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+---
+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
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+---
+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
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+---
+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
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--- /dev/null
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+---
+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
index 0000000..c7f7483
--- /dev/null
+++ b/source/know/concept/bernoullis-theorem/index.md
@@ -0,0 +1,85 @@
+---
+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
new file mode 100644
index 0000000..83ccde2
--- /dev/null
+++ b/source/know/concept/bernstein-vazirani-algorithm/bernstein-vazirani-circuit.png
diff --git a/source/know/concept/bernstein-vazirani-algorithm/index.md b/source/know/concept/bernstein-vazirani-algorithm/index.md
new file mode 100644
index 0000000..8720c81
--- /dev/null
+++ b/source/know/concept/bernstein-vazirani-algorithm/index.md
@@ -0,0 +1,101 @@
+---
+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:
+
+<a href="bernstein-vazirani-circuit.png">
+<img src="bernstein-vazirani-circuit.png" style="width:52%">
+</a>
+
+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
index 0000000..b95788a
--- /dev/null
+++ b/source/know/concept/berry-phase/index.md
@@ -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:
+
+$$\begin{aligned}
+    \boxed{
+        \mu = N p
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-mean"/>
+<label for="proof-mean">Proof</label>
+<div class="hidden">
+<label for="proof-mean">Proof.</label>
+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.
+</div>
+</div>
+
+Meanwhile, we find the following variance $\sigma^2$,
+with $\sigma$ being the standard deviation:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma^2 = N p q
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-var"/>
+<label for="proof-var">Proof</label>
+<div class="hidden">
+<label for="proof-var">Proof.</label>
+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.
+</div>
+</div>
+
+As $N \to \infty$, the binomial distribution
+turns into the continuous normal distribution,
+a fact that is sometimes called the **de Moivre-Laplace theorem**:
+
+$$\begin{aligned}
+    \boxed{
+        \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-normal"/>
+<label for="proof-normal">Proof</label>
+<div class="hidden">
+<label for="proof-normal">Proof.</label>
+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.
+</div>
+</div>
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/blasius-boundary-layer/index.md b/source/know/concept/blasius-boundary-layer/index.md
new file mode 100644
index 0000000..86b0e02
--- /dev/null
+++ b/source/know/concept/blasius-boundary-layer/index.md
@@ -0,0 +1,113 @@
+---
+title: "Blasius boundary layer"
+date: 2021-05-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+In fluid dynamics, the **Blasius boundary layer** is an application of
+the [Prandtl equations](/know/concept/prandtl-equations/),
+which govern the flow of a fluid
+at large Reynolds number $\mathrm{Re} \gg 1$
+close to a surface.
+Specifically, the Blasius layer is the solution
+for a half-plane approached from the edge by a fluid.
+
+A fluid with velocity field $\va{v} = U \vu{e}_x$ flows to the plane,
+which starts at $y = 0$ and exists for $x \ge 0$.
+To describe this, we make an ansatz
+for the *slip-flow* region's $x$-velocity $v_x(x, y)$:
+
+$$\begin{aligned}
+    v_x
+    = U f'(s)
+    \qquad \quad
+    s
+    \equiv \frac{y}{\delta(x)}
+\end{aligned}$$
+
+Note that $f'(s)$ is the derivative of an unknown $f(s)$,
+and that it obeys the boundary conditions $f'(0) = 0$ and $f'(\infty) = 1$.
+Furthermore, $\delta(x)$ is the thickness of the stationary boundary layer at the surface.
+To derive the Prandtl equations,
+the estimate $\delta(x) = \sqrt{\nu x / U}$ was used,
+which we will stick with.
+For later use, it is worth writing the derivatives of $s$:
+
+$$\begin{aligned}
+    \pdv{s}{x}
+    = - y \frac{\delta'}{\delta^2}
+    = - s \frac{\delta'}{\delta}
+    \qquad \quad
+    \pdv{s}{y}
+    = \frac{1}{\delta}
+\end{aligned}$$
+
+Inserting the ansatz for $v_x$ into the incompressibility condition then yields:
+
+$$\begin{aligned}
+    \pdv{v_y}{y}
+    = - \pdv{v_x}{x}
+    = U s f'' \frac{\delta'}{\delta}
+\end{aligned}$$
+
+Which we integrate to get an expression for the $y$-velocity $v_y$, namely:
+
+$$\begin{aligned}
+    v_y
+    = U \frac{\delta'}{\delta} \int s f'' \dd{y}
+    = U \delta' \: (s f' - f)
+\end{aligned}$$
+
+Now, consider the main Prandtl equation,
+assuming that the attack velocity $U$ is constant:
+
+$$\begin{aligned}
+    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    = \nu \pdvn{2}{v_x}{y}
+\end{aligned}$$
+
+Inserting our expressions for $v_x$ and $v_y$ into this leads us to:
+
+$$\begin{aligned}
+    - U^2 \frac{\delta'}{\delta} s f'' f' + U^2 \frac{\delta'}{\delta} f'' (s f' - f)
+    = \nu U \frac{1}{\delta^2} f'''
+\end{aligned}$$
+
+After  multiplying it by $\delta^2 / U$ and cancelling out some terms,
+it reduces to:
+
+$$\begin{aligned}
+    \nu f''' + U \delta' \delta f'' f
+    = 0
+\end{aligned}$$
+
+Then, substituting $\delta(x) = \sqrt{\nu x / U}$ and $\delta'(x) = (1/2) \sqrt{\nu / (U x)}$ yields:
+
+$$\begin{aligned}
+    \nu f''' + U \frac{\nu}{2 U} f'' f
+    = 0
+\end{aligned}$$
+
+Simplifying this leads us to the **Blasius equation**,
+which is a nonlinear ODE for $f(s)$:
+
+$$\begin{aligned}
+    \boxed{
+        2 f''' + f'' f = 0
+    }
+\end{aligned}$$
+
+Unfortunately, this cannot be solved analytically, only numerically.
+Nevertheless, the result shows a boundary layer $\delta(x)$
+exhibiting the expected downstream thickening.
+
+
+
+## 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/bloch-sphere/bloch-small.jpg b/source/know/concept/bloch-sphere/bloch-small.jpg
new file mode 100644
index 0000000..e99c0e1
--- /dev/null
+++ b/source/know/concept/bloch-sphere/bloch-small.jpg
diff --git a/source/know/concept/bloch-sphere/bloch.jpg b/source/know/concept/bloch-sphere/bloch.jpg
new file mode 100644
index 0000000..9515d84
--- /dev/null
+++ b/source/know/concept/bloch-sphere/bloch.jpg
diff --git a/source/know/concept/bloch-sphere/index.md b/source/know/concept/bloch-sphere/index.md
new file mode 100644
index 0000000..d333c50
--- /dev/null
+++ b/source/know/concept/bloch-sphere/index.md
@@ -0,0 +1,133 @@
+---
+title: "Bloch sphere"
+date: 2021-03-09
+categories:
+- Quantum mechanics
+- Quantum information
+- Two-level system
+layout: "concept"
+---
+
+In quantum mechanics, particularly quantum information,
+the **Bloch sphere** is an invaluable tool to visualize qubits.
+All pure qubit states are represented by a point on the sphere's surface:
+
+<a href="bloch.jpg">
+<img src="bloch-small.jpg" style="width:50%">
+</a>
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-moment0"/>
+<label for="proof-moment0">Proof</label>
+<div class="hidden">
+<label for="proof-moment0">Proof.</label>
+We insert $Q = m$ into our prototype,
+and since $m$ is constant, the rest is trivial:
+
+$$\begin{aligned}
+    0
+    &= \pdv{}{t}\big(n \Expval{m}\big) + \nabla \cdot \big(n \Expval{m \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \Expval{\pdv{m}{\vb{v}}} \bigg)
+    \\
+    &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
+\end{aligned}$$
+</div>
+</div>
+
+If we instead choose the momentum $Q = m \vb{v}$,
+we find that the **first moment** of the BTE describes conservation of momentum,
+where $\hat{P}$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/):
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{}{t}\big(\rho \vb{V}\big) + \rho \vb{V} (\nabla \cdot \vb{V}) + \nabla \cdot \hat{P} - n \vb{F}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-moment1"/>
+<label for="proof-moment1">Proof</label>
+<div class="hidden">
+<label for="proof-moment1">Proof.</label>
+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:
+
+$$\begin{aligned}
+    0
+    &= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
+\end{aligned}$$
+</div>
+</div>
+
+Finally, if we choose the kinetic energy $Q = m |\vb{v}|^2 / 2$,
+we find that the **second moment** gives conservation of energy,
+where $U$ is the thermal energy density and $\vb{J}$ is the heat flux:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{}{t}\bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
+        + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
+        - \vb{F} \cdot \big( n \vb{V} \big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-moment2"/>
+<label for="proof-moment2">Proof</label>
+<div class="hidden">
+<label for="proof-moment2">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \nabla \cdot (\vb{V} \cdot \hat{P})
+    &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V}
+    = \nabla \cdot
+    \begin{bmatrix}
+        P_{xx} & P_{xy} & P_{xz} \\
+        P_{yx} & P_{yy} & P_{yz} \\
+        P_{zx} & P_{zy} & P_{zz}
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        V_x \\ V_y \\ V_z
+    \end{bmatrix}
+    \\
+    &=
+    \begin{bmatrix}
+        \displaystyle \pdv{P_{xx}}{x} + \pdv{P_{xy}}{y} + \pdv{P_{xz}}{z} \\
+        \displaystyle \pdv{P_{yx}}{x} + \pdv{P_{yy}}{y} + \pdv{P_{yz}}{z} \\
+        \displaystyle \pdv{P_{zx}}{x} + \pdv{P_{zy}}{y} + \pdv{P_{zz}}{z}
+    \end{bmatrix}^{\top}
+    \cdot
+    \begin{bmatrix}
+        V_x \\ V_y \\ V_z
+    \end{bmatrix}
+    = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
+\end{aligned}$$
+</div>
+</div>
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/source/know/concept/boltzmann-relation/index.md b/source/know/concept/boltzmann-relation/index.md
new file mode 100644
index 0000000..6dd04d2
--- /dev/null
+++ b/source/know/concept/boltzmann-relation/index.md
@@ -0,0 +1,90 @@
+---
+title: "Boltzmann relation"
+date: 2021-10-18
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+In a plasma where the ions and electrons are both in thermal equilibrium,
+and in the absence of short-lived induced electromagnetic fields,
+their densities $n_i$ and $n_e$ can be predicted.
+
+By definition, a particle in an [electric field](/know/concept/electric-field/) $\vb{E}$
+experiences a [Lorentz force](/know/concept/lorentz-force/) $\vb{F}_e$.
+This corresponds to a force density $\vb{f}_e$,
+such that $\vb{F}_e = \vb{f}_e \dd{V}$.
+For the electrons, we thus have:
+
+$$\begin{aligned}
+    \vb{f}_e
+    = q_e n_e \vb{E}
+    = - q_e n_e \nabla \phi
+\end{aligned}$$
+
+Meanwhile, if we treat the electrons as a gas
+obeying the ideal gas law $p_e = k_B T_e n_e$,
+then the pressure $p_e$ leads to another force density $\vb{f}_p$:
+
+$$\begin{aligned}
+    \vb{f}_p
+    = - \nabla p_e
+    = - k_B T_e \nabla n_e
+\end{aligned}$$
+
+At equilibrium, we demand that $\vb{f}_e = - \vb{f}_p$,
+and isolate this equation for $\nabla n_e$, yielding:
+
+$$\begin{aligned}
+    k_B T_e \nabla n_e
+    = - q_e n_e \nabla \phi
+    \quad \implies \quad
+    \nabla n_e
+    = - \frac{q_e \nabla \phi}{k_B T_e} n_e
+    = - \nabla \bigg( \frac{q_e \phi}{k_B T_e} \bigg) n_e
+\end{aligned}$$
+
+This equation is straightforward to integrate,
+leading to the following expression for $n_e$,
+known as the **Boltzmann relation**,
+due to its resemblance to the statistical Boltzmann distribution
+(see [canonical ensemble](/know/concept/canonical-ensemble/)):
+
+$$\begin{aligned}
+    \boxed{
+        n_e(\vb{r})
+        = n_{e0} \exp\!\bigg( \!-\! \frac{q_e \phi(\vb{r})}{k_B T_e} \bigg)
+    }
+\end{aligned}$$
+
+Where the linearity factor $n_{e0}$ represents
+the electron density for $\phi = 0$.
+We can do the same for ions instead of electrons,
+leading to the following ion density $n_i$:
+
+$$\begin{aligned}
+    \boxed{
+        n_i(\vb{r})
+        = n_{i0} \exp\!\bigg( \!-\! \frac{q_i \phi(\vb{r})}{k_B T_i} \bigg)
+    }
+\end{aligned}$$
+
+However, due to their larger mass,
+ions are much slower to respond to fluctuations in the above equilibrium.
+Consequently, after a perturbation,
+the ions spend much more time in a transient non-equilibrium state
+than the electrons, so this formula for $n_i$ is only valid
+if the perturbation is sufficiently slow,
+allowing the ions to keep up.
+Usually, electrons do not suffer the same issue,
+thanks to their small mass and fast response.
+
+
+## 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/bose-einstein-distribution/index.md b/source/know/concept/bose-einstein-distribution/index.md
new file mode 100644
index 0000000..8397a8f
--- /dev/null
+++ b/source/know/concept/bose-einstein-distribution/index.md
@@ -0,0 +1,77 @@
+---
+title: "Bose-Einstein distribution"
+date: 2021-07-11
+categories:
+- Physics
+- Statistics
+- Quantum mechanics
+layout: "concept"
+---
+
+**Bose-Einstein statistics** describe how bosons,
+which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
+will distribute themselves across the available states
+in a system at equilibrium.
+
+Consider a single-particle state $s$,
+which can contain any number of bosons.
+Since the occupation number $N$ is variable,
+we turn to the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
+whose grand partition function $\mathcal{Z}$ is as follows,
+where $\varepsilon$ is the energy per particle,
+and $\mu$ is the chemical potential:
+
+$$\begin{aligned}
+    \mathcal{Z}
+    = \sum_{N = 0}^\infty \Big( \exp(- \beta (\varepsilon - \mu)) \Big)^{N}
+    = \frac{1}{1 - \exp(- \beta (\varepsilon - \mu))}
+\end{aligned}$$
+
+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 $\Expval{N}$
+is found by taking a derivative of $\Omega$:
+
+$$\begin{aligned}
+    \Expval{N}
+    = - \pdv{\Omega}{\mu}
+    = k T \pdv{\ln{\mathcal{Z}}}{\mu}
+    = \frac{\exp(- \beta (\varepsilon - \mu))}{1 - \exp(- \beta (\varepsilon - \mu))}
+\end{aligned}$$
+
+By multitplying both the numerator and the denominator by $\exp(\beta(\varepsilon \!-\! \mu))$,
+we arrive at the standard form of the **Bose-Einstein distribution** $f_B$:
+
+$$\begin{aligned}
+    \boxed{
+        \Expval{N}
+        = f_B(\varepsilon)
+        = \frac{1}{\exp(\beta (\varepsilon - \mu)) - 1}
+    }
+\end{aligned}$$
+
+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/calculus-of-variations/index.md b/source/know/concept/calculus-of-variations/index.md
new file mode 100644
index 0000000..81d9a5f
--- /dev/null
+++ b/source/know/concept/calculus-of-variations/index.md
@@ -0,0 +1,339 @@
+---
+title: "Calculus of variations"
+date: 2021-02-24
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+The **calculus of variations** lays the mathematical groundwork
+for [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
+
+Consider a **functional** $J$, mapping a function $f(x)$ to a scalar value
+by integrating over the so-called **Lagrangian** $L$,
+which represents an expression involving $x$, $f$ and the derivative $f'$:
+
+$$\begin{aligned}
+    J[f] = \int_{x_0}^{x_1} L(f, f', x) \dd{x}
+\end{aligned}$$
+
+If $J$ in some way measures the physical "cost" (e.g. energy) of
+the path $f(x)$ taken by a physical system,
+the **principle of least action** states that $f$ will be a minimum of $J[f]$,
+so for example the expended energy will be minimized.
