$$\begin{aligned}
r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2
@@ -52,8 +52,8 @@ $$\begin{aligned}
= \ell^2
\end{aligned}$$
-This gives an eigenvalue problem for $\Phi$,
-and the well-known *Bessel equation* for $R$:
+This gives an eigenvalue problem for $$\Phi$$,
+and the well-known *Bessel equation* for $$R$$:
$$\begin{aligned}
\boxed{
@@ -65,9 +65,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-We will return to $R$ later; we start with $\Phi$, because it has the
-simplest equation. Since the angle $\phi$ is limited to $[0,2\pi]$,
-$\Phi$ must be $2 \pi$-periodic, so:
+We will return to $$R$$ later; we start with $$\Phi$$, because it has the
+simplest equation. Since the angle $$\phi$$ is limited to $$[0,2\pi]$$,
+$$\Phi$$ must be $$2 \pi$$-periodic, so:
$$\begin{aligned}
\Phi(0) = \Phi(2\pi)
@@ -75,18 +75,18 @@ $$\begin{aligned}
\Phi'(0) = \Phi'(2\pi)
\end{aligned}$$
-The above equation for $\Phi$ with these periodic boundary conditions
+The above equation for $$\Phi$$ with these periodic boundary conditions
is a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
-Consequently, there are infinitely many allowed values of $\ell^2$,
+Consequently, there are infinitely many allowed values of $$\ell^2$$,
all real, and one of them is lowest, known as the *ground state*.
-To find the eigenvalues $\ell^2$ and their corresponding $\Phi$,
-we in turn assume that $\ell^2 < 0$, $\ell^2 = 0$, or $\ell^2 > 0$,
-and check if we can then arrive at a non-trivial $\Phi$ for each case.
+To find the eigenvalues $$\ell^2$$ and their corresponding $$\Phi$$,
+we in turn assume that $$\ell^2 < 0$$, $$\ell^2 = 0$$, or $$\ell^2 > 0$$,
+and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
-* For $\ell^2 < 0$, solutions have the form $\Phi(\phi) = A \sinh(\phi \ell) + B \cosh(\phi \ell)$,
- where $A$ and $B$ are unknown linearity constants.
- At least one of these constants must be nonzero for $\Phi$ to be non-trivial,
+* For $$\ell^2 < 0$$, solutions have the form $$\Phi(\phi) = A \sinh(\phi \ell) + B \cosh(\phi \ell)$$,
+ where $$A$$ and $$B$$ are unknown linearity constants.
+ At least one of these constants must be nonzero for $$\Phi$$ to be non-trivial,
but the challenge is to satisfy the boundary conditions:
$$\begin{alignedat}{3}
@@ -112,12 +112,12 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
= 2 \big( \cosh(2 \pi \ell) - 1 \big)
\end{aligned}$$
- This can only be zero if $\ell = 0$,
- which contradicts the premise that $\ell^2 < 0$,
- so we conclude that $\ell^2$ cannot be negative,
+ This can only be zero if $$\ell = 0$$,
+ which contradicts the premise that $$\ell^2 < 0$$,
+ so we conclude that $$\ell^2$$ cannot be negative,
because no non-trivial solutions exist here.
-* For $\ell^2 = 0$, the solution is $\Phi(\phi) = A \phi + B$.
+* For $$\ell^2 = 0$$, the solution is $$\Phi(\phi) = A \phi + B$$.
Putting this in the boundary conditions:
$$\begin{alignedat}{3}
@@ -130,11 +130,11 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
B &&= B
\end{alignedat}$$
- $B$ can be nonzero, so this a valid solution.
- We conclude that $\ell^2 = 0$ is the ground state.
+ $$B$$ can be nonzero, so this a valid solution.
+ We conclude that $$\ell^2 = 0$$ is the ground state.
-* For $\ell^2 > 0$, all solutions have the form
- $\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$, therefore:
+* For $$\ell^2 > 0$$, all solutions have the form
+ $$\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$$, therefore:
$$\begin{alignedat}{3}
\Phi(0) &= \Phi(2 \pi)
@@ -159,7 +159,7 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
= 2 \big(\cos(2 \pi \ell) - 1\big)
\end{aligned}$$
- Meaning that $\ell$ must be an integer.
+ Meaning that $$\ell$$ must be an integer.
We revisit the boundary conditions and indeed see:
$$\begin{alignedat}{3}
@@ -172,16 +172,16 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
0 &&= 0
\end{alignedat}$$
- So $A$ and $B$ are *both* unconstrained,
- and each integer $\ell$ is a doubly-degenerate eigenvalue.
+ So $$A$$ and $$B$$ are *both* unconstrained,
+ and each integer $$\ell$$ is a doubly-degenerate eigenvalue.
The two linearly independent solutions,
- $\sin(\phi \ell)$ and $\cos(\phi \ell)$,
+ $$\sin(\phi \ell)$$ and $$\cos(\phi \ell)$$,
represent the polarization of light in the mode.
For simplicity, we assume that all light is in a single polarization,
- so only $\cos(\phi \ell)$ will be considered from now on.
+ so only $$\cos(\phi \ell)$$ will be considered from now on.
-By combining our result for $\ell^2 = 0$ and $\ell^2 > 0$,
-we get the following for $\ell = 0, 1, 2, ...$:
+By combining our result for $$\ell^2 = 0$$ and $$\ell^2 > 0$$,
+we get the following for $$\ell = 0, 1, 2, ...$$:
$$\begin{aligned}
\boxed{
@@ -189,28 +189,28 @@ $$\begin{aligned}
}
\end{aligned}$$
-Here, $\ell$ is called the **primary mode index**.
-We exclude $\ell < 0$ because $\cos(x) \propto \cos(-x)$
-and $\sin(x) \propto \sin(-x)$,
-and because $A$ is free to choose thanks to linearity.
+Here, $$\ell$$ is called the **primary mode index**.
+We exclude $$\ell < 0$$ because $$\cos(x) \propto \cos(-x)$$
+and $$\sin(x) \propto \sin(-x)$$,
+and because $$A$$ is free to choose thanks to linearity.
-Let us now revisit the Bessel equation for the radial function $R(r)$,
+Let us now revisit the Bessel equation for the radial function $$R(r)$$,
which should be continuous and differentiable throughout the fiber:
$$\begin{aligned}
r^2 R'' + r R' + \mu r^2 R - \ell^2 R = 0
\end{aligned}$$
-To continue, we need to specify the refractive index $n(r)$, contained in $\mu(r)$.
+To continue, we need to specify the refractive index $$n(r)$$, contained in $$\mu(r)$$.
We choose a **step-index fiber**,
-whose cross-section consists of a **core** with radius $a$,
-surrounded by a **cladding** that extends to infinity $r \to \infty$.
-In the core $r < a$, the index $n$ is a constant $n_i$,
-while in the cladding $r > a$ it is another constant $n_o$.
+whose cross-section consists of a **core** with radius $$a$$,
+surrounded by a **cladding** that extends to infinity $$r \to \infty$$.
+In the core $$r < a$$, the index $$n$$ is a constant $$n_i$$,
+while in the cladding $$r > a$$ it is another constant $$n_o$$.
-Since $\mu$ is different in the core and cladding,
-we will get different solutions $R_i$ and $R_o$ there,
-so we must demand that the field is continuous at the boundary $r = a$:
+Since $$\mu$$ is different in the core and cladding,
+we will get different solutions $$R_i$$ and $$R_o$$ there,
+so we must demand that the field is continuous at the boundary $$r = a$$:
$$\begin{aligned}
R_i(a) = R_o(a)
@@ -219,14 +219,14 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, for a physically plausible solution,
-we require that $R_i$ is finite
-and that $R_o$ decays monotonically to zero when $r \to \infty$.
-These constraints will turn out to restrict $\mu$.
+we require that $$R_i$$ is finite
+and that $$R_o$$ decays monotonically to zero when $$r \to \infty$$.
+These constraints will turn out to restrict $$\mu$$.
-Introducing a new coordinate $\rho \equiv r \sqrt{|\mu|}$
+Introducing a new coordinate $$\rho \equiv r \sqrt{|\mu|}$$
gives the Bessel equation's standard form,
which has well-known solutions called *Bessel functions*, shown below.
-Let $\pm$ be the sign of $\mu$:
+Let $$\pm$$ be the sign of $$\mu$$:
$$\begin{aligned}
\begin{cases}
@@ -244,11 +244,11 @@ $$\begin{aligned}
-Looking at these solutions with our constraints for $R_o$ in mind,
-we see that for $\mu > 0$ none of the solutions decay
-*monotonically* to zero, so we must have $\mu \le 0$ in the cladding.
-Of the remaining candidates, $\ln\!(r)$, $r^\ell$ and $I_\ell(\rho)$ do not decay at all,
-leading to the following $R_o$:
+Looking at these solutions with our constraints for $$R_o$$ in mind,
+we see that for $$\mu > 0$$ none of the solutions decay
+*monotonically* to zero, so we must have $$\mu \le 0$$ in the cladding.
+Of the remaining candidates, $$\ln\!(r)$$, $$r^\ell$$ and $$I_\ell(\rho)$$ do not decay at all,
+leading to the following $$R_o$$:
$$\begin{aligned}
R_{o,\ell}(r) =
@@ -261,12 +261,12 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-Next, for $R_i$, we see that when $\mu < 0$ all solutions are invalid
-since they diverge at $r = 0$,
-and so do $\ln\!(r)$, $r^{-\ell}$ and $Y_\ell(\rho)$.
-Of the remaining candidates, $r^0$ and $r^\ell$ have a non-negative slope
-at the boundary $r = a$, so they can never be continuous with $R_o'$.
-This leaves $J_\ell(\rho)$ for $\mu > 0$:
+Next, for $$R_i$$, we see that when $$\mu < 0$$ all solutions are invalid
+since they diverge at $$r = 0$$,
+and so do $$\ln\!(r)$$, $$r^{-\ell}$$ and $$Y_\ell(\rho)$$.
+Of the remaining candidates, $$r^0$$ and $$r^\ell$$ have a non-negative slope
+at the boundary $$r = a$$, so they can never be continuous with $$R_o'$$.
+This leaves $$J_\ell(\rho)$$ for $$\mu > 0$$:
$$\begin{aligned}
R_{i,\ell}(r) =
@@ -274,7 +274,7 @@ $$\begin{aligned}
\qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,...
\end{aligned}$$
-Putting this all together, we now know what the full solution for $F$ should look like:
+Putting this all together, we now know what the full solution for $$F$$ should look like:
$$\begin{aligned}
F_\ell(r, \phi)
@@ -289,25 +289,25 @@ $$\begin{aligned}
\end{cases}
\end{aligned}$$
-Where $A_\ell$ and $B_\ell$ are constants to be chosen
-based on the light's intensity, and to satisfy the continuity condition at $r = a$.
+Where $$A_\ell$$ and $$B_\ell$$ are constants to be chosen
+based on the light's intensity, and to satisfy the continuity condition at $$r = a$$.
-We found that $\mu \le 0$ in the cladding and $\mu > 0$ in the core.
-Since $\mu \equiv n^2 k^2 \!-\! \beta^2$ by definition,
-this discovery places a constraint on the propagation constant $\beta$:
+We found that $$\mu \le 0$$ in the cladding and $$\mu > 0$$ in the core.
+Since $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ by definition,
+this discovery places a constraint on the propagation constant $$\beta$$:
$$\begin{aligned}
n_i^2 k^2 > \beta^2 \ge n_o^2 k^2
\end{aligned}$$
-Therefore, $n_i > n_o$ in a step-index fiber,
-and there is only a limited range of allowed $\beta$-values;
+Therefore, $$n_i > n_o$$ in a step-index fiber,
+and there is only a limited range of allowed $$\beta$$-values;
the fiber is not able to guide the light outside this range.
-However, not all $\beta$ in this range are created equal for all $k$.
+However, not all $$\beta$$ in this range are created equal for all $$k$$.
To investigate further, let us define the quantities
-$\xi_\mathrm{core}$ and $\xi_\mathrm{clad}$ like so,
-assuming $n_i$ and $n_o$ do not depend on $k$:
+$$\xi_\mathrm{core}$$ and $$\xi_\mathrm{clad}$$ like so,
+assuming $$n_i$$ and $$n_o$$ do not depend on $$k$$:
$$\begin{aligned}
\xi_i(k)
@@ -317,13 +317,13 @@ $$\begin{aligned}
\equiv \sqrt{ \beta^2(k) - n_o^2 k^2 }
\end{aligned}$$
-It is important to note that the sum of their squares is constant with respect to $\beta$:
+It is important to note that the sum of their squares is constant with respect to $$\beta$$:
$$\begin{aligned}
\xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2
\end{aligned}$$
-Where $\mathrm{NA}$ is the so-called **numerical aperture**,
+Where $$\mathrm{NA}$$ is the so-called **numerical aperture**,
often mentioned in papers and datasheets as one of a fiber's key parameters.
It is defined as:
@@ -334,7 +334,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-From this, we define a new fiber parameter: the $V$-**number**,
+From this, we define a new fiber parameter: the $$V$$-**number**,
which is extremely useful:
$$\begin{aligned}
@@ -345,8 +345,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Now, the allowed values of $\beta$ are found
-by fulfilling the boundary conditions (for $\mu \neq 0$):
+Now, the allowed values of $$\beta$$ are found
+by fulfilling the boundary conditions (for $$\mu \neq 0$$):
$$\begin{aligned}
A_\ell J_\ell(a \xi_i)
@@ -356,10 +356,10 @@ $$\begin{aligned}
&= B_\ell \xi_o K_\ell'(a \xi_o)
\end{aligned}$$
-To remove $A_\ell$ and $B_\ell$,
+To remove $$A_\ell$$ and $$B_\ell$$,
we divide the latter equation by the former,
-meanwhile defining $X \equiv a \xi_i$ and $Y \equiv a \xi_o$
-for convenience, such that $X^2 + Y^2 = V^2$:
+meanwhile defining $$X \equiv a \xi_i$$ and $$Y \equiv a \xi_o$$
+for convenience, such that $$X^2 + Y^2 = V^2$$:
$$\begin{aligned}
X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)}
@@ -374,7 +374,7 @@ $$\begin{aligned}
K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x}
\end{aligned}$$
-With this, the transcendental equation for $\beta$
+With this, the transcendental equation for $$\beta$$
takes this convenient form:
$$\begin{aligned}
@@ -383,27 +383,27 @@ $$\begin{aligned}
}
\end{aligned}$$
-All $\beta$ that satisfy this indicate the existence
+All $$\beta$$ that satisfy this indicate the existence
of a **linearly polarized** mode.
-These modes are called $\mathrm{LP}_{\ell m}$,
-where $\ell$ is the primary (azimuthal) mode index,
-and $m$ the secondary (radial) mode index,
-which is needed because multiple $\beta$ may exist for a single $\ell$.
+These modes are called $$\mathrm{LP}_{\ell m}$$,
+where $$\ell$$ is the primary (azimuthal) mode index,
+and $$m$$ the secondary (radial) mode index,
+which is needed because multiple $$\beta$$ may exist for a single $$\ell$$.
An example graphical solution of the transcendental equation
-is illustrated below for a fiber with $V = 5$,
+is illustrated below for a fiber with $$V = 5$$,
where red and blue respectively denote the left and right-hand side:
-This shows that each $\mathrm{LP}_{\ell m}$ has an associated cut-off $V_{\ell m}$,
-so that if $V > V_{\ell m}$ then $\mathrm{LP}_{lm}$ exists,
-as long as $\beta$ stays in the allowed range.
-The cut-offs of the secondary modes for a given $\ell$
-are found as the $m$th roots of $J_{\ell-1}(V_{\ell m}) = 0$.
-In the above figure, they are $V_{01} = 0$, $V_{11} = 2.405$, and $V_{02} = V_{21} = 3.832$.
+This shows that each $$\mathrm{LP}_{\ell m}$$ has an associated cut-off $$V_{\ell m}$$,
+so that if $$V > V_{\ell m}$$ then $$\mathrm{LP}_{lm}$$ exists,
+as long as $$\beta$$ stays in the allowed range.
+The cut-offs of the secondary modes for a given $$\ell$$
+are found as the $$m$$th roots of $$J_{\ell-1}(V_{\ell m}) = 0$$.
+In the above figure, they are $$V_{01} = 0$$, $$V_{11} = 2.405$$, and $$V_{02} = V_{21} = 3.832$$.
All differential equations have been linear,
so a linear combination of these solutions is also valid.
diff --git a/source/know/concept/stochastic-process/index.md b/source/know/concept/stochastic-process/index.md
index dc3d30e..68f028d 100644
--- a/source/know/concept/stochastic-process/index.md
+++ b/source/know/concept/stochastic-process/index.md
@@ -9,24 +9,24 @@ categories:
layout: "concept"
---
-A **stochastic process** $X_t$ is a time-indexed
+A **stochastic process** $$X_t$$ is a time-indexed
[random variable](/know/concept/random-variable/),
-$\{ X_t : t > 0 \}$, i.e. a set of (usually correlated)
-random variables, each labelled with a unique timestamp $t$.
+$$\{ X_t : t > 0 \}$$, i.e. a set of (usually correlated)
+random variables, each labelled with a unique timestamp $$t$$.
Whereas "ordinary" random variables are defined on
-a probability space $(\Omega, \mathcal{F}, P)$,
+a probability space $$(\Omega, \mathcal{F}, P)$$,
stochastic process are defined on
-a **filtered probability space** $(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$.
-As before, $\Omega$ is the sample space,
-$\mathcal{F}$ is the event space,
-and $P$ is the probability measure.
+a **filtered probability space** $$(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$$.
+As before, $$\Omega$$ is the sample space,
+$$\mathcal{F}$$ is the event space,
+and $$P$$ is the probability measure.
-The **filtration** $\{ \mathcal{F}_t : t \ge 0 \}$
-is a time-indexed set of [$\sigma$-algebras](/know/concept/sigma-algebra/) on $\Omega$,
+The **filtration** $$\{ \mathcal{F}_t : t \ge 0 \}$$
+is a time-indexed set of [$$\sigma$$-algebras](/know/concept/sigma-algebra/) on $$\Omega$$,
which contains at least all the information generated
-by $X_t$ up to the current time $t$,
-and is a subset of $\mathcal{F}_t$:
+by $$X_t$$ up to the current time $$t$$,
+and is a subset of $$\mathcal{F}_t$$:
$$\begin{aligned}
\mathcal{F}
@@ -34,18 +34,18 @@ $$\begin{aligned}
\supseteq \sigma(X_s : 0 \le s \le t)
\end{aligned}$$
-In other words, $\mathcal{F}_t$ is the "accumulated" $\sigma$-algebra
-of all information extractable from $X_t$,
-and hence grows with time: $\mathcal{F}_s \subseteq \mathcal{F}_t$ for $s < t$.
