+‐
+{% for letter in by_letter %}
+{{ letter.name }}
+‐
+{% endfor %}
+
+
+{% for letter in by_letter %}
+{{ letter.name }}
+
+ {% assign items = letter.items | sort: "title" %}
+ {% for item in items %}
+ {{ item.title }}
+{% endfor %}
diff --git a/source/know/category/laser-theory/index.md b/source/know/category/laser-theory/index.md
new file mode 100644
index 0000000..f56e7a1
--- /dev/null
+++ b/source/know/category/laser-theory/index.md
@@ -0,0 +1,4 @@
+---
+title: "Laser theory"
+layout: "category"
+---
diff --git a/source/know/category/mathematics/index.md b/source/know/category/mathematics/index.md
new file mode 100644
index 0000000..e40678f
--- /dev/null
+++ b/source/know/category/mathematics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Mathematics"
+layout: "category"
+---
diff --git a/source/know/category/measure-theory/index.md b/source/know/category/measure-theory/index.md
new file mode 100644
index 0000000..2b77528
--- /dev/null
+++ b/source/know/category/measure-theory/index.md
@@ -0,0 +1,4 @@
+---
+title: "Measure theory"
+layout: "category"
+---
diff --git a/source/know/category/nonlinear-optics/index.md b/source/know/category/nonlinear-optics/index.md
new file mode 100644
index 0000000..edf73d4
--- /dev/null
+++ b/source/know/category/nonlinear-optics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Nonlinear optics"
+layout: "category"
+---
diff --git a/source/know/category/numerical-methods/index.md b/source/know/category/numerical-methods/index.md
new file mode 100644
index 0000000..7e7ee13
--- /dev/null
+++ b/source/know/category/numerical-methods/index.md
@@ -0,0 +1,4 @@
+---
+title: "Numerical methods"
+layout: "category"
+---
diff --git a/source/know/category/optics/index.md b/source/know/category/optics/index.md
new file mode 100644
index 0000000..2889f03
--- /dev/null
+++ b/source/know/category/optics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Optics"
+layout: "category"
+---
diff --git a/source/know/category/perturbation/index.md b/source/know/category/perturbation/index.md
new file mode 100644
index 0000000..3901a32
--- /dev/null
+++ b/source/know/category/perturbation/index.md
@@ -0,0 +1,4 @@
+---
+title: "Perturbation"
+layout: "category"
+---
diff --git a/source/know/category/physics/index.md b/source/know/category/physics/index.md
new file mode 100644
index 0000000..5fecc13
--- /dev/null
+++ b/source/know/category/physics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Physics"
+layout: "category"
+---
diff --git a/source/know/category/plasma-physics/index.md b/source/know/category/plasma-physics/index.md
new file mode 100644
index 0000000..8e23912
--- /dev/null
+++ b/source/know/category/plasma-physics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Plasma physics"
+layout: "category"
+---
diff --git a/source/know/category/plasma-waves/index.md b/source/know/category/plasma-waves/index.md
new file mode 100644
index 0000000..4fb7b4a
--- /dev/null
+++ b/source/know/category/plasma-waves/index.md
@@ -0,0 +1,4 @@
+---
+title: "Plasma waves"
+layout: "category"
+---
diff --git a/source/know/category/quantum-information/index.md b/source/know/category/quantum-information/index.md
new file mode 100644
index 0000000..4fb3a26
--- /dev/null
+++ b/source/know/category/quantum-information/index.md
@@ -0,0 +1,4 @@
+---
+title: "Quantum information"
+layout: "category"
+---
diff --git a/source/know/category/quantum-mechanics/index.md b/source/know/category/quantum-mechanics/index.md
new file mode 100644
index 0000000..279062a
--- /dev/null
+++ b/source/know/category/quantum-mechanics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Quantum mechanics"
+layout: "category"
+---
diff --git a/source/know/category/statistics/index.md b/source/know/category/statistics/index.md
new file mode 100644
index 0000000..be0f520
--- /dev/null
+++ b/source/know/category/statistics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Statistics"
+layout: "category"
+---
diff --git a/source/know/category/stochastic-analysis/index.md b/source/know/category/stochastic-analysis/index.md
new file mode 100644
index 0000000..99827b4
--- /dev/null
+++ b/source/know/category/stochastic-analysis/index.md
@@ -0,0 +1,4 @@
+---
+title: "Stochastic analysis"
+layout: "category"
+---
diff --git a/source/know/category/surface-tension/index.md b/source/know/category/surface-tension/index.md
new file mode 100644
index 0000000..a0e8dbb
--- /dev/null
+++ b/source/know/category/surface-tension/index.md
@@ -0,0 +1,4 @@
+---
+title: "Surface tension"
+layout: "category"
+---
diff --git a/source/know/category/thermodynamic-ensembles/index.md b/source/know/category/thermodynamic-ensembles/index.md
new file mode 100644
index 0000000..57fb482
--- /dev/null
+++ b/source/know/category/thermodynamic-ensembles/index.md
@@ -0,0 +1,4 @@
+---
+title: "Thermodynamic ensembles"
+layout: "category"
+---
diff --git a/source/know/category/thermodynamics/index.md b/source/know/category/thermodynamics/index.md
new file mode 100644
index 0000000..1f1fa11
--- /dev/null
+++ b/source/know/category/thermodynamics/index.md
@@ -0,0 +1,4 @@
+---
+title: "Thermodynamics"
+layout: "category"
+---
diff --git a/source/know/category/two-level-system/index.md b/source/know/category/two-level-system/index.md
new file mode 100644
index 0000000..57d03f7
--- /dev/null
+++ b/source/know/category/two-level-system/index.md
@@ -0,0 +1,4 @@
+---
+title: "Two-level system"
+layout: "category"
+---
diff --git a/source/know/concept/alfven-waves/index.md b/source/know/concept/alfven-waves/index.md
new file mode 100644
index 0000000..9c5e460
--- /dev/null
+++ b/source/know/concept/alfven-waves/index.md
@@ -0,0 +1,243 @@
+---
+title: "Alfvén waves"
+date: 2022-01-31
+categories:
+- Physics
+- Plasma physics
+- Plasma waves
+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 $1$):
+
+$$\begin{aligned}
+ \vb{u}
+ = \vb{u}_0 + \vb{u}_1
+ \qquad
+ \vb{J}
+ = \vb{J}_0 + \vb{J}_1
+ \qquad
+ \vb{B}
+ = \vb{B}_0 + \vb{B}_1
+ \qquad
+ \vb{E}
+ = \vb{E}_0 + \vb{E}_1
+\end{aligned}$$
+
+Inserting this decomposition into the ideal form of the generalized Ohm's law
+and keeping only terms that are first-order in the perturbation, we get:
+
+$$\begin{aligned}
+ 0
+ &= (\vb{E}_0 + \vb{E}_1) + (\vb{u}_0 + \vb{u}_1) \cross (\vb{B}_0 + \vb{B}_1)
+ \\
+ &= \vb{E}_1 + \vb{u}_1 \cross \vb{B}_0
+\end{aligned}$$
+
+We do this for the momentum equation too,
+assuming that $\vb{J}_0 \!=\! 0$ (to be justified later).
+Note that the temperature is set to zero, such that the pressure vanishes:
+
+$$\begin{aligned}
+ \rho \pdv{\vb{u}_1}{t}
+ = \vb{J}_1 \cross \vb{B}_0
+\end{aligned}$$
+
+Where $\rho$ is the uniform equilibrium density.
+We would like an equation for $\vb{J}_1$,
+which is provided by the magnetohydrodynamic form of Ampère's law:
+
+$$\begin{aligned}
+ \nabla \cross \vb{B}_1
+ = \mu_0 \vb{J}_1
+ \qquad \implies \quad
+ \vb{J}_1
+ = \frac{1}{\mu_0} \nabla \cross \vb{B}_1
+\end{aligned}$$
+
+Substituting this into the momentum equation,
+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 $\ipdv{\vb{B}_1}{t}$,
+incorporating Ohm's law too:
+
+$$\begin{aligned}
+ \pdv{\vb{B}_1}{t}
+ = - \nabla \cross \vb{E}_1
+ = \nabla \cross (\vb{u}_1 \cross \vb{B}_0)
+\end{aligned}$$
+
+Inserting this into the momentum equation for $\vb{u}_1$
+thus yields its final form:
+
+$$\begin{aligned}
+ \rho \pdvn{2}{\vb{u}_1}{t}
+ = \frac{1}{\mu_0} \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vb{B}_0) \big) \Big) \cross \vb{B}_0 \bigg)
+\end{aligned}$$
+
+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 $\vb{J}_0 = 0$,
+and the equation can be written as:
+
+$$\begin{aligned}
+ \pdvn{2}{\vb{u}_1}{t}
+ = v_A^2 \bigg( \Big( \nabla \cross \big( \nabla \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
+\end{aligned}$$
+
+Where we have defined the so-called **Alfvén velocity** $v_A$ to be given by:
+
+$$\begin{aligned}
+ \boxed{
+ v_A
+ \equiv \sqrt{\frac{B_0^2}{\mu_0 \rho}}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \omega^2 \vb{u}_1
+ = v_A^2 \bigg( \Big( \vb{k} \cross \big( \vb{k} \cross (\vb{u}_1 \cross \vu{e}_z) \big) \Big) \cross \vu{e}_z \bigg)
+\end{aligned}$$
+
+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.
+Calculating the cross products:
+
+$$\begin{aligned}
+ \omega^2 \vb{u}_1
+ &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix}
+ \cross \big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix}
+ \cross ( \begin{bmatrix} u_{1x} \\ u_{1y} \\ u_{1z} \end{bmatrix}
+ \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} ) \big) \Big)
+ \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg)
+ \\
+ &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix}
+ \cross \big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix}
+ \cross \begin{bmatrix} u_{1y} \\ -u_{1x} \\ 0 \end{bmatrix} \big) \Big)
+ \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg)
+ \\
+ &= v_A^2 \bigg( \Big( \begin{bmatrix} 0 \\ k_\perp \\ k_\parallel \end{bmatrix}
+ \cross \begin{bmatrix} k_\parallel u_{1x} \\ k_\parallel u_{1y} \\ -k_\perp u_{1y} \end{bmatrix} \Big)
+ \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg)
+ \\
+ &= v_A^2 \bigg( \begin{bmatrix} -(k_\perp^2 \!+ k_\parallel^2) u_{1y} \\ k_\parallel^2 u_{1x} \\ -k_\perp k_\parallel u_{1x} \end{bmatrix}
+ \cross \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \bigg)
+ \\
+ &= v_A^2 \begin{bmatrix} k_\parallel^2 u_{1x} \\ (k_\perp^2 \!+ k_\parallel^2) u_{1y} \\ 0 \end{bmatrix}
+\end{aligned}$$
+
+We rewrite this equation in matrix form,
+using that $k_\perp^2 \!+ k_\parallel^2 = k^2 \equiv |\vb{k}|^2$:
+
+$$\begin{aligned}
+ \begin{bmatrix}
+ \omega^2 - v_A^2 k_\parallel^2 & 0 & 0 \\
+ 0 & \omega^2 - v_A^2 k^2 & 0 \\
+ 0 & 0 & \omega^2
+ \end{bmatrix}
+ \vb{u}_1
+ = 0
+\end{aligned}$$
+
+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.
+To achieve this, we demand that the matrix' determinant is zero:
+
+$$\begin{aligned}
+ \big(\omega^2 - v_A^2 k_\parallel^2\big) \: \big(\omega^2 - v_A^2 k^2\big) \: \omega^2
+ = 0
+\end{aligned}$$
+
+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,
+because we are looking for waves.
+
+The first interesting case is $\omega^2 = v_A^2 k_\parallel^2$,
+yielding the following dispersion relation:
+
+$$\begin{aligned}
+ \boxed{
+ \omega
+ = \pm v_A k_\parallel
+ }
+\end{aligned}$$
+
+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 $\vb{u}_1 \cdot \vb{k} = 0$:
+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 $\vb{B}_0$:
+
+$$\begin{aligned}
+ v_p
+ = \frac{|\omega|}{k}
+ = v_A \frac{k_\parallel}{k}
+ = v_A \cos(\theta)
+ \qquad \qquad
+ v_g
+ = \pdv{|\omega|}{k}
+ = v_A
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \omega
+ = \pm v_A k
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ v_p
+ = \frac{|\omega|}{k}
+ = v_A
+ \qquad \qquad
+ v_g
+ = \pdv{|\omega|}{k}
+ = v_A
+\end{aligned}$$
+
+The mechanism behind both of these oscillations is magnetic tension:
+the waves are "ripples" in the field lines,
+which get straightened out by Faraday's law,
+but the ions' inertia causes them to overshoot and form ripples again.
+
+
+
+## References
+1. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
+
diff --git a/source/know/concept/archimedes-principle/index.md b/source/know/concept/archimedes-principle/index.md
new file mode 100644
index 0000000..4bab87b
--- /dev/null
+++ b/source/know/concept/archimedes-principle/index.md
@@ -0,0 +1,89 @@
+---
+title: "Archimedes' principle"
+date: 2021-04-10
+categories:
+- Fluid statics
+- Fluid mechanics
+- Physics
+layout: "concept"
+---
+
+Many objects float when placed on a liquid,
+but some float higher than others,
+and some do not float at all, sinking instead.
+**Archimedes' principle** balances the forces,
+and predicts how much of a body is submerged,
+and how much is non-submerged.
+
+In truth, there is no real distinction between
+the submerged and non-submerged parts,
+since the latter is surrounded by another fluid (air),
+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.
+This volume will experience a downward force due to gravity, given by:
+
+$$\begin{aligned}
+ \va{F}_g
+ = \int_V \va{g} \rho_\mathrm{b} \dd{V}
+\end{aligned}$$
+
+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 $V$:
+
+$$\begin{aligned}
+ \va{F}_p
+ = - \oint_S p \dd{\va{S}}
+ = - \int_V \nabla p \dd{V}
+\end{aligned}$$
+
+Where we have used the divergence theorem.
+Assuming [hydrostatic equilibrium](/know/concept/hydrostatic-pressure/),
+we replace $\nabla p$,
+leading to the definition of the **buoyant force**:
+
+$$\begin{aligned}
+ \boxed{
+ \va{F}_p
+ = - \int_V \va{g} \rho_\mathrm{f} \dd{V}
+ }
+\end{aligned}$$
+
+For the body to be at rest, we require $\va{F}_g + \va{F}_p = 0$.
+Concretely, the equilibrium condition is:
+
+$$\begin{aligned}
+ \boxed{
+ \int_V \va{g} (\rho_\mathrm{b} - \rho_\mathrm{f}) \dd{V}
+ = 0
+ }
+\end{aligned}$$
+
+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,
+and zero on the "non-submerged" side, we find:
+
+$$\begin{aligned}
+ 0
+ = \mathrm{g} (m_\mathrm{b} - m_\mathrm{f})
+\end{aligned}$$
+
+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.
+This is the best-known version of Archimedes' principle.
+
+Note that if $\rho_\mathrm{b} > \rho_\mathrm{f}$,
+then the displaced mass $m_\mathrm{f} < m_\mathrm{b}$
+even if the entire body is submerged,
+and the object will therefore continue to sink.
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/bb84-protocol/index.md b/source/know/concept/bb84-protocol/index.md
new file mode 100644
index 0000000..95ba720
--- /dev/null
+++ b/source/know/concept/bb84-protocol/index.md
@@ -0,0 +1,233 @@
+---
+title: "BB84 protocol"
+date: 2021-03-06
+categories:
+- Quantum information
+- Cryptography
+layout: "concept"
+---
+
+The **BB84** or **Bennett-Brassard 1984** protocol is
+a *quantum key distribution* (QKD) protocol,
+whose purpose is to securely transmit a string of random bits
+over a quantum channel for later use as a one-time pad.
+It is provably information-secure, thanks to the fact that
+quantum channels cannot be eavesdropped without interfering with the signal.
+
+Alice wants to send a secret key to Bob.
+Between them, they have one quantum channel and one classical channel.
+Both channels may have eavesdroppers without compromising the security of the BB84 protocol,
+as long as Alice and Bob authenticate all data sent over the classical channel.
+
+First, Alice securely generates a sequence of random (classical) bits.
+Note that the BB84 protocol is only suitable for random (high-entropy) data,
+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 $\hat{\sigma}_x$).
+Using the basis she chose, she then transmits the bits to Bob over the quantum channel,
+encoding them as follows:
+
+$$\begin{aligned}
+ 0 \:\:\rightarrow\:\: \Ket{0} \:\mathrm{or}\: \Ket{+}
+ \qquad \quad
+ 1 \:\:\rightarrow\:\: \Ket{1} \:\mathrm{or}\: \Ket{-}
+\end{aligned}$$
+
+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,
+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:
+
+$$\begin{aligned}
+ | \Inprod{0}{+} |^2 = | \Inprod{0}{-} |^2 = | \Inprod{1}{+} |^2 = | \Inprod{1}{-} |^2 = \frac{1}{2}
+\end{aligned}$$
+
+After Alice has sent all her qubits,
+the next step is **basis reconciliation**:
+over the classical channel, Bob announces, for each bit,
+which basis he chose, and Alice tells him if he was right or wrong.
+Bob discards all bits where he guessed wrongly.
+If their quantum channel did not have any noise or eavesdroppers,
+Alice and Bob now have a perfectly correlated secret string of bits.
+
+
+## Eavesdropper detection
+
+But what if there is actually an eavesdropper?
+Consider a third party, Eve, who wants to intercept Alice' secret string.
+The main advantage of using QKD compared to classical protocols
+is that such an eavesdropper can be detected.
+
+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.
+She records her results and resends the qubits to Bob
+using the basis she chose, which may or may not be what Alice intended.
+
+If Eve guesses Alice' basis correctly, her presence is not revealed,
+and the protocol proceeds as normal.
+However, if she guesses wrongly (which has a probability of 50%),
+her bit value might be incorrect, and she will resend whatever she measured
+to Bob encoded in the wrong basis.
+
+If we assume that Eve chose wrongly,
+Bob might measure using Eve's basis,
+which will yield a 50% error rate due to Eve's mistake.
+Otherwise, Bob might measure using Alice' basis,
+which will also yield a 50% error rate due to the disagreement with Eve.
+
+In the end, the probability is 0.5 that Eve chose correctly,
+which, multiplied by the 0.5 error rate from Bob's choice,
+will result in a 0.25 error rate due to Eve's presence.
+To detect this, after basis reconciliation,
+Bob reveals a part of his secret string as a sacrifice,
+which allows Alice to estimate the error rate.
+If the rate is too high, Eve is detected, and the protocol can be aborted,
+although that is not mandatory.
+
+This was an intercept-resend attack.
+There exist other attacks, but similar logic holds.
+It has been proven by Shor and Preskill in 2000
+that as long as the error rate is below 11%,
+the BB84 protocol is fully secure, i.e. there cannot be any eavesdroppers.
+
+
+## Error correction
+
+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,
+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:
+
+$$\begin{aligned}
+ A = a_n \oplus a_{n+1}
+ \qquad \quad
+ B = b_n \oplus b_{n+1}
+\end{aligned}$$
+
+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.
+
+Given that $A = B$, the probability that $a_n = b_n$,
+which is what we want, is given by:
+
+$$\begin{aligned}
+ P(a_{n} = b_{n} | A = B)
+ &= \frac{P(a_{n} = b_{n} \land A = B)}{P(A = B)}
+ \\
+ &= \frac{P(a_{n} = b_{n} \land a_{n+1} = b_{n+1})}{P(a_{n} = b_{n} \land a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n} \land a_{n+1} \neq b_{n+1})}
+ \\
+ &= \frac{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1})}{P(a_{n} = b_{n}) \: P(a_{n+1} = b_{n+1}) + P(a_{n} \neq b_{n}) \: P(a_{n+1} \neq b_{n+1})}
+\end{aligned}$$
+
+We use the definition of $p$ to get the following inequality,
+which can be verified by plotting:
+
+$$\begin{aligned}
+ P(a_{n} = b_{n} | A = B)
+ = \frac{p^2}{p^2 + (1 - p)^2}
+ > p
+\end{aligned}$$
+
+Alice and Bob can repeat this error correction scheme multiple times,
+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
+after one iteration is given by:
+
+$$\begin{aligned}
+ N_\mathrm{new}
+ = \frac{1}{2} N_\mathrm{old} P(A = B)
+ = \frac{1}{2} N_\mathrm{old} \big( p^2 + (1 - p)^2 \big)
+\end{aligned}$$
+
+More efficient schemes exist, which do not consume so many bits.
+
+
+## Privacy amplification
+
+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 $\{e_1, ..., e_N\}$,
+such that she knows more that 50% of the bits:
+
+$$\begin{aligned}
+ q = P(e_n = a_n) > \frac{1}{2}
+\end{aligned}$$
+
+**Privacy amplification** is an optional final step of the BB84 protocol
+which aims to reduce Eve's $q$.
+Alice and Bob use their existing strings to generate a new one
+$\{a_1', ..., a_M'\}$:
+
+$$\begin{aligned}
+ a_1'
+ = a_1 \oplus a_2 = b_1 \oplus b_2
+ \qquad
+ a_2'
+ = a_3 \oplus a_4 = b_3 \oplus b_4
+ \qquad
+ \cdots
+\end{aligned}$$
+
+Note that this halves the string's length;
+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 $e_m' = e_{2m - 1} \oplus e_{2m}$.
+The probability that Eve's result agrees with
+Alice and Bob's string is given by:
+
+$$\begin{aligned}
+ P(e_m' = a_m')
+ &= P(e_1 = a_1 \land e_2 = a_2) + P(e_1 \neq a_1 \land e_2 \neq a_2)
+ \\
+ &= P(e_1 = a_1) \: P(e_2 = a_2) + P(e_1 \neq a_1) \: P(e_2 \neq a_2)
+\end{aligned}$$
+
+Recognizing $q$ 's definition,
+we find the following inequality,
+which can be verified by plotting:
+
+$$\begin{aligned}
+ P(e_m' = a_m')
+ = q^2 + (1 - q)^2
+ < q
+\end{aligned}$$
+
+After repeating this step several times, $q$ will be close to 1/2,
+which is the ideal value: for $q =$ 0.5,
+Eve would only know 50% of the bits,
+which is equivalent to her guessing at random.
+
+
+## References
+1. N. Brunner,
+ *Quantum information theory: lecture notes*,
+ 2019, unpublished.
+2. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/bell-state/index.md b/source/know/concept/bell-state/index.md
new file mode 100644
index 0000000..15b916d
--- /dev/null
+++ b/source/know/concept/bell-state/index.md
@@ -0,0 +1,93 @@
+---
+title: "Bell state"
+date: 2021-03-09
+categories:
+- Quantum mechanics
+- Quantum information
+layout: "concept"
+---
+
+In quantum information, the **Bell states** are a set of four two-qubit states
+which are simple and useful examples of [quantum entanglement](/know/concept/quantum-entanglement/).
+They are given by:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \ket{\Phi^{\pm}}
+ &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{0}_B \pm \Ket{1}_A \Ket{1}_B \Big)
+ \\
+ \ket{\Psi^{\pm}}
+ &= \frac{1}{\sqrt{2}} \Big( \Ket{0}_A \Ket{1}_B \pm \Ket{1}_A \Ket{0}_B \Big)
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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 $\Ket{1}$.
+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 $\ket{\Phi^{+}}$.
+Consider the following pure [density operator](/know/concept/density-operator/):
+
+$$\begin{aligned}
+ \hat{\rho}
+ = \ket{\Phi^{+}} \bra{\Phi^{+}}
+ &= \frac{1}{2} \Big( \Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big) \Big( \Bra{0}_A \Bra{0}_B + \Bra{1}_A \Bra{1}_B \Big)
+\end{aligned}$$
+
+The reduced density operator $\hat{\rho}_A$ of qubit $A$ is then calculated as follows:
+
+$$\begin{aligned}
+ \hat{\rho}_A
+ &= \Tr_B(\hat{\rho})
+ = \sum_{b = 0, 1} \Bra{b}_B \Big( \ket{\Phi^{+}} \bra{\Phi^{+}} \Big) \Ket{b}_B
+ \\
+ &= \sum_{b = 0, 1} \Big( \Ket{0}_A \Inprod{b}{0}_B + \Ket{1}_A \Inprod{b}{1}_B \Big)
+ \Big( \Bra{0}_A \Inprod{0}{b}_B + \Bra{1}_A \Inprod{1}{b}_B \Big)
+ \\
+ &= \frac{1}{2} \Big( \Ket{0}_A \Bra{0}_A + \Ket{1}_A \Bra{1}_A \Big)
+ = \frac{1}{2} \hat{I}
+\end{aligned}$$
+
+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 $B$.
+
+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 $B$:
+
+$$\begin{aligned}
+ \big| \Bra{0}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+ &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{0}{0}_B + \Inprod{0}{1}_A \Inprod{0}{1}_B \Big)^2
+ = \frac{1}{2}
+ \\
+ \big| \Bra{0}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+ &= \frac{1}{2} \Big( \Inprod{0}{0}_A \Inprod{1}{0}_B + \Inprod{0}{1}_A \Inprod{1}{1}_B \Big)^2
+ = 0
+ \\
+ \big| \Bra{1}_A \! \Bra{0}_B \cdot \ket{\Phi^{+}} \big|^2
+ &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{0}{0}_B + \Inprod{1}{1}_A \Inprod{0}{1}_B \Big)^2
+ = 0
+ \\
+ \big| \Bra{1}_A \! \Bra{1}_B \cdot \ket{\Phi^{+}} \big|^2
+ &= \frac{1}{2} \Big( \Inprod{1}{0}_A \Inprod{1}{0}_B + \Inprod{1}{1}_A \Inprod{1}{1}_B \Big)^2
+ = \frac{1}{2}
+\end{aligned}$$
+
+As an example, if $A$ collapses into $\Ket{0}$ due to a measurement,
+then $B$ instantly also collapses into $\Ket{0}$, never $\Ket{1}$,
+even if it was not measured.
+This was a specific example for $\ket{\Phi^{+}}$,
+but analogous results can be found for the other Bell states.
+
+
+## References
+1. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md
new file mode 100644
index 0000000..1752854
--- /dev/null
+++ b/source/know/concept/bells-theorem/index.md
@@ -0,0 +1,372 @@
+---
+title: "Bell's theorem"
+date: 2021-03-28
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+layout: "concept"
+---
+
+**Bell's theorem** states that the laws of quantum mechanics
+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 $B$,
+in an entangled [Bell state](/know/concept/bell-state/):
+
+$$\begin{aligned}
+ \Ket{\Psi^{-}}
+ = \frac{1}{\sqrt{2}} \Big( \Ket{\uparrow \downarrow} - \Ket{\downarrow \uparrow} \Big)
+\end{aligned}$$
+
+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 $\Ket{\downarrow}$.
+The point is that this collapse is instant,
+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 $B$
+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 $\lambda$,
+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:
+
+$$\begin{aligned}
+ A(\vec{a}, \lambda) = \pm 1
+ \qquad \quad
+ B(\vec{b}, \lambda) = \pm 1
+\end{aligned}$$
+
+Where $\pm 1$ are the eigenvalues of the Pauli matrices
+in the chosen directions $\vec{a}$ and $\vec{b}$:
+
+$$\begin{aligned}
+ \hat{\sigma}_a
+ &= \vec{a} \cdot \vec{\sigma}
+ = a_x \hat{\sigma}_x + a_y \hat{\sigma}_y + a_z \hat{\sigma}_z
+ \\
+ \hat{\sigma}_b
+ &= \vec{b} \cdot \vec{\sigma}
+ = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z
+\end{aligned}$$
+
+Whether $\lambda$ is a scalar or a vector does not matter;
+we simply demand that it follows an unknown probability distribution $\rho(\lambda)$:
+
+$$\begin{aligned}
+ \int \rho(\lambda) \dd{\lambda} = 1
+ \qquad \quad
+ \rho(\lambda) \ge 0
+\end{aligned}$$
+
+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:
+this is what makes it a **local** hidden variable:
+
+$$\begin{aligned}
+ \Expval{A_a B_b}
+ = \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+From this, two inequalities can be derived,
+which both prove Bell's theorem.
+
+
+## Bell inequality
+
+If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins:
+
+$$\begin{aligned}
+ A(\vec{a}, \lambda)
+ = A(\vec{b}, \lambda)
+ = - B(\vec{b}, \lambda)
+\end{aligned}$$
+
+The expectation value of the product can therefore be rewritten as follows:
+
+$$\begin{aligned}
+ \Expval{A_a B_b}
+ = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+Next, we introduce an arbitrary third direction $\vec{c}$,
+and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$:
+
+$$\begin{aligned}
+ \Expval{A_a B_b} - \Expval{A_a B_c}
+ &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda}
+ \\
+ &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+Inside the integral, the only factors that can be negative
+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:
+
+$$\begin{aligned}
+ \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big|
+ &\le \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big)
+ \: \Big| A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \Big| \dd{\lambda}
+ \\
+ &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda}
+\end{aligned}$$
+
+Since $\rho(\lambda)$ is a normalized probability density function,
+we arrive at the **Bell inequality**:
+
+$$\begin{aligned}
+ \boxed{
+ \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big|
+ \le 1 + \Expval{A_b B_c}
+ }
+\end{aligned}$$
+
+Any theory involving an LHV $\lambda$ must obey this inequality.
+The problem, however, is that quantum mechanics dictates the expectation values
+for the state $\Ket{\Psi^{-}}$:
+
+$$\begin{aligned}
+ \Expval{A_a B_b} = - \vec{a} \cdot \vec{b}
+\end{aligned}$$
+
+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 $0.293$,
+which clearly disagrees with the inequality,
+meaning that LHVs are impossible.
+
+
+## CHSH inequality
+
+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 $\vec{b}_2$.
+Let us introduce the following abbreviations:
+
+$$\begin{aligned}
+ A_1 &= A(\vec{a}_1, \lambda)
+ \qquad \quad
+ A_2 = A(\vec{a}_2, \lambda)
+ \\
+ B_1 &= B(\vec{b}_1, \lambda)
+ \qquad \quad
+ B_2 = B(\vec{b}_2, \lambda)
+\end{aligned}$$
+
+From the definition of the expectation value,
+we know that the difference is given by:
+
+$$\begin{aligned}
+ \Expval{A_1 B_1} - \Expval{A_1 B_2}
+ = \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \Big) \dd{\lambda}
+\end{aligned}$$
+
+We introduce some new terms and rearrange the resulting expression:
+
+$$\begin{aligned}
+ \Expval{A_1 B_1} - \Expval{A_1 B_2}
+ &= \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \pm A_1 B_1 A_2 B_2 \mp A_1 B_1 A_2 B_2 \Big) \dd{\lambda}
+ \\
+ &= \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
+\end{aligned}$$
+
+Taking the absolute value of both sides
+and invoking the triangle inequality then yields:
+
+$$\begin{aligned}
+ \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
+ &= \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
+ \\
+ &\le \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} \!\bigg|
+ + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
+\end{aligned}$$
+
+Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$,
+we can reduce this to:
+
+$$\begin{aligned}
+ \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
+ &\le \int \rho(\lambda) \Big| A_1 B_1 \Big| \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ + \!\int \rho(\lambda) \Big| A_1 B_2 \Big| \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
+ \\
+ &\le \int \rho(\lambda) \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
+ + \!\int \rho(\lambda) \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
+\end{aligned}$$
+
+Evaluating these integrals gives us the following inequality,
+which holds for both choices of $\pm$:
+
+$$\begin{aligned}
+ \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
+ &\le 2 \pm \Expval{A_2 B_2} \pm \Expval{A_2 B_1}
+\end{aligned}$$
+
+We should choose the signs such that the right-hand side is as small as possible, that is:
+
+$$\begin{aligned}
+ \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
+ &\le 2 \pm \Big( \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big)
+ \\
+ &\le 2 - \Big| \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
+\end{aligned}$$
+
+Rearranging this and once again using the triangle inequality,
+we get the CHSH inequality:
+
+$$\begin{aligned}
+ 2
+ &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| + \Big| \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
+ \\
+ &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} + \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
+\end{aligned}$$
+
+The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$,
+and measures the correlation between the spins of $A$ and $B$:
+
+$$\begin{aligned}
+ \boxed{
+ S \equiv \Expval{A_2 B_1} + \Expval{A_2 B_2} + \Expval{A_1 B_1} - \Expval{A_1 B_2}
+ }
+\end{aligned}$$
+
+The CHSH inequality places an upper bound on the magnitude of $S$
+for LHV-based theories:
+
+$$\begin{aligned}
+ \boxed{
+ |S| \le 2
+ }
+\end{aligned}$$
+
+
+## Tsirelson's bound
+
+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. $S = \expval{\hat{S}}$:
+
+$$\begin{aligned}
+ \hat{S}
+ = \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2
+\end{aligned}$$
+
+Where $\otimes$ is the tensor product,
+and e.g. $\hat{A}_1$ is the Pauli matrix for the $\vec{a}_1$-direction.
+The square of this operator is then given by:
+
+$$\begin{aligned}
+ \hat{S}^2
+ = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_1 \hat{B}_2
+ + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1^2 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1 \hat{B}_2
+ \\
+ + &\hat{A}_2^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_2^2 \otimes \hat{B}_2^2
+ + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2^2
+ \\
+ + &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_1^2 + \hat{A}_1 \hat{A}_2 \otimes \hat{B}_1 \hat{B}_2
+ + \hat{A}_1^2 \otimes \hat{B}_1^2 - \hat{A}_1^2 \otimes \hat{B}_1 \hat{B}_2
+ \\
+ - &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_1 \hat{A}_2 \otimes \hat{B}_2^2
+ - \hat{A}_1^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_1^2 \otimes \hat{B}_2^2
+ \\
+ = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_2^2 + \hat{A}_1^2 \otimes \hat{B}_1^2 + \hat{A}_1^2 \otimes \hat{B}_2^2
+ \\
+ + &\hat{A}_2^2 \otimes \acomm{\hat{B}_1}{\hat{B}_2} - \hat{A}_1^2 \otimes \acomm{\hat{B}_1}{\hat{B}_2}
+ + \acomm{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_1^2 - \acomm{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_2^2
+ \\
+ + &\hat{A}_1 \hat{A}_2 \otimes \comm{\hat{B}_1}{\hat{B}_2} - \hat{A}_2 \hat{A}_1 \otimes \comm{\hat{B}_1}{\hat{B}_2}
+\end{aligned}$$
+
+Spin operators are unitary, so their square is the identity,
+e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to:
+
+$$\begin{aligned}
+ \hat{S}^2
+ &= 4 \: (\hat{I} \otimes \hat{I}) + \comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}
+\end{aligned}$$
+
+The *norm* $\norm{\hat{S}^2}$ of this operator
+is the largest possible expectation value $\expval{\hat{S}^2}$,
+which is the same as its largest eigenvalue.
+It is given by:
+
+$$\begin{aligned}
+ \Norm{\hat{S}^2}
+ &= 4 + \Norm{\comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}}
+ \\
+ &\le 4 + \Norm{\comm{\hat{A}_1}{\hat{A}_2}} \Norm{\comm{\hat{B}_1}{\hat{B}_2}}
+\end{aligned}$$
+
+We find a bound for the norm of the commutators by using the triangle inequality, such that:
+
+$$\begin{aligned}
+ \Norm{\comm{\hat{A}_1}{\hat{A}_2}}
+ = \Norm{\hat{A}_1 \hat{A}_2 - \hat{A}_2 \hat{A}_1}
+ \le \Norm{\hat{A}_1 \hat{A}_2} + \Norm{\hat{A}_2 \hat{A}_1}
+ \le 2 \Norm{\hat{A}_1 \hat{A}_2}
+ \le 2
+\end{aligned}$$
+
+And $\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason.
+The norm is the largest eigenvalue, therefore:
+
+$$\begin{aligned}
+ \Norm{\hat{S}^2}
+ \le 4 + 2 \cdot 2
+ = 8
+ \quad \implies \quad
+ \Norm{\hat{S}}
+ \le \sqrt{8}
+ = 2 \sqrt{2}
+\end{aligned}$$
+
+We thus arrive at **Tsirelson's bound**,
+which states that quantum mechanics can violate
+the CHSH inequality by a factor of $\sqrt{2}$:
+
+$$\begin{aligned}
+ \boxed{
+ |S|
+ \le 2 \sqrt{2}
+ }
+\end{aligned}$$
+
+Importantly, this is a *tight* bound,
+meaning that there exist certain spin measurement directions
+for which Tsirelson's bound becomes an equality, for example:
+
+$$\begin{aligned}
+ \hat{A}_1 = \hat{\sigma}_z
+ \qquad
+ \hat{A}_2 = \hat{\sigma}_x
+ \qquad
+ \hat{B}_1 = \frac{\hat{\sigma}_z + \hat{\sigma}_x}{\sqrt{2}}
+ \qquad
+ \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}}
+\end{aligned}$$
+
+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.
+
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
+2. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/beltrami-identity/index.md b/source/know/concept/beltrami-identity/index.md
new file mode 100644
index 0000000..e9d7a4c
--- /dev/null
+++ b/source/know/concept/beltrami-identity/index.md
@@ -0,0 +1,134 @@
+---
+title: "Beltrami identity"
+date: 2022-09-17
+categories:
+- Physics
+- Mathematics
+layout: "concept"
+---
+
+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 $J[f]$,
+the [calculus of variations](/know/concept/calculus-of-variations/)
+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 $f'(x)$.
+Using the chain rule:
+
+$$\begin{aligned}
+ \dv{L}{x}
+ = \pdv{L}{f} \dv{f}{x} + \pdv{L}{f'} \dv{f'}{x} + \pdv{L}{x}
+\end{aligned}$$
+
+Substituting the Euler-Lagrange equation into the first term gives us:
+
+$$\begin{aligned}
+ \dv{L}{x}
+ &= f' \dv{}{x} \Big( \pdv{L}{f'} \Big) + \dv{f'}{x} \pdv{L}{f'} + \pdv{L}{x}
+ \\
+ &= \dv{}{x} \bigg( f' \pdv{L}{f'} \bigg) + \pdv{L}{x}
+\end{aligned}$$
+
+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 $x$:
+
+$$\begin{aligned}
+ - \pdv{L}{x}
+ = \dv{}{x} \bigg( f' \pdv{L}{f'} - L \bigg)
+\end{aligned}$$
+
+What if $L$ does not explicitly depend on $x$, i.e. $\ipdv{L}{x} = 0$?
+In that case, the equation can be integrated to give the **Beltrami identity**:
+
+$$\begin{aligned}
+ \boxed{
+ f' \pdv{L}{f'} - L
+ = C
+ }
+\end{aligned}$$
+
+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.
+
+
+## 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:
+
+$$\begin{aligned}
+ J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
+\end{aligned}$$
+
+In which case the Euler-Lagrange equation takes the following form:
+
+$$\begin{aligned}
+ 0 = \pdv{L}{f} - \dv{}{x} \Big( \pdv{L}{f_x} \Big) - \dv{}{y} \Big( \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+Once again, we calculate the hard $x$-derivative of $L$ (the $y$-derivative is analogous):
+
+$$\begin{aligned}
+ \dv{L}{x}
+ &= \pdv{L}{f} \dv{f}{x} + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
+ \\
+ &= \dv{f}{x} \bigg( \dv{}{x} \Big( \pdv{L}{f_x} \Big) + \dv{}{y} \Big( \pdv{L}{f_y} \Big) \bigg)
+ + \pdv{L}{f_x} \dv{f_x}{x} + \pdv{L}{f_y} \dv{f_y}{x} + \pdv{L}{x}
+ \\
+ &= \dv{}{x} \Big( f_x \pdv{L}{f_x} \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) + \pdv{L}{x}
+\end{aligned}$$
+
+This time, we arrive at the following expression for the soft derivative $\ipdv{L}{x}$:
+
+$$\begin{aligned}
+ - \pdv{L}{x}
+ &= \dv{}{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+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.
+For an off-topic demonstration of this fact,
+let us choose $x$ as the transverse coordinate, and integrate over it to get:
+
+$$\begin{aligned}
+ 0
+ &= - \int_{x_0}^{x_1} \pdv{L}{x} \dd{x}
+ \\
+ &= \int_{x_0}^{x_1} \dv{}{x} \Big( f_x \pdv{L}{f_x} - L \Big) + \dv{}{y} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
+ \\
+ &= \Big[ f_x \pdv{L}{f_x} - L \Big]_{x_0}^{x_1} + \dv{}{y} \int_{x_0}^{x_1} \Big( f_x \pdv{L}{f_y} \Big) \dd{x}
+\end{aligned}$$
+
+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 $y$.
+While not as elegant as the 1D Beltrami identity,
+the above 2D counterpart still fulfills the same role.
+
+
+
+## References
+1. O. Bang,
+ *Nonlinear mathematical physics: lecture notes*, 2020,
+ unpublished.
diff --git a/source/know/concept/bernoullis-theorem/index.md b/source/know/concept/bernoullis-theorem/index.md
new file mode 100644
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+---
+title: "Bernoulli's theorem"
+date: 2021-04-02
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+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.
+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/):
+
+$$\begin{aligned}
+ \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+ = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
+ = \va{g} - \frac{\nabla p}{\rho}
+\end{aligned}$$
+
+Assuming that $\va{v}$ is constant in $t$,
+it becomes clear that a higher $\va{v}$ requires a lower $p$.
+
+
+## Simple form
+
+For an incompressible fluid
+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:
+
+$$\begin{aligned}
+ \boxed{
+ H
+ = \frac{1}{2} \va{v}^2 + \Phi + \frac{p}{\rho}
+ }
+\end{aligned}$$
+
+Where $\Phi$ is the gravitational potential, such that $\va{g} = - \nabla \Phi$.
+To prove this theorem, we take the material derivative of $H$:
+
+$$\begin{aligned}
+ \frac{\mathrm{D} H}{\mathrm{D} t}
+ &= \va{v} \cdot \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+ + \frac{\mathrm{D} \Phi}{\mathrm{D} t}
+ + \frac{1}{\rho} \frac{\mathrm{D} p}{\mathrm{D} t}
+\end{aligned}$$
+
+In the first term we insert the Euler equation,
+and in the other two we expand the derivatives:
+
+$$\begin{aligned}
+ \frac{\mathrm{D} H}{\mathrm{D} t}
+ &= \va{v} \cdot \Big( \va{g} - \frac{\nabla p}{\rho} \Big)
+ + \Big( \pdv{\Phi}{t} + (\va{v} \cdot \nabla) \Phi \Big)
+ + \frac{1}{\rho} \Big( \pdv{p}{t} + (\va{v} \cdot \nabla) p \Big)
+ \\
+ &= \pdv{\Phi}{t} + \frac{1}{\rho} \pdv{p}{t}
+ + \va{v} \cdot \big( \va{g} + \nabla \Phi \big) + \va{v} \cdot \Big( \frac{\nabla p}{\rho} - \frac{\nabla p}{\rho} \Big)
+\end{aligned}$$
+
+Using the fact that $\va{g} = - \nabla \Phi$,
+we are left with the following equation:
+
+$$\begin{aligned}
+ \frac{\mathrm{D} H}{\mathrm{D} t}
+ &= \pdv{\Phi}{t} + \frac{1}{\rho} \pdv{p}{t}
+\end{aligned}$$
+
+Assuming that the flow is steady, both derivatives vanish,
+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.
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/bernstein-vazirani-algorithm/bernstein-vazirani-circuit.png b/source/know/concept/bernstein-vazirani-algorithm/bernstein-vazirani-circuit.png
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index 0000000..83ccde2
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diff --git a/source/know/concept/bernstein-vazirani-algorithm/index.md b/source/know/concept/bernstein-vazirani-algorithm/index.md
new file mode 100644
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+++ b/source/know/concept/bernstein-vazirani-algorithm/index.md
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+---
+title: "Bernstein-Vazirani algorithm"
+date: 2021-05-01
+categories:
+- Quantum information
+- Algorithms
+layout: "concept"
+---
+
+In quantum information,
+the **Bernstein-Vazirani algorithm** proves
+the supremacy of quantum computers
+over classical deterministic or probabilistic computers.
+It is extremely similar to the
+[Deutsch-Jozsa algorithm](/know/concept/deutsch-jozsa-algorithm/),
+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,
+which we are promised is the lowest bit of the bitwise dot product
+of $x$ with an unknown $N$-bit string $s$:
+
+$$\begin{aligned}
+ f(x)
+ = s \cdot x \:\:(\bmod \: 2)
+ = (s_1 x_1 + s_2 x_2 + \:...\: + s_N x_N) \:\:(\bmod \: 2)
+\end{aligned}$$
+
+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 $n \in \{ 0, ..., N \!-\! 1\}$.
+However, the Bernstein-Vazirani algorithm
+allows a quantum computer to do it with only a single query.
+It uses the following circuit:
+
+
+
+
+Where $U_f$ is a phase oracle,
+whose action is defined as follows,
+where $\Ket{x} = \Ket{x_1} \cdots \Ket{x_N}$:
+
+$$\begin{aligned}
+ \Ket{x}
+ \quad \to \boxed{U_f} \to \quad
+ (-1)^{f(x)} \Ket{x}
+ = (-1)^{s \cdot x} \Ket{x}
+\end{aligned}$$
+
+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.
+
+Starting from the state $\Ket{0}^{\otimes N}$,
+applying the [Hadamard gate](/know/concept/quantum-gate/) $H$
+to all qubits yields:
+
+$$\begin{aligned}
+ \Ket{0}^{\otimes N}
+ \quad \to \boxed{H^{\otimes N}} \to \quad
+ \Ket{+}^{\otimes N}
+ = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x}
+\end{aligned}$$
+
+This is an equal superposition of all candidates $\Ket{x}$,
+which we feed to the oracle:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x}
+ \quad \to \boxed{U_f} \to \quad
+ \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \Ket{x}
+\end{aligned}$$
+
+Then, thanks to the definition of the Hadamard transform,
+a final set of $H$-gates leads us to:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{s \cdot x} \Ket{x}
+ \quad \to \boxed{H^{\otimes N}} \to \quad
+ \Ket{s}
+ = \Ket{s_1} \cdots \Ket{s_N}
+\end{aligned}$$
+
+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 $s$.
+
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
diff --git a/source/know/concept/berry-phase/index.md b/source/know/concept/berry-phase/index.md
new file mode 100644
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@@ -0,0 +1,213 @@
+---
+title: "Berry phase"
+date: 2021-11-29
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+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:
+
+$$\begin{aligned}
+ i \hbar \dv{}{t}\Ket{\Psi_n(t)}
+ &= \hat{H}(\vb{R}) \Ket{\Psi_n(t)}
+ \\
+ \hat{H}(\vb{R}) \Ket{\psi_n(\vb{R})}
+ &= E_n(\vb{R}) \Ket{\psi_n(\vb{R})}
+\end{aligned}$$
+
+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 $L_n$:
+
+$$\begin{aligned}
+ \Ket{\Psi_n(t)}
+ = \exp(i \gamma_n(t)) \exp(-i L_n(t) / \hbar) \: \Ket{\psi_n(\vb{R}(t))}
+ \qquad
+ L_n(t) \equiv \int_0^t E_n(\vb{R}(t')) \dd{t'}
+\end{aligned}$$
+
+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 $\Ket{\Psi_n}$
+into the time-dependent Schrödinger equation:
+
+$$\begin{aligned}
+ \dv{}{t}\Ket{\Psi_n}
+ &= i \dv{\gamma_n}{t} \Ket{\Psi_n} - \frac{i}{\hbar} \dv{L_n}{t} \Ket{\Psi_n}
+ + \exp(i \gamma_n) \exp(-i L_n / \hbar) \dv{}{t}\Ket{\psi_n}
+ \\
+ &= i \dv{\gamma_n}{t} \Ket{\Psi_n} + \frac{1}{i \hbar} E_n \Ket{\Psi_n}
+ + \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
+ \\
+ &= i \dv{\gamma_n}{t} \Ket{\Psi_n} + \frac{1}{i \hbar} \hat{H} \Ket{\Psi_n}
+ + \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
+\end{aligned}$$
+
+Here we recognize the Schrödinger equation, so those terms cancel.
+We are then left with:
+
+$$\begin{aligned}
+ - i \dv{\gamma_n}{t} \Ket{\Psi_n}
+ &= \exp(i \gamma_n) \exp(-i L_n / \hbar) \: \Ket{\nabla_\vb{R} \psi_n} \cdot \dv{\vb{R}}{t}
+\end{aligned}$$
+
+Front-multiplying by $i \Bra{\Psi_n}$ gives us
+the equation of motion of the geometric phase $\gamma_n$:
+
+$$\begin{aligned}
+ \boxed{
+ \dv{\gamma_n}{t}
+ = - \vb{A}_n(\vb{R}) \cdot \dv{\vb{R}}{t}
+ }
+\end{aligned}$$
+
+Where we have defined the so-called **Berry connection** $\vb{A}_n$ as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{A}_n(\vb{R})
+ \equiv -i \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})}
+ }
+\end{aligned}$$
+
+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 $\nabla_\vb{R} 1 = 0$:
+
+$$\begin{aligned}
+ 0
+ &= \nabla_\vb{R} \Inprod{\psi_n}{\psi_n}
+ = \Inprod{\nabla_\vb{R} \psi_n}{\psi_n} + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}
+ \\
+ &= \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}^* + \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}
+ = 2 \Real\{ - i \vb{A}_n \}
+ = 2 \Imag\{ \vb{A}_n \}
+\end{aligned}$$
+
+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.
+
+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** $\gamma_n(C)$:
+
+$$\begin{aligned}
+ \boxed{
+ \gamma_n(C)
+ = - \oint_C \vb{A}_n(\vb{R}) \cdot \dd{\vb{R}}
+ }
+\end{aligned}$$
+
+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 $\gamma_n(t)$
+without making an immediate physical difference to the state.
+Consider the following general gauge transformation:
+
+$$\begin{aligned}
+ \ket{\tilde{\psi}_n(\vb{R})}
+ \equiv \exp(i f(\vb{R})) \: \Ket{\psi_n(\vb{R})}
+\end{aligned}$$
+
+To find $\vb{A}_n$ for a particular choice of $f$,
+we need to evaluate the inner product
+$\inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}$:
+
+$$\begin{aligned}
+ \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \tilde{\psi}_n}
+ &= \exp(i f) \Big( i \nabla_\vb{R} f \: \inprod{\tilde{\psi}_n}{\psi_n} + \inprod{\tilde{\psi}_n}{\nabla_\vb{R} \psi_n} \Big)
+ \\
+ &= i \nabla_\vb{R} f \: \inprod{\psi_n}{\psi_n} + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
+ \\
+ &= i \nabla_\vb{R} f + \inprod{\psi_n}{\nabla_\vb{R} \psi_n}
+\end{aligned}$$
+
+Unfortunately, $f$ does not vanish as we would have liked,
+so $\vb{A}_n$ depends on our choice of $f$.
+
+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 $\gamma_n(C)$
+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):
+
+$$\begin{aligned}
+ \boxed{
+ \gamma_n(C)
+ = - \iint_{S(C)} \vb{B}_n(\vb{R}) \cdot \dd{\vb{S}}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \vb{B}_n(\vb{R})
+ \equiv \nabla_\vb{R} \cross \vb{A}_n(\vb{R})
+ = \Imag\!\Big\{ \nabla_\vb{R} \cross \Inprod{\psi_n(\vb{R})}{\nabla_\vb{R} \psi_n(\vb{R})} \Big\}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ i \vb{B}_n
+ = \nabla_\vb{R} \cross \Inprod{\psi_n}{\nabla_\vb{R} \psi_n}
+ &= \Inprod{\psi_n}{\nabla_\vb{R} \cross \nabla_\vb{R} \psi_n} + \Bra{\nabla_\vb{R} \psi_n} \cross \Ket{\nabla_\vb{R} \psi_n}
+ \\
+ &= \sum_{m} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
+\end{aligned}$$
+
+The fact that $\Inprod{\psi_n}{\nabla_\vb{R} \psi_n}$ is imaginary
+means it is parallel to its complex conjugate,
+and thus the cross product vanishes, so we exclude $n$ from the sum:
+
+$$\begin{aligned}
+ \vb{B}_n
+ &= \sum_{m \neq n} \Inprod{\nabla_\vb{R} \psi_n}{\psi_m} \cross \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
+\end{aligned}$$
+
+From the [Hellmann-Feynman theorem](/know/concept/hellmann-feynman-theorem/),
+we know that the inner products can be rewritten:
+
+$$\begin{aligned}
+ \Inprod{\psi_m}{\nabla_\vb{R} \psi_n}
+ = \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m}}{E_n - E_m}
+\end{aligned}$$
+
+Where we have assumed that there is no degeneracy.
+This leads to the following result:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{B}_n
+ = \Imag \sum_{m \neq n}
+ \frac{\matrixel{\psi_n}{\nabla_\vb{R} \hat{H}}{\psi_m} \cross \matrixel{\psi_m}{\nabla_\vb{R} \hat{H}}{\psi_n}}{(E_n - E_m)^2}
+ }
+\end{aligned}$$
+
+Which only involves $\nabla_\vb{R} \hat{H}$,
+and is therefore easier to evaluate than any $\Ket{\nabla_\vb{R} \psi_n}$.
+
+
+
+## References
+1. M.V. Berry,
+ [Quantal phase factors accompanying adiabatic changes](https://doi.org/10.1098/rspa.1984.0023),
+ 1984, Royal Society.
+2. G. Grosso, G.P. Parravicini,
+ *Solid state physics*,
+ 2nd edition, Elsevier.
diff --git a/source/know/concept/binomial-distribution/index.md b/source/know/concept/binomial-distribution/index.md
new file mode 100644
index 0000000..c25da3d
--- /dev/null
+++ b/source/know/concept/binomial-distribution/index.md
@@ -0,0 +1,220 @@
+---
+title: "Binomial distribution"
+date: 2021-02-26
+categories:
+- Statistics
+- Mathematics
+layout: "concept"
+---
+
+The **binomial distribution** is a discrete probability distribution
+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 binomial distribution then gives the probability
+that $n$ out of the $N$ trials succeed:
+
+$$\begin{aligned}
+ \boxed{
+ P_N(n) = \binom{N}{n} \: p^n q^{N - n}
+ }
+\end{aligned}$$
+
+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 $(a + b)^N$,
+and can be read off of Pascal's triangle.
+It is defined as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \binom{N}{n} = \frac{N!}{n! (N - n)!}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \mu = N p
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+The trick is to treat $p$ and $q$ as independent until the last moment:
+
+$$\begin{aligned}
+ \mu
+ &= \sum_{n = 0}^N n \binom{N}{n} p^n q^{N - n}
+ = \sum_{n = 0}^N \binom{N}{n} \Big( p \pdv{(p^n)}{p} \Big) q^{N - n}
+ \\
+ &= p \pdv{}{p}\sum_{n = 0}^N \binom{N}{n} p^n q^{N - n}
+ = p \pdv{}{p}(p + q)^N
+ = N p (p + q)^{N - 1}
+\end{aligned}$$
+
+Inserting $q = 1 - p$ then gives the desired result.
+
+
+
+
Proof
+
+Proof.
+We use the same trick to calculate $\overline{n^2}$
+(the mean squared number of successes):
+
+$$\begin{aligned}
+ \overline{n^2}
+ &= \sum_{n = 0}^N n^2 \binom{N}{n} p^n q^{N - n}
+ = \sum_{n = 0}^N n \binom{N}{n} \Big( p \pdv{}{p}\Big)^2 p^n q^{N - n}
+ \\
+ &= \Big( p \pdv{}{p}\Big)^2 \sum_{n = 0}^N \binom{N}{n} p^n q^{N - n}
+ = \Big( p \pdv{}{p}\Big)^2 (p + q)^N
+ \\
+ &= N p \pdv{}{p}p (p + q)^{N - 1}
+ = N p \big( (p + q)^{N - 1} + (N - 1) p (p + q)^{N - 2} \big)
+ \\
+ &= N p + N^2 p^2 - N p^2
+\end{aligned}$$
+
+Using this and the earlier expression $\mu = N p$, we find the variance $\sigma^2$:
+
+$$\begin{aligned}
+ \sigma^2
+ &= \overline{n^2} - \mu^2
+ = N p + N^2 p^2 - N p^2 - N^2 p^2
+ = N p (1 - p)
+\end{aligned}$$
+
+By inserting $q = 1 - p$, we arrive at the desired expression.
+
+
+
+
Proof
+
+Proof.
+We take the Taylor expansion of $\ln\!\big(P_N(n)\big)$
+around the mean $\mu = Np$:
+
+$$\begin{aligned}
+ \ln\!\big(P_N(n)\big)
+ &= \sum_{m = 0}^\infty \frac{(n - \mu)^m}{m!} D_m(\mu)
+ \quad \mathrm{where} \quad
+ D_m(n) = \dvn{m}{\ln\!\big(P_N(n)\big)}{n}
+\end{aligned}$$
+
+We use Stirling's approximation to calculate the factorials in $D_m$:
+
+$$\begin{aligned}
+ \ln\!\big(P_N(n)\big)
+ &= \ln(N!) - \ln(n!) - \ln\!\big((N - n)!\big) + n \ln(p) + (N - n) \ln(q)
+ \\
+ &\approx \ln(N!) - n \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) - (N\!-\!n) \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big)
+\end{aligned}$$
+
+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 $N - N p = N q$:
+
+$$\begin{aligned}
+ D_0(\mu)
+ &= \ln(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln(p) + N q \ln(q)
+ \\
+ &= \Big( N \ln(N) - N + \frac{1}{2} \ln(2\pi N) \Big)
+ - \Big( N p \ln(N p) - N p + \frac{1}{2} \ln(2\pi N p) \Big) \\
+ &\qquad - \Big( N q \ln(N q) - N q + \frac{1}{2} \ln(2\pi N q) \Big)
+ + N p \ln(p) + N q \ln(q)
+ \\
+ &= N \ln(N) - N (p + q) \ln(N) + N (p + q) - N - \frac{1}{2} \ln(2\pi N p q)
+ \\
+ &= - \frac{1}{2} \ln(2\pi N p q)
+ = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big)
+\end{aligned}$$
+
+Next, we expect that $D_1(\mu) = 0$, because $\mu$ is the maximum.
+This is indeed the case:
+
+$$\begin{aligned}
+ D_1(n)
+ &= - \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) + \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big) - 1 + 1
+ \\
+ &= - \ln(n) + \ln(N - n) + \ln(p) - \ln(q)
+ \\
+ D_1(\mu)
+ &= \ln(N q) - \ln(N p) + \ln(p) - \ln(q)
+ = \ln(N p q) - \ln(N p q)
+ = 0
+\end{aligned}$$
+
+For the same reason, we expect that $D_2(\mu)$ is negative.
+We find the following expression:
+
+$$\begin{aligned}
+ D_2(n)
+ &= - \frac{1}{n} - \frac{1}{N - n}
+ \qquad
+ D_2(\mu)
+ = - \frac{1}{Np} - \frac{1}{Nq}
+ = - \frac{p + q}{N p q}
+ = - \frac{1}{\sigma^2}
+\end{aligned}$$
+
+The higher-order derivatives tend to zero for $N \to \infty$, so we discard them:
+
+$$\begin{aligned}
+ D_3(n)
+ = \frac{1}{n^2} - \frac{1}{(N - n)^2}
+ \qquad
+ D_4(n)
+ = - \frac{2}{n^3} - \frac{2}{(N - n)^3}
+ \qquad
+ \cdots
+\end{aligned}$$
+
+Putting everything together, for large $N$,
+the Taylor series approximately becomes:
+
+$$\begin{aligned}
+ \ln\!\big(P_N(n)\big)
+ \approx D_0(\mu) + \frac{(n - \mu)^2}{2} D_2(\mu)
+ = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) - \frac{(n - \mu)^2}{2 \sigma^2}
+\end{aligned}$$
+
+Taking $\exp$ of this expression then yields a normalized Gaussian distribution.
+
+
+
+
+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:
+
+$$\begin{aligned}
+ \hat{\sigma}_z
+ \to \{\Ket{0}, \Ket{1}\}
+ \qquad
+ \hat{\sigma}_x
+ \to \{\Ket{+}, \Ket{-}\}
+ \qquad
+ \hat{\sigma}_y
+ \to \{\Ket{+i}, \Ket{-i}\}
+\end{aligned}$$
+
+Where the latter two states are expressed as follows in the conventional $z$-basis:
+
+$$\begin{aligned}
+ \Ket{\pm}
+ = \frac{\Ket{0} \pm \Ket{1}}{\sqrt{2}}
+ \qquad \quad
+ \Ket{\pm i}
+ = \frac{\Ket{0} \pm i \Ket{1}}{\sqrt{2}}
+\end{aligned}$$
+
+More generally, every point on the surface of the sphere
+describes a pure qubit state in terms of the angles $\theta$ and $\varphi$,
+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 $r \le 1$:
+
+$$\begin{aligned}
+ \boxed{
+ \vec{r}
+ = \begin{bmatrix} r_x \\ r_y \\ r_z \end{bmatrix}
+ = \begin{bmatrix} r \sin\theta \cos\varphi \\ r \sin\theta \sin\varphi \\ r \cos\theta \end{bmatrix}
+ }
+\end{aligned}$$
+
+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
+to describe the qubit using a [density operator](/know/concept/density-operator/):
+
+$$\begin{aligned}
+ \boxed{
+ \hat{\rho}
+ = \frac{1}{2} \Big( \hat{I} + \vec{r} \cdot \vec{\sigma} \Big)
+ }
+\end{aligned}$$
+
+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. $\hat{\rho}^2 = \hat{\rho}$:
+
+$$\begin{aligned}
+ \hat{\rho}^2
+ &= \frac{1}{4} \Big( \hat{I}^2 + 2 \hat{I} (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
+ = \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ (\vec{r} \cdot \vec{\sigma})^2
+ &= (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z)^2
+ \\
+ &= r_x^2 \hat{\sigma}_x^2 + r_x r_y \hat{\sigma}_x \hat{\sigma}_y + r_x r_z \hat{\sigma}_x \hat{\sigma}_z
+ + r_x r_y \hat{\sigma}_y \hat{\sigma}_x + r_y^2 \hat{\sigma}_y^2
+ \\
+ &\quad + r_y r_z \hat{\sigma}_y \hat{\sigma}_z + r_x r_z \hat{\sigma}_z \hat{\sigma}_x
+ + r_y r_z \hat{\sigma}_z \hat{\sigma}_y + r_z^2 \hat{\sigma}_z^2
+ \\
+ &= r_x^2 \hat{I} + r_y^2 \hat{I} + r_z^2 \hat{I}
+ + r_x r_y \{ \hat{\sigma}_x, \hat{\sigma}_y \}
+ + r_y r_z \{ \hat{\sigma}_y, \hat{\sigma}_z \}
+ + r_x r_z \{ \hat{\sigma}_x, \hat{\sigma}_z \}
+ \\
+ &= (r_x^2 + r_y^2 + r_z^2) \hat{I}
+ = r^2 \hat{I}
+\end{aligned}$$
+
+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 $\hat{\sigma}_z$:
+
+$$\begin{aligned}
+ \Expval{\hat{\sigma}_z}
+ &= \Tr(\hat{\rho} \hat{\sigma}_z)
+ = \frac{1}{2} \Tr\!\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
+ = \frac{1}{2} \Tr\!\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
+ \\
+ &= \frac{1}{2} \Tr\!\big( r_x \hat{\sigma}_x \hat{\sigma}_z + r_y \hat{\sigma}_y \hat{\sigma}_z + r_z \hat{\sigma}_z^2 \big)
+ = \frac{1}{2} \Tr\!\big( r_z \hat{I} \big)
+ = r_z
+\end{aligned}$$
+
+
+## References
+1. N. Brunner,
+ *Quantum information theory: lecture notes*,
+ 2019, unpublished.