+In practice, various cost metrics may be used,
+so maxima of $J[f]$ are also interesting to us.
+
+If $f(x, \varepsilon\!=\!0)$ is the optimal route, then a slightly
+different (and therefore worse) path between the same two points can be expressed
+using the parameter $\varepsilon$:
+
+$$\begin{aligned}
+    f(x, \varepsilon) = f(x, 0) + \varepsilon \eta(x)
+    \qquad \mathrm{or} \qquad
+    \delta f = \varepsilon \eta(x)
+\end{aligned}$$
+
+Where $\eta(x)$ is an arbitrary differentiable deviation.
+Since $f(x, \varepsilon)$ must start and end in the same points as $f(x,0)$,
+we have the boundary conditions:
+
+$$\begin{aligned}
+    \eta(x_0) = \eta(x_1) = 0
+\end{aligned}$$
+
+Given $L$, the goal is to find an equation for the optimal path $f(x,0)$.
+Just like when finding the minimum of a real function,
+the minimum $f$ of a functional $J[f]$ is a stationary point
+with respect to the deviation weight $\varepsilon$,
+a condition often written as $\delta J = 0$.
+In the following, the integration limits have been omitted:
+
+$$\begin{aligned}
+    0
+    &= \delta J
+    = \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \int \pdv{L}{\varepsilon} \dd{x}
+    = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f'} \pdv{f'}{\varepsilon} \dd{x}
+    \\
+    &= \int \pdv{L}{f} \eta + \pdv{L}{f'} \eta' \dd{x}
+    = \Big[ \pdv{L}{f'} \eta \Big]_{x_0}^{x_1} + \int \pdv{L}{f} \eta - \dv{}{x}\Big( \pdv{L}{f'} \Big) \eta \dd{x}
+\end{aligned}$$
+
+The boundary term from partial integration vanishes due to the boundary
+conditions for $\eta(x)$. We are thus left with:
+
+$$\begin{aligned}
+    0
+    = \int \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f'} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+This holds for all $\eta$, but $\eta$ is arbitrary, so in fact
+only the parenthesized expression matters:
+
+$$\begin{aligned}
+    \boxed{
+        0 = \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f'} \Big)
+    }
+\end{aligned}$$
+
+This is known as the **Euler-Lagrange equation** of the Lagrangian $L$,
+and its solutions represent the optimal paths $f(x, 0)$.
+
+
+## Multiple functions
+
+Suppose that the Lagrangian $L$ depends on multiple independent functions
+$f_1, f_2, ..., f_N$:
+
+$$\begin{aligned}
+    J[f_1, ..., f_N] = \int_{x_0}^{x_1} L(f_1, ..., f_N, f_1', ..., f_N', x) \dd{x}
+\end{aligned}$$
+
+In this case, every $f_n(x)$ has its own deviation $\eta_n(x)$,
+satisfying $\eta_n(x_0) = \eta_n(x_1) = 0$:
+
+$$\begin{aligned}
+    f_n(x, \varepsilon) = f_n(x, 0) + \varepsilon \eta_n(x)
+\end{aligned}$$
+
+The derivation procedure is identical to the case $N = 1$ from earlier:
+
+$$\begin{aligned}
+    0
+    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \int \pdv{L}{\varepsilon} \dd{x}
+    = \int \sum_{n} \Big( \pdv{L}{f_n} \pdv{f_n}{\varepsilon} + \pdv{L}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x}
+    \\
+    &= \int \sum_{n} \Big( \pdv{L}{f_n} \eta_n + \pdv{L}{f_n'} \eta_n' \Big) \dd{x}
+    \\
+    &= \Big[ \sum_{n} \pdv{L}{f_n'} \eta_n \Big]_{x_0}^{x_1}
+    + \int \sum_{n} \eta_n \bigg( \pdv{L}{f_n} - \dv{}{x}\Big( \pdv{L}{f_n'} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+Once again, $\eta_n(x)$ is arbitrary and disappears at the boundaries,
+so we end up with $N$ equations of the same form as for a single function:
+
+$$\begin{aligned}
+    \boxed{
+        0 = \pdv{L}{f_1} - \dv{}{x}\Big( \pdv{L}{f_1'} \Big)
+        \quad \cdots \quad
+        0 = \pdv{L}{f_N} - \dv{}{x}\Big( \pdv{L}{f_N'} \Big)
+    }
+\end{aligned}$$
+
+
+## Higher-order derivatives
+
+Suppose that the Lagrangian $L$ depends on multiple derivatives of $f(x)$:
+
+$$\begin{aligned}
+    J[f] = \int_{x_0}^{x_1} L(f, f', f'', ..., f^{(N)}, x) \dd{x}
+\end{aligned}$$
+
+Once again, the derivation procedure is the same as before:
+
+$$\begin{aligned}
+    0
+    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \int \pdv{L}{\varepsilon} \dd{x}
+    = \int \pdv{L}{f} \pdv{f}{\varepsilon} + \sum_{n} \pdv{L}{f^{(n)}} \pdv{f^{(n)}}{\varepsilon} \dd{x}
+    \\
+    &= \int \pdv{L}{f} \eta + \sum_{n} \pdv{L}{f^{(n)}} \eta^{(n)} \dd{x}
+\end{aligned}$$
+
+The goal is to turn each $\eta^{(n)}(x)$ into $\eta(x)$, so we need to
+partially integrate the $n$th term of the sum $n$ times. In this case,
+we will need some additional boundary conditions for $\eta(x)$:
+
+$$\begin{aligned}
+    \eta'(x_0) = \eta'(x_1) = 0
+    \qquad \cdots \qquad
+    \eta^{(N-1)}(x_0) = \eta^{(N-1)}(x_1) = 0
+\end{aligned}$$
+
+This eliminates the boundary terms from partial integration, leaving:
+
+$$\begin{aligned}
+    0
+    &= \int \eta \bigg( \pdv{L}{f} + \sum_{n} (-1)^n \dvn{n}{}{x}\Big( \pdv{L}{f^{(n)}} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+Once again, because $\eta(x)$ is arbitrary, the Euler-Lagrange equation becomes:
+
+$$\begin{aligned}
+    \boxed{
+        0 = \pdv{L}{f} + \sum_{n} (-1)^n \dvn{n}{}{x}\Big( \pdv{L}{f^{(n)}} \Big)
+    }
+\end{aligned}$$
+
+
+## Multiple coordinates
+
+Suppose now that $f$ is a function of multiple variables.
+For brevity, we only consider two variables $x$ and $y$,
+but the results generalize effortlessly to larger amounts.
+The Lagrangian now depends on all the partial derivatives of $f(x, y)$:
+
+$$\begin{aligned}
+    J[f] = \iint_{(x_0, y_0)}^{(x_1, y_1)} L(f, f_x, f_y, x, y) \dd{x} \dd{y}
+\end{aligned}$$
+
+The arbitrary deviation $\eta$ is then also a function of multiple variables:
+
+$$\begin{aligned}
+    f(x, y; \varepsilon) = f(x, y; 0) + \varepsilon \eta(x, y)
+\end{aligned}$$
+
+The derivation procedure starts in the exact same way as before:
+
+$$\begin{aligned}
+    0
+    &= \pdv{J}{\varepsilon} \Big|_{\varepsilon = 0}
+    = \iint \pdv{L}{\varepsilon} \dd{x} \dd{y}
+    \\
+    &= \iint \pdv{L}{f} \pdv{f}{\varepsilon} + \pdv{L}{f_x} \pdv{f_x}{\varepsilon} + \pdv{L}{f_y} \pdv{f_y}{\varepsilon} \dd{x} \dd{y}
+    \\
+    &= \iint \pdv{L}{f} \eta + \pdv{L}{f_x} \eta_x + \pdv{L}{f_y} \eta_y \dd{x} \dd{y}
+\end{aligned}$$
+
+We partially integrate for both $\eta_x$ and $\eta_y$, yielding:
+
+$$\begin{aligned}
+    0
+    &= \int \Big[ \pdv{L}{f_x} \eta \Big]_{x_0}^{x_1} \dd{y} + \int \Big[ \pdv{L}{f_y} \eta \Big]_{y_0}^{y_1} \dd{x}
+    \\
+    &\quad + \iint \eta \bigg( \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big) \bigg) \dd{x} \dd{y}
+\end{aligned}$$
+
+But now, to eliminate these boundary terms, we need extra conditions for $\eta$:
+
+$$\begin{aligned}
+    \forall y: \eta(x_0, y) = \eta(x_1, y) = 0
+    \qquad
+    \forall x: \eta(x, y_0) = \eta(x, y_1) = 0
+\end{aligned}$$
+
+In other words, the deviation $\eta$ must be zero on the whole "box".
+Again relying on the fact that $\eta$ is arbitrary, the Euler-Lagrange
+equation is:
+
+$$\begin{aligned}
+    0 = \pdv{L}{f} - \dv{}{x}\Big( \pdv{L}{f_x} \Big) - \dv{}{y}\Big( \pdv{L}{f_y} \Big)
+\end{aligned}$$
+
+This generalizes nicely to functions of even more variables $x_1, x_2, ..., x_N$:
+
+$$\begin{aligned}
+    \boxed{
+        0 = \pdv{L}{f} - \sum_{n} \dv{}{x_n}\Big( \pdv{L}{f_{x_n}} \Big)
+    }
+\end{aligned}$$
+
+
+## Constraints
+
+So far, for multiple functions $f_1, ..., f_N$,
+we have been assuming that all $f_n$ are independent, and by extension all $\eta_n$.
+Suppose that we now have $M < N$ constraints $\phi_m$
+that all $f_n$ need to obey, introducing implicit dependencies between them.
+
+Let us consider constraints $\phi_m$ of the two forms below.
+It is important that they are **holonomic**,
+meaning they do not depend on any derivatives of any $f_n(x)$:
+
+$$\begin{aligned}
+    \phi_m(f_1, ..., f_N, x) = 0
+    \qquad \mathrm{or} \qquad
+    \int_{x_0}^{x_1} \phi_m(f_1, ..., f_N, x) \dd{x} = C_m
+\end{aligned}$$
+
+Where $C_m$ is a constant.
+Note that the first form can also be used for $\phi_m = C_m \neq 0$,
+by simply redefining the constraint as $\phi_m^0 = \phi_m - C_m = 0$.
+
+To solve this constrained optimization problem for $f_n(x)$,
+we introduce [Lagrange multipliers](/know/concept/lagrange-multiplier/) $\lambda_m$.
+In the former case $\lambda_m(x)$ is a function of $x$, while in the
+latter case $\lambda_m$ is constant:
+
+$$\begin{aligned}
+    \int \lambda_m(x) \: \phi_m(\{f_n\}, x) \dd{x} = 0
+    \qquad \mathrm{or} \qquad
+    \lambda_m \int \phi_m(\{f_n\}, x) \dd{x} = \lambda_m C_m
+\end{aligned}$$
+
+The reason for this distinction in $\lambda_m$
+is that we need to find the stationary points with respect to $\varepsilon$
+of both constraint types. Written in the variational form, this is:
+
+$$\begin{aligned}
+    \delta \int \lambda_m \: \phi_m \dd{x} = 0
+\end{aligned}$$
+
+From this, we define a new Lagrangian $\Lambda$ for the functional $J$,
+with the contraints built in:
+
+$$\begin{aligned}
+    J[f_n]
+    &= \int \Lambda(f_1, ..., f_N; f_1', ..., f_N'; \lambda_1, ..., \lambda_M; x) \dd{x}
+    \\
+    &= \int L + \sum_{m} \lambda_m \phi_m \dd{x}
+\end{aligned}$$
+
+Then we derive the Euler-Lagrange equation as usual for $\Lambda$ instead of $L$:
+
+$$\begin{aligned}
+    0
+    &= \delta \int \Lambda \dd{x}
+    = \int \pdv{\Lambda}{\varepsilon} \dd{x}
+    = \int \sum_n \Big( \pdv{\Lambda}{f_n} \pdv{f_n}{\varepsilon} + \pdv{\Lambda}{f_n'} \pdv{f_n'}{\varepsilon} \Big) \dd{x}
+    \\
+    &= \int \sum_n \Big( \pdv{\Lambda}{f_n} \eta_n + \pdv{\Lambda}{f_n'} \eta_n' \Big) \dd{x}
+    \\
+    &= \Big[ \sum_n \pdv{\Lambda}{f_n'} \eta_n \Big]_{x_0}^{x_1}
+    + \int \sum_n \eta_n \bigg( \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big) \bigg) \dd{x}
+\end{aligned}$$
+
+Using the same logic as before, we end up with a set of Euler-Lagrange equations with $\Lambda$:
+
+$$\begin{aligned}
+    0
+    = \pdv{\Lambda}{f_n} - \dv{}{x}\Big( \pdv{\Lambda}{f_n'} \Big)
+\end{aligned}$$
+
+By inserting the definition of $\Lambda$, we then get the following.
+Recall that $\phi_m$ is holonomic, and thus independent of all derivatives $f_n'$:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{L}{f_n} - \dv{}{x}\Big( \pdv{L}{f_n'} \Big) + \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
+    }
+\end{aligned}$$
+
+These are **Lagrange's equations of the first kind**,
+with their second-kind counterparts being the earlier Euler-Lagrange equations.
+Note that there are $N$ separate equations, one for each $f_n$.
+
+Due to the constraints $\phi_m$, the functions $f_n$ are not independent.
+This is solved by choosing $\lambda_m$ such that $M$ of the $N$ equations hold,
+i.e. solving a system of $M$ equations for $\lambda_m$:
+
+$$\begin{aligned}
+    \dv{}{x}\Big( \pdv{L}{f_n'} \Big) - \pdv{L}{f_n}
+    = \sum_{m} \lambda_m \pdv{\phi_m}{f_n}
+\end{aligned}$$
+
+And then the remaining $N - M$ equations can be solved in the normal unconstrained way.
+
+
+
+## References
+1.  G.B. Arfken, H.J. Weber,
+    *Mathematical methods for physicists*, 6th edition, 2005,
+    Elsevier.