-Given $\mathcal{F}_t$, all values $X_s$ for $s \le t$ can be computed,
-i.e. if you know $\mathcal{F}_t$, then the present and past of $X_t$ can be reconstructed.
-
-Given any filtration $\mathcal{H}_t$, a stochastic process $X_t$
-is said to be *"$\mathcal{H}_t$-adapted"*
-if $X_t$'s own filtration $\sigma(X_s : 0 \le s \le t) \subseteq \mathcal{H}_t$,
-meaning $\mathcal{H}_t$ contains enough information
-to determine the current and past values of $X_t$.
-Clearly, $X_t$ is always adapted to its own filtration.
+In other words, $$\mathcal{F}_t$$ is the "accumulated" $$\sigma$$-algebra
+of all information extractable from $$X_t$$,
+and hence grows with time: $$\mathcal{F}_s \subseteq \mathcal{F}_t$$ for $$s < t$$.
+Given $$\mathcal{F}_t$$, all values $$X_s$$ for $$s \le t$$ can be computed,
+i.e. if you know $$\mathcal{F}_t$$, then the present and past of $$X_t$$ can be reconstructed.
+
+Given any filtration $$\mathcal{H}_t$$, a stochastic process $$X_t$$
+is said to be *"$$\mathcal{H}_t$$-adapted"*
+if $$X_t$$'s own filtration $$\sigma(X_s : 0 \le s \le t) \subseteq \mathcal{H}_t$$,
+meaning $$\mathcal{H}_t$$ contains enough information
+to determine the current and past values of $$X_t$$.
+Clearly, $$X_t$$ is always adapted to its own filtration.
Filtration and their adaptations are very useful
for working with stochastic processes,
diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md
index e3b4526..3a02a83 100644
--- a/source/know/concept/stokes-law/index.md
+++ b/source/know/concept/stokes-law/index.md
@@ -9,17 +9,17 @@ categories:
layout: "concept"
---
-**Stokes' law** describes the size of the drag force $D$
-at low [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \ll 1$
-experienced by a spherical object in a steady, uniform flow at velocity $U$.
+**Stokes' law** describes the size of the drag force $$D$$
+at low [Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re} \ll 1$$
+experienced by a spherical object in a steady, uniform flow at velocity $$U$$.
## Flow field
-Imagine a sphere with radius $a$ sinking in a viscous liquid.
+Imagine a sphere with radius $$a$$ sinking in a viscous liquid.
To model this situation, let us pretend that the sphere is fixed instead,
-and the fluid comes from infinity at velocity $U$ along the $z$-axis,
-flows past the sphere, and continues to infinity at the same $U$.
+and the fluid comes from infinity at velocity $$U$$ along the $$z$$-axis,
+flows past the sphere, and continues to infinity at the same $$U$$.
The Reynolds number is:
$$\begin{aligned}
@@ -27,7 +27,7 @@ $$\begin{aligned}
= \frac{2 a U}{\nu}
\end{aligned}$$
-We assume that $\mathrm{Re} \ll 1$, in which case
+We assume that $$\mathrm{Re} \ll 1$$, in which case
the incompressible [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
are reduced to the **steady Stokes equations**:
@@ -39,10 +39,10 @@ $$\begin{aligned}
= 0
\end{aligned}$$
-The goal is to solve for $p$ and $\va{v}$.
+The goal is to solve for $$p$$ and $$\va{v}$$.
We make the following ansatz in
-[spherical coordinates](/know/concept/spherical-coordinates/) $(r, \theta, \phi)$,
-where $q(r)$, $f(r)$ and $g(r)$ are unknown functions:
+[spherical coordinates](/know/concept/spherical-coordinates/) $$(r, \theta, \phi)$$,
+where $$q(r)$$, $$f(r)$$ and $$g(r)$$ are unknown functions:
$$\begin{gathered}
p
@@ -59,12 +59,12 @@ $$\begin{gathered}
\end{gathered}$$
The fluid hits the sphere head on,
-so the solution is taken to be $\phi$-independent due to symmetry.
-Note that $\theta$ is the angle to the positive $z$-axis,
-which is the direction of $\va{U} = U \vu{e}_z$.
-Moreover, note that $\va{U} \cdot \vu{e}_r = U \cos\theta$
-and $\va{U} \cdot \vu{e}_\theta = - U \sin\theta$,
-where $\vu{e}_r$ and $\vu{e}_\theta$ are basis vectors.
+so the solution is taken to be $$\phi$$-independent due to symmetry.
+Note that $$\theta$$ is the angle to the positive $$z$$-axis,
+which is the direction of $$\va{U} = U \vu{e}_z$$.
+Moreover, note that $$\va{U} \cdot \vu{e}_r = U \cos\theta$$
+and $$\va{U} \cdot \vu{e}_\theta = - U \sin\theta$$,
+where $$\vu{e}_r$$ and $$\vu{e}_\theta$$ are basis vectors.
To begin with, we insert this ansatz into the incompressibility condition,
yielding:
@@ -79,7 +79,7 @@ $$\begin{aligned}
&= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big)
\end{aligned}$$
-The parenthesized expression must be zero for all $r$,
+The parenthesized expression must be zero for all $$r$$,
leading us to the following relation:
$$\begin{aligned}
@@ -98,7 +98,7 @@ $$\begin{aligned}
\end{aligned}$$
This is simply the Laplace equation,
-which is as follows for our ansatz $p(r, \theta)$:
+which is as follows for our ansatz $$p(r, \theta)$$:
$$\begin{aligned}
0
@@ -115,8 +115,8 @@ $$\begin{aligned}
&= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big)
\end{aligned}$$
-Again, the parenthesized expression must be zero for all $r$,
-meaning it is an ODE for $q(r)$,
+Again, the parenthesized expression must be zero for all $$r$$,
+meaning it is an ODE for $$q(r)$$,
whose solution is straightforwardly found to be:
$$\begin{aligned}
@@ -124,7 +124,7 @@ $$\begin{aligned}
= \frac{C_3}{r^2} + C_4 r
\end{aligned}$$
-Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later).
+Where $$C_3$$ and $$C_4$$ are linearity constants ($$C_1$$ and $$C_2$$ appear later).
The pressure is therefore:
$$\begin{aligned}
@@ -132,7 +132,7 @@ $$\begin{aligned}
= \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big)
\end{aligned}$$
-Consequently, its gradient $\nabla p$ in spherical coordinates is as follows:
+Consequently, its gradient $$\nabla p$$ in spherical coordinates is as follows:
$$\begin{aligned}
\nabla p
@@ -140,8 +140,8 @@ $$\begin{aligned}
= \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big)
\end{aligned}$$
-According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$.
-Let us look at the $r$-component of $\nabla^2 \va{v}$:
+According to the Stokes equation, this equals $$\eta \nabla^2 \va{v}$$.
+Let us look at the $$r$$-component of $$\nabla^2 \va{v}$$:
$$\begin{aligned}
(\nabla^2 \va{v})_r
@@ -154,22 +154,22 @@ $$\begin{aligned}
&= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big)
\end{aligned}$$
-Substituting $g$ for the expression we found from incompressibility lets us simplify this:
+Substituting $$g$$ for the expression we found from incompressibility lets us simplify this:
$$\begin{aligned}
\eta (\nabla^2 \va{v})_r
&= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
\end{aligned}$$
-The Stokes equation says that this must be equal to the $r$-component of $\nabla p$:
+The Stokes equation says that this must be equal to the $$r$$-component of $$\nabla p$$:
$$\begin{aligned}
\eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
= \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big)
\end{aligned}$$
-Where we have inserted $\idv{q}{r}$.
-Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$,
+Where we have inserted $$\idv{q}{r}$$.
+Dividing out $$\eta U \cos\theta$$ leaves an ODE for $$f(r)$$,
satisfied by:
$$\begin{aligned}
@@ -178,17 +178,17 @@ $$\begin{aligned}
\end{aligned}$$
Then, thanks to our earlier relation again,
-we know that $g(r)$ is as follows:
+we know that $$g(r)$$ is as follows:
$$\begin{aligned}
g(r)
= C_1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r} + \frac{C_4 r^2}{5}
\end{aligned}$$
-So what about $C_1$, $C_2$, $C_3$ and $C_4$?
-For $r\!\to\!\infty$, we expect that $\va{v}\!\to\!\va{U}$,
-meaning that $f(r)\!\to\!1$ and $g(r)\!\to\!1$.
-This implies that $C_4 = 0$ and $C_1 = 1$, leaving:
+So what about $$C_1$$, $$C_2$$, $$C_3$$ and $$C_4$$?
+For $$r\!\to\!\infty$$, we expect that $$\va{v}\!\to\!\va{U}$$,
+meaning that $$f(r)\!\to\!1$$ and $$g(r)\!\to\!1$$.
+This implies that $$C_4 = 0$$ and $$C_1 = 1$$, leaving:
$$\begin{aligned}
f(r)
@@ -199,10 +199,10 @@ $$\begin{aligned}
\end{aligned}$$
Furthermore, the viscous *no-slip* condition demands
-that $\va{v} = 0$ at the sphere's surface $r = a$, so $f(a) = g(a) = 0$ there.
-Inserting $a$ into $f$ and $g$, setting them to zero,
+that $$\va{v} = 0$$ at the sphere's surface $$r = a$$, so $$f(a) = g(a) = 0$$ there.
+Inserting $$a$$ into $$f$$ and $$g$$, setting them to zero,
and solving the resulting system of equations
-yields $C_2 = a^3 / 2$ and $C_3 = -3 a / 2$.
+yields $$C_2 = a^3 / 2$$ and $$C_3 = -3 a / 2$$.
Therefore the full solution is:
$$\begin{gathered}
@@ -225,8 +225,8 @@ $$\begin{gathered}
From the definition of [viscosity](/know/concept/viscosity/),
we know that there must be shear stresses at the sphere surface,
-described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$.
-The drag force $\va{D}$ on the surface is:
+described by the fluid's [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{\sigma}$$.
+The drag force $$\va{D}$$ on the surface is:
$$\begin{aligned}
\va{D}
@@ -234,7 +234,7 @@ $$\begin{aligned}
= \int_0^{2\pi} \!\!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:a^2 \sin\theta \dd{\theta} \dd{\phi}
\end{aligned}$$
-Where $\vu{e}_r$ is the sphere's surface normal vector.
+Where $$\vu{e}_r$$ is the sphere's surface normal vector.
The integrand can be expanded as follows:
$$\begin{aligned}
@@ -242,7 +242,7 @@ $$\begin{aligned}
= \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r}
\end{aligned}$$
-To calculate this, we start by taking the gradient of the velocity field $\va{v}$:
+To calculate this, we start by taking the gradient of the velocity field $$\va{v}$$:
$$\begin{aligned}
\nabla\va{v}
@@ -253,7 +253,7 @@ $$\begin{aligned}
+ \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big)
\end{aligned}$$
-Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$:
+Some of these terms are necessary to calculate the stress elements $$\sigma_{rr}$$ and $$\sigma_{\theta r}$$:
$$\begin{aligned}
\sigma_{rr}
@@ -264,6 +264,7 @@ $$\begin{aligned}
\\
&= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big)
\end{aligned}$$
+
$$\begin{aligned}
\sigma_{\theta r}
&= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big)
@@ -276,7 +277,7 @@ $$\begin{aligned}
&= - \frac{3 \eta U a^3}{2 r^4} \sin\theta
\end{aligned}$$
-At the sphere's surface we set $r = a$, so these expressions reduce to the following:
+At the sphere's surface we set $$r = a$$, so these expressions reduce to the following:
$$\begin{aligned}
\sigma_{rr}
@@ -287,7 +288,7 @@ $$\begin{aligned}
\end{aligned}$$
Now we can finally calculate the effective stress on the surface,
-by converting the basis vectors $\vu{e}_r$ and $\vu{e}_\theta$ to Cartesian coordinates:
+by converting the basis vectors $$\vu{e}_r$$ and $$\vu{e}_\theta$$ to Cartesian coordinates:
$$\begin{aligned}
\hat{\sigma} \cdot \vu{e}_r
@@ -301,11 +302,11 @@ $$\begin{aligned}
= \vu{e}_z \frac{3 \eta U}{2 a}
\end{aligned}$$
-Remarkably, the stress at every point on the sphere is purely in the $z$-direction!
+Remarkably, the stress at every point on the sphere is purely in the $$z$$-direction!
This is not entirely unexpected though: symmetry cancels out all other components.
-With this, we can do the integrals for $\va{D}$,
-which reduce to a surface area factor $4 \pi a^2$:
+With this, we can do the integrals for $$\va{D}$$,
+which reduce to a surface area factor $$4 \pi a^2$$:
$$\begin{aligned}
\va{D}
@@ -315,7 +316,7 @@ $$\begin{aligned}
\end{aligned}$$
At last, we arrive at Stokes' law,
-which simply expresses the magnitude of $\va{D}$:
+which simply expresses the magnitude of $$\va{D}$$:
$$\begin{aligned}
\boxed{
@@ -327,8 +328,8 @@ $$\begin{aligned}
To arrive at this result,
we assumed that the sphere was fixed, and the fluid was flowing past it.
We can equally well let the fluid be at rest,
-with the sphere falling through it at $U$.
-The force of gravity then exerts the following force $G$ on it,
+with the sphere falling through it at $$U$$.
+The force of gravity then exerts the following force $$G$$ on it,
subtracting [buoyancy](/know/concept/archimedes-principle/):
$$\begin{aligned}
@@ -336,9 +337,9 @@ $$\begin{aligned}
= \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
\end{aligned}$$
-Where $\rho_s$ and $\rho_f$ are the sphere's and fluid's densities,
-and $g_0$ is the gravitational acceleration.
-Since $D$ acts in the opposite sense of $G$,
+Where $$\rho_s$$ and $$\rho_f$$ are the sphere's and fluid's densities,
+and $$g_0$$ is the gravitational acceleration.
+Since $$D$$ acts in the opposite sense of $$G$$,
after some time, they cancel out:
$$\begin{aligned}
@@ -346,7 +347,7 @@ $$\begin{aligned}
= \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
\end{aligned}$$
-This is an equation for the **terminal velocity** $U_t$,
+This is an equation for the **terminal velocity** $$U_t$$,
which we find to be as follows:
$$\begin{aligned}
@@ -356,7 +357,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-The falling sphere will accelerate until $U_t$,
+The falling sphere will accelerate until $$U_t$$,
and then continue falling at constant speed.
diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md
index 23922d5..75daae3 100644
--- a/source/know/concept/sturm-liouville-theory/index.md
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -21,23 +21,23 @@ of eigenfunctions.
## General operator
Consider the most general form of a second-order linear
-differential operator $\hat{L}$, where $p_0(x)$, $p_1(x)$, and $p_2(x)$
-are real functions of $x \in [a,b]$ which are non-zero for all $x \in ]a, b[$:
+differential operator $$\hat{L}$$, where $$p_0(x)$$, $$p_1(x)$$, and $$p_2(x)$$
+are real functions of $$x \in [a,b]$$ which are non-zero for all $$x \in ]a, b[$$:
$$\begin{aligned}
\hat{L} \{u(x)\} = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x)
\end{aligned}$$
We now define the **adjoint** or **Hermitian** operator
-$\hat{L}^\dagger$ analogously to matrices:
+$$\hat{L}^\dagger$$ analogously to matrices:
$$\begin{aligned}
\inprod{f}{\hat{L} g}
= \inprod{\hat{L}^\dagger f}{g}
\end{aligned}$$
-What is $\hat{L}^\dagger$, given the above definition of $\hat{L}$?
-We start from the inner product $\inprod{f}{\hat{L} g}$:
+What is $$\hat{L}^\dagger$$, given the above definition of $$\hat{L}$$?
+We start from the inner product $$\inprod{f}{\hat{L} g}$$:
$$\begin{aligned}
\inprod{f}{\hat{L} g}
@@ -51,7 +51,7 @@ $$\begin{aligned}
&= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x}
\end{aligned}$$
-We now have an expression for $\hat{L}^\dagger$, but are left with an
+We now have an expression for $$\hat{L}^\dagger$$, but are left with an
annoying boundary term:
$$\begin{aligned}
@@ -60,8 +60,8 @@ $$\begin{aligned}
\end{aligned}$$
To fix this,
-let us demand that $p_1(x) = p_0'(x)$ and that
-$[p_0(f^* g' - (f^*)' g)]_a^b = 0$, leaving:
+let us demand that $$p_1(x) = p_0'(x)$$ and that
+$$[p_0(f^* g' - (f^*)' g)]_a^b = 0$$, leaving:
$$\begin{aligned}
\inprod{f}{\hat{L} g}
@@ -69,8 +69,8 @@ $$\begin{aligned}
= \inprod{\hat{L}^\dagger f}{g}
\end{aligned}$$
-Using the aforementioned restriction $p_1(x) = p_0'(x)$,
-we then take a look at the definition of $\hat{L}^\dagger$:
+Using the aforementioned restriction $$p_1(x) = p_0'(x)$$,
+we then take a look at the definition of $$\hat{L}^\dagger$$:
$$\begin{aligned}
\hat{L}^\dagger \{f\}
@@ -83,7 +83,7 @@ $$\begin{aligned}
&= (p_0 f')' + p_2 f
\end{aligned}$$
-The original operator $\hat{L}$ reduces to the same form,
+The original operator $$\hat{L}$$ reduces to the same form,
so it is **self-adjoint**:
$$\begin{aligned}
@@ -93,19 +93,19 @@ $$\begin{aligned}
= \hat{L}^\dagger \{f\}
\end{aligned}$$
-Consequently, every such second-order linear operator $\hat{L}$ is self-adjoint,
-as long as it satisfies the constraints $p_1(x) = p_0'(x)$ and $[p_0 (f^* g' - (f^*)' g)]_a^b = 0$.
+Consequently, every such second-order linear operator $$\hat{L}$$ is self-adjoint,
+as long as it satisfies the constraints $$p_1(x) = p_0'(x)$$ and $$[p_0 (f^* g' - (f^*)' g)]_a^b = 0$$.