+2. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
new file mode 100644
index 0000000..496d8d3
--- /dev/null
+++ b/source/know/concept/blochs-theorem/index.md
@@ -0,0 +1,110 @@
+---
+title: "Bloch's theorem"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+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 $\psi(\vb{r})$
+to the time-independent Schrödinger equation
+take the following form,
+where the function $u(\vb{r})$ is periodic on the same lattice,
+i.e. $u(\vb{r}) = u(\vb{r} + \vb{a})$:
+
+$$
+\begin{aligned}
+ \boxed{
+ \psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}}
+ }
+\end{aligned}
+$$
+
+In other words, in a periodic potential,
+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 $\psi(\vb{r} + \vb{a})$
+are eigenstates with the same energy:
+
+$$
+\begin{aligned}
+ \hat{H} \psi(\vb{r}) = E \psi(\vb{r})
+ \qquad
+ \hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a})
+\end{aligned}
+$$
+
+Now define the unitary translation operator $\hat{T}(\vb{a})$ such that
+$\psi(\vb{r} + \vb{a}) = \hat{T}(\vb{a}) \psi(\vb{r})$.
+From the previous equation, we then know that:
+
+$$
+\begin{aligned}
+ \hat{H} \hat{T}(\vb{a}) \psi(\vb{r})
+ = E \hat{T}(\vb{a}) \psi(\vb{r})
+ = \hat{T}(\vb{a}) \big(E \psi(\vb{r})\big)
+ = \hat{T}(\vb{a}) \hat{H} \psi(\vb{r})
+\end{aligned}
+$$
+
+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 $\psi(\vb{r})$:
+
+$$
+\begin{aligned}
+ \hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r})
+ \qquad \qquad
+ \hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r})
+\end{aligned}
+$$
+
+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 $\vb{k}$:
+
+$$
+\begin{aligned}
+ \psi(\vb{r} + \vb{a})
+ = \hat{T}(\vb{a}) \:\psi(\vb{r})
+ = e^{i \theta} \:\psi(\vb{r})
+ = e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r})
+\end{aligned}
+$$
+
+Let us now define the following function,
+keeping our arbitrary choice of $\vb{k}$:
+
+$$
+\begin{aligned}
+ u(\vb{r})
+ = e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
+\end{aligned}
+$$
+
+As it turns out, this function is guaranteed to be lattice-periodic for any $\vb{k}$:
+
+$$
+\begin{aligned}
+ u(\vb{r} + \vb{a})
+ &= e^{- i \vb{k} \cdot (\vb{r} + \vb{a})} \:\psi(\vb{r} + \vb{a})
+ \\
+ &= e^{- i \vb{k} \cdot \vb{r}} e^{- i \vb{k} \cdot \vb{a}} e^{i \vb{k} \cdot \vb{a}} \:\psi(\vb{r})
+ \\
+ &= e^{- i \vb{k} \cdot \vb{r}} \:\psi(\vb{r})
+ \\
+ &= u(\vb{r})
+\end{aligned}
+$$
+
+Then Bloch's theorem follows from
+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
new file mode 100644
index 0000000..6193fe6
--- /dev/null
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -0,0 +1,357 @@
+---
+title: "Boltzmann equation"
+date: 2022-10-02
+categories:
+- Physics
+- Thermodynamics
+- Fluid mechanics
+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:
+
+$$\begin{aligned}
+ N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}}
+\end{aligned}$$
+
+At equilibrium, all processes affecting the particles
+no longer have a net effect, so $f$ is fixed:
+
+$$\begin{aligned}
+ \dv{f}{t}
+ = 0
+\end{aligned}$$
+
+If each particle's momentum only changes due to collisions,
+then a non-equilibrium state can be described as follows, very generally:
+
+$$\begin{aligned}
+ \dv{f}{t}
+ = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
+ &= \pdv{f}{t} + \bigg( \pdv{f}{x} \dv{x}{t} \!+\! \pdv{f}{y} \dv{y}{t} \!+\! \pdv{f}{z} \dv{z}{t} \bigg)
+ + \bigg( \pdv{f}{v_x} \dv{v_x}{t} \!+\! \pdv{f}{v_y} \dv{v_y}{t} \!+\! \pdv{f}{v_z} \dv{v_z}{t} \bigg)
+ \\
+ &= \pdv{f}{t} + \bigg( v_x \pdv{f}{x} \!+\! v_y \pdv{f}{y} \!+\! v_z \pdv{f}{z} \bigg)
+ + \bigg( a_x \pdv{f}{v_x} \!+\! a_y \pdv{f}{v_y} \!+\! a_z \pdv{f}{v_z} \bigg)
+ \\
+ &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}}
+\end{aligned}$$
+
+Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$.
+Inserting Newton's second law $\vb{F} = m \vb{a}$
+leads us to the **Boltzmann equation** or
+**Boltzmann transport equation** (BTE):
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}}
+ = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
+ }
+\end{aligned}$$
+
+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,
+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 $f_0$:
+
+$$\begin{aligned}
+ \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}}
+ = \frac{f_0 - f}{\tau}
+\end{aligned}$$
+
+Where $\tau$ is the average collision period.
+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 $n$:
+
+$$\begin{aligned}
+ n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+\end{aligned}$$
+
+Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so:
+
+$$\begin{aligned}
+ \Expval{Q}
+ = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+\end{aligned}$$
+
+With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate,
+assuming that $\vb{F}$ does not depend on $\vb{v}$:
+
+$$\begin{aligned}
+ 0
+ &= \int_{-\infty}^\infty Q \bigg( \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} \bigg) \dd{\vb{v}}
+ \\
+ &= \int Q \pdv{f}{t} \dd{\vb{v}} + \int (\vb{v} \cdot \nabla f) \: Q \dd{\vb{v}} + \frac{\vb{F}}{m} \cdot \int Q \pdv{f}{\vb{v}} \dd{\vb{v}}
+ \\
+ &= \pdv{}{t}\int Q f \dd{\vb{v}} + \int \Big( \nabla \cdot (\vb{v} f) - f (\nabla \cdot \vb{v}) \Big) Q \dd{\vb{v}}
+ + \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{}{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}}
+\end{aligned}$$
+
+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 $\vb{v} \to \pm\infty$,
+so the first term in the third integral vanishes after it is integrated:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(n \Expval{Q}\big) + \int \nabla \cdot (\vb{v} f) \: Q \dd{\vb{v}}
+ + \frac{\vb{F}}{m} \cdot \bigg( \Big[ Q f \Big]_{-\infty}^\infty - \int f \pdv{Q}{\vb{v}} \dd{\vb{v}} \bigg)
+ \\
+ &= \pdv{}{t}\big(n \Expval{Q}\big) + \nabla \cdot \int Q \vb{v} f \dd{\vb{v}}
+ - \frac{\vb{F}}{m} \cdot \int f \pdv{Q}{\vb{v}} \dd{\vb{v}}
+\end{aligned}$$
+
+We thus arrive at the prototype of the BTE's so-called **moment equations**:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \pdv{}{t}\big(n \Expval{Q}\big) + \nabla \cdot \big(n \Expval{Q \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{Q}{\vb{v}}} \bigg)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big)
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We insert $Q = m$ into our prototype,
+and since $m$ is constant, the rest is trivial:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(n \Expval{m}\big) + \nabla \cdot \big(n \Expval{m \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{m}{\vb{v}}} \bigg)
+ \\
+ &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(n \Expval{m \vb{v}}\big) + \nabla \cdot \big(n \Expval{m \vb{v} \vb{v}}\big)
+ - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{(m \vb{v})}{\vb{v}}} \bigg)
+ \\
+ &= \pdv{}{t}\big(\rho \Expval{\vb{v}}\big) + \nabla \cdot \big(\rho \Expval{\vb{v} \vb{v}}\big)
+ - \vb{F} \cdot \bigg( n \Expval{\pdv{\vb{v}}{\vb{v}}} \bigg)
+\end{aligned}$$
+
+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 $\vb{V}$:
+
+$$\begin{aligned}
+ \Expval{\vb{v} \vb{v}}
+ &= \Expval{(\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})}
+ = \Expval{\vb{V} \vb{V} + 2 \vb{V} \vb{w} + \vb{w} \vb{w}}
+ = \vb{V} \vb{V} + 2 \vb{V} \Expval{\vb{w}} + \Expval{\vb{w} \vb{w}}
+\end{aligned}$$
+
+Since $\vb{w}$ represents a deviation from the mean, $\Expval{\vb{w}} = 0$.
+We define the pressure tensor:
+
+$$\begin{aligned}
+ \hat{P}
+ \equiv \rho \Expval{\vb{w} \vb{w}}
+ = \rho \Expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We insert $Q = m |\vb{v}|^2 / 2$ into our prototype and recognize $\rho$ wherever possible:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\bigg(n \Expval{\frac{m |\vb{v}|^2}{2}}\bigg)
+ + \nabla \cdot \bigg(n \Expval{\frac{m |\vb{v}|^2}{2} \vb{v}}\bigg)
+ - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{}{\vb{v}} \frac{m |\vb{v}|^2}{2}} \bigg)
+ \\
+ &= \pdv{}{t}\bigg(\frac{\rho}{2} \Expval{|\vb{v}|^2}\bigg)
+ + \nabla \cdot \bigg(\frac{\rho}{2} \Expval{|\vb{v}|^2 \vb{v}}\bigg)
+ - \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg)
+\end{aligned}$$
+
+We handle these terms one by one. Substituting $\vb{v} = \vb{V} + \vb{w}$ in the first gives:
+
+$$\begin{aligned}
+ \Expval{|\vb{v}|^2}
+ &= \Expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w})}
+ = \Expval{|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2}
+ \\
+ &= |\vb{V}|^2 + 2 \vb{V} \cdot \Expval{\vb{w}} + \Expval{|\vb{w}|^2}
+ = |\vb{V}|^2 + \Expval{|\vb{w}|^2}
+\end{aligned}$$
+
+And likewise for the second term,
+where we recognize the stress tensor $\Expval{\vb{w} \vb{w}}$:
+
+$$\begin{aligned}
+ \Expval{|\vb{v}|^2 \vb{v}}
+ &= \Expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})}
+ = \Expval{(|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2) (\vb{V} \!+\! \vb{w})}
+ \\
+ &= \Expval{|\vb{V}|^2 \vb{V} + |\vb{V}|^2 \vb{w}
+ + 2 (\vb{V} \cdot \vb{w}) \vb{V} + 2 (\vb{V} \cdot \vb{w}) \vb{w}
+ + |\vb{w}|^2 \vb{V} + |\vb{w}|^2 \vb{w}}
+ \\
+ &= |\vb{V}|^2 \vb{V} + |\vb{V}|^2 \Expval{\vb{w}}
+ + 2 (\vb{V} \cdot \Expval{\vb{w}}) \vb{V} + 2 \Expval{(\vb{V} \cdot \vb{w}) \vb{w}}
+ + \Expval{|\vb{w}|^2} \vb{V} + \Expval{|\vb{w}|^2 \vb{w}}
+ \\
+ &= |\vb{V}|^2 \vb{V} + 0 + 0 + 2 \vb{V} \cdot \Expval{\vb{w} \vb{w}}
+ + \Expval{|\vb{w}|^2} \vb{V} + \Expval{|\vb{w}|^2 \vb{w}}
+\end{aligned}$$
+
+The third term is fairly obvious, but we calculate it rigorously just to be safe:
+
+$$\begin{aligned}
+ \pdv{|\vb{v}|^2}{\vb{v}}
+ &= \pdv{}{\vb{v}} \big( v_x^2 + v_y^2 + v_z^2 \big)
+ = \vu{e}_x \pdv{v_x^2}{v_x} + \vu{e}_y \pdv{v_y^2}{v_y} + \vu{e}_z \pdv{v_z^2}{v_z}
+ = 2 \vb{v}
+\end{aligned}$$
+
+To clarify the physical interpretation,
+we define $U$, $\vb{J}$ and $\hat{P}$ as follows:
+
+$$\begin{aligned}
+ U
+ &\equiv \frac{\rho}{2} \Expval{|\vb{w}|^2}
+ = \frac{\rho}{2} \Expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})}
+ \\
+ \vb{J}
+ &\equiv \frac{\rho}{2} \Expval{|\vb{w}|^2 \vb{w}}
+ = \frac{\rho}{2} \Expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})(\vb{v} \!-\! \vb{V})}
+ \\
+ \hat{P}
+ &\equiv \rho \Expval{\vb{w} \vb{w}}
+ = \rho \Expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})}
+\end{aligned}$$
+
+Putting it all together, we arrive at the expected result, namely:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
+ + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
+ - \vb{F} \cdot \big( n \vb{V} \big)
+\end{aligned}$$
+
+For the sake of clarity, we write out the pressure term, including the outer divergence:
+
+$$\begin{aligned}
+ \nabla \cdot (\vb{V} \cdot \hat{P})
+ &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V}
+ = \nabla \cdot
+ \begin{bmatrix}
+ P_{xx} & P_{xy} & P_{xz} \\
+ P_{yx} & P_{yy} & P_{yz} \\
+ P_{zx} & P_{zy} & P_{zz}
+ \end{bmatrix}
+ \cdot
+ \begin{bmatrix}
+ V_x \\ V_y \\ V_z
+ \end{bmatrix}
+ \\
+ &=
+ \begin{bmatrix}
+ \displaystyle \pdv{P_{xx}}{x} + \pdv{P_{xy}}{y} + \pdv{P_{xz}}{z} \\
+ \displaystyle \pdv{P_{yx}}{x} + \pdv{P_{yy}}{y} + \pdv{P_{yz}}{z} \\
+ \displaystyle \pdv{P_{zx}}{x} + \pdv{P_{zy}}{y} + \pdv{P_{zz}}{z}
+ \end{bmatrix}^{\top}
+ \cdot
+ \begin{bmatrix}
+ V_x \\ V_y \\ V_z
+ \end{bmatrix}
+ = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We expand the right-hand side of the theorem and
+rearrange the integrals:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
+ &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k}
+ \\
+ &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'}
+ \\
+ &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
+ = A \cdot (f * g)(x)
+\end{aligned}$$
+
+Then we do the same again,
+this time starting from a product in the $x$-domain:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}\{f(x) \: g(x)\}
+ &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
+ \\
+ &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
+ \\
+ &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
+ = B \cdot (\tilde{f} * \tilde{g})(k)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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 $t < 0$:
+
+$$\begin{aligned}
+ \hat{\mathcal{L}}\{(f * g)(t)\}
+ &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t}
+ \\
+ &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
+\end{aligned}$$
+
+Then we define a new integration variable $\tau = t - t'$, yielding:
+
+$$\begin{aligned}
+ \hat{\mathcal{L}}\{(f * g)(t)\}
+ &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
+ \\
+ &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
+ \\
+ &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'}
+ = \tilde{f}(s) \: \tilde{g}(s)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+For a direction $\dd{\ell}$, we know that
+$\idv{f}{\ell}$ is the component of $\nabla f$ in that direction:
+
+$$\begin{aligned}
+ \dv{f}{\ell}
+ = \pdv{f}{x} \dv{x}{\ell} + \pdv{f}{y} \dv{y}{\ell} + \pdv{f}{z} \dv{z}{\ell}
+ = \nabla f \cdot \bigg( \dv{x}{\ell}, \dv{y}{\ell}, \dv{z}{\ell} \bigg)
+ = \nabla f \cdot \vu{u}
+\end{aligned}$$
+
+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:
+
+$$\begin{gathered}
+ \nabla f
+ = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1}
+ + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
+ + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
+\end{gathered}$$
+
+
+
+
Proof
+
+Proof.
+As preparation, we rewrite $\vb{V}$ as follows
+to introduce the scale factors:
+
+$$\begin{aligned}
+ \vb{V}
+ &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1)
+ + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2)
+ + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
+\end{aligned}$$
+
+We start by taking only the $\vu{e}_1$-component of this vector,
+and expand its divergence using the following vector identity:
+
+$$\begin{gathered}
+ \nabla \cdot (\vb{U} \: f)
+ = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) f
+\end{gathered}$$
+
+Inserting the scalar $f = h_2 h_3 V_1$
+the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$,
+we arrive at:
+
+$$\begin{gathered}
+ \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} (h_2 h_3 V_1) \Big)
+ = \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
+ + \Big( \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} \Big) (h_2 h_3 V_1)
+\end{gathered}$$
+
+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:
+
+$$\begin{aligned}
+ \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
+ = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1}
+\end{aligned}$$
+
+The second term is more involved.
+First, we use the gradient formula to observe that:
+
+$$\begin{aligned}
+ \nabla x_1
+ = \frac{\vu{e}_1}{h_1}
+ \qquad \quad
+ \nabla x_2
+ = \frac{\vu{e}_2}{h_2}
+ \qquad \quad
+ \nabla x_3
+ = \frac{\vu{e}_3}{h_3}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \nabla x_2 \cross \nabla x_3
+ = \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3}
+ = \frac{\vu{e}_1}{h_2 h_3}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \nabla \cdot \frac{\vu{e}_1}{h_2 h_3}
+ = \nabla \cdot \big( \nabla x_2 \cross \nabla x_3 \big)
+ = \nabla x_3 \cdot (\nabla \cross \nabla x_2) - \nabla x_2 \cdot (\nabla \cross \nabla x_3)
+ = 0
+\end{aligned}$$
+
+After repeating this procedure for the other components of $\vb{V}$,
+we get the desired general expression for the divergence.
+
+
+
+
Proof
+
+Proof.
+The curl is found in a similar way as the divergence.
+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:
+
+$$\begin{gathered}
+ \nabla \cross (\vb{U} \: f)
+ = (\nabla \cross \vb{U}) f - \vb{U} \cross (\nabla f)
+\end{gathered}$$
+
+Inserting the scalar $f = h_1 V_1$
+and the vector $\vb{U} = \vu{e}_1 / h_1$, we arrive at:
+
+$$\begin{gathered}
+ \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
+ = \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big) (h_1 V_1) - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
+\end{gathered}$$
+
+Previously, when proving the divergence,
+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:
+
+$$\begin{aligned}
+ \nabla (h_1 V_1)
+ = \vu{e}_1 \frac{1}{h_1} \pdv{(h_1 V_1)}{x_1}
+ + \vu{e}_2 \frac{1}{h_2} \pdv{(h_1 V_1)}{x_2}
+ + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3}
+\end{aligned}$$
+
+Consequently, the curl of the first component of $\vb{V}$ is as follows,
+using the fact that $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+are related to each other by cross products:
+
+$$\begin{aligned}
+ \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
+ = - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
+ = - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3}
+\end{aligned}$$
+
+If we go through the same process for the other components of $\vb{V}$
+and add up the results, we get the desired expression for the curl.
+
+
+
+
Proof
+
+Proof.
+Consider the following weighted inner product,
+whose weight function is a stationary distribution $\pi$
+satisfying detailed balance,
+where $\hat{L}$ is the Kolmogorov operator:
+
+$$\begin{aligned}
+ \inprod{\hat{L} h}{k}_\pi
+ \equiv \int_{-\infty}^\infty \hat{L}\{h(x)\} \: \pi(x) \: k(x) \dd{x}
+ = \int_{-\infty}^\infty h(x) \: \hat{L}{}^\dagger\{\pi(x) k(x)\} \dd{x}
+\end{aligned}$$
+
+Where we have used the definition of an adjoint operator.
+We would like to rewrite this:
+
+$$\begin{aligned}
+ \hat{L}{}^\dagger \{\pi k\}
+ = -\nabla \cdot \big( \vb{u} \pi k - D \nabla(\pi k) \big)
+ = -\nabla \cdot (\vb{u} \pi k - D k \nabla \pi - D \pi \nabla k)
+\end{aligned}$$
+
+Since $\pi$ is stationary by definition,
+we know that $\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$,
+meaning:
+
+$$\begin{aligned}
+ \hat{L}{}^\dagger \{\pi k\}
+ = \nabla \cdot (D \pi \nabla k)
+ = \nabla \pi \cdot (D \nabla k) + \pi \nabla \cdot (D \nabla k)
+\end{aligned}$$
+
+Detailed balance demands that $\vb{u} \pi = D \nabla \pi$,
+leading to the following:
+
+$$\begin{aligned}
+ \hat{L}{}^\dagger \{\pi k\}
+ &= D \nabla \pi \cdot \nabla k + \pi \nabla \cdot (D \nabla k)
+ = \pi \vb{u} \cdot \nabla k + \pi \nabla \cdot (D \nabla k)
+ \\
+ &= \pi \big( \vb{u} \cdot \nabla k + \nabla \cdot (D \nabla k) \big)
+ = \pi \hat{L}\{k\}
+\end{aligned}$$
+
+Where we recognized the definition of $\hat{L}$
+from the backward Kolmogorov equation.
+Now that we have established that $\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$,
+we return to the inner product:
+
+$$\begin{aligned}
+ \inprod{\hat{L} h}{k}_\pi
+ = \int_{-\infty}^\infty h(x) \: \pi(x) \: \hat{L}\{k(x)\} \dd{x}
+ = \inprod{h}{\hat{L} k}_\pi
+\end{aligned}$$
+
+Consequently, the following weighted inner products must also be equivalent:
+
+$$\begin{aligned}
+ \Inprod{\exp(t \hat{L}) h}{k}_\pi
+ = \Inprod{h}{\exp(t \hat{L}) k}_\pi
+\end{aligned}$$
+
+Now, consider the time evolution of the
+[conditional expectation](/know/concept/conditional-expectation/)
+$\mathbf{E}\big[ k(X_t) | X_0 \big]$:
+
+$$\begin{aligned}
+ \pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big]
+ &= \pdv{}{t}\int_{-\infty}^\infty k(x) \: \phi(t, x) \dd{x}
+ = \int_{-\infty}^\infty k \pdv{\phi}{t} \dd{x}
+ \\
+ &= \int_{-\infty}^\infty k \: \hat{L}{}^\dagger\{\phi\} \dd{x}
+ = \int_{-\infty}^\infty \hat{L}\{k\} \: \phi \dd{x}
+ = \mathbf{E}\big[ \hat{L}\{k(X_t)\} | X_0 \big]
+\end{aligned}$$
+
+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 $X_t$),
+we can naively move the differentiation inside
+(only valid within $\mathbf{E}$):
+
+$$\begin{aligned}
+ \pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big]
+ = \mathbf{E}\bigg[ \pdv{k(X_t)}{t} \bigg| X_0 \bigg]
+ = \mathbf{E}\bigg[ \hat{L}\{k(X_0)\} \bigg| X_0 \bigg]
+\end{aligned}$$
+
+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 $k(t, x) = \exp(t \hat{L})\{k(0,x)\}$,
+therefore:
+
+$$\begin{aligned}
+ \mathbf{E}\big[ k(X_t) \big| X_0 \big]
+ = \mathbf{E}\big[ \exp(t \hat{L})\{k(X_0)\} \big| X_0 \big]
+ = \exp(t \hat{L})\{k(X_0)\}
+\end{aligned}$$
+
+With this, we can evaluate the two weighted inner products from earlier,
+which we know are equal to each other.
+Using the *tower property* of the conditional expectation:
+
+$$\begin{aligned}
+ \Inprod{h}{\exp(t \hat{L}) k}_\pi
+ &= \mathbf{E}\big[ h(X_0) \: \mathbf{E}[k(X_t) | X_0] \big]
+ = \mathbf{E}\big[ h(X_0) \: k(X_t) \big]
+ \\
+ = \Inprod{\exp(t \hat{L}) h}{k}_\pi
+ &= \mathbf{E}\big[ \mathbf{E}[h(X_t) | X_0] \: k(X_0) \big]
+ = \mathbf{E}\big[ h(X_t) \: k(X_0) \big]
+\end{aligned}$$
+
+Where the integral gave the expectation value at $X_0$,
+since $\pi$ does not change in time.
+
+
+
+
+Due to unitarity constraints,
+the action of $U_f$ is defined to be as follows,
+with $\oplus$ meaning XOR:
+
+$$\begin{aligned}
+ \Ket{x} \Ket{y}
+ \quad \to \boxed{U_f} \to \quad
+ \Ket{x} \Ket{y \oplus f(x)}
+\end{aligned}$$
+
+Starting on the left from two qubits $\Ket{0}$ and $\Ket{1}$,
+we apply the [Hadamard gate](/know/concept/quantum-gate/) $H$ to both:
+
+$$\begin{aligned}
+ \Ket{0} \Ket{1}
+ \quad \to \boxed{H^{\otimes 2}} \to \quad
+ \Ket{+} \Ket{-}
+ = \frac{1}{2} \Big( \Ket{0} + \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
+\end{aligned}$$
+
+Feeding this result into the oracle $U_f$ then leads us to:
+
+$$\begin{aligned}
+ \to \boxed{U_f} \to \quad
+ \frac{1}{2} \Ket{0} \Big( \Ket{0 \oplus f(0)} - \Ket{1 \oplus f(0)} \Big)
+ + \frac{1}{2} \Ket{1} \Big( \Ket{0 \oplus f(1)} - \Ket{1 \oplus f(1)} \Big)
+\end{aligned}$$
+
+The parenthesized superpositions can be reduced.
+Assuming that $f(b) = 0$, we notice:
+
+$$\begin{aligned}
+ \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
+ = \Ket{0 \oplus 0} - \Ket{1 \oplus 0}
+ = \Ket{0} - \Ket{1}
+\end{aligned}$$
+
+On the other hand, if we assume that $f(b) = 1$,
+we get the opposite result:
+
+$$\begin{aligned}
+ \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
+ = \Ket{0 \oplus 1} - \Ket{1 \oplus 1}
+ = - \big(\Ket{0} - \Ket{1}\big)
+\end{aligned}$$
+
+We can thus combine both cases, $f(b) = 0$ or $f(b) = 1$,
+into the following single expression:
+
+$$\begin{aligned}
+ \Ket{0 \oplus f(b)} - \Ket{1 \oplus f(b)}
+ = (-1)^{f(b)} \big(\Ket{0} - \Ket{1}\big)
+\end{aligned}$$
+
+Using this, we rewrite the intermediate state of the quantum circuit like so:
+
+$$\begin{aligned}
+ \Ket{0} \Ket{1}
+ \quad \to \boxed{H^{\otimes 2}} \to \boxed{U_f} \to \quad
+ \frac{1}{2} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big) \Big( \Ket{0} - \Ket{1} \Big)
+\end{aligned}$$
+
+The second qubit in state $\Ket{-}$ is garbage; it is no longer of interest.
+The first qubit is given by:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{2}} \Big( (-1)^{f(0)} \Ket{0} + (-1)^{f(1)} \Ket{1} \Big)
+ = \frac{(-1)^{f(0)}}{\sqrt{2}} \Big( \Ket{0} + (-1)^{f(0) \oplus f(1)} \Ket{1} \Big)
+\end{aligned}$$
+
+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 $(-1)^{f(0)} \Ket{-}$.
+Taking the Hadamard transform of this qubit therefore yields:
+
+$$\begin{aligned}
+ \to \boxed{H} \to \quad
+ (-1)^{f(0)} \Ket{f(0) \oplus f(1)}
+\end{aligned}$$
+
+Depending on whether $f$ is constant or balanced,
+the mearurement outcome of this state will be $\Ket{0}$ or $\Ket{1}$
+with 100\% probability. We have solved the problem!
+
+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 $x = 1$.
+
+
+## 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;
+other possibilities are assumed to be impossible.
+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:
+
+$$\begin{aligned}
+ \Ket{x_1} \Ket{x_2} \cdots \Ket{x_N} \Ket{y}
+ \quad \to \boxed{U_f} \to \quad
+ \Ket{x_1} \cdots \Ket{x_N} \Ket{y \oplus f(x_1, ..., x_N)}
+\end{aligned}$$
+
+Applying the $N + 1$ Hadamard gates to the initial state
+yields the following superposition:
+
+$$\begin{aligned}
+ \Ket{0}^{\otimes N} \Ket{1}
+ \quad \to \boxed{H^{\otimes N + 1}} \to \quad
+ \Ket{+}^{\otimes N} \Ket{-}
+ = \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} \Ket{x} \Ket{-}
+\end{aligned}$$
+
+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 $\Ket{x} = \Ket{1} \Ket{0} \Ket{1} \Ket{0}^{\otimes N-3}$
+(from least to most significant).
+
+We give this state to the oracle,
+and, by the same logic as for the Deutsch algorithm,
+get back:
+
+$$\begin{aligned}
+ \to \boxed{U_f} \to \quad
+ \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \Ket{x} \Ket{-}
+\end{aligned}$$
+
+The last qubit $\Ket{-}$ is garbage.
+Next, applying the Hadamard transform to the other $N$ gives:
+
+$$\begin{aligned}
+ \to \boxed{H^{\otimes N}} \to \quad
+ \frac{1}{\sqrt{2^N}} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)}
+ \bigg( \frac{1}{\sqrt{2^N}} \sum_{y = 0}^{2^N - 1} (-1)^{x \cdot y} \Ket{y} \bigg)
+\end{aligned}$$
+
+Where $x \cdot y$ is the bitwise dot product of the binary representations of $x$ and $y$,
+so, for example, if $N = 2$, then $x \cdot y = x_1 y_1 + x_2 y_2$.
+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:
+
+$$\begin{aligned}
+ \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} \sum_{y = 0}^{2^N - 1} (-1)^{f(x) + x \cdot y} \Ket{y}
+ = \sum_{y = 0}^{2^N - 1} \bigg( \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x) + x \cdot y} \bigg) \Ket{y}
+ = \sum_{y = 0}^{2^N - 1} c_y \Ket{y}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ |c_0|^2
+ = \bigg| \frac{1}{2^N} \sum_{x = 0}^{2^N - 1} (-1)^{f(x)} \bigg|^2
+\end{aligned}$$
+
+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 $|c_0|^2 = 0$.
+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.
+
+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;
+a revolutionary discovery.
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
diff --git a/source/know/concept/dielectric-function/index.md b/source/know/concept/dielectric-function/index.md
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+---
+title: "Dielectric function"
+date: 2022-01-24
+categories:
+- Physics
+- Electromagnetism
+- Quantum mechanics
+layout: "concept"
+---
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{D} = \varepsilon_0 \varepsilon_r \vb{E}
+ }
+\end{aligned}$$
+
+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 $\vb{E}$.
+
+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 $\Phi_\mathrm{ind}$
+are the total, external and induced potentials, respectively:
+
+$$\begin{aligned}
+ \vb{E}
+ = -\nabla \Phi_\mathrm{tot}
+ \qquad \qquad
+ \vb{D}
+ = - \varepsilon_0 \nabla \Phi_\mathrm{ext}
+ \qquad \qquad
+ \vb{P}
+ = \varepsilon_0 \nabla \Phi_\mathrm{ind}
+\end{aligned}$$
+
+Such that $\Phi_\mathrm{tot} = \Phi_\mathrm{ext} + \Phi_\mathrm{ind}$.
+Inserting this into $\vb{D} = \varepsilon_0 \varepsilon_r \vb{E}$
+then suggests defining:
+
+$$\begin{aligned}
+ \boxed{
+ \varepsilon_r
+ \equiv \frac{\Phi_\mathrm{ext}}{\Phi_\mathrm{tot}}
+ }
+\end{aligned}$$
+
+
+## From induced charge density
+
+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 $\vb{P}$:
+
+$$\begin{aligned}
+ \nabla \cdot \vb{P}
+ = \varepsilon_0 \nabla^2 \Phi_\mathrm{ind}(\vb{r})
+ = - \rho_\mathrm{ind}(\vb{r})
+\end{aligned}$$
+
+This is Poisson's equation, which has the following well-known
+[Fourier transform](/know/concept/fourier-transform/):
+
+$$\begin{aligned}
+ \Phi_\mathrm{ind}(\vb{q})
+ = \frac{\rho_\mathrm{ind}(\vb{q})}{\varepsilon_0 |\vb{q}|^2}
+ = V(\vb{q}) \: \rho_\mathrm{ind}(\vb{q})
+\end{aligned}$$
+
+Where $V(\vb{q})$ represents Coulomb interactions,
+and $V(0) = 0$ to ensure overall neutrality:
+
+$$\begin{aligned}
+ V(\vb{q})
+ = \frac{1}{\varepsilon_0 |\vb{q}|^2}
+ \qquad \implies \qquad
+ V(\vb{r} - \vb{r}')
+ = \frac{1}{4 \pi \varepsilon_0 |\vb{r} - \vb{r}'|}
+\end{aligned}$$
+
+The [convolution theorem](/know/concept/convolution-theorem/)
+then gives us the solution $\Phi_\mathrm{ind}$ in the $\vb{r}$-domain:
+
+$$\begin{aligned}
+ \Phi_\mathrm{ind}(\vb{r})
+ = (V * \rho_\mathrm{ind})(\vb{r})
+ = \int_{-\infty}^\infty V(\vb{r} - \vb{r}') \: \rho_\mathrm{ind}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+To proceed, we need to find an expression for $\rho_\mathrm{ind}$
+that is proportional to $\Phi_\mathrm{tot}$ or $\Phi_\mathrm{ext}$,
+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:
+
+$$\begin{aligned}
+ \Phi_\mathrm{tot}
+ = (1 + C_\mathrm{ext} V) \Phi_\mathrm{ext}
+ \quad \implies \quad
+ \boxed{
+ \varepsilon_r(\vb{q})
+ = \frac{1}{1 + C_\mathrm{ext}(\vb{q}) V(\vb{q})}
+ }
+\end{aligned}$$
+
+Similarly, suppose we can show that $\rho_\mathrm{ind} = C_\mathrm{tot} \Phi_\mathrm{tot}$,
+for some quantity $C_\mathrm{tot}$, then:
+
+$$\begin{aligned}
+ \Phi_\mathrm{ext}
+ = (1 - C_\mathrm{tot} V) \Phi_\mathrm{tot}
+ \quad \implies \quad
+ \boxed{
+ \varepsilon_r(\vb{q})
+ = 1 - C_\mathrm{tot}(\vb{q}) V(\vb{q})
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
+2. M. Fox,
+ *Optical properties of solids*, 2nd edition,
+ Oxford.
diff --git a/source/know/concept/diffie-hellman-key-exchange/index.md b/source/know/concept/diffie-hellman-key-exchange/index.md
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+---
+title: "Diffie-Hellman key exchange"
+date: 2021-03-06
+categories:
+- Cryptography
+layout: "concept"
+---
+
+In cryptography, the **Diffie-Hellman key exchange** is a method
+for two parties to securely agree on an encryption key,
+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:
+
+$$\begin{aligned}
+ f(n) = g^n \bmod p
+\end{aligned}$$
+
+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 $n$.
+
+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:
+
+$$\begin{aligned}
+ A = g^a \bmod p
+ \qquad \quad
+ B = g^b \bmod p
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ k = A^b \bmod p = B^a \bmod p = g^{ab} \bmod p
+ }
+\end{aligned}$$
+
+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 $b$.
+
+This assumption is just that: an assumption.
+So far, nobody has been able to prove or disprove it
+for classical computation.
+However, for quantum computers,
+it has already been *dis*proven!
+In this case, another method must be used,
+for example the [BB84 protocol](/know/concept/bb84-protocol/).
+
+
+
+## References
+1. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/dirac-delta-function/index.md b/source/know/concept/dirac-delta-function/index.md
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+---
+title: "Dirac delta function"
+date: 2021-02-22
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+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
+defined to be 1:
+
+$$\begin{aligned}
+ \boxed{
+ \delta(x) \equiv
+ \begin{cases}
+ +\infty & \mathrm{if}\: x = 0 \\
+ 0 & \mathrm{if}\: x \neq 0
+ \end{cases}
+ \quad \mathrm{and} \quad
+ \int_{-\varepsilon}^\varepsilon \delta(x) \dd{x} = 1
+ }
+\end{aligned}$$
+
+It is sometimes also called the **sampling function**, thanks to its most
+important property: the so-called **sampling property**:
+
+$$\begin{aligned}
+ \boxed{
+ \int f(x) \: \delta(x - x_0) \: dx = \int f(x) \: \delta(x_0 - x) \: dx = f(x_0)
+ }
+\end{aligned}$$
+
+$\delta(x)$ 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:
+
+$$\begin{aligned}
+ \delta(x)
+ = \lim_{n \to +\infty} \!\Big\{ \frac{n}{\sqrt{\pi}} \exp(- n^2 x^2) \Big\}
+ = \lim_{n \to +\infty} \!\Big\{ \frac{n}{\pi} \frac{1}{1 + n^2 x^2} \Big\}
+ = \lim_{n \to +\infty} \!\Big\{ \frac{\sin(n x)}{\pi x} \Big\}
+\end{aligned}$$
+
+The last one is especially important, since it is equivalent to the
+following integral, which appears very often in the context of
+[Fourier transforms](/know/concept/fourier-transform/):
+
+$$\begin{aligned}
+ \delta(x)
+ = \lim_{n \to +\infty} \!\Big\{\frac{\sin(n x)}{\pi x}\Big\}
+ = \frac{1}{2\pi} \int_{-\infty}^\infty \exp(i k x) \dd{k}
+ \:\:\propto\:\: \hat{\mathcal{F}}\{1\}
+\end{aligned}$$
+
+When the argument of $\delta(x)$ is scaled, the delta function is itself scaled:
+
+$$\begin{aligned}
+ \boxed{
+ \delta(s x) = \frac{1}{|s|} \delta(x)
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+Because it is symmetric, $\delta(s x) = \delta(|s| x)$.
+Then by substituting $\sigma = |s| x$:
+
+$$\begin{aligned}
+ \int \delta(|s| x) \dd{x}
+ &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+Note which variable is used for the
+differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:
+
+$$\begin{aligned}
+ \int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi}
+ &= \dv{}{x}\int f(\xi) \: \delta(x - \xi) \dd{x}
+ = f'(x)
+\end{aligned}$$
+
+
+
+
+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 $A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$:
+
+$$\begin{aligned}
+ \omega_{\mathrm{GVD}}(z,t)
+ = \pdv{}{t}\Big( \frac{\beta_2 z t^2 / (2 T_0^4)}{1 + \beta_2^2 z^2 / T_0^4} \Big)
+ = \frac{\beta_2 z / T_0^2}{1 + \beta_2^2 z^2 / T_0^4} \frac{t}{T_0^2}
+\end{aligned}$$
+
+This expression is linear in time, and depending on the sign of $\beta_2$,
+frequencies on one side of the pulse arrive first,
+and those on the other side arrive last.
+The effect is stronger for smaller $T_0$:
+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** ($\beta_2 < 0$),
+lower frequencies travel more slowly than higher ones,
+and vice versa in the **normal dispersion regime** ($\beta_2 > 0$).
+
+
+
+## References
+1. O. Bang,
+ *Numerical methods in photonics: lecture notes*, 2019,
+ unpublished.
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diff --git a/source/know/concept/drude-model/index.md b/source/know/concept/drude-model/index.md
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+---
+title: "Drude model"
+date: 2021-09-23
+categories:
+- Physics
+- Electromagnetism
+- Optics
+layout: "concept"
+---
+
+The **Drude model** classically predicts
+the dielectric function and electric conductivity of a gas of free charge carriers,
+as found in metals and doped semiconductors.
+
+
+## Metals
+
+An [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+has an oscillating [electric field](/know/concept/electric-field/)
+$E(t) = E_0 \exp(- i \omega t)$
+that exerts a force on the charge carriers,
+which have mass $m$ and charge $q$.
+They thus obey the following equation of motion,
+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:
+
+$$\begin{aligned}
+ - x_0 m \omega^2 - i x_0 m \gamma \omega
+ = q E_0
+ \quad \implies \quad
+ x_0
+ = - \frac{q E_0}{m (\omega^2 + i \gamma \omega)}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ P(t)
+ = N p(t)
+ = N q x(t)
+ = - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} E(t)
+\end{aligned}$$
+
+The electric displacement field $D$ is thus as follows,
+where $\varepsilon_r$ is the unknown relative permittivity of the gas,
+which we will find shortly:
+
+$$\begin{aligned}
+ D
+ = \varepsilon_0 \varepsilon_r E
+ = \varepsilon_0 E + P
+ = \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) E
+\end{aligned}$$
+
+The parenthesized expression is the desired dielectric function $\varepsilon_r$,
+which depends on $\omega$:
+
+$$\begin{aligned}
+ \boxed{
+ \varepsilon_r(\omega)
+ = 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
+ }
+\end{aligned}$$
+
+Where we have defined the important so-called **plasma frequency** like so:
+
+$$\begin{aligned}
+ \boxed{
+ \omega_p
+ \equiv \sqrt{\frac{N q^2}{\varepsilon_0 m}}
+ }
+\end{aligned}$$
+
+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),
+and the possibility of self-sustained plasma oscillations.
+For metals, $\omega_p$ lies in the UV.
+
+We can refine this result for $\varepsilon_r$,
+by recognizing the (mean) velocity $v = \idv{x}{t}$,
+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** $\tau \equiv 1 / \gamma$,
+which represents the average time between collisions,
+where each collision resets the involved particles' momentums to zero.
+Or, more formally:
+
+$$\begin{aligned}
+ \dv{p}{t}
+ = - \frac{p}{\tau} + q E
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ - i m \omega v_0 + \frac{m}{\tau} v_0 = q E_0
+ \quad \implies \quad
+ v_0 = \frac{q \tau}{m (1 - i \omega \tau)} E_0
+\end{aligned}$$
+
+From $v(t)$, we find the resulting average current density $J(t)$ to be as follows:
+
+$$\begin{aligned}
+ J(t)
+ = - N q v(t)
+ = \sigma E(t)
+\end{aligned}$$
+
+Where $\sigma(\omega)$ is the **AC conductivity**,
+which depends on the **DC conductivity** $\sigma_0$:
+
+$$\begin{aligned}
+ \boxed{
+ \sigma
+ = \frac{\sigma_0}{1 - i \omega \tau}
+ }
+ \qquad \quad
+ \boxed{
+ \sigma_0
+ = \frac{N q^2 \tau}{m}
+ }
+\end{aligned}$$
+
+We can use these quantities to rewrite
+the dielectric function $\varepsilon_r$ from earlier:
+
+$$\begin{aligned}
+ \boxed{
+ \varepsilon_r(\omega)
+ = 1 + \frac{i \sigma(\omega)}{\varepsilon_0 \omega}
+ }
+\end{aligned}$$
+
+
+## Doped semiconductors
+
+Doping a semiconductor introduces
+free electrons (n-type)
+or free holes (p-type),
+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 $m^*$.
+Furthermore, semiconductors already have
+a high intrinsic permittivity $\varepsilon_{\mathrm{int}}$
+before the dopant is added,
+so the diplacement field $D$ is:
+
+$$\begin{aligned}
+ D
+ = \varepsilon_0 E + P_{\mathrm{int}} + P_{\mathrm{free}}
+ = \varepsilon_{\mathrm{int}} \varepsilon_0 E - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} E
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \varepsilon_r(\omega)
+ = \varepsilon_{\mathrm{int}} \Big( 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega} \Big)
+ }
+\end{aligned}$$
+
+Where the plasma frequency $\omega_p$ has been redefined as follows
+to include $\varepsilon_\mathrm{int}$:
+
+$$\begin{aligned}
+ \boxed{
+ \omega_p
+ = \sqrt{\frac{N q^2}{\varepsilon_{\mathrm{int}} \varepsilon_0 m^*}}
+ }
+\end{aligned}$$
+
+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.
+
+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:
+
+$$\begin{aligned}
+ \omega^2
+ = \frac{\varepsilon_{\mathrm{int}}}{\varepsilon_{\mathrm{int}} - 1} \omega_p^2
+\end{aligned}$$
+
+This is used to experimentally determine the effective mass $m^*$
+of the doped semiconductor,
+by finding which value of $m^*$ gives the measured $\omega$.
+
+
+
+## References
+1. M. Fox,
+ *Optical properties of solids*, 2nd edition,
+ Oxford.
+2. S.H. Simon,
+ *The Oxford solid state basics*,
+ Oxford.
diff --git a/source/know/concept/dynkins-formula/index.md b/source/know/concept/dynkins-formula/index.md
new file mode 100644
index 0000000..be7fa17
--- /dev/null
+++ b/source/know/concept/dynkins-formula/index.md
@@ -0,0 +1,193 @@
+---
+title: "Dynkin's formula"
+date: 2021-11-28
+categories:
+- Mathematics
+- Stochastic analysis
+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
+[conditional expectation](/know/concept/conditional-expectation/):
+
+$$\begin{aligned}
+ \boxed{
+ \hat{A}\{h(X_0)\}
+ \equiv \lim_{t \to 0^+} \bigg[ \frac{1}{t} \mathbf{E}\Big[ h(X_t) - h(X_0) \Big| X_0 \Big] \bigg]
+ }
+\end{aligned}$$
+
+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 $t$.
+
+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/):
+
+$$\begin{aligned}
+ \boxed{
+ \hat{A}\{h(x)\}
+ = \hat{L}\{h(x)\}
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We define a new process $Y_t \equiv h(X_t)$, and then apply Itō's lemma, leading to:
+
+$$\begin{aligned}
+ \dd{Y_t}
+ &= \bigg( \pdv{h}{x} f(X_t) + \frac{1}{2} \pdvn{2}{h}{x} g^2(X_t) \bigg) \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
+ \\
+ &= \hat{L}\{h(X_t)\} \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
+\end{aligned}$$
+
+Where we have recognized the definition of $\hat{L}$.
+Integrating the above equation yields:
+
+$$\begin{aligned}
+ Y_t
+ = Y_0 + \int_0^t \hat{L}\{h(X_s)\} \dd{s} + \int_0^\tau \pdv{h}{x} g(X_s) \dd{B_s}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \mathbf{E}[Y_t | X_0]
+ = Y_0 + \mathbf{E}\bigg[ \int_0^t \hat{L}\{h(X_s)\} \dd{s} \bigg| X_0 \bigg]
+\end{aligned}$$
+
+For suffiently small $t$, the integral can be replaced by its first-order approximation:
+
+$$\begin{aligned}
+ \mathbf{E}[Y_t | X_0]
+ \approx Y_0 + \hat{L}\{h(X_0)\} \: t
+\end{aligned}$$
+
+Rearranging this gives the following,
+to be understood in the limit $t \to 0^+$:
+
+$$\begin{aligned}
+ \hat{L}\{h(X_0)\}
+ \approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0]
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+The proof is similar to the one above.
+Define $Y_t = h(X_t)$ and use Itō’s lemma:
+
+$$\begin{aligned}
+ \dd{Y_t}
+ &= \bigg( \pdv{h}{x} f(X_t) + \frac{1}{2} \pdvn{2}{h}{x} g^2(X_t) \bigg) \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
+ \\
+ &= \hat{L} \{h(X_t)\} \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
+\end{aligned}$$
+
+And then integrate this from $t = 0$ to the provided stopping time $t = \tau$:
+
+$$\begin{aligned}
+ Y_\tau
+ = Y_0 + \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} + \int_0^\tau \pdv{h}{x} g(X_t) \dd{B_t}
+\end{aligned}$$
+
+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 $X_0$.
+The rest of the above equality is also a martingale:
+
+$$\begin{aligned}
+ 0
+ = \mathbf{E}\bigg[ Y_\tau - Y_0 - \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
+\end{aligned}$$
+
+Isolating this equation for $\mathbf{E}[Y_\tau | X_0]$ then gives Dynkin's formula.
+
+
+
+
Proof
+
+Proof.
+Using the commutator identity for $\comm{A B}{C}$,
+we decompose it like so:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{c}_\nu}
+ &= \sum_{\nu' \nu''} u_{\nu \nu''} \comm{\hat{c}_{\nu'}^\dagger \hat{c}_{\nu''}}{\hat{c}_\nu}
+ = \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{c}_{\nu'}^\dagger \comm{\hat{c}_{\nu''}}{\hat{c}_\nu}
+ + \comm{\hat{c}_{\nu'}^\dagger}{\hat{c}_\nu} \hat{c}_{\nu''} \Big)
+\end{aligned}$$
+
+Bosons have well-known commutation relations,
+so the result follows directly:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{b}_\nu}
+ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{b}_{\nu'}^\dagger \comm{\hat{b}_{\nu''}}{\hat{b}_\nu}
+ + \comm{\hat{b}_{\nu'}^\dagger}{\hat{b}_\nu} \hat{b}_{\nu''} \Big)
+ = - \sum_{\nu''} u_{\nu \nu''} \hat{b}_{\nu''}
+\end{aligned}$$
+
+Fermions only have anticommutation relations,
+so a bit more work is necessary:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{f}_{\!\nu}}
+ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
+ + \comm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
+ \\
+ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
+ - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''}
+ + \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''}
+ - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big)
+ \\
+ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''}
+ - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
+ = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+By integrating the above strains $u_{ii} = \ipdv{u_i}{i}$,
+we get the components of $\va{u}$:
+
+$$\begin{aligned}
+ u_x
+ = \nu \frac{x y}{R} + f_x(y, z)
+ \qquad
+ u_y
+ = \nu \frac{y^2}{2 R} + f_y(x, z)
+ \qquad
+ u_z
+ = - \frac{y z}{R} + f_z(x, y)
+\end{aligned}$$
+
+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 $u_{xz} = 0$:
+
+$$\begin{aligned}
+ 0
+ = u_{xz}
+ = \frac{1}{2} \Big( \pdv{u_x}{z} + \pdv{u_z}{x} \Big)
+ = \frac{1}{2} \Big( \pdv{f_x}{z} + \pdv{f_z}{x} \Big)
+\end{aligned}$$
+
+Here, only $f_x$ may depend on $z$,
+and only $f_z$ may depend on $x$.
+This equation thus tell us:
+
+$$\begin{aligned}
+ f_x(y, z)
+ = z \: g(y)
+ \qquad \quad
+ f_z(x, y)
+ = - x \: g(y)
+\end{aligned}$$
+
+Where $g(y)$ is an unknown integration constant.
+Moving on to $u_{xy} = 0$:
+
+$$\begin{aligned}
+ 0
+ = \frac{1}{2} \Big( \pdv{u_x}{y} + \pdv{u_y}{x} \Big)
+ = \frac{1}{2} \Big( \nu \frac{x}{R} + \pdv{f_x}{y} + \pdv{f_y}{x} \Big)
+\end{aligned}$$
+
+Only $f_x$ may contain $y$,
+so its $y$-derivative must be a constant,
+so $g(y) = C y$. Therefore:
+
+$$\begin{aligned}
+ f_x(y, z)
+ = C y z
+ \qquad
+ f_y(x, z)
+ = - \nu \frac{x^2}{2 R} - C x z + h(z)
+ \qquad
+ f_z(x, y)
+ = - C x y
+\end{aligned}$$
+
+Where $h(z)$ is an unknown integration constant.
+Finally, we put everything in $u_{yz} = 0$:
+
+$$\begin{aligned}
+ 0
+ = \frac{1}{2} \Big( \pdv{u_y}{z} + \pdv{u_z}{y} \Big)
+ = \frac{1}{2} \Big( \pdv{f_y}{z} - \frac{z}{R} + \pdv{f_z}{y} \Big)
+ = \frac{1}{2} \Big( \!-\! 2 C x + \dv{h}{z} - \frac{z}{R} \Big)
+\end{aligned}$$
+
+Only the first term contains $x$, so to satisfy this equation, we must set $C = 0$.
+The remaining terms then tell us that $h(z) = z^2 / (2 R)$.
+Therefore:
+
+$$\begin{aligned}
+ f_x = 0
+ \qquad
+ f_y = - \nu \frac{x^2}{2 R} + \frac{z^2}{2 R}
+ \qquad
+ f_z = 0
+\end{aligned}$$
+
+Inserting this into the components $u_x$, $u_y$ and $u_z$
+then yields the full displacement field.
+
+
+
+
+
+## Microscopic cavity
+
+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:
+
+$$\begin{aligned}
+ E_m(x)
+ = \begin{cases}
+ A_1 e^{- i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
+ A_2 e^{- i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
+ A_4 e^{i k_m n_R x} & \mathrm{for}\; x > \ell/2
+ \end{cases}
+\end{aligned}$$
+
+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 $x = \pm \ell/2$:
+
+$$\begin{aligned}
+ A_1 e^{i k_m n_L \ell/2}
+ &= A_2 e^{i k_m n_C \ell/2} + A_3 e^{- i k_m n_C \ell/2}
+ \\
+ A_4 e^{i k_m n_R \ell/2}
+ &= A_2 e^{- i k_m n_C \ell/2} + A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+$$\begin{aligned}
+ - i k_m n_L A_1 e^{i k_m n_L \ell/2}
+ &= - i k_m n_C A_2 e^{i k_m n_C \ell/2} + i k_m n_C A_3 e^{- i k_m n_C \ell/2}
+ \\
+ i k_m n_R A_4 e^{i k_m n_R \ell/2}
+ &= - i k_m n_C A_2 e^{- i k_m n_C \ell/2} + i k_m n_C A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+
+Rearranging the four equations above yields the following linear system:
+
+$$\begin{aligned}
+ 0
+ &= A_1 - A_2 e^{i k_m (n_C - n_L) \ell/2} - A_3 e^{- i k_m (n_C + n_L) \ell/2}
+ \\
+ 0
+ &= A_2 e^{- i k_m (n_C + n_R) \ell/2} + A_3 e^{i k_m (n_C - n_R) \ell/2} - A_4
+ \\
+ 0
+ &= n_L A_1 + n_C \big( A_3 e^{- i k_m (n_C + n_L) \ell/2} - A_2 e^{i k_m (n_C - n_L) \ell/2} \big)
+ \\
+ 0
+ &= n_C \big( A_3 e^{i k_m (n_C - n_R) \ell/2} - A_2 e^{- i k_m (n_C + n_R) \ell/2} \big) - n_R A_4
+\end{aligned}$$
+
+Which can be rewritten in matrix form as follows, with the system matrix on the left:
+
+$$\begin{aligned}
+ \begin{bmatrix}
+ 1 & -e^{i k_m (n_C - n_L) \ell/2} & -e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
+ 0 & e^{- i k_m (n_C + n_R) \ell/2} & e^{i k_m (n_C - n_R) \ell/2} & -1 \\
+ n_L & -n_C e^{i k_m (n_C - n_L) \ell/2} & n_C e^{- i k_m (n_C + n_L) \ell/2} & 0 \\
+ 0 & -n_C e^{- i k_m (n_C + n_R) \ell/2} & n_C e^{i k_m (n_C - n_R) \ell/2} & -n_R
+ \end{bmatrix}
+ \cdot
+ \begin{bmatrix}
+ A_1 \\ A_2 \\ A_3 \\ A_4
+ \end{bmatrix}
+ =
+ \begin{bmatrix}
+ 0 \\ 0 \\ 0 \\ 0
+ \end{bmatrix}
+\end{aligned}$$
+
+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
+demand that the system matrix is singular, i.e. its determinant is zero:
+
+$$\begin{aligned}
+ 0 =
+ &- n_C (n_L + n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} + e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
+ \\
+ &+ (n_C^2 + n_L n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} - e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big)
+\end{aligned}$$
+
+We multiply by $e^{i k_m (n_L + n_R) \ell / 2}$ and
+decompose the exponentials into sines and cosines:
+
+$$\begin{aligned}
+ 0
+ = i 2 (n_C^2 + n_L n_R) \sin(k_m n_C \ell)
+ - 2 n_C (n_L + n_R) \cos(k_m n_C \ell)
+\end{aligned}$$
+
+Finally, some further rearranging gives a convenient transcendental equation:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \tan(k_m n_C \ell) + i \frac{n_C (n_L + n_R)}{n_C^2 + n_L n_R}
+ }
+\end{aligned}$$
+
+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.
+
+
+## Macroscopic cavity
+
+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 $r_R$.
+We use the same ansatz:
+
+$$\begin{aligned}
+ E_m(x)
+ =
+ \begin{cases}
+ A_1 e^{-i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\
+ A_2 e^{-i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\
+ A_4 e^{i k_m n_R x} & \mathrm{for}\; \ell/2 < x
+ \end{cases}
+\end{aligned}$$
+
+On the left, $A_3$ is the reflection of $A_2$,
+and on the right, $A_2$ is the reflection of $A_3$,
+where the reflected amplitudes are determined
+by the coefficients $r_L$ and $r_R$, respectively:
+
+$$\begin{aligned}
+ A_3 e^{- i k_m n_C \ell/2}
+ &= r_L A_2 e^{i k_m n_C \ell/2}
+ \\
+ A_2 e^{-i k_m n_C \ell/2}
+ &= r_R A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+
+These equations might seem to contradict each other.
+We recast them into matrix form:
+
+$$\begin{aligned}
+ \begin{bmatrix}
+ 1 & - r_R e^{i k_m n_C \ell} \\
+ - r_L e^{i k_m n_C \ell} & 1
+ \end{bmatrix}
+ \cdot
+ \begin{bmatrix}
+ A_2 \\ A_3
+ \end{bmatrix}
+ =
+ \begin{bmatrix}
+ 0 \\ 0
+ \end{bmatrix}
+\end{aligned}$$
+
+Again, we demand that the determinant is zero, in order to get non-trivial solutions:
+
+$$\begin{aligned}
+ 0
+ &= 1 - r_L r_R e^{i 2 k_m n_C \ell}
+\end{aligned}$$
+
+Isolating this for $k_m$ yields the following modes,
+where $m$ is an arbitrary integer:
+
+$$\begin{aligned}
+ \boxed{
+ k_m
+ = - \frac{\ln(r_L r_R) + i 2 \pi m}{i 2 n_C \ell}
+ }
+\end{aligned}$$
+
+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 $1 \!-\! r_R$:
+
+$$\begin{aligned}
+ A_1 e^{i k_m n_L \ell/2}
+ &= (1 - r_L) A_2 e^{i k_m n_C \ell/2}
+ \\
+ A_4 e^{i k_m n_R \ell/2}
+ &= (1 - r_R) A_3 e^{i k_m n_C \ell/2}
+\end{aligned}$$
+
+We simply isolate for $A_1$ and $A_4$ respectively,
+yielding the following amplitudes:
+
+$$\begin{aligned}
+ A_1
+ &= (1 - r_L) A_2 e^{i k_m (n_C - n_L) \ell/2}
+ \\
+ A_4
+ &= (1 - r_R) A_3 e^{i k_m (n_C - n_R) \ell/2}
+\end{aligned}$$
+
+Note that we have not demanded continuity of the electric field.
+This is because the mirrors are infinitely thin "magic" planes;
+had we instead used the full mirror structure,
+then we would have demanded continuity, as you maybe expected.
+
+
+
+## References
+1. P.T. Kristensen, K. Herrmann, F. Intravaia, K. Busch,
+ [Modeling electromagnetic resonators using quasinormal modes](https://doi.org/10.1364/AOP.377940),
+ 2020, Optical Society of America.
diff --git a/source/know/concept/fermi-dirac-distribution/index.md b/source/know/concept/fermi-dirac-distribution/index.md
new file mode 100644
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@@ -0,0 +1,80 @@
+---
+title: "Fermi-Dirac distribution"
+date: 2021-07-11
+categories:
+- Physics
+- Statistics
+- Quantum mechanics
+layout: "concept"
+---
+
+**Fermi-Dirac statistics** describe how identical **fermions**,
+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,
+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 $s$:
+
+$$\begin{aligned}
+ \mathcal{Z}
+ = \sum_{N = 0}^1 \exp(- \beta N (\varepsilon - \mu))
+ = 1 + \exp(- \beta (\varepsilon - \mu))
+\end{aligned}$$
+
+Where $\mu$ is the chemical potential,
+and $\varepsilon$ is the energy contribution per particle in $s$,
+i.e. the total energy of all particles $E = \varepsilon N$.
+
+The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
+is the Landau potential $\Omega$, given by:
+
+$$\begin{aligned}
+ \Omega
+ = - k T \ln{\mathcal{Z}}
+ = - k T \ln\!\Big( 1 + \exp(- \beta (\varepsilon - \mu)) \Big)
+\end{aligned}$$
+
+The average number of particles $\Expval{N}$
+in state $s$ is then found to be as follows:
+
+$$\begin{aligned}
+ \Expval{N}
+ = - \pdv{\Omega}{\mu}
+ = k T \pdv{\ln{\mathcal{Z}}}{\mu}
+ = \frac{\exp(- \beta (\varepsilon - \mu))}{1 + \exp(- \beta (\varepsilon - \mu))}
+\end{aligned}$$
+
+By multiplying both the numerator and the denominator by $\exp(\beta (\varepsilon \!-\! \mu))$,
+we arrive at the standard form of
+the **Fermi-Dirac distribution** or **Fermi function** $f_F$:
+
+$$\begin{aligned}
+ \boxed{
+ \Expval{N}
+ = f_F(\varepsilon)
+ = \frac{1}{\exp(\beta (\varepsilon - \mu)) + 1}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \sigma^2
+ = k T \pdv{\Expval{N}}{\mu}
+ = \Expval{N} \big(1 - \Expval{N}\big)
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Gould, J. Tobochnik,
+ *Statistical and thermal physics*, 2nd edition,
+ Princeton.
diff --git a/source/know/concept/fermis-golden-rule/index.md b/source/know/concept/fermis-golden-rule/index.md
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+---
+title: "Fermi's golden rule"
+date: 2021-07-10
+categories:
+- Physics
+- Quantum mechanics
+- Two-level system
+- Optics
+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
+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 $\omega$:
+
+$$\begin{aligned}
+ P_{ab}
+ = \frac{|V_{ba}|^2}{\hbar^2} \frac{\sin^2\!\big((\omega_{ba} - \omega) t / 2\big)}{(\omega_{ba} - \omega)^2}
+\end{aligned}$$
+
+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 $2 \pi n / (\omega_{ba} \!-\! \omega)$.
+
+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/):
+
+$$\begin{aligned}
+ P_{ab}
+ &= \frac{|V_{ba}|^2}{\hbar^2} \int_{E_b - \Delta E / 2}^{E_b + \Delta E / 2}
+ \frac{\sin^2\!\big((\omega_{xa} - \omega) t / 2\big)}{(\omega_{xa} - \omega)^2} \:\rho(E_x) \dd{E_x}
+\end{aligned}$$
+
+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 $\rho(E_x)$.
+In that case, we also simplify the integration limits.