+2.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/canonical-ensemble/index.md b/source/know/concept/canonical-ensemble/index.md
new file mode 100644
index 0000000..aca2a33
--- /dev/null
+++ b/source/know/concept/canonical-ensemble/index.md
@@ -0,0 +1,239 @@
+---
+title: "Canonical ensemble"
+date: 2021-07-10
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+layout: "concept"
+---
+
+The **canonical ensemble** or **NVT ensemble** builds on
+the [microcanonical ensemble](/know/concept/microcanonical-ensemble/),
+by allowing the system to exchange energy with a very large heat bath,
+such that its temperature $T$ remains constant,
+but internal energy $U$ does not.
+The conserved state functions are
+the temperature $T$, the volume $V$, and the particle count $N$.
+
+We refer to the system of interest as $A$, and the heat bath as $B$.
+The combination $A\!+\!B$ forms a microcanonical ensemble,
+i.e. it has a fixed total energy $U$,
+and eventually reaches an equilibrium
+with a uniform temperature $T$ in both $A$ and $B$.
+
+Assuming that this equilibrium has been reached,
+we want to know which microstates $A$ prefers in that case.
+Specifically, if $A$ has energy $U_A$, and $B$ has $U_B$,
+which $U_A$ does $A$ prefer?
+
+Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
+Then the probability that $A$ is in a specific microstate $s_A$ is as follows,
+where $U_A(s_A)$ is the resulting energy:
+
+$$\begin{aligned}
+    p(s_A)
+    = \frac{c_B(U - U_A(s_A))}{D}
+    \qquad \quad
+    D \equiv \sum_{s_A} c_B(U - U_A(s_A))
+\end{aligned}$$
+
+In other words, we choose an $s_A$,
+and count the number $c_B$ of compatible $B$-microstates.
+
+Since the heat bath is large, let us assume that $U_B \gg U_A$.
+We thus approximate $\ln{p(s_A)}$ by
+Taylor-expanding $\ln{c_B(U_B)}$ around $U_B = U$:
+
+$$\begin{aligned}
+    \ln{p(s_A)}
+    &= -\ln{D} + \ln\!\big(c_B(U - U_A(s_A))\big)
+    \\
+    &\approx - \ln{D} + \ln{c_B(U)} - \bigg( \dv{(\ln{c_B})}{U_B} \bigg) \: U_A(s_A)
+\end{aligned}$$
+
+Here, we use the definition of entropy $S_B \equiv k \ln{c_B}$,
+and that its $U_B$-derivative is $1/T$:
+
+$$\begin{aligned}
+    \ln{p(s_A)}
+    &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k} \Big( \pdv{S_B}{U_B} \Big)
+    \\
+    &\approx - \ln{D} + \ln{c_B(U)} - \frac{U_A(s_A)}{k T}
+\end{aligned}$$
+
+We now define the **partition function** or **Zustandssumme** $Z$ as follows,
+which will act as a normalization factor for the probability:
+
+$$\begin{aligned}
+    \boxed{
+        Z
+        \equiv \sum_{s_A}^{} \exp(- \beta U_A(s_A))
+    }
+    = \frac{D}{c_B(U)}
+\end{aligned}$$
+
+Where $\beta \equiv 1/ (k T)$.
+The probability of finding $A$ in a microstate $s_A$ is thus given by:
+
+$$\begin{aligned}
+    \boxed{
+        p(s_A) = \frac{1}{Z} \exp(- \beta U_A(s_A))
+    }
+\end{aligned}$$
+
+This is the **Boltzmann distribution**,
+which, as it turns out, maximizes the entropy $S_A$
+for a fixed value of the average energy $\Expval{U_A}$,
+i.e. a fixed $T$ and set of microstates $s_A$.
+
+Because $A\!+\!B$ is a microcanonical ensemble,
+we know that its [thermodynamic potential](/know/concept/thermodynamic-potential/)
+is the entropy $S$.
+But what about the canonical ensemble, just $A$?
+
+The solution is a bit backwards.
+Note that the partition function $Z$ is not a constant;
+it depends on $T$ (via $\beta$), $V$ and $N$ (via $s_A$).
+Using the same logic as for the microcanonical ensemble,
+we define "equilibrium" as the set of microstates $s_A$
+that $A$ is most likely to occupy,
+which must be the set (as a function of $T,V,N$) that maximizes $Z$.
+
+However, $T$, $V$ and $N$ are fixed,
+so how can we maximize $Z$?
+Well, as it turns out,
+the Boltzmann distribution has already done it for us!
+We will return to this point later.
+
+Still, $Z$ does not have a clear physical interpretation.
+To find one, we start by showing that the ensemble averages
+of the energy $U_A$, pressure $P_A$ and chemical potential $\mu_A$
+can be calculated by differentiating $Z$.
+As preparation, note that:
+
+$$\begin{aligned}
+    \pdv{Z}{\beta} = - \sum_{s_A} U_A \exp(- \beta U_A)
+\end{aligned}$$
+
+With this, we can find the ensemble averages
+$\Expval{U_A}$, $\Expval{P_A}$ and $\Expval{\mu_A}$ of the system:
+
+$$\begin{aligned}
+    \Expval{U_A}
+    &= \sum_{s_A} p(s_A) \: U_A
+    = \frac{1}{Z} \sum_{s_A} U_A \exp(- \beta U_A)
+    = - \frac{1}{Z} \pdv{Z}{\beta}
+    \\
+    \Expval{P_A}
+    &= - \sum_{s_A} p(s_A) \pdv{U_A}{V}
+    = - \frac{1}{Z} \sum_{s_A} \exp(- \beta U_A) \pdv{U_A}{V}
+    \\
+    &= \frac{1}{Z \beta} \pdv{}{V}\sum_{s_A} \exp(- \beta U_A)
+    = \frac{1}{Z \beta} \pdv{Z}{V}
+    \\
+    \Expval{\mu_A}
+    &= \sum_{s_A} p(s_A) \pdv{U_A}{N}
+    = \frac{1}{Z} \sum_{s_A} \exp(- \beta U_A) \pdv{U_A}{N}
+    \\
+    &= - \frac{1}{Z \beta} \pdv{}{N}\sum_{s_A} \exp(- \beta U_A)
+    = - \frac{1}{Z \beta} \pdv{Z}{N}
+\end{aligned}$$
+
+It will turn out more convenient to use derivatives of $\ln{Z}$ instead,
+in which case:
+
+$$\begin{aligned}
+    \Expval{U_A}
+    = - \pdv{\ln{Z}}{\beta}
+    \qquad \quad
+    \Expval{P_A}
+    = \frac{1}{\beta} \pdv{\ln{Z}}{V}
+    \qquad \quad
+    \Expval{\mu_A}
+    = - \frac{1}{\beta} \pdv{\ln{Z}}{N}
+\end{aligned}$$
+
+Now, to find a physical interpretation for $Z$.
+Consider the quantity $F$, in units of energy,
+whose minimum corresponds to a maximum of $Z$:
+
+$$\begin{aligned}
+    F \equiv - k T \ln{Z}
+\end{aligned}$$
+
+We rearrange the equation to $\beta F = - \ln{Z}$ and take its differential element:
+
+$$\begin{aligned}
+    \dd{(\beta F)}
+    = - \dd{(\ln{Z})}
+    &= - \pdv{\ln{Z}}{\beta} \dd{\beta} - \pdv{\ln{Z}}{V} \dd{V} - \pdv{\ln{Z}}{N} \dd{N}
+    \\
+    &= \Expval{U_A} \dd{\beta} - \beta \Expval{P_A} \dd{V} + \beta \Expval{\mu_A} \dd{N}
+    \\
+    &= \Expval{U_A} \dd{\beta} + \beta \dd{\Expval{U_A}} - \beta \dd{\Expval{U_A}} - \beta \Expval{P_A} \dd{V} + \beta \Expval{\mu_A} \dd{N}
+    \\
+    &= \dd{(\beta \Expval{U_A})} - \beta \: \big( \dd{\Expval{U_A}} + \Expval{P_A} \dd{V} - \Expval{\mu_A} \dd{N} \big)
+\end{aligned}$$
+
+Rearranging and substituting
+the [fundamental thermodynamic relation](/know/concept/fundamental-thermodynamic-relation/)
+then gives:
+
+$$\begin{aligned}
+    \dd{(\beta F - \beta \Expval{U_A})}
+    &= - \beta \: \big( \dd{\Expval{U_A}} + \Expval{P_A} \dd{V} - \Expval{\mu_A} \dd{N} \big)
+    = - \beta T \dd{S_A}
+\end{aligned}$$
+
+We integrate this and ignore the integration constant,
+leading us to the desired result:
+
+$$\begin{aligned}
+    - \beta T S_A
+    &= \beta F - \beta \Expval{U_A}
+    \quad \implies \quad
+    F = \Expval{U_A} -  T S_A
+\end{aligned}$$
+
+As was already suggested by our notation,
+$F$ turns out to be the **Helmholtz free energy**:
+
+$$\begin{aligned}
+    \boxed{
+        F
+        \equiv - k T \ln{Z}
+        = \Expval{U_A} - T S_A
+    }
+\end{aligned}$$
+
+We can therefore reinterpret
+the partition function $Z$ and the Boltzmann distribution $p(s_A)$
+in the following "more physical" way:
+
+$$\begin{aligned}
+    Z
+    = \exp(- \beta F)
+    \qquad \quad
+    p(s_A)
+    = \exp\!\Big( \beta \big( F \!-\! U_A(s_A) \big) \Big)
+\end{aligned}$$
+
+Finally, by rearranging the expressions for $F$,
+we find the entropy $S_A$ to be:
+
+$$\begin{aligned}
+    S_A
+    = k \ln{Z} + \frac{\Expval{U_A}}{T}
+\end{aligned}$$
+
+This is why $Z$ is already maximized:
+the Boltzmann distribution maximizes $S_A$ for fixed values of $T$ and $\Expval{U_A}$,
+leaving $Z$ as the only "variable".
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/capillary-action/index.md b/source/know/concept/capillary-action/index.md
new file mode 100644
index 0000000..b388389
--- /dev/null
+++ b/source/know/concept/capillary-action/index.md
@@ -0,0 +1,125 @@
+---
+title: "Capillary action"
+date: 2021-03-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+- Surface tension
+layout: "concept"
+---
+
+**Capillary action** refers to the movement of liquid
+through narrow spaces due to surface tension, often against gravity.
+It occurs when the [Laplace pressure](/know/concept/young-laplace-law/)
+from surface tension is much larger in magnitude than the
+[hydrostatic pressure](/know/concept/hydrostatic-pressure/) from gravity.
+
+Consider a spherical droplet of liquid with radius $R$.
+The hydrostatic pressure difference
+between the top and bottom of the drop
+is much smaller than the Laplace pressure:
+
+$$\begin{aligned}
+    2 R \rho g \ll 2 \frac{\alpha}{R}
+\end{aligned}$$
+
+Where $\rho$ is the density of the liquid,
+$g$ is the acceleration due to gravity,
+and $\alpha$ is the energy cost per unit surface area.
+Rearranging the inequality yields:
+
+$$\begin{aligned}
+    R^2 \ll \frac{\alpha}{\rho g}
+\end{aligned}$$
+
+From the right-hand side we define the **capillary length** $L_c$,
+so gravity is negligible if $R \ll L_c$:
+
+$$\begin{aligned}
+    \boxed{
+        L_c
+        \equiv \sqrt{\frac{\alpha}{\rho g}}
+    }
+\end{aligned}$$
+
+In general, for a system with characteristic length $L$,
+the relative strength of gravity compared to surface tension
+is described by the **Bond number** $\mathrm{Bo}$
+or **Eötvös number** $\mathrm{Eo}$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{Bo}
+        \equiv \mathrm{Eo}
+        \equiv \frac{L^2}{L_c^2}
+        = \frac{m g}{\alpha L}
+    }
+\end{aligned}$$
+
+The right-most side gives an alternative way of understanding $\mathrm{Bo}$:
+$m$ is the mass of a cube with side $L$, such that the numerator is the weight force,
+and the denominator is the tension force of the surface.
+In any case, capillary action can be observed when $\mathrm{Bo \ll 1}$.
+
+The most famous example of capillary action is **capillary rise**,
+where a liquid "climbs" upwards in a narrow vertical tube with radius $R$,
+apparently defying gravity.
+Assuming the liquid-air interface is a spherical cap
+with constant [curvature](/know/concept/curvature/) radius $R_c$,
+then we know that the liquid is at rest
+when the hydrostatic pressure equals the Laplace pressure:
+
+$$\begin{aligned}
+    \rho g h
+    \approx \alpha \frac{2}{R_c}
+    = 2 \alpha \frac{\cos\theta}{R}
+\end{aligned}$$
+
+Where $\theta$ is the liquid-tube contact angle,
+and we are neglecting variations of the height $h$ due to the curvature
+(i.e. the [meniscus](/know/concept/meniscus/)).
+By isolating the above equation for $h$,
+we arrive at **Jurin's law**,
+which predicts the height climbed by a liquid in a tube with radius $R$:
+
+$$\begin{aligned}
+    \boxed{
+        h
+        = 2 \frac{L_c^2}{R} \cos\theta
+    }
+\end{aligned}$$
+
+Depending on $\theta$, $h$ can be negative,
+i.e. the liquid might descend below the ambient level.
+
+
+An alternative derivation of Jurin's law balances the forces instead of the pressures.
+On the right, we have the gravitational force
+(i.e. the energy-per-distance to lift the liquid),
+and on the left, the surface tension force
+(i.e. the energy-per-distance of the liquid-tube interface):
+
+$$\begin{aligned}
+     \pi R^2 \rho g h
+     \approx 2 \pi R (\alpha_{sg} - \alpha_{sl})
+\end{aligned}$$
+
+Where $\alpha_{sg}$ and $\alpha_{sl}$ are the energy costs
+of the solid-gas and solid-liquid interfaces.
+Thanks to the [Young-Dupré relation](/know/concept/young-dupre-relation/),
+we can rewrite this as follows:
+
+$$\begin{aligned}
+     R \rho g h
+     = 2 \alpha \cos\theta
+\end{aligned}$$
+
+Isolating this for $h$ simply yields Jurin's law again, as expected.
+
+
+
+## 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/cauchy-principal-value/index.md b/source/know/concept/cauchy-principal-value/index.md
new file mode 100644
index 0000000..fec0826
--- /dev/null
+++ b/source/know/concept/cauchy-principal-value/index.md
@@ -0,0 +1,52 @@
+---
+title: "Cauchy principal value"
+date: 2021-11-01
+categories:
+- Mathematics
+layout: "concept"
+---
+
+The **Cauchy principal value** $\mathcal{P}$,
+or just **principal value**,
+is a method for integrating problematic functions,
+i.e. functions with singularities,
+whose integrals would otherwise diverge.
+
+Consider a function $f(x)$ with a singularity at some finite $x = b$,
+which is hampering attempts at integrating it.
