-{% 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 devia
-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}