+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 $t > 0$:
+
+$$\begin{aligned}
+ P_{ab}
+ &= \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b) \: t
+\end{aligned}$$
+
+The transition rate $R_{ab}$,
+i.e. the number of particles per unit time,
+then takes this form:
+
+$$\begin{aligned}
+ \boxed{
+ R_{ab}
+ = \pdv{P_{ab}}{t}
+ = \frac{2 \pi}{\hbar} |V_{ba}|^2 \rho(E_b)
+ }
+\end{aligned}$$
+
+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.
+
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
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+---
+title: "Feynman diagram"
+date: 2021-11-18
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+When doing calculations in the context of condensed matter physics and quantum field theory,
+**Feynman diagrams** graphically represent expressions
+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 $\hat{H}_1$
+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 $\Expval{}$.
+
+Below, we go through the most notable components of Feynman diagrams
+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 $\hat{H}_0$.
+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
+[interaction picture](/know/concept/interaction-picture/),
+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 $\hat{\Psi}$.
+The dots at the ends are called **vertices**,
+which represent points in space and time with a spin.
+Vertices can be
+**internal** (one Green's function entering AND one leaving)
+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/):
+
+
+
+$$\begin{aligned}
+ = i \hbar G_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
+ = \Expval{\mathcal{T} \Big\{ \hat{\Psi}_{s_2 H}(\vb{r}_2, t_2) \hat{\Psi}_{s_1 H}^\dagger(\vb{r}_1, t_1) \Big\}}
+\end{aligned}$$
+
+Next, an **interaction line** or **boson line** represents
+a two-body interaction operator $\hat{W}$ (in $\hat{H}_1$),
+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,
+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 $W$
+has a time-dependence coming only from the interaction picture,
+and is to be used as follows to get the full two-body operator $\hat{W}$:
+
+$$\begin{aligned}
+ \hat{W}
+ = \frac{1}{2} \sum_{s_1 s_2} \iint_{-\infty}^\infty \hat{\Psi}_{s_1}^\dagger(\vb{r}_1) \hat{\Psi}_{s_2}^\dagger(\vb{r}_2)
+ W(\vb{r}_1, \vb{r}_2) \hat{\Psi}_{s_2}(\vb{r}_2) \hat{\Psi}_{s_1}(\vb{r}_1) \dd{\vb{r}_1} \dd{\vb{r}_2}
+\end{aligned}$$
+
+One-body (time-dependent) operators $\hat{V}$ in $\hat{H}_1$
+are instead represented by a special vertex:
+
+
+
+$$\begin{aligned}
+ = \frac{1}{i \hbar} V_s(\vb{r}, t)
+\end{aligned}$$
+
+Other graphical components exist representing
+more complicated operators and quantities,
+but these deserve their own articles.
+
+In order for a given Feynman diagram to be valid,
+it must satisfy the following criteria:
+
+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 $W$.
+
+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
+ 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,
+ but not external ones.
+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 $W$;
+it simply turns out to be nicer to do it this way
+when using Feynman diagrams in the wild.
+
+The combination of rules 2 and 3 means that spin
+belongs to lines rather than vertices,
+so that a particle with a given spin propagates
+from vertex to vertex without getting flipped.
+
+
+## Fourier space
+
+If the system is time-independent and spatially uniform,
+meaning it has continuous translational symmetry in time and space,
+then it is useful to work in [Fourier space](/know/concept/fourier-transform/):
+
+$$\begin{aligned}
+ G_{s_2 s_1}^0(\vb{r}_2, t_2; \vb{r}_1, t_1)
+ &= G_{s_1}^0(\vb{r}_2 - \vb{r}_1, t_2 - t_1) \: \delta_{s_2 s_1}
+ \\
+ &= \frac{\delta_{s_2 s_1}}{(2 \pi)^4} \iint_{-\infty}^\infty G_{s_1}^0(\vb{k}, \omega) \:
+ e^{i \vb{k} \cdot (\vb{r}_2 - \vb{r}_1)} e^{- i \omega (t_2 - t_1)} \dd{\vb{k}} \dd{\omega}
+ \\
+ W_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
+ &= W(\vb{r}_2 - \vb{r}_1) \: \delta(t_2 - t_1)
+ \\
+ &= \frac{1}{(2 \pi)^4} \iint_{\infty}^\infty W(\vb{k}) \:
+ e^{i \vb{k} \cdot (\vb{r}_2 - \vb{r}_1)} e^{- i \omega (t_2 - t_1)} \dd{\vb{k}} \dd{\omega}
+\end{aligned}$$
+
+Where we have used an integral representation of
+the [Dirac delta function](/know/concept/dirac-delta-function/).
+Note the inconsistent sign of the exponent
+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 $\tilde{\vb{k}} = (\vb{k}, \omega)$:
+
+
+
+$$\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'}
+ W(\tilde{\vb{r}}, \tilde{\vb{r}}')
+ G_{s_2' s'}^0(\tilde{\vb{r}}_2', \tilde{\vb{r}}') G_{s' s_2}^0(\tilde{\vb{r}}', \tilde{\vb{r}}_2) \delta_{s_2 s_2'}
+ \\
+ &= \frac{-i \hbar^3}{(2 \pi)^{20}}
+ \sum_{s_1 s_2} \!\!\iint \dd{\tilde{\vb{r}}} \dd{\tilde{\vb{r}}'}
+ \bigg(\! \int \dd{\tilde{\vb{k}}_2} G_{s_1}^0(\tilde{\vb{k}}_2) e^{i \tilde{\vb{k}}_2 \cdot (\tilde{\vb{r}}_1' - \tilde{\vb{r}})} \!\bigg)
+ \bigg(\! \int \dd{\tilde{\vb{k}}_1} G_{s_1}^0(\tilde{\vb{k}}_1) e^{i \tilde{\vb{k}}_1 \cdot (\tilde{\vb{r}} - \tilde{\vb{r}}_1)} \!\bigg)
+ \\
+ &\qquad\times \bigg(\! \int \dd{\tilde{\vb{p}}} W(\tilde{\vb{p}}) e^{i \tilde{\vb{p}} \cdot (\tilde{\vb{r}}' - \tilde{\vb{r}})} \!\bigg)
+ \bigg(\! \int \dd{\tilde{\vb{q}}_2} G_{s_2}^0(\tilde{\vb{q}}_2) e^{i \tilde{\vb{q}}_2 \cdot (\tilde{\vb{r}}_2' - \tilde{\vb{r}}')} \!\bigg)
+ \bigg(\! \int \dd{\tilde{\vb{q}}_1} G_{s_2}^0(\tilde{\vb{q}}_1) e^{i \tilde{\vb{q}}_1 \cdot (\tilde{\vb{r}}' - \tilde{\vb{r}}_2)} \!\bigg)
+ \\
+ &= \frac{-i \hbar^3}{(2 \pi)^{12}}
+ \sum_{s_1 s_2} \!\!\iint \dd{\tilde{\vb{k}}_1} \dd{\tilde{\vb{k}}_2}
+ G_{s_1}^0(\tilde{\vb{k}}_2) G_{s_1}^0(\tilde{\vb{k}}_1)
+ \iint \dd{\tilde{\vb{q}}_1} \dd{\tilde{\vb{q}}_2}
+ G_{s_2}^0(\tilde{\vb{q}}_2) G_{s_2}^0(\tilde{\vb{q}}_1)
+ \\
+ &\qquad\times
+ e^{i \tilde{\vb{k}}_2 \cdot \tilde{\vb{r}}_1' - i \tilde{\vb{k}}_1 \cdot \tilde{\vb{r}}_1
+ + i \tilde{\vb{q}}_2 \cdot \tilde{\vb{r}}_2' - i \tilde{\vb{q}}_1 \cdot \tilde{\vb{r}}_2}
+ \!\!\int \dd{\tilde{\vb{p}}} W(\tilde{\vb{p}})
+ \bigg( \frac{1}{(2 \pi)^8} \!\!\iint \dd{\tilde{\vb{r}}} \dd{\tilde{\vb{r}}'}
+ e^{i (\tilde{\vb{k}}_1 - \tilde{\vb{k}}_2 - \tilde{\vb{p}}) \cdot \tilde{\vb{r}}}
+ e^{i (\tilde{\vb{q}}_1 - \tilde{\vb{q}}_2 + \tilde{\vb{p}}) \cdot \tilde{\vb{r}}'} \bigg)
+ \\
+ &= \frac{-i \hbar^3}{(2 \pi)^{12}}
+ \sum_{s_1 s_2} \!\!\iint \dd{\tilde{\vb{k}}_1} \dd{\tilde{\vb{k}}_2}
+ G_{s_1}^0(\tilde{\vb{k}}_2) G_{s_1}^0(\tilde{\vb{k}}_1)
+ \iint \dd{\tilde{\vb{q}}_1} \dd{\tilde{\vb{q}}_2}
+ G_{s_2}^0(\tilde{\vb{q}}_2) G_{s_2}^0(\tilde{\vb{q}}_1)
+ \\
+ &\qquad\times
+ e^{i \tilde{\vb{k}}_2 \cdot \tilde{\vb{r}}_1' - i \tilde{\vb{k}}_1 \cdot \tilde{\vb{r}}_1
+ + i \tilde{\vb{q}}_2 \cdot \tilde{\vb{r}}_2' - i \tilde{\vb{q}}_1 \cdot \tilde{\vb{r}}_2}
+ \!\!\int \dd{\tilde{\vb{p}}} W(\tilde{\vb{p}})
+ \: \delta(\tilde{\vb{k}}_1 \!-\! \tilde{\vb{k}}_2 \!-\! \tilde{\vb{p}})
+ \: \delta(\tilde{\vb{q}}_1 \!-\! \tilde{\vb{q}}_2 \!+\! \tilde{\vb{p}})
+ \\
+ &= \frac{-i \hbar^3}{(2 \pi)^{12}}
+ \sum_{s_1 s_2} \!\!\int \dd{\tilde{\vb{p}}} W(\tilde{\vb{p}})
+ \int \dd{\tilde{\vb{k}}_1} G_{s_1}^0(\tilde{\vb{k}}_1 \!-\! \tilde{\vb{p}}) G_{s_1}^0(\tilde{\vb{k}}_1)
+ \int \dd{\tilde{\vb{q}}_1} G_{s_2}^0(\tilde{\vb{q}}_1 \!+\! \tilde{\vb{p}}) G_{s_2}^0(\tilde{\vb{q}}_1)
+ \\
+ &\qquad\times
+ e^{i \tilde{\vb{k}}_1 \cdot (\tilde{\vb{r}}_1' - \tilde{\vb{r}}_1)}
+ e^{i \tilde{\vb{q}}_1 \cdot (\tilde{\vb{r}}_2' - \tilde{\vb{r}}_2)}
+ e^{i \tilde{\vb{p}} \cdot (\tilde{\vb{r}}_2' - \tilde{\vb{r}}_1')}
+\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 $\hbar \omega$).
+
+In Fourier space, it makes more sense
+to regard the incoming energies and momenta and spins as given,
+and only integrate over the internal quantities.
+We thus modify the Feynman diagram rules
+such that we end up with the following result:
+
+$$\begin{aligned}
+ \equiv \frac{-i \hbar^3}{(2 \pi)^4}
+ \sum_{s} \!\!\int \dd{\tilde{\vb{p}}} W(\tilde{\vb{p}})
+ \: G_{s_1}^0(\tilde{\vb{k}}_1 \!-\! \tilde{\vb{p}}) \: G_{s_1}^0(\tilde{\vb{k}}_1)
+ \: G_{s_2}^0(\tilde{\vb{q}}_1 \!+\! \tilde{\vb{p}}) \: G_{s_2}^0(\tilde{\vb{q}}_1)
+\end{aligned}$$
+
+Therefore, we say that fermion lines represent $i \hbar G_{s}^0(\vb{k}, \omega)$,
+interaction lines $W(\vb{k}) / i \hbar$, etc.,
+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 $s$;
+ 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.
+5. Depending on the context, additional constant factors may be required;
+ sometimes they are changed on-the-fly during a calculation.
+
+Note that if the diagram is linear (i.e. does not contain interactions),
+then conservation removes all internal variables,
+so no integrals would be needed.
+
+
+## Imaginary time
+
+Feynman diagrams are also useful when working with
+[imaginary time](/know/concept/imaginary-time/).
+In that case, the meaning of fermion lines is changed as follows,
+involving the [Matsubara Green's function](/know/concept/matsubara-greens-function/):
+
+$$\begin{aligned}
+ i \hbar G_{s_2 s_1}^0(\vb{r}_2, t_2; \vb{r}_1, t_1)
+ \:\: &\longrightarrow \:\:
+ \hbar G_{s_2 s_1}^0(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+ = \Expval{\mathcal{T} \Big\{ \hat{\Psi}_I(\vb{r}_2, \tau_2) \hat{\Psi}_I^\dagger(\vb{r}_1, \tau_1) \Big\}}
+ \\
+ i \hbar G_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
+ \:\: &\longrightarrow \:\:
+ \hbar G_{s_2 s_1}(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+ = \Expval{\mathcal{T} \Big\{ \hat{\Psi}_H(\vb{r}_2, \tau_2) \hat{\Psi}_H^\dagger(\vb{r}_1, \tau_1) \Big\}}
+\end{aligned}$$
+
+Where the time-ordering is with respect to $\tau$.
+Interaction lines are modified like so:
+
+$$\begin{aligned}
+ \frac{1}{i \hbar} W_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
+ \:\: &\longrightarrow \:\:
+ -\frac{1}{\hbar} W_{s_2 s_1}(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+ = -\frac{1}{\hbar} W(\vb{r}_2, \vb{r}_1; \tau_1) \delta(\tau_2 \!-\! \tau_1)
+\end{aligned}$$
+
+One-body $V$-vertices are usually not used,
+because they are intended for real-time-dependent operators,
+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 $W$).
+This distinction is compatible with frequency conservation,
+since a sum of two fermionic frequencies is always bosonic.
+We have:
+
+$$\begin{aligned}
+ G_{s_2 s_1}^0(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+ &= \frac{\delta_{s_2 s_1}}{(2 \pi)^3} \int_{-\infty}^\infty \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty
+ G_{s_1}^0(\vb{k}, i \omega_n^f) e^{i \vb{k} \cdot (\vb{r}_2 - \vb{r}_1)} e^{- i \omega_n^f (\tau_2 - \tau_1)} \dd{\vb{k}}
+ \\
+ W_{s_2 s_1}(\vb{r}_2, \tau_2; \vb{r}_1, \tau_1)
+ &= \frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty
+ W(\vb{k}) e^{i \vb{k} \cdot (\vb{r}_2 - \vb{r}_1)} e^{- i \omega_n^b (\tau_2 - \tau_1)} \dd{\vb{k}}
+\end{aligned}$$
+
+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 $e^{i \omega_n^f \eta}$.
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
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diff --git a/source/know/concept/ficks-laws/index.md b/source/know/concept/ficks-laws/index.md
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+---
+title: "Fick's laws"
+date: 2021-09-05
+categories:
+- Physics
+- Mathematics
+layout: "concept"
+---
+
+**Fick's laws of diffusion** govern the majority of diffusion processes,
+where a certain "impurity" substance redistributes itself through a medium over time.
+A diffusion process that obeys Fick's laws is called **Fickian**,
+as opposed to **non-Fickian** or **anomalous diffusion**.
+
+
+## Fick's first law
+
+**Fick's first law** states that diffusing matter
+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 $\mathrm{m}^{-2} \mathrm{s}^{-1}$),
+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 $\mathrm{m}^{-3}$):
+
+$$\begin{aligned}
+ \boxed{
+ \vec{J} = - D \: \nabla C
+ }
+\end{aligned}$$
+
+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.
+
+Fick's first law is a general physical principle,
+which was discovered experimentally,
+and thus does not have a general derivation.
+Proofs for specific systems do exist,
+but they say more about those systems
+than about diffusion in general.
+
+
+## Fick's second law
+
+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:
+
+$$\begin{aligned}
+ M
+ \equiv \int_V C \dd{V}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \dv{M}{t}
+ = - \int_S \vec{J} \cdot \dd{\vec{S}}
+ = - \int_V \nabla \cdot \vec{J} \dd{V}
+\end{aligned}$$
+
+For comparison, we differentiate the definition of $M$,
+and exploit that the integral ignores $t$:
+
+$$\begin{aligned}
+ \dv{M}{t}
+ = \dv{}{t}\int_V C \dd{V}
+ = \int_V \pdv{C}{t} \dd{V}
+\end{aligned}$$
+
+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 $\vec{J}$.
+Substituting this into the continuity equation yields
+the general form of Fick's second law:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{C}{t}
+ = \nabla \cdot \Big( D \: \nabla C \Big)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \pdv{C}{t} = D \: \nabla^2 C
+\end{aligned}$$
+
+
+## Fundamental solution
+
+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
+a [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+ C(x, 0) = \delta(x - x_0)
+\end{aligned}$$
+
+According to Fick's second law,
+the concentration's time evolution of $C$ turns out to be:
+
+$$\begin{aligned}
+ H(x - x_0, t)
+ \equiv C(x, t)
+ = \frac{1}{\sqrt{4 \pi D t}} \exp\!\Big( \!-\!\frac{(x - x_0)^2}{4 D t} \Big)
+\end{aligned}$$
+
+This result is a normalized Gaussian,
+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 $\sqrt{t}$.
+
+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:
+
+$$\begin{aligned}
+ C(x, 0)
+ = (C * \delta)(x)
+ = \int_{-\infty}^\infty C(x_0, 0) \: \delta(x - x_0) \dd{x_0}
+\end{aligned}$$
+
+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 $H$:
+
+$$\begin{aligned}
+ C(x, t)
+ = (C * H)(x)
+ &= \int_{-\infty}^\infty C(x_0, 0) \: H(x - x_0, t) \dd{x_0}
+ \\
+ &= \int_{-\infty}^\infty \frac{1}{\sqrt{4 \pi D t}} \exp\!\Big( \!-\!\frac{(x - x_0)^2}{4 D t} \Big) \: C(x_0, 0) \dd{x_0}
+\end{aligned}$$
+
+This technique is analogous to using
+the [impulse response](/know/concept/impulse-response/)
+of a linear operator to extrapolate all its inhomogeneous solutions.
+The difference is that here, we used the initial condition
+instead of the forcing function.
+
+
+
+## References
+1. U.F. Thygesen,
+ *Lecture notes on diffusions and stochastic differential equations*,
+ 2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md
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@@ -0,0 +1,245 @@
+---
+title: "Fourier transform"
+date: 2021-02-22
+categories:
+- Mathematics
+- Physics
+- Optics
+layout: "concept"
+---
+
+The **Fourier transform** (FT) is an integral transform which converts a
+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
+(for now):
+
+$$\begin{aligned}
+ \boxed{
+ \tilde{f}(k)
+ \equiv \hat{\mathcal{F}}\{f(x)\}
+ \equiv A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+ }
+\end{aligned}$$
+
+The **inverse Fourier transform** (iFT) undoes the forward FT operation:
+
+$$\begin{aligned}
+ \boxed{
+ f(x)
+ \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}
+ \equiv B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k}
+ }
+\end{aligned}$$
+
+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/) $\delta(x)$:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\}
+ &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k}
+ \\
+ &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'}
+ \\
+ &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'}
+ = \frac{2 \pi A B}{|s|} f(x)
+\end{aligned}$$
+
+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 $B$
+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.
+
+
+## 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 $x \to \pm \infty$:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}\{f'(x)\}
+ &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x}
+ \\
+ &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+ \\
+ &= (- i s k) \tilde{f}(k)
+\end{aligned}$$
+
+Therefore, as long as $f(x)$ is localized, the FT eliminates derivatives
+of the transformed variable, which makes it useful against PDEs:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{\mathcal{F}}\{f'(x)\} = (- i s k) \tilde{f}(k)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{\mathcal{F}} \Big\{ \dvn{n}{f}{x} \Big\}
+ = (- i s k)^n \tilde{f}(k)
+ }
+\end{aligned}$$
+
+Derivatives in the frequency domain have an analogous property:
+
+$$\begin{aligned}
+ \dvn{n}{\tilde{f}}{k}
+ &= A \dvn{n}{}{k}\int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+ \\
+ &= A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x}
+ = \hat{\mathcal{F}}\{ (i s x)^n f(x) \}
+\end{aligned}$$
+
+
+## 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:
+
+$$\begin{aligned}
+ \boxed{
+ \tilde{f}(\vb{k})
+ \equiv \hat{\mathcal{F}}\{f(\vb{x})\}
+ \equiv A \int_{-\infty}^\infty f(\vb{x}) \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
+ }
+\end{aligned}$$
+
+Where the wavevector $\vb{k} = (k_1, ..., k_N)$.
+Likewise, the inverse FT is given by:
+
+$$\begin{aligned}
+ \boxed{
+ f(\vb{x})
+ \equiv \hat{\mathcal{F}}^{-1}\{\tilde{f}(\vb{k})\}
+ \equiv B \int_{-\infty}^\infty \tilde{f}(\vb{k}) \exp(- i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{k}}
+ }
+\end{aligned}$$
+
+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 $s = \pm 1$.
+Any choice will do, as long as:
+
+$$\begin{aligned}
+ \boxed{
+ A B
+ = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N}
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+The inverse FT of the forward FT of $f(\vb{x})$ must be equal to $f(\vb{x})$ again, so:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
+ &= A B \int \exp(- i s \vb{k} \cdot \vb{x})
+ \int f(\vb{x}') \exp(i s \vb{k} \cdot \vb{x}') \ddn{N}{\vb{x}'} \ddn{N}{\vb{k}}
+ \\
+ &= (2 \pi)^N A B \int f(\vb{x}')
+ \Big( \frac{1}{(2 \pi)^N} \int \exp(i s \vb{k} \cdot (\vb{x}' - \vb{x})) \ddn{N}{\vb{k}} \Big) \ddn{N}{\vb{x}'}
+ \\
+ &= (2 \pi)^N A B \int f(\vb{x}')
+ \Big( \prod_{n = 1}^N \frac{1}{2 \pi} \int \exp(i s k_n (x_n' - x_n)) \dd{k_n} \Big) \ddn{N}{\vb{x}'}
+\end{aligned}$$
+
+Here, we recognize the definition of the Dirac delta function again,
+leading to:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
+ &= (2 \pi)^N A B \int f(\vb{x}')
+ \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \ddn{N}{\vb{x}'}
+ \\
+ &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
+ = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We insert $\nabla^2 f$ into the FT,
+decompose the exponential and the Laplacian,
+and then integrate by parts (limits $\pm \infty$ omitted):
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}\{\nabla^2 f\}
+ &= A \int \big( \nabla^2 f \big) \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
+ \\
+ &= A \int \Big( \sum_{n = 1}^N \pdv{ {}^2 f}{x_n^2} \Big) \Big( \prod_{m = 1}^N \exp(i s k_m x_m) \Big) \ddn{N}{\vb{x}}
+ \\
+ &= A \sum_{n = 1}^N \bigg[ \pdv{f}{x_n} \exp(i s \vb{k} \cdot \vb{x}) \bigg]
+ - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
+\end{aligned}$$
+
+Just like in 1D, we get rid of the boundary term
+by assuming that all derivatives $\idv{f}{x_n}$ are nicely localized.
+To proceed, we then integrate by parts again:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}\{\nabla^2 f\}
+ &= - A \sum_{n = 1}^N i s k_n \int \pdv{f}{x_n} \Big( \prod_{m = 1}^N \exp(i s k_m x_m) \Big) \ddn{N}{\vb{x}}
+ \\
+ &= - A \sum_{n = 1}^N i s k_n \bigg[ f \exp(i s \vb{k} \cdot \vb{x}) \bigg]
+ + A \sum_{n = 1}^N (i s k_n)^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
+\end{aligned}$$
+
+Once again, we remove the boundary term
+by assuming that $f$ is localized, yielding:
+
+$$\begin{aligned}
+ \hat{\mathcal{F}}\{\nabla^2 f\}
+ &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
+ = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+$\hat{L}$ 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:
+
+$$\begin{aligned}
+ A f(x') \: \delta(x - x')
+ = f(x') \hat{L}\{ G(x, x') \}
+ = \hat{L}\{ f(x') \: G(x, x') \}
+\end{aligned}$$
+
+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 $x$:
+
+$$\begin{aligned}
+ A \int_a^b f(x') \: \delta(x - x') \dd{x'}
+ &= \int_a^b \hat{L}\{ f(x') \: G(x, x') \} \dd{x'}
+ \\
+ A f(x)
+ &= \hat{L} \int_a^b f(x') \: G(x, x') \dd{x'}
+\end{aligned}$$
+
+By definition, $\hat{L}$'s response $u(x)$ to $f(x)$
+satisfies $\hat{L}\{ u(x) \} = f(x)$, recognizable here.
+
+
+
+
Proof
+
+Proof.
+Consider two parameters $x_1'$ and $x_2'$.
+The self-adjointness of $\hat{L}$ means that:
+
+$$\begin{aligned}
+ \int_a^b G^*(x, x_1') \Big( \hat{L} \{ G(x, x_2') \} \Big) \dd{x}
+ &= \int_a^b \Big( \hat{L} \{ G(x, x_1') \} \Big)^* G(x, x_2') \dd{x}
+ \\
+ \int_a^b G^*(x, x_1') \: \delta(x - x_2') \dd{x}
+ &= \int_a^b \delta^*(x - x_1') \: G(x, x_2') \dd{x}
+ \\
+ G^*(x_2', x_1')
+ &= G(x_1', x_2')
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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 $\hat{B}$,
+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:
+
+$$\begin{aligned}
+ \hat{\rho} = \frac{1}{Z} \exp(- \beta \hat{H})
+\end{aligned}$$
+
+The expected value of the product
+of the time-independent operators $\hat{A}$ and $\hat{B}$ is then:
+
+$$\begin{aligned}
+ \expval{\hat{A}(t) \hat{B}(t')}
+ &= \frac{1}{Z} \Tr\!\big( \hat{\rho} \hat{A}(t) \hat{B}(t') \big)
+ \\
+ &= \frac{1}{Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{i t \hat{H} / \hbar} \hat{A} e^{-i t \hat{H} / \hbar}
+ e^{i t' \hat{H} / \hbar} \hat{B} e^{-i t' \hat{H} / \hbar} \Big)
+\end{aligned}$$
+
+Using that the trace $\Tr$ is invariant
+under cyclic permutations of its argument,
+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.
+Note that thermodynamic equilibrium is crucial:
+intuitively, if the system is not in equilibrium,
+then it evolves in some transient time-dependent way.
+
+
+
+
Proof
+
+Proof.
+In the second quantization,
+the Hamiltonian $\hat{H}_0$ is written like so:
+
+$$\begin{aligned}
+ \hat{H}_0
+ &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \Inprod{\psi_\nu}{\nabla^2 \psi_{\nu'}}
+ \\
+ &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_\nu^\dagger \hat{c}_{\nu'} \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+ \\
+ &= - \frac{\hbar^2}{2 m}
+ \int \Big( \sum_{\nu} \psi_\nu^*(\vb{r}') \hat{c}_\nu^\dagger \Big) \Big( \nabla^2 \sum_{\nu'} \psi_{\nu'}(\vb{r}') \hat{c}_{\nu'} \Big) \dd{\vb{r}'}
+ \\
+ &= - \frac{\hbar^2}{2 m}
+ \int \hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+We then insert this into the commutator that we want to prove, yielding:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+ &= - \frac{\hbar^2}{2 m} \int \Comm{\hat{\Psi}^\dagger(\vb{r}') \: \nabla^2 \hat{\Psi}(\vb{r}')}{\hat{\Psi}(\vb{r})} \dd{\vb{r}'}
+ \\
+ &= - \frac{\hbar^2}{2 m} \int \hat{\Psi}^\dagger(\vb{r}') \Comm{\nabla^2 \hat{\Psi}(\vb{r}')}{\hat{\Psi}(\vb{r})}
+ + \Comm{\hat{\Psi}^\dagger(\vb{r}')}{\hat{\Psi}(\vb{r})} \nabla^2 \hat{\Psi}(\vb{r}') \dd{\vb{r}'}
+ \\
+ &= - \frac{\hbar^2}{2 m} \sum_{\nu \nu' \nu''}
+ \Big( \hat{c}_\nu^\dagger \comm{\hat{c}_{\nu''}}{\hat{c}_{\nu'}} + \comm{\hat{c}_\nu^\dagger}{\hat{c}_{\nu'}} \hat{c}_{\nu''} \Big)
+ \psi_{\nu'}(\vb{r}) \int \psi_\nu^*(\vb{r}') \: \nabla^2 \psi_{\nu''}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+When deriving equation-of-motion theory,
+we already showed that the following identity
+holds for both bosons and fermions:
+
+$$\begin{aligned}
+ \hat{c}_\nu^\dagger \comm{\hat{c}_{\nu''}}{\hat{c}_{\nu'}} + \comm{\hat{c}_\nu^\dagger}{\hat{c}_{\nu'}} \hat{c}_{\nu''}
+ = - \delta_{\nu \nu'} \hat{c}_{\nu''}
+\end{aligned}$$
+
+Such that the commutator can be significantly simplified to:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+ &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
+ \int \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r}) \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+\end{aligned}$$
+
+We know that the $\psi_\nu$ form a *complete* basis,
+which implies (see [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)):
+
+$$\begin{aligned}
+ \sum_{\nu} \psi_\nu^*(\vb{r}') \: \psi_\nu(\vb{r})
+ = \delta(\vb{r} - \vb{r}')
+\end{aligned}$$
+
+With this, the commutator can be reduced even further as follows:
+
+$$\begin{aligned}
+ \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+ &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
+ \int \delta(\vb{r} - \vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+ \\
+ &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
+ = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We define $w(t)$ to equal the upper bounds above
+on both $w'(t)$ and $w(t)$ itself:
+
+$$\begin{aligned}
+ w(t)
+ \equiv u(0) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
+ \quad \implies \quad
+ w'(t)
+ = \beta(t) \: w(t)
+\end{aligned}$$
+
+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 $u(t)$
+in order to satisfy the inequality.
+We thus calculate:
+
+$$\begin{aligned}
+ \dv{}{t}\bigg( \frac{u}{w} \bigg)
+ = \frac{u' w - u w'}{w^2}
+ = \frac{u' w - u \beta w}{w^2}
+ = \frac{u' - u \beta}{w}
+\end{aligned}$$
+
+Since $u' \le \beta u$ as a condition,
+the above derivative is always negative.
+
+
+
+
Proof
+
+Proof.
+We start by defining $w(t)$ as follows,
+which will act as shorthand:
+
+$$\begin{aligned}
+ w(t)
+ \equiv \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg) \bigg( \int_0^t \beta(s) \: u(s) \dd{s} \bigg)
+\end{aligned}$$
+
+Its derivative $w'(t)$ is then straightforwardly calculated to be given by:
+
+$$\begin{aligned}
+ w'(t)
+ &= \bigg( \dv{}{t} \int_0^t \beta(s) \: u(s) \dd{s} - \beta(t)\int_0^t \beta(s) \: u(s) \dd{s} \bigg)
+ \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
+ \\
+ &= \beta(t) \bigg( u(t) - \int_0^t \beta(s) \: u(s) \dd{s} \bigg)
+ \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
+\end{aligned}$$
+
+The parenthesized expression it bounded from above by $\alpha(t)$,
+thanks to the condition that $u(t)$ is assumed to satisfy,
+for the Grönwall-Bellman inequality to be true:
+
+$$\begin{aligned}
+ w'(t)
+ \le \alpha(t) \: \beta(t) \exp\!\bigg( \!-\!\! \int_0^t \beta(s) \dd{s} \bigg)
+\end{aligned}$$
+
+Integrating this to find $w(t)$ yields the following result:
+
+$$\begin{aligned}
+ w(t)
+ \le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s}
+\end{aligned}$$
+
+In the initial definition of $w(t)$,
+we now move the exponential to the other side,
+and rewrite it using the above inequality for $w(t)$:
+
+$$\begin{aligned}
+ \int_0^t \beta(s) \: u(s) \dd{s}
+ &= w(t) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
+ \\
+ &\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg) \exp\!\bigg( \!-\!\! \int_0^s \beta(r) \dd{r} \bigg) \dd{s}
+ \\
+ &\le \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
+\end{aligned}$$
+
+Insert this into the condition under which the Grönwall-Bellman inequality holds.
+
+
+
+
Proof
+
+Proof.
+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:
+
+$$\begin{aligned}
+ u(t)
+ &\le \alpha(t) + \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
+ \\
+ &\le \alpha(t) + \alpha(t) \int_0^t \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
+\end{aligned}$$
+
+Now, consider the following straightfoward identity, involving the exponential:
+
+$$\begin{aligned}
+ \dv{}{s}\exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
+ &= - \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg)
+\end{aligned}$$
+
+By inserting this into Grönwall-Bellman inequality, we arrive at:
+
+$$\begin{aligned}
+ u(t)
+ &\le \alpha(t) - \alpha(t) \int_0^t \dv{}{s}\exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
+ \\
+ &\le \alpha(t) - \alpha(t) \bigg[ \int \dv{}{s}\exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s} \bigg]_{s = 0}^{s = t}
+\end{aligned}$$
+
+Where we have converted the outer integral from definite to indefinite.
+Continuing:
+
+$$\begin{aligned}
+ u(t)
+ &\le \alpha(t) - \alpha(t) \bigg[ \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \bigg]_{s = 0}^{s = t}
+ \\
+ &\le \alpha(t) - \alpha(t) \exp\!\bigg( \int_t^t \beta(r) \dd{r} \bigg) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
+ \\
+ &\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We know what $\vb{x}_L$ is,
+so we can write out $(\vb{x}_L \cdot \nabla) \vb{B}$
+for $\vb{B} = (B_x, B_y, B_z)$:
+
+$$\begin{aligned}
+ (\vb{x}_L \cdot \nabla) \vb{B}
+ = \frac{u_L}{\omega_c}
+ \begin{pmatrix}
+ \displaystyle \sin(\omega_c t) \pdv{B_x}{x} + \cos(\omega_c t) \pdv{B_x}{y} \\
+ \displaystyle \sin(\omega_c t) \pdv{B_y}{x} + \cos(\omega_c t) \pdv{B_y}{y} \\
+ \displaystyle \sin(\omega_c t) \pdv{B_z}{x} + \cos(\omega_c t) \pdv{B_z}{y}
+ \end{pmatrix}
+\end{aligned}$$
+
+Integrating $\sin$ and $\cos$ over their period yields zero,
+so the average vanishes:
+
+$$\begin{aligned}
+ \Expval{ (\vb{x}_L \cdot \nabla) \vb{B} }
+ = 0
+\end{aligned}$$
+
+Moving on, we write out $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$,
+suppressing the arguments of $\sin$ and $\cos$:
+
+$$\begin{aligned}
+ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}
+ &= \frac{u_L^2}{\omega_c}
+ \begin{pmatrix}
+ \cos \\
+ - \sin \\
+ 0
+ \end{pmatrix}
+ \cross
+ \begin{pmatrix}
+ \displaystyle \pdv{B_x}{x} \sin + \pdv{B_x}{y} \cos \\
+ \displaystyle \pdv{B_y}{x} \sin + \pdv{B_y}{y} \cos \\
+ \displaystyle \pdv{B_z}{x} \sin + \pdv{B_z}{y} \cos
+ \end{pmatrix}
+ \\
+ &= \frac{u_L^2}{\omega_c}
+ \begin{pmatrix}
+ \displaystyle - \pdv{B_z}{x} \sin^2 - \pdv{B_z}{y} \sin \cos \\
+ \displaystyle - \pdv{B_z}{x} \sin \cos - \pdv{B_z}{y} \cos^2 \\
+ \displaystyle \pdv{B_y}{x} \sin \cos + \pdv{B_y}{y} \cos^2
+ \displaystyle + \pdv{B_x}{x} \sin^2 + \pdv{B_x}{y} \sin \cos
+ \end{pmatrix}
+\end{aligned}$$
+
+Integrating products of $\sin$ and $\cos$ over their period gives us the following:
+
+$$\begin{aligned}
+ \Expval{\cos^2} = \Expval{\sin^2} = \frac{1}{2}
+ \qquad \quad
+ \Expval{\sin \cos} = 0
+\end{aligned}$$
+
+Inserting this tells us that the average
+of $\vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B}$ is given by:
+
+$$\begin{aligned}
+ \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+ &= \frac{u_L^2}{2 \omega_c}
+ \begin{pmatrix}
+ \displaystyle - \pdv{B_z}{x} \\
+ \displaystyle - \pdv{B_z}{y} \\
+ \displaystyle \pdv{B_y}{y} + \pdv{B_x}{x}
+ \end{pmatrix}
+\end{aligned}$$
+
+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 $B \equiv |\vb{B}| \approx B_z$:
+
+$$\begin{aligned}
+ \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+ &= - \frac{u_L^2}{2 \omega_c}
+ \begin{pmatrix}
+ \displaystyle \pdv{B_z}{x} \\
+ \displaystyle \pdv{B_z}{y} \\
+ \displaystyle \pdv{B_z}{z}
+ \end{pmatrix}
+ \approx - \frac{u_L^2}{2 \omega_c}
+ \begin{pmatrix}
+ \displaystyle \pdv{B}{x} \\
+ \displaystyle \pdv{B}{y} \\
+ \displaystyle \pdv{B}{z}
+ \end{pmatrix}
+ = - \frac{u_L^2}{2 \omega_c} \nabla B
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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:
+
+$$\begin{aligned}
+ \dv{\vu{b}}{t}
+ = \pdv{\vu{b}}{s} \dv{s}{t}
+ = u_{gc\parallel} \dv{\vu{b}}{s}
+\end{aligned}$$
+
+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 $\dd{\theta}$:
+
+$$\begin{aligned}
+ \dd{s}
+ = R_c \dd{\theta}
+\end{aligned}$$
+
+Meanwhile, across this arc, $\vu{b}$ rotates by $\dd{\theta}$,
+such that the tip travels a distance $|\dd{\vu{b}}|$:
+
+$$\begin{aligned}
+ |\!\dd{\vu{b}}\!|
+ = |\vu{b}| \dd{\theta}
+ = \dd{\theta}
+\end{aligned}$$
+
+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 $\vu{b}$:
+
+$$\begin{aligned}
+ \dd{\vu{b}}
+ = - \vu{R}_c \dd{\theta}
+\end{aligned}$$
+
+Combining these expressions for $\dd{s}$ and $\dd{\vu{b}}$,
+we find the following derivative:
+
+$$\begin{aligned}
+ \dv{\vu{b}}{s}
+ = - \frac{\vu{R}_c \dd{\theta}}{R_c \dd{\theta}}
+ = - \frac{\vu{R}_c}{R_c}
+ = - \frac{\vb{R}_c}{R_c^2}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+For the first equation,
+we differentiate $H$ with respect to $q_n$,
+and use the chain rule:
+
+$$\begin{aligned}
+ \pdv{H}{q_n}
+ &= \pdv{}{q_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
+ \\
+ &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{q_n} + p_j \pdv{\dot{q}_j}{q_n} \Big)
+ - \Big( \pdv{L}{q_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{q_n} \Big) \bigg)
+ \\
+ &= \sum_{j} \Big( p_j \pdv{\dot{q}_j}{q_n} - \pdv{L}{q_n} - p_j \pdv{\dot{q}_j}{q_n} \Big)
+ = - \pdv{L}{q_n}
+\end{aligned}$$
+
+We use the Euler-Lagrange equation here,
+leading to the desired equation:
+
+$$\begin{aligned}
+ - \pdv{L}{q_n} = - \dv{}{t}\Big( \pdv{L}{\dot{q}_n} \Big) = - \dv{p_n}{t} = - \dot{p}_n
+\end{aligned}$$
+
+The second equation is somewhat trivial,
+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:
+
+$$\begin{aligned}
+ \pdv{H}{p_n}
+ &= \pdv{}{p_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
+ \\
+ &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big)
+ - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg)
+ \\
+ &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n}
+ - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
+ = \dot{q}_n
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We differentiate via the multivariate chain rule,
+insert the canonical equations,
+and eventually recognize the PB definition:
+
+$$\begin{aligned}
+ \dv{A}{t}
+ &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t}
+ \\
+ &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t}
+ \\
+ &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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:
+
+$$\begin{aligned}
+ \dot{Q}_n
+ &= \{Q_n, H\}
+ = \sum_{n} \bigg( \pdv{Q_n}{q_n} \pdv{H}{p_n} - \pdv{Q_n}{p_n} \pdv{H}{q_n} \bigg)
+ \\
+ &= \sum_{n, j} \bigg( \pdv{Q_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big)
+ - \pdv{Q_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg)
+ \\
+ &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{Q_n}{q_n} \pdv{Q_j}{p_n} - \pdv{Q_n}{p_n} \pdv{Q_j}{q_n} \Big)
+ + \pdv{H}{P_j} \Big( \pdv{Q_n}{q_n} \pdv{P_j}{p_n} - \pdv{Q_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg)
+ \\
+ &= \sum_{j} \bigg( \pdv{H}{Q_j} \{Q_n, Q_j\} + \pdv{H}{P_j} \{Q_n, P_j\} \bigg)
+\end{aligned}$$
+
+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 $\{Q_n, P_j\} = \delta_{nj}$.
+
+Next, we do the exact same thing with $P_n$ instead of $Q_n$,
+giving an analogous result:
+
+$$\begin{aligned}
+ \dot{P}_n
+ &= \{P_n, H\}
+ = \sum_{n} \bigg( \pdv{P_n}{q_n} \pdv{H}{p_n} - \pdv{P_n}{p_n} \pdv{H}{q_n} \bigg)
+ \\
+ &= \sum_{n, j} \bigg( \pdv{P_n}{q_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{p_n} + \pdv{H}{P_j} \pdv{P_j}{p_n} \Big)
+ - \pdv{P_n}{p_n} \Big( \pdv{H}{Q_j} \pdv{Q_j}{q_n} + \pdv{H}{P_j} \pdv{P_j}{q_n} \Big) \bigg)
+ \\
+ &= \sum_{n, j} \bigg( \pdv{H}{Q_j} \Big( \pdv{P_n}{q_n} \pdv{Q_j}{p_n} - \pdv{P_n}{p_n} \pdv{Q_j}{q_n} \Big)
+ + \pdv{H}{P_j} \Big( \pdv{P_n}{q_n} \pdv{P_j}{p_n} - \pdv{P_n}{p_n} \pdv{P_j}{q_n} \Big) \bigg)
+ \\
+ &= \sum_{j} \bigg( \pdv{H}{Q_j} \{P_n, Q_j\} + \pdv{H}{P_j} \{P_n, P_j\} \bigg)
+\end{aligned}$$
+
+Which is equivalent to Hamilton's equation $\dot{P}_n = -\ipdv{H}{Q_n}$
+if and only if $\{P_n, P_j\} = 0$,
+and $\{Q_n, P_j\} = - \delta_{nj}$.
+The PB is anticommutative,
+i.e. $\{A, B\} = - \{B, A\}$.
+
+
+
+
Proof
+
+Proof.
+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 $\mathrm{sgn}(t)$:
+
+$$\begin{aligned}
+ \Theta(t) = \frac{1}{2} + \frac{\mathrm{sgn}(t)}{2}
+\end{aligned}$$
+
+We then take the Fourier transform,
+where $A$ and $s$ are constants from its definition:
+
+$$\begin{aligned}
+ \tilde{\Theta}(\omega)
+ = \hat{\mathcal{F}}\{\Theta(t)\}
+ = \frac{A}{2} \Big( \int_{-\infty}^\infty \exp(i s \omega t) \dd{t} + \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t} \Big)
+\end{aligned}$$
+
+The first term is proportional to the Dirac delta function.
+The second integral is problematic, so we take the Cauchy principal value $\pv{}$
+and look up the integral:
+
+$$\begin{aligned}
+ \tilde{\Theta}(\omega)
+ &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
+ = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+Just like before, we decompose $f(z)$ into its real and imaginary parts:
+
+$$\begin{aligned}
+ \oint_C f(z) \dd{z}
+ &= \oint_C (u + i v) \dd{(x + i y)}
+ = \oint_C (u + i v) \:(\dd{x} + i \dd{y})
+ \\
+ &= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
+\end{aligned}$$
+
+Using Green's theorem, we integrate over the area $A$ enclosed by $C$:
+
+$$\begin{aligned}
+ \oint_C f(z) \dd{z}
+ &= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
+\end{aligned}$$
+
+Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+equations, such that the integrands disappear and the final result is zero.
+
+
+
+
Proof
+
+Proof.
+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
+we integrate by substitution:
+
+$$\begin{aligned}
+ \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
+ &= \frac{1}{2 \pi i} \int_0^{2 \pi} f(z) \frac{i r e^{i \theta}}{r e^{i \theta}} \dd{\theta}
+ = \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
+\end{aligned}$$
+
+We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:
+
+$$\begin{aligned}
+ \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
+ &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
+ = f(z_0)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+By definition, the first derivative $f'(z)$ of a
+holomorphic function exists and is:
+
+$$\begin{aligned}
+ f'(z_0)
+ = \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
+\end{aligned}$$
+
+We evaluate the numerator using Cauchy's integral theorem as follows:
+
+$$\begin{aligned}
+ f'(z_0)
+ &= \lim_{z \to z_0} \frac{1}{z - z_0}
+ \bigg( \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} - \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta} \bigg)
+ \\
+ &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0}
+ \oint_C \frac{f(\zeta)}{\zeta - z} - \frac{f(\zeta)}{\zeta - z_0} \dd{\zeta}
+ \\
+ &= \frac{1}{2 \pi i} \lim_{z \to z_0} \frac{1}{z - z_0}
+ \oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta}
+\end{aligned}$$
+
+This contour integral converges uniformly, so we may apply the limit on the inside:
+
+$$\begin{aligned}
+ f'(z_0)
+ &= \frac{1}{2 \pi i} \oint_C \Big( \lim_{z \to z_0} \frac{f(\zeta)}{(\zeta - z)(\zeta - z_0)} \Big) \dd{\zeta}
+ = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta}
+\end{aligned}$$
+
+Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
+this proof works inductively for all higher orders $n$.
+
+
+
+
Proof
+
+Proof.
+Starting from the definition of $u_p(t)$,
+we shift the argument by some constant $\tau$,
+and multiply both sides by the constant $f(\tau)$:
+
+$$\begin{aligned}
+ \hat{L} \{ u_p(t - \tau) \} &= \delta(t - \tau)
+ \\
+ \hat{L} \{ f(\tau) \: u_p(t - \tau) \} &= f(\tau) \: \delta(t - \tau)
+\end{aligned}$$
+
+Where $f(\tau)$ can be moved inside using the
+linearity of $\hat{L}$. Integrating over $\tau$ then gives us:
+
+$$\begin{aligned}
+ \int_0^\infty \hat{L} \{ f(\tau) \: u_p(t - \tau) \} \dd{\tau}
+ &= \int_0^\infty f(\tau) \: \delta(t - \tau) \dd{\tau}
+ = f(t)
+\end{aligned}$$
+
+The integral and $\hat{L}$ are operators of different variables, so we reorder them:
+
+$$\begin{aligned}
+ \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
+ &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
+\end{aligned}$$
+
+
{{ letter.name }}
+
+ {% assign items = letter.items | sort: "title" %}
+ {% for item in items %}
+ {{ item.title }}
+{% endfor %}
diff --git a/source/know/concept/interaction-picture/index.md b/source/know/concept/interaction-picture/index.md
new file mode 100644
index 0000000..3912b46
--- /dev/null
+++ b/source/know/concept/interaction-picture/index.md
@@ -0,0 +1,211 @@
+---
+title: "Interaction picture"
+date: 2021-09-13
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+The **interaction picture** or **Dirac picture**
+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 $\hat{L}_H(t)$.
+
+However, in the interaction picture,
+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 $\hat{H}_{0,S}$:
+
+$$\begin{aligned}
+ \hat{H}_S(t)
+ = \hat{H}_{0,S} + \hat{H}_{1,S}(t)
+\end{aligned}$$
+
+With $\hat{H}_S(t)$ the full Schrödinger Hamiltonian.
+We define the unitary conversion operator:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{U}(t)
+ \equiv \exp\!\bigg( i \frac{\hat{H}_{0,S} t}{\hbar} \bigg)
+ }
+\end{aligned}$$
+
+The interaction-picture states $\Ket{\psi_I(t)}$ and operators $\hat{L}_I(t)$
+are then defined to be:
+
+$$\begin{aligned}
+ \boxed{
+ \Ket{\psi_I(t)}
+ \equiv \hat{U}(t) \Ket{\psi_S(t)}
+ \qquad
+ \hat{L}_I(t)
+ \equiv \hat{U}(t) \: \hat{L}_S(t) \: \hat{U}{}^\dagger(t)
+ }
+\end{aligned}$$
+
+
+## Equations of motion
+
+To find the equation of motion for $\Ket{\psi_I(t)}$,
+we differentiate it and multiply by $i \hbar$:
+
+$$\begin{aligned}
+ i \hbar \dv{}{t}\Ket{\psi_I}
+ &= i \hbar \Big( \dv{\hat{U}}{t} \Ket{\psi_S} + \hat{U} \dv{}{t}\Ket{\psi_S} \Big)
+ \\
+ &= i \hbar \Big( i \frac{\hat{H}_{0,S}}{\hbar} \Big) \hat{U} \Ket{\psi_S} + \hat{U} \Big( i \hbar \dv{}{t}\Ket{\psi_S} \Big)
+\end{aligned}$$
+
+We insert the Schrödinger equation into the second term,
+and use $\comm{\hat{U}}{\hat{H}_{0,S}} = 0$:
+
+$$\begin{aligned}
+ i \hbar \dv{}{t}\Ket{\psi_I}
+ &= - \hat{H}_{0,S} \hat{U} \Ket{\psi_S} + \hat{U} \hat{H}_S \Ket{\psi_S}
+ \\
+ &= \hat{U} \big( \!-\! \hat{H}_{0,S} + \hat{H}_S \big) \Ket{\psi_S}
+ \\
+ &= \hat{U} \big( \hat{H}_{1,S} \big) \hat{U}{}^\dagger \hat{U} \Ket{\psi_S}
+\end{aligned}$$
+
+Which leads to an analogue of the Schrödinger equation,
+with $\hat{H}_{1,I} = \hat{U} \hat{H}_{1,S} \hat{U}{}^\dagger$:
+
+$$\begin{aligned}
+ \boxed{
+ i \hbar \dv{}{t}\Ket{\psi_I(t)}
+ = \hat{H}_{1,I}(t) \Ket{\psi_I(t)}
+ }
+\end{aligned}$$
+
+Next, we do the same with an operator $\hat{L}_I$
+to find a description of its evolution in time:
+
+$$\begin{aligned}
+ \dv{}{t}\hat{L}_I
+ &= \dv{\hat{U}}{t} \hat{L}_S \hat{U}{}^\dagger + \hat{U} \hat{L}_S \dv{\hat{U}{}^\dagger}{t} + \hat{U} \dv{\hat{L}_S}{t} \hat{U}{}^\dagger
+ \\
+ &= \frac{i}{\hbar} \hat{U} \hat{H}_{0,S} \big( \hat{U}{}^\dagger \hat{U} \big) \hat{L}_S \hat{U}{}^\dagger
+ - \frac{i}{\hbar} \hat{U} \hat{L}_S \big( \hat{U}{}^\dagger \hat{U} \big) \hat{H}_{0,S} \hat{U}{}^\dagger
+ + \Big( \dv{\hat{L}_S}{t} \Big)_I
+ \\
+ &= \frac{i}{\hbar} \hat{H}_{0,I} \hat{L}_I
+ - \frac{i}{\hbar} \hat{L}_I \hat{H}_{0,I}
+ + \Big( \dv{\hat{L}_S}{t} \Big)_I
+ = \frac{i}{\hbar} \comm{\hat{H}_{0,I}}{\hat{L}_I} + \Big( \dv{\hat{L}_S}{t} \Big)_I
+\end{aligned}$$
+
+The result is analogous to the equation of motion in the Heisenberg picture:
+
+$$\begin{aligned}
+ \boxed{
+ \dv{}{t}\hat{L}_I(t)
+ = \frac{i}{\hbar} \comm{\hat{H}_{0,I}(t)}{\hat{L}_I(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_I
+ }
+\end{aligned}$$
+
+
+## Time evolution operator
+
+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 $t$:
+
+$$\begin{aligned}
+ \Ket{\psi_S(t)}
+ = \hat{K}_S(t, t_0) \Ket{\psi_S(t_0)}
+ \qquad \quad
+ \hat{K}_S(t, t_0)
+ \equiv \exp\!\Big( \!-\! i \frac{\hat{H}_S (t - t_0)}{\hbar} \Big)
+\end{aligned}$$
+
+We want to find an analogous operator in the interaction picture, satisfying:
+
+$$\begin{aligned}
+ \Ket{\psi_I(t)}
+ \equiv \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)}
+\end{aligned}$$
+
+Inserting this definition into the equation of motion for $\Ket{\psi_I}$ yields
+an equation for $\hat{K}_I$, with the logical boundary condition $\hat{K}_I(t_0, t_0) = 1$:
+
+$$\begin{aligned}
+ i \hbar \dv{}{t}\Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big)
+ &= \hat{H}_{1,I}(t) \Big( \hat{K}_I(t, t_0) \Ket{\psi_I(t_0)} \Big)
+ \\
+ i \hbar \dv{}{t}\hat{K}_I(t, t_0)
+ &= \hat{H}_{1,I}(t) \hat{K}_I(t, t_0)
+\end{aligned}$$
+
+We turn this into an integral equation
+by integrating both sides from $t_0$ to $t$:
+
+$$\begin{aligned}
+ i \hbar \int_{t_0}^t \dv{}{t'}K_I(t', t_0) \dd{t'}
+ = \int_{t_0}^t \hat{H}_{1,I}(t') \hat{K}_I(t', t_0) \dd{t'}
+\end{aligned}$$
+
+After evaluating the left integral,
+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:
+
+$$\begin{aligned}
+ K_I(t, t_0)
+ = 1 + \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'}
+ + \frac{1}{(i \hbar)^2} \int_{t_0}^t \hat{H}_{1,I}(t') \int_{t_0}^{t'} \hat{H}_{1,I}(t'') \hat{K}_I(t'', t_0) \dd{t''} \dd{t'}
+\end{aligned}$$
+
+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 $\hat{H}_{1,I}$:
+
+$$\begin{aligned}
+ \hat{K}_I(t, t_0)
+ &= 1 + \int_{t_0}^t \hat{H}_{1,I}(t_1) \dd{t_1}
+ + \frac{1}{2} \int_{t_0}^{t} \int_{t_0}^{t_1} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \hat{H}_{1,I}(t_2) \Big\} \dd{t_1} \dd{t_2}
+ + \: ...
+ \\
+ &= 1 + \sum_{n = 1}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
+ \int_{t_0}^{t} \cdots \int_{t_0}^{t_n} \mathcal{T} \Big\{ \hat{H}_{1,I}(t_1) \cdots \hat{H}_{1,I}(t_n) \Big\} \dd{t_1} \cdots \dd{t_n}
+ \\
+ &= \sum_{n = 0}^\infty \frac{1}{n!} \frac{1}{(i \hbar)^n}
+ \mathcal{T} \bigg\{ \bigg( \int_{t_0}^{t} \hat{H}_{1,I}(t') \dd{t'} \bigg)^n \bigg\}
+\end{aligned}$$
+
+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 $\hat{K}_I$:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{K}_I(t, t_0)
+ = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
+
diff --git a/source/know/concept/ion-sound-wave/index.md b/source/know/concept/ion-sound-wave/index.md
new file mode 100644
index 0000000..48a727d
--- /dev/null
+++ b/source/know/concept/ion-sound-wave/index.md
@@ -0,0 +1,261 @@
+---
+title: "Ion sound wave"
+date: 2021-10-31
+categories:
+- Physics
+- Plasma physics
+- Plasma waves
+- Perturbation
+layout: "concept"
+---
+
+In a plasma, electromagnetic interactions allow
+compressional longitudinal waves to propagate
+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 $m_e \ll m_i$:
+
+$$\begin{aligned}
+ m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+ &= - q_i n_i \nabla \phi - \gamma_i k_B T_i \nabla n_i
+ \\
+ 0
+ &= - q_e n_e \nabla \phi - \gamma_e k_B T_e \nabla n_e
+\end{aligned}$$
+
+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 $1$):
+
+$$\begin{aligned}
+ n_i
+ = n_{i0} + n_{i1}
+ \qquad
+ n_e
+ = n_{e0} + n_{e1}
+ \qquad
+ \vb{u}_i
+ = \vb{u}_{i0} + \vb{u}_{i1}
+ \qquad
+ \phi
+ = \phi_0 + \phi_1
+\end{aligned}$$
+
+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 $\phi_0$
+by definition satisfy:
+
+$$\begin{aligned}
+ \pdv{n_{i0}}{t} = 0
+ \qquad
+ \frac{\mathrm{D} \vb{u}_{i0}}{\mathrm{D} t} = 0
+ \qquad
+ \nabla n_{i0} = \nabla n_{e0} = 0
+ \qquad
+ \vb{u}_{i0} = 0
+ \qquad
+ \phi_0 = 0
+\end{aligned}$$
+
+Inserting this decomposition into the momentum equations
+yields new equations.
+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:
+
+$$\begin{aligned}
+ m_i (n_{i0} \!+\! n_{i1}) \frac{\mathrm{D} (\vb{u}_{i0} \!+\! \vb{u}_{i1})}{\mathrm{D} t}
+ &= - q_i (n_{i0} \!+\! n_{i1}) \nabla (\phi_0 \!+\! \phi_1) - \gamma_i k_B T_i \nabla (n_{i0} \!+\! n_{i1})
+ \\
+ 0
+ &= - q_e (n_{e0} \!+\! n_{e1}) \nabla (\phi_0 \!+\! \phi_1) - \gamma_e k_B T_e \nabla (n_{e0} \!+\! n_{e1})
+\end{aligned}$$
+
+Using the defined properties of the equilibrium components
+$n_{i0}$, $n_{e0}$, $\vb{u}_{i0}$ and $\phi_0$,
+and neglecting all products of perturbations for being small,
+this reduces to:
+
+$$\begin{aligned}
+ m_i n_{i0} \pdv{\vb{u}_{i1}}{t}
+ &= - q_i n_{i0} \nabla \phi_1 - \gamma_i k_B T_i \nabla n_{i1}
+ \\
+ 0
+ &= - q_e n_{e0} \nabla \phi_1 - \gamma_e k_B T_e \nabla n_{e1}
+\end{aligned}$$
+
+Because we are interested in linear waves,
+we make the following plane-wave ansatz:
+
+$$\begin{aligned}
+ n_{i1}(\vb{r}, t)
+ &= n_{i1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ \\
+ n_{e1}(\vb{r}, t)
+ &= n_{e1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ \\
+ \vb{u}_{i1}(\vb{r}, t)
+ &= \vb{u}_{i1} \exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+ \\
+ \phi_1(\vb{r}, t)
+ &= \phi_1 \,\,\exp\!(i \vb{k} \cdot \vb{r} - i \omega t)
+\end{aligned}$$
+
+Which we then insert into the momentum equations for the ions and electrons:
+
+$$\begin{aligned}
+ - i \omega m_i n_{i0} \vb{u}_{i1}
+ &= - i \vb{k} q_i n_{i0} \phi_1 - i \vb{k} \gamma_i k_B T_i n_{i1}
+ \\
+ 0
+ &= - i \vb{k} q_e n_{e0} \phi_1 - i \vb{k} \gamma_e k_B T_e n_{e1}
+\end{aligned}$$
+
+The electron equation can easily be rearranged
+to get a relation between $n_{e1}$ and $n_{e0}$:
+
+$$\begin{aligned}
+ i \vb{k} \gamma_e k_B T_e n_{e1}
+ = - i \vb{k} q_e n_{e0} \phi_1
+ \quad \implies \quad
+ n_{e1}
+ = - \frac{q_e \phi_1}{\gamma_e k_B T_e} n_{e0}
+\end{aligned}$$
+
+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 $\gamma_e = 1$.
+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 $n_0 = - k_B T_e n_{e1} / (q_e \phi_1)$,
+which we insert into the ion equation to get:
+
+$$\begin{gathered}
+ i \omega m_i \frac{k_B T_e n_{e1}}{q_e \phi_1} \vb{u}_{i1}
+ = - i q_i \frac{k_B T_e n_{e1}}{q_e \phi_1} \phi_1 \vb{k} - i \gamma_i k_B T_i n_{i1} \vb{k}
+ \\
+ \implies \qquad
+ \omega m_i \frac{T_e n_{e1}}{q_e \phi_1} \vb{k} \cdot \vb{u}_{i1}
+ = T_e n_{e1} |\vb{k}|^2 - \gamma_i T_i n_{i1} |\vb{k}|^2
+\end{gathered}$$
+
+Where we have taken the dot product with $\vb{k}$,
+and used that $q_i / q_e = -1$.
+In order to simplify this equation,
+we turn to the two-fluid ion continuity relation:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{(n_{i0} \!+\! n_{i1})}{t} + \nabla \cdot \Big( (n_{i0} \!+\! n_{i1}) (\vb{u}_{i0} \!+\! \vb{u}_{i1}) \Big)
+ \approx \pdv{n_{i1}}{t} + n_{i0} \nabla \cdot \vb{u}_{i1}
+\end{aligned}$$
+
+Then we insert our plane-wave ansatz,
+and substitute $n_{i0} = n_0$ as before, yielding:
+
+$$\begin{aligned}
+ 0
+ = - i \omega n_{i1} + i n_{i0} \vb{k} \cdot \vb{u}_{i1}
+ \quad \implies \quad
+ \vb{k} \cdot \vb{u}_{i1}
+ = \omega \frac{n_{i1}}{n_{i0}}
+ = \omega \frac{q_e n_{i1} \phi_1}{k_B T_e n_{e1}}
+\end{aligned}$$
+
+Substituting this in the ion momentum equation
+leads us to a dispersion relation $\omega(\vb{k})$:
+
+$$\begin{gathered}
+ \omega^2 m_i \frac{T_e n_{e1}}{q_e \phi_1} \frac{q_e n_{i1} \phi_1}{k_B T_e n_{e1}}
+ = \omega^2 m_i \frac{n_{i1}}{k_B}
+ = |\vb{k}|^2 \big( T_e n_{e1} - \gamma_i T_i n_{i1} \big)
+ \\
+ \implies \qquad
+ \omega^2
+ = \frac{|\vb{k}|^2}{m_i} \Big( k_B T_e \frac{n_{e1}}{n_{i1}} - \gamma_i k_B T_i \Big)
+\end{gathered}$$
+
+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,
+according to [Gauss' law](/know/concept/maxwells-equations/).
+Nevertheless, authors often ignore this fact,
+thereby making the so-called **plasma approximation**.