Let us ignore the latter constraint for now (it will return later),
-and focus on the former: what if $\hat{L}$ does not satisfy $p_0' \neq p_1$?
-We multiply it by an unknown $p(x) \neq 0$, and divide by $p_0(x) \neq 0$:
+and focus on the former: what if $$\hat{L}$$ does not satisfy $$p_0' \neq p_1$$?
+We multiply it by an unknown $$p(x) \neq 0$$, and divide by $$p_0(x) \neq 0$$:
$$\begin{aligned}
\frac{p(x)}{p_0(x)} \hat{L} \{u\} = p(x) u'' + p(x) \frac{p_1(x)}{p_0(x)} u' + p(x) \frac{p_2(x)}{p_0(x)} u
\end{aligned}$$
-We now define $q(x)$,
-and demand that the derivative $p'(x)$ of the unknown $p(x)$ satisfies:
+We now define $$q(x)$$,
+and demand that the derivative $$p'(x)$$ of the unknown $$p(x)$$ satisfies:
$$\begin{aligned}
q(x) = p(x) \frac{p_2(x)}{p_0(x)}
@@ -113,7 +113,7 @@ $$\begin{aligned}
p'(x) = p(x) \frac{p_1(x)}{p_0(x)}
\end{aligned}$$
-The latter is a differential equation for $p(x)$, which we solve by integration:
+The latter is a differential equation for $$p(x)$$, which we solve by integration:
$$\begin{gathered}
\frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x}
@@ -128,7 +128,7 @@ $$\begin{gathered}
p(x) = p(a) \exp\!\Big( \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} \Big)
\end{gathered}$$
-Now that we have $p(x)$ and $q(x)$, we can define a new operator $\hat{L}_p$ as follows:
+Now that we have $$p(x)$$ and $$q(x)$$, we can define a new operator $$\hat{L}_p$$ as follows:
$$\begin{aligned}
\hat{L}_p \{u\}
@@ -138,11 +138,11 @@ $$\begin{aligned}
\end{aligned}$$
This is the self-adjoint form from earlier!
-So even if $p_0' \neq p_1$, any second-order linear operator with $p_0(x) \neq 0$
+So even if $$p_0' \neq p_1$$, any second-order linear operator with $$p_0(x) \neq 0$$
can easily be put in self-adjoint form.
-This general form is known as the **Sturm-Liouville operator** $\hat{L}_{SL}$,
-where $p(x)$ and $q(x)$ are non-zero real functions of the variable $x \in [a,b]$:
+This general form is known as the **Sturm-Liouville operator** $$\hat{L}_{SL}$$,
+where $$p(x)$$ and $$q(x)$$ are non-zero real functions of the variable $$x \in [a,b]$$:
$$\begin{aligned}
\boxed{
@@ -156,8 +156,8 @@ $$\begin{aligned}
## Eigenvalue problem
A **Sturm-Liouville problem** (SLP) is analogous to a matrix eigenvalue problem,
-where $w(x)$ is a real weight function, $\lambda$ is the **eigenvalue**,
-and $u(x)$ is the corresponding **eigenfunction**:
+where $$w(x)$$ is a real weight function, $$\lambda$$ is the **eigenvalue**,
+and $$u(x)$$ is the corresponding **eigenfunction**:
$$\begin{aligned}
\boxed{
@@ -165,14 +165,14 @@ $$\begin{aligned}
}
\end{aligned}$$
-Necessarily, $w(x) > 0$ except in isolated points, where $w(x) = 0$ is allowed;
-the point is that any inner product $\Inprod{f}{w g}$ may never be zero due to $w$'s fault.
-Furthermore, the convention is that $u(x)$ cannot be trivially zero.
+Necessarily, $$w(x) > 0$$ except in isolated points, where $$w(x) = 0$$ is allowed;
+the point is that any inner product $$\Inprod{f}{w g}$$ may never be zero due to $$w$$'s fault.
+Furthermore, the convention is that $$u(x)$$ cannot be trivially zero.
-In our derivation of $\hat{L}_{SL}$,
+In our derivation of $$\hat{L}_{SL}$$,
we removed a boundary term to get self-adjointness.
Consequently, to have a valid SLP, the boundary conditions for
-$u(x)$ must be as follows, otherwise the operator cannot be self-adjoint:
+$$u(x)$$ must be as follows, otherwise the operator cannot be self-adjoint:
$$\begin{aligned}
\Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0
@@ -181,16 +181,16 @@ $$\begin{aligned}
There are many boundary conditions (BCs) which satisfy this requirement.
Some notable ones are listed here non-exhaustively:
-+ **Dirichlet BCs**: $u(a) = u(b) = 0$
-+ **Neumann BCs**: $u'(a) = u'(b) = 0$
-+ **Robin BCs**: $\alpha_1 u(a) + \beta_1 u'(a) = \alpha_2 u(b) + \beta_2 u'(b) = 0$ with $\alpha_{1,2}, \beta_{1,2} \in \mathbb{R}$
-+ **Periodic BCs**: $p(a) = p(b)$, $u(a) = u(b)$, and $u'(a) = u'(b)$
-+ **Legendre "BCs"**: $p(a) = p(b) = 0$
++ **Dirichlet BCs**: $$u(a) = u(b) = 0$$
++ **Neumann BCs**: $$u'(a) = u'(b) = 0$$
++ **Robin BCs**: $$\alpha_1 u(a) + \beta_1 u'(a) = \alpha_2 u(b) + \beta_2 u'(b) = 0$$ with $$\alpha_{1,2}, \beta_{1,2} \in \mathbb{R}$$
++ **Periodic BCs**: $$p(a) = p(b)$$, $$u(a) = u(b)$$, and $$u'(a) = u'(b)$$
++ **Legendre "BCs"**: $$p(a) = p(b) = 0$$
Once this requirement is satisfied, Sturm-Liouville theory gives us
-some very useful information about $\lambda$ and $u(x)$.
+some very useful information about $$\lambda$$ and $$u(x)$$.
From the definition of an SLP, we know that, given two arbitrary (and possibly identical)
-eigenfunctions $u_n$ and $u_m$, the following must be satisfied:
+eigenfunctions $$u_n$$ and $$u_m$$, the following must be satisfied:
$$\begin{aligned}
0 = \hat{L}_{SL}\{u_n\} + \lambda_n w u_n = \hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*
@@ -215,7 +215,7 @@ $$\begin{aligned}
&= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
\end{aligned}$$
-The operator $\hat{L}_{SL}$ is self-adjoint by definition,
+The operator $$\hat{L}_{SL}$$ is self-adjoint by definition,
so the left-hand side vanishes, leaving us with:
$$\begin{aligned}
@@ -223,20 +223,20 @@ $$\begin{aligned}
&= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
\end{aligned}$$
-When $m = n$, the inner product $\Inprod{u_n}{w u_n}$ is real and positive
-(assuming $u_n$ is not trivially zero, in which case it would be disqualified anyway).
-In this case we thus know that $\lambda_n^* = \lambda_n$,
-i.e. the eigenvalue $\lambda_n$ is real for any $n$.
+When $$m = n$$, the inner product $$\Inprod{u_n}{w u_n}$$ is real and positive
+(assuming $$u_n$$ is not trivially zero, in which case it would be disqualified anyway).
+In this case we thus know that $$\lambda_n^* = \lambda_n$$,
+i.e. the eigenvalue $$\lambda_n$$ is real for any $$n$$.
-When $m \neq n$, then $\lambda_m^* - \lambda_n$ may or may not be zero,
+When $$m \neq n$$, then $$\lambda_m^* - \lambda_n$$ may or may not be zero,
depending on the degeneracy. If there is no degeneracy, we
-see that $\Inprod{u_m}{w u_n} = 0$, i.e. the eigenfunctions are orthogonal.
+see that $$\Inprod{u_m}{w u_n} = 0$$, i.e. the eigenfunctions are orthogonal.
In case of degeneracy, manual orthogonalization is needed, but as it turns out,
this is guaranteed to be doable, using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method/).
-In conclusion, **a Sturm-Liouville problem has real eigenvalues $\lambda$,
-and all the corresponding eigenfunctions $u(x)$ are mutually orthogonal**:
+In conclusion, **a Sturm-Liouville problem has real eigenvalues $$\lambda$$,
+and all the corresponding eigenfunctions $$u(x)$$ are mutually orthogonal**:
$$\begin{aligned}
\boxed{
@@ -252,16 +252,16 @@ so it is always worth checking whether you are dealing with an SLP.
Another useful fact of SLPs is that they always
have an infinite number of discrete eigenvalues.
-Furthermore, the eigenvalues always ascend to $+\infty$;
-in other words, there always exists a *lowest* eigenvalue $\lambda_0 > -\infty$,
+Furthermore, the eigenvalues always ascend to $$+\infty$$;
+in other words, there always exists a *lowest* eigenvalue $$\lambda_0 > -\infty$$,
known as the **ground state**.
## Completeness
-Not only are the eigenfunctions $u_n(x)$ of an SLP orthogonal, they
-also form a **complete basis**, meaning that any well-behaved function $f(x)$ can be
-expanded as a **generalized Fourier series** with coefficients $a_n$:
+Not only are the eigenfunctions $$u_n(x)$$ of an SLP orthogonal, they
+also form a **complete basis**, meaning that any well-behaved function $$f(x)$$ can be
+expanded as a **generalized Fourier series** with coefficients $$a_n$$:
$$\begin{aligned}
\boxed{
@@ -271,12 +271,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-This series will converge significantly faster if $f(x)$
-satisfies the same BCs as $u_n(x)$. In that case the
-expansion will even be valid for the inclusive interval $x \in [a, b]$.
+This series will converge significantly faster if $$f(x)$$
+satisfies the same BCs as $$u_n(x)$$. In that case the
+expansion will even be valid for the inclusive interval $$x \in [a, b]$$.
-To find an expression for the coefficients $a_n$,
-we multiply the above generalized Fourier series by $w(x) u_m^*(x)$ for an arbitrary $m$:
+To find an expression for the coefficients $$a_n$$,
+we multiply the above generalized Fourier series by $$w(x) u_m^*(x)$$ for an arbitrary $$m$$:
$$\begin{aligned}
f(x) w(x) u_m^*(x)
@@ -303,9 +303,9 @@ $$\begin{aligned}
= a_m A_m
\end{aligned}$$
-After isolating this for $a_n$, we see that
+After isolating this for $$a_n$$, we see that
the coefficients are given by the projection of the target
-function $f(x)$ onto the normalized eigenfunctions $u_n(x) / A_n$:
+function $$f(x)$$ onto the normalized eigenfunctions $$u_n(x) / A_n$$:
$$\begin{aligned}
\boxed{
@@ -317,7 +317,7 @@ $$\begin{aligned}
As a final remark, we can see something interesting
by rearranging the generalized Fourier series
-after inserting the expression for $a_n$:
+after inserting the expression for $$a_n$$:
$$\begin{aligned}
f(x)
@@ -328,7 +328,7 @@ $$\begin{aligned}
\end{aligned}$$
Upon closer inspection, the parenthesized summation
-must be the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(x)$
+must be the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$
for the integral to work out.
This is in fact the underlying requirement for completeness:
diff --git a/source/know/concept/superdense-coding/index.md b/source/know/concept/superdense-coding/index.md
index 5c1e4ca..ba6e898 100644
--- a/source/know/concept/superdense-coding/index.md
+++ b/source/know/concept/superdense-coding/index.md
@@ -20,48 +20,78 @@ She could send a qubit, which has a larger state space than a classical bit,
but it can only be measured once, thereby yielding only one bit of data.
However, they are already sharing an entangled pair of qubits
-in the [Bell state](/know/concept/bell-state/) $\ket{\Phi^{+}}_{AB}$,
-where $A$ and $B$ are qubits belonging to Alice and Bob, respectively.
+in the [Bell state](/know/concept/bell-state/) $$\ket{\Phi^{+}}_{AB}$$,
+where $$A$$ and $$B$$ are qubits belonging to Alice and Bob, respectively.
-Based on the values of the two classical bits $(a_1, a_2)$,
-Alice performs the following operations on her side $A$
+Based on the values of the two classical bits $$(a_1, a_2)$$,
+Alice performs the following operations on her side $$A$$
of the Bell state:
-This gate is reversible, because $A$ and $B$ are preserved,
-and are all you need to reconstruct to $C$.
+This gate is reversible, because $$A$$ and $$B$$ are preserved,
+and are all you need to reconstruct to $$C$$.
Moreover, this gate is universal,
because we can make a NAND gate from it:
@@ -47,8 +47,8 @@ Note that NOT is a special case of NAND:
Using these, we can, as an example, make an OR gate
from three Toffoli gates,
-thanks to the fact that $A \lor B = \neg (\neg A \land \neg B)$,
-i.e. OR is NAND of NOT $A$ and NOT $B$:
+thanks to the fact that $$A \lor B = \neg (\neg A \land \neg B)$$,
+i.e. OR is NAND of NOT $$A$$ and NOT $$B$$:
@@ -57,7 +57,7 @@ i.e. OR is NAND of NOT $A$ and NOT $B$:
Thanks to its reversibility and universality,
the Toffoli gate is interesting for quantum computing.
Its [quantum gate](/know/concept/quantum-gate/) form is often called **CCNOT**.
-In the basis $\Ket{A} \Ket{B} \Ket{C}$, its matrix is:
+In the basis $$\Ket{A} \Ket{B} \Ket{C}$$, its matrix is:
$$\begin{aligned}
\boxed{
@@ -75,7 +75,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-If we apply this gate to an arbitrary three-qubit state $\Ket{\psi}$,
+If we apply this gate to an arbitrary three-qubit state $$\Ket{\psi}$$,
it swaps the last two coefficients:
$$\begin{aligned}
diff --git a/source/know/concept/two-fluid-equations/index.md b/source/know/concept/two-fluid-equations/index.md
index 425d50d..e224e3e 100644
--- a/source/know/concept/two-fluid-equations/index.md
+++ b/source/know/concept/two-fluid-equations/index.md
@@ -11,13 +11,13 @@ layout: "concept"
The **two-fluid model** describes a plasma as two separate but overlapping fluids,
one for ions and one for electrons.
Instead of tracking individual particles,
-it gives the dynamics of fluid elements $\dd{V}$ (i.e. small "blobs").
+it gives the dynamics of fluid elements $$\dd{V}$$ (i.e. small "blobs").
These blobs are assumed to be much larger than
the [Debye length](/know/concept/debye-length/),
such that electromagnetic interactions between nearby blobs can be ignored.
-From Newton's second law, we know that the velocity $\vb{v}$
-of a particle with mass $m$ and charge $q$ is as follows,
+From Newton's second law, we know that the velocity $$\vb{v}$$
+of a particle with mass $$m$$ and charge $$q$$ is as follows,
when subjected only to the [Lorentz force](/know/concept/lorentz-force/):
$$\begin{aligned}
@@ -27,19 +27,19 @@ $$\begin{aligned}
From here, the derivation is similar to that of the
[Navier-Stokes equations](/know/concept/navier-stokes-equations/).
-We replace $\idv{}{t}$ with a
-[material derivative](/know/concept/material-derivative/) $\mathrm{D}/\mathrm{D}t$,
-and define $\vb{u}$ as the blob's center-of-mass velocity:
+We replace $$\idv{}{t}$$ with a
+[material derivative](/know/concept/material-derivative/) $$\mathrm{D}/\mathrm{D}t$$,
+and define $$\vb{u}$$ as the blob's center-of-mass velocity:
$$\begin{aligned}
m n \frac{\mathrm{D} \vb{u}}{\mathrm{D} t}
= q n (\vb{E} + \vb{u} \cross \vb{B})
\end{aligned}$$
-Where we have multiplied by the number density $n$ of the particles.
+Where we have multiplied by the number density $$n$$ of the particles.
Due to particle collisions in the fluid,
stresses become important. Therefore, we include
-the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{P}$,
+the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $$\hat{P}$$,
leading to the following two equations:
$$\begin{aligned}
@@ -50,7 +50,7 @@ $$\begin{aligned}
&= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) + \nabla \cdot \hat{P}_e{}^\top
\end{aligned}$$
-Where the subscripts $i$ and $e$ refer to ions and electrons, respectively.
+Where the subscripts $$i$$ and $$e$$ refer to ions and electrons, respectively.
Finally, we also account for momentum transfer between ions and electrons
due to [Rutherford scattering](/know/concept/rutherford-scattering/),
leading to these **two-fluid momentum equations**:
@@ -67,12 +67,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-Where $f_{ie}$ is the mean frequency at which an ion collides with electrons,
-and vice versa for $f_{ei}$.
+Where $$f_{ie}$$ is the mean frequency at which an ion collides with electrons,
+and vice versa for $$f_{ei}$$.
For simplicity, we assume that the plasma is isotropic
and that shear stresses are negligible,
in which case the stress term can be replaced
-by the gradient $- \nabla p$ of a scalar pressure $p$:
+by the gradient $$- \nabla p$$ of a scalar pressure $$p$$:
$$\begin{aligned}
m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
@@ -83,8 +83,8 @@ $$\begin{aligned}
\end{aligned}$$
Next, we demand that matter is conserved.
-In other words, the rate at which particles enter/leave a volume $V$
-must be equal to the flux through the enclosing surface $S$:
+In other words, the rate at which particles enter/leave a volume $$V$$
+must be equal to the flux through the enclosing surface $$S$$:
$$\begin{aligned}
0
@@ -93,7 +93,7 @@ $$\begin{aligned}
\end{aligned}$$
Where we have used the divergence theorem.
-Since $V$ is arbitrary, we can remove the integrals,
+Since $$V$$ is arbitrary, we can remove the integrals,
leading to the following **continuity equations**:
$$\begin{aligned}
@@ -107,7 +107,7 @@ $$\begin{aligned}
\end{aligned}$$
These are 8 equations (2 scalar continuity, 2 vector momentum),
-but 16 unknowns $\vb{u}_i$, $\vb{u}_e$, $\vb{E}$, $\vb{B}$, $n_i$, $n_e$, $p_i$ and $p_e$.
+but 16 unknowns $$\vb{u}_i$$, $$\vb{u}_e$$, $$\vb{E}$$, $$\vb{B}$$, $$n_i$$, $$n_e$$, $$p_i$$ and $$p_e$$.
We would like to close this system, so we need 8 more.