+To resolve this, we define the Cauchy principal value $\mathcal{P}$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \mathcal{P} \int_a^c f(x) \dd{x}
+        = \lim_{\varepsilon \to 0^{+}} \!\bigg( \int_a^{b - \varepsilon} f(x) \dd{x} + \int_{b + \varepsilon}^c f(x) \dd{x} \bigg)
+    }
+\end{aligned}$$
+
+If $f(x)$ instead has a singularity at postive infinity $+\infty$,
+then we define $\mathcal{P}$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \mathcal{P} \int_{a}^\infty f(x) \dd{x}
+        = \lim_{c \to \infty} \!\bigg( \int_{a}^c f(x) \dd{x} \bigg)
+    }
+\end{aligned}$$
+
+And analogously for $-\infty$.
+If $f(x)$ has singularities both at $+\infty$ and at $b$,
+then we simply combine the two previous cases,
+such that $\mathcal{P}$ is given by:
+
+$$\begin{aligned}
+    \mathcal{P} \int_{a}^\infty f(x) \:dx
+    = \lim_{c \to \infty} \lim_{\varepsilon \to 0^{+}}
+    \!\bigg( \int_{a}^{b - \varepsilon} f(x) \:dx + \int_{b + \varepsilon}^{c} f(x) \:dx \bigg)
+\end{aligned}$$
+
+And so on, until all problematic singularities have been dealt with.
+
+In some situations, for example involving
+the [Sokhotski-Plemelj theorem](/know/concept/sokhotski-plemelj-theorem/),
+the symbol $\mathcal{P}$ is written without an integral,
+in which case the calculations are implicitly integrated.
diff --git a/source/know/concept/cauchy-strain-tensor/index.md b/source/know/concept/cauchy-strain-tensor/index.md
new file mode 100644
index 0000000..b5840e7
--- /dev/null
+++ b/source/know/concept/cauchy-strain-tensor/index.md
@@ -0,0 +1,325 @@
+---
+title: "Cauchy strain tensor"
+date: 2021-03-31
+categories:
+- Physics
+- Continuum physics
+layout: "concept"
+---
+
+**Strain** quantifies the deformation of a solid object.
+If the body has been deformed, e.g. by pulling or bending,
+its constituent particles have moved a bit.
+Let $\va{X}$ be the original location of a particle,
+and $\va{x}$ its new location after the deformation.
+We can thus define the **displacement field** $\va{u}$:
+
+$$\begin{aligned}
+    \va{u}
+    \equiv \va{x} - \va{X}
+\end{aligned}$$
+
+We restrict ourselves to **infinitesimal strain**,
+where $\va{u}$ is so tiny that the material's properties are unchanged,
+and a **slowly-varying strain**,
+where the particle's neighbourhood has been distorted,
+but not completely changed.
+
+A key challenge when quantifying deformation
+is that we need to somehow exclude movements of the *entire* body:
+for example, you can bend a twig in your hands while walking or dancing,
+but we are only interested in the twig's shape change,
+not in your movements.
+The above definition of $\vu{u}$ includes both,
+so we should be careful how we extract the strain from it.
+
+
+## Definition
+
+We use the **Eulerian description** of deformation,
+where the new position $\va{x}$ is the reference,
+and the old position $\va{X}$ is expressed as a function of $\va{x}$:
+
+$$\begin{aligned}
+    \va{u}(\va{x})
+    \equiv \va{x} - \va{X}(\va{x})
+\end{aligned}$$
+
+Let us choose two nearby points in the deformed solid,
+and call them $\va{x}$ and $\va{x} + \va{a}$,
+where $\va{a}$ is a tiny vector pointing from one to the other.
+Before the displacement, those points respectively had these positions,
+where we define $\va{A}$ as the "old" version of $\va{a}$:
+
+$$\begin{aligned}
+    \va{X} = \va{X}(\va{x})
+    \qquad
+    \va{X} + \va{A} = \va{X}(\va{x} + \va{a})
+\end{aligned}$$
+
+Because the new positions $\va{x}$ are our reference,
+we would like to write $\va{A}$ without $\va{X}$.
+To do so, we use the definition of $\va{u}(\va{x})$, yielding:
+
+$$\begin{aligned}
+    \va{A}
+    &= \va{X}(\va{x} + \va{a}) - \va{X}(\va{x})
+    \\
+    &= \big( \va{x} + \va{a} - \va{u}(\va{x} + \va{a}) \big) - \big( \va{x} - \va{u}(\va{x}) \big)
+    \\
+    &= \va{a} - \va{u}(\va{x} + \va{a}) - \va{u}(\va{x})
+\end{aligned}$$
+
+Using the fact that $\va{a}$ is tiny by definition,
+we expand the middle term to first order in $\va{a}$:
+
+$$\begin{aligned}
+    \va{u}(\va{x} + \va{a})
+    \approx \va{u}(\va{x}) + a_x \pdv{\va{u}}{x} + a_y \pdv{\va{u}}{y} + a_z \pdv{\va{u}}{z}
+    = \va{u}(\va{x}) + (\va{a} \cdot \nabla) \va{u}(\va{x})
+\end{aligned}$$
+
+With this, we can now define the "shift" $\delta\va{a}$
+as the difference between $\va{a}$ and $\va{A}$ like so:
+
+$$\begin{aligned}
+    \delta{\va{a}}
+    \equiv \va{a} - \va{A}
+    = (\va{a} \cdot \nabla) \va{u}(\va{x})
+\end{aligned}$$
+
+In index notation, we write this expression as follows,
+with $\nabla_j \equiv \ipdv{}{x_j}$ simply being the partial derivative
+with respect to the $j$th coordinate:
+
+$$\begin{aligned}
+    \delta a_i
+    = \sum_{j} a_j \nabla_j u_i
+\end{aligned}$$
+
+Where $\nabla_j u_i$ are called the **displacement gradients**,
+and are just one step away from the desired definition of strain.
+Note that these gradients are dimensionless,
+so we can more formally define a *slowly-varying* displacement $\va{u}(\va{x})$
+as one where $|\nabla_j u_i| \ll 1$.
+
+Now, to solve the problem of macroscopic movements,
+we take another tiny vector $\va{b}$ starting in the same point $\va{x}$ as $\va{a}$.
+Here is the trick: if the whole body is uniformly translated or rotated,
+the scalar product $\va{a} \cdot \va{b}$ is unchanged,
+but if there is a non-uniform distortion, it changes.
+We thus define the scalar product's difference like so:
+
+$$\begin{aligned}
+    \delta(\va{a} \cdot \va{b})
+    \equiv \va{a} \cdot \va{b} - \va{A} \cdot \va{B}
+\end{aligned}$$
+
+Where $\va{B}$ is the old version of $\va{b}$.
+Since these vectors are all tiny, we apply the product rule:
+
+$$\begin{aligned}
+    \delta(\va{a} \cdot \va{b})
+    &= \delta\va{a} \cdot \va{b} + \va{a} \cdot \delta\va{b}
+\end{aligned}$$
+
+It is more informative to switch to index notation here.
+Inserting $\delta\va{a}$ and $\delta\va{b}$ yields:
+
+$$\begin{aligned}
+    \delta(\va{a} \cdot \va{b})
+    &= \sum_{i} \delta{a}_i \: b_i + \sum_{i} \delta{b}_i \: a_i
+    \\
+    &= \sum_{ij} \nabla_j u_i \: a_j b_i + \sum_{ij} \nabla_j u_i \: a_i b_j
+    \\
+    &= \sum_{ij} \big( \nabla_i u_j + \nabla_j u_i \big) \: a_i b_j
+\end{aligned}$$
+
+At last, we define the **Cauchy infinitesimal strain tensor** $\hat{u}$
+such that it has $u_{ij}$ as components:
+
+$$\begin{aligned}
+    \boxed{
+        u_{ij}
+        \equiv \frac{1}{2} \big( \nabla_j u_i + \nabla_i u_j \big)
+    }
+\end{aligned}$$
+
+Which allows us to rewrite the shift of the scalar product in the following compact way:
+
+$$\begin{aligned}
+    \delta(\va{a} \cdot \va{b})
+    &= 2 \sum_{ij} u_{ij} a_i b_j
+    = 2 \va{a} \cdot \hat{u} \cdot \va{b}
+\end{aligned}$$
+
+The Cauchy strain tensor $\hat{u}$ is a second-rank tensor,
+and can alternatively be expressed like so:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{u}
+        \equiv \frac{1}{2} \big( \nabla \va{u} + (\nabla \va{u})^\top \big)
+    }
+\end{aligned}$$
+
+Where $\top$ is the transpose. Being defined from the scalar product,
+all macroscopic movements of the body are removed from the tensor,
+which turns out to make it symmetric, i.e. $u_{ij} = u_{ji}$.
+
+
+## Geometry
+
+So far we have used Cartesian coordinates,
+but we can choose any three vectors $\va{a}$, $\va{b}$ and $\va{c}$,
+and **project** $\hat{u}$ onto this basis.
+For example, the component $u_{ab}$ then becomes:
+
+$$\begin{aligned}
+    \boxed{
+        u_{ab}
+        = \frac{\va{a} \cdot \hat{u} \cdot \va{b}}{\big|\va{a}\big| \big|\va{b}\big|}
+    }
+\end{aligned}$$
+
+And so forth, for the other eight components.
+The basis in which $\hat{u}$ is diagonal is the one formed by its eigenvectors,
+and their directions are the **principal axes of strain**
+at that point in the solid.
+Because $\hat{u}$ is symmetric, such a basis always exists.
+
+Given a vector $\va{a}$, its relative length change
+due to the deformation is simply given by:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\delta|\va{a}|}{|\va{a}|}
+        = u_{aa}
+    }
+\end{aligned}$$
+
+To find the angle change $\delta\theta$
+between two vectors $\va{a}$ and $\va{b}$,
+we start with the product rule:
+
+$$\begin{aligned}
+    \delta(\va{a} \cdot \va{b})
+    = \delta(\big|\va{a}\big| \big|\va{b}\big| \cos\theta)
+    = \delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta
+    + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta
+    - \big|\va{a}\big| \big|\va{b}\big| \sin\theta \: \delta\theta
+\end{aligned}$$
+
+We isolate this for $\delta\theta$, using the fact that
+$\delta(\va{a} \cdot \va{b}) = 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}$
+thanks to the projection $u_{ab}$:
+
+$$\begin{aligned}
+    \delta\theta
+    = \frac{\delta\big|\va{a}\big| \big|\va{b}\big| \cos\theta
+    + \big|\va{a}\big| \delta\big|\va{b}\big| \cos\theta
+    - 2 \big|\va{a}\big| \big|\va{b}\big| u_{ab}}
+    {\big|\va{a}\big| \big|\va{b}\big| \sin\theta}
+\end{aligned}$$
+
+By recognizing the length change $\delta|\va{a}|/|\va{a}| = u_{aa}$,
+we arrive at the following expression:
+
+$$\begin{aligned}
+    \boxed{
+        \delta\theta
+        = \frac{(u_{aa} + u_{bb}) \cos\theta - u_{ab}}{\sin\theta}
+    }
+\end{aligned}$$
+
+Now, everything so far has been about tiny vectors,
+so the change of the line element $\dd{\va{l}}$
+is easy to express using the displacement field $\va{u}$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(\dd{\va{l}})
+        = (\dd{\va{l}} \cdot \nabla) \va{u}
+    }
+\end{aligned}$$
+
+Next, we calculate the change of the differential volume element $\dd{V}$
+by treating it as the volume of a tiny parallelepiped
+spanned by $\va{a}$, $\va{b}$ and $\va{c}$:
+
+$$\begin{aligned}
+    \delta(\dd{V})
+    = \delta(\va{a} \cross \va{b} \cdot \va{c})
+    &= \delta\va{a} \cross \va{b} \cdot \va{c} + \va{a} \cross \delta\va{b} \cdot \va{c} + \va{a} \cross \va{b} \cdot \delta\va{c}
+    \\
+    &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c}
+    + \va{a} \cross (\va{b} \cdot \nabla )\va{u} \cdot \va{c}
+    + \va{a} \cross \va{b} \cdot (\va{c} \cdot \nabla) \va{u}
+\end{aligned}$$
+
+We can reorder the factors like so
+(write it out in index notation if you are not convinced):
+
+$$\begin{aligned}
+    \delta(\dd{V})
+    &= (\va{a} \cdot \nabla) \va{u} \cross \va{b} \cdot \va{c}
+    + (\va{b} \cdot \nabla) \va{a} \cross \va{u} \cdot \va{c}
+    + (\va{c} \cdot \nabla) \va{a} \cross \va{b} \cdot \va{u}
+\end{aligned}$$
+
+By applying a couple of vector identities,
+we can rewrite this more compactly as follows:
+
+$$\begin{aligned}
+    \delta(\dd{V})
+    &= \Big( \va{b} \cross \va{c} (\va{a} \cdot \nabla) \cross \va{b}
+    + \va{c} \cross \va{a} (\va{b} \cdot \nabla)
+    + \va{a} \cross \va{b} (\va{c} \cdot \nabla) \Big) \cdot \va{u}
+    \\
+    &= (\va{a} \cross \va{b} \cdot \va{c}) (\nabla \cdot \va{u})
+\end{aligned}$$
+
+Here, we recognize the definition of $\dd{V}$,
+leading to the following infinitesimal volume change:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(\dd{V})
+        = \nabla \cdot \va{u} \dd{V}
+    }
+\end{aligned}$$
+
+Finally, for the surface element $\dd{\va{S}} = \va{a} \cross \va{b}$,
+we use that the volume element $\dd{V} = \va{c} \cdot \dd{\va{S}}$:
+
+$$\begin{aligned}
+    \delta(\dd{V})
+    = \delta(\va{c} \cdot \dd{\va{S}})
+    = \delta\va{c} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
+    = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
+\end{aligned}$$
+
+By comparing this to the previous result for $\delta(\dd{V})$,
+we arrive at the following equation:
+
+$$\begin{aligned}
+    \nabla \cdot \va{u} (\va{c} \cdot \dd{\va{S}})
+    = (\va{c} \cdot \nabla) \va{u} \cdot \dd{\va{S}} + \va{c} \cdot \delta(\dd{\va{S}})
+\end{aligned}$$
+
+Since $\va{c}$ is dot-multiplied at the front of each term,
+we remove it, and isolate the rest for $\delta(\dd{\va{S}})$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(\dd{\va{S}})
+        = \big( (\nabla \cdot \va{u}) \hat{1} - \nabla \va{u} \big) \cdot \dd{\va{S}}
+    }
+\end{aligned}$$
+
+
+
+## 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/cauchy-stress-tensor/index.md b/source/know/concept/cauchy-stress-tensor/index.md
new file mode 100644
index 0000000..c56d997
--- /dev/null
+++ b/source/know/concept/cauchy-stress-tensor/index.md
@@ -0,0 +1,238 @@
+---
+title: "Cauchy stress tensor"
+date: 2021-03-31
+categories:
+- Physics
+- Continuum physics
+layout: "concept"
+---
+
+Roughly speaking, **stress** is the solid equivalent of fluid pressure:
+it describes the net force acting on an imaginary partition surface inside a solid.
+However, unlike fluids at rest,
+where the pressure is always perpendicular to such a surface,
+solid stress is usually much more complicated.
+
+Formally, the concept of stress can be applied to any continuum
+(not just solids), including fluids,
+but it is arguably most intuitive for solids.