+We will not, and therefore turn to Gauss' law:
+
+$$\begin{aligned}
+ \varepsilon_0 \nabla \cdot \vb{E}
+ = - \varepsilon_0 \nabla^2 \phi_1
+ = q_i n_i - q_e n_e
+ = - q_e (n_{i1} - n_{e1})
+\end{aligned}$$
+
+One final time, we insert our plane-wave ansatz,
+and use our Boltzmann-like relation between $n_{e1}$ and $n_{e0}$
+to substitute $\phi_1 = - k_B T_e n_{e1} / (q_e n_{e0})$:
+
+$$\begin{gathered}
+ q_e (n_{e1} - n_{i1})
+ = |\vb{k}|^2 \varepsilon_0 \phi_1
+ = - |\vb{k}|^2 \varepsilon_0 \frac{k_B T_e n_{e1}}{q_e n_{e0}}
+ \\
+ \implies \qquad
+ n_{i1}
+ = n_{e1} + |\vb{k}|^2 \varepsilon_0 \frac{k_B T_e n_{e1}}{q_e^2 n_{e0}}
+ = n_{e1} \big( 1 + |\vb{k}|^2 \lambda_{De}^2 \big)
+\end{gathered}$$
+
+Where $\lambda_{De}$ is the electron [Debye length](/know/concept/debye-length/).
+We thus reach the following dispersion relation,
+which governs **ion sound waves** or **ion acoustic waves**:
+
+$$\begin{aligned}
+ \boxed{
+ \omega^2
+ = \frac{|\vb{k}|^2}{m_i} \bigg( \frac{k_B T_e}{1 + |\vb{k}|^2 \lambda_{De}^2} + \gamma_i k_B T_i \bigg)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \omega^2
+ = \frac{|\vb{k}|^2}{m_i} \bigg( k_B T_e + \gamma_i k_B T_i \bigg)
+\end{aligned}$$
+
+The phase velocity $v_s$ of these waves,
+i.e. the speed of sound, is then given by:
+
+$$\begin{aligned}
+ \boxed{
+ v_s
+ = \frac{\omega}{k}
+ = \sqrt{\frac{k_B T_e}{m_i} + \frac{\gamma_i k_B T_i}{m_i}}
+ }
+\end{aligned}$$
+
+Curiously, unlike a neutral gas,
+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 $T_e \gg T_i$,
+even though the main feature of these waves
+is that they involve ion density fluctuations $n_{i1}$.
+
+
+
+## References
+1. F.F. Chen,
+ *Introduction to plasma physics and controlled fusion*,
+ 3rd edition, Springer.
+2. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/ito-integral/index.md b/source/know/concept/ito-integral/index.md
new file mode 100644
index 0000000..da3c706
--- /dev/null
+++ b/source/know/concept/ito-integral/index.md
@@ -0,0 +1,268 @@
+---
+title: "Itō integral"
+date: 2021-11-06
+categories:
+- Mathematics
+- Stochastic analysis
+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/) $B_t$,
+which is also a stochastic process.
+The Itō integral $I_t$ of $G_t$ is defined as follows:
+
+$$\begin{aligned}
+ \boxed{
+ I_t
+ \equiv \int_a^b G_t \dd{B_t}
+ \equiv \lim_{h \to 0} \sum_{t = a}^{t = b} G_t \big(B_{t + h} - B_t\big)
+ }
+\end{aligned}$$
+
+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 $B_t$.
+
+
+## Motivation
+
+Consider the following simple first-order differential equation for $X_t$,
+for some function $f$:
+
+$$\begin{aligned}
+ \dv{X_t}{t}
+ = f(X_t)
+\end{aligned}$$
+
+This can be solved numerically using the explicit Euler scheme
+by discretizing it with step size $h$,
+which can be applied recursively, leading to:
+
+$$\begin{aligned}
+ X_{t+h}
+ \approx X_{t} + f(X_t) \: h
+ \quad \implies \quad
+ X_t
+ \approx X_0 + \sum_{s = 0}^{s = t} f(X_s) \: h
+\end{aligned}$$
+
+In the limit $h \to 0$, this leads to the following unsurprising integral for $X_t$:
+
+$$\begin{aligned}
+ \int_0^t f(X_s) \dd{s}
+ = \lim_{h \to 0} \sum_{s = 0}^{s = t} f(X_s) \: h
+\end{aligned}$$
+
+In contrast, consider the *stochastic differential equation* below,
+where $\xi_t$ represents white noise,
+which is informally the $t$-derivative
+of the Wiener process $\xi_t = \idv{B_t}{t}$:
+
+$$\begin{aligned}
+ \dv{X_t}{t}
+ = g(X_t) \: \xi_t
+\end{aligned}$$
+
+Now $X_t$ is not deterministic,
+since $\xi_t$ is derived from a random variable $B_t$.
+If $g = 1$, we expect $X_t = X_0 + B_t$.
+With this in mind, we introduce the **Euler-Maruyama scheme**:
+
+$$\begin{aligned}
+ X_{t+h}
+ &= X_t + g(X_t) \: (\xi_{t+h} - \xi_t) \: h
+ \\
+ &= X_t + g(X_t) \: (B_{t+h} - B_t)
+\end{aligned}$$
+
+We would like to turn this into an integral for $X_t$, as we did above.
+Therefore, we state:
+
+$$\begin{aligned}
+ X_t
+ = X_0 + \int_0^t g(X_s) \dd{B_s}
+\end{aligned}$$
+
+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.
+This is an Itō integral:
+
+$$\begin{aligned}
+ \int_0^t g(X_s) \dd{B_s}
+ \equiv \lim_{h \to 0} \sum_{s = 0}^{s = t} g(X_s) \big(B_{s + h} - B_s\big)
+\end{aligned}$$
+
+For more information about applying the Itō integral in this way,
+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.
+
+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 $v, w \in \mathbb{R}$:
+
+$$\begin{aligned}
+ \int_a^b v G_t + w H_t \dd{B_t}
+ = v\! \int_a^b G_t \dd{B_t} +\: w\! \int_a^b H_t \dd{B_t}
+\end{aligned}$$
+
+By adding multiple summations,
+the Itō integral clearly satisfies, for $a < b < c$:
+
+$$\begin{aligned}
+ \int_a^c G_t \dd{B_t}
+ = \int_a^b G_t \dd{B_t} + \int_b^c G_t \dd{B_t}
+\end{aligned}$$
+
+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 exists by the definition of Itō-integrability):
+
+$$\begin{aligned}
+ \boxed{
+ \mathbf{E} \bigg( \int_a^b G_t \dd{B_t} \bigg)^2
+ = \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We write out the left-hand side of the Itō isometry,
+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 $t \ge s \!+\! h$,
+a given term of this summation can be rewritten
+as follows using the *law of total expectation*
+(see [conditional expectation](/know/concept/conditional-expectation/)):
+
+$$\begin{aligned}
+ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+ = \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big| \mathcal{F}_t \Big] \bigg]
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+ = \mathbf{E} \bigg[ G_t G_s (B_{s + h} \!-\! B_s) \: \mathbf{E} \Big[ (B_{t + h} \!-\! B_t) \Big| \mathcal{F}_t \Big] \bigg]
+\end{aligned}$$
+
+However, $\mathcal{F}_t$ says nothing about
+the increment $(B_{t + h} \!-\! B_t) \sim \mathcal{N}(0, h)$,
+meaning that the conditional expectation is zero:
+
+$$\begin{aligned}
+ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+ = 0
+ \qquad \mathrm{for}\; t \ge s + h
+\end{aligned}$$
+
+By swapping $s$ and $t$, the exact same result can be obtained for $s \ge t \!+\! h$:
+
+$$\begin{aligned}
+ \mathbf{E} \Big[ G_t G_s (B_{t + h} \!-\! B_t) (B_{s + h} \!-\! B_s) \Big]
+ = 0
+ \qquad \mathrm{for}\; s \ge t + h
+\end{aligned}$$
+
+This leaves only one case which can be nonzero: $[t, t\!+\!h] = [s, s\!+\!h]$.
+Applying the law of total expectation again yields:
+
+$$\begin{aligned}
+ \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
+ &= \sum_{t = a}^{t = b} \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big]
+ \\
+ &= \sum_{t = a}^{t = b} \mathbf{E} \bigg[ \mathbf{E} \Big[ G_t^2 (B_{t + h} \!-\! B_t)^2 \Big| \mathcal{F}_t \Big] \bigg]
+\end{aligned}$$
+
+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 $h$:
+
+$$\begin{aligned}
+ \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
+ &= \sum_{t = a}^{t = b} \mathbf{E} \big[ G_t^2 \big] h
+ \longrightarrow
+ \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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 $\mathcal{F}_s$:
+
+$$\begin{aligned}
+ \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+ = \mathbf{E} \bigg[ \int_s^t G_u \dd{B_u} \bigg| \mathcal{F}_s \bigg]
+ = \lim_{h \to 0} \sum_{u = s}^{u = t} \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_s \Big]
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+ &= \lim_{h \to 0} \sum_{u = s}^{u = t}
+ \mathbf{E} \bigg[ \mathbf{E} \Big[ G_u (B_{u + h} \!-\! B_u) \Big| \mathcal{F}_u \Big] \bigg| \mathcal{F}_s \bigg]
+ = 0
+\end{aligned}$$
+
+We now have everything we need to calculate $\mathbf{E} [ I_t | \mathcal{F_s} ]$,
+giving the martingale property:
+
+$$\begin{aligned}
+ \mathbf{E} \big[ I_t | \mathcal{F}_s \big]
+ = \mathbf{E} \big[ I_s | \mathcal{F}_s \big] + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+ = I_s + \mathbf{E} \big[ I_t \!-\! I_s | \mathcal{F}_s \big]
+ = I_s
+\end{aligned}$$
+
+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.
+
+
+
+
Proof
+
+Proof.
+We start by applying the classical chain rule,
+but we go to second order in $x$.
+This is also valid classically,
+but there we would neglect all higher-order infinitesimals:
+
+$$\begin{aligned}
+ \dd{Y_t}
+ = \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{X_t} + \frac{1}{2} \pdvn{2}{h}{x} \dd{X_t}^2
+\end{aligned}$$
+
+But here we cannot neglect $\dd{X_t}^2$.
+We insert the definition of an Itō process:
+
+$$\begin{aligned}
+ \dd{Y_t}
+ &= \pdv{h}{t} \dd{t} + \pdv{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big) + \frac{1}{2} \pdvn{2}{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big)^2
+ \\
+ &= \pdv{h}{t} \dd{t} + \pdv{h}{x} \Big( F_t \dd{t} + G_t \dd{B_t} \Big)
+ + \frac{1}{2} \pdvn{2}{h}{x} \Big( F_t^2 \dd{t}^2 + 2 F_t G_t \dd{t} \dd{B_t} + G_t^2 \dd{B_t}^2 \Big)
+\end{aligned}$$
+
+In the limit of small $\dd{t}$, we can neglect $\dd{t}^2$,
+and as it turns out, $\dd{t} \dd{B_t}$ too:
+
+$$\begin{aligned}
+ \dd{t} \dd{B_t}
+ &= (B_{t + \dd{t}} - B_t) \dd{t}
+ \sim \dd{t} \mathcal{N}(0, \dd{t})
+ \sim \mathcal{N}(0, \dd{t}^3)
+ \longrightarrow 0
+\end{aligned}$$
+
+However, due to the scaling property of $B_t$,
+we cannot ignore $\dd{B_t}^2$, which has order $\dd{t}$:
+
+$$\begin{aligned}
+ \dd{B_t}^2
+ &= (B_{t + \dd{t}} - B_t)^2
+ \sim \big( \mathcal{N}(0, \dd{t}) \big)^2
+ \sim \chi^2_1(\dd{t})
+ \longrightarrow \dd{t}
+\end{aligned}$$
+
+Where $\chi_1^2(\dd{t})$ is the generalized chi-squared distribution
+with one term of variance $\dd{t}$.
+
+
+
+
Proof
+
+Proof.
+If we define $Y_t \equiv X_t^2$,
+then Itō's lemma tells us that the following holds:
+
+$$\begin{aligned}
+ \dd{Y_t}
+ = \big( 2 X_t \: f(X_t) + g^2(X_t) \big) \dd{t} + 2 X_t \: g(X_t) \dd{B_t}
+\end{aligned}$$
+
+Integrating and taking the expectation value
+removes the Wiener term, leaving:
+
+$$\begin{aligned}
+ \mathbf{E}[Y_t]
+ = Y_0 + \mathbf{E}\! \int_0^t 2 X_s f(X_s) + g^2(X_s) \dd{s}
+\end{aligned}$$
+
+Given that $K (1 \!+\! x^2)$ is an upper bound of $x f(x)$ and $g^2(x)$,
+we get an inequality:
+
+$$\begin{aligned}
+ \mathbf{E}[Y_t]
+ &\le Y_0 + \mathbf{E}\! \int_0^t 2 K (1 \!+\! X_s^2) + K (1 \!+\! X_s^2) \dd{s}
+ \\
+ &\le Y_0 + \int_0^t 3 K (1 + \mathbf{E}[Y_s]) \dd{s}
+ \\
+ &\le Y_0 + 3 K t + \int_0^t 3 K \big( \mathbf{E}[Y_s] \big) \dd{s}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \mathbf{E}[Y_t]
+ &\le (Y_0 + 3 K t) \exp\!\bigg( \int_0^t 3 K \dd{s} \bigg)
+ \\
+ &\le (Y_0 + 3 K t) \exp\!\big(3 K t\big)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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:
+
+$$\begin{aligned}
+ \dd{Z_t}
+ = \big( 2 D_t F_t + G_t^2 \big) \dd{t} + 2 D_t G_t \dd{B_t}
+\end{aligned}$$
+
+Integrating and taking the expectation value
+removes the Wiener term, leaving:
+
+$$\begin{aligned}
+ \mathbf{E}[Z_t]
+ = Z_0 + \mathbf{E}\! \int_0^t 2 D_s F_s + G_s^2 \dd{s}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \mathbf{E}[Z_t]
+ &\le Z_0 + \mathbf{E}\! \int_0^t 2 K D_s^2 + K^2 D_s^2 \dd{s}
+ \\
+ &\le Z_0 + \int_0^t (2 K \!+\! K^2) \: \mathbf{E}[Z_s] \dd{s}
+\end{aligned}$$
+
+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.
+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):
+
+$$\begin{aligned}
+ \mathbf{E}[Z_t]
+ &\le Z_0 \exp\!\bigg( \int_0^t 2 K \!+\! K^2 \dd{s} \bigg)
+ \\
+ &\le Z_0 \exp\!\Big( \big( 2 K \!+\! K^2 \big) t \Big)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+The exponential $\exp(- s t)$ is the only thing that depends on $s$ here:
+
+$$\begin{aligned}
+ \dvn{n}{\tilde{f}}{s}
+ &= \dvn{n}{}{s}\int_0^\infty f(t) \exp(- s t) \dd{t}
+ \\
+ &= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t}
+ = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We integrate by parts and use the fact that $\lim_{x \to \infty} \exp(-x) = 0$:
+
+$$\begin{aligned}
+ \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
+ &= \int_0^\infty f^{(n)}(t) \exp(- s t) \dd{t}
+ \\
+ &= \Big[ f^{(n - 1)}(t) \exp(- s t) \Big]_0^\infty + s \int_0^\infty f^{(n-1)}(t) \exp(- s t) \dd{t}
+ \\
+ &= - f^{(n - 1)}(0) + s \Big[ f^{(n - 2)}(t) \exp(- s t) \Big]_0^\infty + s^2 \int_0^\infty f^{(n-2)}(t) \exp(- s t) \dd{t}
+\end{aligned}$$
+
+And so on.
+By partially integrating $n$ times in total we arrive at the conclusion.
+
+
+
+
Proof
+
+Proof.
+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
+$\psi_\vb{k}(\vb{r}) = e^{i \vb{k} \cdot \vb{r}} / \sqrt{V}$:
+
+$$\begin{aligned}
+ \hat{n}(\vb{r})
+ \equiv \hat{\Psi}{}^\dagger(\vb{r}) \hat{\Psi}(\vb{r})
+ = \sum_{\vb{k} \vb{k}'} \psi_{\vb{k}}^*(\vb{r}) \: \psi_{\vb{k}'}(\vb{r})\: \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'}
+ = \frac{1}{V} \sum_{\vb{k} \vb{k}'} e^{i (\vb{k}' - \vb{k}) \cdot \vb{r}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'}
+\end{aligned}$$
+
+Taking the Fourier transfom yields a Dirac delta function $\delta$:
+
+$$\begin{aligned}
+ \hat{n}(\vb{q})
+ = \frac{1}{V} \int_{-\infty}^\infty
+ \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: e^{i (\vb{k}' - \vb{k} - \vb{q})\cdot \vb{r}} \dd{\vb{r}}
+ = \frac{(2 \pi)^D}{V} \sum_{\vb{k} \vb{k}'} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q})
+\end{aligned}$$
+
+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,
+which is what we want for jellium.
+
+We apply this standard trick from condensed matter physics to $\hat{n}$,
+and $V$ cancels out:
+
+$$\begin{aligned}
+ \hat{n}(\vb{q})
+ &= \frac{(2 \pi)^D}{V} \frac{V}{(2 \pi)^D} \sum_{\vb{k}} \int_{-\infty}^\infty
+ \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}'} \: \delta(\vb{k}' \!-\! \vb{k} \!-\! \vb{q}) \dd{\vb{k}'}
+ = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \hat{n}(-\vb{q})
+ = \sum_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} - \vb{q}}
+ = \sum_{\vb{k}'} \hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}
+ = \hat{n}^\dagger(\vb{q})
+\end{aligned}$$
+
+The summation variable $\vb{k}$ has an associated spin $\sigma$,
+and $\hat{n}$ does not carry any spin.
+
+
+
+
Proof
+
+Proof.
+In general, for any single-particle states labeled by $m$, $n$, $o$ and $p$, we have:
+$$\begin{aligned}
+ \comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
+ &= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n
+ \\
+ &= \hat{c}_m^\dagger \big( \acomm{\hat{c}_n}{\hat{c}_o^\dagger} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p
+ - \hat{c}_o^\dagger \big( \acomm{\hat{c}_p}{\hat{c}_m^\dagger} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n
+\end{aligned}$$
+
+Using the standard fermion anticommutation relations, this becomes:
+
+$$\begin{aligned}
+ \comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
+ &= \hat{c}_m^\dagger \big( \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \big) \hat{c}_p
+ - \hat{c}_o^\dagger \big( \delta_{pm} - \hat{c}_m^\dagger \hat{c}_p \big) \hat{c}_n
+ \\
+ &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_m^\dagger \hat{c}_o^\dagger \hat{c}_n \hat{c}_p
+ - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm} + \hat{c}_o^\dagger \hat{c}_m^\dagger \hat{c}_p \hat{c}_n
+ \\
+ &= \hat{c}_m^\dagger \hat{c}_p \: \delta_{no} - \hat{c}_o^\dagger \hat{c}_n \: \delta_{pm}
+\end{aligned}$$
+
+In this case, $m = p = \vb{k}$ and $n = o = \vb{k} \!+\! \vb{q}$,
+so the Kronecker deltas are unnecessary.
+
+
+
+
Proof
+
+Proof.
+First $\tau \!-\! \tau' < 0$.
+We insert the argument $\tau \!-\! \tau' \!+\! \hbar \beta$,
+and use the cyclic property:
+
+$$\begin{aligned}
+ C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
+ &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau' + \hbar \beta) \hat{H} / \hbar}
+ \hat{A} e^{-(\tau - \tau' + \hbar \beta) \hat{H} / \hbar} \hat{B} \Big)
+ \\
+ &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{B} \Big)
+ \\
+ &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar}
+ e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar} \Big)
+ \\
+ &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big)
+\end{aligned}$$
+
+Since $\tau < \tau'$ by assumption,
+we can bring back the time-ordered product $\mathcal{T}$:
+
+$$\begin{aligned}
+ C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
+ &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big)
+ \\
+ &= \pm C_{AB}(\tau \!-\! \tau')
+\end{aligned}$$
+
+Moving on to $\tau \!-\! \tau' > 0$, the proof is perfectly analogous:
+
+$$\begin{aligned}
+ C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta)
+ &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau' - \hbar \beta) \hat{H} / \hbar}
+ \hat{B} e^{(\tau - \tau' - \hbar \beta) \hat{H} / \hbar} \hat{A} \Big)
+ \\
+ &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{A} \Big)
+ \\
+ &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar}
+ e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big)
+ \\
+ &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big)
+ \\
+ &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big)
+ \\
+ &= \pm C_{AB}(\tau \!-\! \tau')
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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:
+
+$$\begin{aligned}
+ C_{AB}(\tau)
+ &= \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty
+ \bigg( \frac{1}{2} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: e^{i k_n \tau'} \dd{\tau'} \bigg) e^{-i k_n \tau}
+ \\
+ &= \frac{1}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau')
+ \bigg( \frac{1}{2} \sum_{n = -\infty}^\infty e^{i k_n (\tau' - \tau)} \bigg) \dd{\tau'}
+ \\
+ &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau')
+ \bigg( \frac{1}{2 \pi} \sum_{n = -\infty}^\infty e^{i \pi n (\tau' - \tau) / \hbar \beta} \bigg) \dd{\tau'}
+\end{aligned}$$
+
+Here, the inner expression turns out to be
+a [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+ \frac{1}{2 \pi} \sum_{n = -\infty}^\infty e^{i n x}
+ = \delta(x)
+\end{aligned}$$
+
+From which the rest of the proof follows straightforwardly:
+
+$$\begin{aligned}
+ C_{AB}(\tau)
+ &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta\big( (\tau' \!-\! \tau) \pi / \hbar \beta \big) \dd{\tau'}
+ \\
+ &= \frac{\pi \hbar \beta}{\pi \hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'}
+ \\
+ &= \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'}
+ \\
+ &= C_{AB}(\tau)
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We split the integral, shift its limits,
+and use the (anti)periodicity of $C_{AB}$:
+
+$$\begin{aligned}
+ C_{AB}(i k_n)
+ &= \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+ + \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+ \\
+ &= \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+ + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau \!-\! \hbar \beta) \: e^{i k_n (\tau - \hbar \beta)} \dd{\tau}
+ \\
+ &= \frac{1}{2} \int_0^{\hbar \beta} \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{-i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau}
+ \\
+ &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+\end{aligned}$$
+
+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 $n$,
+yielding the desired result.
+
+For the other case, we simply shift the first integral's limits instead of the seconds':
+
+$$\begin{aligned}
+ C_{AB}(i k_n)
+ &= \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau \!+\! \hbar \beta) \: e^{i k_n (\tau + \hbar \beta)} \dd{\tau}
+ + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+ \\
+ &= \frac{1}{2} \int_{-\hbar \beta}^0 \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau}
+ \\
+ &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+For $\tau \!-\! \tau' > 0$, we start by expanding
+in the many-particle eigenstates $\Ket{n}$:
+
+$$\begin{aligned}
+ C_{AB}(\tau \!-\! \tau')
+ &= - \frac{1}{\hbar Z} \sum_{n}
+ \Matrixel{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n}
+ \\
+ &= - \frac{1}{\hbar Z} \sum_{n n'} \Matrixel{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n'}
+ \Matrixel{n'}{e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n}
+ \\
+ &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{A}}{n'}
+ \matrixel{n'}{\hat{B}}{n} e^{(E_n - E_{n'})(\tau - \tau') / \hbar}
+\end{aligned}$$
+
+We take the Fourier transform by integrating over $[0, \hbar \beta]$:
+
+$$\begin{aligned}
+ C_{AB}(i \omega_m)
+ &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{A}}{n'}
+ \matrixel{n'}{\hat{B}}{n} \int_0^{\hbar \beta} e^{(E_n - E_{n'}) \tau / \hbar} e^{i \omega_m \tau} \dd{\tau}
+ \\
+ &= - \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}
+ \bigg[ \frac{\hbar e^{(i \hbar \omega_m + E_n - E_{n'}) \tau / \hbar}}{i \hbar \omega_m + E_n - E_{n'}} \bigg]_0^{\hbar \beta}
+ \\
+ &= - \frac{1}{Z} \sum_{n n'} e^{-\beta E_n} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+ \Big( e^{(i \hbar \omega_m + E_n - E_{n'}) \beta} - 1 \Big)
+ \\
+ &= - \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+ \Big( e^{i \hbar \omega_m \beta} e^{-\beta E_{n'}} - e^{-\beta E_n} \Big)
+ \\
+ &= \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+ \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+\end{aligned}$$
+
+Moving on to $\tau \!-\! \tau' < 0$,
+we again expand in the many-particle eigenstates $\Ket{n}$:
+
+$$\begin{aligned}
+ C_{AB}(\tau \!-\! \tau')
+ &= \mp \frac{1}{\hbar Z} \sum_{n}
+ \Matrixel{n}{e^{-\beta \hat{H}} e^{- (\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n}
+ \\
+ &= \mp \frac{1}{\hbar Z} \sum_{n n'} \Matrixel{n}{e^{-\beta \hat{H}} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n'}
+ \Matrixel{n'}{e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n}
+ \\
+ &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{B}}{n'}
+ \matrixel{n'}{\hat{A}}{n} e^{-(E_n - E_{n'})(\tau - \tau') / \hbar}
+\end{aligned}$$
+
+Since $\tau \!-\! \tau' < 0$ this time,
+we take the Fourier transform over $[-\hbar \beta, 0]$:
+
+$$\begin{aligned}
+ C_{AB}(i \omega_m)
+ &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{B}}{n'}
+ \matrixel{n'}{\hat{A}}{n} \int_{-\hbar \beta}^0 e^{-(E_n - E_{n'}) \tau / \hbar} e^{i \omega_m \tau} \dd{\tau}
+ \\
+ &= \mp \frac{1}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}
+ \bigg[ \frac{\hbar e^{(i \hbar \omega_m - E_n + E_{n'}) \tau / \hbar}}{i \hbar \omega_m - E_n + E_{n'}} \bigg]_{-\hbar \beta}^0
+ \\
+ &= \mp \frac{1}{Z} \sum_{n n'} e^{-\beta E_n} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+ \Big( 1 - e^{(-i \hbar \omega_m + E_n - E_{n'}) \beta} \Big)
+ \\
+ &= \mp \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+ \Big( e^{-\beta E_n} - e^{-i \hbar \omega_m \beta} e^{-\beta E_{n'}} \Big)
+ \\
+ &= \mp \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+ \Big( e^{- \beta E_n} \pm e^{-\beta E_{n'}} \Big)
+ \\
+ &= \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{B}}{n'} \matrixel{n'}{\hat{A}}{n}}{i \hbar \omega_m - E_n + E_{n'}}
+ \Big( e^{- \beta E_{n'}} \mp e^{-\beta E_n} \Big)
+\end{aligned}$$
+
+Where swapping $n$ and $n'$ gives the desired result.
+
+
+
+
Proof
+
+Proof.
+We introduce some new terms, and reorganize the expression:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ &= 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ + \gamma_g \rho_{ee} - \gamma_g \rho_{ee}
+ + \gamma_e \rho_{gg} - \gamma_e \rho_{gg}
+ \\
+ &= \gamma_g (\rho_{gg} + \rho_{ee}) - \gamma_e (\rho_{gg} + \rho_{ee})
+ + \gamma_g (\rho_{gg} - \rho_{ee}) + \gamma_e (\rho_{gg} - \rho_{ee})
+\end{aligned}$$
+
+Since the total probability $\rho_{gg} + \rho_{ee} = 1$,
+and $d \equiv \rho_{ee} - \rho_{gg}$, this reduces to:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ &= \gamma_g - \gamma_e - (\gamma_g + \gamma_e) d
+ \\
+ &= (\gamma_g + \gamma_e) \Big( \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e} - d \Big)
+ \\
+ &= \gamma_\parallel ( d_0 - d )
+\end{aligned}$$
+
+
+
+
+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 $\vb{F}_G$
+have equal magnitudes in opposite directions,
+and $B$ is directly above or below $G$,
+or in other words, $x_B = x_G$ and $y_B = y_G$,
+which are calculated as follows:
+
+$$\begin{aligned}
+ (x_G, y_G, z_G)
+ &\equiv \frac{1}{V_{boat}} \iiint_{boat} (x, y, z) \dd{V}
+ \\
+ (x_B, y_B, z_B)
+ &\equiv \frac{1}{V_{disp}} \iiint_{disp} (x, y, z) \dd{V}
+\end{aligned}$$
+
+Where $V_{boat}$ is the volume of the whole boat,
+and $V_{disp}$ is the volume of liquid it displaces.
+
+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,
+gets shifted to a new plane, namely:
+
+$$\begin{aligned}
+ z
+ = \sin(\theta) \: y
+ \approx \theta y
+\end{aligned}$$
+
+Then $V_{disp}$ changes by $\Delta V_{disp}$, which is estimated below.
+If a point of the old waterline is raised by $z$,
+then the displaced liquid underneath it is reduced proportionally,
+hence the sign:
+
+$$\begin{aligned}
+ \Delta V_{disp}
+ \approx - \iint_{wl} z \dd{A}
+ \approx - \theta \iint_{wl} y \dd{A}
+ = 0
+\end{aligned}$$
+
+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 $\theta$:
+
+$$\begin{aligned}
+ y_B'
+ = y_B - \theta z_B + \Delta y_B
+\end{aligned}$$
+
+We find $\Delta y_B$ by calculating the virtual buoyancy center of the shape difference:
+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:
+
+$$\begin{aligned}
+ \Delta y_B
+ = - \frac{1}{V_{disp}} \iint_{wl} y z \dd{A}
+ \approx - \frac{\theta}{V_{disp}} \iint_{wl} y^2 \dd{A}
+ = - \frac{\theta I}{V_{disp}}
+\end{aligned}$$
+
+Where we have defined the so-called **area moment** $I$ of the waterline as follows:
+
+$$\begin{aligned}
+ \boxed{
+ I
+ \equiv \iint_{wl} y^2 \dd{A}
+ }
+\end{aligned}$$
+
+Now that we have an expression for $\Delta y_B$,
+the new center's position $y_B'$ is found to be:
+
+$$\begin{aligned}
+ y_B'
+ = y_B - \theta \Big( z_B + \frac{I}{V_{disp}} \Big)
+ \approx y_B - \sin(\theta) \: \Big( z_B + \frac{I}{V_{disp}} \Big)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ z_M
+ \equiv z_B + \frac{I}{V_{disp}}
+ }
+\end{aligned}$$
+
+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 $M$.
+
+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:
+
+$$\begin{aligned}
+ \vb{T}
+ = (\vb{r}_G' - \vb{r}_B') \cross m \vb{g}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ T_x
+ = - (y_G' - y_B') m \mathrm{g}
+ = - \big((y_G - \theta z_G) - (y_B - \theta z_M)\big) m \mathrm{g}
+\end{aligned}$$
+
+With $y_G' = y_G - \theta z_G$ being a simple rotation of $G$.
+At the initial equilibrium $y_G = y_B$, so:
+
+$$\begin{aligned}
+ T_x
+ = \theta (z_G - z_M) m \mathrm{g}
+\end{aligned}$$
+
+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,
+and the boat returns to the original equilibrium,
+leading us to the following stability condition:
+
+$$\begin{aligned}
+ \boxed{
+ z_M > z_G
+ }
+\end{aligned}$$
+
+In other words, for a given boat design (or general shape)
+$z_G$ 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).
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
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+---
+title: "Microcanonical ensemble"
+date: 2021-07-09
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+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 $N$.
+
+Consider a box with those properties.
+We now put an imaginary rigid wall inside the box,
+thus dividing it into two subsystems $A$ and $B$,
+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 $U = U_A \!+\! U_B$.
+
+The particles in each subsystem are in a certain **microstate** (configuration).
+For a given $U$, there is a certain number $c$
+of possible whole-box microstates with that energy, given by:
+
+$$\begin{aligned}
+ c(U)
+ = \sum_{U_A \le U} c_A(U_A) \: c_B(U - U_A)
+\end{aligned}$$
+
+Where $c_A$ and $c_B$ are the numbers of subsystem microstates
+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:
+
+$$\begin{aligned}
+ p_A(U_A)
+ = \frac{c_A(U_A) \: c_B(U - U_A)}{c(U)}
+\end{aligned}$$
+
+If a certain $U_A$ has a higher probability,
+then there are more $A$-microstates with that energy,
+so, statistically, for an *ensemble* of many boxes,
+we expect that $U_A$ is more common.
+
+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 $p_B$.
+
+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 $\ln{c_A}$,
+we recognize the definition of
+the entropy $S_A \equiv k \ln{c_A}$,
+where $k$ is Boltzmann's constant.
+We thus multiply by $k$:
+
+$$\begin{aligned}
+ k \ln p_A(U_A)
+ = S_A(U_A) + S_B(U - U_A) - k \ln{c(U)}
+\end{aligned}$$
+
+Since entropy is additive over subsystems,
+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/)
+that corresponds to the microcanonical ensemble.
+
+For our example, maximizing gives the following,
+more concrete, equilibrium condition:
+
+$$\begin{aligned}
+ 0
+ = k \dv{(\ln{p_A})}{U_A}
+ = \pdv{S_A}{U_A} + \pdv{S_B}{U_A}
+ = \pdv{S_A}{U_A} - \pdv{S_B}{U_B}
+\end{aligned}$$
+
+By definition, the energy-derivative of the entropy
+is the reciprocal temperature $1 / T$.
+In other words,
+equilibrium is reached when both subsystems
+are at the same temperature:
+
+$$\begin{aligned}
+ \frac{1}{T_A}
+ = \pdv{S_A}{U_A}
+ = \pdv{S_B}{U_B}
+ = \frac{1}{T_B}
+\end{aligned}$$
+
+Recall that our partitioning into $A$ and $B$ was arbitrary,
+meaning that, in fact, the temperature $T$ must be uniform in the whole box.
+We get this specific result because
+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 $S$.
+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 $N_A$.
+
+
+
+## References
+1. H. Gould, J. Tobochnik,
+ *Statistical and thermal physics*, 2nd edition,
+ Princeton.
diff --git a/source/know/concept/modulational-instability/index.md b/source/know/concept/modulational-instability/index.md
new file mode 100644
index 0000000..255a385
--- /dev/null
+++ b/source/know/concept/modulational-instability/index.md
@@ -0,0 +1,202 @@
+---
+title: "Modulational instability"
+date: 2021-02-26
+categories:
+- Physics
+- Fiber optics
+- Optics
+- Perturbation
+- Nonlinear optics
+layout: "concept"
+---
+
+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.
+
+Consider the following simple solution to the nonlinear Schrödinger equation:
+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,
+representing background noise:
+
+$$\begin{aligned}
+ A(z,t) = \big(\sqrt{P_0} + \varepsilon(z,t)\big) \exp( i \gamma P_0 z)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ 0
+ &= - P_0 \sqrt{P_0} \gamma - P_0 \gamma \varepsilon + i \pdv{\varepsilon}{z}
+ - \frac{\beta_2}{2} \pdvn{2}{\varepsilon}{t}
+ + \gamma \big(\sqrt{P_0} + \varepsilon\big)^2 \big(\sqrt{P_0} + \varepsilon\big)^*
+ \\
+ &= i \pdv{\varepsilon}{z}
+ - \frac{\beta_2}{2} \pdvn{2}{\varepsilon}{t}
+ + \gamma \big( P_0 (\varepsilon + \varepsilon^*) + \sqrt{P_0} |\varepsilon|^2
+ + \sqrt{P_0} \varepsilon (\varepsilon + \varepsilon^*) + \varepsilon |\varepsilon|^2 \big)
+ \\
+ &= i \pdv{\varepsilon}{z} - \frac{\beta_2}{2} \pdvn{2}{\varepsilon}{t} + \gamma P_0 (\varepsilon + \varepsilon^*)
+\end{aligned}$$
+
+We split the perturbation into real and imaginary parts
+$\varepsilon(z,t) = \varepsilon_r(z,t) + i \varepsilon_i(z,t)$,
+which we fill in in this equation.
+The point is that $\varepsilon_r$ and $\varepsilon_i$ are real functions:
+
+$$\begin{aligned}
+ 0
+ &= i \pdv{\varepsilon_r}{z} - \pdv{\varepsilon_i}{z}
+ - \frac{\beta_2}{2} \pdvn{2}{\varepsilon_r}{t} - i \frac{\beta_2}{2} \pdvn{2}{\varepsilon_i}{t}
+ + 2 \gamma P_0 \varepsilon_r
+\end{aligned}$$
+
+Splitting this into its real and imaginary parts gives two PDEs
+relating $\varepsilon_r$ and $\varepsilon_i$:
+
+$$\begin{aligned}
+ \pdv{\varepsilon_r}{z} = \frac{\beta_2}{2} \pdvn{2}{\varepsilon_i}{t}
+ \qquad \quad
+ \pdv{\varepsilon_i}{z} = - \frac{\beta_2}{2} \pdvn{2}{\varepsilon_r}{t} + 2 \gamma P_0 \varepsilon_r
+\end{aligned}$$
+
+We [Fourier transform](/know/concept/fourier-transform/)
+these in $t$ to turn them into ODEs relating
+$\tilde{\varepsilon}_r(z,\omega)$ and $\tilde{\varepsilon}_i(z,\omega)$:
+
+$$\begin{aligned}
+ \pdv{\tilde{\varepsilon}_r}{z} = - \frac{\beta_2}{2} \omega^2 \tilde{\varepsilon}_i
+ \qquad \quad
+ \pdv{\tilde{\varepsilon}_i}{z} = \Big(\frac{\beta_2}{2} \omega^2 + 2 \gamma P_0 \Big) \tilde{\varepsilon}_r
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \tilde{\varepsilon}_r(z, \omega) = \tilde{\varepsilon}_r(0, \omega) \exp(k z)
+ \qquad \quad
+ \tilde{\varepsilon}_i(z, \omega) = \tilde{\varepsilon}_i(0, \omega) \exp(k z)
+\end{aligned}$$
+
+With this, we can write the system of ODEs for
+$\tilde{\varepsilon}_r(z,\omega)$ and $\tilde{\varepsilon}_i(z,\omega)$
+in matrix form:
+
+$$\begin{aligned}
+ \begin{bmatrix}
+ k & \beta_2 \omega^2 / 2 \\
+ \beta_2 \omega^2 / 2 \!+\! 2 \gamma P_0 & - k
+ \end{bmatrix}
+ \cdot
+ \begin{bmatrix} \tilde{\varepsilon}_r(0, \omega) \\ \tilde{\varepsilon}_i(0, \omega) \end{bmatrix}
+ =
+ \begin{bmatrix} 0 \\ 0 \end{bmatrix}
+\end{aligned}$$
+
+This has non-zero solutions if the system matrix' determinant is zero,
+which is true when:
+
+$$\begin{aligned}
+ k = \pm \sqrt{ - \frac{\beta_2}{2} \omega^2 \Big( \frac{\beta_2}{2} \omega^2 + 2 \gamma P_0 \Big) }
+\end{aligned}$$
+
+To get exponential growth, it is essential that $\mathrm{Re}\{k\} > 0$,
+so we discard the negative sign,
+and get the following condition for MI:
+
+$$\begin{aligned}
+ - \frac{\beta_2}{2} \omega^2 \Big( \frac{\beta_2}{2} \omega^2 + 2 \gamma P_0 \Big) > 0
+ \quad \implies \quad
+ \boxed{
+ \omega^2 < -\frac{4 \gamma P_0}{\beta_2}
+ }
+\end{aligned}$$
+
+Since $\omega^2$ is positive, $\beta_2$ must be negative,
+so MI can only occur in the ADR.
+It is worth noting that $\beta_2 = \beta_2(\omega_0)$,
+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 $|\varepsilon|^2 \ll P_0$.
+Over time, the noise gets amplified so greatly
+that this approximation breaks down.
+
+Next, we define the **gain** $g(\omega)$,
+which expresses how quickly the
+perturbation grows as a function of the frequency offset $\omega$:
+
+$$\begin{aligned}
+ \boxed{
+ g(\omega)
+ = \mathrm{Re}\{k\}
+ = \mathrm{Re} \bigg\{ \sqrt{ - \frac{\beta_2}{2} \omega^2 \Big( \frac{\beta_2}{2} \omega^2 + 2 \gamma P_0 \Big) } \bigg\}
+ }
+\end{aligned}$$
+
+The frequencies with maximum gain are then found as extrema of $g(\omega)$,
+which satisfy:
+
+$$\begin{aligned}
+ g'(\omega_\mathrm{max}) = 0
+ \qquad \implies \qquad
+ \boxed{
+ \omega_\mathrm{max} = \pm \sqrt{\frac{2 \gamma P_0}{-\beta_2}}
+ }
+\end{aligned}$$
+
+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 $\gamma = 0.1/\mathrm{W}/\mathrm{m}$,
+whose peak is approximately flat, so our derivation is valid there,
+hence it "wrinkles" in the $t$-domain:
+
+$$\begin{aligned}
+ A(0, t)
+ = \sqrt{P_0} \sech\!\Big(\frac{t}{T_0}\Big)
+\end{aligned}$$
+
+
+
+
+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),
+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 $z > 30 L_\mathrm{NL}$.
+
+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 $\tilde{g}_R(\omega)$,
+even when the parent pulse is in the NDR.
+This is an example of stimulated Raman scattering (SRS).
+
+
+
+## References
+1. O. Bang,
+ *Numerical methods in photonics: lecture notes*, 2019,
+ unpublished.
+2. O. Bang,
+ *Nonlinear mathematical physics: lecture notes*, 2020,
+ unpublished.
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diff --git a/source/know/concept/multi-photon-absorption/index.md b/source/know/concept/multi-photon-absorption/index.md
new file mode 100644
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@@ -0,0 +1,353 @@
+---
+title: "Multi-photon absorption"
+date: 2022-01-30
+categories:
+- Physics
+- Optics
+- Quantum mechanics
+- Nonlinear optics
+- Perturbation
+layout: "concept"
+---
+
+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
+[electric dipole approximation](/know/concept/electric-dipole-approximation/) by:
+
+$$\begin{aligned}
+ \hat{H}_1(t)
+ = -\vu{p} \cdot \vb{E} \cos(\omega t)
+ \approx -\vu{p} \cdot \vb{E} e^{-i \omega t}
+\end{aligned}$$
+
+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.
+Here, we have made the
+[rotating wave approximation](/know/concept/rotating-wave-approximation/)
+to neglect the $e^{i \omega t}$ term,
+because it turns out to be irrelevant in this discussion.
+
+
+We call the ground state $\Ket{0}$,
+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 $\Ket{u'}$:
+
+$$\begin{aligned}
+ \matrixel{e}{\hat{H}_1}{e'} = \matrixel{u}{\hat{H}_1}{u'} = 0
+ \qquad \quad
+ \matrixel{e}{\hat{H}_1}{u} \neq 0
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ c_e^{(N)}(t)
+ &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(N-1)}(\tau) \: e^{i \omega_{eu} \tau} \dd{\tau}
+ \\
+ c_u^{(N)}(t)
+ &= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(N-1)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau}
+\end{aligned}$$
+
+Where $\omega_{eu} = (E_e \!-\! E_u) / \hbar$.
+For simplicity, the electron starts in the lowest-energy state $\Ket{0}$:
+
+$$\begin{aligned}
+ c_0^{(0)} = 1
+ \qquad \qquad
+ c_u^{(0)} = c_{e \neq 0}^{(0)} = 0
+\end{aligned}$$
+
+Finally, we prove the following useful relation for large $t$,
+involving a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
+
+$$\begin{aligned}
+ \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
+ = 2 \pi \: \delta(x) \: t
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+First, observe that we can rewrite the fraction using an integral:
+
+$$\begin{aligned}
+ \frac{e^{i x t} - 1}{x}
+ = e^{i x t / 2} \frac{e^{i x t / 2} - e^{-i x t / 2}}{x}
+ = i e^{i x t / 2} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
+\end{aligned}$$
+
+By taking the limit $t \to \infty$,
+it can be turned into a nascent Dirac delta function:
+
+$$\begin{aligned}
+ \lim_{t \to \infty} \frac{e^{i x t} - 1}{x}
+ = \lim_{t \to \infty} i e^{i x t / 2} \frac{2 \pi}{2 \pi} \int_{-\infty}^{\infty} e^{i x \tau} \dd{\tau}
+ = \lim_{t \to \infty} i 2 \pi e^{i x t / 2} \: \delta(x)
+\end{aligned}$$
+
+Consequently, the absolute value squared is as follows:
+
+$$\begin{aligned}
+ \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
+ = 4 \pi^2 \delta^2(x)
+\end{aligned}$$
+
+However, a squared delta function $\delta^2$ is not ideal,
+so we take a step back:
+
+$$\begin{aligned}
+ \delta^2(x)
+ = \delta(x) \lim_{t \to \infty} \frac{1}{2 \pi} \int_{-t/2}^{t/2} e^{i x \tau} \dd{\tau}
+ = \delta(x) \lim_{t \to \infty} \frac{t}{2 \pi}
+\end{aligned}$$
+
+Where we have set $x = 0$ according to the first delta function.
+This gives the target:
+
+$$\begin{aligned}
+ \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
+ = 4 \pi^2 \delta^2(x)
+ = 2 \pi \: \delta(x) \: t
+\end{aligned}$$
+
+
+
+
+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
+(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.
+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.
+Dispersive broadening then continues normally:
+
+
+
+
+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 $\omega_i$,
+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
+the instantaneous frequencies for these separate effects:
+
+$$\begin{aligned}
+ \omega_i(z,t)
+ &\approx \omega_\mathrm{GVD}(z,t) + \omega_\mathrm{SPM}(z,t)
+ = \frac{tz}{T_0^2} \bigg( \frac{\beta_2 / T_0^2}{1 + \beta_2^2 z^2 / T_0^4}
+ + 2\gamma P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
+\end{aligned}$$
+
+Assuming that $z$ is small enough such that $z^2 \approx 0$, this
+expression can be reduced to:
+
+$$\begin{aligned}
+ \omega_i(z,t)
+ \approx \frac{\beta_2 tz}{T_0^4} \bigg( 1 + 2\frac{\gamma P_0 T_0^2}{\beta_2} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
+ = \frac{\beta_2 t z}{T_0^4} \bigg( 1 + 2 N_\mathrm{sol}^2 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
+\end{aligned}$$
+
+Where we have assumed $\beta_2 > 0$,
+and $N_\mathrm{sol}$ is the **soliton number**,
+which is defined as:
+
+$$\begin{aligned}
+ N_\mathrm{sol}^2 \equiv \frac{L_D}{L_N} = \frac{\gamma P_0 T_0^2}{|\beta_2|}
+\end{aligned}$$
+
+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 $t^2$:
+
+$$\begin{aligned}
+ 0
+ = \xi_i(z,t)
+ = \pdv{\omega_i}{t}
+ &= \frac{\beta_2 z}{T_0^4} \bigg( 1 + 2 N_\mathrm{sol}^2 \Big( 1 - \frac{2 t^2}{T_0^2} \Big) \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
+ = \frac{\beta_2 z}{T_0^4} \: f\Big(\frac{t^2}{T_0^2}\Big)
+\end{aligned}$$
+
+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 $N_\mathrm{sol}^2 > N_\mathrm{min}^2$:
+
+$$\begin{aligned}
+ \boxed{
+ N_\mathrm{min}^2 = \frac{1}{4} \exp\!\Big(\frac{3}{2}\Big) \approx 1.12
+ }
+\end{aligned}$$
+
+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.
+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 $\omega_i$
+will cause them to displace relative to each other
+after a short distance $z$ by some amount $\Delta t$,
+estimated by:
+
+$$\begin{aligned}
+ \Delta t
+ &\approx z \Delta\beta_1
+ \qquad
+ &&\Delta\beta_1 \equiv \beta_1(\omega_i(z,t_2)) - \beta_1(\omega_i(z,t_1))
+ \\
+ &\approx z \beta_2 \Delta\omega_i
+ \qquad
+ &&\Delta\omega_i \equiv \omega_i(z,t_2) - \omega_i(z,t_1)
+ \\
+ &\approx z \beta_2 \Delta\xi_i \,(t_2 - t_1)
+ \qquad \quad
+ &&\Delta\xi_i \equiv \xi_i(z,t_2) - \xi_i(z,t_1)
+\end{aligned}$$
+
+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, $z = L_\mathrm{WB}$,
+must therefore satisfy the following condition
+for a particular value of $t$:
+
+$$\begin{aligned}
+ L_\mathrm{WB} \, \beta_2 \, \xi_i(L_\mathrm{WB}, t) = -1
+ \qquad \implies \qquad
+ L_\mathrm{WB}^2 = - \frac{T_0^4}{\beta_2^2 \, f(t^2/T_0^2)}
+\end{aligned}$$
+
+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 $f(3/2)$:
+
+$$\begin{aligned}
+ f_\mathrm{min} = f(3/2)
+ = 1 - 4 N_\mathrm{sol}^2 \exp(-3/2)
+ = 1 - N_\mathrm{sol}^2 / N_\mathrm{min}^2
+\end{aligned}$$
+
+Clearly, $f_\mathrm{min} \ge 0$ when $N_\mathrm{sol}^2 \le N_\mathrm{min}^2$,
+which, when inserted above, leads to an imaginary $L_\mathrm{WB}$,
+confirming that OWB cannot occur in that case.
+Otherwise, if $N_\mathrm{sol}^2 > N_\mathrm{min}^2$, then:
+
+$$\begin{aligned}
+ \boxed{
+ L_\mathrm{WB}
+ = \frac{T_0^2}{\beta_2 \, \sqrt{- f_\mathrm{min}}}
+ = \frac{L_D}{\sqrt{N_\mathrm{sol}^2 / N_\mathrm{min}^2 - 1}}
+ }
+\end{aligned}$$
+
+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,
+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:
+
+$$\begin{aligned}
+ \omega_{\mathrm{SPM}}(L_\mathrm{WB},t)
+ = \frac{2 \gamma P_0 t}{\beta_2 \sqrt{4 N_\mathrm{sol}^2 \exp(-3/2) - 1}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
+\end{aligned}$$
+
+Assuming that $N_\mathrm{sol}^2$ is large in the denominator, this can
+be approximately reduced to:
+
+$$\begin{aligned}
+ \omega_\mathrm{SPM}(L_\mathrm{WB}, t)
+ \approx \frac{2 \gamma P_0 t}{\beta_2 N_\mathrm{sol}} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
+ = 2 \sqrt{\frac{\gamma P_0}{\beta_2}} \frac{t}{T_0} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \omega_\mathrm{max} = \sqrt{\frac{2 \gamma P_0}{e \beta_2}}
+\end{aligned}$$
+
+Interestingly, this expression does not contain $T_0$ at all,
+so the achieved spectrum when SPM is halted by OWB
+is independent of the pulse width,
+for sufficiently large $N_\mathrm{sol}$.
+
+
+## References
+1. D. Anderson, M. Desaix, M. Lisak, M.L. Quiroga-Teixeiro,
+ [Wave breaking in nonlinear-optical fibers](https://doi.org/10.1364/JOSAB.9.001358),
+ 1992, Optical Society of America.
+2. A.M. Heidt, A. Hartung, H. Bartelt,
+ [Generation of ultrashort and coherent supercontinuum light pulses in all-normal dispersion fibers](https://doi.org/10.1007/978-1-4939-3326-6_6),
+ 2016, Springer Media.
+
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+---
+title: "Parseval's theorem"
+date: 2021-02-22
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**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 $\tilde{g}(k)$.
+There are two equivalent ways of stating it,
+where $A$, $B$, and $s$ are constants from the FT's definition:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \Inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+ \\
+ \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)}
+ \end{aligned}
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We insert the inverse FT into the defintion of the inner product:
+
+$$\begin{aligned}
+ \Inprod{f}{g}
+ &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
+ \\
+ &= B^2 \int
+ \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
+ \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
+ \dd{x}
+ \\
+ &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
+ \\
+ &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
+ \\
+ &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
+ = \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}}{\tilde{g}}
+\end{aligned}$$
+
+Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Note that we can equally well do this proof in the opposite direction,
+which yields an equivalent result:
+
+$$\begin{aligned}
+ \inprod{\tilde{f}}{\tilde{g}}
+ &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
+ \\
+ &= A^2 \int
+ \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
+ \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
+ \dd{k}
+ \\
+ &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
+ \\
+ &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
+ \\
+ &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
+ = \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
+\end{aligned}$$
+
+
+
+
+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:
+
+
+
+
+For each of the $n$ qubits, $\mathcal{O}(n)$ gates are applied,
+so overall the QFT algorithm is $\mathcal{O}(n^2)$.
+
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
+
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+---
+title: "Quantum gate"
+date: 2021-03-29
+categories:
+- Quantum information
+layout: "concept"
+---
+
+In quantum computing, **quantum gates** are the equivalent
+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.
+
+
+## One-qubit gates
+
+As an example, consider the following must general single-qubit state $\Ket{\psi}$:
+
+$$\begin{aligned}
+ \Ket{\psi}
+ = \alpha \Ket{0} + \beta \Ket{1}
+ = \begin{bmatrix} \alpha \\ \beta \end{bmatrix}
+\end{aligned}$$
+
+Arguably the most famous and/or most fundamental quantum gates are the **Pauli matrices**:
+
+$$\begin{aligned}
+ \boxed{
+ X =
+ \begin{bmatrix}
+ 0 & 1 \\
+ 1 & 0
+ \end{bmatrix}
+ }
+ \qquad
+ \boxed{
+ Y =
+ \begin{bmatrix}
+ 0 & -i \\
+ i & 0
+ \end{bmatrix}
+ }
+ \qquad
+ \boxed{
+ Z =
+ \begin{bmatrix}
+ 1 & 0 \\
+ 0 & -1
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+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**:
+
+$$\begin{aligned}
+ X \Ket{\psi}
+ = \begin{bmatrix} \beta \\ \alpha \end{bmatrix}
+ \qquad
+ Y \Ket{\psi}
+ = \begin{bmatrix} -i \beta \\ i \alpha \end{bmatrix}
+ \qquad
+ Z \Ket{\psi}
+ = \begin{bmatrix} \alpha \\ -\beta \end{bmatrix}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ R_\phi =
+ \begin{bmatrix}
+ 1 & 0 \\
+ 0 & e^{i \phi}
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+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 $T$:
+
+$$\begin{aligned}
+ \boxed{
+ S = R_{\pi/2} =
+ \begin{bmatrix}
+ 1 & 0 \\
+ 0 & i
+ \end{bmatrix}
+ }
+ \qquad \quad
+ \boxed{
+ T = R_{\pi/4} =
+ \frac{1}{\sqrt{2}}
+ \begin{bmatrix}
+ \sqrt{2} & 0 \\
+ 0 & 1 + i
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+Finally, we have the **Hadamard gate** $H$,
+which is defined as follows:
+
+$$\begin{aligned}
+ \boxed{
+ H = \frac{1}{\sqrt{2}}
+ \begin{bmatrix}
+ 1 & 1 \\
+ 1 & -1
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+Its action consists of rotating the qubit
+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,
+and is its own inverse (i.e. unitary):
+
+$$\begin{aligned}
+ H \Ket{0} = \Ket{+}
+ \qquad
+ H \Ket{1} = \Ket{-}
+ \qquad
+ H \Ket{+} = \Ket{0}
+ \qquad
+ H \Ket{-} = \Ket{1}
+\end{aligned}$$
+
+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.
+
+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 $\varepsilon$.
+This is the definition of universality:
+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 $\Ket{\psi_2}$:
+
+$$\begin{aligned}
+ \Ket{\psi_1}
+ = \alpha_1 \Ket{0} + \beta_1 \Ket{1}
+ = \begin{bmatrix} \alpha_1 \\ \beta_1 \end{bmatrix}
+ \qquad \quad
+ \Ket{\psi_2}
+ = \alpha_2 \Ket{0} + \beta_2 \Ket{1}
+ = \begin{bmatrix} \alpha_2 \\ \beta_2 \end{bmatrix}
+\end{aligned}$$
+
+The composite state of both qubits, assuming they are pure,
+is then their tensor product $\otimes$:
+
+$$\begin{aligned}
+ \Ket{\psi_1 \psi_2}
+ = \Ket{\psi_1} \otimes \Ket{\psi_2}
+ &= \alpha_1 \alpha_2 \Ket{00} + \alpha_1 \beta_2 \Ket{01} + \beta_1 \alpha_2 \Ket{10} + \beta_1 \beta_2 \Ket{11}
+ \\
+ &= c_{00} \Ket{00} + c_{01} \Ket{01} + c_{10} \Ket{10} + c_{11} \Ket{11}
+\end{aligned}$$
+
+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 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 $\Ket{1}_2$.
+
+With that said, the first two-qubit gate is $\mathrm{SWAP}$,
+which simply swaps $\Ket{\psi_1}$ and $\Ket{\psi_2}$:
+
+
+
+
+$$\begin{aligned}
+ \boxed{
+ \mathrm{SWAP} =
+ \begin{bmatrix}
+ 1 & 0 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 0 & 1
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+This matrix is given in the basis of $\Ket{00}$, $\Ket{01}$, $\Ket{10}$ and $\Ket{11}$.
+Note that $\mathrm{SWAP}$ cannot generate entanglement,
+so if its input is separable, its output is too.
+In any case, its effect is clear:
+
+$$\begin{aligned}
+ \mathrm{SWAP} \Ket{\psi_1 \psi_2}
+ &= c_{00} \Ket{00} + c_{10} \Ket{01} + c_{01} \Ket{10} + c_{11} \Ket{11}
+\end{aligned}$$
+
+Next, there is the **controlled NOT gate** $\mathrm{CNOT}$,
+which "flips" (applies $X$ to) $\Ket{\psi_2}$ if $\Ket{\psi_1}$ is true:
+
+
+
+
+$$\begin{aligned}
+ \boxed{
+ \mathrm{CNOT} =
+ \begin{bmatrix}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 0 & 1 \\
+ 0 & 0 & 1 & 0
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+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:
+
+
+
+
+$$\begin{aligned}
+ \boxed{
+ \mathrm{CU} =
+ \begin{bmatrix}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & u_{00} & u_{01} \\
+ 0 & 0 & u_{10} & u_{11}
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+Where the lower-right 2x2 block is simply $U$.
+The general action of this gate is given by:
+
+$$\begin{aligned}
+ \mathrm{CU} \Ket{\psi_1 \psi_2}
+ &= c_{00} \Ket{00} + c_{01} \Ket{01} + (c_{10} u_{00} + c_{11} u_{01}) \Ket{10} + (c_{10} u_{10} + c_{11} u_{11}) \Ket{11}
+\end{aligned}$$
+
+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 $\{\mathrm{CNOT}, T, S\}$,
+and there exist many others.
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
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+---
+title: "Quantum teleportation"
+date: 2021-03-07
+categories:
+- Quantum information
+layout: "concept"
+---
+
+**Quantum teleportation** is a method to transfer quantum information
+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 $\beta$;
+she just wants Bob to get the same qubit:
+
+$$\begin{aligned}
+ \Ket{q}
+ = \alpha \Ket{0}_{A'} + \beta \Ket{1}_{A'}
+\end{aligned}$$
+
+She can only directly communicate with Bob over a classical channel.
+This is not enough: even if Alice did know $\alpha$ and $\beta$ exactly
+(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)
+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:
+
+$$\begin{aligned}
+ \Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB}
+ &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \Big( \Ket{00} + \Ket{11} \Big)_{AB}
+ \\
+ &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{000} + \beta \Ket{100}
+ + \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB}
+\end{aligned}$$
+
+Now, observe that we can write any combination of $\Ket{0}$ and $\Ket{1}$
+in the Bell basis like so:
+
+$$\begin{aligned}
+ \Ket{00}
+ &= \frac{\Ket{\Phi^{+}} + \Ket{\Phi^{-}}}{\sqrt{2}}
+ \qquad \quad
+ \Ket{11}
+ = \frac{\Ket{\Phi^{+}} - \Ket{\Phi^{-}}}{\sqrt{2}}
+ \\
+ \Ket{01}
+ &= \frac{\Ket{\Psi^{+}} + \Ket{\Psi^{-}}}{\sqrt{2}}
+ \qquad \quad
+ \Ket{10}
+ = \frac{\Ket{\Psi^{+}} - \Ket{\Psi^{-}}}{\sqrt{2}}
+\end{aligned}$$
+
+Using this, we can rewrite our previous result in terms of the Bell states as follows:
+
+$$\begin{aligned}
+ \Ket{q}_{A'} \ket{\Phi^+}_{AB}
+ &= \frac{\alpha}{2} \Big( \ket{\Phi^{+}} + \ket{\Phi^{-}} \Big)_{A'A} \Ket{0}_B
+ + \frac{\beta}{2} \Big( \ket{\Psi^{+}} - \ket{\Psi^{-}} \Big)_{A'A} \Ket{0}_B
+ \\
+ &+ \frac{\alpha}{2} \Big( \ket{\Psi^{+}} + \ket{\Psi^{-}} \Big)_{A'A} \Ket{1}_B
+ + \frac{\beta}{2} \Big( \ket{\Phi^{+}} - \ket{\Phi^{-}} \Big)_{A'A} \Ket{1}_B
+\end{aligned}$$
+
+If we group all terms according to the Bell states,
+we end up with an interesting expression:
+
+$$\begin{aligned}
+ \Ket{q}_{A'} \ket{\Phi^+}_{AB}
+ = \frac{1}{2} \bigg( &\ket{\Phi^{+}}_{A'A} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{B}
+ + \ket{\Phi^{-}}_{A'A} \Big( \alpha \Ket{0} - \beta \Ket{1} \Big)_{B}
+ \\
+ + &\ket{\Psi^{+}}_{A'A} \Big( \alpha \Ket{1} + \beta \Ket{0} \Big)_{B}
+ + \ket{\Psi^{-}}_{A'A} \Big( \alpha \Ket{1} - \beta \Ket{0} \Big)_{B} \bigg)
+\end{aligned}$$
+
+Thus, purely due to entanglement,
+Bob's qubit $B$ is in a superposition of the following states:
+
+$$\begin{aligned}
+ \Ket{q}
+ &= \alpha \Ket{0} + \beta \Ket{1}
+ \qquad \quad
+ \quad \hat{\sigma}_z \Ket{q}
+ = \alpha \Ket{0} - \beta \Ket{1}
+ \\
+ \hat{\sigma}_x \Ket{q}
+ &= \alpha \Ket{1} + \beta \Ket{0}
+ \qquad \quad
+ \hat{\sigma}_x \hat{\sigma}_z \Ket{q}
+ = \alpha \Ket{1} - \beta \Ket{0}
+\end{aligned}$$
+
+Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of)
+the following entangled three-qubit state:
+
+$$\begin{aligned}
+ \Ket{q}_{A'} \ket{\Phi^+}_{AB}
+ = \frac{1}{2} \bigg( &\ket{\Phi^{+}}_{A'A} \Big( \Ket{q} \Big)_B \quad\, + \ket{\Phi^{-}}_{A'A} \Big( \hat{\sigma}_z \Ket{q} \Big)_B
+ \\
+ + &\ket{\Psi^{+}}_{A'A} \Big( \hat{\sigma}_x \Ket{q} \Big)_B + \ket{\Psi^{-}}_{A'A} \Big( \hat{\sigma}_x \hat{\sigma}_z \Ket{q} \Big)_B \bigg)
+\end{aligned}$$
+
+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,
+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
+$\ket{\Phi^{+}}$, $\ket{\Phi^{-}}$, $\ket{\Psi^{+}}$, $\ket{\Psi^{-}}$
+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 $A$.