An obvious choice is [Maxwell's equations](/know/concept/maxwells-equations/),
in particular Faraday's and Ampère's law
@@ -121,7 +121,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-Now we have 14 equations, so we need 2 more, for the pressures $p_i$ and $p_e$.
+Now we have 14 equations, so we need 2 more, for the pressures $$p_i$$ and $$p_e$$.
This turns out to be the thermodynamic **equation of state**:
for quasistatic, reversible, adiabatic compression
of a gas with constant heat capacity (i.e. a *calorically perfect* gas),
@@ -135,14 +135,14 @@ $$\begin{aligned}
= \frac{N + 2}{N}
\end{aligned}$$
-Where $\gamma$ is the *heat capacity ratio*,
-and can be calculated from the number of degrees of freedom $N$
+Where $$\gamma$$ is the *heat capacity ratio*,
+and can be calculated from the number of degrees of freedom $$N$$
of each particle in the gas.
-In a fully ionized plasma, $N = 3$.
+In a fully ionized plasma, $$N = 3$$.
-The density $n \propto 1/V$,
-so since $p V^\gamma$ is constant in time,
-for some constant $C$:
+The density $$n \propto 1/V$$,
+so since $$p V^\gamma$$ is constant in time,
+for some constant $$C$$:
$$\begin{aligned}
\frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p}{n^\gamma} \Big) = 0
@@ -163,8 +163,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Note that from the relation $p = C n^\gamma$,
-we can calculate the $\nabla p$ term in the momentum equation,
+Note that from the relation $$p = C n^\gamma$$,
+we can calculate the $$\nabla p$$ term in the momentum equation,
using simple differentiation and the ideal gas law:
$$\begin{aligned}
@@ -177,13 +177,13 @@ $$\begin{aligned}
\end{aligned}$$
Note that the ideal gas law was not used immediately,
-to allow for $\gamma \neq 1$.
+to allow for $$\gamma \neq 1$$.
## Fluid drifts
The momentum equations reduce to the following
-if we assume the flow is steady $\ipdv{\vb{u}}{t} = 0$,
+if we assume the flow is steady $$\ipdv{\vb{u}}{t} = 0$$,
and neglect electron-ion momentum transfer on the right:
$$\begin{aligned}
@@ -194,9 +194,9 @@ $$\begin{aligned}
&\approx q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e
\end{aligned}$$
-We take the cross product with $\vb{B}$,
-which leaves only the component $\vb{u}_\perp$ of $\vb{u}$
-perpendicular to $\vb{B}$ in the Lorentz term:
+We take the cross product with $$\vb{B}$$,
+which leaves only the component $$\vb{u}_\perp$$ of $$\vb{u}$$
+perpendicular to $$\vb{B}$$ in the Lorentz term:
$$\begin{aligned}
0
@@ -205,9 +205,9 @@ $$\begin{aligned}
&= q n (\vb{E} \cross \vb{B} - \vb{u}_\perp B^2) - \nabla p \cross \vb{B} - m n \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}
\end{aligned}$$
-Isolating for $\vb{u}_\perp$ tells us
-that the fluids drifts perpendicularly to $\vb{B}$,
-with velocity $\vb{u}_\perp$:
+Isolating for $$\vb{u}_\perp$$ tells us
+that the fluids drifts perpendicularly to $$\vb{B}$$,
+with velocity $$\vb{u}_\perp$$:
$$\begin{aligned}
\vb{u}_\perp
@@ -216,11 +216,11 @@ $$\begin{aligned}
\end{aligned}$$
The last term is often neglected,
-which turns out to be a valid approximation if $\vb{E} = 0$,
-or if $\vb{E}$ is parallel to $\nabla p$.
-The first term is the familiar $\vb{E} \cross \vb{B}$ drift $\vb{v}_E$
+which turns out to be a valid approximation if $$\vb{E} = 0$$,
+or if $$\vb{E}$$ is parallel to $$\nabla p$$.
+The first term is the familiar $$\vb{E} \cross \vb{B}$$ drift $$\vb{v}_E$$
from [guiding center theory](/know/concept/guiding-center-theory/),
-and the second term is called the **diamagnetic drift** $\vb{v}_D$:
+and the second term is called the **diamagnetic drift** $$\vb{v}_D$$:
$$\begin{aligned}
\boxed{
@@ -236,9 +236,9 @@ $$\begin{aligned}
It is called *diamagnetic* because
it creates a current that induces
-a magnetic field opposite to the original $\vb{B}$.
-In a quasi-neutral plasma $q_e n_e = - q_i n_i$,
-the current density $\vb{J}$ is given by:
+a magnetic field opposite to the original $$\vb{B}$$.
+In a quasi-neutral plasma $$q_e n_e = - q_i n_i$$,
+the current density $$\vb{J}$$ is given by:
$$\begin{aligned}
\vb{J}
@@ -247,7 +247,7 @@ $$\begin{aligned}
= \frac{\vb{B} \cross \nabla (p_i + p_e)}{B^2}
\end{aligned}$$
-Using the ideal gas law $p = k_B T n$,
+Using the ideal gas law $$p = k_B T n$$,
this can be rewritten as follows:
$$\begin{aligned}
@@ -255,7 +255,7 @@ $$\begin{aligned}
= k_B \frac{\vb{B} \cross \nabla (T_i n_i + T_e n_e)}{B^2}
\end{aligned}$$
-Curiously, $\vb{v}_D$ does not involve any net movement of particles,
+Curiously, $$\vb{v}_D$$ does not involve any net movement of particles,
because a pressure gradient does not necessarily cause particles to move.
Instead, there is a higher density of gyration paths
in the high-pressure region,
diff --git a/source/know/concept/viscosity/index.md b/source/know/concept/viscosity/index.md
index 829b186..31d5336 100644
--- a/source/know/concept/viscosity/index.md
+++ b/source/know/concept/viscosity/index.md
@@ -15,19 +15,19 @@ when one part of the fluid moves, it "drags"
neighbouring parts by an amount proportional to the viscosity.
Imagine a liquid in a canal,
-flowing in the $x$-direction at a velocity $v(z)$
-as a function of depth $z$.
+flowing in the $$x$$-direction at a velocity $$v(z)$$
+as a function of depth $$z$$.
Due to the liquid's viscosity,
-its molecules are "stuck" to the bottom of the canal $z = 0$,
-such that it is stationary there $v(0) = 0$.
-However, at the surface $z = z_s$, there is a flow at $v(z_s) = v_s$.
+its molecules are "stuck" to the bottom of the canal $$z = 0$$,
+such that it is stationary there $$v(0) = 0$$.
+However, at the surface $$z = z_s$$, there is a flow at $$v(z_s) = v_s$$.
-This difference in $v$ means that there is a velocity gradient across $z$.
+This difference in $$v$$ means that there is a velocity gradient across $$z$$.
Each infinitesimal layer of the liquid
is dragging on the layers above and below it,
-meaning there is a nonzero shear stress $\sigma_{xz}$
+meaning there is a nonzero shear stress $$\sigma_{xz}$$
(see [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/)).
-Formally, the **dynamic viscosity** $\eta$ is defined as follows:
+Formally, the **dynamic viscosity** $$\eta$$ is defined as follows:
$$\begin{aligned}
\boxed{
@@ -39,16 +39,16 @@ $$\begin{aligned}
This is **Newton's law of viscosity**,
and fluids obeying it are known as **Newtonian**.
In a Newtonian fluid *at rest*, there are no such shear stresses,
-and the Cauchy stress tensor $\hat{\sigma}$ is diagonal:
+and the Cauchy stress tensor $$\hat{\sigma}$$ is diagonal:
$$\begin{aligned}
\sigma_{ij} = - p \delta_{ij}
\end{aligned}$$
-Where $p$ is the pressure, and $\delta_{ij}$ is the Kronecker delta.
-If the fluid flows according to a velocity field $\va{v}$,
-then a more general definition of $\eta$ is as follows,
-in index notation with $\nabla_i \!=\! \ipdv{}{x_i}$:
+Where $$p$$ is the pressure, and $$\delta_{ij}$$ is the Kronecker delta.
+If the fluid flows according to a velocity field $$\va{v}$$,
+then a more general definition of $$\eta$$ is as follows,
+in index notation with $$\nabla_i \!=\! \ipdv{}{x_i}$$:
$$\begin{aligned}
\boxed{
@@ -57,12 +57,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-The double term $\nabla_i v_j + \nabla_j v_i$ comes from the fact that
+The double term $$\nabla_i v_j + \nabla_j v_i$$ comes from the fact that
the stress tensor of a Newtonian fluid is always symmetric;
-this definition of $\sigma_{ij}$ enforces that.
+this definition of $$\sigma_{ij}$$ enforces that.
-Another quantity is the **kinematic viscosity** $\nu$,
-which is simply $\eta$ divided by the density $\rho$:
+Another quantity is the **kinematic viscosity** $$\nu$$,
+which is simply $$\eta$$ divided by the density $$\rho$$:
$$\begin{aligned}
\boxed{
@@ -72,7 +72,7 @@ $$\begin{aligned}
\end{aligned}$$
With this, Newton's law of viscosity is written
-using the momentum density $P = \rho v$:
+using the momentum density $$P = \rho v$$:
$$\begin{aligned}
\sigma_{xz}
@@ -80,10 +80,10 @@ $$\begin{aligned}
\end{aligned}$$
Because momentum is "more fundamental" than velocity,
-is $\nu$ often more useful than $\eta$.
+is $$\nu$$ often more useful than $$\eta$$.
However, this comes at the cost of our intuition:
-for example, as you would expect, $\eta_\mathrm{water} > \eta_\mathrm{air}$,
-but you may be surprised that $\nu_\mathrm{water} < \nu_\mathrm{air}$.
+for example, as you would expect, $$\eta_\mathrm{water} > \eta_\mathrm{air}$$,
+but you may be surprised that $$\nu_\mathrm{water} < \nu_\mathrm{air}$$.
Since air is less dense, it is easier to set in motion,
hence we expect it to be less viscous than water,
but in fact air's molecules are stickier than water's.
diff --git a/source/know/concept/von-neumann-extractor/index.md b/source/know/concept/von-neumann-extractor/index.md
index 59566f1..3c54196 100644
--- a/source/know/concept/von-neumann-extractor/index.md
+++ b/source/know/concept/von-neumann-extractor/index.md
@@ -14,20 +14,20 @@ it extracts the entropy, and outputs a "perfectly random" stream.
As input, the Von Neumann extractor expects
a stream of independent (uncorrelated) bits,
i.e. the result of a [Bernoulli process](/know/concept/binomial-distribution/),
-where each bit is $0$ with probability $p$,
-and $1$ with probability $1 \!-\! p$.
-Crucially, $p$ does not need to be $1/2$;
+where each bit is $$0$$ with probability $$p$$,
+and $$1$$ with probability $$1 \!-\! p$$.
+Crucially, $$p$$ does not need to be $$1/2$$;
there may be a bias.
-The extractor will output a uniformly random stream with $p = 1/2$.
-Given input bits $a_1, a_2, ...$, it achieves this
-by looking at the bits in pairs $(a_1, a_2)$, $(a_3, a_4)$, etc.
+The extractor will output a uniformly random stream with $$p = 1/2$$.
+Given input bits $$a_1, a_2, ...$$, it achieves this
+by looking at the bits in pairs $$(a_1, a_2)$$, $$(a_3, a_4)$$, etc.
Then:
-+ If $a_n = a_{n+1}$, it discards both bits.
-+ If $a_n \neq a_{n+1}$, it keeps the first bit $a_n$, and discards $a_{n+1}$.
++ If $$a_n = a_{n+1}$$, it discards both bits.
++ If $$a_n \neq a_{n+1}$$, it keeps the first bit $$a_n$$, and discards $$a_{n+1}$$.
-Evidently, the first case $a_n = a_{n+1}$ occurs with the following probabilities:
+Evidently, the first case $$a_n = a_{n+1}$$ occurs with the following probabilities:
$$\begin{aligned}
P(0, 0)
@@ -37,7 +37,7 @@ $$\begin{aligned}
= (1 - p)^2
\end{aligned}$$
-Meanwhile, the second case $a_n \neq a_{n+1}$ occurs with probabilities given by:
+Meanwhile, the second case $$a_n \neq a_{n+1}$$ occurs with probabilities given by:
$$\begin{aligned}
P(0, 1)
@@ -47,18 +47,18 @@ $$\begin{aligned}
= (p - 1) p
\end{aligned}$$
-Crucially, they are equal; $P(0, 1) = P(1, 0)$.
-Therefore, if the extractor encounters an input pair satisfying $a_n \neq a_{n+1}$,
-the first bit $a_n$ is $0$ or $1$ with a 50-50 probability,
-regardless of $p$.
+Crucially, they are equal; $$P(0, 1) = P(1, 0)$$.
+Therefore, if the extractor encounters an input pair satisfying $$a_n \neq a_{n+1}$$,
+the first bit $$a_n$$ is $$0$$ or $$1$$ with a 50-50 probability,
+regardless of $$p$$.
Since the extractor only keeps those bits,
its output is guaranteed to be "perfectly random".
Clearly, because it discards many of the bits,
-the output stream will have a length $N_\mathrm{out} < N_\mathrm{in}$.
-The exact value of $N_\mathrm{out}$ is as follows,
-where $P(0, 1) + P(1, 0)$ is the probability that we keep a bit,
-and the factor $1/2$ is due to us discarding half of the pair even in that case:
+the output stream will have a length $$N_\mathrm{out} < N_\mathrm{in}$$.
+The exact value of $$N_\mathrm{out}$$ is as follows,
+where $$P(0, 1) + P(1, 0)$$ is the probability that we keep a bit,
+and the factor $$1/2$$ is due to us discarding half of the pair even in that case:
$$\begin{aligned}
N_\mathrm{out}
diff --git a/source/know/concept/vorticity/index.md b/source/know/concept/vorticity/index.md
index da1205b..f7b5c6e 100644
--- a/source/know/concept/vorticity/index.md
+++ b/source/know/concept/vorticity/index.md
@@ -9,9 +9,9 @@ categories:
layout: "concept"
---
-In fluid mechanics, the **vorticity** $\va{\omega}$
+In fluid mechanics, the **vorticity** $$\va{\omega}$$
is a measure of the local circulation in a fluid.
-It is defined as the curl of the flow velocity field $\va{v}$:
+It is defined as the curl of the flow velocity field $$\va{v}$$:
$$\begin{aligned}
\boxed{
@@ -20,13 +20,13 @@ $$\begin{aligned}
}
\end{aligned}$$
-Just as curves tangent to $\va{v}$ are called *streamlines*,
-curves tangent to $\va{\omega}$ are **vortex lines**,
-which are to be interpreted as the "axes" that $\va{v}$ is circulating around.
+Just as curves tangent to $$\va{v}$$ are called *streamlines*,
+curves tangent to $$\va{\omega}$$ are **vortex lines**,
+which are to be interpreted as the "axes" that $$\va{v}$$ is circulating around.
The vorticity is a local quantity,
-and the corresponding global quantity is the **circulation** $\Gamma$,
-which is defined as the projection of $\va{v}$ onto a close curve $C$.
+and the corresponding global quantity is the **circulation** $$\Gamma$$,
+which is defined as the projection of $$\va{v}$$ onto a close curve $$C$$.
Then, by Stokes' theorem:
$$\begin{aligned}
@@ -41,15 +41,15 @@ $$\begin{aligned}
## Ideal fluids
For an inviscid, incompressible fluid,
-consider the *Bernoulli field* $H$, which is defined as:
+consider the *Bernoulli field* $$H$$, which is defined as:
$$\begin{aligned}
H
\equiv \frac{1}{2} \va{v}^2 + \Phi + \frac{p}{\rho}
\end{aligned}$$
-Where $\Phi$ is the gravitational potential,
-$p$ is the pressure, and $\rho$ is the (constant) density.
+Where $$\Phi$$ is the gravitational potential,
+$$p$$ is the pressure, and $$\rho$$ is the (constant) density.
We then take the gradient of this scalar field:
$$\begin{aligned}
@@ -59,7 +59,7 @@ $$\begin{aligned}
&= \va{v} \cdot (\nabla \va{v}) - \Big( \!-\! \nabla \Phi - \frac{\nabla p}{\rho} \Big)
\end{aligned}$$
-Since $-\nabla \Phi = \va{g}$,
+Since $$-\nabla \Phi = \va{g}$$,
the rightmost term is the right-hand side of
the [Euler equation](/know/concept/euler-equations/).
We substitute the other side of said equation, yielding:
@@ -70,7 +70,7 @@ $$\begin{aligned}
= \va{v} \cdot (\nabla \va{v}) - \pdv{\va{v}}{t} - (\va{v} \cdot \nabla) \va{v}
\end{aligned}$$
-We isolate this equation for $\ipdv{\va{v}}{t}$,
+We isolate this equation for $$\ipdv{\va{v}}{t}$$,
and apply a vector identity to reduce it to the following:
$$\begin{aligned}
@@ -79,8 +79,8 @@ $$\begin{aligned}
= \va{v} \cross (\nabla \cross \va{v}) - \nabla H
\end{aligned}$$
-Here, the definition of the vorticity $\va{\omega}$ is clear to see,
-leading us to an equation of motion for $\va{v}$:
+Here, the definition of the vorticity $$\va{\omega}$$ is clear to see,
+leading us to an equation of motion for $$\va{v}$$:
$$\begin{aligned}
\boxed{
@@ -97,9 +97,9 @@ $$\begin{aligned}
= \nabla \cross (\va{v} \cross \va{\omega}) - \nabla \cross (\nabla H)
\end{aligned}$$
-On the left, we swap $\nabla$ with $\ipdv{}{t}$,
+On the left, we swap $$\nabla$$ with $$\ipdv{}{t}$$,
and on the right, the curl of a gradient is always zero.
-We are thus left with the equation of motion of the vorticity $\va{\omega}$:
+We are thus left with the equation of motion of the vorticity $$\va{\omega}$$:
$$\begin{aligned}
\boxed{
@@ -108,8 +108,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-Let us now return to the equation of motion for $\va{v}$.
-For *steady* flows where $\ipdv{\va{v}}{t} = 0$, in which case
+Let us now return to the equation of motion for $$\va{v}$$.