+
+
+## Definition
+
+In the solid, imagine an infinitesimal cube
+whose sides, $\dd{S}_x$, $\dd{S}_y$ and $\dd{S}_z$,
+are orthogonal to the $x$, $y$ and $z$ axes, respectively.
+There is a force $\dd{\va{F}}_1$ acting on $\dd{S}_x$,
+$\dd{\va{F}}_2$ on $\dd{S}_y$, and $\dd{\va{F}}_3$ on $\dd{S}_z$.
+Then we can decompose each of these forces, for example:
+
+$$\begin{aligned}
+    \dd{\va{F}}_1
+    = \va{e}_x F_{x1} + \va{e}_y F_{y1} + \va{e}_z F_{z1}
+\end{aligned}$$
+
+Where $\va{e}_x$, $\va{e}_y$ and $\va{e}_z$ are the basis unit vectors.
+If we divide each of the force components by the area $\dd{S}_x$
+(like in a fluid, in order to get the pressure),
+we find the stresses $\sigma_{xx}$, $\sigma_{yx}$ and $\sigma_{zx}$
+that are being "felt" by the $x$ surface element $\dd{S}_x$:
+
+$$\begin{aligned}
+    \dd{\va{F}}_1
+    = \big( \va{e}_x \sigma_{xx}  + \va{e}_y \sigma_{yx} + \va{e}_z \sigma_{zx} \big) \dd{S}_x
+\end{aligned}$$
+
+The perpendicular component $\sigma_{xx}$ is called a **tensile stress**,
+and its sign is always chosen so that a positive value corresponds to a tension,
+i.e. the $x$-side is pulled away from the rest of the cube.
+The tangential components $\sigma_{yx}$ and $\sigma_{zx}$
+are called **shear stresses**.
+
+Evidently, the other two forces $\dd{\va{F}}_2$ and $\dd{\va{F}}_3$
+can be decomposed in the exact same way,
+yielding nine stress components in total:
+
+$$\begin{aligned}
+    \dd{\va{F}}_2
+    &= \va{e}_x F_{x2} + \va{e}_y F_{y2} + \va{e}_z F_{z2}
+    = \big( \va{e}_x \sigma_{xy}  + \va{e}_y \sigma_{yy} + \va{e}_z \sigma_{zy} \big) \dd{S}_y
+    \\
+    \dd{\va{F}}_3
+    &= \va{e}_x F_{x3} + \va{e}_y F_{y3} + \va{e}_z F_{z3}
+    = \big( \va{e}_x \sigma_{xz}  + \va{e}_y \sigma_{yz} + \va{e}_z \sigma_{zz} \big) \dd{S}_z
+\end{aligned}$$
+
+The total force $\dd{\va{F}}$ on the entire infinitesimal cube
+is simply the sum of the previous three:
+
+$$\begin{aligned}
+    \dd{\va{F}}
+    = \dd{\va{F}}_1 + \dd{\va{F}}_2 + \dd{\va{F}}_3
+\end{aligned}$$
+
+We can then decompose $\dd{\va{F}}$ into its net components
+along the $x$, $y$ and $z$ axes:
+
+$$\begin{aligned}
+    \dd{\va{F}}
+    = \va{e}_x \dd{F}_x + \va{e}_y \dd{F}_y + \va{e}_z \dd{F}_z
+\end{aligned}$$
+
+From the preceding equations, we find that these components are given by:
+
+$$\begin{aligned}
+    \dd{F}_x
+    &= \sigma_{xx} \dd{S}_x + \sigma_{xy} \dd{S}_y + \sigma_{xz} \dd{S}_z
+    \\
+    \dd{F}_y
+    &= \sigma_{yx} \dd{S}_x + \sigma_{yy} \dd{S}_y + \sigma_{yz} \dd{S}_z
+    \\
+    \dd{F}_z
+    &= \sigma_{zx} \dd{S}_x + \sigma_{zy} \dd{S}_y + \sigma_{zz} \dd{S}_z
+\end{aligned}$$
+
+We can write this much more compactly using index notation,
+where $i, j \in \{x, y, z\}$:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{F}_i
+        = \sum_{j} \sigma_{ij} \dd{S}_j
+    }
+\end{aligned}$$
+
+The stress components $\sigma_{ij}$ can be written as a second-rank tensor
+(i.e. a matrix that transforms in a certain way),
+called the **Cauchy stress tensor** $\hat{\sigma}$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\sigma} \equiv
+        \{ \sigma_{ij} \} =
+        \begin{pmatrix}
+            \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\
+            \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\
+            \sigma_{zx} & \sigma_{zy} & \sigma_{zz}
+        \end{pmatrix}
+    }
+\end{aligned}$$
+
+Then $\dd{\va{F}}$ is written even more compactly
+using the dot product, with $\dd{\va{S}} = (\dd{S}_x, \dd{S}_y, \dd{S}_z)$:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\va{F}}
+        = \hat{\sigma} \cdot \dd{\va{S}}
+    }
+\end{aligned}$$
+
+All forces on the cube's sides can be written in this form.
+**Cauchy's stress theorem** states that the force on *any*
+surface element inside the solid can be written like this,
+simply by projecting it onto the $x$, $y$ and $z$ zero-planes
+to get the areas $\dd{S}_x$, $\dd{S}_y$ and $\dd{S}_z$.
+
+Note that for fluids, the pressure $p$ was defined
+such that $\dd{\va{F}} = - p \dd{\va{S}}$.
+If we wanted to define $p$ for solids in the same way,
+we would need $\hat{\sigma}$ to be diagonal *and*
+all of its diagonal elements to be identical.
+Since this is almost never the case,
+the scalar pressure is ill-defined in solids.
+
+
+## Equilibrium
+
+The total force $\va{F}$ acting on a (non-infinitesimal) volume $V$ of the solid
+is given by the sum of the total body force $\va{F}_b$ and total surface force $\va{F}_s$,
+where $\vec{f}$ is the body force density:
+
+$$\begin{aligned}
+    \va{F}
+    = \va{F}_b + \va{F}_s
+    = \int_V \va{f} \dd{V} + \oint_S \hat{\sigma} \cdot \dd{\va{S}}
+\end{aligned}$$
+
+We can rewrite the surface term using the divergence theorem,
+where $\top$ is the transpose:
+
+$$\begin{aligned}
+    \va{F}_s
+    = \oint_S \hat{\sigma} \cdot \dd{\va{S}}
+    = \int_V \nabla \cdot \hat{\sigma}^{\top} \dd{V}
+\end{aligned}$$
+
+For some people, this equation may be more enlightening in index notation,
+where $\nabla_j \equiv \ipdv{}{x_j}$ is the partial derivative with respect to the $j$th coordinate:
+
+$$\begin{aligned}
+    F_{s, i}
+    = \oint_S \sum_j \sigma_{ij} \dd{S_j}
+    = \int_V \sum_{j} \nabla_{\!j} \sigma_{ij} \dd{V}
+\end{aligned}$$
+
+In any case, the total force $\va{F}$ can then be expressed
+as a single volume integral over $V$:
+
+$$\begin{aligned}
+    \va{F}
+    = \int_V \va{f} \dd{V} + \int_V \nabla \cdot \hat{\sigma}^{\top} \dd{V}
+    = \int_V \va{f^*} \dd{V}
+\end{aligned}$$
+
+Where we have defined the **effective force density** $\va{f^*}$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \va{f^*}
+        = \va{f} + \nabla \cdot \hat{\sigma}^{\top}
+    }
+\end{aligned}$$
+
+The volume $V$ is in **mechanical equilibrium** if the net force acting on it amounts to zero:
+
+$$\begin{aligned}
+    \va{F}
+    = 0
+\end{aligned}$$
+
+However, because $V$ is abritrary, the equilibrium condition for the whole solid is in fact:
+
+$$\begin{aligned}
+    \boxed{
+        \va{f^*}
+        = 0
+    }
+\end{aligned}$$
+
+This is reminiscent of the equilibrium condition of a fluid
+(see [hydrostatic pressure](/know/concept/hydrostatic-pressure/)).
+Note that it is a set of coupled differential equations,
+which needs boundary conditions at the object's surface.
+Newton's third law states that the two sides of the boundary
+exert opposite forces on each other,
+so the boundary condition is continuity of the **stress vector**
+$\hat{\sigma} \cdot \va{n}$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\sigma}_{\mathrm{outer}} \cdot \va{n}
+        = - \hat{\sigma}_{\mathrm{inner}} \cdot \va{n}
+    }
+\end{aligned}$$
+
+Where the normal of the outer surface is $\va{n}$,
+and the normal of the inner surface is $-\va{n}$.
+Note that the above equation does *not* mean
+that $-\hat{\sigma}_{\mathrm{inner}}$ equals $\hat{\sigma}_{\mathrm{outer}}$:
+the tensors are allowed to be very different,
+as long as the stress vector's three components are equal.
+
+
+## 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/cavitation/index.md b/source/know/concept/cavitation/index.md
new file mode 100644
index 0000000..95822ae
--- /dev/null
+++ b/source/know/concept/cavitation/index.md
@@ -0,0 +1,113 @@
+---
+title: "Cavitation"
+date: 2021-04-09
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+In a liquid, **cavitation** is the spontaneous appearance of bubbles,
+occurring when the pressure in a part of the liquid drops
+below its vapour pressure, e.g. due to the fast movements.
+When such a bubble is subjected to a higher pressure
+by the surrounding liquid, it quickly implodes.
+
+To model this case, we use the simple form of
+the [Rayleigh-Plesset equation](/know/concept/rayleigh-plesset-equation/)
+for an inviscid liquid without surface tension.
+Note that the RP equation assumes incompressibility.
+
+We assume that the whole liquid is at a constant pressure $p_\infty$,
+and the bubble is empty, such that the interface pressure $P = 0$,
+meaning $\Delta p = - p_\infty$.
+At first, the radius is stationary $R'(0) = 0$,
+and given by a constant $R(0) = a$.
+The simple Rayleigh-Plesset equation is then:
+
+$$\begin{aligned}
+    R \dvn{2}{R}{t} + \frac{3}{2} \bigg( \dv{R}{t} \bigg)^2
+    = - \frac{p_\infty}{\rho}
+\end{aligned}$$
+
+To solve it, we multiply both sides by $R^2 R'$
+and rewrite it in the following way:
+
+$$\begin{aligned}
+    - 2 \frac{p_\infty}{\rho} R^2 R'
+    &= 2 R^3 R' R'' + 3 R^2 (R')^3
+    \\
+    - \frac{2 p_\infty}{3 \rho} \dv{}{t}\Big( R^3 \Big)
+    &= \dv{}{t}\Big( R^3 (R')^2 \Big)
+\end{aligned}$$
+
+It is then straightforward to integrate both sides
+with respect to time $\tau$, from $0$ to $t$:
+
+$$\begin{aligned}
+    - \frac{2 p_\infty}{3 \rho} \int_0^t \dv{}{\tau}\Big( R^3 \Big) \dd{\tau}
+    &= \int_0^t \dv{}{\tau}\Big( R^3 (R')^2 \Big) \dd{\tau}
+    \\
+    - \frac{2 p_\infty}{3 \rho} \Big[ R^3 \Big]_0^t
+    &= \Big[ R^3 (R')^2 \Big]_0^t
+    \\
+    - \frac{2 p_\infty}{3 \rho} \Big( R^3(t) - a^3 \Big)
+    &= \Big( R^3(t) \: \big(R'(t)\big)^2 \Big)
+\end{aligned}$$
+
+Rearranging this equation yields the following expression
+for the derivative $R'$:
+
+$$\begin{aligned}
+    (R')^2
+    = \frac{2 p_\infty}{3 \rho} \Big( \frac{a^3}{R^3} - 1 \Big)
+\end{aligned}$$
+
+This equation is nasty to integrate.
+The trick is to invert $R(t)$ into $t(R)$,
+and, because we are only interested in collapse,
+we just need to consider the case $R' < 0$.
+The time of a given radius $R$ is then as follows,
+where we are using slightly sloppy notation:
+
+$$\begin{aligned}
+    t
+    = \int_0^t \dd{\tau}
+    = - \int_{a}^{R} \frac{\dd{R}}{R'}
+    = \int_{R}^{a} \frac{\dd{R}}{R'}
+\end{aligned}$$
+
+The minus comes from the constraint that $R' < 0$, but $t \ge 0$.
+We insert the expression for $R'$:
+
+$$\begin{aligned}
+    t
+    = \sqrt{\frac{3 \rho}{2 p_\infty}} \int_{R}^{a} \Bigg( \sqrt{ \frac{a^3}{R^3} - 1 } \Bigg)^{-1} \dd{R}
+    = \sqrt{\frac{3 \rho a^2}{2 p_\infty}} \int_{R/a}^{1} \frac{1}{\sqrt{x^{-3} - 1}} \dd{x}
+\end{aligned}$$
+
+This integral needs to be looked up,
+and involves the hypergeometric function ${}_2 F_1$.
+However, we only care about *collapse*, which is when $R = 0$.
+The time $t_0$ at which this occurs is:
+
+$$\begin{aligned}
+    t_0
+    = \sqrt{\frac{3 \rho a^2}{2 p_\infty}} \sqrt{\frac{3 \pi}{2}} \frac{\Gamma(5/6)}{\Gamma(1/3)}
+    \approx \sqrt{\frac{3 \rho a^2}{2 p_\infty}} 0.915 \:\mathrm{s}
+\end{aligned}$$
+
+With our assumptions, a bubble will always collapse.
+However, unsurprisingly, reality turns out to be more complicated:
+as $R \to 0$, the interface velocity $R' \to \infty$.
+By looking at the derivation of the Rayleigh-Plesset equation,
+it can be shown that the pressure just outside the bubble diverges due to $R'$.
+This drastically changes the liquid's properties, and breaks our assumptions.
+
+
+
+## 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/central-limit-theorem/index.md b/source/know/concept/central-limit-theorem/index.md
new file mode 100644
index 0000000..126ff3b
--- /dev/null
+++ b/source/know/concept/central-limit-theorem/index.md
@@ -0,0 +1,203 @@
+---
+title: "Central limit theorem"
+date: 2021-03-09
+categories:
+- Statistics
+- Mathematics
+layout: "concept"
+---
+
+In statistics, the **central limit theorem** states that
+the sum of many independent variables tends towards a normal distribution,
+even if the individual variables $x_n$ follow different distributions.
+
+For example, by taking $M$ samples of size $N$ from a population,
+and calculating $M$ averages $\mu_m$ (which involves summing over $N$),
+the resulting means $\mu_m$ are normally distributed
+across the $M$ samples if $N$ is sufficiently large.
+
+More formally, for $N$ independent variables $x_n$ with probability distributions $p(x_n)$,
+the central limit theorem states the following,
+where we define the sum $S$:
+
+$$\begin{aligned}
+    S = \sum_{n = 1}^N x_n
+    \qquad
+    \mu_S = \sum_{n = 1}^N \mu_n
+    \qquad
+    \sigma_S^2 = \sum_{n = 1}^N \sigma_n^2
+\end{aligned}$$
+
+And crucially, it states that the probability distribution $p_N(S)$ of $S$ for $N$ variables
+will become a normal distribution when $N$ goes to infinity:
+
+$$\begin{aligned}
+    \boxed{
+        \lim_{N \to \infty} \!\big(p_N(S)\big)
+        = \frac{1}{\sigma_S \sqrt{2 \pi}} \exp\!\Big( -\frac{(\mu_S - S)^2}{2 \sigma_S^2} \Big)
+    }
+\end{aligned}$$
+
+We prove this below,
+but first we need to introduce some tools.