+
+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 $B$.
+
+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.
+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,
+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 $\Ket{q}$.
+
+
+## References
+1. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/rabi-oscillation/index.md b/source/know/concept/rabi-oscillation/index.md
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+---
+title: "Rabi oscillation"
+date: 2021-09-22
+categories:
+- Physics
+- Quantum mechanics
+- Two-level system
+- Optics
+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
+affects the state as follows,
+where $c_n(t)$ are the coefficients of the linear combination
+of basis states $\Ket{n} \exp(-i E_n t / \hbar)$:
+
+$$\begin{aligned}
+ i \hbar \dv{c_m}{t}
+ = \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1}{n} \exp(i \omega_{mn} t)
+\end{aligned}$$
+
+Where $\omega_{mn} \equiv (E_m \!-\! E_n) / \hbar$
+for energies $E_m$ and $E_n$.
+Note that this equation is exact,
+despite being used for deriving perturbation theory.
+Consider a two-level system where $n \in \{a, b\}$,
+in which case the above equation can be expanded to the following:
+
+$$\begin{aligned}
+ \dv{c_a}{t}
+ &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp(- i \omega_0 t) \: c_b - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{a} \: c_a
+ \\
+ \dv{c_b}{t}
+ &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{b} \: c_b
+\end{aligned}$$
+
+Where $\omega_0 \equiv \omega_{ba}$ is positive.
+We assume that $\hat{H}_1$ has odd spatial parity,
+in which case [Laporte's selection rule](/know/concept/selection-rules/)
+states that the diagonal matrix elements vanish, leaving:
+
+$$\begin{aligned}
+ \dv{c_a}{t}
+ &= - \frac{i}{\hbar} \matrixel{a}{\hat{H}_1}{b} \exp(- i \omega_0 t) \: c_b
+ \\
+ \dv{c_b}{t}
+ &= - \frac{i}{\hbar} \matrixel{b}{\hat{H}_1}{a} \exp(i \omega_0 t) \: c_a
+\end{aligned}$$
+
+We now choose $\hat{H}_1$ to be as follows,
+sinusoidally oscillating with a spatially odd $V(\vec{r})$:
+
+$$\begin{aligned}
+ \hat{H}_1(t)
+ = V \cos(\omega t)
+ = \frac{V}{2} \Big( \exp(i \omega t) + \exp(-i \omega t) \Big)
+\end{aligned}$$
+
+We insert this into the equations for $c_a$ and $c_b$,
+and define $V_{ab} \equiv \matrixel{a}{V}{b}$, leading us to:
+
+$$\begin{aligned}
+ \dv{c_a}{t}
+ &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!-\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!+\! \omega_0) t\big) \Big) \: c_b
+ \\
+ \dv{c_b}{t}
+ &= - i \frac{V_{ab}}{2 \hbar} \Big( \exp\!\big(i (\omega \!+\! \omega_0) t\big) + \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t\big) \Big) \: c_a
+\end{aligned}$$
+
+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 $\exp(i (\omega \!+\! \omega_0) t)$
+oscillates so fast that its effect is negligible
+when the system is observed over a reasonable time interval.
+Dropping those terms leaves us with:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \dv{c_a}{t}
+ &= - i \frac{V_{ab}}{2 \hbar} \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b
+ \\
+ \dv{c_b}{t}
+ &= - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \dvn{2}{c_a}{t}
+ &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b + \dv{c_b}{t} \bigg) \exp\!\big(i (\omega \!-\! \omega_0) t \big)
+ \\
+ &= - i \frac{V_{ab}}{2 \hbar} \bigg( i (\omega - \omega_0) \: c_b
+ - i \frac{V_{ba}}{2 \hbar} \exp\!\big(\!-\! i (\omega \!-\! \omega_0) t \big) \: c_a \bigg)
+ \exp\!\big(i (\omega \!-\! \omega_0) t \big)
+ \\
+ &= \frac{V_{ab}}{2 \hbar} (\omega - \omega_0) \exp\!\big(i (\omega \!-\! \omega_0) t \big) \: c_b - \frac{|V_{ab}|^2}{(2 \hbar)^2} c_a
+\end{aligned}$$
+
+In the first term, we recognize $\idv{c_a}{t}$,
+which we insert to arrive at an equation for $c_a(t)$:
+
+$$\begin{aligned}
+ 0
+ = \dvn{2}{c_a}{t} - i (\omega - \omega_0) \dv{c_a}{t} + \frac{|V_{ab}|^2}{(2 \hbar)^2} \: c_a
+\end{aligned}$$
+
+To solve this, we make the ansatz $c_a(t) = \exp(\lambda t)$,
+which, upon insertion, gives us:
+
+$$\begin{aligned}
+ 0
+ = \lambda^2 - i (\omega - \omega_0) \lambda + \frac{|V_{ab}|^2}{(2 \hbar)^2}
+\end{aligned}$$
+
+This quadratic equation has two complex roots $\lambda_1$ and $\lambda_2$,
+which are found to be:
+
+$$\begin{aligned}
+ \lambda_1
+ = i \frac{\omega - \omega_0 + \tilde{\Omega}}{2}
+ \qquad \quad
+ \lambda_2
+ = i \frac{\omega - \omega_0 - \tilde{\Omega}}{2}
+\end{aligned}$$
+
+Where we have defined the **generalized Rabi frequency** $\tilde{\Omega}$ to be given by:
+
+$$\begin{aligned}
+ \boxed{
+ \tilde{\Omega}
+ \equiv \sqrt{(\omega - \omega_0)^2 + \frac{|V_{ab}|^2}{\hbar^2}}
+ }
+\end{aligned}$$
+
+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):
+
+$$\begin{aligned}
+ \boxed{
+ c_a(t)
+ = \Big( A \sin(\tilde{\Omega} t / 2) + B \cos(\tilde{\Omega} t / 2) \Big) \exp\!\big(i (\omega \!-\! \omega_0) t / 2 \big)
+ }
+\end{aligned}$$
+
+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 $B = 1$,
+we get the following probabilities
+
+$$\begin{aligned}
+ |c_a(t)|^2
+ &= \cos^2(\tilde{\Omega} t / 2)
+ = \frac{1}{2} \Big( 1 + \cos(\tilde{\Omega} t) \Big)
+ \\
+ |c_b(t)|^2
+ &= \sin^2(\tilde{\Omega} t / 2)
+ = \frac{1}{2} \Big( 1 - \cos(\tilde{\Omega} t) \Big)
+\end{aligned}$$
+
+Note that the period was halved by squaring.
+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 $\omega = \omega_0$:
+
+$$\begin{aligned}
+ \Omega
+ \equiv \frac{V_{ba}}{\hbar}
+\end{aligned}$$
+
+As an example, Rabi oscillation arises
+in the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
+where $\hat{H}_1$ is:
+
+$$\begin{aligned}
+ \hat{H}_1(t)
+ = - q \vec{r} \cdot \vec{E}_0 \cos(\omega t)
+\end{aligned}$$
+
+After making the rotating wave approximation,
+the resulting Rabi frequency is given by:
+
+$$\begin{aligned}
+ \Omega
+ = - \frac{\vec{d} \cdot \vec{E}_0}{\hbar}
+\end{aligned}$$
+
+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,
+thereby departing from its classical interpretation.
+
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
diff --git a/source/know/concept/random-phase-approximation/dyson.png b/source/know/concept/random-phase-approximation/dyson.png
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+---
+title: "Random phase approximation"
+date: 2021-12-01
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+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.
+We consider the self-energy in the context of [jellium](/know/concept/jellium/),
+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 $1 / \beta$:
+
+$$\begin{aligned}
+ \frac{1}{(2 \pi)^3} \int_{-\infty}^\infty \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty \cdots \:\dd{\vb{k}}
+ \:\:\sim\:\:
+ k_F^5
+\end{aligned}$$
+
+Meanwhile, every line gives a factor $1 / k_F^2$.
+The [Matsubara Green's function](/know/concept/matsubara-greens-function/) $G^0$
+for a system with continuous translational symmetry
+is found from [equation-of-motion theory](/know/concept/equation-of-motion-theory/):
+
+$$\begin{aligned}
+ W(\vb{k}) = \frac{e^2}{\varepsilon_0 |\vb{k}|^2}
+ \:\:\sim\:\:
+ \frac{1}{k_F^2}
+ \qquad \qquad
+ G_s^0(\vb{k}, i \omega_n^F)
+ = \frac{1}{i \hbar \omega_n^F - \varepsilon_\vb{k}}
+ \:\:\sim\:\:
+ \frac{1}{k_F^2}
+\end{aligned}$$
+
+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 $\Sigma$
+converge to zero faster than lower-order terms.
+
+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 $\vb{k}$,
+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,
+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
+carry the same momentum and energy:
+
+
+
+
+Where we have defined the **screened interaction** $W^\mathrm{RPA}$,
+denoted by a double wavy line:
+
+
+
+
+Rearranging the above sequence of diagrams quickly leads to the following
+[Dyson equation](/know/concept/dyson-equation/):
+
+
+
+
+In Fourier space, this equation's linear shape
+means it is algebraic, so we can write it out:
+
+$$\begin{aligned}
+ \boxed{
+ W^\mathrm{RPA}
+ = W + W \Pi_0 W^\mathrm{RPA}
+ }
+\end{aligned}$$
+
+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 $\tilde{\vb{q}} \equiv (\vb{q}, i \omega_n^F)$:
+
+
+
+
+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 $\Pi_0$:
+
+$$\begin{aligned}
+ W^\mathrm{RPA}(\vb{k}, i \omega_n^B)
+ = \frac{W(\vb{k})}{1 - W(\vb{k}) \: \Pi_0(\vb{k}, i \omega_n^B)}
+ = \frac{e^2}{\varepsilon_0 |\vb{k}|^2 - e^2 \Pi_0(\vb{k}, i \omega_n^B)}
+\end{aligned}$$
+
+Let us evaluate the pair-bubble $\Pi_0$ more concretely.
+The Feynman diagram translates to:
+
+$$\begin{aligned}
+ -\hbar \Pi_0(\vb{k}, i \omega_n^B)
+ &= - \sum_{s} \frac{1}{(2 \pi)^3} \int \frac{1}{\hbar \beta} \sum_{m = -\infty}^\infty
+ \hbar G_s(\vb{k} \!+\! \vb{q}, i \omega_n^B \!+\! i \omega_m^F) \: \hbar G_s(\vb{q}, i \omega_m^F) \dd{\vb{q}}
+ \\
+ &= - \frac{2 \hbar}{(2 \pi)^3} \int \frac{1}{\beta} \sum_{m = -\infty}^\infty
+ \frac{1}{i \hbar \omega_n^B + i \hbar \omega_m^F - \varepsilon_{\vb{k}+\vb{q}}} \: \frac{1}{i \hbar \omega_m^F - \varepsilon_{\vb{q}}} \dd{\vb{q}}
+\end{aligned}$$
+
+Here we recognize a [Matsubara sum](/know/concept/matsubara-sum/),
+and rewrite accordingly.
+Note that the residues of $n_F$ are $1 / (\hbar \beta)$
+when it is a function of frequency,
+and $1 / \beta$ when it is a function of energy, so:
+
+$$\begin{aligned}
+ \Pi_0(\vb{k}, i \omega_n^B)
+ &= \frac{2}{(2 \pi)^3} \int
+ \frac{n_F(\varepsilon_{\vb{k}+\vb{q}} - i \hbar \omega_n^B)}{(\varepsilon_{\vb{k}+\vb{q}} - i \hbar \omega_n^B) - \varepsilon_{\vb{q}}}
+ + \frac{n_F(\varepsilon_{\vb{q}})}{i \hbar \omega_n^B + (\varepsilon_{\vb{q}}) - \varepsilon_{\vb{k}+\vb{q}}} \dd{\vb{q}}
+ \\
+ &= \frac{2}{(2 \pi)^3} \int \frac{n_F(\varepsilon_{\vb{q}}) - n_F(\varepsilon_{\vb{k}+\vb{q}})}
+ {i \hbar \omega_n^B + \varepsilon_{\vb{q}} - \varepsilon_{\vb{k}+\vb{q}}} \dd{\vb{q}}
+\end{aligned}$$
+
+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 $\Pi_0^R$:
+
+$$\begin{aligned}
+ \boxed{
+ \Pi_0^R(\vb{k}, \omega)
+ = \frac{2}{(2 \pi)^3} \int \frac{n_F(\varepsilon_{\vb{q}}) - n_F(\varepsilon_{\vb{k}+\vb{q}})}
+ {\hbar (\omega + i \eta) + \varepsilon_{\vb{q}} - \varepsilon_{\vb{k}+\vb{q}}} \dd{\vb{q}}
+ }
+\end{aligned}$$
+
+This is as far as we can go before making simplifying assumptions.
+Therefore, we leave it at:
+
+$$\begin{aligned}
+ \boxed{
+ W^\mathrm{RPA}(\vb{k}, \omega)
+ = \frac{e^2}{\varepsilon_0 |\vb{k}|^2 - e^2 \Pi_0(\vb{k}, \omega)}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
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+---
+title: "Random variable"
+date: 2021-10-22
+categories:
+- Mathematics
+- Statistics
+- Measure theory
+layout: "concept"
+---
+
+**Random variables** are the bread and butter
+of probability theory and statistics,
+and are simply variables whose value depends
+on the outcome of a random experiment.
+Here, we will describe the formal mathematical definition
+of a random variable.
+
+
+## Probability space
+
+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 **event space** $\mathcal{F}$ is a set of events $A$
+that are interesting to us,
+i.e. we have subjectively chosen $\mathcal{F}$
+based on the problem at hand.
+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/).
+
+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 $P(\Omega) = 1$.
+
+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 $\omega$
+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 $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$:
+
+$$\begin{aligned}
+ \{ \omega \in \Omega : X(\omega) \in B \} \in \mathcal{F}
+ \quad \mathrm{for\:any\:} B \in \mathcal{B}(\mathbb{R}^n)
+\end{aligned}$$
+
+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 $\mathcal{F}$.
+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 $\mathcal{F}$.
+
+Given such an $X$, and a set $B \subseteq \mathbb{R}$,
+the **preimage** or **inverse image** $X^{-1}$ is defined as:
+
+$$\begin{aligned}
+ X^{-1}(B)
+ = \{ \omega \in \Omega : X(\omega) \in B \}
+\end{aligned}$$
+
+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 $X$,
+or, more formally:
+
+$$\begin{aligned}
+ \sigma(X)
+ = X^{-1}(\mathcal{B}(\mathbb{R}^n))
+ = \{ A \in \mathcal{F} : A = X^{-1}(B) \mathrm{\:for\:some\:} B \in \mathcal{B}(\mathbb{R}^n) \}
+\end{aligned}$$
+
+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 $B$)
+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 $Y$.
+
+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 $\omega \in \Omega$.
+
+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 $x \in \mathbb{R}$:
+
+$$\begin{aligned}
+ F_X(x)
+ = P(X \le x)
+ = P(\{ \omega \in \Omega : X(\omega) \le x \})
+ = P(X^{-1}(]\!-\!\infty, x]))
+\end{aligned}$$
+
+If $F_X(x)$ is differentiable,
+then the **probability density function** $f_X(x)$ is defined as:
+
+$$\begin{aligned}
+ f_X(x)
+ = \dv{F_X}{x}
+\end{aligned}$$
+
+
+## Expectation value
+
+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:
+
+$$\begin{aligned}
+ \mathbf{E}[X]
+ = \int_{-\infty}^\infty x \: f_X(x) \dd{x}
+ \qquad \mathrm{or} \qquad
+ \mathbf{E}[X]
+ = \sum_{i = 1}^N x_i \: P(X \!=\! x_i)
+\end{aligned}$$
+
+However, $f_X(x)$ is not guaranteed to exist,
+and the distinction between continuous and discrete is cumbersome.
+A more general definition of $\mathbf{E}[X]$
+is the following Lebesgue-Stieltjes integral,
+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 $\Omega$.
+Or, equivalently, a Lebesgue integral can be used:
+
+$$\begin{aligned}
+ \mathbf{E}[X]
+ = \int_\Omega X(\omega) \dd{P(\omega)}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \mathbf{V}[X]
+ = \mathbf{E}\big[ (X - \mathbf{E}[X])^2 \big]
+ = \mathbf{E}[X^2] - \big(\mathbf{E}[X]\big)^2
+\end{aligned}$$
+
+It is also possible to calculate expectation values and variances
+adjusted to some given event information:
+see [conditional expectation](/know/concept/conditional-expectation/).
+
+
+
+## References
+1. U.H. Thygesen,
+ *Lecture notes on diffusions and stochastic differential equations*,
+ 2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/rayleigh-plateau-instability/index.md b/source/know/concept/rayleigh-plateau-instability/index.md
new file mode 100644
index 0000000..38973ae
--- /dev/null
+++ b/source/know/concept/rayleigh-plateau-instability/index.md
@@ -0,0 +1,282 @@
+---
+title: "Rayleigh-Plateau instability"
+date: 2021-03-10
+categories:
+- Physics
+- Fluid mechanics
+- Perturbation
+- Surface tension
+layout: "concept"
+---
+
+In fluid mechanics, the **Rayleigh-Plateau instability** causes
+a column of liquid to break up due to surface tension.
+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 $\alpha$.
+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:
+
+$$\begin{aligned}
+ p_i
+ = p_o + \frac{\alpha}{R_0}
+\end{aligned}$$
+
+We assume that the liquid is at rest.
+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,
+to the internal pressure, which we allow to vary with time and space.
+We use cylindrical coordinates:
+
+$$\begin{aligned}
+ p(r, \phi, z, t) = p_i + p_\epsilon(r, \phi, z, t)
+\end{aligned}$$
+
+This internal pressure difference will cause the liquid to start to flow.
+We express the flow velocity as a vector $\vec{u} = (u_r, u_\phi, u_z)$,
+which obeys the following Euler equations:
+
+$$\begin{aligned}
+ \pdv{\vec{u}}{t} + (\vec{u} \cdot \nabla) \vec{u}
+ = - \frac{1}{\rho} \nabla p
+ \qquad \qquad
+ \nabla \cdot \vec{u} = 0
+\end{aligned}$$
+
+The latter equation states that the fluid is incompressible.
+We assume that $\vec{u}$ is so small that we can ignore
+the quadratic term in the former equation, leaving:
+
+$$\begin{aligned}
+ \pdv{\vec{u}}{t}
+ = - \frac{1}{\rho} \nabla p_\epsilon
+\end{aligned}$$
+
+Taking the divergence and using incompressibility
+yields the Laplace equation for $p_\epsilon$:
+
+$$\begin{aligned}
+ - \frac{1}{\rho} \nabla^2 p_\epsilon
+ = \pdv{}{t}(\nabla \cdot \vec{u})
+ = 0
+ \qquad \implies \qquad
+ \nabla^2 p_\epsilon = 0
+\end{aligned}$$
+
+We write out the Laplacian in cylindrical coordinates
+to get the following problem:
+
+$$\begin{aligned}
+ \nabla^2 p_\epsilon
+ = \pdvn{2}{p_\epsilon}{r} + \frac{1}{r} \pdv{p_\epsilon}{r} + \pdvn{2}{p_\epsilon}{z} + \frac{1}{r^2} \pdvn{2}{p_\epsilon}{\phi}
+ = 0
+\end{aligned}$$
+
+Finally, we add a perturbation $R_\epsilon \ll R_0$
+to the radius of the surface of the liquid column:
+
+$$\begin{aligned}
+ R(z, t)
+ = R_0 + R_\epsilon(z, t)
+\end{aligned}$$
+
+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 $\phi$.
+The Laplace equation then reduces to:
+
+$$\begin{aligned}
+ \nabla^2 p_\epsilon
+ = \pdvn{2}{p_\epsilon}{r} + \frac{1}{r} \pdv{p_\epsilon}{r} + \pdvn{2}{p_\epsilon}{z}
+ = 0
+\end{aligned}$$
+
+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 $R_\epsilon$:
+
+$$\begin{aligned}
+ u_r(r\!=\!R)
+ = \frac{\mathrm{D} R_\epsilon}{\mathrm{D} t}
+ = \pdv{R_\epsilon}{t} + u_z(r\!=\!R) \pdv{R_\epsilon}{z}
+\end{aligned}$$
+
+We linearize this by assuming that the deformation $R_\epsilon$
+varies slowly with respect to $z$:
+
+$$\begin{aligned}
+ u_r(r\!=\!R)
+ \approx \pdv{R_\epsilon}{t}
+\end{aligned}$$
+
+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 $p_\epsilon$:
+
+$$\begin{aligned}
+ p(r\!=\!R)
+ = p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+ \qquad \quad
+ p(r\!=\!R)
+ = p_i + p_\epsilon(r\!=\!R)
+\end{aligned}$$
+
+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 $p_i = p_o + \alpha / R_0$:
+
+$$\begin{aligned}
+ p_o + \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+ = p_o + \frac{\alpha}{R_0} + p_\epsilon(r\!=\!R)
+\end{aligned}$$
+
+Isolating this equation for $p_\epsilon$ yields the desired boundary condition:
+
+$$\begin{aligned}
+ p_\epsilon(r\!=\!R)
+ = \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big) - \frac{\alpha}{R_0}
+\end{aligned}$$
+
+The principal radius around the circumference is $R_0 + R_\epsilon$,
+while the curvature along the length can be approximated
+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:
+
+$$\begin{aligned}
+ p_\epsilon(r\!=\!R)
+ \approx - \alpha \Big( \frac{R_\epsilon}{R_0^2 + R_\epsilon} + \pdvn{2}{R_\epsilon}{z} \Big)
+ \approx - \alpha \Big( \frac{R_\epsilon}{R_0^2} + \pdvn{2}{R_\epsilon}{z} \Big)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \vec{u}(r, z, t)
+ &= \vec{u}(r) \exp(\sigma t) \cos(k z)
+ \\
+ p_\epsilon(r, z, t)
+ &= p_\epsilon(r) \exp(\sigma t) \cos(k z)
+ \\
+ R_\epsilon(z, t)
+ &= R_\epsilon \exp(\sigma t) \cos(k z)
+\end{aligned}$$
+
+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
+Bessel's modified equation of order zero:
+
+$$\begin{aligned}
+ \dvn{2}{p_\epsilon}{r} + \frac{1}{r} \dv{p_\epsilon}{r} - k^2 p_\epsilon
+ = 0
+\end{aligned}$$
+
+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 $B = 0$:
+
+$$\begin{aligned}
+ p_\epsilon(r)
+ = A I_0(kr) + B K_0(kr)
+ = A I_0(kr)
+\end{aligned}$$
+
+Inserting the ansatz into the boundary condition for $p_\epsilon$
+gives us the following relation:
+
+$$\begin{aligned}
+ p_\epsilon(r\!=\!R)
+ = - \alpha R_\epsilon \Big( \frac{1}{R_0^2} + k^2 \Big)
+ = A I_0(k R)
+\end{aligned}$$
+
+Meanwhile, the linearized Euler equation governing $\vec{u}$
+states that $u_r$ is given by:
+
+$$\begin{aligned}
+ \sigma u_r
+ = - \frac{1}{\rho} \dv{p_\epsilon}{r}
+ = - \frac{A k}{\rho} I_0'(kr)
+\end{aligned}$$
+
+Now that we have an expression for $u_r$,
+we can revisit its boundary condition:
+
+$$\begin{aligned}
+ u_r(r\!=\!R)
+ = - \frac{A k}{\rho \sigma} I_0'(k R)
+ = \sigma R_\epsilon
+\end{aligned}$$
+
+Isolating this for $R_\epsilon$ and inserting it
+into the boundary condition for $p_\epsilon$ yields:
+
+$$\begin{aligned}
+ p_\epsilon(r\!=\!R)
+ = A I_0(kR)
+ = \alpha \Big( \frac{1}{R_0^2} + k^2 \Big) \Big( \frac{A k}{\rho \sigma^2} I_0'(k R) \Big)
+\end{aligned}$$
+
+Isolating this for the exponential growth/decay parameter $\sigma$
+gives us the desired result,
+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** $\lambda_c$:
+
+$$\begin{aligned}
+ \boxed{
+ \lambda_c
+ = \frac{2 \pi}{k}
+ = 2 \pi R_0
+ }
+\end{aligned}$$
+
+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
+dominate the action of surface tension,
+which will then try to smoothen the surface by shrinking the bulges
+and widening the constrictions.
+In other words, the liquid column is stable in this case.
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
+2. T. Bohr, A. Andersen,
+ *The Rayleigh-Plateau instability of a liquid column*, 2020,
+ unpublished.
diff --git a/source/know/concept/rayleigh-plesset-equation/index.md b/source/know/concept/rayleigh-plesset-equation/index.md
new file mode 100644
index 0000000..9108d51
--- /dev/null
+++ b/source/know/concept/rayleigh-plesset-equation/index.md
@@ -0,0 +1,132 @@
+---
+title: "Rayleigh-Plesset equation"
+date: 2021-04-06
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+In fluid dynamics, the **Rayleigh-Plesset equation**
+describes how the radius of a spherical bubble evolves in time
+inside an incompressible liquid.
+Notably, it leads to [cavitation](/know/concept/cavitation/).
+
+Consider the main
+[Navier-Stokes equation](/know/concept/navier-stokes-equations/)
+for the velocity field $\va{v}$:
+
+$$\begin{aligned}
+ \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+ = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
+ = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+\end{aligned}$$
+
+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 $r$.
+The above equation then becomes:
+
+$$\begin{aligned}
+ \pdv{v}{t} + v \pdv{v}{r}
+ = - \frac{1}{\rho} \pdv{p}{r}
+ + \nu \bigg( \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{v}{r} \Big) - \frac{2}{r^2} v \bigg)
+\end{aligned}$$
+
+Meanwhile, the incompressibility condition
+in [spherical coordinates](/know/concept/spherical-coordinates/) yields:
+
+$$\begin{aligned}
+ \nabla \cdot \va{v}
+ = \frac{1}{r^2} \pdv{(r^2 v)}{r}
+ = 0
+\end{aligned}$$
+
+This is only satisfied if $r^2 v$ is constant with respect to $r$,
+leading us to a solution $v(r)$ given by:
+
+$$\begin{aligned}
+ v(r)
+ = \frac{C(t)}{r^2}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \pdv{p}{r}
+ = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2
+ - \nu \Big( \frac{2}{r^4} C - \frac{2}{r^4} C \Big) \bigg)
+ = - \rho \bigg( \frac{1}{r^2} C' - \frac{2}{r^5} C^2 \bigg)
+\end{aligned}$$
+
+Integrating this with respect to $r$ yields the following expression for $p$,
+where $p_\infty(t)$ is the (possibly time-dependent) pressure at $r = \infty$:
+
+$$\begin{aligned}
+ p(r)
+ = p_\infty + \rho \bigg( \frac{1}{r} C' - \frac{1}{2 r^4} C^2 \bigg)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \sigma_{rr}(r)
+ = - p(r) + 2 \rho \nu \pdv{v(r)}{r}
+\end{aligned}$$
+
+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 $\sigma_{rr}(r)$.
+Furthermore, to include the effects of surface tension, we simply add
+the [Young-Laplace law](/know/concept/young-laplace-law/) to $P$:
+
+$$\begin{aligned}
+ P
+ = - \sigma_{rr}(R) + \alpha \frac{2}{R}
+ = p(R) - 2 \rho \nu \Big( \frac{-2}{R^3} C \Big) + \alpha \frac{2}{R}
+\end{aligned}$$
+
+We isolate this for $p(R)$, and equate it to
+our expression for $p(r)$
+at the surface $r\!=\!R$:
+
+$$\begin{aligned}
+ P - \rho \nu \frac{4}{R^3} C - \alpha \frac{2}{R}
+ = p_\infty + \rho \bigg( \frac{1}{R} C' - \frac{1}{2 R^4} C^2 \bigg)
+\end{aligned}$$
+
+Isolating for $P$,
+and inserting the fact that $R'(t) = v(t)$,
+such that $C = r^2 v = R^2 R'$,
+yields:
+
+$$\begin{aligned}
+ P
+ &= p_\infty + \rho \bigg( \frac{1}{R} \dv{(R^2 R')}{t} - \frac{1}{2 R^4} (R^2 R')^2
+ + \nu \frac{4}{R^3} (R^2 R') \bigg) + \alpha \frac{2}{R}
+ \\
+ &= p_\infty + \rho \bigg( 2 (R')^2 + R R'' - \frac{1}{2} (R')^2 + \nu \frac{4}{R} R' \bigg) + \alpha \frac{2}{R}
+\end{aligned}$$
+
+Rearranging this and defining $\Delta p \equiv P - p_\infty$
+leads to the Rayleigh-Plesset equation:
+
+$$\begin{aligned}
+ \boxed{
+ \frac{\Delta p}{\rho}
+ = R \dvn{2}{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2 + \nu \frac{4}{R} \dv{R}{t} + \frac{\alpha}{\rho} \frac{2}{R}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/reduced-mass/index.md b/source/know/concept/reduced-mass/index.md
new file mode 100644
index 0000000..12c2ce5
--- /dev/null
+++ b/source/know/concept/reduced-mass/index.md
@@ -0,0 +1,134 @@
+---
+title: "Reduced mass"
+date: 2021-07-05
+categories:
+- Physics
+layout: "concept"
+---
+
+Problems with two interacting objects can be simplified
+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 $m_2$:
+
+$$\begin{aligned}
+ \boxed{
+ \mu \equiv \frac{m_1 m_2}{m_1 + m_2}
+ }
+\end{aligned}$$
+
+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 $\va{a}_r$:
+
+$$\begin{aligned}
+ \va{x}_r
+ \equiv \va{x}_1 - \va{x}_2
+ \qquad
+ \va{v}_r
+ \equiv \va{v}_1 - \va{v}_2
+ = \dv{\va{x}_r}{t}
+ \qquad \quad
+ \va{a}_r
+ \equiv \va{a}_1 - \va{a}_2
+ = \dvn{2}{\va{x}_r}{t}
+\end{aligned}$$
+
+We now choose the coordinate system's origin
+to be the center of mass of both objects:
+
+$$\begin{aligned}
+ m_1 \va{x}_1 + m_2 \va{x}_2 = 0
+\end{aligned}$$
+
+Rearranging and differentiating then yields the following useful equations:
+
+$$\begin{aligned}
+ \va{x}_2 = - \frac{m_1}{m_2} \va{x}_1
+ \qquad \quad
+ \va{v}_2 = - \frac{m_1}{m_2} \va{v}_1
+ \qquad \quad
+ \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1
+\end{aligned}$$
+
+Using these relations, we can rewrite the relative quantities we defined earlier:
+
+$$\begin{aligned}
+ \va{x}_r
+ = \Big( 1 + \frac{m_1}{m_2} \Big) \va{x}_1
+ = \frac{m_1 + m_2}{m_2} \va{x}_1
+ \qquad
+ \va{v}_r
+ = \frac{m_1 + m_2}{m_2} \va{v}_1
+ \qquad
+ \va{a}_r
+ = \frac{m_1 + m_2}{m_2} \va{a}_1
+\end{aligned}$$
+
+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 $\va{a}_1$
+boils down to Newton's third law:
+
+$$\begin{aligned}
+ \va{F}_2 = m_2 \va{a}_2 = - m_1 \va{a}_1 = - \va{F}_1
+ \quad \implies \quad
+ \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1
+\end{aligned}$$
+
+With all that in mind, let us take a closer look at the relative acceleration $\va{a}_r$:
+
+$$\begin{aligned}
+ \va{a}_r
+ = \frac{m_1 + m_2}{m_2} \Big( \frac{m_1}{m_1} \Big) \va{a}_1
+ = \frac{m_1 + m_2}{m_1 m_2} \big( m_1 \va{a}_1 \big)
+ = \frac{\va{F}_1}{\mu}
+ = - \frac{\va{F}_2}{\mu}
+\end{aligned}$$
+
+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 $m_1 / \mu$.
+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:
+
+$$\begin{aligned}
+ \va{x}_1
+ = \frac{\mu}{m_1} \va{x}_r
+ = \frac{m_2}{m_1 + m_2} \va{x}_r
+ \qquad \quad
+ \va{x}_2
+ = \frac{\mu}{m_2} \va{x}_r
+ = - \frac{m_1}{m_1 + m_2} \va{x}_r
+\end{aligned}$$
+
+With this, we can rewrite the total kinetic energy $T$ in an elegant way:
+
+$$\begin{aligned}
+ T
+ &= \frac{1}{2} m_1 \va{v}_1^2 + \frac{1}{2} m_2 \va{v}_2^2
+ = \frac{1}{2} m_1 \Big( \frac{\mu}{m_1} \va{v}_r \Big)^2 + \frac{1}{2} m_2 \Big( \frac{\mu}{m_2} \va{v}_r \Big)^2
+ \\
+ &= \frac{1}{2} \frac{\mu^2}{m_1} \va{v}_r^2 + \frac{1}{2} \frac{\mu^2}{m_2} \va{v}_r^2
+ = \frac{1}{2} \Big( \frac{m_2 \mu^2}{m_1 m_2} + \frac{m_1 \mu^2}{m_1 m_2} \Big) \va{v}_r^2
+ \\
+ &= \frac{1}{2} \frac{(m_1 + m_2) \mu^2}{m_1 m_2} \va{v}_r^2
+ = \frac{1}{2} \frac{\mu^2}{\mu} \va{v}_r^2
+ = \frac{1}{2} \mu \va{v}_r^2
+\end{aligned}$$
+
+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.
+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.
diff --git a/source/know/concept/renyi-entropy/index.md b/source/know/concept/renyi-entropy/index.md
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+---
+title: "Rényi entropy"
+date: 2021-04-11
+categories:
+- Cryptography
+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 $X$.
+It is defined as follows:
+
+$$\begin{aligned}
+ \boxed{
+ H_\alpha(X)
+ = \frac{1}{1 - \alpha} \log\!\bigg( \sum_{i = 1}^N p_i^\alpha \bigg)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ H_0(X)
+ = \log N
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ H_1(X)
+ = \lim_{\alpha \to 1} H_\alpha(X)
+ = \lim_{\alpha \to 1} \frac{\log\!\left( \sum_i p_i^\alpha \right)}{1 - \alpha}
+\end{aligned}$$
+
+We then apply L'Hôpital's rule to evaluate this limit,
+and use the fact that all $p_i$ sum to $1$:
+
+$$\begin{aligned}
+ H_1(X)
+ = \lim_{\alpha \to 1} \frac{\displaystyle \dv{}{\alpha}\log\!\left( \sum_i p_i^\alpha \right)}{\displaystyle \dv{}{\alpha}(1 - \alpha)}
+ = \lim_{\alpha \to 1} \frac{\sum_i p_i^\alpha \log p_i}{- \sum_i p_i^\alpha}
+ = - \sum_{i = 1}^N p_i \log p_i
+\end{aligned}$$
+
+This quantity is the **Shannon entropy**,
+which is the most general measure of "surprise":
+
+$$\begin{aligned}
+ \boxed{
+ H_1(X)
+ = \lim_{\alpha \to 1} H_\alpha(X)
+ = - \sum_{i = 1}^N p_i \log p_i
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ H_2(X)
+ = - \log\!\bigg( \sum_{i = 1}^N p_i^2 \bigg)
+ = - \log P(X = Y)
+ }
+\end{aligned}$$
+
+Finally, in the limit $\alpha \to \infty$,
+the largest probability dominates the sum,
+leading to the definition of the **min-entropy** $H_\infty$,
+describing the surprise of the most likely event:
+
+$$\begin{aligned}
+ \boxed{
+ H_\infty(X)
+ = \lim_{\alpha \to \infty} H_\alpha(x)
+ = - \log\!\big( \max_{i} p_i \big)
+ }
+\end{aligned}$$
+
+It is straightforward to convince yourself that these entropies
+are ordered in the following way:
+
+$$\begin{aligned}
+ H_0 \ge H_1 \ge H_2 \ge H_\infty
+\end{aligned}$$
+
+In other words, from left to right,
+they go from permissive to conservative, roughly speaking.
+
+
+## References
+1. P.A. Bromiley, N.A. Thacker, E. Bouhova-Thacker,
+ [Shannon entropy, Rényi entropy, and information](https://www.researchgate.net/publication/253537416_Shannon_Entropy_Renyi_Entropy_and_Information),
+ 2010, University of Manchester.
+2. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
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+---
+title: "Repetition code"
+date: 2021-05-07
+categories:
+- Quantum information
+layout: "concept"
+---
+
+A **repetition code** is a simple approach to error correction:
+to protect a bit $x$, make two copies:
+
+$$\begin{aligned}
+ 0 \to 000
+ \qquad \quad
+ 1 \to 111
+\end{aligned}$$
+
+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.
+
+In quantum computing, where error correction is much more important,
+repetition codes can also be used,
+albeit with some complications,
+as discussed below.
+
+
+## Bit flip code
+
+Suppose that we want to detect errors in
+the following arbitrary qubit state $\Ket{\psi}$:
+
+$$\begin{aligned}
+ \Ket{\psi}
+ = \alpha \Ket{0} + \beta \Ket{1}
+\end{aligned}$$
+
+For now, let us limit ourselves to detecting **bit flips**,
+where $\alpha$ and $\beta$ get switched:
+
+$$\begin{aligned}
+ \alpha \Ket{0} + \beta \Ket{1}
+ \quad \to \boxed{\mathrm{Error}} \to \quad
+ \beta \Ket{0} \!+\! \alpha \Ket{1}
+\end{aligned}$$
+
+One way to defend against this is
+the quantum version of a classical repetition code:
+
+$$\begin{aligned}
+ \Ket{\psi}
+ \quad \to \boxed{\mathrm{Encoder}} \to \quad
+ \ket{\overline{\psi}}
+ = \alpha \Ket{000} + \beta \Ket{111}
+\end{aligned}$$
+
+In other words, a *logical* $\Ket{0}$ (written $\ket{\overline{0}}$)
+is represented by 3 *physical* qubits, and vice versa:
+
+$$\begin{aligned}
+ \boxed{
+ \Ket{0}
+ \to
+ \ket{\overline{0}}
+ = \Ket{000}
+ \qquad \quad
+ \Ket{1}
+ \to
+ \ket{\overline{1}}
+ = \Ket{111}
+ }
+\end{aligned}$$
+
+Such a transformation is easy to achieve with the following sequence
+of [quantum gates](/know/concept/quantum-gate/):
+
+
+
+
+So, a little while after encoding the state $\Ket{\psi}$ like that,
+a bit flip occurs on the 2nd qubit:
+
+$$\begin{aligned}
+ \ket{\overline{\psi}}
+ \quad \to \boxed{\mathrm{Error}} \to \quad
+ \alpha \Ket{010} + \beta \Ket{101}
+\end{aligned}$$
+
+But now there is a problem: how do we detect this error?
+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.
+The 3-qubit basis states are its eigenvectors:
+
+$$\begin{alignedat}{2}
+ ZZI \Ket{000}
+ &= + \Ket{000}
+ \qquad
+ ZZI \Ket{001}
+ &&= + \Ket{001}
+ \\
+ ZZI \Ket{010}
+ &= - \Ket{010}
+ \qquad
+ ZZI \Ket{011}
+ &&= - \Ket{011}
+ \\
+ ZZI \Ket{100}
+ &= - \Ket{100}
+ \qquad
+ ZZI \Ket{101}
+ &&= - \Ket{101}
+ \\
+ ZZI \Ket{110}
+ &= + \Ket{110}
+ \qquad
+ ZZI \Ket{111}
+ &&= + \Ket{111}
+\end{alignedat}$$
+
+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).
+
+These false negatives are fixed by including another stabilizer $IZZ$,
+with these eigenvectors:
+
+$$\begin{alignedat}{2}
+ IZZ \Ket{000}
+ &= + \Ket{000}
+ \qquad
+ IZZ \Ket{001}
+ &&= - \Ket{001}
+ \\
+ IZZ \Ket{010}
+ &= - \Ket{010}
+ \qquad
+ IZZ \Ket{011}
+ &&= + \Ket{011}
+ \\
+ IZZ \Ket{100}
+ &= + \Ket{100}
+ \qquad
+ IZZ \Ket{101}
+ &&= - \Ket{101}
+ \\
+ IZZ \Ket{110}
+ &= - \Ket{110}
+ \qquad
+ IZZ \Ket{111}
+ &&= + \Ket{111}
+\end{alignedat}$$
+
+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.
+But by using both, we know exactly which qubit was flipped
+thanks to the eigenvalues:
+
+
+
+ Error
+ $ZZI$
+ $IZZ$
+
+
+ $I$
+ $+1$
+ $+1$
+
+
+ $X_1$
+ $-1$
+ $+1$
+
+
+ $X_2$
+ $-1$
+ $-1$
+
+
+ $X_1$
+ $+1$
+ $-1$
+
+
+
+Where e.g. $X_3$ denotes that the 3rd qubit was flipped.
+The measurement outcomes on the last three rows are called **error syndromes**,
+and are obtained by a **syndrome measurement**.
+
+Fortunately, we can measure $ZZI$ and $IZZ$
+without affecting $\ket{\overline{\psi}}$ itself,
+by applying $\mathrm{CNOT}$s to some ancillary qubits
+and then measuring those:
+
+
+
+
+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.
+
+
+## Phase flip code
+
+The above system protects us against all single-qubit bit flips.
+Unfortunately, that is not enough:
+qubits can also experience a **phase flip**:
+
+$$\begin{aligned}
+ \alpha \Ket{0} + \beta \Ket{1}
+ \quad \to \boxed{\mathrm{Error}} \to \quad
+ \alpha \Ket{0} - \beta \Ket{1}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \Ket{0}
+ \to
+ \ket{\overline{0}}
+ = \Ket{+\!+\!+}
+ \qquad \quad
+ \Ket{1}
+ \to
+ \ket{\overline{1}}
+ = \Ket{-\!-\!-}
+ }
+\end{aligned}$$
+
+Such that an arbitrary state $\Ket{\psi}$ is encoded as follows,
+by the circuit shown below:
+
+$$\begin{aligned}
+ \Ket{\psi}
+ \quad \to \boxed{\mathrm{Encoder}} \to \quad
+ \ket{\overline{\psi}}
+ = \alpha \Ket{+\!+\!+} + \beta \Ket{-\!-\!-}
+\end{aligned}$$
+
+
+
+
+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 $IXX$,
+and the error detection circuit is as follows:
+
+
+
+
+This system protects us against all single-qubit phase flips,
+but not against bit flips.
+
+
+## Shor code
+
+What kind of repetition code would we need
+if we want to detect both bit flips *and* phase flips?
+The most straightforward option is the **Shor code**.
+Starting from a phase flip encoding:
+
+$$\begin{aligned}
+ \Ket{0} \to
+ \ket{\overline{0}}
+ &= \Ket{+\!+\!+}
+ = \bigg( \frac{\Ket{0} + \Ket{1}}{\sqrt{2}} \bigg)^{\otimes 3}
+ \\
+ \Ket{1} \to
+ \ket{\overline{1}}
+ &= \Ket{-\!-\!-}
+ = \bigg( \frac{\Ket{0} - \Ket{1}}{\sqrt{2}} \bigg)^{\otimes 3}
+\end{aligned}$$
+
+We add protection against bit flips
+by using a repetition code for each physical qubit:
+
+$$\begin{aligned}
+ \boxed{
+ \ket{\overline{0}}
+ = \bigg( \frac{\Ket{000} + \Ket{111}}{\sqrt{2}} \bigg)^{\otimes 3}
+ \qquad \quad
+ \ket{\overline{1}}
+ = \bigg( \frac{\Ket{000} - \Ket{111}}{\sqrt{2}} \bigg)^{\otimes 3}
+ }
+\end{aligned}$$
+
+This encoding is achieved by the following quantum circuit,
+which simply consists of the phase flip encoder,
+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 $IZZ$
+are applied on a per-block basis, like so:
+
+$$\begin{aligned}
+ ZZI \: III \: III \qquad\quad III \: ZZI \: III \qquad\quad III \: III \: ZZI
+ \\
+ IZZ \: III \: III \qquad\quad III \: IZZ \: III \qquad\quad III \: III \: IZZ
+\end{aligned}$$
+
+Whereas the phase flip stabilizers $XXI$ and $IXX$
+are applied to entire blocks at once:
+
+$$\begin{aligned}
+ XXX \: XXX \: III
+ \qquad \quad
+ III \: XXX \: XXX
+\end{aligned}$$
+
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
+
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+---
+title: "Residue theorem"
+date: 2021-11-13
+categories:
+- Mathematics
+- Complex analysis
+layout: "concept"
+---
+
+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 $z_p$:
+
+$$\begin{aligned}
+ f(z)
+ \approx \frac{R_p}{z - z_p}
+\end{aligned}$$
+
+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 $z_p$:
+
+$$\begin{aligned}
+ \boxed{
+ R_p = \lim_{z \to z_p} (z - z_p) f(z)
+ }
+\end{aligned}$$
+
+**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 $C$:
+
+$$\begin{aligned}
+ \boxed{
+ \oint_C f(z) \dd{z} = i 2 \pi \sum_{z_p} R_p
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+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:
+
+$$\begin{aligned}
+ f(z) = h(z) + \sum_{z_p} \frac{R_p}{z - z_p}
+\end{aligned}$$
+
+We integrate this over a contour $C$ which contains all poles, and apply
+both Cauchy's integral theorem and Cauchy's integral formula to get:
+
+$$\begin{aligned}
+ \oint_C f(z) \dd{z}
+ &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
+ = \sum_{p} R_p \: 2 \pi i
+\end{aligned}$$
+
+
+
+
+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:
+
+$$\begin{aligned}
+ m_1 \dv{\vb{v}_1}{t}
+ = \vb{F}_1
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}_1 - \vb{r}_2}{r^3}
+ \qquad \quad
+ m_2 \dv{\vb{v}_2}{t}
+ = \vb{F}_2
+ = - \vb{F}_1
+\end{aligned}$$
+
+Using the [reduced mass](/know/concept/reduced-mass/)
+$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$,
+we turn this into a one-body problem:
+
+$$\begin{aligned}
+ \mu \dv{\vb{v}}{t}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3}
+\end{aligned}$$
+
+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.
+The latter is as follows in
+[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/)
+$(r, \varphi, z)$:
+
+$$\begin{aligned}
+ \vb{r}
+ = r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
+ = r \:\vu{e}_r + z \:\vu{e}_z
+\end{aligned}$$
+
+These new coordinates are sketched below,
+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:
+
+
+
+
+From our expression for $\vb{r}$,
+we can find $\vb{v}$ by differentiating with respect to time:
+
+$$\begin{aligned}
+ \vb{v}
+ &= \big( r' \cos{\varphi} - r \varphi' \sin{\varphi} \big) \:\vu{e}_x
+ + \big( r' \sin{\varphi} + r \varphi' \cos{\varphi} \big) \:\vu{e}_y + z' \:\vu{e}_z
+ \\
+ &= r' \: \big( \cos{\varphi} \:\vu{e}_x + \sin{\varphi} \:\vu{e}_y \big)
+ + r \varphi' \: \big( \!-\! \sin{\varphi} \:\vu{e}_x + \cos{\varphi} \:\vu{e}_y \big) + z' \:\vu{e}_z
+ \\
+ &= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi + z' \:\vu{e}_z
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \vb{r}
+ = r \: \vu{e}_r
+ \qquad \qquad
+ \vb{v}
+ = r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi
+\end{aligned}$$
+
+Consequently, the angular momentum $\vb{L}$ is as follows,
+pointing purely in the $z$-direction:
+
+$$\begin{aligned}
+ \vb{L}(t)
+ = \mu \vb{r} \cross \vb{v}
+ = \mu \big( r \vu{e}_r \cross r \varphi' \vu{e}_\varphi \big)
+ = \mu r^2 \varphi' \:\vu{e}_z
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \frac{b}{r(\pm \infty)}
+ = \sin{\chi(\pm \infty)}
+ \qquad \quad
+ \chi(t)
+ \equiv \measuredangle(\vb{r}, \vb{v})
+\end{aligned}$$
+
+With this, we can rewrite
+the magnitude of the angular momentum $\vb{L}$ as follows,
+where the total velocity $|\vb{v}|$ is a constant,
+thanks to conservation of energy:
+
+$$\begin{aligned}
+ \big| \vb{L}(\pm \infty) \big|
+ = \mu \big| \vb{r} \cross \vb{v} \big|
+ = \mu r |\vb{v}| \sin{\chi}
+ = \mu b |\vb{v}|
+\end{aligned}$$
+
+However, conveniently,
+angular momentum is also conserved, i.e. $\vb{L}$ is constant in time:
+
+$$\begin{aligned}
+ \vb{L}'(t)
+ &= \mu \big( \vb{r} \cross \vb{v}' + \vb{v} \cross \vb{v} \big)
+ = \vb{r} \cross (\mu \vb{v}')
+ = \vb{r} \cross \Big( \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \Big)
+ = 0
+\end{aligned}$$
+
+Where we have replaced $\mu \vb{v}'$ with the equation of motion.
+Thanks to this, we can equate the two preceding expressions for $\vb{L}$,
+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 $t$:
+
+$$\begin{aligned}
+ - \mu r^2 \dv{\varphi}{t}
+ = \mu b |\vb{v}|
+ \quad \implies \quad
+ \dd{t}
+ = - \frac{r^2}{b |\vb{v}|} \dd{\varphi}
+\end{aligned}$$
+
+Now, at last, we turn to the main equation of motion.
+Its $y$-component is given by:
+
+$$\begin{aligned}
+ \mu \dv{v_y}{t}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3}
+ \quad \implies \quad
+ \mu \dd{v_y}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \dd{t}
+\end{aligned}$$
+
+We replace $\dd{t}$ with our earlier relation,
+and recognize geometrically that $y/r = \sin{\varphi}$:
+
+$$\begin{aligned}
+ \mu \dd{v_y}
+ = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \frac{y}{r} \dd{\varphi}
+ = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \sin{\varphi} \dd{\varphi}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})}
+\end{aligned}$$
+
+Integrating this from the initial state $i$ at $t = -\infty$
+to the final state $f$ at $t = \infty$ yields:
+
+$$\begin{aligned}
+ \Delta v_y
+ \equiv \int_{i}^{f} \dd{v_y}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ 2 v_{y,f}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos(\pi \!-\! \varphi_f) \big)
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big)
+\end{aligned}$$
+
+Furthermore, geometrically, at $t = \infty$
+we notice that $v_{y,f} = |\vb{v}| \sin{\varphi_f}$,
+leading to:
+
+$$\begin{aligned}
+ 2 |\vb{v}| \sin{\varphi_f}
+ = \frac{q_1 q_2}{2 \pi \varepsilon_0 b |\vb{v}| \mu} \cos{\varphi_f}
+\end{aligned}$$
+
+Rearranging this yields the following equation
+for the final polar angle $\varphi_f \equiv \varphi(\infty)$:
+
+$$\begin{aligned}
+ \tan{\varphi_f}
+ = \frac{\sin{\varphi_f}}{\cos{\varphi_f}}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
+\end{aligned}$$
+
+However, we want $\theta$, not $\varphi_f$.
+One last use of symmetry and geometry
+tells us that $\theta = 2 \varphi_f$,
+and we thus arrive at the celebrated **Rutherford scattering formula**:
+
+$$\begin{aligned}
+ \boxed{
+ \tan\!\Big( \frac{\theta}{2} \Big)
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
+ }
+\end{aligned}$$
+
+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.
+
+
+
+## References
+1. P.M. Bellan,
+ *Fundamentals of plasma physics*,
+ 1st edition, Cambridge.
+2. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
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diff --git a/source/know/concept/salt-equation/index.md b/source/know/concept/salt-equation/index.md
new file mode 100644
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+---
+title: "SALT equation"
+date: 2022-02-07
+categories:
+- Physics
+- Optics
+- Laser theory
+layout: "concept"
+---
+
+The **steady-state *ab initio* laser theory** (SALT) is
+a theoretical description of lasers, whose mode-centric approach
+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
+active atoms (or quantum dots) embedded in a passive linear background
+material with refractive index $c / v$.
+The system is affected by a driving [electric field](/know/concept/electric-field/)
+$\vb{E}^{+}(t) = \vb{E}_0^{+} e^{-i \omega t}$,
+such that the set of equations is:
+
+$$\begin{aligned}
+ - \mu_0 \pdvn{2}{\vb{P}^{+}}{t}
+ &= \nabla \cross \nabla \cross \vb{E}^{+} + \frac{1}{v^2} \pdvn{2}{\vb{E}^{+}}{t}
+ \\
+ \pdv{\vb{P}^{+}}{t}
+ &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+}
+ - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} D
+ \\
+ \pdv{D}{t}
+ &= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
+\end{aligned}$$
+
+Where $\hbar \omega_0$ is the band gap of the active atoms,
+and $\gamma_\perp$ and $\gamma_\parallel$ are relaxation rates
+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:
+
+$$\begin{aligned}
+ \vb{p}_0^{-}
+ \equiv q \matrixel{e}{\vu{x}}{g}
+ \qquad \qquad
+ \vb{p}_0^{+}
+ \equiv q \matrixel{g}{\vu{x}}{e}
+ = (\vb{p}_0^{-})^*
+\end{aligned}$$
+
+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.
+Note that this ansatz already suggests
+that the interactions between the modes are limited:
+
+$$\begin{aligned}
+ \vb{E}^{+}(\vb{x}, t)
+ = \sum_{n = 1}^N \Psi_n(\vb{x}) \: e^{- i \omega_n t}
+ \qquad \qquad
+ \vb{P}^{+}(\vb{x}, t)
+ = \sum_{n = 1}^N \vb{p}_n(\vb{x}) \: e^{- i \omega_n t}
+\end{aligned}$$
+
+Using the modes' linear independence to treat each term of the summation individually,
+the first two Maxwell-Bloch equations turn into, respectively:
+
+$$\begin{aligned}
+ \mu_0 \omega_n^2 \vb{p}_n
+ &= \nabla \cross \nabla \cross \Psi_n - \frac{1}{v^2} \omega_n^2 \Psi_n
+ \\
+ i \omega_n \vb{p}_n
+ &= \big( i \omega_0 + \gamma_\perp \big) \vb{p}_n
+ + \frac{i}{\hbar} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n \: D
+\end{aligned}$$
+
+With being $\vb{p}_0^{+} \vb{p}_0^{-}$ a dyadic product.
+Isolating the latter equation for $\vb{p}_n$ gives us:
+
+$$\begin{aligned}
+ \vb{p}_n
+ &= \frac{\big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n \: D}{\hbar \big((\omega_n - \omega_0) + i \gamma_\perp\big)}
+ = \frac{\gamma(\omega_n) D}{\hbar \gamma_\perp} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n
+\end{aligned}$$
+
+Where we have defined the Lorentzian gain curve $\gamma(\omega_n)$ as follows,
+which represents the laser's preferred frequencies for amplification:
+
+$$\begin{aligned}
+ \gamma(\omega_n)
+ \equiv \frac{\gamma_\perp}{(\omega_n - \omega_0) + i \gamma_\perp}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ 0
+ &= \bigg( \nabla \cross \nabla \cross - \, \omega_n^2 \frac{1}{v^2}
+ - \omega_n^2 \frac{\mu_0 \gamma(\omega_n) D}{\hbar \gamma_\perp} (\vb{p}_0^{+} \vb{p}_0^{-}) \cdot \bigg) \Psi_n
+\end{aligned}$$
+
+To rewrite $D$, we turn to its (Maxwell-Bloch) equation of motion,
+making the crucial **stationary inversion approximation** $\ipdv{D}{t} = 0$:
+
+$$\begin{aligned}
+ D
+ &= D_0 + \frac{i 2}{\hbar \gamma_\parallel} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
+\end{aligned}$$
+
+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 $D$:
+
+$$\begin{aligned}
+ D
+ &= D_0 + \frac{i 2}{\hbar \gamma_\parallel} \sum_{\nu, \mu}^\mathrm{active}
+ \bigg( \vb{p}_\nu^* \cdot \Psi_\mu e^{i (\omega_\nu - \omega_\mu) t}
+ - \vb{p}_\nu \cdot \Psi_\mu^* e^{i (\omega_\mu - \omega_\nu) t} \bigg)
+\end{aligned}$$
+
+Here, we make the [rotating wave approximation](/know/concept/rotating-wave-approximation/)
+to neglect all terms where $\nu \neq \mu$
+on the basis that they oscillate too quickly,
+leaving only $\nu = \mu$:
+
+$$\begin{aligned}
+ D
+ &= D_0 + \frac{i 2}{\hbar \gamma_\parallel} \sum_{\nu}^\mathrm{act.}
+ \bigg( \vb{p}_\nu^* \cdot \Psi_\nu - \vb{p}_\nu \cdot \Psi_\nu^* \bigg)
+\end{aligned}$$
+
+Inserting our earlier equation for $\vb{p}_n$
+and using the fact that $\vb{p}_0^{+} = (\vb{p}_0^{-})^*$ leads us to:
+
+$$\begin{aligned}
+ D
+ &= D_0 + \frac{i 2 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.}
+ \bigg( \gamma^*(\omega_\nu) \big(\vb{p}_0^{+} \vb{p}_0^{-}\big)^* \!\cdot\! \Psi_\nu^* \cdot \Psi_\nu
+ - \gamma(\omega_\nu) \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \!\cdot\! \Psi_\nu \cdot \Psi_\nu^* \bigg)
+ \\
+ &= D_0 + \frac{i 2 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.}
+ \bigg( \gamma^*(\omega_\nu) \big(\vb{p}_0^{+} \cdot \Psi_\nu^*\big) \vb{p}_0^{-} \cdot \Psi_\nu
+ - \gamma(\omega_\nu) \big(\vb{p}_0^{-} \cdot \Psi_\nu\big) \vb{p}_0^{+} \cdot \Psi_\nu^* \bigg)
+ \\
+ &= D_0 + \frac{i 2 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.}
+ \Big( \gamma^*(\omega_\nu) - \gamma(\omega_\nu) \Big) \big|\vb{p}_0^{-} \cdot \Psi_\nu\big|^2
+\end{aligned}$$
+
+By putting the terms on a common denominator, it is easily shown that:
+
+$$\begin{aligned}
+ \gamma^*(\omega_\nu) - \gamma(\omega_\nu)
+ &= \frac{\gamma_\perp ((\omega_\nu - \omega_0) + i \gamma_\perp)}{(\omega_\nu - \omega_0)^2 + \gamma_\perp^2}
+ - \frac{\gamma_\perp ((\omega_\nu - \omega_0) - i \gamma_\perp)}{(\omega_\nu - \omega_0)^2 + \gamma_\perp^2}
+ \\
+ &= \frac{\gamma_\perp (i \gamma_\perp + i \gamma_\perp)}{(\omega_\nu - \omega_0)^2 + \gamma_\perp^2}
+ = i 2 \big|\gamma(\omega_\nu)\big|^2
+\end{aligned}$$
+
+Inserting this into our equation for $D$ gives the following expression:
+
+$$\begin{aligned}
+ D
+ &= D_0 - \frac{4 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.}
+ \Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu\Big|^2
+\end{aligned}$$
+
+We then properly isolate this for $D$ to get its final form, namely:
+
+$$\begin{aligned}
+ D
+ &= D_0 \bigg( 1 + \frac{4}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.}
+ \Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu\Big|^2 \bigg)^{-1}
+\end{aligned}$$
+
+Substituting this into the prototypical SALT equation from earlier
+yields the most general form of the **SALT equation**,
+upon which the theory is built:
+
+$$\begin{aligned}
+ \boxed{
+ 0
+ = \bigg( \nabla \cross \nabla \cross
+ -\,\omega_n^2 \bigg[ \frac{1}{v^2(\vb{x})} + \frac{\mu_0 \gamma(\omega_n)}{\hbar \gamma_\perp}
+ \frac{D_0(\vb{x})}{1 + h(\vb{x})} (\vb{p}_0^{+} \vb{p}_0^{-}) \cdot \bigg] \bigg) \Psi_n(\vb{x})
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ h(\vb{x})
+ \equiv \frac{4}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.}
+ \Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu(\vb{x})\Big|^2
+ }
+\end{aligned}$$
+
+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 $e_c$:
+
+$$\begin{aligned}
+ \tilde{\Psi}_n
+ \equiv \frac{\Psi_n}{e_c}
+ \qquad
+ e_c
+ \equiv \frac{\hbar \sqrt{\gamma_\parallel \gamma_\perp}}{2 |g|}
+ \qquad \qquad
+ \tilde{D}
+ \equiv \frac{D}{d_c}
+ \qquad
+ d_c
+ \equiv \frac{\varepsilon_0 \hbar \gamma_\perp}{|g|^2}
+\end{aligned}$$
+
+And then the SALT equation and hole burning function $h$ are reduced to the following,
+where the vacuum wavenumber $k_n = \omega_n / c$:
+
+$$\begin{aligned}
+ 0
+ = \bigg( \nabla \cross \nabla \cross -\,k_n^2 \bigg[ \varepsilon_r
+ + \gamma(c k_n) \frac{\tilde{D}_0}{1 + h} \bigg] \bigg) \tilde{\Psi}_n
+ \qquad
+ h(\vb{x})
+ = \sum_{\nu}^\mathrm{act.} \Big|\gamma(c k_\nu) \tilde{\Psi}_\nu(\vb{x})\Big|^2
+\end{aligned}$$
+
+
+In addition, some papers only consider 1D or 2D *transverse magnetic* (TM) modes,
+in which case the fields are scalars. Using the vector identity
+
+$$\begin{aligned}
+ \nabla \cross \nabla \cross \Psi
+ = \nabla (\nabla \cdot \Psi) - \nabla^2 \Psi
+\end{aligned}$$
+
+Where $\nabla \cdot \Psi = 0$ thanks to [Gauss' law](/know/concept/maxwells-equations/),
+so we get an even further simplified SALT equation:
+
+$$\begin{aligned}
+ 0
+ = \bigg( \nabla^2 +\,k_n^2 \bigg[ \varepsilon_r
+ + \gamma(c k_n) \frac{\tilde{D}_0}{1 + h} \bigg] \bigg) \tilde{\Psi}_n
+\end{aligned}$$
+
+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,
+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,
+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.
+
+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 $D_0$.
+
+
+## References
+1. L. Ge, Y.D. Chong, A.D. Stone,
+ [Steady-state *ab initio* laser theory: generalizations and analytic results](http://dx.doi.org/10.1103/PhysRevA.82.063824),
+ 2010, American Physical Society.