+For *steady* flows where $$\ipdv{\va{v}}{t} = 0$$, in which case
[Bernoulli's theorem](/know/concept/bernoullis-theorem/) applies,
it reduces to:
@@ -118,13 +118,13 @@ $$\begin{aligned}
= \va{v} \cross \va{\omega}
\end{aligned}$$
-If a fluid has $\va{\omega} = 0$ in some regions, it is known as **irrotational**.
-From this equation, we see that, in that case, $\nabla H = 0$,
-meaning that $H$ is a constant in those regions,
+If a fluid has $$\va{\omega} = 0$$ in some regions, it is known as **irrotational**.
+From this equation, we see that, in that case, $$\nabla H = 0$$,
+meaning that $$H$$ is a constant in those regions,
a fact sometimes referred to as **Bernoulli's stronger theorem**.
-Furthermore, irrotationality $\va{\omega} = 0$
-implies that $\va{v}$ is the gradient of a potential $\Psi$:
+Furthermore, irrotationality $$\va{\omega} = 0$$
+implies that $$\va{v}$$ is the gradient of a potential $$\Psi$$:
$$\begin{aligned}
\va{v}
@@ -140,8 +140,8 @@ $$\begin{aligned}
= \nabla^2 \Psi
\end{aligned}$$
-And second, the main equation of motion for $\va{v}$ states
-that the quantity $H + \ipdv{\Psi}{t}$ is spatially constant
+And second, the main equation of motion for $$\va{v}$$ states
+that the quantity $$H + \ipdv{\Psi}{t}$$ is spatially constant
in the irrotational region:
$$\begin{aligned}
diff --git a/source/know/concept/wetting/index.md b/source/know/concept/wetting/index.md
index db40521..8c51acf 100644
--- a/source/know/concept/wetting/index.md
+++ b/source/know/concept/wetting/index.md
@@ -13,18 +13,18 @@ layout: "concept"
In fluid statics, **wetting** is the ability
of a given liquid to touch a given surface.
When a droplet of the liquid is placed on the surface,
-the **wettability** determines the contact angle $\theta$.
+the **wettability** determines the contact angle $$\theta$$.
-If $\theta = 0$, we have **perfect** or **complete wetting**:
+If $$\theta = 0$$, we have **perfect** or **complete wetting**:
the droplet spreads out over the entire surface.
The other extreme is **dewetting** or **non-wetting**,
-where $\theta = \pi$, such that the droplet "floats" on the surface,
+where $$\theta = \pi$$, such that the droplet "floats" on the surface,
which in the specific case of water is called **hydrophobia**.
-Furthermore, $\theta < \pi/2$ is **high wettability**,
-and $\pi/2 < \theta < \pi$ is **low wettability**.
+Furthermore, $$\theta < \pi/2$$ is **high wettability**,
+and $$\pi/2 < \theta < \pi$$ is **low wettability**.
For a perfectly smooth homogeneous surface,
-$\theta$ is determined by
+$$\theta$$ is determined by
the [Young-Dupré relation](/know/concept/young-dupre-relation/):
$$\begin{aligned}
@@ -40,15 +40,15 @@ There are two options:
either the droplet fills those gaps (a **Wenzel state**),
or it floats over them (a **Cassie-Baxter state**).
-For a Wenzel state, we define the **roughness ratio** $r$
+For a Wenzel state, we define the **roughness ratio** $$r$$
as the relative increase of the surface's area due to its rough structure,
-where $A_{real}$ and $A_{app}$ are the real and apparent areas:
+where $$A_{real}$$ and $$A_{app}$$ are the real and apparent areas:
$$\begin{aligned}
r = \frac{A_{real}}{A_{app}}
\end{aligned}$$
-The net energy cost $E$ of spreading the droplet over the surface is then given by:
+The net energy cost $$E$$ of spreading the droplet over the surface is then given by:
$$\begin{aligned}
E_{sl}
@@ -59,8 +59,8 @@ $$\begin{aligned}
= \alpha_{gl} A_{app} \cos\theta^*
\end{aligned}$$
-Where we have defined the **apparent contact angle** $\theta^*$
-as the correction to $\theta$ to account for the roughness.
+Where we have defined the **apparent contact angle** $$\theta^*$$
+as the correction to $$\theta$$ to account for the roughness.
It is expressed as follows:
$$\begin{aligned}
@@ -71,7 +71,7 @@ $$\begin{aligned}
\end{aligned}$$
For Cassie-Baxter states, where the gaps remain air-filled,
-we define $f$ as the "non-gap" fraction of the apparent surface, such that:
+we define $$f$$ as the "non-gap" fraction of the apparent surface, such that:
$$\begin{aligned}
E
@@ -81,10 +81,10 @@ $$\begin{aligned}
\end{aligned}$$
Note the signs: for the solid-liquid interface,
-we "spend" $\alpha_{sg}$ and "get back" $\alpha_{sl}$,
+we "spend" $$\alpha_{sg}$$ and "get back" $$\alpha_{sl}$$,
while for the gas-liquid interface, we spend nothing,
-but get $\alpha_{gl}$.
-The apparent angle $\theta^*$ is therefore:
+but get $$\alpha_{gl}$$.
+The apparent angle $$\theta^*$$ is therefore:
$$\begin{aligned}
\boxed{
@@ -94,7 +94,7 @@ $$\begin{aligned}
\end{aligned}$$
We generalize this equation to inhomogeneous surfaces
-consisting of two materials with contact angles $\theta_1$ and $\theta_2$.
+consisting of two materials with contact angles $$\theta_1$$ and $$\theta_2$$.
The energy cost of the interface is then given by:
$$\begin{aligned}
@@ -104,7 +104,7 @@ $$\begin{aligned}
&= A \alpha_{gl} \big( f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2 \big)
\end{aligned}$$
-Such that $\theta^*$ for an inhomogeneous surface is given by this equation,
+Such that $$\theta^*$$ for an inhomogeneous surface is given by this equation,
called **Cassie's law**:
$$\begin{aligned}
@@ -117,7 +117,7 @@ $$\begin{aligned}
Note that the materials need not be solids,
for example, if one is air, we recover the previous case for rough surfaces.
Cassie's law can also easily be generalized to three or more materials,
-and to include Wenzel-style roughness ratios $r_1$, $r_2$, etc.
+and to include Wenzel-style roughness ratios $$r_1$$, $$r_2$$, etc.
diff --git a/source/know/concept/wicks-theorem/index.md b/source/know/concept/wicks-theorem/index.md
index 42a1bec..35be8fa 100644
--- a/source/know/concept/wicks-theorem/index.md
+++ b/source/know/concept/wicks-theorem/index.md
@@ -38,9 +38,9 @@ The normal product of three or more operators works in the same way,
but might not be unique depending,
on how many of each type there are.
-Next, the **contraction** of the operators $A$ and $B$
+Next, the **contraction** of the operators $$A$$ and $$B$$
is defined as the vacuum matrix element,
-i.e. the expectation value of $\Ket{0}$:
+i.e. the expectation value of $$\Ket{0}$$:
$$\begin{aligned}
\Expval{A B}_0
@@ -48,8 +48,8 @@ $$\begin{aligned}
\end{aligned}$$
Unsurprisingly, a contraction can only be nonzero if
-$A = \hat{c}_\alpha$ is an annihilation and $B = \hat{c}_\alpha^\dagger$
-a creation for the same state $\alpha$.
+$$A = \hat{c}_\alpha$$ is an annihilation and $$B = \hat{c}_\alpha^\dagger$$
+a creation for the same state $$\alpha$$.
Wick's theorem states:
**any product of second quantization operators can be
@@ -71,7 +71,7 @@ $$\begin{aligned}
\end{aligned}$$
Where the negative signs apply to fermions only.
-We take the normal product with 0 contractions removed ($\underline{ABCD}$),
+We take the normal product with 0 contractions removed ($$\underline{ABCD}$$),
then with 1 contraction removed in every possible way (first two lines),
then with 2 contractions removed in every possible way (last line), and so on.
@@ -94,9 +94,9 @@ $$\begin{aligned}
= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}
\end{aligned}$$
-This anticommutator is known to be $\delta_{\alpha\beta}$,
+This anticommutator is known to be $$\delta_{\alpha\beta}$$,
so we can inconsequentially take
-its inner product with the vacuum state $\Ket{0}$:
+its inner product with the vacuum state $$\Ket{0}$$:
$$\begin{aligned}
\hat{f}_\alpha \hat{f}_\beta^\dagger
@@ -114,8 +114,8 @@ $$\begin{aligned}
= \hat{b}_\beta^\dagger \hat{b}_\alpha + [\hat{b}_\alpha, \hat{b}_\beta^\dagger]
\end{aligned}$$
-This commutator is known to be $\delta_{\alpha\beta}$,
-so we take the inner product with $\Ket{0}$, like before:
+This commutator is known to be $$\delta_{\alpha\beta}$$,
+so we take the inner product with $$\Ket{0}$$, like before:
$$\begin{aligned}
\hat{b}_\alpha \hat{b}_\beta^\dagger
@@ -127,9 +127,9 @@ $$\begin{aligned}
\end{aligned}$$
Which again agrees with Wick's theorem.
-Next, we prove that if it holds for $N$ operators, then it also holds for $N + 1$.
+Next, we prove that if it holds for $$N$$ operators, then it also holds for $$N + 1$$.
To begin with, consider the following statement about right-multiplying
-by an extra $A_{N+1}$, with $s = 1$ for bosons and $s = -1$ for fermions:
+by an extra $$A_{N+1}$$, with $$s = 1$$ for bosons and $$s = -1$$ for fermions:
$$\begin{aligned}
\underline{A_1 ... A_N} A_{N+1}
@@ -137,11 +137,11 @@ $$\begin{aligned}
+ \sum_{n = 1}^N s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
\end{aligned}$$
-If $A_{N + 1}$ is an annihilation operator, then this is trivial:
+If $$A_{N + 1}$$ is an annihilation operator, then this is trivial:
appending it does not break the existing normal order,
-and $\Expval{A_n A_{N+1}}_0 = 0$ for all $A_n$.
+and $$\Expval{A_n A_{N+1}}_0 = 0$$ for all $$A_n$$.
-However, if $A_{N + 1}$ is a creation operator,
+However, if $$A_{N + 1}$$ is a creation operator,
then to restore the normal order,
we move it to the front by swapping,
which introduces a bunch of (anti)commutators:
@@ -155,12 +155,12 @@ $$\begin{aligned}
+ \sum_{n} s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
\end{aligned}$$
-Where $\{[]\}$ is the anticommutator or commutator,
+Where $$\{[]\}$$ is the anticommutator or commutator,
respectively for fermions or bosons.
-If we take Wick's theorem for $N$ operators $A_1 ... A_N$,
-and right-multiply it by $A_{N + 1}$,
-then each term will contain a product of the form $\underline{A_{v} ... A_{w}} A_{N+1}$.
+If we take Wick's theorem for $$N$$ operators $$A_1 ... A_N$$,
+and right-multiply it by $$A_{N + 1}$$,
+then each term will contain a product of the form $$\underline{A_{v} ... A_{w}} A_{N+1}$$.
Using the relation that we just proved,
each such product can be rewritten as follows:
@@ -171,15 +171,15 @@ $$\begin{aligned}
\end{aligned}$$
Inserting this back into Wick's theorem,
-we get new terms with contractions of $A_{N+1}$.
+we get new terms with contractions of $$A_{N+1}$$.
After a lot of rearranging,
-the result turns out to just be Wick's theorem for $N\!+\!1$ operators.
+the result turns out to just be Wick's theorem for $$N\!+\!1$$ operators.
Therefore,
-if Wick's theorem holds for $N$ operators,
-it also holds for $N\!+\!1$.
+if Wick's theorem holds for $$N$$ operators,
+it also holds for $$N\!+\!1$$.
-We showed that Wick's theorem holds for $N = 2$,
-so, by induction, it holds for all $N \ge 2$.
+We showed that Wick's theorem holds for $$N = 2$$,
+so, by induction, it holds for all $$N \ge 2$$.
diff --git a/source/know/concept/wiener-process/index.md b/source/know/concept/wiener-process/index.md
index 629da20..c6fcf8e 100644
--- a/source/know/concept/wiener-process/index.md
+++ b/source/know/concept/wiener-process/index.md
@@ -14,34 +14,34 @@ that provides a pure mathematical definition
of the physical phenomenon of **Brownian motion**,
and hence is also called *Brownian motion*.
-A Wiener process $B_t$ is defined as any
-stochastic process $\{B_t: t \ge 0\}$ that satisfies:
-
-1. Initial condition $B_0 = 0$.
-2. Each **increment** of $B_t$ is independent of the past:
- given $0 \le s < t \le u < v$, then
- $B_t \!-\! B_s$ and $B_v \!-\! B_u$ are independent random variables.
-3. The increments of $B_t$ are Gaussian with mean $0$
- and variance $h$, where $h$ is the time step,
- such that $B_{t+h} \!-\! B_t \sim \mathcal{N}(0, h)$.
-4. $B_t$ is a continuous function of $t$.
+A Wiener process $$B_t$$ is defined as any
+stochastic process $$\{B_t: t \ge 0\}$$ that satisfies:
+
+1. Initial condition $$B_0 = 0$$.
+2. Each **increment** of $$B_t$$ is independent of the past:
+ given $$0 \le s < t \le u < v$$, then
+ $$B_t \!-\! B_s$$ and $$B_v \!-\! B_u$$ are independent random variables.
+3. The increments of $$B_t$$ are Gaussian with mean $$0$$
+ and variance $$h$$, where $$h$$ is the time step,
+ such that $$B_{t+h} \!-\! B_t \sim \mathcal{N}(0, h)$$.
+4. $$B_t$$ is a continuous function of $$t$$.
There exist stochastic processes that satisfy these requirements,
infinitely many in fact.
In other words, Brownian motion exists,
and can be constructed in various ways.
-Since the variance of an increment is expressed in units of time $t$,
-the physical unit of the Wiener process is the square root of time $\sqrt{t}$.
+Since the variance of an increment is expressed in units of time $$t$$,
+the physical unit of the Wiener process is the square root of time $$\sqrt{t}$$.
Brownian motion is **self-similar**:
-if we define a rescaled $W_t = \sqrt{\alpha} B_{t/\alpha}$ for some $\alpha$,
-then $W_t$ is also a valid Wiener process,
+if we define a rescaled $$W_t = \sqrt{\alpha} B_{t/\alpha}$$ for some $$\alpha$$,
+then $$W_t$$ is also a valid Wiener process,
meaning that there are no fundemental scales.
A consequence of this is that:
-$\mathbf{E}|B_t|^p = \mathbf{E}|\sqrt{t} B_1|^p = t^{p/2} \mathbf{E}|B_1|^p$.
+$$\mathbf{E}|B_t|^p = \mathbf{E}|\sqrt{t} B_1|^p = t^{p/2} \mathbf{E}|B_1|^p$$.
Another consequence is invariance under "time inversion",
-by defining $\sqrt{\alpha} = t$, such that $W_t = t B_{1/t}$.
+by defining $$\sqrt{\alpha} = t$$, such that $$W_t = t B_{1/t}$$.
Despite being continuous by definition,
the Wiener process is not differentiable in general,
@@ -68,13 +68,13 @@ and all increments are independent (so it is a Markov process).
An important question about the Wiener process
is whether it is **recurrent** or **transient**:
given a hypersphere (interval in 1D, circle in 2D, sphere in 3D)
-away from the origin, will $B_t$ visit it after a finite time $\tau\!<\!\infty$?
-It is *recurrent* if yes, i.e. $P(\tau \!<\! \infty) = 1$, or *transient* otherwise.
+away from the origin, will $$B_t$$ visit it after a finite time $$\tau\!<\!\infty$$?
+It is *recurrent* if yes, i.e. $$P(\tau \!<\! \infty) = 1$$, or *transient* otherwise.
The answer to this question turns out to depend on the number of dimenions.
-To demonstrate this, we model the $d$-dimensional Wiener process
-as an [Itō diffusion](/know/concept/ito-calculus/) $X_t$,
-which also allows us to shift the initial condition $X_0$
+To demonstrate this, we model the $$d$$-dimensional Wiener process
+as an [Itō diffusion](/know/concept/ito-calculus/) $$X_t$$,
+which also allows us to shift the initial condition $$X_0$$
(or resume a "paused" process):
$$\begin{aligned}
@@ -82,10 +82,10 @@ $$\begin{aligned}
= X_0 + \int_0^t \dd{B_s}
\end{aligned}$$
-Consider two hyperspheres, the inner with radius $R_i$,
-and the outer with $R_o > R_i$.
-Let the initial condition $|X_0| \in \: ]R_i, R_o[$,
-then we define the stopping times $\tau_i$, $\tau_o$ and $\tau$ like so:
+Consider two hyperspheres, the inner with radius $$R_i$$,
+and the outer with $$R_o > R_i$$.
+Let the initial condition $$|X_0| \in \: ]R_i, R_o[$$,
+then we define the stopping times $$\tau_i$$, $$\tau_o$$ and $$\tau$$ like so:
$$\begin{aligned}
\tau_i
@@ -98,15 +98,15 @@ $$\begin{aligned}
\equiv \min\{\tau_i, \tau_o\}
\end{aligned}$$
-We stop when the inner or outer hypersphere is touched by $X_t$,
+We stop when the inner or outer hypersphere is touched by $$X_t$$,
whichever happens first.