+Given a probability density $p(x)$, its [Fourier transform](/know/concept/fourier-transform/)
+is called the **characteristic function** $\phi(k)$:
+
+$$\begin{aligned}
+    \phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x}
+\end{aligned}$$
+
+Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$.
+We take its Taylor expansion in two separate ways,
+where an overline denotes the mean:
+
+$$\begin{aligned}
+    \phi(k)
+    = \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0)
+    \qquad
+    \phi(k)
+    = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
+\end{aligned}$$
+
+By comparing the coefficients of these two power series,
+we get a useful relation:
+
+$$\begin{aligned}
+    \phi^{(n)}(0) = i^n \: \overline{x^n}
+\end{aligned}$$
+
+Next, the **cumulants** $C^{(n)}$ are defined from the Taylor expansion of $\ln\!\big(\phi(k)\big)$:
+
+$$\begin{aligned}
+    \ln\!\big( \phi(k) \big)
+    = \sum_{n = 1}^\infty \frac{(ik)^n}{n!} C^{(n)}
+    \quad \mathrm{where} \quad
+    C^{(n)} = \frac{1}{i^n} \: \dvn{n}{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0}
+\end{aligned}$$
+
+The first two cumulants $C^{(1)}$ and $C^{(2)}$ are of particular interest,
+since they turn out to be the mean and the variance respectively,
+using our earlier relation:
+
+$$\begin{aligned}
+    C^{(1)}
+    &= - i \dv{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0}
+    = - i \frac{\phi'(0)}{\exp(0)}
+    = \overline{x}
+    \\
+    C^{(2)}
+    &= - \dvn{2}{}{k} \Big(\ln\!\big(\phi(k)\big)\Big) \Big|_{k = 0}
+    = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)}
+    = - \overline{x}^2 + \overline{x^2} = \sigma^2
+\end{aligned}$$
+
+Let us now define $S$ as the sum of $N$ independent variables $x_n$, in other words:
+
+$$\begin{aligned}
+    S = \sum_{n = 1}^N x_n = x_1 + x_2 + ... + x_N
+\end{aligned}$$
+
+The probability density of $S$ is then as follows, where $p(x_n)$ are
+the densities of all the individual variables and $\delta$ is
+the [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+    p(S)
+    &= \int\cdots\int_{-\infty}^\infty \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( S - \sum_{n = 1}^N x_n \Big) \dd{x_1} \cdots \dd{x_N}
+    \\
+    &= \Big( p_1 * \big( p_2 * ( ... * (p_N * \delta))\big)\Big)(S)
+\end{aligned}$$
+
+In other words, the integrals pick out all combinations of $x_n$ which
+add up to the desired $S$-value, and multiply the probabilities
+$p(x_1) p(x_2) \cdots p(x_N)$ of each such case. This is a convolution,
+so the [convolution theorem](/know/concept/convolution-theorem/)
+states that it is a product in the Fourier domain:
+
+$$\begin{aligned}
+    \phi_S(k) = \prod_{n = 1}^N \phi_n(k)
+\end{aligned}$$
+
+By taking the logarithm of both sides, the product becomes a sum,
+which we further expand:
+
+$$\begin{aligned}
+    \ln\!\big(\phi_S(k)\big)
+    = \sum_{n = 1}^N \ln\!\big(\phi_n(k)\big)
+    = \sum_{n = 1}^N \sum_{m = 1}^{\infty} \frac{(ik)^m}{m!} C_n^{(m)}
+\end{aligned}$$
+
+Consequently, the cumulants $C^{(m)}$ stack additively for the sum $S$
+of independent variables $x_m$, and therefore
+the means $C^{(1)}$ and variances $C^{(2)}$ do too:
+
+$$\begin{aligned}
+    C_S^{(m)} = \sum_{n = 1}^N C_n^{(m)} = C_1^{(m)} + C_2^{(m)} + ... + C_N^{(m)}
+\end{aligned}$$
+
+We now introduce the scaled sum $z$ as the new combined variable:
+
+$$\begin{aligned}
+    z = \frac{S}{\sqrt{N}} = \frac{1}{\sqrt{N}} (x_1 + x_2 + ... + x_N)
+\end{aligned}$$
+
+Its characteristic function $\phi_z(k)$ is then as follows,
+with $\sqrt{N}$ appearing in the arguments of $\phi_n$:
+
+$$\begin{aligned}
+    \phi_z(k)
+    &= \int\cdots\int
+    \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z)
+    \dd{x_1} \cdots \dd{x_N}
+    \\
+    &= \int\cdots\int
+    \Big( \prod_{n = 1}^N p(x_n) \Big) \exp\!\Big( i \frac{k}{\sqrt{N}} \sum_{n = 1}^N x_n \Big)
+    \dd{x_1} \cdots \dd{x_N}
+    \\
+    &= \prod_{n = 1}^N \phi_n\Big(\frac{k}{\sqrt{N}}\Big)
+\end{aligned}$$
+
+By expanding $\ln\!\big(\phi_z(k)\big)$ in terms of its cumulants $C^{(m)}$
+and introducing $\kappa = k / \sqrt{N}$, we see that the higher-order terms
+become smaller for larger $N$:
+
+$$\begin{gathered}
+    \ln\!\big( \phi_z(k) \big)
+    = \sum_{m = 1}^\infty \frac{(ik)^m}{m!} C^{(m)}
+    \\
+    C^{(m)}
+    = \frac{1}{i^m} \dvn{m}{}{k} \sum_{n = 1}^N \ln\!\bigg( \phi_n\Big(\frac{k}{\sqrt{N}}\Big) \bigg)
+    = \frac{1}{i^m N^{m/2}} \dvn{m}{}{\kappa} \sum_{n = 1}^N \ln\!\big( \phi_n(\kappa) \big)
+\end{gathered}$$
+
+For sufficiently large $N$, we can therefore approximate it using just the first two terms:
+
+$$\begin{aligned}
+    \ln\!\big( \phi_z(k) \big)
+    &\approx i k C^{(1)} - \frac{k^2}{2} C^{(2)}
+    = i k \overline{z} - \frac{k^2}{2} \sigma_z^2
+    \\
+    \phi_z(k)
+    &\approx \exp(i k \overline{z}) \exp(- k^2 \sigma_z^2 / 2)
+\end{aligned}$$
+
+We take its inverse Fourier transform to get the density $p(z)$,
+which turns out to be a Gaussian normal distribution,
+which is even already normalized:
+
+$$\begin{aligned}
+    p(z)
+    = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\}
+    &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k}
+    \\
+    &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big)
+\end{aligned}$$
+
+Therefore, the sum of many independent variables tends to a normal distribution,
+regardless of the densities of the individual variables.
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/conditional-expectation/index.md b/source/know/concept/conditional-expectation/index.md
new file mode 100644
index 0000000..c545cef
--- /dev/null
+++ b/source/know/concept/conditional-expectation/index.md
@@ -0,0 +1,172 @@
+---
+title: "Conditional expectation"
+date: 2021-10-23
+categories:
+- Mathematics
+- Statistics
+- Measure theory
+- Stochastic analysis
+layout: "concept"
+---
+
+Recall that the expectation value $\mathbf{E}[X]$
+of a [random variable](/know/concept/random-variable/) $X$
+is a function of the probability space $(\Omega, \mathcal{F}, P)$
+on which $X$ is defined, and the definition of $X$ itself.
+
+The **conditional expectation** $\mathbf{E}[X|A]$
+is the expectation value of $X$ given that an event $A$ has occurred,
+i.e. only the outcomes $\omega \in \Omega$
+satisfying $\omega \in A$ should be considered.
+If $A$ is obtained by observing a variable,
+then $\mathbf{E}[X|A]$ is a random variable in its own right.
+
+Consider two random variables $X$ and $Y$
+on the same probability space $(\Omega, \mathcal{F}, P)$,
+and suppose that $\Omega$ is discrete.
+If $Y = y$ has been observed,
+then the conditional expectation of $X$
+given the event $Y = y$ is as follows:
+
+$$\begin{aligned}
+    \mathbf{E}[X | Y \!=\! y]
+    = \sum_{x} x \: Q(X \!=\! x)
+    \qquad \quad
+    Q(X \!=\! x)
+    = \frac{P(X \!=\! x \cap Y \!=\! y)}{P(Y \!=\! y)}
+\end{aligned}$$
+
+Where $Q$ is a renormalized probability function,
+which assigns zero to all events incompatible with $Y = y$.
+If we allow $\Omega$ to be continuous,
+then from the definition $\mathbf{E}[X]$,
+we know that the following Lebesgue integral can be used,
+which we call $f(y)$:
+
+$$\begin{aligned}
+    \mathbf{E}[X | Y \!=\! y]
+    = f(y)
+    = \int_\Omega X(\omega) \dd{Q(\omega)}
+\end{aligned}$$
+
+However, this is only valid if $P(Y \!=\! y) > 0$,
+which is a problem for continuous sample spaces $\Omega$.
+Sticking with the assumption $P(Y \!=\! y) > 0$, notice that:
+
+$$\begin{aligned}
+    f(y)
+    = \frac{1}{P(Y \!=\! y)} \int_\Omega X(\omega) \dd{P(\omega \cap Y \!=\! y)}
+    = \frac{\mathbf{E}[X \cdot I(Y \!=\! y)]}{P(Y \!=\! y)}
+\end{aligned}$$
+
+Where $I$ is the indicator function,
+equal to $1$ if its argument is true, and $0$ if not.
+Multiplying the definition of $f(y)$ by $P(Y \!=\! y)$ then leads us to:
+
+$$\begin{aligned}
+    \mathbf{E}[X \cdot I(Y \!=\! y)]
+    &= f(y) \cdot P(Y \!=\! y)
+    \\
+    &= \mathbf{E}[f(Y) \cdot I(Y \!=\! y)]
+\end{aligned}$$
+
+Recall that because $Y$ is a random variable,
+$\mathbf{E}[X|Y] = f(Y)$ is too.
+In other words, $f$ maps $Y$ to another random variable,
+which, thanks to the *Doob-Dynkin lemma*
+(see [random variable](/know/concept/random-variable/)),
+means that $\mathbf{E}[X|Y]$ is measurable with respect to $\sigma(Y)$.
+Intuitively, this makes sense:
+$\mathbf{E}[X|Y]$ cannot contain more information about events
+than the $Y$ it was calculated from.
+
+This suggests a straightforward generalization of the above:
+instead of a specific value $Y = y$,
+we can condition on *any* information from $Y$.
+If $\mathcal{H} = \sigma(Y)$ is the information generated by $Y$,
+then the conditional expectation $\mathbf{E}[X|\mathcal{H}] = Z$
+is $\mathcal{H}$-measurable, and given by a $Z$ satisfying:
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E}\big[X \cdot I(H)\big]
+        = \mathbf{E}\big[Z \cdot I(H)\big]
+    }
+\end{aligned}$$
+
+For any $H \in \mathcal{H}$. Note that $Z$ is almost surely unique:
+*almost* because it could take any value
+for an event $A$ with zero probability $P(A) = 0$.
+Fortunately, if there exists a continuous $f$
+such that $\mathbf{E}[X | \sigma(Y)] = f(Y)$,
+then $Z = \mathbf{E}[X | \sigma(Y)]$ is unique.
+
+
+## Properties
+
+A conditional expectation defined in this way has many useful properties,
+most notably linearity:
+$\mathbf{E}[aX \!+\! bY | \mathcal{H}] = a \mathbf{E}[X|\mathcal{H}] + b \mathbf{E}[Y|\mathcal{H}]$
+for any $a, b \in \mathbb{R}$.
+
+The **tower property** states that if $\mathcal{F} \supset \mathcal{G} \supset \mathcal{H}$,
+then $\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}] = \mathbf{E}[X|\mathcal{H}]$.
+Intuitively, this works as follows:
+suppose person $G$ knows more about $X$ than person $H$,
+then $\mathbf{E}[X | \mathcal{H}]$ is $H$'s expectation,
+$\mathbf{E}[X | \mathcal{G}]$ is $G$'s "better" expectation,
+and then $\mathbf{E}[\mathbf{E}[X|\mathcal{G}]|\mathcal{H}]$
+is $H$'s prediction about what $G$'s expectation will be.
+However, $H$ does not have access to $G$'s extra information,
+so $H$'s best prediction is simply $\mathbf{E}[X | \mathcal{H}]$.
+
+The **law of total expectation** says that
+$\mathbf{E}[\mathbf{E}[X | \mathcal{G}]] = \mathbf{E}[X]$,
+and follows from the above tower property
+by choosing $\mathcal{H}$ to contain no information:
+$\mathcal{H} = \{ \varnothing, \Omega \}$.
+
+Another useful property is that $\mathbf{E}[X | \mathcal{H}] = X$
+if $X$ is $\mathcal{H}$-measurable.
+In other words, if $\mathcal{H}$ already contains
+all the information extractable from $X$,
+then we know $X$'s exact value.
+Conveniently, this can easily be generalized to products:
+$\mathbf{E}[XY | \mathcal{H}] = X \mathbf{E}[Y | \mathcal{H}]$
+if $X$ is $\mathcal{H}$-measurable:
+since $X$'s value is known, it can simply be factored out.
+
+Armed with this definition of conditional expectation,
+we can define other conditional quantities,
+such as the **conditional variance** $\mathbf{V}[X | \mathcal{H}]$:
+
+$$\begin{aligned}
+    \mathbf{V}[X | \mathcal{H}]
+    = \mathbf{E}[X^2 | \mathcal{H}] - \big[\mathbf{E}[X | \mathcal{H}]\big]^2
+\end{aligned}$$
+
+The **law of total variance** then states that
+$\mathbf{V}[X] = \mathbf{E}[\mathbf{V}[X | \mathcal{H}]] + \mathbf{V}[\mathbf{E}[X | \mathcal{H}]]$.
+
+Likewise, we can define the **conditional probability** $P$,
+**conditional distribution function** $F_{X|\mathcal{H}}$,
+and **conditional density function** $f_{X|\mathcal{H}}$
+like their non-conditional counterparts:
+
+$$\begin{aligned}
+    P(A | \mathcal{H})
+    = \mathbf{E}[I(A) | \mathcal{H}]
+    \qquad
+    F_{X|\mathcal{H}}(x)
+    = P(X \le x | \mathcal{H})
+    \qquad
+    f_{X|\mathcal{H}}(x)
+    = \dv{F_{X|\mathcal{H}}}{x}
+\end{aligned}$$
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
new file mode 100644
index 0000000..d6c578b
--- /dev/null
+++ b/source/know/concept/convolution-theorem/index.md
@@ -0,0 +1,116 @@
+---
+title: "Convolution theorem"
+date: 2021-02-22
+categories:
+- Mathematics
+layout: "concept"
+---
+
+The **convolution theorem** states that a convolution in the direct domain
+is equal to a product in the frequency domain. This is especially useful
+for computation, replacing an $\mathcal{O}(n^2)$ convolution with an
+$\mathcal{O}(n \log(n))$ transform and product.