+
diff --git a/source/know/concept/schwartz-distribution/index.md b/source/know/concept/schwartz-distribution/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/schwartz-distribution/index.md
@@ -0,0 +1,120 @@
+---
+title: "Schwartz distribution"
+date: 2021-02-25
+categories:
+- Mathematics
+layout: "concept"
+---
+
+A **Schwartz distribution**, also known as a **generalized function**,
+is a generalization of a function,
+allowing us to work with otherwise pathological definitions.
+
+Notable examples of distributions are
+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,
+whose members are often called **test functions**.
+Every such $\phi(x) \in \mathcal{S}$ must satisfy
+the following constraint for any $p, q \in \mathbb{N}$:
+
+$$\begin{aligned}
+ \mathrm{max} \big| x^p \phi^{(q)}(x) \big| < \infty
+\end{aligned}$$
+
+In other words, a test function and its derivatives
+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 $\Inprod{f}{\phi}$.
+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:
+
+$$\begin{aligned}
+ f[\phi]
+ = \Inprod{f}{\phi}
+ = \int_{-\infty}^\infty f(x) \: \phi(x) \dd{x}
+\end{aligned}$$
+
+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 $f$.
+
+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 $\mathcal{S}$,
+we can work out what the derivative of a distribution is:
+
+$$\begin{aligned}
+ \Inprod{f'}{\phi}
+ = \int_{-\infty}^\infty f'(x) \: \phi(x) \dd{x}
+ = \Big[ f(x) \: \phi(x) \Big]_{-\infty}^\infty - \int_{-\infty}^\infty f(x) \: \phi'(x) \dd{x}
+\end{aligned}$$
+
+The test function removes the boundary term, yielding the result
+$- \Inprod{f}{\phi'}$. Although this was an example for a specific $f(x)$,
+we use it to define the derivative of any distribution:
+
+$$\begin{aligned}
+ \boxed{
+ \Inprod{f'}{\phi} = - \Inprod{f}{\phi'}
+ }
+\end{aligned}$$
+
+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 $x$:
+
+$$\begin{aligned}
+ \tilde{\phi}(x)
+ = \int_{-\infty}^\infty \phi(k) \exp(- i k x) \dd{k}
+\end{aligned}$$
+
+The FT of a Schwartz distribution $f$ then turns out to be as follows:
+
+$$\begin{aligned}
+ \inprod{\tilde{f}}{\phi}
+ &= \int_{-\infty}^\infty \tilde{f}(k) \: \phi(k) \dd{k}
+ = \iint_{-\infty}^\infty f(x) \exp(- i k x) \: \phi(k) \dd{x} \dd{k}
+ \\
+ &= \int_{-\infty}^\infty f(x) \: \tilde{\phi}(x) \dd{x}
+ = \inprod{f}{\tilde{\phi}}
+\end{aligned}$$
+
+Note that the ordinary FT $\tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\}$ is
+already a 1:1 mapping of test functions $\phi \leftrightarrow \tilde{\phi}$.
+As it turns out,
+in this generalization it is also a 1:1 mapping of distributions in $\mathcal{S}'$,
+defined as:
+
+$$\begin{aligned}
+ \boxed{
+ \inprod{\tilde{f}}{\phi}
+ = \inprod{f}{\tilde{\phi}}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. K.W. Jacobsen,
+ *Note on generalized functions (distributions)*, 2020,
+ unpublished.
diff --git a/source/know/concept/screw-pinch/index.md b/source/know/concept/screw-pinch/index.md
new file mode 100644
index 0000000..a279b98
--- /dev/null
+++ b/source/know/concept/screw-pinch/index.md
@@ -0,0 +1,203 @@
+---
+title: "Screw pinch"
+date: 2022-03-06
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+A **pinch** is a type of plasma confinement,
+which relies on [magnetic fields](/know/concept/magnetic-field/)
+to squeeze the plasma into the desired area.
+Examples include tokamaks and stellarators,
+although the term *pinch* is typically introduced for simpler 1D confinement.
+
+Suppose that we want to pinch a plasma into a cylindrical shape.
+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/)
+$(r, \theta, z)$.
+
+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.
+From the (ideal) generalized Ohm's law, it then follows
+that the [electric field](/know/concept/electric-field/) $\vb{E} = 0$:
+
+$$\begin{aligned}
+ \vb{u}
+ = 0
+ \qquad \qquad
+ \pdv{\vb{u}}{t}
+ = 0
+ \qquad \qquad
+ \pdv{\vb{B}}{t}
+ = 0
+ \qquad \qquad
+ \vb{E}
+ = 0
+\end{aligned}$$
+
+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 $p$.
+Symmetries should be used whenever possible to reduce these equations:
+
+$$\begin{aligned}
+ \nabla \cross \vb{B}
+ = \mu_0 \vb{J}
+ \qquad \qquad
+ \vb{J} \cross \vb{B}
+ = \nabla p
+\end{aligned}$$
+
+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 $r$.
+Concretely, we have:
+
+$$\begin{aligned}
+ \vb{B}
+ = B_z(r) \: \vu{e}_z
+\end{aligned}$$
+
+Where $\vu{e}_z$ is the basis vector of the $z$-axis.
+This $\vb{B}$ confines the plasma thanks to
+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 $(r, \theta, z)$:
+
+$$\begin{aligned}
+ \vb{J}
+ = \frac{1}{\mu_0} \nabla \cross \vb{B}
+ = \frac{1}{\mu_0}
+ \begin{bmatrix}
+ \displaystyle \frac{1}{r} \pdv{B_z}{\theta} - \pdv{B_\theta}{z} \\
+ \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\
+ \displaystyle \frac{1}{r} \Big( \pdv{(r B_\theta)}{r} - \pdv{B_r}{\theta} \Big)
+ \end{bmatrix}
+ = -\frac{1}{\mu_0} \pdv{B_z}{r} \: \vu{e}_\theta
+\end{aligned}$$
+
+Where we have used that only $B_z$ is nonzero,
+and that it only depends on $r$.
+This yields a circular current parallel to $\vu{e}_\theta$,
+hence the name *ϴ-pinch*.
+
+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 $\vu{e}_\theta$:
+
+$$\begin{aligned}
+ \nabla p
+ &= \vb{J} \cross \vb{B}
+ = J_\theta \vu{e}_\theta \cross B_z \vu{e}_z
+ = J_\theta B_z \vu{e}_r
+ = - \frac{1}{\mu_0} \pdv{B_z}{r} B_z \: \vu{e}_r
+\end{aligned}$$
+
+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
+into the following equilibrium condition:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{}{r}\bigg( p + \frac{B_z^2}{2 \mu_0} \bigg)
+ = 0
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \vb{J}
+ = J_z(r) \: \vu{e}_z
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \vb{J}
+ = \frac{1}{\mu_0} \nabla \cross \vb{B}
+ = \frac{1}{\mu_0}
+ \begin{bmatrix}
+ \displaystyle \frac{1}{r} \pdv{B_z}{\theta} - \pdv{B_\theta}{z} \\
+ \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\
+ \displaystyle \frac{1}{r} \Big( \pdv{(r B_\theta)}{r} - \pdv{B_r}{\theta} \Big)
+ \end{bmatrix}
+ = \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} \: \vu{e}_z
+\end{aligned}$$
+
+Therefore, $\vb{J}$ induces a circular $\vb{B} = B_\theta(r) \: \vu{e}_\theta$,
+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:
+
+$$\begin{aligned}
+ \nabla p
+ &= \vb{J} \cross \vb{B}
+ = J_z \vb{e}_z \cross B_\theta \vb{e}_\theta
+ = - J_z B_\theta \vu{e}_r
+ = - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta \: \vu{e}_r
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{}{r}\bigg( p + \frac{B_\theta^2}{2 \mu_0} \bigg) + \frac{B_\theta^2}{\mu_0 r}
+ = 0
+ }
+\end{aligned}$$
+
+
+## Screw pinch
+
+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.
+By performing the above procedure again,
+the following equilibrium condition is obtained:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{}{r}\bigg( p + \frac{B_z^2}{2 \mu_0} + \frac{B_\theta^2}{2 \mu_0} \bigg) + \frac{B_\theta^2}{\mu_0 r}
+ = 0
+ }
+\end{aligned}$$
+
+Which simply combines the terms of the preceding equations.
+Indirectly, this result is relevant for certain types of nuclear fusion reactor,
+e.g. the tokamak, which basically consists of a screw pinch bent into a torus.
+The resulting equilibrium is given by
+the [Grad-Shafranov equation](/know/concept/grad-shafranov-equation/).
+
+
+
+## References
+1. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/second-quantization/index.md b/source/know/concept/second-quantization/index.md
new file mode 100644
index 0000000..0df70e0
--- /dev/null
+++ b/source/know/concept/second-quantization/index.md
@@ -0,0 +1,326 @@
+---
+title: "Second quantization"
+date: 2021-02-26
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+The **second quantization** is a technique to deal with quantum systems
+containing a large and/or variable number of identical particles.
+Its exact formulation depends on
+whether it is fermions or bosons that are being considered
+(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)).
+
+Regardless of whether the system is fermionic or bosonic,
+the idea is to change basis to a set of certain many-particle wave functions,
+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 $N$.
+
+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**
+and **particle annihilation operators**,
+which respectively add/remove a particle to/from a given state.
+In other words, these operators relate the Fock basis vectors
+to one another, and are very useful.
+
+The point is to express the system's state in such a way that the
+fermionic/bosonic constraints are automatically satisfied, and the
+formulae look the same regardless of the number of particles.
+
+
+## Fermions
+
+Fermions need to obey the Pauli exclusion principle, so each state can only
+contain one particle. In this case, the Fock states are given by:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ n &= 0:
+ \qquad \Ket{0, 0, 0, ...}
+ \\
+ n &= 1:
+ \qquad \Ket{1, 0, 0, ...} \quad \Ket{0, 1, 0, ...} \quad \Ket{0, 0, 1, ...} \quad \cdots
+ \\
+ n &= 2:
+ \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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 $c$
+in states 2, 4 and 5:
+
+$$\begin{aligned}
+ \Ket{0, 1, 0, 1, 1}
+ = \Psi(x_a, x_b, x_c)
+ = \frac{1}{\sqrt{3!}} \det\!
+ \begin{bmatrix}
+ \psi_2(x_a) & \psi_4(x_a) & \psi_5(x_a) \\
+ \psi_2(x_b) & \psi_4(x_b) & \psi_5(x_b) \\
+ \psi_2(x_c) & \psi_4(x_c) & \psi_5(x_c)
+ \end{bmatrix}
+\end{aligned}$$
+
+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 $\psi_\alpha$:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \hat{c}_\alpha^\dagger \Ket{... (N_\alpha\!=\!0) ...}
+ &= J_\alpha \Ket{... (N_\alpha\!=\!1) ...}
+ \\
+ \hat{c}_\alpha \Ket{... (N_\alpha\!=\!1) ...}
+ &= J_\alpha \Ket{... (N_\alpha\!=\!0) ...}
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ J_\alpha = (-1)^{\sum_{j < \alpha} N_j}
+\end{aligned}$$
+
+So, for example, when creating a particle in state 4
+of $\Ket{0, 1, 1, 0, 1}$, we get the following:
+
+$$\begin{aligned}
+ \hat{c}_4^\dagger \Ket{0, 1, 1, 0, 1}
+ = (-1)^{0 + 1 + 1} \Ket{0, 1, 1, 1, 1}
+\end{aligned}$$
+
+The point of the Jordan-Wigner string
+is that the order matters when applying the creation and annihilation operators:
+
+$$\begin{aligned}
+ \hat{c}_1^\dagger \hat{c}_2 \Ket{0, 1}
+ &= \hat{c}_1^\dagger \Ket{0, 0}
+ = \Ket{1, 0}
+ \\
+ \hat{c}_2 \hat{c}_1^\dagger \Ket{0, 1}
+ &= \hat{c}_2 \Ket{1, 1}
+ = - \Ket{1, 0}
+\end{aligned}$$
+
+In other words, $\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$,
+meaning that the anticommutator $\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$.
+You can verify for youself that
+the general anticommutators of these operators are given by:
+
+$$\begin{aligned}
+ \boxed{
+ \{\hat{c}_\alpha, \hat{c}_\beta\} = \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\} = 0
+ \qquad \quad
+ \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\} = \delta_{\alpha\beta}
+ }
+\end{aligned}$$
+
+Each single-particle state can only contain 0 or 1 fermions,
+so these operators **quench** states that would violate this rule.
+Note that these are *scalar* zeros:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{c}_\alpha^\dagger \Ket{... (N_\alpha\!=\!1) ...} = 0
+ \qquad \quad
+ \hat{c}_\alpha \Ket{... (N_\alpha\!=\!0) ...} = 0
+ }
+\end{aligned}$$
+
+Finally, as has already been suggested by the notation, they are each other's adjoint:
+
+$$\begin{aligned}
+ \matrixel{... (N_\alpha\!=\!1) ...}{\hat{c}_\alpha^\dagger}{... (N_\alpha\!=\!0) ...}
+ = \matrixel{...(N_\alpha\!=\!0) ...}{\hat{c}_\alpha}{... (N_\alpha\!=\!1) ...}
+\end{aligned}$$
+
+Let us now use these operators to define the **number operator** $\hat{N}_\alpha$ as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
+ }
+\end{aligned}$$
+
+Its eigenvalue is the number of particles residing in state $\psi_\alpha$
+(look at the hats):
+
+$$\begin{aligned}
+ \hat{N}_\alpha \Ket{... N_\alpha ...}
+ = N_\alpha \Ket{... N_\alpha ...}
+\end{aligned}$$
+
+
+## Bosons
+
+Bosons do not need to obey the Pauli exclusion principle, so multiple can occupy a single state.
+The Fock states are therefore as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ n &= 0:
+ \qquad \Ket{0, 0, 0, ...}
+ \\
+ n &= 1:
+ \qquad \Ket{1, 0, 0, ...} \quad \Ket{0, 1, 0, ...} \quad \Ket{0, 0, 1, ...} \quad \cdots
+ \\
+ n &= 2:
+ \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots
+ \\
+ &\qquad\:\:\:
+ \qquad \Ket{2, 0, 0, ...} \quad \Ket{0, 2, 0, ...} \quad \Ket{0, 0, 2, ...} \quad \cdots
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{gathered}
+ \boxed{
+ \begin{aligned}
+ \hat{c}_\alpha^\dagger \Ket{... N_\alpha ...}
+ &= \sqrt{N_\alpha + 1} \: \Ket{... (N_\alpha \!+\! 1) ...}
+ \\
+ \hat{c}_\alpha \Ket{... N_\alpha ...}
+ &= \sqrt{N_\alpha} \: \Ket{... (N_\alpha \!-\! 1) ...}
+ \end{aligned}
+}\end{gathered}$$
+
+Applying the annihilation operator $\hat{c}_\alpha$ when there are zero
+particles in $\alpha$ will quench the state:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{c}_\alpha \Ket{... (N_\alpha\!=\!0) ...} = 0
+ }
+\end{aligned}$$
+
+There is no Jordan-Wigner string, and therefore no sign change when commuting.
+Consequently, these operators therefore satisfy the following:
+
+$$\begin{aligned}
+ \boxed{
+ [\hat{c}_\alpha, \hat{c}_\beta] = [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] = 0
+ \qquad
+ [\hat{c}_\alpha, \hat{c}_\beta^\dagger] = \delta_{\alpha\beta}
+ }
+\end{aligned}$$
+
+The constant factors applied by $\hat{c}_\alpha^\dagger$ and $\hat{c}_\alpha$
+ensure that $\hat{N}_\alpha$ keeps the same nice form:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
+ }
+\end{aligned}$$
+
+
+## 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:
+
+$$\begin{aligned}
+ \hat{V}
+ = \sum_{n = 1}^N \hat{V}_1
+\end{aligned}$$
+
+This can be rewritten using the second quantization operators as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{V}
+ = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}_1}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta
+ }
+\end{aligned}$$
+
+Where the matrix element $\matrixel{\alpha}{\hat{V}_1}{\beta}$ is to be
+evaluated in the normal way:
+
+$$\begin{aligned}
+ \matrixel{\alpha}{\hat{V}_1}{\beta}
+ = \int \psi_\alpha^*(\vec{r}) \: \hat{V}_1(\vec{r}) \: \psi_\beta(\vec{r}) \dd{\vec{r}}
+\end{aligned}$$
+
+Similarly, given some two-particle operator $\hat{V}$ in first-quantized form:
+
+$$\begin{aligned}
+ \hat{V}
+ = \sum_{n \neq m} v(\vec{r}_n, \vec{r}_m)
+\end{aligned}$$
+
+We can rewrite this in second-quantized form as follows.
+Note the ordering of the subscripts:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{V}
+ = \sum_{\alpha, \beta, \gamma, \delta}
+ v_{\alpha \beta \gamma \delta} \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma
+ }
+\end{aligned}$$
+
+Where the constant $v_{\alpha \beta \gamma \delta}$ is defined from the
+single-particle wave functions:
+
+$$\begin{aligned}
+ v_{\alpha \beta \gamma \delta}
+ = \iint \psi_\alpha^*(\vec{r}_1) \: \psi_\beta^*(\vec{r}_2)
+ \: v(\vec{r}_1, \vec{r}_2) \: \psi_\gamma(\vec{r}_1)
+ \: \psi_\delta(\vec{r}_2) \dd{\vec{r}_1} \dd{\vec{r}_2}
+\end{aligned}$$
+
+Finally, in the second quantization, changing basis is done in the usual way:
+
+$$\begin{aligned}
+ \hat{c}_b^\dagger \Ket{0}
+ = \Ket{b}
+ = \sum_{\alpha} \Ket{\alpha} \Inprod{\alpha}{b}
+ = \sum_{\alpha} \Inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0}
+\end{aligned}$$
+
+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 $\vec{r}$:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{\Psi}^\dagger(\vec{r})
+ = \sum_{\alpha} \Inprod{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger
+ \qquad \quad
+ \hat{\Psi}(\vec{r})
+ = \sum_{\alpha} \Inprod{\vec{r}}{\alpha} \hat{c}_\alpha
+ }
+\end{aligned}$$
+
+
+## References
+1. L.E. Ballentine,
+ *Quantum mechanics: a modern development*, 2nd edition,
+ World Scientific.
diff --git a/source/know/concept/selection-rules/index.md b/source/know/concept/selection-rules/index.md
new file mode 100644
index 0000000..2ce5748
--- /dev/null
+++ b/source/know/concept/selection-rules/index.md
@@ -0,0 +1,698 @@
+---
+title: "Selection rules"
+date: 2021-06-02
+categories:
+- Physics
+- Quantum mechanics
+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:
+
+$$\begin{aligned}
+ \matrixel{f}{\hat{O}}{i}
+ = \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
+\end{aligned}$$
+
+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 $\Ket{f}$
+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,
+guarantee that the above matrix element is zero.
+
+
+## Parity rules
+
+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 $\Ket{\ell_i m_i}$:
+
+$$\begin{aligned}
+ \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
+ &= - \matrixel{\ell_f m_f}{\hat{\Pi}^\dagger \hat{O} \hat{\Pi}}{\ell_i m_i}
+ \\
+ &= - \matrixel{\ell_f m_f}{(-1)^{\ell_f} \hat{O} (-1)^{\ell_i}}{\ell_i m_i}
+ \\
+ &= (-1)^{\ell_f + \ell_i + 1} \matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i}
+\end{aligned}$$
+
+Which clearly can only be true if the exponent is even,
+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**:
+
+$$\begin{aligned}
+ \boxed{
+ \Delta \ell \:\:\text{is odd}
+ }
+\end{aligned}$$
+
+If this is not the case,
+then the only possible way that the above equation can be satisfied
+is if the matrix element vanishes $\matrixel{\ell_f m_f}{\hat{O}}{\ell_i m_i} = 0$.
+We can derive an analogous rule for
+any operator $\hat{E}$ which is even under parity:
+
+$$\begin{aligned}
+ \hat{\Pi}^\dagger \hat{E} \hat{\Pi} = \hat{E}
+ \quad \implies \quad
+ \boxed{
+ \Delta \ell \:\:\text{is even}
+ }
+\end{aligned}$$
+
+
+## Dipole rules
+
+Arguably the most common operator found in such matrix elements
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \Delta m = 0 \:\:\mathrm{or}\: \pm 1
+ }
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+We know that the angular momentum $z$-component operator $\hat{L}_z$ satisfies:
+
+$$\begin{aligned}
+ \comm{\hat{L}_z}{\hat{x}} = i \hbar \hat{y}
+ \qquad
+ \comm{\hat{L}_z}{\hat{y}} = - i \hbar \hat{x}
+ \qquad
+ \comm{\hat{L}_z}{\hat{z}} = 0
+\end{aligned}$$
+
+We take the first relation,
+and wrap it in $\Bra{\ell_f m_f}$ and $\Ket{\ell_i m_i}$, giving:
+
+$$\begin{aligned}
+ i \hbar \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+ &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{x}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{x} \hat{L}_z}{\ell_i m_i}
+ \\
+ &= \hbar m_f \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
+ \\
+ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
+\end{aligned}$$
+
+Next, we do the same thing with the second relation, for $[\hat{L}_z, \hat{y}]$, giving:
+
+$$\begin{aligned}
+ - i \hbar \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
+ &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{y}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{y} \hat{L}_z}{\ell_i m_i}
+ \\
+ &= \hbar m_f \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+ \\
+ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+\end{aligned}$$
+
+Respectively isolating the two above results for $\hat{x}$ and $\hat{y}$,
+we arrive at these equations:
+
+$$\begin{aligned}
+ \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
+ &= i (m_f - m_i) \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+ \\
+ \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+ &= - i (m_f - m_i) \matrixel{\ell_f m_f}{\hat{x}}{\ell_i m_i}
+\end{aligned}$$
+
+By inserting the first into the second,
+we find (part of) the selection rule:
+
+$$\begin{aligned}
+ \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+ &= (m_f - m_i)^2 \matrixel{\ell_f m_f}{\hat{y}}{\ell_i m_i}
+\end{aligned}$$
+
+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.
+Assuming the latter, we take the inner product of
+the commutator $\comm{\hat{L}_z}{\hat{z}} = 0$, and find:
+
+$$\begin{aligned}
+ 0
+ &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{z}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{z} \hat{L}_z}{\ell_i m_i}
+ \\
+ &= \hbar m_f \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i} - \hbar m_i \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i}
+ \\
+ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{z}}{\ell_i m_i}
+\end{aligned}$$
+
+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.
+
+
+
+
Proof
+
+
Proof.
+We start from the following relation
+(which is already quite a chore to prove):
+
+$$\begin{aligned}
+ \Comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}
+ = 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r})
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+To begin with, we want to find the commutator of $\hat{L}^2$ and $\hat{x}$:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\hat{x}}
+ &= \comm{\hat{L}_x^2}{\hat{x}} + \comm{\hat{L}_y^2}{\hat{x}} + \comm{\hat{L}_z^2}{\hat{x}}
+ = \comm{\hat{L}_y^2}{\hat{x}} + \comm{\hat{L}_z^2}{\hat{x}}
+ \\
+ &= \hat{L}_y \comm{\hat{L}_y}{\hat{x}} + \comm{\hat{L}_y}{\hat{x}} \hat{L}_y
+ + \hat{L}_z \comm{\hat{L}_z}{\hat{x}} + \comm{\hat{L}_z}{\hat{x}} \hat{L}_z
+\end{aligned}$$
+
+Evaluating these commutators gives us:
+
+$$\begin{aligned}
+ \comm{\hat{L}_y}{\hat{x}}
+ &= \comm{\hat{z} \hat{p}_x}{\hat{x}} - \comm{\hat{x} \hat{p}_z}{\hat{x}}
+ = \hat{z} \comm{\hat{p}_x}{\hat{x}} + \comm{\hat{z}}{\hat{x}} \hat{p}_x
+ - \hat{x} \comm{\hat{p}_z}{\hat{x}} - \comm{\hat{x}}{\hat{x}} \hat{p}_z
+ = - i \hbar \hat{z}
+ \\
+ \comm{\hat{L}_z}{\hat{x}}
+ &= \comm{\hat{x} \hat{p}_y}{\hat{x}} - \comm{\hat{y} \hat{p}_x}{\hat{x}}
+ = \hat{x} \comm{\hat{p}_y}{\hat{x}} + \comm{\hat{x}}{\hat{x}} \hat{p}_y
+ - \hat{y} \comm{\hat{p}_x}{\hat{x}} - \comm{\hat{y}}{\hat{x}} \hat{p}_x
+ = i \hbar \hat{y}
+\end{aligned}$$
+
+Which we then insert back into the original equation, yielding:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\hat{x}}
+ &= i \hbar (- \hat{L}_y \hat{z} - \hat{z} \hat{L}_y + \hat{L}_z \hat{y} + \hat{y} \hat{L}_z)
+\end{aligned}$$
+
+This can be simplified by introducing some more commutators:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\hat{x}}
+ &= i \hbar \big( \!-\! ( \comm{\hat{L}_y}{\hat{z}} + \hat{z} \hat{L}_y ) - \hat{z} \hat{L}_y
+ + ( \comm{\hat{L}_z}{\hat{y}} + \hat{y} \hat{L}_z ) + \hat{y} \hat{L}_z \big)
+\end{aligned}$$
+
+Evaluating these commutators gives us:
+
+$$\begin{aligned}
+ \comm{\hat{L}_y}{\hat{z}}
+ &= \comm{\hat{z} \hat{p}_x}{\hat{z}} - \comm{\hat{x} \hat{p}_z}{\hat{z}}
+ = \hat{z} \comm{\hat{p}_x}{\hat{z}} + \comm{\hat{z}}{\hat{z}} \hat{p}_x
+ - \hat{x} \comm{\hat{p}_z}{\hat{z}} - \comm{\hat{x}}{\hat{z}} \hat{p}_z
+ = i \hbar \hat{x}
+ \\
+ \comm{\hat{L}_z}{\hat{y}}
+ &= \comm{\hat{x} \hat{p}_y}{\hat{y}} - \comm{\hat{y} \hat{p}_x}{\hat{y}}
+ = \hat{x} \comm{\hat{p}_y}{\hat{y}} + \comm{\hat{x}}{\hat{y}} \hat{p}_y
+ - \hat{y} \comm{\hat{p}_x}{\hat{y}} - \comm{\hat{y}}{\hat{y}} \hat{p}_x
+ = - i \hbar \hat{x}
+\end{aligned}$$
+
+Substituting these then leads us to the first milestone of this proof:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\hat{x}}
+ &= i \hbar \big( \!-\! i \hbar \hat{x} - \hat{z} \hat{L}_y - \hat{z} \hat{L}_y
+ - i \hbar \hat{x} + \hat{y} \hat{L}_z + \hat{y} \hat{L}_z \big)
+ \\
+ &= 2 i \hbar (\hat{y} \hat{L}_z - \hat{z} \hat{L}_y - i \hbar \hat{x})
+\end{aligned}$$
+
+Repeating this process for $\comm{\hat{L}^2}{\hat{y}}$ and $\comm{\hat{L}^2}{\hat{z}}$,
+we find analogous expressions:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\hat{y}}
+ &= 2 i \hbar (\hat{z} \hat{L}_x - \hat{x} \hat{L}_z - i \hbar \hat{y})
+ \\
+ \comm{\hat{L}^2}{\hat{z}}
+ &= 2 i \hbar (\hat{x} \hat{L}_y - \hat{y} \hat{L}_x - i \hbar \hat{z})
+\end{aligned}$$
+
+Next, we take the commutator with $\hat{L}^2$ of the commutator we just found:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
+ &= 2 i \hbar \big(\comm{\hat{L}^2}{\hat{y} \hat{L}_z} - \comm{\hat{L}^2}{\hat{z} \hat{L}_y} - i \hbar \comm{\hat{L}^2}{\hat{x}}\big)
+ \\
+ &= 2 i \hbar \big( \hat{y} \comm{\hat{L}^2}{\hat{L}_z} + \comm{\hat{L}^2}{\hat{y}} \hat{L}_z
+ - \hat{z} \comm{\hat{L}^2}{\hat{L}_y} - \comm{\hat{L}^2}{\hat{z}} \hat{L}_y
+ - i \hbar \comm{\hat{L}^2}{\hat{x}} \big)
+\end{aligned}$$
+
+Where we used that $\comm{\hat{L}^2}{\hat{L}_y} = \comm{\hat{L}^2}{\hat{L}_z} = 0$.
+The other commutators look familiar:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
+ &= 2 i \hbar \big( \comm{\hat{L}^2}{\hat{y}} \hat{L}_z
+ - \comm{\hat{L}^2}{\hat{z}} \hat{L}_y
+ - i \hbar \comm{\hat{L}^2}{\hat{x}} \big)
+\end{aligned}$$
+
+By inserting the expressions we found earlier for these commutators, we get:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
+ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z - \hat{x} \hat{L}_z^2 - i \hbar \hat{y} \hat{L}_z
+ + \hat{y} \hat{L}_x \hat{L}_y - \hat{x} \hat{L}_y^2 + i \hbar \hat{z} \hat{L}_y \big) \\
+ &\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
+\end{aligned}$$
+
+Substituting the well-known commutators
+$i \hbar \hat{L}_y = \comm{\hat{L}_z}{\hat{L}_x}$ and
+$i \hbar \hat{L}_z = \comm{\hat{L}_x}{\hat{L}_y}$:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
+ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_x \hat{L}_z + \hat{y} \hat{L}_x \hat{L}_y
+ - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2
+ + \hat{z} \comm{\hat{L}_z}{\hat{L}_x} - \hat{y} \comm{\hat{L}_x}{\hat{L}_y} \big) \\
+ &\qquad\qquad + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
+ \\
+ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x
+ - \hat{x} \hat{L}_y^2 - \hat{x} \hat{L}_z^2 \big)
+ + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
+\end{aligned}$$
+
+By definition, $\hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 = \hat{L}^2$,
+which we use to arrive at:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
+ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 - \hat{x} \hat{L}^2 \big)
+ + 2 \hbar^2 \big( \hat{L}^2 \hat{x} - \hat{x} \hat{L}^2 \big)
+ \\
+ &= - 4 \hbar^2 \big( \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2 \big)
+ + 2 \hbar^2 \big( \hat{L}^2 \hat{x} + \hat{x} \hat{L}^2 \big)
+\end{aligned}$$
+
+The second term is what we want to prove,
+so the first term must vanish:
+
+$$\begin{aligned}
+ \hat{z} \hat{L}_z \hat{L}_x + \hat{y} \hat{L}_y \hat{L}_x + \hat{x} \hat{L}_x^2
+ = (\vu{r} \cdot \vu{L}) \hat{L}_x
+ = (\vu{r} \cdot (\vu{r} \cross \vu{p})) \hat{L}_x
+ = (\vu{p} \cdot (\vu{r} \cross \vu{r})) \hat{L}_x
+ = 0
+\end{aligned}$$
+
+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
+$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}$,
+leading us to:
+
+$$\begin{aligned}
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{x}}}
+ &= 2 \hbar^2 (\hat{x} \hat{L}^2 + \hat{L}^2 \hat{x})
+ \\
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{y}}}
+ &= 2 \hbar^2 (\hat{y} \hat{L}^2 + \hat{L}^2 \hat{y})
+ \\
+ \comm{\hat{L}^2}{\comm{\hat{L}^2}{\hat{z}}}
+ &= 2 \hbar^2 (\hat{z} \hat{L}^2 + \hat{L}^2 \hat{z})
+\end{aligned}$$
+
+At last, this brings us to the desired equation for $\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}$,
+with $\vu{r} = (\hat{x}, \hat{y}, \hat{z})$.
+
+
+
+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:
+
+$$\begin{aligned}
+ 2 \hbar^2 \matrixel{f}{\vu{r} \hat{L}^2 \!\!+\! \hat{L}^2 \!\vu{r}}{i}
+ &= 2 \hbar^2 \big( \matrixel{f}{\vu{r} \hat{L}^2}{i} + \matrixel{f}{\hat{L}^2 \vu{r}}{i} \big)
+ \\
+ &= 2 \hbar^2 \big( \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\vu{r}}{i}
+ + \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\vu{r}}{i} \big)
+ \\
+ &= 2 \hbar^4 \big(\ell_f (\ell_f \!+\! 1) + \ell_i (\ell_i \!+\! 1)\big) \matrixel{f}{\vu{r}}{i}
+\end{aligned}$$
+
+And, likewise, the left-hand side becomes:
+
+$$\begin{aligned}
+ \matrixel{f}{\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}}{i}
+ &= \matrixel{f}{\hat{L}^2 \comm{\hat{L}^2}{\vu{r}}}{i}
+ - \matrixel{f}{\comm{\hat{L}^2}{\vu{r}} \hat{L}^2}{i}
+ \\
+ &= \hbar^2 \ell_f (\ell_f \!+\! 1) \matrixel{f}{\comm{\hat{L}^2}{\vu{r}}}{i}
+ - \hbar^2 \ell_i (\ell_i \!+\! 1) \matrixel{f}{\comm{\hat{L}^2}{\vu{r}}}{i}
+ \\
+ &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{f}{\comm{\hat{L}^2}{\vu{r}}}{i}
+ \\
+ &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)
+ \big( \matrixel{f}{\hat{L}^2 \vu{r}}{i} - \matrixel{f}{\vu{r} \hat{L}^2}{i} \big)
+ \\
+ &= \hbar^4 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2 \matrixel{f}{\vu{r}}{i}
+\end{aligned}$$
+
+Obviously, both sides are equal to each other,
+leading to the following equation:
+
+$$\begin{aligned}
+ 2 \ell_f (\ell_f \!+\! 1) + 2 \ell_i (\ell_i \!+\! 1)
+ &= \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2
+\end{aligned}$$
+
+To proceed, we rewrite the right-hand side like so:
+
+$$\begin{aligned}
+ \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big)^2
+ &= \big( \ell_f^2 - \ell_i^2 + \ell_f - \ell_i \big)^2
+ \\
+ &= \big( (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i) \big)^2
+ \\
+ &= (\ell_f + \ell_i)^2 (\ell_f - \ell_i)^2 + 2 (\ell_f + \ell_i) (\ell_f - \ell_i)^2 + (\ell_f - \ell_i)^2
+ \\
+ &= \big( (\ell_f + \ell_i)^2 + 2 (\ell_f + \ell_i) + 1 \big) (\ell_f - \ell_i)^2
+ \\
+ &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2
+\end{aligned}$$
+
+And then we do the same to the left-hand side, yielding:
+
+$$\begin{aligned}
+ 2 (\ell_f^2 + \ell_i^2 + \ell_f + \ell_i)
+ &= 2 \ell_f^2 + 2 \ell_i^2 + 2 \ell_f \ell_i - 2 \ell_f \ell_i + 2 \ell_f + 2 \ell_i + 1 - 1
+ \\
+ &= (\ell_f + \ell_i + 1)^2 + \ell_f^2 + \ell_i^2 - 2 \ell_f \ell_i - 1
+ \\
+ &= (\ell_f + \ell_i + 1)^2 + (\ell_f - \ell_i)^2 - 1
+\end{aligned}$$
+
+The equation above has thus been simplified to the following form:
+
+$$\begin{aligned}
+ (\ell_f + \ell_i + 1)^2 + (\ell_f - \ell_i)^2 - 1
+ &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2
+\end{aligned}$$
+
+Rearranging yields a product equal to zero,
+so one or both of the factors must vanish:
+
+$$\begin{aligned}
+ 0
+ &= (\ell_f + \ell_i + 1)^2 (\ell_f - \ell_i)^2 - (\ell_f + \ell_i + 1)^2 - (\ell_f - \ell_i)^2 + 1
+ \\
+ &= \big( (\ell_f + \ell_i + 1)^2 - 1 \big) \big( (\ell_f - \ell_i)^2 - 1 \big)
+\end{aligned}$$
+
+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 other, non-trivial option is therefore:
+
+$$\begin{aligned}
+ (\ell_f - \ell_i)^2
+ = 1
+\end{aligned}$$
+
+
+
+
Proof
+
+Proof.
+Firstly, we look at the commutator of $\hat{s}$ with the $z$-component $\hat{L}_z$:
+$$\begin{aligned}
+ 0
+ = \matrixel{\ell_f m_f}{\comm{\hat{L}_z}{\hat{s}}}{\ell_i m_i}
+ &= \matrixel{\ell_f m_f}{\hat{L}_z \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_z}{\ell_i m_i}
+ \\
+ &= \hbar (m_f - m_i) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+\end{aligned}$$
+
+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.
+
+Secondly, we look at the commutator of $\hat{s}$ with the total angular momentum $\hat{L}^2$:
+
+$$\begin{aligned}
+ 0
+ = \matrixel{\ell_f m_f}{\comm{\hat{L}^2}{\hat{s}}}{\ell_i m_i}
+ &= \matrixel{\ell_f m_f}{\hat{L}^2 \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}^2}{\ell_i m_i}
+ \\
+ &= \hbar^2 \big( \ell_f (\ell_f \!+\! 1) - \ell_i (\ell_i \!+\! 1) \big) \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+\end{aligned}$$
+
+Assuming $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i} \neq 0$,
+this can only be satisfied if the following holds:
+
+$$\begin{aligned}
+ 0
+ = \ell_f^2 + \ell_f - \ell_i^2 - \ell_i
+ = (\ell_f + \ell_i) (\ell_f - \ell_i) + (\ell_f - \ell_i)
+\end{aligned}$$
+
+If $\ell_f = \ell_i = 0$ this equation is trivially satisfied.
+Otherwise, the only option is $\ell_f \!-\! \ell_i = 0$,
+which is another part of the selection rule.
+
+Thirdly, we look at the commutator of $\hat{s}$ with the ladder operators $\hat{L}_\pm$:
+
+$$\begin{aligned}
+ 0
+ = \matrixel{\ell_f m_f}{\comm{\hat{L}_\pm}{\hat{s}}}{\ell_i m_i}
+ &= \matrixel{\ell_f m_f}{\hat{L}_\pm \hat{s}}{\ell_i m_i} - \matrixel{\ell_f m_f}{\hat{s} \hat{L}_\pm}{\ell_i m_i}
+ \\
+ &= C_f \matrixel{\ell_f (m_f\!\mp\!1)}{\hat{s}}{\ell_i m_i} - C_i \matrixel{\ell_f m_f}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
+\end{aligned}$$
+
+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 $m_f = m_i \pm 1$.
+Therefore:
+
+$$\begin{aligned}
+ C_i
+ &= \hbar \sqrt{\ell_i (\ell_i + 1) - m_i (m_i \pm 1)}
+ \\
+ C_f
+ &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_f (m_f \!\mp\! 1)}
+ \\
+ &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - (m_i \!\pm\! 1) (m_i \!\pm\! 1 \!\mp\! 1)}
+ \\
+ &= \hbar \sqrt{\ell_f (\ell_f \!+\! 1) - m_i (m_i \!\pm\! 1)}
+\end{aligned}$$
+
+In other words, $C_f = C_i$. The above equation therefore reduces to:
+
+$$\begin{aligned}
+ \matrixel{\ell_f m_i}{\hat{s}}{\ell_i m_i}
+ &= \matrixel{\ell_f (m_i \!\pm\! 1)}{\hat{s}}{\ell_i (m_i\!\pm\!1)}
+\end{aligned}$$
+
+Which means that the value of the matrix element
+does not depend on $m_i$ (or $m_f$) at all.
+
+
+
+
+A **reducible diagram** is a Feynman diagram
+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 $b$:
+
+
+
+
+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,
+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 $b$.
+
+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 $\Sigma(y, x)$:
+
+
+
+
+This makes sense: in the "normal" Dyson equation
+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:
+
+$$\begin{aligned}
+ \boxed{
+ G(b, a)
+ = G^0(b, a) + \iint G^0(b, y) \: \Sigma(y, x) \: G(x, a) \dd{x} \dd{y}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ G_{s}(\tilde{\vb{k}})
+ = G_{s}^0(\tilde{\vb{k}}) + G_{s}^0(\tilde{\vb{k}}) \: \Sigma_{s}(\tilde{\vb{k}}) \: G_{s}(\tilde{\vb{k}})
+ }
+\end{aligned}$$
+
+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 $\omega_n$,
+together with the linear structure of the Dyson equation,
+makes $\Sigma$ diagonal in all of those quantities.
+Isolating for $G$:
+
+$$\begin{aligned}
+ G_{s}(\tilde{\vb{k}})
+ = \frac{G_{s}^0(\tilde{\vb{k}})}{1 - G_{s}^0(\tilde{\vb{k}}) \: \Sigma_{s}(\tilde{\vb{k}})}
+ = \frac{1}{1 / G_{s}^0(\tilde{\vb{k}}) - \Sigma_{s}(\tilde{\vb{k}})}
+\end{aligned}$$
+
+From [equation-of-motion theory](/know/concept/equation-of-motion-theory/),
+we already know an expression for $G$ in diagonal $\vb{k}$-space:
+
+$$\begin{aligned}
+ G_s^0(\vb{k}, i \omega_n)
+ = \frac{1}{i \hbar \omega_n - \varepsilon_\vb{k}}
+ \quad \implies \quad
+ G_{s}(\vb{k}, i \omega_n)
+ = \frac{1}{i \hbar \omega_n - \varepsilon_\vb{k} - \Sigma_{s}(\vb{k}, i \omega_n)}
+\end{aligned}$$
+
+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
+in the translationally-invariant example above;
+in fact, it does not even need to be Hermitian,
+i.e. $\Sigma(y,x) \neq \Sigma^*(x,y)$,
+in which case it resists the standard techniques for analysis.
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
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+---
+title: "Self-phase modulation"
+date: 2021-02-26
+categories:
+- Physics
+- Optics
+- Fiber optics
+- Nonlinear optics
+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,
+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
+control dispersion and nonlinearity:
+
+$$\begin{aligned}
+ 0
+ = i \pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma |A|^2 A
+\end{aligned}$$
+
+By setting $\beta_2 = 0$ to neglect dispersion,
+solving this equation becomes trivial.
+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,
+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** $L_N$,
+which gives the distance after which the phase of the
+peak has increased by exactly 1 radian:
+
+$$\begin{aligned}
+ \gamma P_0 L_N = 1
+ \qquad \implies \qquad
+ \boxed{
+ L_N = \frac{1}{\gamma P_0}
+ }
+\end{aligned}$$
+
+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 $\gamma = 0.1/\mathrm{W}/\mathrm{m}$:
+
+$$\begin{aligned}
+ A(0, t)
+ = \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big)
+\end{aligned}$$
+
+From earlier, we then know the analytical solution for the $z$-evolution:
+
+$$\begin{aligned}
+ A(z, t) = \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big) \exp\!\bigg( i \gamma z P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
+\end{aligned}$$
+
+
+
+
+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 $A(z, t) = \sqrt{P(z, t)} \exp(i \phi(z, t))$:
+
+$$\begin{aligned}
+ \omega_{\mathrm{SPM}}(z,t)
+ = - \pdv{\phi}{t}
+ = 2 \gamma z P_0 \frac{t}{T_0^2} \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big)
+\end{aligned}$$
+
+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 interaction between self-phase modulation
+and [dispersion](/know/concept/dispersive-broadening/)
+leads to many interesting effects,
+such as [modulational instability](/know/concept/modulational-instability/)
+and [optical wave breaking](/know/concept/optical-wave-breaking/).
+
+
+
+## References
+1. O. Bang,
+ *Numerical methods in photonics: lecture notes*, 2019,
+ unpublished.
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diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
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+---
+title: "Self-steepening"
+date: 2021-02-26
+categories:
+- Physics
+- Optics
+- Fiber optics
+- Nonlinear optics
+layout: "concept"
+---
+
+For a laser pulse travelling through an optical fiber,
+its intensity is highest at its peak, so the Kerr effect will be strongest there.
+This means that the peak travels slightly slower
+than the rest of the pulse, leading to **self-steepening** of its trailing edge.
+Mathematically, this is described by adding a new term to the
+nonlinear Schrödinger equation:
+
+$$\begin{aligned}
+ 0
+ = i\pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{}{t} \Big) \big(|A|^2 A\big)
+\end{aligned}$$
+
+Where $\omega_0$ is the angular frequency of the pump.
+We will use the following ansatz,
+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 $\varepsilon = \gamma / \omega_0$:
+
+$$\begin{aligned}
+ 0 &= i \frac{1}{2} \frac{P_z}{\sqrt{P}} - \sqrt{P} \phi_z + \gamma P \sqrt{P} + i \varepsilon \frac{3}{2} P_t \sqrt{P} - \varepsilon P \sqrt{P} \phi_t
+\end{aligned}$$
+
+This results in two equations, respectively corresponding to the real
+and imaginary parts:
+
+$$\begin{aligned}
+ 0 &= - \phi_z - \varepsilon P \phi_t + \gamma P
+ \\
+ 0 &= P_z + \varepsilon 3 P_t P
+\end{aligned}$$
+
+The phase $\phi$ is not so interesting, so we focus on the latter equation for $P$.
+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:
+
+$$\begin{aligned}
+ P(z,t) = f(t - 3 \varepsilon z P)
+\end{aligned}$$
+
+Where $f$ is the initial power profile: $f(t) = P(0,t)$.
+The derivatives $P_t$ and $P_z$ are then given by:
+
+$$\begin{aligned}
+ P_t
+ &= (1 - 3 \varepsilon z P_t) \: f'
+ \qquad \quad \implies \quad
+ P_t
+ = \frac{f'}{1 + 3 \varepsilon z f'}
+ \\
+ P_z
+ &= (-3 \varepsilon P - 3 \varepsilon z P_z) \: f'
+ \quad \implies \quad
+ P_z
+ = \frac{- 3 \varepsilon P f'}{1 + 3 \varepsilon z f'}
+\end{aligned}$$
+
+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 is located on the trailing edge of the pulse.
+At the propagation distance where this occurs, $L_\mathrm{shock}$,
+the pulse will "tip over", creating a discontinuous shock:
+
+$$\begin{aligned}
+ \boxed{
+ L_\mathrm{shock} = -\frac{1}{3 \varepsilon f_\mathrm{min}'}
+ }
+\end{aligned}$$
+
+In practice, however, this will never actually happen, because by the time
+$L_\mathrm{shock}$ is reached, the pulse spectrum will have become so
+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 $\gamma = 0.1/\mathrm{W}/\mathrm{m}$:
+
+$$\begin{aligned}
+ f(t) = P(0,t) = P_0 \exp\!\Big(\! -\!\frac{t^2}{T_0^2} \Big)
+\end{aligned}$$
+
+
+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:
+
+$$\begin{aligned}
+ f_\mathrm{min}' = - \frac{\sqrt{2} P_0}{T_0} \exp\!\Big(\!-\!\frac{1}{2}\Big)
+ \quad \implies \quad
+ L_\mathrm{shock} = \frac{T_0}{3 \sqrt{2} \varepsilon P_0} \exp\!\Big(\frac{1}{2}\Big)
+\end{aligned}$$
+
+This example Gaussian pulse therefore has a theoretical
+$L_\mathrm{shock} = 0.847\,\mathrm{m}$,
+which turns out to be accurate,
+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
+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:
+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:
+
+$$\begin{aligned}
+ \boxed{
+ N_\mathrm{ph}(z) = \int_0^\infty \frac{|\tilde{A}(z,\omega)|^2}{\omega} \dd{\omega}
+ }
+\end{aligned}$$
+
+
+## References
+1. B.R. Suydam, [Self-steepening of optical pulses](https://doi.org/10.1007/0-387-25097-2_6), 2006, Springer.
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diff --git a/source/know/concept/shors-algorithm/index.md b/source/know/concept/shors-algorithm/index.md
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+---
+title: "Shor's algorithm"
+date: 2021-04-13
+categories:
+- Quantum information
+- Cryptography
+- Algorithms
+layout: "concept"
+---
+
+**Shor's algorithm** was the first truly useful quantum algorithm.
+It can solve important problems,
+most notably integer factorization,
+much more efficiently than any classical algorithm.
+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:
+
+$$\begin{aligned}
+ f(x_1, ..., x_A)
+ = f(x_1 + s_1, ..., x_A + s_A)
+\end{aligned}$$
+
+This is a so-called *hidden subgroup problem* for a *finite Abelian group*.
+With minimal modifications,
+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):
+
+$$\begin{aligned}
+ f(x)
+ = a^x \bmod N
+\end{aligned}$$
+
+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 $N^2 \le 2^q < 2 N^2$:
+
+
+
+
+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 $N$.
+It is an XOR oracle, working as follows:
+
+$$\begin{aligned}
+ \Ket{x} \Ket{y}
+ \quad \to \boxed{U_f(a, N)} \to \quad
+ \Ket{x} \Ket{y \oplus f(x)}
+\end{aligned}$$
+
+Execution starts by applying the [Hadamard gate](/know/concept/quantum-gate/) $H$
+to the first $q$ qubits, yielding:
+
+$$\begin{aligned}
+ \Ket{0}^{\otimes q} \Ket{0}^{\otimes q}
+ \quad \to \boxed{H^{\otimes q}} \to \quad
+ \Ket{+}^{\otimes q} \Ket{0}^{\otimes q}
+ = \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
+\end{aligned}$$
+
+Where $Q = 2^q$, and $\Ket{x}$ is the computational basis state $\Ket{x_1} \cdots \Ket{x_q}$.
+Moving on to $U_f$:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
+ \quad \to \boxed{U_f(a, N)} \to \quad
+ \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{f(x)}
+\end{aligned}$$
+
+Then we measure $f(x)$, causing it collapse as follows,
+for an unknown arbitrary value of $x_0$:
+
+$$\begin{aligned}
+ f(x_0) = f(x_0 + s) = f(x_0 + 2s) = \cdots = f(x_0 + (L-1) s)
+\end{aligned}$$
+
+Due to [entanglement](/know/concept/quantum-entanglement/),
+the unmeasured (top $q$) qubits change state into a superposition:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{L}} \sum_{\ell = 0}^{L - 1} \Ket{x_0 + \ell s}
+\end{aligned}$$
+
+Clearly, there is a periodic structure here,
+but we cannot measure it directly,
+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:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{QL}} \sum_{k = 0}^{Q - 1} \bigg( \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg) \Ket{k}
+\end{aligned}$$
+
+Where $\omega_Q$ is a $Q$th root of unity.
+Measuring this state yields a $\Ket{k}$, with a probability $P(k)$:
+
+$$\begin{aligned}
+ P(k)
+ = \frac{1}{QL} \bigg| \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg|^2
+ = \frac{1}{QL} \bigg| \omega_Q^{x_0 k} \sum_{\ell = 0}^{L - 1} \omega_Q^{\ell s k} \bigg|^2
+ = \frac{1}{QL} \bigg| \sum_{\ell = 0}^{L - 1} \omega_Q^{\ell s k} \bigg|^2
+\end{aligned}$$
+
+The last step holds because $|\omega_Q| = 1$.
+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 $s$?
+
+So, what does the above probability $P(k)$ work out to?
+There are two cases:
+
+$$\begin{alignedat}{2}
+ &\mathrm{if} \: \omega_Q^{sk} = 1: \qquad
+ &&P(k) = \frac{1}{QL} |L|^2 = \frac{L}{Q}
+ \\
+ &\mathrm{if} \: \omega_Q^{sk} \neq 1: \qquad
+ &&P(k) = \frac{1}{QL} \Bigg| \frac{1 - \omega_Q^{sk L}}{1 - \omega_Q^{sk}} \Bigg|^2
+\end{alignedat}$$
+
+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 $c$.
+
+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 $P(k) = L / Q$,
+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:
+
+$$\begin{aligned}
+ \mathrm{if} \: (Q/s) \in \mathbb{N}: \qquad
+ P(k)
+ = \frac{L}{Q}
+ = \frac{1}{s}
+\end{aligned}$$
+
+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,
+leaving only the "good" ones.
+
+**So, to summarize: if** $Q/s$ **is an integer**,
+then measuring only yields $k$-values that are multiples of $L\!=\!Q/s$.
+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 $s = Q/L$.
+
+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:
+
+$$\begin{aligned}
+ \mathrm{if} \: (Q/s) \not\in \mathbb{N}: \qquad
+ P(k)
+ = \frac{1}{QL} \Bigg| \frac{1 - \omega_Q^{sk L}}{1 - \omega_Q^{sk}} \Bigg|^2
+ = \frac{1}{QL} \Bigg| \frac{\sin(\pi s k L / Q)}{\sin(\pi s k / Q)} \Bigg|^2
+\end{aligned}$$
+
+This function peaks if $s k$ is close to a multiple of $Q$, i.e. $s k \approx c Q$,
+which we rearrange:
+
+$$\begin{aligned}
+ \frac{k}{Q} \approx \frac{c}{s}
+\end{aligned}$$
+
+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 $k < Q$.
+
+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 $c/s\!=\!1/3$.
+
+This can be done rigorously using the **continued fractions algorithm**:
+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 $c/s$.
+
+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.
+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
+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,
+but fortunately, for most applications of this algorithm,
+that does not actually matter,
+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 $\mathcal{O}(q^2 (\log{q}) \log{\log{q}})$
+and therefore worse than the QFT,
+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:
+
+$$\begin{aligned}
+ f(x)
+ = a^x \bmod N
+\end{aligned}$$
+
+$N$ 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:
+
+$$\begin{aligned}
+ a^x
+ = a^{x + s} \bmod N
+ \quad \implies \quad
+ 1
+ = a^s \bmod N
+\end{aligned}$$
+
+Suppose that $s$ is even. In that case,
+we can rewrite the above equation as follows:
+
+$$\begin{aligned}
+ (a^{s/2})^2 - 1
+ = 0 \bmod N
+\end{aligned}$$
+
+In other words, $(a^{s/2})^2 \!-\! 1$ is a multiple of $N$.
+We then use that $(a\!-\!b) (a\!+\!b) = a^2\!-\!b^2$:
+
+$$\begin{aligned}
+ \big( a^{s/2} - 1 \big) \big( a^{s/2} + 1 \big)
+ = 0 \bmod N
+\end{aligned}$$
+
+Because $s$ is even by assumption, the two factors on the left are integers,
+and as just mentioned, their product is a multiple of $N$.
+Then we only need to calculate:
+
+$$\begin{aligned}
+ \gcd\!\big( a^{s/2}\!-\!1, N \big) > 1
+ \quad\:\: \mathrm{and} \quad\:\:
+ \gcd\!\big( a^{s/2}\!+\!1, N \big) > 1
+\end{aligned}$$
+
+And there we have the factors of $N$!
+The $\gcd$ can be calculated efficiently in $\mathcal{O}(q^2)$ time.
+
+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 $N$.
+
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
+
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+---
+title: "Sigma-algebra"
+date: 2021-10-22
+categories:
+- Mathematics
+- Measure theory
+layout: "concept"
+---
+
+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.
+
+This forms a Boolean algebra:
+property (1) represents TRUE,
+(2) is NOT, and (3) is AND,
+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.
+
+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 $\mathcal{F}$
+(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 $\mathcal{B}(\mathbb{R})$
+is defined as the family of all open intervals of the real line,
+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**.
+
+
+
+## References
+1. U.H. Thygesen,
+ *Lecture notes on diffusions and stochastic differential equations*,
+ 2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/simons-algorithm/index.md b/source/know/concept/simons-algorithm/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/simons-algorithm/index.md
@@ -0,0 +1,184 @@
+---
+title: "Simon's algorithm"
+date: 2021-05-01
+categories:
+- Quantum information
+- Algorithms
+layout: "concept"
+---
+
+**Simon's algorithm** was the first proof that quantum computers
+are able to solve some problems *exponentially* faster
+than classical computers.
+In the same spirit as
+the [Deutsch-Jozsa algorithm](/know/concept/deutsch-jozsa-algorithm/)
+and the [Bernstein-Vazirani algorithm](/know/concept/bernstein-vazirani-algorithm/),
+the problem it solves, known as **Simon's problem**,
+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 $x_2$:
+
+$$\begin{aligned}
+ f(x_1)
+ = f(x_2)
+ \quad \Leftrightarrow \quad
+ x_2 = s \oplus x_1
+\end{aligned}$$
+
+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, $x_1 = s \oplus x_2$.
+
+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 $x_2 = s \oplus x_1$.
+
+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
+(the square root is due to the birthday paradox).
+
+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:
+
+
+
+
+The XOR oracle $U_f$ implements $f$,
+and has the following action for $n$-bit $a$ and $b$:
+
+$$\begin{aligned}
+ \Ket{a} \Ket{b}
+ \quad \to \boxed{U_f} \to \quad
+ \Ket{a} \Ket{b \oplus f(a)}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \Ket{0}^{\otimes n} \Ket{0}^{\otimes n}
+ \quad \to \boxed{H^{\otimes n}} \to \quad
+ \Ket{+}^{\otimes n} \Ket{0}^{\otimes n}
+ = \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{0}^{\otimes n}
+\end{aligned}$$
+
+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 $U_f$:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{0}^{\otimes n}
+ \quad \to \boxed{U_f} \to \quad
+ \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{f(x)}
+\end{aligned}$$
+
+Then we apply $H^{\otimes n}$ to the first $n$ qubits again,
+which, thanks to the definition of the Hadamard transform,
+yields the following,
+where $x \cdot y$ is the bitwise dot product:
+
+$$\begin{aligned}
+ \frac{1}{\sqrt{2^n}} \sum_{x = 0}^{2^n - 1} \Ket{x} \Ket{f(x)}
+ \quad \to \boxed{H^{\otimes n}} \to \quad
+ &\frac{1}{2^n} \sum_{x = 0}^{2^n - 1} \bigg( \sum_{y = 0}^{2^n - 1} (-1)^{x \cdot y} \Ket{y} \bigg) \Ket{f(x)}
+\end{aligned}$$
+
+
+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 $x_2 = s \oplus x_1$:
+
+$$\begin{alignedat}{2}
+ &\mathrm{if} \: s = 0: \qquad
+ &&\frac{1}{\sqrt{2^{n}}} \sum_{y = 0}^{2^n - 1} (-1)^{x_1 \cdot y} \Ket{y} \Ket{f(x_1)}
+ \\
+ &\mathrm{if} \: s \neq 0: \qquad
+ &&\frac{1}{\sqrt{2^{n+1}}} \sum_{y = 0}^{2^n - 1} \Big( (-1)^{x_1 \cdot y} + (-1)^{x_2 \cdot y} \Big) \Ket{y} \Ket{f(x_1)}
+\end{alignedat}$$
+
+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 $s \neq 0$, 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,
+we can rewrite this as follows:
+
+$$\begin{aligned}
+ (-1)^{x_1 \cdot y} + (-1)^{x_1 \cdot y \oplus s \cdot y}
+ = (-1)^{x_1 \cdot y} + (-1)^{x_1 \cdot y} (-1)^{s \cdot y}
+ \neq 0
+\end{aligned}$$
+
+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.
+
+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 $y_1, ..., y_N$,
+from which we can build a system of linear equations:
+
+$$\begin{aligned}
+ s \cdot y_1 &= 0 \:\:(\bmod 2)
+ \\
+ s \cdot y_2 &= 0 \:\:(\bmod 2)
+ \\
+ &\:\:\vdots
+ \\
+ s \cdot y_N &= 0 \:\:(\bmod 2)
+\end{aligned}$$
+
+This can be solved efficiently by a classical computer.
+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 $\mathcal{O}(n)$.
+
+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,
+which we did successfully.
+Querying an oracle might be a very expensive operation,
+so that is a big improvement!
+That said, Simon's algorithm currently has no known practical uses.
+
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
+2. S. Aaronson,
+ *Introduction to quantum information science: lecture notes*,
+ 2018, unpublished.
diff --git a/source/know/concept/simons-algorithm/simons-circuit.png b/source/know/concept/simons-algorithm/simons-circuit.png
new file mode 100644
index 0000000..b1f5ffb
Binary files /dev/null and b/source/know/concept/simons-algorithm/simons-circuit.png differ
diff --git a/source/know/concept/slater-determinant/index.md b/source/know/concept/slater-determinant/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/slater-determinant/index.md
@@ -0,0 +1,48 @@
+---
+title: "Slater determinant"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In quantum mechanics, the **Slater determinant** is a trick
+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
+goal is to find an expression for an overall many-particle wave
+function $\Psi(x_1, ..., x_N)$ that satisfies the
+[Pauli exclusion principle](/know/concept/pauli-exclusion-principle/).
+Enter the Slater determinant:
+
+$$\begin{aligned}
+ \boxed{
+ \Psi(x_1, ..., x_N)
+ = \frac{1}{\sqrt{N!}} \det\!
+ \begin{bmatrix}
+ \psi_1(x_1) & \cdots & \psi_N(x_1) \\
+ \vdots & \ddots & \vdots \\
+ \psi_1(x_N) & \cdots & \psi_N(x_N)
+ \end{bmatrix}
+ }\end{aligned}$$
+
+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.
+
+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 $\Psi$.
+
+In fact, there exists a similar trick for bosons, where the goal is to
+create a symmetric wave function which allows multiple particles to
+occupy the same state. In this case, one needs to take the **Slater
+permanent** of the same matrix, which is simply the determinant, but with
+all minuses replaced by pluses.
diff --git a/source/know/concept/sokhotski-plemelj-theorem/index.md b/source/know/concept/sokhotski-plemelj-theorem/index.md
new file mode 100644
index 0000000..b5c0b1b
--- /dev/null
+++ b/source/know/concept/sokhotski-plemelj-theorem/index.md
@@ -0,0 +1,109 @@
+---
+title: "Sokhotski-Plemelj theorem"
+date: 2021-11-01
+categories:
+- Mathematics
+- Complex analysis
+- Quantum mechanics
+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 $[a, b]$:
+
+$$\begin{aligned}
+ \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
+\end{aligned}$$
+
+To do so, we start by splitting the integrand
+into its real and imaginary parts (limit hidden):
+
+$$\begin{aligned}
+ \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
+ &= \int_a^b \frac{x - i \eta}{x^2 + \eta^2} f(x) \dd{x}
+ = \int_a^b \bigg( \frac{x}{x^2 + \eta^2} - i \frac{\eta}{x^2 + \eta^2} \bigg) f(x) \dd{x}
+\end{aligned}$$
+
+To evaluate the real part,
+we notice that for $\eta \to 0^+$ the integrand diverges for $x \to 0$,
+and thus split the integral as follows:
+
+$$\begin{aligned}
+ \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x}
+ &= \lim_{\eta \to 0^+} \bigg( \int_a^{-\eta} \frac{x f(x)}{x^2 + \eta^2} \dd{x} + \int_\eta^b \frac{x f(x)}{x^2 + \eta^2} \dd{x} \bigg)
+\end{aligned}$$
+
+This is simply the definition of the
+[Cauchy principal value](/know/concept/cauchy-principal-value/) $\mathcal{P}$,
+so the real part is given by:
+
+$$\begin{aligned}
+ \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x}
+ &= \mathcal{P} \int_a^b \frac{x f(x)}{x^2} \dd{x}
+ = \mathcal{P} \int_a^b \frac{f(x)}{x} \dd{x}
+\end{aligned}$$
+
+Meanwhile, in the imaginary part,
+we substitute $\eta$ for $1 / m$, and introduce $\pi$:
+
+$$\begin{aligned}
+ \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x}
+ &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{1/m}{x^2 + 1/m^2} f(x) \dd{x}
+ \\
+ &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{m}{1 + m^2 x^2} f(x) \dd{x}
+\end{aligned}$$
+
+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 [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+ \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x}
+ &= \pi \int_a^b \delta(x) \: f(x) \dd{x}
+ = \pi f(0)
+\end{aligned}$$
+
+By combining the real and imaginary parts,
+we thus arrive at the (real version of the)
+so-called **Sokhotski-Plemelj theorem** of complex analysis:
+
+$$\begin{aligned}
+ \boxed{
+ \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
+ = \mathcal{P} \int_a^b \frac{f(x)}{x} \dd{x} - i \pi f(0)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \frac{1}{x + i \eta}
+ = \mathcal{P} \Big( \frac{1}{x} \Big) - i \pi \delta(x)
+\end{aligned}$$
+
+The full, complex version of the Sokhotski-Plemelj theorem
+evaluates integrals of the following form
+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/).