[Dynkin's formula](/know/concept/dynkins-formula/)
-is applicable to this situation, if we define $h(x)$ as follows,
-where the *terminal reward* $\Gamma$ equals $1$ for $|X_\tau| = R_i$,
-and $0$ for $|X_\tau| = R_o$,
-such that $h(X_0)$ equals the probability
-that we touch $R_i$ before $R_o$ for a given $X_0$:
+is applicable to this situation, if we define $$h(x)$$ as follows,
+where the *terminal reward* $$\Gamma$$ equals $$1$$ for $$|X_\tau| = R_i$$,
+and $$0$$ for $$|X_\tau| = R_o$$,
+such that $$h(X_0)$$ equals the probability
+that we touch $$R_i$$ before $$R_o$$ for a given $$X_0$$:
$$\begin{aligned}
h(X_0)
@@ -114,8 +114,8 @@ $$\begin{aligned}
= P\Big[|X_\tau| \!=\! R_i \:\Big|\: X_0\Big]
\end{aligned}$$
-Dynkin's formula then tells us that $h(x)$ is given by the following equation,
-with the boundary conditions $h(R_i) = 1$ and $h(R_o) = 0$:
+Dynkin's formula then tells us that $$h(x)$$ is given by the following equation,
+with the boundary conditions $$h(R_i) = 1$$ and $$h(R_o) = 0$$:
$$\begin{aligned}
0
@@ -124,9 +124,9 @@ $$\begin{aligned}
\end{aligned}$$
Thanks to this problem's spherical symmetry,
-$h$ only depends on the radial coodinate $r$,
-so the Laplacian $\nabla^2$ can be written as follows
-in $d$-dimensional [spherical coordinates](/know/concept/spherical-coordinates/):
+$$h$$ only depends on the radial coodinate $$r$$,
+so the Laplacian $$\nabla^2$$ can be written as follows
+in $$d$$-dimensional [spherical coordinates](/know/concept/spherical-coordinates/):
$$\begin{aligned}
0
@@ -134,8 +134,8 @@ $$\begin{aligned}
= \pdvn{2}{h}{r} + \frac{d - 1}{r} \pdv{h}{r}
\end{aligned}$$
-For $d = 1$, the solution $h_1(r)$ is as follows,
-of which we take the limit for $R_o \to \infty$:
+For $$d = 1$$, the solution $$h_1(r)$$ is as follows,
+of which we take the limit for $$R_o \to \infty$$:
$$\begin{aligned}
h_1(r)
@@ -144,14 +144,14 @@ $$\begin{aligned}
1
\end{aligned}$$
-The outer hypersphere becomes harder to reach for larger $R_o$,
-and for $R_o \to \infty$ we are left with
-the probability of hitting $R_i$ only.
-This turns out to be $1$, so in 1D the Wiener process is recurrent:
+The outer hypersphere becomes harder to reach for larger $$R_o$$,
+and for $$R_o \to \infty$$ we are left with
+the probability of hitting $$R_i$$ only.
+This turns out to be $$1$$, so in 1D the Wiener process is recurrent:
it always comes close to the origin in finite time.
-For $d = 2$, the solution $h_2(r)$ is as follows,
-whose limit turns out to be $1$,
+For $$d = 2$$, the solution $$h_2(r)$$ is as follows,
+whose limit turns out to be $$1$$,
so the Wiener process is also recurrent in 2D:
$$\begin{aligned}
@@ -161,8 +161,8 @@ $$\begin{aligned}
1
\end{aligned}$$
-However, for $d \ge 3$, the solution $h_d(r)$
-does not converge to $1$ for $R_o \to \infty$,
+However, for $$d \ge 3$$, the solution $$h_d(r)$$
+does not converge to $$1$$ for $$R_o \to \infty$$,
meaning the Wiener process is transient in 3D or higher:
$$\begin{aligned}
diff --git a/source/know/concept/wkb-approximation/index.md b/source/know/concept/wkb-approximation/index.md
index efeeca6..ef57a3b 100644
--- a/source/know/concept/wkb-approximation/index.md
+++ b/source/know/concept/wkb-approximation/index.md
@@ -9,13 +9,13 @@ layout: "concept"
---
In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
-approximation** is a technique to approximate the wave function $\psi(x)$ of
+approximation** is a technique to approximate the wave function $$\psi(x)$$ of
the one-dimensional time-independent Schrödinger equation. It is an example
of a **semiclassical approximation**, because it tries to find a
balance between classical and quantum physics.
-In classical mechanics, a particle travelling in a potential $V(x)$
-along a path $x(t)$ has a total energy $E$ as follows, which we
+In classical mechanics, a particle travelling in a potential $$V(x)$$
+along a path $$x(t)$$ has a total energy $$E$$ as follows, which we
rearrange:
$$\begin{aligned}
@@ -25,13 +25,13 @@ $$\begin{aligned}
\end{aligned}$$
The left-hand side of the rearranged version is simply the momentum squared,
-so we define the magnitude of the momentum $p(x)$ accordingly:
+so we define the magnitude of the momentum $$p(x)$$ accordingly:
$$\begin{aligned}
p(x) = \sqrt{2 m (E - V(x))}
\end{aligned}$$
-Note that this is under the assumption that $E > V$,
+Note that this is under the assumption that $$E > V$$,
which is always true in classical mechanics,
but not necessarily in quantum mechanics.
We rewrite the Schrödinger equation:
@@ -42,7 +42,7 @@ $$\begin{aligned}
= \dvn{2}{\psi}{x} + \frac{p^2}{\hbar^2} \psi
\end{aligned}$$
-If $V(x)$ were constant, and by extension $p(x)$ too, then the solution
+If $$V(x)$$ were constant, and by extension $$p(x)$$ too, then the solution
is easy:
$$\begin{aligned}
@@ -51,9 +51,9 @@ $$\begin{aligned}
\end{aligned}$$
This form is reminiscent of the generator of translations. In practice,
-$V(x)$ and $p(x)$ vary with $x$, but we can still salvage this solution
-by assuming that $V(x)$ varies slowly compared to the wavelength
-$\lambda(x) = 2 \pi / k(x)$, where $k(x) = p(x) / \hbar$ is the
+$$V(x)$$ and $$p(x)$$ vary with $$x$$, but we can still salvage this solution
+by assuming that $$V(x)$$ varies slowly compared to the wavelength
+$$\lambda(x) = 2 \pi / k(x)$$, where $$k(x) = p(x) / \hbar$$ is the
wavenumber. The solution then takes the following form:
$$\begin{aligned}
@@ -61,9 +61,9 @@ $$\begin{aligned}
= \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big)
\end{aligned}$$
-$\chi(\xi)$ is an unknown function, which intuitively should be related
-to $p(x)$. The purpose of the integral is to accumulate the change of
-$\chi$ from the initial point $0$ to the current position $x$.
+$$\chi(\xi)$$ is an unknown function, which intuitively should be related
+to $$p(x)$$. The purpose of the integral is to accumulate the change of
+$$\chi$$ from the initial point $$0$$ to the current position $$x$$.
Let us write this as an indefinite integral for convenience:
$$\begin{aligned}
@@ -71,8 +71,8 @@ $$\begin{aligned}
= \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg)
\end{aligned}$$
-Where $C = \int \chi(x) \dd{x} |_{x = 0}$ is the initial point of the definite integral.
-For simplicity, we absorb the constant $C$ into $\psi(0)$.
+Where $$C = \int \chi(x) \dd{x} |_{x = 0}$$ is the initial point of the definite integral.
+For simplicity, we absorb the constant $$C$$ into $$\psi(0)$$.
We can now clearly see that:
$$\begin{aligned}
@@ -81,7 +81,7 @@ $$\begin{aligned}
\chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)}
\end{aligned}$$
-Next, we insert this ansatz for $\psi(x)$ into the Schrödinger equation
+Next, we insert this ansatz for $$\psi(x)$$ into the Schrödinger equation
to get:
$$\begin{aligned}
@@ -91,7 +91,7 @@ $$\begin{aligned}
= \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
\end{aligned}$$
-Dividing out $\psi$ and rearranging gives us the following, which is
+Dividing out $$\psi$$ and rearranging gives us the following, which is
still exact:
$$\begin{aligned}
@@ -99,9 +99,9 @@ $$\begin{aligned}
= p^2 - \chi^2
\end{aligned}$$
-Next, we expand this as a power series of $\hbar$. This is why it is
+Next, we expand this as a power series of $$\hbar$$. This is why it is
called *semiclassical*: so far we have been using full quantum mechanics,
-but now we are treating $\hbar$ as a parameter which controls the
+but now we are treating $$\hbar$$ as a parameter which controls the
strength of quantum effects:
$$\begin{aligned}
@@ -109,9 +109,9 @@ $$\begin{aligned}
\end{aligned}$$
The heart of the WKB approximation is its assumption that quantum effects are
-sufficiently weak (i.e. $\hbar$ is small enough) that we only need to
+sufficiently weak (i.e. $$\hbar$$ is small enough) that we only need to
consider the first two terms, or, more specifically, that we only go up to
-$\hbar$, not $\hbar^2$ or higher. Inserting the first two terms of this
+$$\hbar$$, not $$\hbar^2$$ or higher. Inserting the first two terms of this
expansion into the equation:
$$\begin{aligned}
@@ -119,8 +119,8 @@ $$\begin{aligned}
&= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1
\end{aligned}$$
-Where we have discarded all terms containing $\hbar^2$. At order
-$\hbar^0$, we then get the expected classical result for $\chi_0(x)$:
+Where we have discarded all terms containing $$\hbar^2$$. At order
+$$\hbar^0$$, we then get the expected classical result for $$\chi_0(x)$$:
$$\begin{aligned}
0 = p^2 - \chi_0^2
@@ -128,7 +128,7 @@ $$\begin{aligned}
\chi_0(x) = p(x)
\end{aligned}$$
-While at order $\hbar$, we get the following quantum-mechanical
+While at order $$\hbar$$, we get the following quantum-mechanical
correction:
$$\begin{aligned}
@@ -138,7 +138,7 @@ $$\begin{aligned}
\chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
\end{aligned}$$
-Therefore, our approximated wave function $\psi(x)$ currently looks like
+Therefore, our approximated wave function $$\psi(x)$$ currently looks like
this:
$$\begin{aligned}
@@ -158,7 +158,7 @@ $$\begin{aligned}
= \frac{1}{\sqrt{p(x)}}
\end{aligned}$$
-In the WKB approximation for $E > V$, the solution $\psi(x)$ is thus
+In the WKB approximation for $$E > V$$, the solution $$\psi(x)$$ is thus
given by:
$$\begin{aligned}
@@ -167,7 +167,7 @@ $$\begin{aligned}
}
\end{aligned}$$
-What if $E < V$? In classical mechanics, this is just not allowed; a ball
+What if $$E < V$$? In classical mechanics, this is just not allowed; a ball
cannot simply go through a potential bump without the necessary energy.
On the other hand, in quantum physics, particles can **tunnel** through barriers.
@@ -178,7 +178,7 @@ $$\begin{aligned}
p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
\end{aligned}$$
-And then take the absolute value in the appropriate place in front of $\psi(x)$:
+And then take the absolute value in the appropriate place in front of $$\psi(x)$$:
$$\begin{aligned}
\boxed{
@@ -186,10 +186,10 @@ $$\begin{aligned}
}
\end{aligned}$$
-In the classical region ($E > V$), the wave function oscillates, and
-in the quantum-physical region ($E < V$) it is exponential.
-Note that for $E \approx V$ the approximation breaks down,
-because of the appearance of $p(x)$ in the denominator.
+In the classical region ($$E > V$$), the wave function oscillates, and
+in the quantum-physical region ($$E < V$$) it is exponential.
+Note that for $$E \approx V$$ the approximation breaks down,
+because of the appearance of $$p(x)$$ in the denominator.
## References
diff --git a/source/know/concept/young-dupre-relation/index.md b/source/know/concept/young-dupre-relation/index.md
index 686d2a8..ed41aee 100644
--- a/source/know/concept/young-dupre-relation/index.md
+++ b/source/know/concept/young-dupre-relation/index.md
@@ -12,7 +12,7 @@ layout: "concept"
In fluid mechanics, the **Young-Dupré relation** relates the contact
angle of a droplet at rest on a surface to the surface tensions of the interfaces.
-Let $\alpha_{gl}$, $\alpha_{sl}$ and $\alpha_{sg}$ respectively be
+Let $$\alpha_{gl}$$, $$\alpha_{sl}$$ and $$\alpha_{sg}$$ respectively be
the energy costs of the liquid-gas, solid-liquid and solid-gas interfaces:
$$\begin{aligned}
@@ -28,8 +28,8 @@ when you account for the surface tension force pulling along each interface.
A more general derivation is possible by using the
[calculus of variations](/know/concept/calculus-of-variations/).
-In 2D, the upper surface of the droplet is denoted by $y(x)$.
-Consider the following Lagrangian $\mathcal{L}$,
+In 2D, the upper surface of the droplet is denoted by $$y(x)$$.
+Consider the following Lagrangian $$\mathcal{L}$$,
with the two first terms respectively being the energy costs
of the top and bottom surfaces:
@@ -39,24 +39,24 @@ $$\begin{aligned}
\end{aligned}$$
And the last term comes from the constraint
-that the volume $V$ of the droplet must be constant:
+that the volume $$V$$ of the droplet must be constant:
$$\begin{aligned}
V = \int_0^L y \dd{x}
\end{aligned}$$
The total energy to be minimized is thus given by the following functional,
-where the endpoints of the droplet are $x = 0$ and $x = L$:
+where the endpoints of the droplet are $$x = 0$$ and $$x = L$$:
$$\begin{aligned}
E[y(x)]
= \int_0^L \Big( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \Big) \dd{x}
\end{aligned}$$
-In this optimization problem, the endpoint $L$ is a free parameter,
-i.e. the $L$-value of the optimum is unknown and must be found.
-In such cases, the optimum $y(x)$ needs to satisfy the so-called *transversality condition*
-at the variable endpoint, in this case $x = L$:
+In this optimization problem, the endpoint $$L$$ is a free parameter,
+i.e. the $$L$$-value of the optimum is unknown and must be found.
+In such cases, the optimum $$y(x)$$ needs to satisfy the so-called *transversality condition*
+at the variable endpoint, in this case $$x = L$$:
$$\begin{aligned}
0
@@ -67,7 +67,7 @@ $$\begin{aligned}
&= \bigg( \alpha_{gl} \frac{1}{\sqrt{1 + (y')^2}} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \bigg)_{x = L}
\end{aligned}$$
-Due to the droplet's shape, we have the boundary condition $y(L) = 0$,
+Due to the droplet's shape, we have the boundary condition $$y(L) = 0$$,
so the last term vanishes.
We are thus left with the following equation:
@@ -77,9 +77,9 @@ $$\begin{aligned}
\end{aligned}$$
At the edge of the droplet, imagine a small right-angled triangle
-with one side $\dd{x}$ on the $x$-axis,
-the hypotenuse on $y(x)$ having length $\dd{x} \sqrt{1 + (y')^2}$,
-and the corner between them being the contact point with angle $\theta$.
+with one side $$\dd{x}$$ on the $$x$$-axis,
+the hypotenuse on $$y(x)$$ having length $$\dd{x} \sqrt{1 + (y')^2}$$,
+and the corner between them being the contact point with angle $$\theta$$.
Then, from the definition of the cosine:
$$\begin{aligned}
diff --git a/source/know/concept/young-laplace-law/index.md b/source/know/concept/young-laplace-law/index.md
index 28c3026..f890beb 100644
--- a/source/know/concept/young-laplace-law/index.md
+++ b/source/know/concept/young-laplace-law/index.md
@@ -13,21 +13,21 @@ In liquids, the **Young-Laplace law** governs surface tension:
it describes the tension forces on a surface
as a pressure difference between the two sides of the liquid.
-Consider a small rectangle on the surface with sides $\dd{\ell_1}$ and $\dd{\ell_2}$,
+Consider a small rectangle on the surface with sides $$\dd{\ell_1}$$ and $$\dd{\ell_2}$$,
orientated such that the sides are parallel to the (orthogonal)
principal directions of the surface' [curvature](/know/concept/curvature/).
Surface tension then pulls at the sides with a force
-of magnitude $\alpha \dd{\ell_2}$ and $\alpha \dd{\ell_2}$,
-where $\alpha$ is the energy cost per unit of area,
+of magnitude $$\alpha \dd{\ell_2}$$ and $$\alpha \dd{\ell_2}$$,
+where $$\alpha$$ is the energy cost per unit of area,
which is the same as the force per unit of distance.
However, due to the surface' curvature,
those forces are not quite in the same plane as the rectangle.
Along both principal directions,
if we treat this portion of the surface as a small arc of a circle
-with a radius equal to the principal radius of curvature $R_1$ or $R_2$,
-then the tension forces are at angles $\theta_1$ and $\theta_2$
+with a radius equal to the principal radius of curvature $$R_1$$ or $$R_2$$,
+then the tension forces are at angles $$\theta_1$$ and $$\theta_2$$
calculated from the arc length:
$$\begin{aligned}
@@ -38,14 +38,14 @@ $$\begin{aligned}
= \frac{1}{2} \dd{\ell_1}
\end{aligned}$$
-Pay attention to the indices $1$ and $2$:
-to get the angle of the force pulling at $\dd{\ell_1}$,
-we need to treat $\dd{\ell_2} / 2$ as an arc,
+Pay attention to the indices $$1$$ and $$2$$:
+to get the angle of the force pulling at $$\dd{\ell_1}$$,
+we need to treat $$\dd{\ell_2} / 2$$ as an arc,
and vice versa.
Since the forces are not quite in the plane,
they have a small component acting *perpendicular* to the surface,
-with the following magnitudes $\dd{F_1}$ and $\dd{F_2}$
+with the following magnitudes $$\dd{F_1}$$ and $$\dd{F_2}$$
along the principal axes:
$$\begin{aligned}
@@ -62,13 +62,13 @@ $$\begin{aligned}
= \frac{\alpha}{R_2} \dd{A}
\end{aligned}$$
-The initial factor of $2$ is there since
+The initial factor of $$2$$ is there since
the same force is pulling at opposide sides of the rectangle.
-We end up with $\alpha / R_{1,2}$ multiplied by
-the surface area $\dd{A} = \dd{\ell_1} \dd{\ell_2}$ of the rectangle.
+We end up with $$\alpha / R_{1,2}$$ multiplied by
+the surface area $$\dd{A} = \dd{\ell_1} \dd{\ell_2}$$ of the rectangle.
-Adding together $\dd{F_1}$ and $\dd{F_2}$ and
-dividing out $\dd{A}$ gives us the force-per-area (i.e. the pressure)
+Adding together $$\dd{F_1}$$ and $$\dd{F_2}$$ and
+dividing out $$\dd{A}$$ gives us the force-per-area (i.e. the pressure)
added by surface tension,
which is given by the **Young-Laplace law**:
@@ -79,9 +79,9 @@ $$\begin{aligned}
}
\end{aligned}$$
-The total excess pressure $\Delta p$ is called the **Laplace pressure**,
+The total excess pressure $$\Delta p$$ is called the **Laplace pressure**,
and fully determines the effects of surface tension:
-a certain interface shape leads to a certain $\Delta p$,
+a certain interface shape leads to a certain $$\Delta p$$,
and the liquid will flow (i.e. the surface will move)
to try to reach an equilibrium.