+
+## Fourier transform
+
+The convolution theorem is usually expressed as follows, where
+$\hat{\mathcal{F}}$ is the [Fourier transform](/know/concept/fourier-transform/),
+and $A$ and $B$ are constants from its definition:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
+            B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-fourier"/>
+<label for="proof-fourier">Proof</label>
+<div class="hidden">
+<label for="proof-fourier">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{f(x) \: g(x)\}
+    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
+    \\
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
+    \\
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
+    = B \cdot (\tilde{f} * \tilde{g})(k)
+\end{aligned}$$
+</div>
+</div>
+
+
+## Laplace transform
+
+For functions $f(t)$ and $g(t)$ which are only defined for $t \ge 0$,
+the convolution theorem can also be stated using
+the [Laplace transform](/know/concept/laplace-transform/):
+
+$$\begin{aligned}
+    \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
+\end{aligned}$$
+
+Because the inverse Laplace transform $\hat{\mathcal{L}}{}^{-1}$ is
+unpleasant, the theorem is often stated using the forward transform
+instead:
+
+$$\begin{aligned}
+    \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-laplace"/>
+<label for="proof-laplace">Proof</label>
+<div class="hidden">
+<label for="proof-laplace">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \hat{\mathcal{L}}\{(f * g)(t)\}
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
+    \\
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
+    \\
+    &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'}
+    = \tilde{f}(s) \: \tilde{g}(s)
+\end{aligned}$$
+</div>
+</div>
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/coulomb-logarithm/index.md b/source/know/concept/coulomb-logarithm/index.md
new file mode 100644
index 0000000..3bed159
--- /dev/null
+++ b/source/know/concept/coulomb-logarithm/index.md
@@ -0,0 +1,197 @@
+---
+title: "Coulomb logarithm"
+date: 2021-10-03
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+In a plasma, particles often appear to collide,
+although actually it is caused by Coulomb forces,
+i.e. the "collision" is in fact [Rutherford scattering](/know/concept/rutherford-scattering/).
+In any case, the particles' paths are deflected,
+and it would be nice to know
+whether those deflections are usually large or small.
+
+Let us choose $\pi/2$ as an example of a large deflection angle.
+Then Rutherford predicts:
+
+$$\begin{aligned}
+    \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu b_\mathrm{large}}
+    = \tan\!\Big( \frac{\pi}{4} \Big)
+    = 1
+\end{aligned}$$
+
+Isolating this for the impact parameter $b_\mathrm{large}$
+then yields an effective radius of a particle:
+
+$$\begin{aligned}
+    b_\mathrm{large}
+    = \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu}
+\end{aligned}$$
+
+Therefore, the collision cross-section $\sigma_\mathrm{large}$
+for large deflections can be roughly estimated as
+the area of a disc with radius $b_\mathrm{large}$:
+
+$$\begin{aligned}
+    \sigma_\mathrm{large}
+    = \pi b_\mathrm{large}^2
+    = \frac{q_1^2 q_2^2}{16 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
+\end{aligned}$$
+
+Next, we want to find the cross-section for small deflections.
+For sufficiently small angles $\theta$,
+we can Taylor-expand the Rutherford scattering formula to first order:
+
+$$\begin{aligned}
+    \frac{q_1 q_2}{4 \pi \varepsilon_0 |\vb{v}|^2 \mu b}
+    = \tan\!\Big( \frac{\theta}{2} \Big)
+    \approx \frac{\theta}{2}
+    \quad \implies \quad
+    \theta
+    \approx \frac{q_1 q_2}{2 \pi \varepsilon_0 |\vb{v}|^2 \mu b}
+\end{aligned}$$
+
+Clearly, $\theta$ is inversely proportional to $b$.
+Intuitively, we know that a given particle in a uniform plasma
+always has more "distant" neighbours than "close" neighbours,
+so we expect that small deflections (large $b$)
+are more common than large deflections.
+
+That said, many small deflections can add up to a large total.
+They can also add up to zero,
+so we should use random walk statistics.
+We now ask: how many $N$ small deflections $\theta_n$
+are needed to get a large total of, say, $1$ radian?
+
+$$\begin{aligned}
+    \sum_{n = 1}^N \theta_n^2 \approx 1
+\end{aligned}$$
+
+Traditionally, $1$ is chosen instead of $\pi/2$ for convenience.
+We are only making rough estimates,
+so those two angles are close enough for our purposes.
+Furthermore, the end result will turn out to be logarithmic,
+and is thus barely affected by this inconsistency.
+
+You can easily convince yourself
+that the average time $\tau$ between "collisions"
+is related like so to the cross-section $\sigma$,
+the total density $n$ of charged particles,
+and the relative velocity $|\vb{v}|$:
+
+$$\begin{aligned}
+    \frac{1}{\tau}
+    = n |\vb{v}| \sigma
+    \qquad \implies \qquad
+    1
+    = n |\vb{v}| \tau \sigma
+\end{aligned}$$
+
+Therefore, in a given time interval $t$,
+the expected number of collision $N_b$
+for impact parameters between $b$ and $b\!+\!\dd{b}$
+(imagine a ring with these inner and outer radii)
+is given by:
+
+$$\begin{aligned}
+    N_b
+    = n |\vb{v}| t \: \sigma_b
+    = n |\vb{v}| t \:(2 \pi b \dd{b})
+\end{aligned}$$
+
+In this time interval $t$,
+we can thus turn our earlier sum
+into an integral of $N_b$ over $b$:
+
+$$\begin{aligned}
+    1
+    \approx \sum_{n = 1}^N \theta_n^2
+    = \int N_b \:\theta^2 \dd{b}
+    = n |\vb{v}| t \int 2 \pi \theta^2 b \dd{b}
+\end{aligned}$$
+
+Using the formula $n |\vb{v}| \tau \sigma = 1$,
+we thus define $\sigma_{small}$ as the effective cross-section
+needed to get a large deflection (of $1$ radian),
+with an average period $t$:
+
+$$\begin{aligned}
+    \sigma_\mathrm{small}
+    = \int 2 \pi \theta^2 b \dd{b}
+    = \int \frac{2 \pi q_1^2 q_2^2}{4 \pi^2 \varepsilon_0^2 |\vb{v}|^4 \mu^2 b^2} b \dd{b}
+\end{aligned}$$
+
+Where we have replaced $\theta$ with our earlier Taylor expansion.
+Here, we recognize $\sigma_\mathrm{large}$:
+
+$$\begin{aligned}
+    \sigma_\mathrm{small}
+    = \frac{q_1^2 q_2^2}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2} \int \frac{1}{b} \dd{b}
+    = 8 \sigma_\mathrm{large} \int \frac{1}{b} \dd{b}
+\end{aligned}$$
+
+But what are the integration limits?
+We know that the deflection grows for smaller $b$,
+so it would be reasonable to choose $b_\mathrm{large}$ as the lower limit.
+For very large $b$, the plasma shields the particles from each other,
+thereby nullifying the deflection,
+so as upper limit we choose
+the [Debye length](/know/concept/debye-length/) $\lambda_D$,
+i.e. the plasma's self-shielding length.
+We thus find:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_\mathrm{small}
+        = 8 \ln(\Lambda) \sigma_\mathrm{large}
+        = \frac{q_1^2 q_2^2 \ln\!(\Lambda)}{2 \pi \varepsilon_0^2 |\vb{v}|^4 \mu^2}
+    }
+\end{aligned}$$
+
+Here, $\ln\!(\Lambda)$ is known as the **Coulomb logarithm**,
+with the **plasma parameter** $\Lambda$ defined below,
+equal to $9/2$ times the number of particles
+in a sphere with radius $\lambda_D$:
+
+$$\begin{aligned}
+    \boxed{
+        \Lambda
+        \equiv \frac{\lambda_D}{b_\mathrm{large}}
+        = 6 \pi n \lambda_D^3
+    }
+\end{aligned}$$
+
+The above relation between $\sigma_\mathrm{small}$ and $\sigma_\mathrm{large}$
+gives us an estimate of how much more often
+small deflections occur, compared to large ones.
+In a typical plasma, $\ln\!(\Lambda)$ is between 6 and 25,
+such that $\sigma_\mathrm{small}$ is 2-3 orders of magnitude larger than $\sigma_\mathrm{large}$.
+
+Note that $t$ is now fixed as the period
+for small deflections to add up to $1$ radian.
+In more useful words, it is the time scale
+for significant energy transfer between partices:
+
+$$\begin{aligned}
+    \frac{1}{t}
+    = n |\vb{v}| \sigma_\mathrm{small}
+    = \frac{q_1^2 q_2^2 \ln\!(\Lambda) \: n}{2 \pi \varepsilon_0^2 \mu^2 |\vb{v}|^3}
+    \sim \frac{n}{T^{3/2}}
+\end{aligned}$$
+
+Where we have used that $|\vb{v}| \propto \sqrt{T}$, for some temperature $T$.
+Consequently, in hotter plasmas, there is less energy transfer,
+meaning that a hot plasma is hard to heat up further.
+
+
+
+## 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/coupled-mode-theory/index.md b/source/know/concept/coupled-mode-theory/index.md
new file mode 100644
index 0000000..acb4710
--- /dev/null
+++ b/source/know/concept/coupled-mode-theory/index.md
@@ -0,0 +1,230 @@
+---
+title: "Coupled mode theory"
+date: 2022-03-31
+categories:
+- Physics
+- Optics
+layout: "concept"
+---
+
+Given an optical resonator (e.g. a photonic crystal cavity),
+consider one of its quasinormal modes
+with frequency $\omega_0$ and decay rate $1 / \tau_0$.
+Its complex amplitude $A$ is governed by:
+
+$$\begin{aligned}
+    \dv{A}{t}
+    &= \bigg( \!-\! i \omega_0 - \frac{1}{\tau_0} \bigg) A
+\end{aligned}$$
+
+We choose to normalize $A$ so that $|A(t)|^2$
+is the total energy inside the resonator at time $t$.
+
+Suppose that $N$ waveguides are now "connected" to this resonator,
+meaning that the resonator mode $A$ and the outgoing waveguide mode $S_\ell^\mathrm{out}$
+overlap sufficiently for $A$ to leak into $S_\ell^\mathrm{out}$ at a rate $1 / \tau_\ell$.
+Conversely, the incoming mode $S_\ell^\mathrm{in}$ brings energy to $A$.
+Therefore, we can write up the following general set of equations:
+
+$$\begin{aligned}
+    \dv{A}{t}
+    &= \bigg( \!-\! i \omega_0 - \frac{1}{\tau_0} \bigg) A
+    - \sum_{\ell = 1}^N \frac{1}{\tau_\ell} A + \sum_{\ell = 1}^N \alpha_\ell S_\ell^\mathrm{in}
+    \\
+    S_\ell^\mathrm{out}
+    &= \beta_\ell S_\ell^\mathrm{in} + \gamma_\ell A
+\end{aligned}$$
+
+Where $\alpha_\ell$ and $\gamma_\ell$ are unknown coupling constants,
+and $\beta_\ell$ represents reflection.
+We normalize $S_\ell^\mathrm{in}$
+so that $|S_\ell^\mathrm{in}(t)|^2$ is the power flowing towards $A$ at time $t$,
+and likewise for $S_\ell^\mathrm{out}$.
+
+Note that we have made a subtle approximation here:
+by adding new damping mechanisms,
+we are in fact modifying $\omega_0$;
+see the [harmonic oscillator](/know/concept/harmonic-oscillator/) for a demonstration.
+However, the frequency shift is second-order in the decay rate,
+so by assuming that all $\tau_\ell$ are large,
+we only need to keep the first-order terms, as we did.
+This is called **weak coupling**.
+
+If we also assume that $\tau_0$ is large
+(its effect is already included in $\omega_0$),
+then we can treat the decay mechanisms separately:
+to analyze the decay into a certain waveguide $\ell$,
+it is first-order accurate to neglect all other waveguides and $\tau_0$:
+
+$$\begin{aligned}
+    \dv{A}{t}
+    \approx \bigg( \!-\! i \omega_0 - \frac{1}{\tau_\ell} \bigg) A + \sum_{\ell' = 1}^N \alpha_\ell S_{\ell'}^\mathrm{in}
+\end{aligned}$$
+
+To determine $\gamma_\ell$, we use energy conservation.
+If all $S_{\ell'}^\mathrm{in} = 0$,
+then the energy in $A$ decays as:
+
+$$\begin{aligned}
+    \dv{|A|^2}{t}
+    &= \dv{A}{t} A^* + A \dv{A^*}{t}
+    \\
+    &= \bigg( \!-\! i \omega_0 - \frac{1}{\tau_\ell} \bigg) |A|^2
+    + \bigg( i \omega_0 - \frac{1}{\tau_\ell} \bigg) |A|^2
+    \\
+    &= - \frac{2}{\tau_\ell} |A|^2
+\end{aligned}$$
+
+Since all other mechanisms are neglected,
+all this energy must go into $S_\ell^\mathrm{out}$, meaning:
+
+$$\begin{aligned}
+    |S_\ell^\mathrm{out}|^2
+    = - \dv{|A|^2}{t}
+    = \frac{2}{\tau_\ell} |A|^2
+\end{aligned}$$
+
+Taking the square root, we clearly see that $|\gamma_\ell| = \sqrt{2 / \tau_\ell}$.
+Because the phase of $S_\ell^\mathrm{out}$ is arbitrarily defined,
+$\gamma_\ell$ need not be complex, so we choose $\gamma_\ell = \sqrt{2 / \tau_\ell}$.
+
+Next, to find $\alpha_\ell$, we exploit the time-reversal symmetry
+of [Maxwell's equations](/know/concept/maxwells-equations/),
+which govern the light in the resonator and the waveguides.
+In the above calculation of $\gamma_\ell$, $A$ evolved as follows,
+with the lost energy ending up in $S_\ell^\mathrm{out}$:
+
+$$\begin{aligned}
+    A(t)
+    = A e^{-i \omega_0 t - t / \tau_\ell}
+\end{aligned}$$
+
+After reversing time, $A$ evolves like so,
+where we have taken the complex conjugate
+to preserve the meanings of the symbols
+$A$, $S_\ell^\mathrm{out}$, and $S_\ell^\mathrm{in}$:
+
+$$\begin{aligned}
+    A(t)
+    = A e^{-i \omega_0 t + t / \tau_\ell}
+\end{aligned}$$
+
+We insert this expression for $A(t)$ into its original differential equation, yielding:
+
+$$\begin{aligned}
+    \dv{A}{t}
+    = \bigg( \!-\! i \omega_0 + \frac{1}{\tau_\ell} \bigg) A
+    = \bigg( \!-\! i \omega_0 - \frac{1}{\tau_\ell} \bigg) A + \alpha_\ell S_\ell^\mathrm{in}
+\end{aligned}$$
+
+Isolating this for $A$ leads us to the following power balance equation:
+
+$$\begin{aligned}
+    A
+    = \frac{\alpha_\ell \tau_\ell}{2} S_\ell^\mathrm{in}
+    \qquad \implies \qquad
+    |\alpha_\ell|^2 |S_\ell^\mathrm{in}|^2
+    = \frac{4}{\tau_\ell^2} |A|^2
+\end{aligned}$$
+
+But thanks to energy conservation,
+all power delivered by $S_\ell^\mathrm{in}$ ends up in $A$, so we know:
+
+$$\begin{aligned}
+    |S_\ell^\mathrm{in}|^2
+    = \dv{|A|^2}{t}
+    = \frac{2}{\tau_\ell} |A|^2
+\end{aligned}$$
+
+To reconcile the two equations above,
+we need $|\alpha_\ell| = \sqrt{2 / \tau_\ell}$.