+The Sokhotski-Plemelj theorem then states:
+
+$$\begin{aligned}
+ \boxed{
+ \lim_{w \to z} \phi(w)
+ = \frac{1}{2 \pi i} \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \frac{f(z)}{2}
+ }
+\end{aligned}$$
+
+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,
+so that the integrand vanishes away from the real axis,
+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
new file mode 100644
index 0000000..fb56ce0
--- /dev/null
+++ b/source/know/concept/spherical-coordinates/index.md
@@ -0,0 +1,205 @@
+---
+title: "Spherical coordinates"
+date: 2021-03-04
+categories:
+- Mathematics
+- Physics
+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:
+
+* $r$: 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:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ x &= r \sin\theta \cos\varphi \\
+ y &= r \sin\theta \sin\varphi \\
+ z &= r \cos\theta
+ \end{aligned}
+ }
+\end{aligned}$$
+
+Conversely, a point given in $(x, y, z)$
+can be converted to $(r, \theta, \varphi)$
+using these formulae:
+
+$$\begin{aligned}
+ \boxed{
+ r = \sqrt{x^2 + y^2 + z^2}
+ \qquad
+ \theta = \arccos(z / r)
+ \qquad
+ \varphi = \mathtt{atan2}(y, x)
+ }
+\end{aligned}$$
+
+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.
+To do so, we calculate the differentials of the Cartesian coordinates:
+
+$$\begin{aligned}
+ \dd{x} &= \dd{r} \sin\theta \cos\varphi + \dd{\theta} r \cos\theta \cos\varphi - \dd{\varphi} r \sin\theta \sin\varphi
+ \\
+ \dd{y} &= \dd{r} \sin\theta \sin\varphi + \dd{\theta} r \cos\theta \sin\varphi + \dd{\varphi} r \sin\theta \cos\varphi
+ \\
+ \dd{z} &= \dd{r} \cos\theta - \dd{\theta} r \sin\theta
+\end{aligned}$$
+
+And then we calculate the line element $\dd{\ell}^2$,
+skipping many terms thanks to orthogonality:
+
+$$\begin{aligned}
+ \dd{\ell}^2
+ &= \:\:\:\: \dd{r}^2 \big( \sin^2(\theta) \cos^2(\varphi) + \sin^2(\theta) \sin^2(\varphi) + \cos^2(\theta) \big)
+ \\
+ &\quad + \dd{\theta}^2 \big( r^2 \cos^2(\theta) \cos^2(\varphi) + r^2 \cos^2(\theta) \sin^2(\varphi) + r^2 \sin^2(\theta) \big)
+ \\
+ &\quad + \dd{\varphi}^2 \big( r^2 \sin^2(\theta) \sin^2(\varphi) + r^2 \sin^2(\theta) \cos^2(\varphi) \big)
+ \\
+ &= \dd{r}^2 + r^2 \: \dd{\theta}^2 + r^2 \sin^2(\theta) \: \dd{\varphi}^2
+\end{aligned}$$
+
+Finally, we can simply read off
+the squares of the desired scale factors
+$h_r^2$, $h_\theta^2$ and $h_\varphi^2$:
+
+$$\begin{aligned}
+ \boxed{
+ h_r = 1
+ \qquad
+ h_\theta = r
+ \qquad
+ h_\varphi = r \sin\theta
+ }
+\end{aligned}$$
+
+With these factors, we can easily convert things from the Cartesian system
+using the standard formulae for orthogonal curvilinear coordinates.
+The basis vectors are:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \vu{e}_r
+ &= \sin\theta \cos\varphi \:\vu{e}_x + \sin\theta \sin\varphi \:\vu{e}_y + \cos\theta \:\vu{e}_z
+ \\
+ \vu{e}_\theta
+ &= \cos\theta \cos\varphi \:\vu{e}_x + \cos\theta \sin\varphi \:\vu{e}_y - \sin\theta \:\vu{e}_z
+ \\
+ \vu{e}_\varphi
+ &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y
+ \end{aligned}
+ }
+\end{aligned}$$
+
+The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
+
+$$\begin{aligned}
+ \boxed{
+ \nabla f
+ = \vu{e}_r \pdv{f}{r}
+ + \vu{e}_\theta \frac{1}{r} \pdv{f}{\theta} + \mathbf{e}_\varphi \frac{1}{r \sin\theta} \pdv{f}{\varphi}
+ }
+\end{aligned}$$
+
+$$\begin{aligned}
+ \boxed{
+ \nabla \cdot \vb{V}
+ = \frac{1}{r^2} \pdv{(r^2 V_r)}{r}
+ + \frac{1}{r \sin\theta} \pdv{(\sin\theta V_\theta)}{\theta}
+ + \frac{1}{r \sin\theta} \pdv{V_\varphi}{\varphi}
+ }
+\end{aligned}$$
+
+$$\begin{aligned}
+ \boxed{
+ \nabla^2 f
+ = \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{f}{r} \Big)
+ + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{f}{\theta} \Big)
+ + \frac{1}{r^2 \sin^2(\theta)} \pdvn{2}{f}{\varphi}
+ }
+\end{aligned}$$
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \nabla \times \vb{V}
+ &= \frac{\vu{e}_r}{r \sin\theta} \Big( \pdv{(\sin\theta V_\varphi)}{\theta} - \pdv{V_\theta}{\varphi} \Big)
+ \\
+ &+ \frac{\vu{e}_\theta}{r} \Big( \frac{1}{\sin\theta} \pdv{V_r}{\varphi} - \pdv{(r V_\varphi)}{r} \Big)
+ \\
+ &+ \frac{\vu{e}_\varphi}{r} \Big( \pdv{(r V_\theta)}{r} - \pdv{V_r}{\theta} \Big)
+ \end{aligned}
+ }
+\end{aligned}$$
+
+The differential element of volume $\dd{V}$
+takes the following form:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{V}
+ = r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
+ }
+\end{aligned}$$
+
+So, for example, an integral over all of space is converted like so:
+
+$$\begin{aligned}
+ \iiint_{-\infty}^\infty f(x, y, z) \dd{V}
+ = \int_0^{2\pi} \int_0^\pi \int_0^\infty f(r, \theta, \varphi) \: r^2 \sin\theta \dd{r} \dd{\theta} \dd{\varphi}
+\end{aligned}$$
+
+The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ \dd{S}_r = r^2 \sin\theta \dd{\theta} \dd{\varphi}
+ \qquad
+ \dd{S}_\theta = r \sin\theta \dd{r} \dd{\varphi}
+ \qquad
+ \dd{S}_\varphi = r \dd{r} \dd{\theta}
+ \end{aligned}
+ }
+\end{aligned}$$
+
+Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{\vu{S}}
+ = \vu{e}_r \: r^2 \sin\theta \dd{\theta} \dd{\varphi}
+ + \vu{e}_\theta \: r \sin\theta \dd{r} \dd{\varphi}
+ + \vu{e}_\varphi \: r \dd{r} \dd{\theta}
+ }
+\end{aligned}$$
+
+And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{\vu{\ell}}
+ = \vu{e}_r \: \dd{r}
+ + \vu{e}_\theta \: r \dd{\theta}
+ + \vu{e}_\varphi \: r \sin\theta \dd{\varphi}
+ }
+\end{aligned}$$
+
+
+## References
+1. M.L. Boas,
+ *Mathematical methods in the physical sciences*, 2nd edition,
+ Wiley.
diff --git a/source/know/concept/spitzer-resistivity/index.md b/source/know/concept/spitzer-resistivity/index.md
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+---
+title: "Spitzer resistivity"
+date: 2021-10-05
+categories:
+- Physics
+- Plasma physics
+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/) $q_e E$
+(we neglect the ions due to their mass),
+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.
+Balancing the two forces yields the following relation:
+
+$$\begin{aligned}
+ q_e E
+ = f_{ei} m_e v_e
+\end{aligned}$$
+
+Using that the current density $J = q_e n_e v_e$,
+we can rearrange this like so:
+
+$$\begin{aligned}
+ E
+ = f_{ei} m_e \frac{J}{n_e q_e^2}
+ = \frac{m_e f_{ei}}{n_e q_e^2} J
+ = \eta J
+\end{aligned}$$
+
+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/)
+of the electron-ion system:
+
+$$\begin{aligned}
+ f_{ei}
+ = n_i \sigma v_e
+ = \frac{1}{2 \pi} \Big( \frac{q_e q_i}{\varepsilon_0 \mu} \Big)^2 \frac{n_i}{v_e^3} \ln(\Lambda)
+ \approx \frac{1}{2 \pi} \frac{Z q_e^4}{\varepsilon_0^2 m_e^2} \frac{n_e}{v_e^3} \ln(\Lambda)
+\end{aligned}$$
+
+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 $f_{ei}$;
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \eta
+ = \frac{m_e f_{ei}}{n_e q_e^2}
+ = \frac{1}{2 \pi} \frac{Z q_e^2}{\varepsilon_0^2 m_e} \frac{1}{v_e^3} \ln(\Lambda)
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \frac{1}{2} m_e v_e^2
+ = \frac{3}{2} k_B T_e
+ \quad \implies \quad
+ v_e
+ = \sqrt{\frac{3 k_B T_e}{m_e}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \eta
+ = \frac{1}{6 \pi \sqrt{3}} \frac{Z q_e^2 \sqrt{m_e}}{\varepsilon_0^2 (k_B T_e)^{3/2}} \ln(\Lambda)
+\end{aligned}$$
+
+
+
+## References
+1. P.M. Bellan,
+ *Fundamentals of plasma physics*,
+ 1st edition, Cambridge.
+2. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/step-index-fiber/bessel.jpg b/source/know/concept/step-index-fiber/bessel.jpg
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diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
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+---
+title: "Step-index fiber"
+date: 2022-02-11
+categories:
+- Physics
+- Optics
+- Fiber optics
+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/)
+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.
+In [polar coordinates](/know/concept/cylindrical-polar-coordinates/)
+$(r,\phi)$ 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 $r$.
+
+This Helmholtz equation can be solved by *separation of variables*:
+we assume that there exist two functions $R(r)$ and $\Phi(\phi)$
+such that $F(r,\phi) = R(r) \, \Phi(\phi)$.
+Inserting this ansatz:
+
+$$\begin{aligned}
+ R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi = 0
+\end{aligned}$$
+
+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 $\ell^2$:
+
+$$\begin{aligned}
+ r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2
+ = -\frac{\Phi''}{\Phi}
+ = \ell^2
+\end{aligned}$$
+
+This gives an eigenvalue problem for $\Phi$,
+and the well-known *Bessel equation* for $R$:
+
+$$\begin{aligned}
+ \boxed{
+ \Phi'' + \ell^2 \Phi = 0
+ }
+ \qquad \qquad
+ \boxed{
+ r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R = 0
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \Phi(0) = \Phi(2\pi)
+ \qquad \qquad
+ \Phi'(0) = \Phi'(2\pi)
+\end{aligned}$$
+
+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 $\ell^2$,
+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.
+
+* 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:
+
+ $$\begin{alignedat}{3}
+ \Phi(0) &= \Phi(2 \pi)
+ \:\quad &&\implies \quad\:\:
+ 0 &&= A \sinh(2 \pi \ell) + B \big( \cosh(2 \pi \ell) - 1 \big)
+ \\
+ \Phi'(0) &= \Phi'(2 \pi)
+ \: \quad &&\implies \quad \:\:
+ 0 &&= A \ell \big( \cosh(2 \pi \ell) - 1 \big) + B \ell \sinh(2 \pi \ell)
+ \end{alignedat}$$
+
+ This only has non-trivial solutions
+ if the determinant of the system matrix is zero:
+
+ $$\begin{aligned}
+ 0
+ &= \mathrm{det}
+ \begin{bmatrix}
+ \sinh(2 \pi \ell) & \cosh(2 \pi \ell) - 1 \\
+ \cosh(2 \pi \ell) - 1 & \sinh(2 \pi \ell)
+ \end{bmatrix}
+ = 2 \big( \cosh(2 \pi \ell) - 1 \big)
+ \end{aligned}$$
+
+ 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,
+ because no non-trivial solutions exist here.
+
+* For $\ell^2 = 0$, the solution is $\Phi(\phi) = A \phi + B$.
+ Putting this in the boundary conditions:
+
+ $$\begin{alignedat}{3}
+ \Phi(0) &= \Phi(2 \pi)
+ \qquad &&\implies \qquad
+ A &&= 0
+ \\
+ \Phi'(0) &= \Phi'(2 \pi)
+ \qquad &&\implies \qquad
+ B &&= B
+ \end{alignedat}$$
+
+ $B$ 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:
+
+ $$\begin{alignedat}{3}
+ \Phi(0) &= \Phi(2 \pi)
+ \quad &&\implies \quad
+ 0 &&= A \sin(2 \pi \ell) + B \big(\cos(2\pi \ell) - 1\big)
+ \\
+ \Phi'(0) &= \Phi'(2 \pi)
+ \quad &&\implies \quad
+ 0 &&= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell)
+ \end{alignedat}$$
+
+ This system only has nontrivial solutions
+ if the determinant of its matrix is zero:
+
+ $$\begin{aligned}
+ 0
+ &= \mathrm{det}
+ \begin{bmatrix}
+ \sin(2 \pi \ell) & \cos(2 \pi \ell) - 1 \\
+ \cos(2 \pi \ell) - 1 & -\sin(2 \pi \ell)
+ \end{bmatrix}
+ = 2 \big(\cos(2 \pi \ell) - 1\big)
+ \end{aligned}$$
+
+ Meaning that $\ell$ must be an integer.
+ We revisit the boundary conditions and indeed see:
+
+ $$\begin{alignedat}{3}
+ 0 &= A \sin(2 \pi \ell) + B \big(\cos(2 \pi \ell) - 1\big)
+ \qquad &&\implies \qquad
+ 0 &&= 0
+ \\
+ 0 &= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell)
+ \qquad &&\implies \qquad
+ 0 &&= 0
+ \end{alignedat}$$
+
+ 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 $\cos(\phi \ell)$,
+ 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.
+
+By combining our result for $\ell^2 = 0$ and $\ell^2 > 0$,
+we get the following for $\ell = 0, 1, 2, ...$:
+
+$$\begin{aligned}
+ \boxed{
+ \Phi_\ell(\phi) = A \cos(\phi \ell)
+ }
+\end{aligned}$$
+
+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.
+
+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 $\mu(r)$.
+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 $n_o$.
+
+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 $r = a$:
+
+$$\begin{aligned}
+ R_i(a) = R_o(a)
+ \qquad \qquad
+ R_i'(a) = R_o'(a)
+\end{aligned}$$
+
+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 $\mu$.
+
+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 $\mu$:
+
+$$\begin{aligned}
+ \begin{cases}
+ \displaystyle
+ 0 = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R
+ & \mathrm{for}\; \mu \neq 0
+ \\
+ \displaystyle
+ 0 = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R
+ & \mathrm{for}\; \mu = 0
+ \end{cases}
+\end{aligned}$$
+
+
+
+
+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 $R_o$:
+
+$$\begin{aligned}
+ R_{o,\ell}(r) =
+ \begin{cases}
+ r^{-\ell}
+ & \mathrm{for}\; \mu = 0 \;\mathrm{and}\; \ell = 1,2,3,...
+ \\
+ K_\ell(\rho) = K_\ell(r \sqrt{-\mu})
+ & \mathrm{for}\; \mu < 0 \;\mathrm{and}\; \ell = 0,1,2,...
+ \end{cases}
+\end{aligned}$$
+
+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 $\mu > 0$:
+
+$$\begin{aligned}
+ R_{i,\ell}(r) =
+ J_\ell(\rho) = J_\ell(r \sqrt{\mu})
+ \qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,...
+\end{aligned}$$
+
+Putting this all together, we now know what the full solution for $F$ should look like:
+
+$$\begin{aligned}
+ F_\ell(r, \phi)
+ = R_\ell(r) \, \Phi_\ell(\phi)
+ =
+ \begin{cases}
+ A_\ell \: R_{i,\ell}(r) \, \cos(\phi \ell)
+ & \mathrm{for}\; r \le a
+ \\
+ B_\ell \: R_{o,\ell}(r) \, \cos(\phi l)
+ & \mathrm{for}\; r \ge a
+ \end{cases}
+\end{aligned}$$
+
+Where $A_\ell$ and $B_\ell$ are constants to be chosen
+based on the light's intensity, and to satisfy the continuity condition at $r = a$.
+
+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;
+the fiber is not able to guide the light outside this range.
+
+However, not all $\beta$ in this range are created equal for all $k$.
+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 $k$:
+
+$$\begin{aligned}
+ \xi_i(k)
+ \equiv \sqrt{ n_i^2 k^2 - \beta^2(k) }
+ \qquad \qquad
+ \xi_o(k)
+ \equiv \sqrt{ \beta^2(k) - n_o^2 k^2 }
+\end{aligned}$$
+
+It is important to note that the sum of their squares is constant with respect to $\beta$:
+
+$$\begin{aligned}
+ \xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \mathrm{NA}
+ \equiv \sqrt{n_i^2 - n_o^2}
+ }
+\end{aligned}$$
+
+From this, we define a new fiber parameter: the $V$-**number**,
+which is extremely useful:
+
+$$\begin{aligned}
+ \boxed{
+ V
+ \equiv a \sqrt{\xi_i^2 + \xi_o^2}
+ = a k \: \mathrm{NA}
+ }
+\end{aligned}$$
+
+Now, the allowed values of $\beta$ are found
+by fulfilling the boundary conditions (for $\mu \neq 0$):
+
+$$\begin{aligned}
+ A_\ell J_\ell(a \xi_i)
+ &= B_\ell K_\ell(a \xi_o)
+ \\
+ A_\ell \xi_i J_\ell'(a \xi_i)
+ &= B_\ell \xi_o K_\ell'(a \xi_o)
+\end{aligned}$$
+
+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 $X^2 + Y^2 = V^2$:
+
+$$\begin{aligned}
+ X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)}
+\end{aligned}$$
+
+We can turn this result into something a bit nicer
+by using the following identities:
+
+$$\begin{aligned}
+ J_\ell'(x) = -J_{\ell+1}(x) + \ell \frac{J_\ell(x)}{x}
+ \qquad \quad
+ K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x}
+\end{aligned}$$
+
+With this, the transcendental equation for $\beta$
+takes this convenient form:
+
+$$\begin{aligned}
+ \boxed{
+ X \frac{J_{\ell+1}(X)}{J_\ell(X)} = Y \frac{K_{\ell+1}(Y)}{K_\ell(Y)}
+ }
+\end{aligned}$$
+
+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 $\ell$.
+
+An example graphical solution of the transcendental equation
+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 $V_{02} = V_{21} = 3.832$.
+
+All differential equations have been linear,
+so a linear combination of these solutions is also valid.
+Therefore, the fiber modes represent independent "channels" of light.
+However, in practice, they can interact nonlinearly,
+and light can scatter between them, and between polarizations.
+
+
+
+## References
+1. O. Bang,
+ *Applied mathematics for physicists: lecture notes*, 2019,
+ unpublished.
+2. B.E.A. Saleh, M.C. Teich,
+ *Fundamentals of photonics*, 1st edition, 1991,
+ Wiley.
diff --git a/source/know/concept/step-index-fiber/modes.jpg b/source/know/concept/step-index-fiber/modes.jpg
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diff --git a/source/know/concept/stochastic-process/index.md b/source/know/concept/stochastic-process/index.md
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+---
+title: "Stochastic process"
+date: 2021-11-07
+categories:
+- Mathematics
+- Stochastic analysis
+- Measure theory
+layout: "concept"
+---
+
+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 $t$.
+
+Whereas "ordinary" random variables are defined on
+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.
+
+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 $\mathcal{F}_t$:
+
+$$\begin{aligned}
+ \mathcal{F}
+ \supseteq \mathcal{F}_t
+ \supseteq \sigma(X_s : 0 \le s \le t)
+\end{aligned}$$
+
+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.
+
+Filtration and their adaptations are very useful
+for working with stochastic processes,
+most notably for calculating [conditional expectations](/know/concept/conditional-expectation/).
+
+
+
+## References
+1. U.H. Thygesen,
+ *Lecture notes on diffusions and stochastic differential equations*,
+ 2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/stokes-law/index.md b/source/know/concept/stokes-law/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/stokes-law/index.md
@@ -0,0 +1,366 @@
+---
+title: "Stokes' law"
+date: 2021-05-04
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+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 $U$.
+
+
+## Flow field
+
+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 $U$.
+The Reynolds number is:
+
+$$\begin{aligned}
+ \mathrm{Re}
+ = \frac{2 a U}{\nu}
+\end{aligned}$$
+
+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**:
+
+$$\begin{aligned}
+ \nabla p
+ = \eta \nabla^2 \va{v}
+ \qquad \quad
+ \nabla \cdot \va{v}
+ = 0
+\end{aligned}$$
+
+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:
+
+$$\begin{gathered}
+ p
+ = \eta U q(r) \cos\theta
+ \\
+ v_r
+ = U f(r) \cos\theta
+ \qquad
+ v_\theta
+ = - U g(r) \sin\theta
+ \qquad
+ v_\phi
+ = 0
+\end{gathered}$$
+
+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.
+
+To begin with, we insert this ansatz into the incompressibility condition,
+yielding:
+
+$$\begin{aligned}
+ 0
+ = \nabla \cdot \va{v}
+ &= \pdv{v_r}{r} + \frac{1}{r} \pdv{v_\theta}{\theta} + \frac{2 v_r}{r} + \frac{v_\theta}{r \tan \theta}
+ \\
+ &= U \dv{f}{r} \cos\theta - \frac{U g}{r} \cos\theta + \frac{2 U f}{r} \cos\theta - \frac{U g}{r} \cos\theta
+ \\
+ &= U \cos\theta \Big( \dv{f}{r} + \frac{2}{r} f - \frac{2}{r} g \Big)
+\end{aligned}$$
+
+The parenthesized expression must be zero for all $r$,
+leading us to the following relation:
+
+$$\begin{aligned}
+ g(r)
+ = f + \frac{r}{2} \dv{f}{r}
+\end{aligned}$$
+
+Next, we take the divergence of the first Stokes equation,
+and insert incompressibility:
+
+$$\begin{aligned}
+ \nabla^2 p
+ = \eta \nabla \cdot (\nabla^2 \va{v})
+ = \eta \nabla^2 (\nabla \cdot \va{v})
+ = 0
+\end{aligned}$$
+
+This is simply the Laplace equation,
+which is as follows for our ansatz $p(r, \theta)$:
+
+$$\begin{aligned}
+ 0
+ = \nabla^2 p
+ &= \frac{1}{r^2} \pdv{}{r}\Big( r^2 \pdv{p}{r} \Big) + \frac{1}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin\theta \pdv{p}{\theta} \Big)
+ \\
+ 0
+ &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
+ - \frac{\eta U q}{r^2 \sin\theta} \pdv{}{\theta}\Big( \sin^2\theta \Big)
+ \\
+ &= \frac{\eta U \cos\theta}{r^2} \dv{}{r}\Big( r^2 \dv{q}{r} \Big)
+ - \frac{2 \eta U q}{r^2 \sin\theta} \sin\theta \cos\theta
+ \\
+ &= \eta U \cos\theta \Big( \dvn{2}{q}{r} + \frac{2}{r} \dv{q}{r} - \frac{2}{r^2} q \Big)
+\end{aligned}$$
+
+Again, the parenthesized expression must be zero for all $r$,
+meaning it is an ODE for $q(r)$,
+whose solution is straightforwardly found to be:
+
+$$\begin{aligned}
+ q(r)
+ = \frac{C_3}{r^2} + C_4 r
+\end{aligned}$$
+
+Where $C_3$ and $C_4$ are linearity constants ($C_1$ and $C_2$ appear later).
+The pressure is therefore:
+
+$$\begin{aligned}
+ p
+ = \eta U \cos\theta \Big( \frac{C_3}{r^2} + C_4 r \Big)
+\end{aligned}$$
+
+Consequently, its gradient $\nabla p$ in spherical coordinates is as follows:
+
+$$\begin{aligned}
+ \nabla p
+ = \vu{e}_r \pdv{p}{r} + \vu{e}_\theta \frac{1}{r} \pdv{p}{\theta}
+ = \vu{e}_r \Big( \eta U \cos\theta \dv{q}{r} \Big) - \vu{e}_\theta \Big( \eta U \sin\theta \frac{q}{r} \Big)
+\end{aligned}$$
+
+According to the Stokes equation, this equals $\eta \nabla^2 \va{v}$.
+Let us look at the $r$-component of $\nabla^2 \va{v}$:
+
+$$\begin{aligned}
+ (\nabla^2 \va{v})_r
+ &= \pdvn{2}{v_r}{r} + \frac{1}{r^2} \pdvn{2}{v_r}{\theta} + \frac{2}{r} \pdv{v_r}{r}
+ + \frac{\cot\theta}{r^2} \pdv{v_r}{\theta} - \frac{2}{r^2} \pdv{v_\theta}{\theta} - \frac{2}{r^2} v_r - \frac{2 \cot\theta}{r^2} v_\theta
+ \\
+ &= U \cos\theta \Big( \dvn{2}{f}{r} - \frac{1}{r^2} f + \frac{2}{r} \dv{f}{r} - \frac{1}{r^2} f
+ + \frac{2}{r^2} g - \frac{2}{r^2} f + \frac{2}{r^2} g \Big)
+ \\
+ &= U \cos\theta \Big( \dvn{2}{f}{r} + \frac{2}{r} \dv{f}{r} - \frac{4}{r^2} f + \frac{4}{r^2} g \Big)
+\end{aligned}$$
+
+Substituting $g$ for the expression we found from incompressibility lets us simplify this:
+
+$$\begin{aligned}
+ \eta (\nabla^2 \va{v})_r
+ &= \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
+\end{aligned}$$
+
+The Stokes equation says that this must be equal to the $r$-component of $\nabla p$:
+
+$$\begin{aligned}
+ \eta U \cos\theta \Big( \dvn{2}{f}{r} + \frac{4}{r} \dv{f}{r} \Big)
+ = \eta U \cos\theta \Big( \!-\! \frac{2 C_3}{r^3} + C_4 \Big)
+\end{aligned}$$
+
+Where we have inserted $\idv{q}{r}$.
+Dividing out $\eta U \cos\theta$ leaves an ODE for $f(r)$,
+satisfied by:
+
+$$\begin{aligned}
+ f(r)
+ = C_1 + \frac{C_2}{r^3} + \frac{C_3}{r} + \frac{C_4 r^2}{10}
+\end{aligned}$$
+
+Then, thanks to our earlier relation again,
+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:
+
+$$\begin{aligned}
+ f(r)
+ = 1 + \frac{C_2}{r^3} + \frac{C_3}{r}
+ \qquad \quad
+ g(r)
+ = 1 - \frac{C_2}{2 r^3} + \frac{C_3}{2 r}
+\end{aligned}$$
+
+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,
+and solving the resulting system of equations
+yields $C_2 = a^3 / 2$ and $C_3 = -3 a / 2$.
+Therefore the full solution is:
+
+$$\begin{gathered}
+ \boxed{
+ p
+ = - \frac{3 \eta U a}{2 r^2} \cos\theta
+ }
+ \\
+ \boxed{
+ v_r
+ = U \cos\theta \Big( 1 + \frac{a^3}{2 r^3} - \frac{3 a}{2 r} \Big)
+ \qquad
+ v_\theta
+ = - U \sin\theta \Big( 1 - \frac{a^3}{4 r^3} - \frac{3 a}{4 r} \Big)
+ }
+\end{gathered}$$
+
+
+## Drag force
+
+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:
+
+$$\begin{aligned}
+ \va{D}
+ = \oint \hat{\sigma} \cdot \dd{\va{S}}
+ = \int_0^{2\pi} \!\!\!\! \int_0^\pi \big( \hat{\sigma} \cdot \vu{e}_r \big) \:a^2 \sin\theta \dd{\theta} \dd{\phi}
+\end{aligned}$$
+
+Where $\vu{e}_r$ is the sphere's surface normal vector.
+The integrand can be expanded as follows:
+
+$$\begin{aligned}
+ \hat{\sigma} \cdot \vu{e}_r
+ = \vu{e}_r \sigma_{rr} + \vu{e}_\theta \sigma_{\theta r}
+\end{aligned}$$
+
+To calculate this, we start by taking the gradient of the velocity field $\va{v}$:
+
+$$\begin{aligned}
+ \nabla\va{v}
+ &= \vu{e}_r \vu{e}_r \pdv{v_r}{r} + \vu{e}_r \vu{e}_\theta \pdv{v_\theta}{r}
+ + \vu{e}_\theta \vu{e}_r \Big( \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big)
+ \\
+ &\qquad + \vu{e}_\theta \vu{e}_\theta \Big( \frac{1}{r} \pdv{v_\theta}{\theta} - \frac{v_r}{r} \Big)
+ + \vu{e}_\phi \vu{e}_\phi \Big( \frac{v_\theta}{r \tan\theta} + \frac{v_r}{r} \Big)
+\end{aligned}$$
+
+Some of these terms are necessary to calculate the stress elements $\sigma_{rr}$ and $\sigma_{\theta r}$:
+
+$$\begin{aligned}
+ \sigma_{rr}
+ &= - p + 2 \eta (\nabla\va{v})_{rr}
+ = - p + 2 \eta \pdv{v_r}{r}
+ \\
+ &= \frac{3 \eta U a}{2 r^2} \cos\theta + 2 \eta U \cos\theta \: \Big( \!-\! \frac{3 a^3}{2 r^4} + \frac{3 a}{2 r^2} \Big)
+ \\
+ &= \frac{3 \eta U a}{2 r^2} \cos\theta \: \Big( 3 - 2 \frac{a^2}{r^2} \Big)
+\end{aligned}$$
+$$\begin{aligned}
+ \sigma_{\theta r}
+ &= \eta \big( (\nabla\va{v})_{\theta r} + (\nabla\va{v})_{r \theta} \big)
+ = \eta \: \Big( \pdv{v_\theta}{r} + \frac{1}{r} \pdv{v_r}{\theta} - \frac{v_\theta}{r} \Big)
+ \\
+ &= \eta U \sin\theta \: \Big( \!-\! \frac{3 a^3}{4 r^4} - \frac{3 a}{4 r^2}
+ - \frac{1}{r} - \frac{a^3}{2 r^4} + \frac{3 a}{2 r^2}
+ + \frac{1}{r} - \frac{a^3}{4 r^4} - \frac{3 a}{4 r^2} \Big)
+ \\
+ &= - \frac{3 \eta U a^3}{2 r^4} \sin\theta
+\end{aligned}$$
+
+At the sphere's surface we set $r = a$, so these expressions reduce to the following:
+
+$$\begin{aligned}
+ \sigma_{rr}
+ = \frac{3 \eta U}{2 a} \cos\theta
+ \qquad \quad
+ \sigma_{\theta r}
+ = - \frac{3 \eta U}{2 a} \sin\theta
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \hat{\sigma} \cdot \vu{e}_r
+ &= \vu{e}_r \frac{3 \eta U}{2 a} \cos\theta - \vu{e}_\theta \frac{3 \eta U}{2 a} \sin\theta
+ \\
+ &= \Big( \vu{e}_x \sin\theta \cos\phi + \vu{e}_y \sin\theta \sin\phi + \vu{e}_z \cos\theta \Big) \frac{3 \eta U}{2 a} \cos\theta
+ \\
+ &\qquad - \Big( \vu{e}_x \cos\theta \cos\phi + \vu{e}_y \cos\theta \sin\phi - \vu{e}_z \sin\theta \Big) \frac{3 \eta U}{2 a} \sin\theta
+ \\
+ &= \Big( \vu{e}_z \cos^2\theta + \vu{e}_z \sin^2\theta \Big) \frac{3 \eta U}{2 a}
+ = \vu{e}_z \frac{3 \eta U}{2 a}
+\end{aligned}$$
+
+Remarkably, the stress at every point on the sphere is purely in the $z$-direction!
+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 $4 \pi a^2$:
+
+$$\begin{aligned}
+ \va{D}
+ = \vu{e}_z \frac{3 \eta U}{2 a} \int_0^{2\pi} \!\!\!\! \int_0^\pi a^2 \sin\theta \dd{\theta} \dd{\phi}
+ = \vu{e}_z \frac{3 \eta U}{2 a} 2 \pi a^2 \int_0^\pi \sin\theta \dd{\theta}
+ = \vu{e}_z \: 6 \pi \eta U a
+\end{aligned}$$
+
+At last, we arrive at Stokes' law,
+which simply expresses the magnitude of $\va{D}$:
+
+$$\begin{aligned}
+ \boxed{
+ D
+ = 6 \pi \eta U a
+ }
+\end{aligned}$$
+
+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,
+subtracting [buoyancy](/know/concept/archimedes-principle/):
+
+$$\begin{aligned}
+ G
+ = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
+\end{aligned}$$
+
+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 $G$,
+after some time, they cancel out:
+
+$$\begin{aligned}
+ 6 \pi \eta U a
+ = \frac{4 \pi a^3}{3} (\rho_s - \rho_f) g_0
+\end{aligned}$$
+
+This is an equation for the **terminal velocity** $U_t$,
+which we find to be as follows:
+
+$$\begin{aligned}
+ \boxed{
+ U_t
+ = \frac{2 a^2 (\rho_s - \rho_f) g_0}{9 \eta}
+ }
+\end{aligned}$$
+
+The falling sphere will accelerate until $U_t$,
+and then continue falling at constant speed.
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md
new file mode 100644
index 0000000..f9cc6b2
--- /dev/null
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -0,0 +1,345 @@
+---
+title: "Sturm-Liouville theory"
+date: 2021-02-23
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**Sturm-Liouville theory** defines the analogue of Hermitian matrix
+eigenvalue problems for linear second-order ODEs.
+
+It states that, given suitable boundary conditions, any linear
+second-order ODE can be rewritten using the **Sturm-Liouville operator**,
+and that the corresponding eigenvalue problem, known as a
+**Sturm-Liouville problem**, will give real eigenvalues and a complete set
+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 $x \in ]a, b[$:
+
+$$\begin{aligned}
+ \hat{L} \{u(x)\} = p_0(x) u''(x) + p_1(x) u'(x) + p_2(x) u(x)
+\end{aligned}$$
+
+We now define the **adjoint** or **Hermitian** operator
+$\hat{L}^\dagger$ analogously to matrices:
+
+$$\begin{aligned}
+ \inprod{f}{\hat{L} g}
+ = \inprod{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+What is $\hat{L}^\dagger$, given the above definition of $\hat{L}$?
+We start from the inner product $\inprod{f}{\hat{L} g}$:
+
+$$\begin{aligned}
+ \inprod{f}{\hat{L} g}
+ &= \int_a^b f^*(x) \hat{L}\{g(x)\} \dd{x}
+ = \int_a^b (f^* p_0) g'' + (f^* p_1) g' + (f^* p_2) g \dd{x}
+ \\
+ &= \big[ (f^* p_0) g' + (f^* p_1) g \big]_a^b - \int_a^b (f^* p_0)' g' + (f^* p_1)' g - (f^* p_2) g \dd{x}
+ \\
+ &= \big[ f^* \big( p_0 g' \!+\! p_1 g \big) \!-\! (f^* p_0)' g \big]_a^b + \int_a^b \! \big( (f p_0)'' - (f p_1)' + (f p_2) \big)^* g \dd{x}
+ \\
+ &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x}
+\end{aligned}$$
+
+We now have an expression for $\hat{L}^\dagger$, but are left with an
+annoying boundary term:
+
+$$\begin{aligned}
+ \inprod{f}{\hat{L} g}
+ &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \inprod{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \inprod{f}{\hat{L} g}
+ &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \Inprod{\hat{L}^\dagger f}{g}
+ = \inprod{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+Using the aforementioned restriction $p_1(x) = p_0'(x)$,
+we then take a look at the definition of $\hat{L}^\dagger$:
+
+$$\begin{aligned}
+ \hat{L}^\dagger \{f\}
+ &= (p_0 f)'' - (p_1 f)' + (p_2 f)
+ \\
+ &= p_0 f'' + (2 p_0' - p_1) f' + (p_0'' - p_1' + p_2) f
+ \\
+ &= p_0 f'' + p_0' f' + p_2 f
+ \\
+ &= (p_0 f')' + p_2 f
+\end{aligned}$$
+
+The original operator $\hat{L}$ reduces to the same form,
+so it is **self-adjoint**:
+
+$$\begin{aligned}
+ \hat{L} \{f\}
+ &= p_0 f'' + p_0' f' + p_2 f
+ = (p_0 f')' + p_2 f
+ = \hat{L}^\dagger \{f\}
+\end{aligned}$$
+
+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 $[p_0 (f^* g' - (f^*)' g)]_a^b = 0$.
+
+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 $p_0(x) \neq 0$:
+
+$$\begin{aligned}
+ \frac{p(x)}{p_0(x)} \hat{L} \{u\} = p(x) u'' + p(x) \frac{p_1(x)}{p_0(x)} u' + p(x) \frac{p_2(x)}{p_0(x)} u
+\end{aligned}$$
+
+We now define $q(x)$,
+and demand that the derivative $p'(x)$ of the unknown $p(x)$ satisfies:
+
+$$\begin{aligned}
+ q(x) = p(x) \frac{p_2(x)}{p_0(x)}
+ \qquad
+ p'(x) = p(x) \frac{p_1(x)}{p_0(x)}
+\end{aligned}$$
+
+The latter is a differential equation for $p(x)$, which we solve by integration:
+
+$$\begin{gathered}
+ \frac{p_1(x)}{p_0(x)} = \frac{1}{p(x)} \dv{p}{x}
+ \quad \implies \quad
+ \frac{p_1(x)}{p_0(x)} \dd{x} = \frac{1}{p(x)} \dd{p}
+ \\
+ \implies \quad
+ \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} = \int_{p(a)}^{p(x)} \frac{1}{f} \dd{f}
+ = \ln\!\Big( \frac{p(x)}{p(a)} \Big)
+ \\
+ \implies \quad
+ p(x) = p(a) \exp\!\Big( \int_a^x \frac{p_1(\xi)}{p_0(\xi)} \dd{\xi} \Big)
+\end{gathered}$$
+
+Now that we have $p(x)$ and $q(x)$, we can define a new operator $\hat{L}_p$ as follows:
+
+$$\begin{aligned}
+ \hat{L}_p \{u\}
+ = \frac{p}{p_0} \hat{L} \{u\}
+ = p u'' + p' u' + q u
+ = (p u')' + q u
+\end{aligned}$$
+
+This is the self-adjoint form from earlier!
+So even if $p_0' \neq p_1$, any second-order linear operator with $p_0(x) \neq 0$
+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 $x \in [a,b]$:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{L}_{SL} \{u(x)\}
+ = \frac{d}{dx}\Big( p(x) \frac{du}{dx} \Big) + q(x) u(x)
+ = \hat{L}_{SL}^\dagger \{u(x)\}
+ }
+\end{aligned}$$
+
+
+## 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**:
+
+$$\begin{aligned}
+ \boxed{
+ \hat{L}_{SL}\{u(x)\} = - \lambda w(x) u(x)
+ }
+\end{aligned}$$
+
+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.
+
+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:
+
+$$\begin{aligned}
+ \Big[ p(x) \big( u^*(x) u'(x) - (u'(x))^* u(x) \big) \Big]_a^b = 0
+\end{aligned}$$
+
+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"**: $p(a) = p(b) = 0$
+
+Once this requirement is satisfied, Sturm-Liouville theory gives us
+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:
+
+$$\begin{aligned}
+ 0 = \hat{L}_{SL}\{u_n\} + \lambda_n w u_n = \hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*
+\end{aligned}$$
+
+We subtract these expressions, multiply by the eigenfunctions, and integrate:
+
+$$\begin{aligned}
+ 0
+ &= \int_a^b u_m^* \big(\hat{L}_{SL}\{u_n\} + \lambda_n w u_n\big) - u_n \big(\hat{L}_{SL}\{u_m^*\} + \lambda_m^* w u_m^*\big) \:dx
+ \\
+ &= \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} + u_n u_m^* w (\lambda_n - \lambda_m^*) \:dx
+\end{aligned}$$
+
+Rearranging this a bit reveals that these are in fact three inner products:
+
+$$\begin{aligned}
+ \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} \:dx
+ &= (\lambda_m^* - \lambda_n) \int_a^b u_n u_m^* w \:dx
+ \\
+ \inprod{u_m}{\hat{L}_{SL} u_n} - \inprod{\hat{L}_{SL} u_m}{u_n}
+ &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
+\end{aligned}$$
+
+The operator $\hat{L}_{SL}$ is self-adjoint by definition,
+so the left-hand side vanishes, leaving us with:
+
+$$\begin{aligned}
+ 0
+ &= (\lambda_m^* - \lambda_n) \Inprod{u_m}{w u_n}
+\end{aligned}$$
+
+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 $m \neq n$, 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.
+
+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**:
+
+$$\begin{aligned}
+ \boxed{
+ \Inprod{u_m(x)}{w(x) u_n(x)}
+ = \Inprod{u_n}{w u_n} \delta_{nm}
+ = A_n \delta_{nm}
+ }
+\end{aligned}$$
+
+When you're solving a differential eigenvalue problem,
+knowing that all eigenvalues are real is a *huge* simplification,
+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 $\lambda_0 > -\infty$,
+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 $a_n$:
+
+$$\begin{aligned}
+ \boxed{
+ f(x)
+ = \sum_{n = 0}^\infty a_n u_n(x)
+ \quad \mathrm{for}\: x \in ]a, b[
+ }
+\end{aligned}$$
+
+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 $x \in [a, b]$.
+
+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 $m$:
+
+$$\begin{aligned}
+ f(x) w(x) u_m^*(x)
+ &= \sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)
+\end{aligned}$$
+
+By integrating we get inner products on both the left and the right:
+
+$$\begin{aligned}
+ \int_a^b f(x) w(x) u_m^*(x) \dd{x}
+ &= \int_a^b \Big(\sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)\Big) \dd{x}
+ \\
+ \Inprod{u_m}{w f}
+ &= \sum_{n = 0}^\infty a_n \Inprod{u_m}{w u_n}
+\end{aligned}$$
+
+Because the eigenfunctions of an SLP are mutually orthogonal,
+the summation disappears:
+
+$$\begin{aligned}
+ \Inprod{u_m}{w f}
+ &= \sum_{n = 0}^\infty a_n \Inprod{u_m}{w u_n}
+ = \sum_{n = 0}^\infty a_n A_n \delta_{nm}
+ = a_m A_m
+\end{aligned}$$
+
+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 $u_n(x) / A_n$:
+
+$$\begin{aligned}
+ \boxed{
+ a_n
+ = \frac{\Inprod{u_n}{w f}}{A_n}
+ = \frac{\Inprod{u_n}{w f}}{\Inprod{u_n}{w u_n}}
+ }
+\end{aligned}$$
+
+As a final remark, we can see something interesting
+by rearranging the generalized Fourier series
+after inserting the expression for $a_n$:
+
+$$\begin{aligned}
+ f(x)
+ &= \sum_{n = 0}^\infty \frac{1}{A_n} \Inprod{u_n}{w f} u_n(x)
+ = \int_a^b \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) f(\xi) u_n(x) \Big) \dd{\xi}
+ \\
+ &= \int_a^b f(\xi) \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) \Big) \dd{\xi}
+\end{aligned}$$
+
+Upon closer inspection, the parenthesized summation
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) = \delta(x - \xi)
+ }
+\end{aligned}$$
+
+
+
+## References
+1. O. Bang,
+ *Applied mathematics for physicists: lecture notes*, 2019,
+ unpublished.
diff --git a/source/know/concept/superdense-coding/index.md b/source/know/concept/superdense-coding/index.md
new file mode 100644
index 0000000..f9ffbc1
--- /dev/null
+++ b/source/know/concept/superdense-coding/index.md
@@ -0,0 +1,71 @@
+---
+title: "Superdense coding"
+date: 2021-03-07
+categories:
+- Quantum information
+layout: "concept"
+---
+
+In quantum information, **(super)dense coding**
+is a protocol to enhance classical communication.
+It uses a quantum communication channel and
+[entanglement](/know/concept/quantum-entanglement/)
+to send two bits of classical data with just one qubit.
+It is conceptually similar to [quantum teleportation](/know/concept/quantum-teleportation/).
+
+Suppose that Alice wants to send two bits of classical data to Bob,
+but she can only communicate with him over a quantum channel.
+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.
+
+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:
+
+
+
+ $(a_1, a_2)$
+ Operator
+ Result
+
+
+ $00$
+ $\hat{I}$
+ $\ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big)$
+
+
+ $01$
+ $\hat{\sigma}_z$
+ $\ket{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B - \Ket{1}_A \Ket{1}_B \Big)$
+
+
+ $10$
+ $\hat{\sigma}_x$
+ $\ket{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B + \Ket{1}_A \Ket{0}_B \Big)$
+
+
+ $11$
+ $\hat{\sigma}_x \hat{\sigma}_z$
+ $\ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B - \Ket{1}_A \Ket{0}_B \Big)$
+
+
+
+Her actions affect the state on Bob's side $B$ due to entanglement.
+Alice then sends her qubit $A$ to Bob over the quantum channel,
+so he has both sides of the entangled pair.
+
+Finally, Bob performs a measurement of his pair in the Bell basis,
+which will yield a Bell state that he can then look up in the table above
+to recover the values of the bits $(a_1, a_2)$.
+In the end, Alice only sent a single qubit,
+and the rest of the information transfer was via entanglement.
+
+
+## References
+1. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/thermodynamic-potential/index.md b/source/know/concept/thermodynamic-potential/index.md
new file mode 100644
index 0000000..fe81731
--- /dev/null
+++ b/source/know/concept/thermodynamic-potential/index.md
@@ -0,0 +1,273 @@
+---
+title: "Thermodynamic potential"
+date: 2021-07-07
+categories:
+- Physics
+- Thermodynamics
+layout: "concept"
+---
+
+**Thermodynamic potentials** are state functions
+whose minima or maxima represent equilibrium states of a system.
+Such functions are either energies (hence *potential*) or entropies.
+
+Which potential (of many) decides the equilibrium states for a given system?
+That depends which variables are assumed to already be in automatic equilibrium.
+Such variables are known as the **natural variables** of that potential.
+For example, if a system can freely exchange heat with its surroundings,
+and is consequently assumed to be at the same temperature $T = T_{\mathrm{sur}}$,
+then $T$ must be a natural variable.
+
+The link from natural variables to potentials
+is established by [thermodynamic ensembles](/know/category/thermodynamic-ensembles/).
+
+Once enough natural variables have been found,
+the appropriate potential can be selected from the list below.
+All non-natural variables can then be calculated
+by taking partial derivatives of the potential
+with respect to the natural variables.
+
+Mathematically, the potentials are related to each other
+by [Legendre transformation](/know/concept/legendre-transform/).
+
+
+## Internal energy
+
+The **internal energy** $U$ represents
+the capacity to do both mechanical and non-mechanical work,
+and to release heat.
+It is simply the integral
+of the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/):
+
+$$\begin{aligned}
+ \boxed{
+ U(S, V, N) \equiv T S - P V + \mu N
+ }
+\end{aligned}$$
+
+It is a function of the entropy $S$, volume $V$, and particle count $N$:
+these are its natural variables.
+An infinitesimal change $\dd{U}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+temperature $T$, pressure $P$, and chemical potential $\mu$.
+They can be recovered by differentiating $U$
+with respect to the natural variables $S$, $V$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ T = \Big( \pdv{U}{S} \Big)_{V,N}
+ \qquad
+ P = - \Big( \pdv{U}{V} \Big)_{S,N}
+ \qquad
+ \mu = \Big( \pdv{U}{N} \Big)_{S,V}
+ }
+\end{aligned}$$
+
+It is convention to write those subscripts,
+to help keep track of which function depends on which variables.
+They are meaningless; these are normal partial derivatives.
+
+
+## Enthalpy
+
+The **enthalpy** $H$ of a system, in units of energy,
+represents its capacity to do non-mechanical work,
+plus its capacity to release heat.
+It is given by:
+
+$$\begin{aligned}
+ \boxed{
+ H(S, P, N) \equiv U + P V
+ }
+\end{aligned}$$
+
+It is a function of the entropy $S$, pressure $P$, and particle count $N$:
+these are its natural variables.
+An infinitesimal change $\dd{H}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{H} = T \dd{S} + V \dd{P} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+temperature $T$, volume $V$, and chemical potential $\mu$.
+They can be recovered by differentiating $H$
+with respect to the natural variables $S$, $P$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ T = \Big( \pdv{H}{S} \Big)_{P,N}
+ \qquad
+ V = \Big( \pdv{H}{P} \Big)_{S,N}
+ \qquad
+ \mu = \Big( \pdv{H}{N} \Big)_{S,P}
+ }
+\end{aligned}$$
+
+
+## Helmholtz free energy
+
+The **Helmholtz free energy** $F$ represents
+the capacity of a system to
+do both mechanical and non-mechanical work,
+and is given by:
+
+$$\begin{aligned}
+ \boxed{
+ F(T, V, N) \equiv U - T S
+ }
+\end{aligned}$$
+
+It depends on the temperature $T$, volume $V$, and particle count $N$:
+these are natural variables.
+An infinitesimal change $\dd{H}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{F} = - P \dd{V} - S \dd{T} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+entropy $S$, pressure $P$, and chemical potential $\mu$.
+They can be recovered by differentiating $F$
+with respect to the natural variables $T$, $V$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ S = - \Big( \pdv{F}{T} \Big)_{V,N}
+ \qquad
+ P = - \Big( \pdv{F}{V} \Big)_{T,N}
+ \qquad
+ \mu = \Big( \pdv{F}{N} \Big)_{T,V}
+ }
+\end{aligned}$$
+
+
+## Gibbs free energy
+
+The **Gibbs free energy** $G$ represents
+the capacity of a system to do non-mechanical work:
+
+$$\begin{aligned}
+ \boxed{
+ G(T, P, N)
+ \equiv U + P V - T S
+ }
+\end{aligned}$$
+
+It depends on the temperature $T$, pressure $P$, and particle count $N$:
+they are natural variables.
+An infinitesimal change $\dd{G}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{G} = V \dd{P} - S \dd{T} + \mu \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+entropy $S$, volume $V$, and chemical potential $\mu$.
+These can be recovered by differentiating $G$
+with respect to the natural variables $T$, $P$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ S = - \Big( \pdv{G}{T} \Big)_{P,N}
+ \qquad
+ V = \Big( \pdv{G}{P} \Big)_{T,N}
+ \qquad
+ \mu = \Big( \pdv{G}{N} \Big)_{T,P}
+ }
+\end{aligned}$$
+
+
+## Landau potential
+
+The **Landau potential** or **grand potential** $\Omega$, in units of energy,
+represents the capacity of a system to do mechanical work,
+and is given by:
+
+$$\begin{aligned}
+ \boxed{
+ \Omega(T, V, \mu) \equiv U - T S - \mu N
+ }
+\end{aligned}$$
+
+It depends on temperature $T$, volume $V$, and chemical potential $\mu$:
+these are natural variables.
+An infinitesimal change $\dd{\Omega}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{\Omega} = - P \dd{V} - S \dd{T} - N \dd{\mu}
+ }
+\end{aligned}$$
+
+The non-natural variables are
+entropy $S$, pressure $P$, and particle count $N$.
+These can be recovered by differentiating $\Omega$
+with respect to the natural variables $T$, $V$, and $\mu$:
+
+$$\begin{aligned}
+ \boxed{
+ S = - \Big( \pdv{\Omega}{T} \Big)_{V,\mu}
+ \qquad
+ P = - \Big( \pdv{\Omega}{V} \Big)_{T,\mu}
+ \qquad
+ N = - \Big( \pdv{\Omega}{\mu} \Big)_{T,V}
+ }
+\end{aligned}$$
+
+
+## Entropy
+
+The **entropy** $S$, in units of energy over temperature,
+is an odd duck, but nevertheless used as a thermodynamic potential.
+It is given by:
+
+$$\begin{aligned}
+ \boxed{
+ S(U, V, N) \equiv \frac{1}{T} U + \frac{P}{T} V - \frac{\mu}{T} N
+ }
+\end{aligned}$$
+
+It depends on the internal energy $U$, volume $V$, and particle count $N$:
+they are natural variables.
+An infinitesimal change $\dd{S}$ is as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \dd{S} = \frac{1}{T} \dd{U} + \frac{P}{T} \dd{V} - \frac{\mu}{T} \dd{N}
+ }
+\end{aligned}$$
+
+The non-natural variables are $1/T$, $P/T$, and $\mu/T$.
+These can be recovered by differentiating $S$
+with respect to the natural variables $U$, $V$, and $N$:
+
+$$\begin{aligned}
+ \boxed{
+ \frac{1}{T} = \Big( \pdv{S}{U} \Big)_{V,N}
+ \qquad
+ \frac{P}{T} = \Big( \pdv{S}{V} \Big)_{U,N}
+ \qquad
+ \frac{\mu}{T} = - \Big( \pdv{S}{N} \Big)_{U,V}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Gould, J. Tobochnik,
+ *Statistical and thermal physics*, 2nd edition,
+ Princeton.
diff --git a/source/know/concept/time-dependent-perturbation-theory/index.md b/source/know/concept/time-dependent-perturbation-theory/index.md
new file mode 100644
index 0000000..2b80316
--- /dev/null
+++ b/source/know/concept/time-dependent-perturbation-theory/index.md
@@ -0,0 +1,202 @@
+---
+title: "Time-dependent perturbation theory"
+date: 2021-03-07
+categories:
+- Physics
+- Quantum mechanics
+- Perturbation
+layout: "concept"
+---
+
+In quantum mechanics, **time-dependent perturbation theory** exists to deal
+with time-varying perturbations to the Schrödinger equation.
+This is in contrast to [time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/),
+where the perturbation is stationary.
+
+Let $\hat{H}_0$ be the base time-independent
+Hamiltonian, and $\hat{H}_1$ be a time-varying perturbation, with
+"bookkeeping" parameter $\lambda$:
+
+$$\begin{aligned}
+ \hat{H}(t) = \hat{H}_0 + \lambda \hat{H}_1(t)
+\end{aligned}$$
+
+We assume that the unperturbed time-independent problem
+$\hat{H}_0 \Ket{n} = E_n \Ket{n}$ has already been solved, such that the
+full solution is:
+
+$$\begin{aligned}
+ \Ket{\Psi_0(t)} = \sum_{n} c_n \Ket{n} \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+Since these $\Ket{n}$ form a complete basis, the perturbed wave function
+can be written in the same form, but with time-dependent coefficients $c_n(t)$:
+
+$$\begin{aligned}
+ \Ket{\Psi(t)} = \sum_{n} c_n(t) \Ket{n} \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+We insert this ansatz in the time-dependent Schrödinger equation, and
+reduce it using the known unperturbed time-independent problem:
+
+$$\begin{aligned}
+ 0
+ &= \hat{H}_0 \Ket{\Psi(t)} + \lambda \hat{H}_1 \Ket{\Psi(t)} - i \hbar \dv{}{t}\Ket{\Psi(t)}
+ \\
+ &= \sum_{n}
+ \Big( c_n \hat{H}_0 \Ket{n} + \lambda c_n \hat{H}_1 \Ket{n} - c_n E_n \Ket{n} - i \hbar \dv{c_n}{t} \Ket{n} \Big) \exp(- i E_n t / \hbar)
+ \\
+ &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \Ket{n} - i \hbar \dv{c_n}{t} \Ket{n} \Big) \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+We then take the inner product with an arbitrary stationary basis state $\Ket{m}$:
+
+$$\begin{aligned}
+ 0
+ &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \dv{c_n}{t} \Inprod{m}{n} \Big) \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+Thanks to orthonormality, this removes the latter term from the summation:
+
+$$\begin{aligned}
+ i \hbar \dv{c_m}{t} \exp(- i E_m t / \hbar)
+ &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp(- i E_n t / \hbar)
+\end{aligned}$$
+
+We divide by the left-hand exponential and define
+$\omega_{mn} \equiv (E_m - E_n) / \hbar$ to get:
+
+$$\begin{aligned}
+ \boxed{
+ i \hbar \dv{c_m}{t}
+ = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp(i \omega_{mn} t)
+ }
+\end{aligned}$$
+
+So far, we have not invoked any approximation,
+so we can analytically find $c_n(t)$ for some simple systems.
+Furthermore, it is useful to write this equation in integral form instead:
+
+$$\begin{aligned}
+ c_m(t)
+ = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+\end{aligned}$$
+
+If this cannot be solved exactly, we must approximate it. We expand
+$c_m(t)$ in the usual way, with the initial condition $c_m^{(j)}(0) = 0$
+for $j > 0$:
+
+$$\begin{aligned}
+ c_m(t) = c_m^{(0)} + \lambda c_m^{(1)}(t) + \lambda^2 c_m^{(2)}(t) + ...
+\end{aligned}$$
+
+We then insert this into the integral and collect the non-zero orders of $\lambda$:
+
+$$\begin{aligned}
+ c_m^{(1)}(t)
+ &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+ \\
+ c_m^{(2)}(t)
+ &= - \frac{i}{\hbar} \sum_{n}
+ \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+ \\
+ c_m^{(3)}(t)
+ &= - \frac{i}{\hbar} \sum_{n}
+ \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+\end{aligned}$$
+
+And so forth. The pattern here is clear: we can calculate the $(j\!+\!1)$th
+correction using only our previous result for the $j$th correction.
+We cannot go any further than this without considering a specific perturbation $\hat{H}_1(t)$.
+
+
+## Sinusoidal perturbation
+
+Arguably the most important perturbation
+is a sinusoidally-varying potential, which represents
+e.g. incoming electromagnetic waves,
+or an AC voltage being applied to the system.
+In this case, $\hat{H}_1$ has the following form:
+
+$$\begin{aligned}
+ \hat{H}_1(\vec{r}, t)
+ \equiv V(\vec{r}) \sin(\omega t)
+ = \frac{1}{2 i} V(\vec{r}) \: \big( \exp(i \omega t) - \exp(-i \omega t) \big)
+\end{aligned}$$
+
+We abbreviate $V_{mn} = \matrixel{m}{V}{n}$,
+and take the first-order correction formula:
+
+$$\begin{aligned}
+ c_m^{(1)}(t)
+ &= - \frac{1}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
+ \int_0^t \exp\!\big(i \tau (\omega_{mn} \!+\! \omega)\big) - \exp\!\big(i \tau (\omega_{mn} \!-\! \omega)\big) \dd{\tau}
+ \\
+ &= \frac{i}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
+ \bigg( \frac{\exp\!\big(i t (\omega_{mn} \!+\! \omega) \big) - 1}{\omega_{mn} + \omega}
+ + \frac{\exp\!\big(i t (\omega_{mn} \!-\! \omega) \big) - 1}{\omega_{mn} - \omega} \bigg)
+\end{aligned}$$
+
+For simplicity, we let the system start in a known state $\Ket{a}$,
+such that $c_n^{(0)} = \delta_{na}$,
+and we assume that the driving frequency is close to resonance $\omega \approx \omega_{ma}$,
+such that the second term dominates the first, which can then be neglected.
+We thus get:
+
+$$\begin{aligned}
+ c_m^{(1)}(t)
+ &= i \frac{V_{ma}}{2 \hbar} \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) \big) - 1}{\omega_{ma} - \omega}
+ \\
+ &= i \frac{V_{ma}}{2 \hbar}
+ \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) - \exp\!\big(\!-\! i t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
+ \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
+ \\
+ &= - \frac{V_{ma}}{\hbar}
+ \frac{\sin\!\big( t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
+ \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
+\end{aligned}$$
+
+Taking the norm squared yields the **transition probability**:
+the probability that a particle that started in state $\Ket{a}$
+will be found in $\Ket{m}$ at time $t$:
+
+$$\begin{aligned}
+ \boxed{
+ P_{a \to m}
+ = |c_m^{(1)}(t)|^2
+ = \frac{|V_{ma}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ma} - \omega) t / 2 \big)}{(\omega_{ma} - \omega)^2}
+ }
+\end{aligned}$$
+
+The result would be the same if $\hat{H}_1 \equiv V \cos(\omega t)$.
+However, if instead $\hat{H}_1 \equiv V \exp(- i \omega t)$,
+the result is larger by a factor of $4$,
+which can cause confusion when comparing literature.
+
+In any case, the probability oscillates as a function of $t$
+with period $T = 2 \pi / (\omega_{ma} \!-\! \omega)$,
+so after one period the particle is back in $\Ket{a}$,
+and after $T/2$ the particle is in $\Ket{b}$.
+See [Rabi oscillation](/know/concept/rabi-oscillation/)
+for a more accurate treatment of this "flopping" behaviour.
+
+However, when regarded as a function of $\omega$,
+the probability takes the form of
+a sinc-function centred around $(\omega_{ma} \!-\! \omega)$,
+so it is highest for transitions with energy $\hbar \omega = E_m \!-\! E_a$.
+
+Also note that the sinc-distribution becomes narrower over time,
+which roughly means that it takes some time
+for the system to "notice" that
+it is being driven periodically.
+In other words, there is some "inertia" to it.
+
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
+2. R. Shankar,
+ *Principles of quantum mechanics*, 2nd edition,
+ Springer.
diff --git a/source/know/concept/time-independent-perturbation-theory/index.md b/source/know/concept/time-independent-perturbation-theory/index.md
new file mode 100644
index 0000000..94aae4e
--- /dev/null
+++ b/source/know/concept/time-independent-perturbation-theory/index.md
@@ -0,0 +1,331 @@
+---
+title: "Time-independent perturbation theory"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+- Perturbation
+- Physics
+layout: "concept"
+---
+
+**Time-independent perturbation theory**, also known as
+**stationary state perturbation theory**, is a specific application of
+perturbation theory to the time-independent Schrödinger
+equation in quantum physics, for
+Hamiltonians of the following form:
+
+$$\begin{aligned}
+ \hat{H} = \hat{H}_0 + \lambda \hat{H}_1
+\end{aligned}$$
+
+Where $\hat{H}_0$ is a Hamiltonian for which the time-independent
+Schrödinger equation has a known solution, and $\hat{H}_1$ is a small
+perturbing Hamiltonian. The eigenenergies $E_n$ and eigenstates
+$\Ket{\psi_n}$ of the composite problem are expanded in the
+perturbation "bookkeeping" parameter $\lambda$:
+
+$$\begin{aligned}
+ \Ket{\psi_n}
+ &= \ket{\psi_n^{(0)}} + \lambda \ket{\psi_n^{(1)}} + \lambda^2 \ket{\psi_n^{(2)}} + ...