--
cgit v1.3
 except in a finite number of **simple poles**, -which are points $z_p$ where $f(z_p)$ diverges, -but where the product $(z - z_p) f(z)$ is non-zero -and still holomorphic close to $z_p$. -In other words, $f(z)$ can be approximated close to $z_p$: +which are points $$z_p$$ where $$f(z_p)$$ diverges, +but where the product $$(z - z_p) f(z)$$ is non-zero +and still holomorphic close to $$z_p$$. +In other words, $$f(z)$$ can be approximated close to $$z_p$$: $$\begin{aligned} f(z) \approx \frac{R_p}{z - z_p} \end{aligned}$$ -Where the **residue** $R_p$ of a simple pole $z_p$ is defined as follows, and -represents the rate at which $f(z)$ diverges close to $z_p$: +Where the **residue** $$R_p$$ of a simple pole $$z_p$$ is defined as follows, and +represents the rate at which $$f(z)$$ diverges close to $$z_p$$: $$\begin{aligned} \boxed{ @@ -32,8 +32,8 @@ $$\begin{aligned} **Cauchy's residue theorem** for meromorphic functions is a generalization of Cauchy's integral theorem for holomorphic functions, -and states that the integral on a contour $C$ -purely depends on the simple poles $z_p$ enclosed by $C$: +and states that the integral on a contour $$C$$ +purely depends on the simple poles $$z_p$$ enclosed by $$C$$: $$\begin{aligned} \boxed{ @@ -47,14 +47,14 @@ $$\begin{aligned} From the definition of a meromorphic function, -we know that we can decompose $f(z)$ like so, -where $h(z)$ is holomorphic and $z_p$ are all its poles: +we know that we can decompose $$f(z)$$ like so, +where $$h(z)$$ is holomorphic and $$z_p$$ are all its poles: $$\begin{aligned} f(z) = h(z) + \sum_{z_p} \frac{R_p}{z - z_p} \end{aligned}$$ -We integrate this over a contour $C$ which contains all poles, and apply +We integrate this over a contour $$C$$ which contains all poles, and apply both Cauchy's integral theorem and Cauchy's integral formula to get: $$\begin{aligned} @@ -62,11 +62,12 @@ $$\begin{aligned} &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z} = \sum_{p} R_p \: 2 \pi i \end{aligned}$$ + This theorem might not seem very useful, -but in fact, by cleverly choosing the contour $C$, +but in fact, by cleverly choosing the contour $$C$$, it lets us evaluate many integrals along the real axis, most notably [Fourier transforms](/know/concept/fourier-transform/). It can also be used to derive the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations). diff --git a/source/know/concept/reynolds-number/index.md b/source/know/concept/reynolds-number/index.md index aa9559c..9ae4f4b 100644 --- a/source/know/concept/reynolds-number/index.md +++ b/source/know/concept/reynolds-number/index.md @@ -12,7 +12,7 @@ layout: "concept" The [Navier-Stokes equations](/know/concept/navier-stokes-equations/) are infamously tricky to solve, so we would like a way to qualitatively predict -the behaviour of a fluid without needing the flow $\va{v}$. +the behaviour of a fluid without needing the flow $$\va{v}$$. Consider the main equation: $$\begin{aligned} @@ -20,14 +20,14 @@ $$\begin{aligned} = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v} \end{aligned}$$ -In this case, the gravity term $\va{g}$ +In this case, the gravity term $$\va{g}$$ has been absorbed into the pressure term: -$p \to p\!+\!\rho \Phi$, -where $\Phi$ is the gravitational scalar potential, -i.e. $\va{g} = - \nabla \Phi$. +$$p \to p\!+\!\rho \Phi$$, +where $$\Phi$$ is the gravitational scalar potential, +i.e. $$\va{g} = - \nabla \Phi$$. -Let us introduce the dimensionless variables $\va{v}'$, $\va{r}'$, $t'$ and $p'$, -where $U$ and $L$ are respectively a characteristic velocity and length +Let us introduce the dimensionless variables $$\va{v}'$$, $$\va{r}'$$, $$t'$$ and $$p'$$, +where $$U$$ and $$L$$ are respectively a characteristic velocity and length of the system at hand: $$\begin{aligned} @@ -57,7 +57,7 @@ $$\begin{aligned} = - \frac{U^2}{L} \nabla' p' + \frac{U \nu}{L^2} \nabla'^2 \va{v}' \end{aligned}$$ -After dividing out $U^2/L$, +After dividing out $$U^2/L$$, we arrive at the form of the original equation again: $$\begin{aligned} @@ -66,7 +66,7 @@ $$\begin{aligned} \end{aligned}$$ The constant factor of the last term -leads to the definition of the **Reynolds number** $\mathrm{Re}$: +leads to the definition of the **Reynolds number** $$\mathrm{Re}$$: $$\begin{aligned} \boxed{ @@ -75,17 +75,17 @@ $$\begin{aligned} } \end{aligned}$$ -If we choose $U$ and $L$ appropriately for a given system, +If we choose $$U$$ and $$L$$ appropriately for a given system, the Reynolds number allows us to predict the general trends. It can be regarded as the inverse of an "effective viscosity": -when $\mathrm{Re}$ is large, viscosity only has a minor role, -but when $\mathrm{Re}$ is small, it dominates the dynamics. +when $$\mathrm{Re}$$ is large, viscosity only has a minor role, +but when $$\mathrm{Re}$$ is small, it dominates the dynamics. Another way is thus to see the Reynolds number as the characteristic ratio between the advective term (see [material derivative](/know/concept/material-derivative/)) to the [viscosity](/know/concept/viscosity/) term, -since $\va{v} \sim U$: +since $$\va{v} \sim U$$: $$\begin{aligned} \mathrm{Re} @@ -94,7 +94,7 @@ $$\begin{aligned} = \frac{U L}{\nu} \end{aligned}$$ -In other words, $\mathrm{Re}$ +In other words, $$\mathrm{Re}$$ describes the relative strength of intertial and viscous forces. Returning to the dimensionless Navier-Stokes equation: @@ -103,7 +103,7 @@ $$\begin{aligned} = - \nabla' p' + \frac{1}{\mathrm{Re}} \nabla'^2 \va{v}' \end{aligned}$$ -For large $\mathrm{Re} \gg 1$, +For large $$\mathrm{Re} \gg 1$$, we can neglect the latter term, such that redimensionalizing yields: @@ -120,10 +120,10 @@ for an ideal fluid, i.e. a fluid without viscosity. ## Stokes flow A notable case is so-called **Stokes flow** or **creeping flow**, -meaning flow at $\mathrm{Re} \ll 1$. +meaning flow at $$\mathrm{Re} \ll 1$$. In this limit, the Navier-Stokes equations can be linearized: -since $\mathrm{Re}$ is the advective-to-viscous ratio, -$\mathrm{Re} \ll 1$ implies that we can ignore the advective term, leaving: +since $$\mathrm{Re}$$ is the advective-to-viscous ratio, +$$\mathrm{Re} \ll 1$$ implies that we can ignore the advective term, leaving: $$\begin{aligned} \boxed{ @@ -135,7 +135,7 @@ $$\begin{aligned} This equation is called the **unsteady Stokes equation**. Usually, however, such flows are assumed to be steady (i.e. time-invariant), leading to the **steady Stokes equation**, -with $\eta = \rho \nu$: +with $$\eta = \rho \nu$$: $$\begin{aligned} \boxed{ diff --git a/source/know/concept/ritz-method/index.md b/source/know/concept/ritz-method/index.md index ff1faae..902b7cf 100644 --- a/source/know/concept/ritz-method/index.md +++ b/source/know/concept/ritz-method/index.md @@ -28,10 +28,10 @@ $$\begin{aligned} = \frac{1}{S} \int_a^b p(x) \big|u_x(x)\big|^2 - q(x) \big|u(x)\big|^2 \dd{x} \end{aligned}$$ -Where $u(x) \in \mathbb{C}$ is the unknown function, -and $p(x), q(x) \in \mathbb{R}$ are given. -In addition, $S$ is the norm of $u$, which we demand be constant -with respect to a weight function $w(x) \in \mathbb{R}$: +Where $$u(x) \in \mathbb{C}$$ is the unknown function, +and $$p(x), q(x) \in \mathbb{R}$$ are given. +In addition, $$S$$ is the norm of $$u$$, which we demand be constant +with respect to a weight function $$w(x) \in \mathbb{R}$$: $$\begin{aligned} S @@ -39,8 +39,8 @@ $$\begin{aligned} \end{aligned}$$ To handle this normalization requirement, -we introduce a [Lagrange multiplier](/know/concept/lagrange-multiplier/) $\lambda$, -and define the Lagrangian $\Lambda$ for the full constrained optimization problem as: +we introduce a [Lagrange multiplier](/know/concept/lagrange-multiplier/) $$\lambda$$, +and define the Lagrangian $$\Lambda$$ for the full constrained optimization problem as: $$\begin{aligned} \Lambda @@ -64,13 +64,13 @@ $$\begin{aligned} \end{aligned}$$ This has the familiar form of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/) (SLP), -with $\lambda$ representing an eigenvalue. +with $$\lambda$$ representing an eigenvalue. SLPs have useful properties, but before we can take advantage of those, -we need to handle an important detail: the boundary conditions (BCs) on $u$. +we need to handle an important detail: the boundary conditions (BCs) on $$u$$. The above equation is only a valid SLP for certain BCs, as seen in the derivation of Sturm-Liouville theory. -Let us return to the definition of $R[u]$, +Let us return to the definition of $$R[u]$$, and integrate it by parts: $$\begin{aligned} @@ -81,7 +81,7 @@ $$\begin{aligned} \end{aligned}$$ The boundary term vanishes for a subset of the BCs that make a valid SLP, -including Dirichlet BCs $u(a) = u(b) = 0$, Neumann BCs $u_x(a) = u_x(b) = 0$, and periodic BCs. +including Dirichlet BCs $$u(a) = u(b) = 0$$, Neumann BCs $$u_x(a) = u_x(b) = 0$$, and periodic BCs. Therefore, we assume that this term does indeed vanish, such that we can use Sturm-Liouville theory later: @@ -91,14 +91,14 @@ $$\begin{aligned} \equiv - \frac{1}{S} \int_a^b u^* \hat{H} u \dd{x} \end{aligned}$$ -Where $\hat{H}$ is the self-adjoint Sturm-Liouville operator. +Where $$\hat{H}$$ is the self-adjoint Sturm-Liouville operator. Because the constrained Euler-Lagrange equation is now an SLP, -we know that it has an infinite number of real discrete eigenvalues $\lambda_n$ with a lower bound, -corresponding to mutually orthogonal eigenfunctions $u_n(x)$. +we know that it has an infinite number of real discrete eigenvalues $$\lambda_n$$ with a lower bound, +corresponding to mutually orthogonal eigenfunctions $$u_n(x)$$. To understand the significance of this result, suppose we have solved the SLP, -and now insert one of the eigenfunctions $u_n$ into $R$: +and now insert one of the eigenfunctions $$u_n$$ into $$R$$: $$\begin{aligned} R[u_n] @@ -109,9 +109,9 @@ $$\begin{aligned} = \frac{S_n}{S_n} \lambda_n \end{aligned}$$ -Where $S_n$ is the normalization of $u_n$. -In other words, when given $u_n$, -the functional $R$ yields the corresponding eigenvalue $\lambda_n$: +Where $$S_n$$ is the normalization of $$u_n$$. +In other words, when given $$u_n$$, +the functional $$R$$ yields the corresponding eigenvalue $$\lambda_n$$: $$\begin{aligned} \boxed{ @@ -120,15 +120,15 @@ $$\begin{aligned} } \end{aligned}$$ -This powerful result was not at all clear from $R$'s initial definition. +This powerful result was not at all clear from $$R$$'s initial definition. ## Justification -But what if we do not know the eigenfunctions? Is $R$ still useful? -Yes, as we shall see. Suppose we make an educated guess $u(x)$ -for the ground state (i.e. lowest-eigenvalue) solution $u_0(x)$: +But what if we do not know the eigenfunctions? Is $$R$$ still useful? +Yes, as we shall see. Suppose we make an educated guess $$u(x)$$ +for the ground state (i.e. lowest-eigenvalue) solution $$u_0(x)$$: $$\begin{aligned} u(x) @@ -136,10 +136,10 @@ $$\begin{aligned} \end{aligned}$$ Here, we are using the fact that the eigenfunctions of an SLP form a complete set, -so our (known) guess $u$ can be expanded in the true (unknown) eigenfunctions $u_n$. -We are assuming that $u$ is already quite close to its target $u_0$, -such that the (unknown) expansion coefficients $c_n$ are small; -specifically $|c_n|^2 \ll 1$. +so our (known) guess $$u$$ can be expanded in the true (unknown) eigenfunctions $$u_n$$. +We are assuming that $$u$$ is already quite close to its target $$u_0$$, +such that the (unknown) expansion coefficients $$c_n$$ are small; +specifically $$|c_n|^2 \ll 1$$. Let us start from what we know: $$\begin{aligned} @@ -150,8 +150,8 @@ $$\begin{aligned} \end{aligned}$$ This quantity is known as the **Rayleigh quotient**. -Inserting our ansatz $u$, -and using that the true $u_n$ have corresponding eigenvalues $\lambda_n$: +Inserting our ansatz $$u$$, +and using that the true $$u_n$$ have corresponding eigenvalues $$\lambda_n$$: $$\begin{aligned} R[u] @@ -176,8 +176,8 @@ $$\begin{aligned} + \sum_{n = 1}^\infty c_n \Inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \Inprod{u_m}{w u_n}} \end{aligned}$$ -Using orthogonality $\Inprod{u_m}{w u_n} = S_n \delta_{mn}$, -and the fact that $n \neq 0$ by definition, we find: +Using orthogonality $$\Inprod{u_m}{w u_n} = S_n \delta_{mn}$$, +and the fact that $$n \neq 0$$ by definition, we find: $$\begin{aligned} R @@ -191,7 +191,7 @@ $$\begin{aligned} {\displaystyle S_0 + \sum_{n} |c_n|^2 S_n} \end{aligned}$$ -It is always possible to choose our normalizations such that $S_n = S$ for all $u_n$, leaving: +It is always possible to choose our normalizations such that $$S_n = S$$ for all $$u_n$$, leaving: $$\begin{aligned} R @@ -211,9 +211,9 @@ $$\begin{aligned} {\displaystyle 1 + \sum_{n} |c_n|^2} \end{aligned}$$ -Thus, if we improve our guess $u$, -then $R[u]$ approaches the true eigenvalue $\lambda_0$. -For numerically finding $u_0$ and $\lambda_0$, this gives us a clear goal: minimize $R$, because: +Thus, if we improve our guess $$u$$, +then $$R[u]$$ approaches the true eigenvalue $$\lambda_0$$. +For numerically finding $$u_0$$ and $$\lambda_0$$, this gives us a clear goal: minimize $$R$$, because: $$\begin{aligned} \boxed{ @@ -228,19 +228,19 @@ In the context of quantum mechanics, this is not surprising, since any superposition of multiple states is guaranteed to have a higher energy than the ground state. -Note that the convergence to $\lambda_0$ goes as $|c_n|^2$, -while $u$ converges to $u_0$ as $|c_n|$ by definition, -so even a fairly bad guess $u$ will give a decent estimate for $\lambda_0$. +Note that the convergence to $$\lambda_0$$ goes as $$|c_n|^2$$, +while $$u$$ converges to $$u_0$$ as $$|c_n|$$ by definition, +so even a fairly bad guess $$u$$ will give a decent estimate for $$\lambda_0$$. ## The method In the following, we stick to Dirac notation, -since the results hold for both continuous functions $u(x)$ and discrete vectors $\vb{u}$, -as long as the operator $\hat{H}$ is self-adjoint. -Suppose we express our guess $\Ket{u}$ as a linear combination -of *known* basis vectors $\Ket{f_n}$ with weights $a_n \in \mathbb{C}$: +since the results hold for both continuous functions $$u(x)$$ and discrete vectors $$\vb{u}$$, +as long as the operator $$\hat{H}$$ is self-adjoint. +Suppose we express our guess $$\Ket{u}$$ as a linear combination +of *known* basis vectors $$\Ket{f_n}$$ with weights $$a_n \in \mathbb{C}$$: $$\begin{aligned} \Ket{u} @@ -249,16 +249,16 @@ $$\begin{aligned} \approx \sum_{n = 0}^{N - 1} a_n \Ket{f_n} \end{aligned}$$ -For numerical tractability, we truncate the sum at $N$ terms, -and for generality, we allow $\Ket{f_n}$ to be non-orthogonal, -as described by an *overlap matrix* with elements $S_{mn}$: +For numerical tractability, we truncate the sum at $$N$$ terms, +and for generality, we allow $$\Ket{f_n}$$ to be non-orthogonal, +as described by an *overlap matrix* with elements $$S_{mn}$$: $$\begin{aligned} \Inprod{f_m}{w f_n} = S_{m n} \end{aligned}$$ From the discussion above, -we know that the ground-state eigenvalue $\lambda_0$ is estimated by: +we know that the ground-state eigenvalue $$\lambda_0$$ is estimated by: $$\begin{aligned} \lambda_0 @@ -269,8 +269,8 @@ $$\begin{aligned} \equiv \frac{\displaystyle \sum_{m n} a_m^* a_n H_{m n}}{\displaystyle \sum_{m n} a_m^* a_n S_{mn}} \end{aligned}$$ -And we also know that our goal is to minimize $R[u]$, -so we vary $a_k^*$ to find its extremum: +And we also know that our goal is to minimize $$R[u]$$, +so we vary $$a_k^*$$ to find its extremum: $$\begin{aligned} 0 @@ -283,7 +283,7 @@ $$\begin{aligned} = \frac{\displaystyle \sum_{n} a_n \big(H_{k n} - \lambda S_{k n}\big)}{\Inprod{u}{w u}} \end{aligned}$$ -Clearly, this is only satisfied if the following holds for all $k = 0, 1, ..., N\!-\!1$: +Clearly, this is only satisfied if the following holds for all $$k = 0, 1, ..., N\!-\!1$$: $$\begin{aligned} 0 @@ -292,7 +292,7 @@ $$\begin{aligned} For illustrative purposes, we can write this as a matrix equation -with $M_{k n} \equiv H_{k n} - \lambda S_{k n}$: +with $$M_{k n} \equiv H_{k n} - \lambda S_{k n}$$: $$\begin{aligned} \begin{bmatrix} @@ -311,10 +311,10 @@ $$\begin{aligned} \end{bmatrix} \end{aligned}$$ -Note that this looks like an eigenvalue problem for $\lambda$. -Indeed, demanding that $\overline{M}$ cannot simply be inverted +Note that this looks like an eigenvalue problem for $$\lambda$$. +Indeed, demanding that $$\overline{M}$$ cannot simply be inverted (i.e. the solution is non-trivial) -yields a characteristic polynomial for $\lambda$: +yields a characteristic polynomial for $$\lambda$$: $$\begin{aligned} 0 @@ -322,39 +322,39 @@ $$\begin{aligned} = \det\!\Big[ \overline{H} - \lambda \overline{S} \Big] \end{aligned}$$ -This gives a set of $\lambda$, -which are the exact eigenvalues of $\overline{H}$, -and the estimated eigenvalues of $\hat{H}$ -(recall that $\overline{H}$ is $\hat{H}$ expressed in a truncated basis). -The eigenvector $\big[ a_0, a_1, ..., a_{N-1} \big]$ of the lowest $\lambda$ -gives the optimal weights to approximate $\Ket{u_0}$ in the basis $\{\Ket{f_n}\}$. -Likewise, the higher $\lambda$'s eigenvectors approximate -excited (i.e. non-ground) eigenstates of $\hat{H}$, +This gives a set of $$\lambda$$, +which are the exact eigenvalues of $$\overline{H}$$, +and the estimated eigenvalues of $$\hat{H}$$ +(recall that $$\overline{H}$$ is $$\hat{H}$$ expressed in a truncated basis). +The eigenvector $$\big[ a_0, a_1, ..., a_{N-1} \big]$$ of the lowest $$\lambda$$ +gives the optimal weights to approximate $$\Ket{u_0}$$ in the basis $$\{\Ket{f_n}\}$$. +Likewise, the higher $$\lambda$$'s eigenvectors approximate +excited (i.e. non-ground) eigenstates of $$\hat{H}$$, although in practice the results are less accurate the higher we go. The overall accuracy is determined by how good our truncated basis is, i.e. how large a subspace it spans -of the [Hilbert space](/know/concept/hilbert-space/) in which the true $\Ket{u_0}$ resides. +of the [Hilbert space](/know/concept/hilbert-space/) in which the true $$\Ket{u_0}$$ resides. Clearly, adding more basis vectors will improve the results, at the cost of computation. -For example, if $\hat{H}$ represents a helium atom, -a good choice for $\{\Ket{f_n}\}$ would be hydrogen orbitals, +For example, if $$\hat{H}$$ represents a helium atom, +a good choice for $$\{\Ket{f_n}\}$$ would be hydrogen orbitals, since those are qualitatively similar. You may find this result unsurprising; -it makes some intuitive sense that approximating $\hat{H}$ -in a limited basis would yield a matrix $\overline{H}$ giving rough eigenvalues. +it makes some intuitive sense that approximating $$\hat{H}$$ +in a limited basis would yield a matrix $$\overline{H}$$ giving rough eigenvalues. The point of this discussion is to rigorously show the validity of this approach. If we only care about the ground state, -then we already know $\lambda$ from $R[u]$, -so all we need to do is solve the above matrix equation for $a_n$. -Keep in mind that $\overline{M}$ is singular, -and $a_n$ are only defined up to a constant factor. +then we already know $$\lambda$$ from $$R[u]$$, +so all we need to do is solve the above matrix equation for $$a_n$$. +Keep in mind that $$\overline{M}$$ is singular, +and $$a_n$$ are only defined up to a constant factor. Nowadays, there exist many other methods to calculate eigenvalues -of complicated operators $\hat{H}$, +of complicated operators $$\hat{H}$$, but an attractive feature of the Ritz method is that it is single-step, whereas its competitors tend to be iterative. That said, the Ritz method cannot recover from a poorly chosen basis. diff --git a/source/know/concept/rotating-wave-approximation/index.md b/source/know/concept/rotating-wave-approximation/index.md index 7066f37..63efc9c 100644 --- a/source/know/concept/rotating-wave-approximation/index.md +++ b/source/know/concept/rotating-wave-approximation/index.md @@ -10,7 +10,7 @@ categories: layout: "concept" --- -Consider the following periodic perturbation $\hat{H}_1$ to a quantum system, +Consider the following periodic perturbation $$\hat{H}_1$$ to a quantum system, which represents e.g. an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/) in the [electric dipole approximation](/know/concept/electric-dipole-approximation/): @@ -20,16 +20,16 @@ $$\begin{aligned} = \frac{\hat{V}}{2} \Big( e^{i \omega t} + e^{-i \omega t} \Big) \end{aligned}$$ -Where $\hat{V}$ is some operator, and we assume that $\omega$ -is fairly close to a resonance frequency $\omega_0$ -of the system that is getting perturbed by $\hat{H}_1$. +Where $$\hat{V}$$ is some operator, and we assume that $$\omega$$ +is fairly close to a resonance frequency $$\omega_0$$ +of the system that is getting perturbed by $$\hat{H}_1$$. As an example, consider a two-level system -consisting of states $\Ket{g}$ and $\Ket{e}$, -with a resonance frequency $\omega_0 = (E_e \!-\! E_g) / \hbar$. +consisting of states $$\Ket{g}$$ and $$\Ket{e}$$, +with a resonance frequency $$\omega_0 = (E_e \!-\! E_g) / \hbar$$. From the derivation of [time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/), -we know that the state $\Ket{\Psi} = c_g \Ket{g} + c_e \Ket{e}$ evolves as: +we know that the state $$\Ket{\Psi} = c_g \Ket{g} + c_e \Ket{e}$$ evolves as: $$\begin{aligned} i \hbar \dv{c_g}{t} @@ -39,7 +39,7 @@ $$\begin{aligned} &= \matrixel{e}{\hat{H}_1(t)}{g} \: c_g(t) \: e^{i \omega_0 t} + \matrixel{e}{\hat{H}_1(t)}{e} \: c_e(t) \end{aligned}$$ -Typically, $\hat{V}$ has odd spatial parity, in which case +Typically, $$\hat{V}$$ has odd spatial parity, in which case [Laporte's selection rule](/know/concept/selection-rules/) reduces this to: @@ -51,8 +51,8 @@ $$\begin{aligned} &= \frac{1}{i \hbar} \matrixel{e}{\hat{H}_1}{g} \: c_g \: e^{i \omega_0 t} \end{aligned}$$ -We now insert the general $\hat{H}_1$ defined above, -and define $V_{eg} \equiv \matrixel{e}{\hat{V}}{g}$ to get: +We now insert the general $$\hat{H}_1$$ defined above, +and define $$V_{eg} \equiv \matrixel{e}{\hat{V}}{g}$$ to get: $$\begin{aligned} \dv{c_g}{t} @@ -65,14 +65,14 @@ $$\begin{aligned} \end{aligned}$$ At last, here we make the **rotating wave approximation**: -since $\omega$ is assumed to be close to $\omega_0$, -we argue that $\omega \!+\! \omega_0$ is so much larger than $\omega \!-\! \omega_0$ +since $$\omega$$ is assumed to be close to $$\omega_0$$, +we argue that $$\omega \!+\! \omega_0$$ is so much larger than $$\omega \!-\! \omega_0$$ that those oscillations turn out negligible if the system is observed over a reasonable time interval. Specifically, since both exponentials have the same weight, -the fast ($\omega \!+\! \omega_0$) oscillations -have a tiny amplitude compared to the slow ($\omega \!-\! \omega_0$) ones. +the fast ($$\omega \!+\! \omega_0$$) oscillations +have a tiny amplitude compared to the slow ($$\omega \!-\! \omega_0$$) ones. Furthermore, since they average out to zero over most realistic time intervals, the fast terms can be dropped, leaving: @@ -103,11 +103,11 @@ $$\begin{aligned} This approximation's name is a bit confusing: the idea is that going from the Schrödinger to the [interaction picture](/know/concept/interaction-picture/) -has the effect of removing the exponentials of $\omega_0$ from the above equations, -i.e. multiplying them by $e^{i \omega_0 t}$ and $e^{- i \omega_0 t}$ +has the effect of removing the exponentials of $$\omega_0$$ from the above equations, +i.e. multiplying them by $$e^{i \omega_0 t}$$ and $$e^{- i \omega_0 t}$$ respectively, which can be regarded as a rotation. -Relative to this rotation, when we split the wave $\cos(\omega t)$ +Relative to this rotation, when we split the wave $$\cos(\omega t)$$ into two exponentials, one co-rotates, and the other counter-rotates. We keep only the co-rotating waves, hence the name. diff --git a/source/know/concept/runge-kutta-method/index.md b/source/know/concept/runge-kutta-method/index.md index 270cb91..4c3dacf 100644 --- a/source/know/concept/runge-kutta-method/index.md +++ b/source/know/concept/runge-kutta-method/index.md @@ -10,16 +10,16 @@ layout: "concept" A **Runge-Kutta method** (RKM) is a popular approach to numerically solving systems of ordinary differential equations. -Let $\vb{x}(t)$ be the vector we want to find, -governed by $\vb{f}(t, \vb{x})$: +Let $$\vb{x}(t)$$ be the vector we want to find, +governed by $$\vb{f}(t, \vb{x})$$: $$\begin{aligned} \vb{x}'(t) = \vb{f}\big(t, \vb{x}(t)\big) \end{aligned}$$ -Like in all numerical methods, the $t$-axis is split into discrete steps. -If a step has size $h$, then as long as $h$ is small enough, +Like in all numerical methods, the $$t$$-axis is split into discrete steps. +If a step has size $$h$$, then as long as $$h$$ is small enough, we can make the following approximation: $$\begin{aligned} @@ -31,9 +31,9 @@ $$\begin{aligned} &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t) \!+\! a h \vb{x}'(t) \big) \end{aligned}$$ -For sufficiently small $h$, +For sufficiently small $$h$$, higher-order derivates can also be included, -albeit still at $t \!+\! a h$: +albeit still at $$t \!+\! a h$$: $$\begin{aligned} \vb{x}'(t) + a h \vb{x}''(t) + b h^2 \vb{x}'''(t) @@ -43,8 +43,8 @@ $$\begin{aligned} Although these approximations might seem innocent, they actually make it quite complicated to determine the error order of a given RKM. -Now, consider a Taylor expansion around the current $t$, -truncated at a chosen order $n$: +Now, consider a Taylor expansion around the current $$t$$, +truncated at a chosen order $$n$$: $$\begin{aligned} \vb{x}(t \!+\! h) @@ -54,7 +54,7 @@ $$\begin{aligned} \end{aligned}$$ We are free to split the terms as follows, -choosing real factors $\omega_{mj}$ subject to $\sum_{j} \omega_{mj} = 1$: +choosing real factors $$\omega_{mj}$$ subject to $$\sum_{j} \omega_{mj} = 1$$: $$\begin{aligned} \vb{x}(t \!+\! h) @@ -63,9 +63,9 @@ $$\begin{aligned} + \:...\, + \frac{h^{n-1}}{n!} \sum_{j = 1}^{N_n} \omega_{nj} \, \vb{x}^{(n)} \bigg] \end{aligned}$$ -Where the integers $N_1,...,N_n$ are also free to choose, +Where the integers $$N_1,...,N_n$$ are also free to choose, but for reasons that will become clear later, -the most general choice for an RKM is $N_1 = n$, $N_n = 1$, and: +the most general choice for an RKM is $$N_1 = n$$, $$N_n = 1$$, and: $$\begin{aligned} N_{n-1} @@ -82,7 +82,7 @@ $$\begin{aligned} = N_3 \!+\! n \!-\! 1 \end{aligned}$$ -In other words, $N_{n-m}$ is the $m$th triangular number. +In other words, $$N_{n-m}$$ is the $$m$$th triangular number. This is not so important, since this is not a practical way to describe RKMs, but it is helpful to understand how they work. @@ -90,11 +90,11 @@ but it is helpful to understand how they work. ## Example derivation -For example, let us truncate at $n = 3$, -such that $N_1 = 3$, $N_2 = 3$ and $N_3 = 1$. +For example, let us truncate at $$n = 3$$, +such that $$N_1 = 3$$, $$N_2 = 3$$ and $$N_3 = 1$$. The following derivation is very general, -except it requires all $\alpha_j \neq 0$. -Renaming $\omega_{mj}$, we start from: +except it requires all $$\alpha_j \neq 0$$. +Renaming $$\omega_{mj}$$, we start from: $$\begin{aligned} \vb{x}(t \!+\! h) @@ -108,7 +108,7 @@ $$\begin{aligned} \end{aligned}$$ As discussed earlier, the parenthesized expressions -can be approximately rewritten with $\vb{f}$: +can be approximately rewritten with $$\vb{f}$$: $$\begin{aligned} \vb{x}(t \!+\! h) @@ -131,7 +131,7 @@ $$\begin{aligned} \end{aligned}$$ Here, we can see an opportunity to save some computational time -by reusing an evaluation of $\vb{f}$. +by reusing an evaluation of $$\vb{f}$$. Technically, this is optional, but it would be madness not to, so we choose: @@ -140,8 +140,8 @@ $$\begin{aligned} = \frac{\gamma_3}{3 \beta_{32}} \end{aligned}$$ -Such that the next step of $\vb{x}$'s numerical solution is as follows, -recalling that $\sum_{j} \alpha_j = 1$: +Such that the next step of $$\vb{x}$$'s numerical solution is as follows, +recalling that $$\sum_{j} \alpha_j = 1$$: $$\begin{aligned} \boxed{ @@ -150,9 +150,9 @@ $$\begin{aligned} } \end{aligned}$$ -Where $\vb{k}_1$, $\vb{k}_2$ and $\vb{k}_3$ are different estimates -of the average slope $\vb{x}'$ between $t$ and $t \!+\! h$, -whose weighted average is used to make the $t$-step. +Where $$\vb{k}_1$$, $$\vb{k}_2$$ and $$\vb{k}_3$$ are different estimates +of the average slope $$\vb{x}'$$ between $$t$$ and $$t \!+\! h$$, +whose weighted average is used to make the $$t$$-step. They are given by: $$\begin{aligned} @@ -172,7 +172,7 @@ $$\begin{aligned} } \end{aligned}$$ -Despite the contraints on $\alpha_j$ and $\beta_j$, +Despite the contraints on $$\alpha_j$$ and $$\beta_j$$, there is an enormous freedom of choice here, all leading to valid RKMs, although not necessarily good ones. @@ -180,9 +180,9 @@ all leading to valid RKMs, although not necessarily good ones. ## General form A more practical description goes as follows: -in an $s$-stage RKM, a weighted average is taken -of up to $s$ slope estimates $\vb{k}_j$ with weights $b_j$. -Let $\sum_{j} b_j = 1$, then: +in an $$s$$-stage RKM, a weighted average is taken +of up to $$s$$ slope estimates $$\vb{k}_j$$ with weights $$b_j$$. +Let $$\sum_{j} b_j = 1$$, then: $$\begin{aligned} \boxed{ @@ -191,7 +191,7 @@ $$\begin{aligned} } \end{aligned}$$ -Where the estimates $\vb{k}_1, ..., \vb{k}_s$ +Where the estimates $$\vb{k}_1, ..., \vb{k}_s$$ depend on each other, and are calculated one by one as: $$\begin{aligned} @@ -201,8 +201,8 @@ $$\begin{aligned} } \end{aligned}$$ -With $c_1 = 1$ and $\sum_{j = 1} a_{mj} = c_m$. -Writing this out for the first few $m$, the pattern is clear: +With $$c_1 = 1$$ and $$\sum_{j = 1} a_{mj} = c_m$$. +Writing this out for the first few $$m$$, the pattern is clear: $$\begin{aligned} \vb{k}_1 @@ -233,26 +233,26 @@ $$\begin{aligned} \end{array} \end{aligned}$$ -Each RKM has an **order** $p$, -such that the global truncation error is $\mathcal{O}(h^p)$, +Each RKM has an **order** $$p$$, +such that the global truncation error is $$\mathcal{O}(h^p)$$, i.e. the accumulated difference between the numerical -and the exact solutions is proportional to $h^p$. +and the exact solutions is proportional to $$h^p$$. -The surprise is that $p$ need not be equal to the Taylor expansion order $n$, -nor the stage count $s$. -Typically, $s = n$ for computational efficiency, but $s \ge n$ is possible in theory. +The surprise is that $$p$$ need not be equal to the Taylor expansion order $$n$$, +nor the stage count $$s$$. +Typically, $$s = n$$ for computational efficiency, but $$s \ge n$$ is possible in theory. -The order $p$ of a given RKM is determined by +The order $$p$$ of a given RKM is determined by a complicated set of equations on the coefficients, -and the lowest possible $s$ for a desired $p$ +and the lowest possible $$s$$ for a desired $$p$$ is in fact only partially known. -For $p \le 4$ the bound is $s \ge p$, -whereas for $p \ge 5$ the only proven bound is $s \ge p \!+\! 1$, -but for $p \ge 7$ no such efficient methods have been found so far. +For $$p \le 4$$ the bound is $$s \ge p$$, +whereas for $$p \ge 5$$ the only proven bound is $$s \ge p \!+\! 1$$, +but for $$p \ge 7$$ no such efficient methods have been found so far. If you need an RKM with a certain order, look it up. -There exist many efficient methods for $p \le 4$ where $s = p$, -and although less popular, higher $p$ are also available. +There exist many efficient methods for $$p \le 4$$ where $$s = p$$, +and although less popular, higher $$p$$ are also available. diff --git a/source/know/concept/rutherford-scattering/index.md b/source/know/concept/rutherford-scattering/index.md index 7ce15ac..a7375d5 100644 --- a/source/know/concept/rutherford-scattering/index.md +++ b/source/know/concept/rutherford-scattering/index.md @@ -15,20 +15,20 @@ It is not a true collision, and is caused by Coulomb repulsion. The general idea is illustrated below. Consider two particles 1 and 2, with the same charge sign. -Let 2 be initially at rest, and 1 approach it with velocity $\vb{v}_1$. -Coulomb repulsion causes 1 to deflect by an angle $\theta$, +Let 2 be initially at rest, and 1 approach it with velocity $$\vb{v}_1$$. +Coulomb repulsion causes 1 to deflect by an angle $$\theta$$, and pushes 2 away in the process:](phase-flip-detect.png){kind=link}