+ \\
+ E_n
+ &= E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + ...
+\end{aligned}$$
+
+Where $E_n^{(1)}$ and $\ket{\psi_n^{(1)}}$ are called the **first-order
+corrections**, and so on for higher orders. We insert this into the
+Schrödinger equation:
+
+$$\begin{aligned}
+ \hat{H} \Ket{\psi_n}
+ &= \hat{H}_0 \ket{\psi_n^{(0)}}
+ + \lambda \big( \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}} \big) \\
+ &\qquad + \lambda^2 \big( \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}} \big) + ...
+ \\
+ E_n \Ket{\psi_n}
+ &= E_n^{(0)} \ket{\psi_n^{(0)}}
+ + \lambda \big( E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}} \big) \\
+ &\qquad + \lambda^2 \big( E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}} \big) + ...
+\end{aligned}$$
+
+If we collect the terms according to the order of $\lambda$, we arrive
+at the following endless series of equations, of which in practice only
+the first three are typically used:
+
+$$\begin{aligned}
+ \hat{H}_0 \ket{\psi_n^{(0)}}
+ &= E_n^{(0)} \ket{\psi_n^{(0)}}
+ \\
+ \hat{H}_1 \ket{\psi_n^{(0)}} + \hat{H}_0 \ket{\psi_n^{(1)}}
+ &= E_n^{(1)} \ket{\psi_n^{(0)}} + E_n^{(0)} \ket{\psi_n^{(1)}}
+ \\
+ \hat{H}_1 \ket{\psi_n^{(1)}} + \hat{H}_0 \ket{\psi_n^{(2)}}
+ &= E_n^{(2)} \ket{\psi_n^{(0)}} + E_n^{(1)} \ket{\psi_n^{(1)}} + E_n^{(0)} \ket{\psi_n^{(2)}}
+ \\
+ ...
+ &= ...
+\end{aligned}$$
+
+The first equation is the unperturbed problem, which we assume has
+already been solved, with eigenvalues $E_n^{(0)} = \varepsilon_n$ and
+eigenvectors $\ket{\psi_n^{(0)}} = \Ket{n}$:
+
+$$\begin{aligned}
+ \hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}
+\end{aligned}$$
+
+The approach to solving the other two equations varies depending on
+whether this $\hat{H}_0$ has a degenerate spectrum or not.
+
+
+## Without degeneracy
+
+We start by assuming that there is no degeneracy, in other words, each
+$\varepsilon_n$ corresponds to one $\Ket{n}$. At order $\lambda^1$, we
+rewrite the equation as follows:
+
+$$\begin{aligned}
+ (\hat{H}_1 - E_n^{(1)}) \Ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} = 0
+\end{aligned}$$
+
+Since $\Ket{n}$ form a complete basis, we can express
+$\ket{\psi_n^{(1)}}$ in terms of them:
+
+$$\begin{aligned}
+ \ket{\psi_n^{(1)}} = \sum_{m \neq n} c_m \Ket{m}
+\end{aligned}$$
+
+Importantly, $n$ has been removed from the summation to prevent dividing
+by zero later. We are allowed to do this, because
+$\ket{\psi_n^{(1)}} - c_n \Ket{n}$ also satisfies the order-$\lambda^1$
+equation for any value of $c_n$, as demonstrated here:
+
+$$\begin{aligned}
+ (\hat{H}_1 - E_n^{(1)}) \Ket{n} + (\hat{H}_0 - \varepsilon_n) \ket{\psi_n^{(1)}} - (\varepsilon_n - \varepsilon_n) c_n \Ket{n} = 0
+\end{aligned}$$
+
+Where we used $\hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}$.
+We insert the series form of $\ket{\psi_n^{(1)}}$ into the $\lambda^1$-equation:
+
+$$\begin{aligned}
+ (\hat{H}_1 - E_n^{(1)}) \Ket{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \Ket{m} = 0
+\end{aligned}$$
+
+We then put an arbitrary basis vector $\Bra{k}$ in front of this
+equation to get:
+
+$$\begin{aligned}
+ \matrixel{k}{\hat{H}_1}{n} - E_n^{(1)} \Inprod{k}{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \Inprod{k}{m} = 0
+\end{aligned}$$
+
+Suppose that $k = n$. Since $\Ket{n}$ form an orthonormal basis, we end
+up with:
+
+$$\begin{aligned}
+ \boxed{
+ E_n^{(1)} = \matrixel{n}{\hat{H}_1}{n}
+ }
+\end{aligned}$$
+
+In other words, the first-order energy correction $E_n^{(1)}$ is the
+expectation value of the perturbation $\hat{H}_1$ for the unperturbed
+state $\Ket{n}$.
+
+Suppose now that $k \neq n$, then only one term of the summation
+survives, and we are left with the following equation, which tells us
+$c_l$:
+
+$$\begin{aligned}
+ \matrixel{k}{\hat{H}_1}{n} + c_k (\varepsilon_k - \varepsilon_n) = 0
+\end{aligned}$$
+
+We isolate this result for $c_k$ and insert it into the series form of
+$\ket{\psi_n^{(1)}}$ to get the full first-order correction to the wave
+function:
+
+$$\begin{aligned}
+ \boxed{
+ \ket{\psi_n^{(1)}}
+ = \sum_{m \neq n} \frac{\matrixel{m}{\hat{H}_1}{n}}{\varepsilon_n - \varepsilon_m} \Ket{m}
+ }
+\end{aligned}$$
+
+Here it is clear why this is only valid in the non-degenerate case:
+otherwise we would divide by zero in the denominator.
+
+Next, to find the second-order energy correction $E_n^{(2)}$,
+we take the corresponding equation and put $\Bra{n}$ in front of it:
+
+$$\begin{aligned}
+ \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}}
+ &= E_n^{(2)} \Inprod{n}{n} + E_n^{(1)} \inprod{n}{\psi_n^{(1)}} + \varepsilon_n \inprod{n}{\psi_n^{(2)}}
+\end{aligned}$$
+
+Because $\hat{H}_0$ is Hermitian, we know that
+$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \inprod{n}{\psi_n^{(2)}}$,
+i.e. we apply it to the bra, which lets us eliminate two terms. Also,
+since $\Ket{n}$ is normalized, we find:
+
+$$\begin{aligned}
+ E_n^{(2)}
+ = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \inprod{n}{\psi_n^{(1)}}
+\end{aligned}$$
+
+We explicitly removed the $\Ket{n}$-dependence of $\ket{\psi_n^{(1)}}$,
+so the last term is zero. By simply inserting our result for
+$\ket{\psi_n^{(1)}}$, we thus arrive at:
+
+$$\begin{aligned}
+ \boxed{
+ E_n^{(2)}
+ = \sum_{m \neq n} \frac{\big| \matrixel{m}{\hat{H}_1}{n} \big|^2}{\varepsilon_n - \varepsilon_m}
+ }
+\end{aligned}$$
+
+In practice, it is not particulary useful to calculate more corrections.
+
+
+## With degeneracy
+
+If $\varepsilon_n$ is $D$-fold degenerate, then its eigenstate could be
+any vector $\Ket{n, d}$ from the corresponding $D$-dimensional
+eigenspace:
+
+$$\begin{aligned}
+ \hat{H}_0 \Ket{n} = \varepsilon_n \Ket{n}
+ \quad \mathrm{where} \quad
+ \Ket{n}
+ = \sum_{d = 1}^{D} c_{d} \Ket{n, d}
+\end{aligned}$$
+
+In general, adding the perturbation $\hat{H}_1$ will *lift* the
+degeneracy, meaning the perturbed states will be non-degenerate. In the
+limit $\lambda \to 0$, these $D$ perturbed states change into $D$
+orthogonal states which are all valid $\Ket{n}$.
+
+However, the $\Ket{n}$ that they converge to are not arbitrary: only
+certain unperturbed eigenstates are "good" states. Without $\hat{H}_1$,
+this distinction is irrelevant, but in the perturbed case it will turn
+out to be important.
+
+For now, we write $\Ket{n, d}$ to refer to any orthonormal set of
+vectors in the eigenspace of $\varepsilon_n$ (not necessarily the "good"
+ones), and $\Ket{n}$ to denote any linear combination of these. We then
+take the equation at order $\lambda^1$ and prepend an arbitrary
+eigenspace basis vector $\Bra{n, \delta}$:
+
+$$\begin{aligned}
+ \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
+ &= E_n^{(1)} \Inprod{n, \delta}{n} + \varepsilon_n \inprod{n, \delta}{\psi_n^{(1)}}
+\end{aligned}$$
+
+Since $\hat{H}_0$ is Hermitian, we use the same trick as before to
+reduce the problem to:
+
+$$\begin{aligned}
+ \matrixel{n, \delta}{\hat{H}_1}{n}
+ &= E_n^{(1)} \Inprod{n, \delta}{n}
+\end{aligned}$$
+
+We express $\Ket{n}$ as a linear combination of the eigenbasis vectors
+$\Ket{n, d}$ to get:
+
+$$\begin{aligned}
+ \sum_{d = 1}^{D} c_d \matrixel{n, \delta}{\hat{H}_1}{n, d}
+ = E_n^{(1)} \sum_{d = 1}^{D} c_d \Inprod{n, \delta}{n, d}
+ = c_{\delta} E_n^{(1)}
+\end{aligned}$$
+
+Let us now interpret the summation terms as matrix elements
+$M_{\delta, d}$:
+
+$$\begin{aligned}
+ M_{\delta, d} = \matrixel{n, \delta}{\hat{H}_1}{n, d}
+\end{aligned}$$
+
+By varying the value of $\delta$ from $1$ to $D$, we end up with
+equations of the form:
+
+$$\begin{aligned}
+ \begin{bmatrix}
+ M_{1, 1} & \cdots & M_{1, D} \\
+ \vdots & \ddots & \vdots \\
+ M_{D, 1} & \cdots & M_{D, D}
+ \end{bmatrix}
+ \begin{bmatrix}
+ c_1 \\ \vdots \\ c_D
+ \end{bmatrix}
+ = E_n^{(1)}
+ \begin{bmatrix}
+ c_1 \\ \vdots \\ c_D
+ \end{bmatrix}
+\end{aligned}$$
+
+This is an eigenvalue problem for $E_n^{(1)}$, where $c_d$ are the
+components of the eigenvectors which represent the "good" states.
+After solving this, let $\Ket{n, g}$ be the resulting "good" states.
+Then, as long as $E_n^{(1)}$ is a non-degenerate eigenvalue of $M$:
+
+$$\begin{aligned}
+ \boxed{
+ E_{n, g}^{(1)} = \matrixel{n, g}{\hat{H}_1}{n, g}
+ }
+\end{aligned}$$
+
+Which is the same as in the non-degenerate case! Even better, the
+first-order wave function correction is also unchanged:
+
+$$\begin{aligned}
+ \boxed{
+ \ket{\psi_{n,g}^{(1)}}
+ = \sum_{m \neq (n, g)} \frac{\matrixel{m}{\hat{H}_1}{n, g}}{\varepsilon_n - \varepsilon_m} \Ket{m}
+ }
+\end{aligned}$$
+
+This works because the matrix $M$ is diagonal in the $\Ket{n, g}$-basis,
+such that when $\Ket{m}$ is any vector $\Ket{n, \gamma}$ in the
+$\Ket{n}$-eigenspace (except for $\Ket{n,g}$, which is
+explicitly excluded), then the corresponding numerator
+$\matrixel{n, \gamma}{\hat{H}_1}{n, g} = M_{\gamma, g} = 0$, so the term
+does not contribute.
+
+If any of the eigenvalues $E_n^{(1)}$ of $M$ are degenerate, then there
+is still information missing about the components $c_d$ of the
+"good" states, in which case we must find them some other way.
+
+Such an alternative way of determining these "good" states is also of
+interest even if there is no degeneracy in $M$, since such a shortcut would
+allow us to use the formulae from non-degenerate perturbation theory
+straight away.
+
+The trick is to find a Hermitian operator $\hat{L}$ (usually using
+symmetries of the system) which commutes with both $\hat{H}_0$ and $\hat{H}_1$:
+
+$$\begin{aligned}
+ \comm{\hat{L}}{\hat{H}_0} = \comm{\hat{L}}{\hat{H}_1} = 0
+\end{aligned}$$
+
+So that it shares its eigenstates with $\hat{H}_0$ (and $\hat{H}_1$),
+meaning all the vectors of the $D$-dimensional
+$\Ket{n}$-eigenspace are also eigenvectors of $\hat{L}$.
+
+The crucial part, however, is that $\hat{L}$ must be chosen such that
+$\Ket{n, d_1}$ and $\Ket{n, d_2}$ have distinct eigenvalues
+$\ell_1 \neq \ell_2$ for $d_1 \neq d_2$:
+
+$$\begin{aligned}
+ \hat{L} \Ket{n, d_1} = \ell_1 \Ket{n, d_1}
+ \qquad
+ \hat{L} \Ket{n, d_2} = \ell_2 \Ket{n, d_2}
+\end{aligned}$$
+
+When this condition holds for any orthogonal choice of $\Ket{n, d_1}$ and
+$\Ket{n, d_2}$, then these specific eigenvectors of $\hat{L}$ are the
+"good states", for any valid choice of $\hat{L}$.
+
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
diff --git a/source/know/concept/time-ordered-product/index.md b/source/know/concept/time-ordered-product/index.md
new file mode 100644
index 0000000..37f47b7
--- /dev/null
+++ b/source/know/concept/time-ordered-product/index.md
@@ -0,0 +1,118 @@
+---
+title: "Time-ordered product"
+date: 2021-09-13
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, especially quantum field theory,
+a **time-ordered product** is a product of
+explicitly time-dependent operators,
+subject to certain ordering constraints.
+
+Let us start with an unusual motivation.
+Suppose that some time-dependent operator $\hat{A}(t)$ is defined like so,
+as a product of $N$ time-dependent sub-operators $\hat{a}_n(t)$:
+
+$$\begin{aligned}
+ \hat{A}(t)
+ \equiv \int_0^{t} \hat{a}_1(t_1) \bigg( \int_0^{t_1} \hat{a}_2(t_2) \bigg( \int_0^{t_2} \hat{a}_3(t_3) \bigg( \cdots \bigg)
+ \dd{t_3} \bigg) \dd{t_2} \bigg) \dd{t_1}
+\end{aligned}$$
+
+Crucially, the upper limits of the inner integrals
+depend on the surrounding variables,
+meaning that these integrals cannot simply be reordered.
+
+An interpretation is that the rightmost $\hat{a}_N(t_N)$ is applied first,
+and then $\hat{a}_{N-1}(t_{N-1})$ secondly with $t_{N-1} > t_N$,
+and so on.
+This suggests there is a form of "time-ordering" here:
+the integrals sweep across all relative timings of $\hat{a}_n$,
+but preserve the ordering.
+Indeed, this could be rewritten as a time-ordered product
+(see the [interaction picture](/know/concept/interaction-picture/) for an example).
+
+A more general and intuitive motivation goes as follows.
+Suppose we have a product of $N$ time-dependent operators $\hat{a}_n(t)$,
+each representing a certain event.
+Clearly, we would want to apply them in chronological order:
+
+$$\begin{aligned}
+ \hat{a}_N(t_N) \: \hat{a}_{N-1}(t_{N-1}) \: \cdots \: \hat{a}_2(t_2) \: \hat{a}_1(t_1)
+ \qquad \mathrm{where} \qquad
+ t_N > t_{N-1} > ... > \: t_2 > t_1
+\end{aligned}$$
+
+But what if the ordering of the arguments $t_N, ..., t_1$
+is not known in advance?
+We thus define the **time-ordering meta-operator** $\mathcal{T}$,
+which reorders the operators based on the $t$-values
+such that they are always in chronological order.
+For example:
+
+$$\begin{aligned}
+ \mathcal{T} \big\{ \hat{a}_1(t_1) \: \hat{a}_2(t_2) \big\}
+ \equiv
+ \begin{cases}
+ \hat{a}_1(t_1) \: \hat{a}_2(t_2) & \mathrm{if} \; t_2 < t_1 \\
+ \hat{a}_2(t_2) \: \hat{a}_1(t_1) & \mathrm{if} \; t_1 < t_2
+ \end{cases}
+\end{aligned}$$
+
+This example suggests a general algorithm for $\mathcal{T}$:
+we need to consider every permutation of the operators $\hat{a}_n(t_n)$,
+and leave only the single one that satisfies our demands.
+
+Mathematically, we do this by summing up all permutations,
+and multiplying each term with a product of
+[Heaviside step functions](/know/concept/heaviside-step-function/) $\Theta$,
+which remove the term if the ordering is wrong:
+
+$$\begin{aligned}
+ \mathcal{T} \big\{ \hat{a}_1 \cdots \hat{a}_N \big\}
+ \equiv \sum_{p \in P_N}^{}
+ \Theta\big(t_{p_1} \!\!-\! t_{p_2}\big) \cdots \Theta\big(t_{p_{N-1}} \!\!-\! t_{p_N}\big)
+ \: \hat{a}_{p_1}(t_{p_1}) \: \cdots \: \hat{a}_{p_N}(t_{p_N})
+\end{aligned}$$
+
+With this, our earlier example for two operators $\hat{a}_1$ and $\hat{a}_2$
+takes the following form:
+
+$$\begin{aligned}
+ \mathcal{T} \big\{ \hat{a}_1(t_1) \: \hat{a}_2(t_2) \big\}
+ = \Theta(t_1 - t_2) \: \hat{a}_1(t_1) \: \hat{a}_2(t_2) + \Theta(t_2 - t_1) \: \hat{a}_2(t_2) \: \hat{a}_1(t_1)
+\end{aligned}$$
+
+However, we are still missing an important detail:
+so far, we have quietly been assuming that the operators are bosonic
+(see [second quantization](/know/concept/second-quantization/)).
+To include fermionic operators,
+we must allow the sign of each term to change,
+based on whether the permutation is even or odd:
+
+$$\begin{aligned}
+ \mathcal{T} \big\{ \hat{a}_1(t_1) \: \hat{a}_2(t_2) \big\}
+ = \Theta(t_1 - t_2) \: \hat{a}_1(t_1) \: \hat{a}_2(t_2) \pm \Theta(t_2 - t_1) \: \hat{a}_2(t_2) \: \hat{a}_1(t_1)
+\end{aligned}$$
+
+Where $\pm$ is $+$ for bosons, and $-$ for fermions in this case.
+The general definition of $\mathcal{T}$ is:
+
+$$\begin{aligned}
+ \boxed{
+ \mathcal{T} \big\{ \hat{a}_1 \cdots \hat{a}_N \big\}
+ \equiv \sum_{p \in P_N}^{} (\pm 1)^p
+ \bigg( \prod_{j = 1}^{N-1} \Theta\big(t_{p_j} \!-\! t_{p_{j+1}}\big) \bigg)
+ \bigg( \prod_{k = 1}^N \hat{a}_{p_k}(t_{p_k}) \bigg)
+ }
+\end{aligned}$$
+
+
+
+## References
+1. H. Bruus, K. Flensberg,
+ *Many-body quantum theory in condensed matter physics*,
+ 2016, Oxford.
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+---
+title: "Toffoli gate"
+date: 2021-04-09
+categories:
+- Quantum information
+layout: "concept"
+---
+
+The **Toffoli gate** or **controlled-controlled-NOT (CCNOT) gate**
+is a logic gate that is *reversible* (no information is lost)
+and *universal* (all reversible logic circuits can be built using Toffoli gates).
+
+It takes three input bits $A$, $B$ and $C$,
+of which it returns $A$ and $B$ unchanged,
+and flips $C$ if both $A$ and $B$ are true.
+In circuit diagrams, its representation is:
+
+
+
+
+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:
+
+
+
+
+A NAND is enough to implement every conceivable circuit.
+That said, we can efficiently implement NOT, AND, and XOR using a single Toffoli gate too.
+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 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:
+
+$$\begin{aligned}
+ \boxed{
+ \mathrm{CCNOT} =
+ \begin{bmatrix}
+ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
+ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
+ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
+ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
+ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0
+ \end{bmatrix}
+ }
+\end{aligned}$$
+
+If we apply this gate to an arbitrary three-qubit state $\Ket{\psi}$,
+it swaps the last two coefficients:
+
+$$\begin{aligned}
+ \mathrm{CCNOT} \Ket{\psi}
+ &= \mathrm{CCNOT} \big( c_{000} \Ket{000} + c_{001} \Ket{001} + c_{010} \Ket{010} + c_{011} \Ket{011} \\
+ &\qquad\qquad\quad\:\; c_{100} \Ket{100} + c_{101} \Ket{101} + c_{110} \Ket{110} + c_{111} \Ket{111} \big)
+ \\
+ &= c_{000} \Ket{000} + c_{001} \Ket{001} + c_{010} \Ket{010} + c_{011} \Ket{011} \\
+ &\quad\,\, c_{100} \Ket{100} + c_{101} \Ket{101} + c_{111} \Ket{110} + c_{110} \Ket{111}
+\end{aligned}$$
+
+
+
+## References
+1. J.S. Neergaard-Nielsen,
+ *Quantum information: lectures notes*,
+ 2021, unpublished.
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diff --git a/source/know/concept/two-fluid-equations/index.md b/source/know/concept/two-fluid-equations/index.md
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+---
+title: "Two-fluid equations"
+date: 2021-10-19
+categories:
+- Physics
+- Plasma physics
+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").
+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,
+when subjected only to the [Lorentz force](/know/concept/lorentz-force/):
+
+$$\begin{aligned}
+ m \dv{\vb{v}}{t}
+ = q (\vb{E} + \vb{v} \cross \vb{B})
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ m n \frac{\mathrm{D} \vb{u}}{\mathrm{D} t}
+ = q n (\vb{E} + \vb{u} \cross \vb{B})
+\end{aligned}$$
+
+Where we have multiplied by the number density $n$ of the particles.
+Due to particle collisions in the fluid,
+stresses become important. Therefore, we include
+the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{P}$,
+leading to the following two equations:
+
+$$\begin{aligned}
+ m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+ &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) + \nabla \cdot \hat{P}_i{}^\top
+ \\
+ m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
+ &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) + \nabla \cdot \hat{P}_e{}^\top
+\end{aligned}$$
+
+Where the subscripts $i$ and $e$ refer to ions and electrons, respectively.
+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**:
+
+$$\begin{aligned}
+ \boxed{
+ \begin{aligned}
+ m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+ &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) + \nabla \cdot \hat{P}_i{}^\top - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
+ \\
+ m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
+ &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) + \nabla \cdot \hat{P}_e{}^\top - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
+ \end{aligned}
+ }
+\end{aligned}$$
+
+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 $p$:
+
+$$\begin{aligned}
+ m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+ &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
+ \\
+ m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
+ &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
+\end{aligned}$$
+
+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 $S$:
+
+$$\begin{aligned}
+ 0
+ &= \pdv{}{t}\int_V n \dd{V} + \oint_S n \vb{u} \cdot \dd{\vb{S}}
+ = \int_V \Big( \pdv{n}{t} + \nabla \cdot (n \vb{u}) \Big) \dd{V}
+\end{aligned}$$
+
+Where we have used the divergence theorem.
+Since $V$ is arbitrary, we can remove the integrals,
+leading to the following **continuity equations**:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{n_i}{t} + \nabla \cdot (n_i \vb{u}_i)
+ = 0
+ \qquad \quad
+ \pdv{n_e}{t} + \nabla \cdot (n_e \vb{u}_e)
+ = 0
+ }
+\end{aligned}$$
+
+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 $p_e$.
+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
+(since Gauss' laws are redundant; see the article on Maxwell's equations):
+
+$$\begin{aligned}
+ \boxed{
+ \nabla \cross \vb{E} = - \pdv{\vb{B}}{t}
+ \qquad \quad
+ \nabla \cross \vb{B} = \mu_0 \Big( n_i q_i \vb{u}_i + n_e q_e \vb{u}_e + \varepsilon_0 \pdv{\vb{E}}{t} \Big)
+ }
+\end{aligned}$$
+
+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),
+it turns out that:
+
+$$\begin{aligned}
+ \frac{\mathrm{D}}{\mathrm{D} t} \big( p V^\gamma \big) = 0
+ \qquad \quad
+ \gamma
+ \equiv \frac{C_P}{C_V}
+ = \frac{N + 2}{N}
+\end{aligned}$$
+
+Where $\gamma$ is the *heat capacity ratio*,
+and can be calculated from the number of degrees of freedom $N$
+of each particle in the gas.
+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 $C$:
+
+$$\begin{aligned}
+ \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p}{n^\gamma} \Big) = 0
+ \quad \implies \quad
+ p = C n^\gamma
+\end{aligned}$$
+
+In the two-fluid model, we thus have the following two equations of state,
+giving us a set of 16 equations for 16 unknowns:
+
+$$\begin{aligned}
+ \boxed{
+ \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p_i}{n_i^\gamma} \Big)
+ = 0
+ \qquad \quad
+ \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p_e}{n_e^\gamma} \Big)
+ = 0
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ p = C n^\gamma
+ \quad \implies \quad
+ \nabla p
+ = \gamma \frac{C n^{\gamma}}{n} \nabla n
+ = \gamma p \frac{\nabla n}{n}
+ = \gamma k_B T \nabla n
+\end{aligned}$$
+
+Note that the ideal gas law was not used immediately,
+to allow for $\gamma \neq 1$.
+
+
+## Fluid drifts
+
+The momentum equations reduce to the following
+if we assume the flow is steady $\ipdv{\vb{u}}{t} = 0$,
+and neglect electron-ion momentum transfer on the right:
+
+$$\begin{aligned}
+ m_i n_i (\vb{u}_i \cdot \nabla) \vb{u}_i
+ &\approx q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i
+ \\
+ m_e n_e (\vb{u}_e \cdot \nabla) \vb{u}_e
+ &\approx q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ 0
+ &= q n (\vb{E} + \vb{u}_\perp \cross \vb{B}) \cross \vb{B} - \nabla p \cross \vb{B} - m n \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}
+ \\
+ &= q n (\vb{E} \cross \vb{B} - \vb{u}_\perp B^2) - \nabla p \cross \vb{B} - m n \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}
+\end{aligned}$$
+
+Isolating for $\vb{u}_\perp$ tells us
+that the fluids drifts perpendicularly to $\vb{B}$,
+with velocity $\vb{u}_\perp$:
+
+$$\begin{aligned}
+ \vb{u}_\perp
+ = \frac{\vb{E} \cross \vb{B}}{B^2} - \frac{\nabla p \cross \vb{B}}{q n B^2}
+ - \frac{m \big( (\vb{u} \cdot \nabla) \vb{u} \big) \cross \vb{B}}{q B^2}
+\end{aligned}$$
+
+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 $\vb{v}_E$
+from [guiding center theory](/know/concept/guiding-center-theory/),
+and the second term is called the **diamagnetic drift** $\vb{v}_D$:
+
+$$\begin{aligned}
+ \boxed{
+ \vb{v}_E
+ = \frac{\vb{E} \cross \vb{B}}{B^2}
+ }
+ \qquad \quad
+ \boxed{
+ \vb{v}_D
+ = - \frac{\nabla p \cross \vb{B}}{q n B^2}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \vb{J}
+ = q_e n_e (\vb{v}_{De} - \vb{v}_{Di})
+ = q_e n_e \Big( \frac{\nabla p_i \cross \vb{B}}{q_i n_i B^2} - \frac{\nabla p_e \cross \vb{B}}{q_e n_e B^2} \Big)
+ = \frac{\vb{B} \cross \nabla (p_i + p_e)}{B^2}
+\end{aligned}$$
+
+Using the ideal gas law $p = k_B T n$,
+this can be rewritten as follows:
+
+$$\begin{aligned}
+ \vb{J}
+ = k_B \frac{\vb{B} \cross \nabla (T_i n_i + T_e n_e)}{B^2}
+\end{aligned}$$
+
+Curiously, $\vb{v}_D$ does not involve any net movement of particles,
+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,
+so that the particle flux through a reference plane is higher.
+This causes the fluid elements to drift,
+but not the guiding centers.
+
+
+
+## References
+1. F.F. Chen,
+ *Introduction to plasma physics and controlled fusion*,
+ 3rd edition, Springer.
+2. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/viscosity/index.md b/source/know/concept/viscosity/index.md
new file mode 100644
index 0000000..026fc8f
--- /dev/null
+++ b/source/know/concept/viscosity/index.md
@@ -0,0 +1,94 @@
+---
+title: "Viscosity"
+date: 2021-04-12
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+The **viscosity** of a fluid describes how
+"sticky" its constituent molecules are;
+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 $z$.
+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 $v(z_s) = v_s$.
+
+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 $\sigma_{xz}$
+(see [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/)).
+Formally, the **dynamic viscosity** $\eta$ is defined as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \sigma_{xz}
+ = \eta \dv{v}{z}
+ }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \sigma_{ij} = - p \delta_{ij}
+\end{aligned}$$
+
+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 $\nabla_i \!=\! \ipdv{}{x_i}$:
+
+$$\begin{aligned}
+ \boxed{
+ \sigma_{ij}
+ = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
+ }
+\end{aligned}$$
+
+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.
+
+Another quantity is the **kinematic viscosity** $\nu$,
+which is simply $\eta$ divided by the density $\rho$:
+
+$$\begin{aligned}
+ \boxed{
+ \nu
+ \equiv \frac{\eta}{\rho}
+ }
+\end{aligned}$$
+
+With this, Newton's law of viscosity is written
+using the momentum density $P = \rho v$:
+
+$$\begin{aligned}
+ \sigma_{xz}
+ = \nu \dv{P}{z}
+\end{aligned}$$
+
+Because momentum is "more fundamental" than velocity,
+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 $\nu_\mathrm{water} < \nu_\mathrm{air}$.
+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.
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/von-neumann-extractor/index.md b/source/know/concept/von-neumann-extractor/index.md
new file mode 100644
index 0000000..5cc13a6
--- /dev/null
+++ b/source/know/concept/von-neumann-extractor/index.md
@@ -0,0 +1,79 @@
+---
+title: "Von Neumann extractor"
+date: 2021-04-09
+categories:
+- Cryptography
+layout: "concept"
+---
+
+The **Von Neumann extractor** is a simple example of a **randomness extractor**:
+given a stream of "imperfectly random" bits,
+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 $1/2$;
+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.
+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 $a_{n+1}$.
+
+Evidently, the first case $a_n = a_{n+1}$ occurs with the following probabilities:
+
+$$\begin{aligned}
+ P(0, 0)
+ = p^2
+ \qquad \qquad
+ P(1, 1)
+ = (1 - p)^2
+\end{aligned}$$
+
+Meanwhile, the second case $a_n \neq a_{n+1}$ occurs with probabilities given by:
+
+$$\begin{aligned}
+ P(0, 1)
+ = p (p - 1)
+ \qquad \qquad
+ P(1, 0)
+ = (p - 1) p
+\end{aligned}$$
+
+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 $p$.
+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:
+
+$$\begin{aligned}
+ N_\mathrm{out}
+ = \frac{1}{2} N_\mathrm{in} \Big( P(0, 1) + P(1, 0) \Big)
+ = \frac{1}{2} N_\mathrm{in} \Big( 2 p (p - 1) \Big)
+ = N_\mathrm{in} p (p - 1)
+\end{aligned}$$
+
+The key assumption that allows the Von Neumann extractor to work
+is that there is no correlation at all between the bits.
+In practice, this may be difficult to achieve,
+in which case a more complex randomness extraction scheme is needed.
+
+
+
+## References
+1. J.B. Brask,
+ *Quantum information: lecture notes*,
+ 2021, unpublished.
diff --git a/source/know/concept/vorticity/index.md b/source/know/concept/vorticity/index.md
new file mode 100644
index 0000000..518025c
--- /dev/null
+++ b/source/know/concept/vorticity/index.md
@@ -0,0 +1,159 @@
+---
+title: "Vorticity"
+date: 2021-04-03
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+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 $\va{v}$:
+
+$$\begin{aligned}
+ \boxed{
+ \va{\omega}
+ \equiv \nabla \cross \va{v}
+ }
+\end{aligned}$$
+
+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.
+
+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 $C$.
+Then, by Stokes' theorem:
+
+$$\begin{aligned}
+ \boxed{
+ \Gamma(C, t)
+ \equiv \oint_C \va{v} \cdot \dd{\va{l}}
+ = \int_S \va{\omega} \cdot \dd{\va{S}}
+ }
+\end{aligned}$$
+
+
+## Ideal fluids
+
+For an inviscid, incompressible fluid,
+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.
+We then take the gradient of this scalar field:
+
+$$\begin{aligned}
+ \nabla H
+ &= \frac{1}{2} \nabla \va{v}^2 + \nabla \Phi + \frac{\nabla p}{\rho}
+ \\
+ &= \va{v} \cdot (\nabla \va{v}) - \Big( \!-\! \nabla \Phi - \frac{\nabla p}{\rho} \Big)
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \nabla H
+ &= \va{v} \cdot (\nabla \va{v}) - \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+ = \va{v} \cdot (\nabla \va{v}) - \pdv{\va{v}}{t} - (\va{v} \cdot \nabla) \va{v}
+\end{aligned}$$
+
+We isolate this equation for $\ipdv{\va{v}}{t}$,
+and apply a vector identity to reduce it to the following:
+
+$$\begin{aligned}
+ \pdv{\va{v}}{t}
+ = \va{v} \cdot (\nabla \va{v}) - (\va{v} \cdot \nabla) \va{v} - \nabla H
+ = \va{v} \cross (\nabla \cross \va{v}) - \nabla H
+\end{aligned}$$
+
+Here, the definition of the vorticity $\va{\omega}$ is clear to see,
+leading us to an equation of motion for $\va{v}$:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{\va{v}}{t}
+ = \va{v} \cross \va{\omega} - \nabla H
+ }
+\end{aligned}$$
+
+More about this later.
+Now, we take the curl of both sides of this equation, giving us:
+
+$$\begin{aligned}
+ \nabla \cross \pdv{\va{v}}{t}
+ = \nabla \cross (\va{v} \cross \va{\omega}) - \nabla \cross (\nabla H)
+\end{aligned}$$
+
+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 $\va{\omega}$:
+
+$$\begin{aligned}
+ \boxed{
+ \pdv{\va{\omega}}{t}
+ = \nabla \cross (\va{v} \cross \va{\omega})
+ }
+\end{aligned}$$
+
+Let us now return to the equation of motion for $\va{v}$.
+For *steady* flows where $\ipdv{\va{v}}{t} = 0$, in which case
+[Bernoulli's theorem](/know/concept/bernoullis-theorem/) applies,
+it reduces to:
+
+$$\begin{aligned}
+ \nabla H
+ = \va{v} \cross \va{\omega}
+\end{aligned}$$
+
+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,
+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 $\Psi$:
+
+$$\begin{aligned}
+ \va{v}
+ = \nabla \Psi
+\end{aligned}$$
+
+This fact allows us to rewrite the Euler equations in a particularly simple way.
+Firstly, the condition of incompressibility becomes the well-known Laplace equation:
+
+$$\begin{aligned}
+ 0
+ = \nabla \cdot \va{v}
+ = \nabla^2 \Psi
+\end{aligned}$$
+
+And second, the main equation of motion for $\va{v}$ states
+that the quantity $H + \ipdv{\Psi}{t}$ is spatially constant
+in the irrotational region:
+
+$$\begin{aligned}
+ \pdv{\va{v}}{t}
+ = \nabla \pdv{\Psi}{t}
+ = - \nabla H
+ \quad \implies \quad
+ \nabla \Big( H + \pdv{\Psi}{t} \Big)
+ = 0
+\end{aligned}$$
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/wetting/index.md b/source/know/concept/wetting/index.md
new file mode 100644
index 0000000..61f31ad
--- /dev/null
+++ b/source/know/concept/wetting/index.md
@@ -0,0 +1,127 @@
+---
+title: "Wetting"
+date: 2021-03-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+- Surface tension
+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 $\theta$.
+
+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,
+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**.
+
+For a perfectly smooth homogeneous surface,
+$\theta$ is determined by
+the [Young-Dupré relation](/know/concept/young-dupre-relation/):
+
+$$\begin{aligned}
+ \alpha_{sg} - \alpha_{sl}
+ = \alpha_{gl} \cos\theta
+\end{aligned}$$
+
+In practice, however, surfaces can be rough and/or inhomogeneous.
+We start with the former.
+
+A rough surface has some structure, which may contain "gaps".
+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** $r$
+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:
+
+$$\begin{aligned}
+ r = \frac{A_{real}}{A_{app}}
+\end{aligned}$$
+
+The net energy cost $E$ of spreading the droplet over the surface is then given by:
+
+$$\begin{aligned}
+ E_{sl}
+ &= (\alpha_{sg} - \alpha_{sl}) A_{real}
+ = \alpha_{gl} A_{real} \cos\theta
+ \\
+ &= \alpha_{gl} A_{app} r \cos\theta
+ = \alpha_{gl} A_{app} \cos\theta^*
+\end{aligned}$$
+
+Where we have defined the **apparent contact angle** $\theta^*$
+as the correction to $\theta$ to account for the roughness.
+It is expressed as follows:
+
+$$\begin{aligned}
+ \boxed{
+ \cos\theta^*
+ = r \cos\theta
+ }
+\end{aligned}$$
+
+For Cassie-Baxter states, where the gaps remain air-filled,
+we define $f$ as the "non-gap" fraction of the apparent surface, such that:
+
+$$\begin{aligned}
+ E
+ &= A_{app} \big( f (\alpha_{sg} - \alpha_{sl}) - (1 - f) \alpha_{gl} \big)
+ \\
+ &= A_{app} \alpha_{gl} \big( f \cos\theta + f - 1 \big)
+\end{aligned}$$
+
+Note the signs: for the solid-liquid interface,
+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:
+
+$$\begin{aligned}
+ \boxed{
+ \cos\theta^*
+ = f (\cos\theta + 1) - 1
+ }
+\end{aligned}$$
+
+We generalize this equation to inhomogeneous surfaces
+consisting of two materials with contact angles $\theta_1$ and $\theta_2$.
+The energy cost of the interface is then given by:
+
+$$\begin{aligned}
+ E
+ &= A \big( f_1 (\alpha_{s1g} - \alpha_{s1l}) + (1 - f_1) (\alpha_{s2g} - \alpha_{s2l}) \big)
+ \\
+ &= A \alpha_{gl} \big( f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2 \big)
+\end{aligned}$$
+
+Such that $\theta^*$ for an inhomogeneous surface is given by this equation,
+called **Cassie's law**:
+
+$$\begin{aligned}
+ \boxed{
+ \cos\theta^*
+ = f_1 \cos\theta_1 + (1 - f_1) \cos\theta_2
+ }
+\end{aligned}$$
+
+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.
+
+
+
+## References
+1. T. Bohr,
+ *Continuum physics: lecture notes*, 2021,
+ unpublished.
+
diff --git a/source/know/concept/wicks-theorem/index.md b/source/know/concept/wicks-theorem/index.md
new file mode 100644
index 0000000..f1a7357
--- /dev/null
+++ b/source/know/concept/wicks-theorem/index.md
@@ -0,0 +1,188 @@
+---
+title: "Wick's theorem"
+date: 2021-05-29
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In the [second quantization](/know/concept/second-quantization/) formalism,
+**Wick's theorem** helps to evaluate products
+of creation and annihilation operators by
+breaking them down into smaller products.
+
+Firstly, let us define the **normal product** or **normal order** as
+a product of second quantization operators
+reordered such that
+all creation operators are on the left of
+all annihilation operators.
+For two operators this is written as follows,
+at least in the case of bosons:
+
+$$\begin{aligned}
+ \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger}
+ \equiv \hat{b}_\beta^\dagger \hat{b}_\alpha
+\end{aligned}$$
+
+For fermions, the result must be negated for each swapping of adjacent operators
+(and every reordering of operators can be treated as a sequence of such swaps):
+
+$$\begin{aligned}
+ \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger}
+ \equiv - \hat{f}_\beta^\dagger \hat{f}_\alpha
+\end{aligned}$$
+
+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 $B$
+is defined as the vacuum matrix element,
+i.e. the expectation value of $\Ket{0}$:
+
+$$\begin{aligned}
+ \Expval{A B}_0
+ \equiv \matrixel{0}{A B}{0}
+\end{aligned}$$
+
+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 $\alpha$.
+
+Wick's theorem states:
+**any product of second quantization operators can be
+rewritten as a sum of normal products,
+from which 0, 1, 2, etc. contractions have been removed
+in every possible way.**
+For fermions, the sign of a term must also be swapped
+every time two adjacent operators are swapped.
+As an example, for four operators:
+
+$$\begin{aligned}
+ A B C D
+ = \underline{A B C D}
+ &+ \underline{A B} \Expval{C D}_0 \pm \underline{A C} \Expval{B D}_0 + \underline{A D} \Expval{B C}_0
+ \\
+ &+ \underline{B C} \Expval{A D}_0 \pm \underline{B D} \Expval{A C}_0 + \underline{C D} \Expval{A B}_0
+ \\
+ &+ \Expval{A B}_0 \Expval{C D}_0 \pm \Expval{A C}_0 \Expval{B D}_0 + \Expval{A D}_0 \Expval{B C}_0
+\end{aligned}$$
+
+Where the negative signs apply to fermions only.
+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.
+
+
+## Proof
+
+We will prove this by induction, with the base case being two operators,
+where Wick's theorem becomes as follows:
+
+$$\begin{aligned}
+ A B
+ = \underline{AB} + \Expval{A B}_0
+\end{aligned}$$
+
+This must be proven separately for fermions and bosons.
+For fermions, a general consequence of the definition of the anticommutator is:
+
+$$\begin{aligned}
+ \hat{f}_\alpha \hat{f}_\beta^\dagger
+ = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}
+\end{aligned}$$
+
+This anticommutator is known to be $\delta_{\alpha\beta}$,
+so we can inconsequentially take
+its inner product with the vacuum state $\Ket{0}$:
+
+$$\begin{aligned}
+ \hat{f}_\alpha \hat{f}_\beta^\dagger
+ &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\{\hat{f}_\alpha, \hat{f}_\beta^\dagger\}}{0}
+ = - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger + \hat{f}_\beta^\dagger \hat{f}_\alpha}{0}
+ \\
+ &= - \hat{f}_\beta^\dagger \hat{f}_\alpha + \matrixel{0}{\hat{f}_\alpha \hat{f}_\beta^\dagger}{0}
+ = \underline{\hat{f}_\alpha \hat{f}_\beta^\dagger} + \expval{\hat{f}_\alpha \hat{f}_\beta^\dagger}_0
+\end{aligned}$$
+
+Which agrees with Wick's theorem. For bosons, we use the commutator:
+
+$$\begin{aligned}
+ \hat{b}_\alpha \hat{b}_\beta^\dagger
+ = \hat{b}_\beta^\dagger \hat{b}_\alpha + [\hat{b}_\alpha, \hat{b}_\beta^\dagger]
+\end{aligned}$$
+
+This commutator is known to be $\delta_{\alpha\beta}$,
+so we take the inner product with $\Ket{0}$, like before:
+
+$$\begin{aligned}
+ \hat{b}_\alpha \hat{b}_\beta^\dagger
+ &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{[\hat{b}_\alpha, \hat{b}_\beta^\dagger]}{0}
+ = \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger - \hat{b}_\beta^\dagger \hat{b}_\alpha}{0}
+ \\
+ &= \hat{b}_\beta^\dagger \hat{b}_\alpha + \matrixel{0}{\hat{b}_\alpha \hat{b}_\beta^\dagger}{0}
+ = \underline{\hat{b}_\alpha \hat{b}_\beta^\dagger} + \expval{\hat{b}_\alpha \hat{b}_\beta^\dagger}_0
+\end{aligned}$$
+
+Which again agrees with Wick's theorem.
+Next, we prove that if it holds for $N$ operators, then it also holds for $N + 1$.
+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:
+
+$$\begin{aligned}
+ \underline{A_1 ... A_N} A_{N+1}
+ = \underline{A_1 ... A_N A_{N+1}}
+ + \sum_{n = 1}^N s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
+\end{aligned}$$
+
+If $A_{N + 1}$ is an annihilation operator, then this is trivial:
+appending it does not break the existing normal order,
+and $\Expval{A_n A_{N+1}}_0 = 0$ for all $A_n$.
+
+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:
+
+$$\begin{aligned}
+ \underline{A_1 ... A_N} A_{N+1}
+ &= s^N A_{N+1} \underline{A_1 ... A_N}
+ + \sum_{n} s^{n + N} \{[A_n, A_{N+1}]\} \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
+ \\
+ &= \underline{A_1 ... A_N A_{N+1}}
+ + \sum_{n} s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_1 ... A_{n-1} A_{n+1} ... A_N}
+\end{aligned}$$
+
+Where $\{[]\}$ is the anticommutator or commutator,
+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 $\underline{A_{v} ... A_{w}} A_{N+1}$.
+Using the relation that we just proved,
+each such product can be rewritten as follows:
+
+$$\begin{aligned}
+ \underline{A_v ... A_w} A_{N+1}
+ &= \underline{A_v ... A_w A_{N+1}}
+ + \sum_{n} s^{n + N} \Expval{A_n A_{N+1}}_0 \underline{A_v ... A_{n-1} A_{n+1} ... A_w}
+\end{aligned}$$
+
+Inserting this back into Wick's theorem,
+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.
+Therefore,
+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 $N \ge 2$.
+
+
+
+## References
+1. L.E. Ballentine,
+ *Quantum mechanics: a modern development*, 2nd edition,
+ World Scientific.
diff --git a/source/know/concept/wiener-process/index.md b/source/know/concept/wiener-process/index.md
new file mode 100644
index 0000000..09d82e7
--- /dev/null
+++ b/source/know/concept/wiener-process/index.md
@@ -0,0 +1,186 @@
+---
+title: "Wiener process"
+date: 2021-10-29
+categories:
+- Physics
+- Mathematics
+- Stochastic analysis
+layout: "concept"
+---
+
+The **Wiener process** is a [stochastic process](/know/concept/stochastic-process/)
+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 $t$.
+
+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 $\sqrt{t}$.
+
+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,
+meaning that there are no fundemental scales.
+A consequence of this is that:
+$\mathbf{E}|B_t|^p = \mathbf{E}|\sqrt{t} B_1|^p = t^{p/2} \mathbf{E}|B_1|^p$.
+Another consequence is invariance under "time inversion",
+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,
+not even in the mean square, because:
+
+$$\begin{aligned}
+ \frac{B_{t+h} - B_t}{h}
+ \sim \frac{1}{h} \mathcal{N}(0, h)
+ \sim \mathcal{N}\Big(0, \frac{1}{h}\Big)
+ \qquad \quad
+ \lim_{h \to 0} \mathbf{E} \bigg|\mathcal{N}\Big(0, \frac{1}{h}\Big) \bigg|^2
+ = \infty
+\end{aligned}$$
+
+Furthermore, the Wiener process is a good example
+of both a [martingale](/know/concept/martingale/)
+and a [Markov process](/know/concept/markov-process/),
+since each increment has mean zero (so it is a martingale),
+and all increments are independent (so it is a Markov process).
+
+
+## Recurrence
+
+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.
+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 $X_0$
+(or resume a "paused" process):
+
+$$\begin{aligned}
+ X_t
+ = X_0 + \int_0^t \dd{B_s}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ \tau_i
+ \equiv \inf\{ t : |X_t| \le R_i \}
+ \qquad
+ \tau_o
+ \equiv \inf\{ t : |X_t| \ge R_o \}
+ \qquad
+ \tau
+ \equiv \min\{\tau_i, \tau_o\}
+\end{aligned}$$
+
+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 $X_0$:
+
+$$\begin{aligned}
+ h(X_0)
+ = \mathbf{E}\Big[ \Gamma(X_\tau) \Big| X_0 \Big]
+ = P\Big[|X_\tau| \!=\! R_i \:\Big|\: X_0\Big]
+\end{aligned}$$
+
+Dynkin's formula then tells us that $h(x)$ is given by the following equation,
+with the boundary conditions $h(R_i) = 1$ and $h(R_o) = 0$:
+
+$$\begin{aligned}
+ 0
+ = \hat{L}\{h(x)\}
+ = \frac{1}{2} \nabla^2 h(x)
+\end{aligned}$$
+
+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/):
+
+$$\begin{aligned}
+ 0
+ = \nabla^2 h(r)
+ = \pdvn{2}{h}{r} + \frac{d - 1}{r} \pdv{h}{r}
+\end{aligned}$$
+
+For $d = 1$, the solution $h_1(r)$ is as follows,
+of which we take the limit for $R_o \to \infty$:
+
+$$\begin{aligned}
+ h_1(r)
+ = \frac{r - R_o}{R_i - R_o}
+ \quad\underset{R_o \to \infty}{\longrightarrow}\quad
+ 1
+\end{aligned}$$
+
+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:
+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 $1$,
+so the Wiener process is also recurrent in 2D:
+
+$$\begin{aligned}
+ h_2(r)
+ = 1 - \frac{\log(r/R_i)}{\log(R_o/R_i)}
+ \quad\underset{R_o \to \infty}{\longrightarrow}\quad
+ 1
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ h_d(r)
+ = \frac{R_o^{2 - d} - r^{2 - d}}{R_o^{2 - d} - R_i^{2 - d}}
+ \quad\underset{R_o \to \infty}{\longrightarrow}\quad
+ \frac{R_i^{d - 2}}{r^{d - 2}}
+ < 1
+\end{aligned}$$
+
+This is a major qualitative difference. For example, consider a situation
+where some substance is diffusing from a localized infinite source:
+in 3D, the substance can escape and therefore a steady state can exist,
+while in 2D, the substance never strays far from the source,
+so no steady state is ever reached as long as the source continues to emit.
+
+
+
+## References
+1. U.H. Thygesen,
+ *Lecture notes on diffusions and stochastic differential equations*,
+ 2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/wkb-approximation/index.md b/source/know/concept/wkb-approximation/index.md
new file mode 100644
index 0000000..ad9b8e0
--- /dev/null
+++ b/source/know/concept/wkb-approximation/index.md
@@ -0,0 +1,200 @@
+---
+title: "WKB approximation"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+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
+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
+rearrange:
+
+$$\begin{aligned}
+ E = \frac{1}{2} m \dot{x}^2 + V(x)
+ \quad \implies \quad
+ m^2 (x')^2 = 2 m (E - V(x))
+\end{aligned}$$
+
+The left-hand side of the rearranged version is simply the momentum squared,
+so we define the magnitude of the momentum $p(x)$ accordingly:
+
+$$\begin{aligned}
+ p(x) = \sqrt{2 m (E - V(x))}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+ 0
+ = \dvn{2}{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
+ = \dvn{2}{\psi}{x} + \frac{p^2}{\hbar^2} \psi
+\end{aligned}$$
+
+If $V(x)$ were constant, and by extension $p(x)$ too, then the solution
+is easy:
+
+$$\begin{aligned}
+ \psi(x)
+ = \psi(0) \exp(\pm i p x / \hbar)
+\end{aligned}$$
+
+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
+wavenumber. The solution then takes the following form:
+
+$$\begin{aligned}
+ \psi(x)
+ = \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big)
+\end{aligned}$$
+
+$\chi(\xi)$ 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 $x$.
+Let us write this as an indefinite integral for convenience:
+
+$$\begin{aligned}
+ \psi(x)
+ = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg)
+\end{aligned}$$
+
+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 $\psi(0)$.
+We can now clearly see that:
+
+$$\begin{aligned}
+ \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x)
+ \quad \implies \quad
+ \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)}
+\end{aligned}$$
+
+Next, we insert this ansatz for $\psi(x)$ into the Schrödinger equation
+to get:
+
+$$\begin{aligned}
+ 0
+ &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi
+ = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi
+ = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
+\end{aligned}$$
+
+Dividing out $\psi$ and rearranging gives us the following, which is
+still exact:
+
+$$\begin{aligned}
+ \pm \frac{\hbar}{i} \chi'
+ = p^2 - \chi^2
+\end{aligned}$$
+
+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
+strength of quantum effects:
+
+$$\begin{aligned}
+ \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ...
+\end{aligned}$$
+
+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
+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
+expansion into the equation:
+
+$$\begin{aligned}
+ \pm \frac{\hbar}{i} \chi_0'
+ &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1
+\end{aligned}$$
+
+Where we have discarded all terms containing $\hbar^2$. At order
+$\hbar^0$, we then get the expected classical result for $\chi_0(x)$:
+
+$$\begin{aligned}
+ 0 = p^2 - \chi_0^2
+ \quad \implies \quad
+ \chi_0(x) = p(x)
+\end{aligned}$$
+
+While at order $\hbar$, we get the following quantum-mechanical
+correction:
+
+$$\begin{aligned}
+ \pm \frac{\hbar}{i} \chi_0'
+ = - 2 \frac{\hbar}{i} \chi_0 \chi_1
+ \quad \implies \quad
+ \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
+\end{aligned}$$
+
+Therefore, our approximated wave function $\psi(x)$ currently looks like
+this:
+
+$$\begin{aligned}
+ \psi(x)
+ &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
+\end{aligned}$$
+
+We can reduce the latter exponential using integration by substitution:
+
+$$\begin{aligned}
+ \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
+ &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big)
+ = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big)
+ \\
+ &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big)
+ = \frac{1}{\sqrt{\chi_0(x)}}
+ = \frac{1}{\sqrt{p(x)}}
+\end{aligned}$$
+
+In the WKB approximation for $E > V$, the solution $\psi(x)$ is thus
+given by:
+
+$$\begin{aligned}
+ \boxed{
+ \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
+ }
+\end{aligned}$$
+
+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.
+
+Luckily, the only thing we need to change for the WKB approximation
+is to let the momentum take imaginary values:
+
+$$\begin{aligned}
+ p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
+\end{aligned}$$
+
+And then take the absolute value in the appropriate place in front of $\psi(x)$:
+
+$$\begin{aligned}
+ \boxed{
+ \psi(x) \approx \frac{A}{\sqrt{|p(x)|}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
+ }
+\end{aligned}$$
+
+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.
+
+
+## References
+1. D.J. Griffiths, D.F. Schroeter,
+ *Introduction to quantum mechanics*, 3rd edition,
+ Cambridge.
+2. R. Shankar,
+ *Principles of quantum mechanics*, 2nd edition,
+ Springer.
diff --git a/source/know/concept/young-dupre-relation/index.md b/source/know/concept/young-dupre-relation/index.md
new file mode 100644
index 0000000..b87e19b
--- /dev/null
+++ b/source/know/concept/young-dupre-relation/index.md
@@ -0,0 +1,98 @@
+---
+title: "Young-Dupré relation"
+date: 2021-03-07
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+- Surface tension
+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
+the energy costs of the liquid-gas, solid-liquid and solid-gas interfaces:
+
+$$\begin{aligned}
+ \boxed{
+ \alpha_{sg} - \alpha_{sl}
+ = \alpha_{gl} \cos\theta
+ }
+\end{aligned}$$
+
+The derivation is simple:
+this is the only expression that maintains the droplet's boundaries
+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 $\mathcal{L}$,
+with the two first terms respectively being the energy costs
+of the top and bottom surfaces:
+
+$$\begin{aligned}
+ \mathcal{L}
+ = \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y
+\end{aligned}$$
+
+And the last term comes from the constraint
+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 $x = L$:
+
+$$\begin{aligned}
+ E[y(x)]
+ = \int_0^L \Big( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \Big) \dd{x}
+\end{aligned}$$
+
+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 $x = L$:
+
+$$\begin{aligned}
+ 0
+ &= \Big( \mathcal{L} - y' \pdv{\mathcal{L}}{y'} \Big)_{x = L}
+ \\
+ &= \bigg( \alpha_{gl} \sqrt{1 + (y')^2} + (\alpha_{sl} - \alpha_{sg}) + \lambda y - \frac{(y')^2}{\sqrt{1 + (y')^2}} \bigg)_{x = L}
+ \\
+ &= \bigg( \alpha_{gl} \frac{1}{\sqrt{1 + (y')^2}} + (\alpha_{sl} - \alpha_{sg}) + \lambda y \bigg)_{x = L}
+\end{aligned}$$
+
+Due to the droplet's shape, we have the boundary condition $y(L) = 0$,
+so the last term vanishes.
+We are thus left with the following equation:
+
+$$\begin{aligned}
+ \alpha_{gl} \frac{1}{\sqrt{1 + (y'(L))^2}}
+ = \alpha_{sg} - \alpha_{sl}
+\end{aligned}$$
+
+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 $\theta$.
+Then, from the definition of the cosine:
+
+$$\begin{aligned}
+ \cos\theta
+ = \frac{\dd{x}}{\dd{x} \sqrt{1 + (y'(L))^2}}
+ = \frac{1}{\sqrt{1 + (y'(L))^2}}
+\end{aligned}$$
+
+When inserted into the above transversality condition,
+this yields the Young-Dupré relation.
+
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
diff --git a/source/know/concept/young-laplace-law/index.md b/source/know/concept/young-laplace-law/index.md
new file mode 100644
index 0000000..445700e
--- /dev/null
+++ b/source/know/concept/young-laplace-law/index.md
@@ -0,0 +1,94 @@
+---
+title: "Young-Laplace law"
+date: 2021-03-11
+categories:
+- Physics
+- Fluid mechanics
+- Surface tension
+layout: "concept"
+---
+
+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 $\dd{\ell_2}$,
+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,
+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 $\theta_2$
+calculated from the arc length:
+
+$$\begin{aligned}
+ \theta_1 R_1
+ = \frac{1}{2} \dd{\ell_2}
+ \qquad \qquad
+ \theta_2 R_2
+ = \frac{1}{2} \dd{\ell_1}
+\end{aligned}$$
+
+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,
+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 $\dd{F_2}$
+along the principal axes:
+
+$$\begin{aligned}
+ \dd{F_1}
+ &= 2 \alpha \dd{\ell_1} \sin\theta_1
+ \approx 2 \alpha \dd{\ell_1} \theta_1
+ = \alpha \dd{\ell_1} \frac{\dd{\ell_2}}{R_1}
+ = \frac{\alpha}{R_1} \dd{A}
+ \\
+ \dd{F_2}
+ &= 2 \alpha \dd{\ell_2} \sin\theta_2
+ \approx 2 \alpha \dd{\ell_2} \theta_2
+ = \alpha \dd{\ell_2} \frac{\dd{\ell_1}}{R_2}
+ = \frac{\alpha}{R_2} \dd{A}
+\end{aligned}$$
+
+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.
+
+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**:
+
+$$\begin{aligned}
+ \boxed{
+ \Delta p
+ = \alpha \Big( \frac{1}{R_1} + \frac{1}{R_2} \Big)
+ }
+\end{aligned}$$
+
+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 $\Delta p$,
+and the liquid will flow (i.e. the surface will move)
+to try to reach an equilibrium.
+
+
+## References
+1. B. Lautrup,
+ *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+ CRC Press.
+2. T. Bohr,
+ *Surface tension and Laplace pressure*, 2021,
+ unpublished.
diff --git a/source/know/index.md b/source/know/index.md
new file mode 100644
index 0000000..f1b2c51
--- /dev/null
+++ b/source/know/index.md
@@ -0,0 +1,48 @@
+---
+title: "Knowledge base"
+date: 2021-02-22
+layout: "default"
+---
+
+# Knowledge base
+
+Welcome to my knowledge base!
+
+Over the years, I've learned a lot of physics, mathematics, and computing.
+These are vast, difficult subjects, and I've spent many hours of my life
+trying to make sense of obscure, poorly explained concepts.
+To help me remember what I learn, I write notes in LaTeX.
+
+This knowledge base is based on my notes, and
+is freely available to anyone who needs it.
+I hope it helps you in your studies or work.
+
+Keep in mind that I'm only human, so there are probably mistakes here.
+I take no responsibility for any injuries.
+If you're doing something important,
+you should check things yourself!
+
+{% assign concepts = site.pages
+ | where_exp: "item", "item.layout == 'concept'"
+%}
+{% assign categories = site.pages
+ | where_exp: "item", "item.layout == 'category'"
+%}
+
+There are currently
+{{ concepts.size }} articles
+in this knowledge base, in
+{{ categories.size }} categories .
+
+Most recent articles:
+
+{% assign newest = concepts
+ | sort: "date"
+ | reverse
+ | slice: 0, 10
+%}
+{% for page in newest %}
+ {{ page.title }}
+
diff --git a/source/robots.txt b/source/robots.txt
new file mode 100644
index 0000000..4738d6a
--- /dev/null
+++ b/source/robots.txt
@@ -0,0 +1,3 @@
+User-agent: *
+Disallow:
+
--
cgit v1.3
+
+
+To their credit, they seem to be offering a way out.
+This approach is reasonable: preventively ban high-risk IP ranges,
+and allow "trustworthy" servers at the owner's request.
+I got error 5.7.511, asking me to send an email to a support address.
+If you're lucky, you may have a different error,
+and get the opportunity to use the slick [delist portal](https://sender.office.com/) instead.
+The URL in the bounce message links to [this list](https://go.microsoft.com/fwlink/?LinkId=526653) of error codes.
+
+I confess, I never actually bothered to forward the message to the provided address:
+my initial email was time-sensitive,
+so I couldn't afford to wait for Microsoft's response.
+Also, their customer support's stellar reputation precedes them,
+so I chose to use my time more wisely.
+Even if they would've resolved it nicely,
+there's nothing preventing Microsoft (or any other provider)
+from breaking my deliverability again in the future.
+Instead, I opted for a compromise.
+
+As a result of providers' IP reputation systems,
+a whole new business has appeared: SMTP relays.
+They offer to take the issue out of your hands:
+you send your emails through their servers,
+and they do their best to deliver them to large providers.
+SMTP relays are mostly used for sending marketing emails in bulk,
+but are also useful to avoid small-scale problems as described above.
+There are many SMTP relay services to choose from, at various prices.
+
+Using an SMTP relay results in more reliable delivery of your messages to large providers,
+but an obvious concern is privacy:
+the relay server can read all your outgoing emails (but not incoming),
+so you'll have to trust the service you choose.
+But it's no worse than using a major provider,
+and if you're sending sensitive material, why use email in the first place?
+Personally, I use [SMTP2GO](https://www.smtp2go.com/),
+but I can't say how good they are; do your own research.
+
+
+
+### OpenSMTPD settings
+
+Once you've chosen an SMTP relay service, let's say `relay.com`,
+and set up your account,
+they'll let you create credentials to use their SMTP servers.
+Suppose these credentials are `