+Discarding the phase thanks to our choice of $\gamma_\ell$,
+we conclude that $\alpha_\ell = \sqrt{2 / \tau_\ell} = \gamma_\ell$.
+
+Finally, $\beta_\ell$ can also be determined using energy conservation.
+Again using our weak coupling assumption,
+if energy is only entering and leaving $A$ through waveguide $\ell$, we have:
+
+$$\begin{aligned}
+    |S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
+    = \dv{|A|^2}{t}
+\end{aligned}$$
+
+Meanwhile, using the differential equation for $A$,
+we find the following relation:
+
+$$\begin{aligned}
+    \dv{|A|^2}{t}
+    &= \dv{A}{t} A^* + A \dv{A^*}{t}
+    \\
+    &= - \frac{2}{\tau_\ell} |A|^2 + \alpha_\ell \Big( S_\ell^\mathrm{in} A^* + (S_\ell^\mathrm{in})^* A \Big)
+\end{aligned}$$
+
+By isolating both of the above relations for $\idv{|A|^2}{t}$
+and equating them, we arrive at:
+
+$$\begin{aligned}
+    |S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
+    &= - \frac{2}{\tau_\ell} |A|^2 + \alpha_\ell \Big( S_\ell^\mathrm{in} A^* + (S_\ell^\mathrm{in})^* A \Big)
+\end{aligned}$$
+
+We insert the definition of $\gamma_\ell$ and $\beta_\ell$,
+namely $\gamma_\ell A = S_\ell^\mathrm{out} - \beta_\ell S_\ell^\mathrm{in}$,
+and use $\alpha_\ell = \gamma_\ell$:
+
+$$\begin{aligned}
+    |S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
+    &= - \Big( S_\ell^\mathrm{out} - \beta_\ell S_\ell^\mathrm{in} \Big) \Big( (S_\ell^\mathrm{out})^* - \beta_\ell^* (S_\ell^\mathrm{in})^* \Big)
+    \\
+    &\quad\; + S_\ell^\mathrm{in} \Big( (S_\ell^\mathrm{out})^* - \beta_\ell^* (S_\ell^\mathrm{in})^* \Big)
+    + (S_\ell^\mathrm{in})^* \Big( S_\ell^\mathrm{out} - \beta_\ell S_\ell^\mathrm{in} \Big)
+    \\
+    &= - |\beta_\ell|^2 |S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
+    + \beta_\ell S_\ell^\mathrm{in} (S_\ell^\mathrm{out})^* + \beta_\ell^* (S_\ell^\mathrm{in})^* S_\ell^\mathrm{out}
+    \\
+    &\quad\; + S_\ell^\mathrm{in} (S_\ell^\mathrm{out})^* - \beta_\ell^* |S_\ell^\mathrm{in}|^2
+    + (S_\ell^\mathrm{in})^* S_\ell^\mathrm{out} - \beta_\ell |S_\ell^\mathrm{in}|^2
+    \\
+    &= - (|\beta_\ell|^2 + \beta_\ell + \beta_\ell^*) |S_\ell^\mathrm{in}|^2 - |S_\ell^\mathrm{out}|^2
+    \\
+    &\quad\; + (1 - \beta_\ell) S_\ell^\mathrm{in} (S_\ell^\mathrm{out})^* + (1 - \beta_\ell^*) (S_\ell^\mathrm{in})^* S_\ell^\mathrm{out}
+\end{aligned}$$
+
+This equation is only satisfied if $\beta_\ell = -1$.
+Combined with $\alpha_\ell = \gamma_\ell = \sqrt{2 / \tau_\ell}$,
+the **coupled-mode equations** take the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dv{A}{t}
+            &= \bigg( \!-\! i \omega_0 - \frac{1}{\tau_0} \bigg) A
+            - \sum_{\ell = 1}^N \frac{1}{\tau_\ell} A
+            + \sum_{\ell = 1}^N \sqrt{\frac{2}{\tau_\ell}} S_\ell^\mathrm{in}
+            \\
+            S_\ell^\mathrm{out}
+            &= - S_\ell^\mathrm{in} + \sqrt{\frac{2}{\tau_\ell}} A
+        \end{aligned}
+    }
+\end{aligned}$$
+
+By connecting multiple resonators with waveguides,
+optical networks can be created,
+whose dynamics are described by these equations.
+
+The coupled-mode equations are extremely general,
+since we have only used weak coupling,
+conservation of energy, and time-reversal symmetry.
+Even if the decay rates are quite large,
+coupled mode theory still tends to give qualitatively correct answers.
+
+
+
+## References
+1.  H.A. Haus,
+    *Waves and fields in optoelectronics*,
+    1984, Prentice-Hall.
+2.  J.D. Joannopoulos, S.G. Johnson, J.N. Winn, R.D. Meade,
+    *Photonic crystals: molding the flow of light*,
+    2nd edition, Princeton.
+
diff --git a/source/know/concept/curvature/index.md b/source/know/concept/curvature/index.md
new file mode 100644
index 0000000..40bd106
--- /dev/null
+++ b/source/know/concept/curvature/index.md
@@ -0,0 +1,389 @@
+---
+title: "Curvature"
+date: 2021-03-07
+categories:
+- Mathematics
+layout: "concept"
+---
+
+Given a curve or surface, its **curvature** $\kappa$
+describes how sharply it is bending at a given point.
+It is defined as the inverse of the **radius of curvature** $R$,
+which is the radius of the tangent circle
+that **osculates** (i.e. best approximates)
+the curve/surface at that point:
+
+$$\begin{aligned}
+    \kappa = \frac{1}{R}
+\end{aligned}$$
+
+Typically, $\kappa$ is positive for convex curves/surfaces,
+and negative for concave ones, although this distinction is somewhat arbitrary.
+Below, we calculate the curvature in several general cases.
+
+
+## 2D height functions
+
+We start with a specialized case: height functions,
+where one coordinate is a function of the other one (2D) or two (3D).
+In this case, we can use the
+[calculus of variations](/know/concept/calculus-of-variations/)
+to find the curvature.
+
+This approach relies on the fact that a circle
+has the highest area-perimeter ratio of any 2D shape,
+and a sphere has the highest volume-surface ratio of any 3D body.
+By the definition of curvature, these shapes have constant $\kappa$.
+
+We will thus minimize the perimeter/surface while keeping the area/volume fixed,
+which will give us a shape with constant curvature,
+and from that we can extrapolate an expression for $\kappa$.
+
+In 2D, for a single-variable height function $h(x)$,
+the length of a small segment of the curve is:
+
+$$\begin{aligned}
+    \sqrt{\dd{x}^2 + \dd{h}^2}
+    = \dd{x} \sqrt{\Big( \dv{x}{x} \Big)^2 + \Big( \dv{h}{x} \Big)^2}
+    = \dd{x} \sqrt{1 + h_x^2}
+\end{aligned}$$
+
+Which leads us to define the following Lagrangian $\mathcal{L}$
+describing the "energy cost" of the curve:
+
+$$\begin{aligned}
+    \mathcal{L}
+    = \sqrt{1 + h_x^2}
+\end{aligned}$$
+
+Furthermore,
+we demand that the area under the curve (i.e. the "volume") is constant:
+
+$$\begin{aligned}
+    V
+    = \int_{x_0}^{x_1} h(x) \dd{x}
+\end{aligned}$$
+
+By putting these things together,
+we arrive at the following energy functional $E[h]$,
+where $\kappa$ is an ominously-named [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+
+$$\begin{aligned}
+    E[h]
+    = \int (\mathcal{L} + \kappa h) \dd{x}
+\end{aligned}$$
+
+Minimizing this functional leads to the following
+Lagrange equation of the first kind:
+
+$$\begin{aligned}
+    0
+    = \pdv{\mathcal{L}}{h} - \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) + \kappa
+\end{aligned}$$
+
+We evaluate the terms of this equation
+to arrive at an expression for the curvature $\kappa$:
+
+$$\begin{aligned}
+    \boxed{
+        \kappa
+        = \frac{h_{xx}}{\big(1 + h_x^2\big)^{3/2}}
+    }
+\end{aligned}$$
+
+In this optimization problem, $\kappa$ is a constant,
+but in fact the statement above is valid for variable curvatures too,
+in which case $\kappa$ is a function of $x$.
+
+
+## 2D in general
+
+We can parametrically describe an arbitrary plane curve
+as a function of the arc length $s$:
+
+$$\begin{aligned}
+    \big( x(s), y(s) \big)
+    \qquad \mathrm{where} \qquad
+    \dd{s}^2 = \dd{x}^2 + \dd{y}^2
+\end{aligned}$$
+
+If we choose the horizontal $x$-axis as a reference,
+we can furthermore define the **elevation angle** $\theta(s)$
+as the angle between the reference and the curve's tangent vector $\vu{t}$:
+
+$$\begin{aligned}
+    \vu{t}
+    = \big( x_s(s), y_s(s) \big)
+    = \big( \cos\theta(s), \sin\theta(s) \big)
+\end{aligned}$$
+
+Where $x_s(s) = \idv{x}{s}$.
+The curvature $\kappa$ is defined as
+the $s$-derivative of this elevation angle:
+
+$$\begin{aligned}
+    \kappa
+    = \dv{\theta}{s}
+    = \theta_s(s)
+\end{aligned}$$
+
+We have two ways of writing $\vu{t}$:
+using the derivatives $x_s$ and $y_s$,
+or the elevation angle $\theta$.
+Now, let us take the $s$-derivative of both expressions,
+and equate them:
+
+$$\begin{aligned}
+    \big( x_{ss}, y_{ss} \big)
+    = \dv{\vu{t}}{s}
+    = \theta_s \: \big( \!-\!\sin\theta, \cos\theta \big)
+    = \kappa \big( \!-\!y_s, x_s \big)
+\end{aligned}$$
+
+$$\begin{aligned}
+    x_{ss} = - \kappa y_s
+    \qquad
+    y_{ss} = \kappa x_s
+\end{aligned}$$
+
+We multiply these equation by $y_s$ and $x_s$, respectively,
+and subtract the first from the last:
+
+$$\begin{aligned}
+    y_{ss} x_s - x_{ss} y_s = \kappa x_s^2 + \kappa y_s^2
+\end{aligned}$$
+
+Isolating this for $\kappa$ and using the fact that $x_s^2 + y_s^2 = 1$
+thanks to $s$ being the arc length:
+
+$$\begin{aligned}
+    \kappa
+    = \frac{y_{ss} x_s - x_{ss} y_s}{x_s^2 + y_s^2}
+    = y_{ss} x_s - x_{ss} y_s
+\end{aligned}$$
+
+While this result is correct,
+we would like to generalize it to cases where the curve
+is parametrized by some other $t$, not necessarily the arc length.
+Let prime denote the $t$-derivative:
+
+$$\begin{aligned}
+    x_s
+    = x' t_s
+    \qquad
+    x_{ss}
+    = x'' t_s^2 + x' t_{ss}
+    \\
+    y_s
+    = y' t_s
+    \qquad \:
+    y_{ss}
+    = y'' t_s^2 + x' t_{ss}
+\end{aligned}$$
+
+By inserting these expression into the earlier formula for $\kappa$, we find:
+
+$$\begin{aligned}
+    \kappa
+    = y_{ss} x_s - x_{ss} y_s
+    &= x' t_s (y'' t_s^2 + y' t_{ss}) - y' t_s (x'' t_s^2 + x' t_{ss})
+    \\
+    &= t_s t_{ss} (x' y' - y' x') + t_s^3 (x' y'' - y' x'')
+    \\
+    &= t_s^3 (x' y'' - y' x'')
+\end{aligned}$$
+
+Since $x_s^2 + y_s^2 = 1$, we know that $(x')^2 + (y')^2 = 1 / t_s^2$,
+which leads us to the following general expression for
+the curvature $\kappa$ of a plane curve:
+
+$$\begin{aligned}
+    \boxed{
+        \kappa
+        = \frac{y'' x' - x'' y'}{\big((x')^2 + (y')^2\big)^{3/2}}
+    }
+\end{aligned}$$
+
+If the curve happens to be a height function, i.e. $y(x)$,
+then $x' = 1$ and $x'' = 0$, and we arrive at our previous result again.
+
+
+## 3D height functions
+
+The generalization to a 3D height function $h(x, y)$ is straightforward:
+the cost of an infinitesimal portion of the surface is as follows,
+using the same reasoning as before:
+
+$$\begin{aligned}
+    \mathcal{L}
+    = \sqrt{1 + h_x^2 + h_y^2}
+\end{aligned}$$
+
+Keeping the volume $V$ constant,
+we get the following energy functional $E$ to minimize:
+
+$$\begin{aligned}
+    E[h]
+    = \iint (\mathcal{L} + \lambda h) \dd{x} \dd{y}
+\end{aligned}$$
+
+Which gives us an Euler-Lagrange equation
+involving the Lagrange multiplier $\lambda$:
+
+$$\begin{aligned}
+    0
+    = \pdv{\mathcal{L}}{h} - \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) - \dv{}{y}\Big( \pdv{\mathcal{L}}{h_y} \Big) + \lambda
+\end{aligned}$$
+
+Inserting $\mathcal{L}$ into this and evaluating all the derivatives
+yields a result for the (variable) curvature:
+
+$$\begin{aligned}
+    \boxed{
+        \lambda
+        = \kappa_1 + \kappa_2
+        = \frac{(1 + h_y^2) h_{xx} - 2 h_x h_y h_{xy} + (1 + h_x^2) h_{yy}}{\big(1 + h_x^2 + h_y^2\big)^{3/2}}
+    }
+\end{aligned}$$
+
+What are $\kappa_1$ and $\kappa_2$?
+Well, the problem in 3D is that the curvature of an osculating circle
+depends on the orientation of that circle.
+The **principal curvatures** $\kappa_1$ and $\kappa_2$
+are the largest and smallest curvatures at a given point,
+but finding their values and the corresponding **principal directions** is not so easy.
+Fortunately, in practice, we are often only interested in their sum:
+
+$$\begin{aligned}
+    \lambda
+    = \kappa_1 + \kappa_2
+    = \frac{1}{R_1} + \frac{1}{R_2}
+\end{aligned}$$
+
+These **principal radii** $R_1$ and $R_2$ are important
+for e.g. the [Young-Laplace law](/know/concept/young-laplace-law/).
+
+
+## 3D in general
+
+To find a general expression for the mean curvature of an arbitrary surface,
+we "cut off" a small part of the surface that we can regard as a height function.
+We call the "cutting" reference plane $(x, y)$,
+and the surface it describes $h(x, y)$.
+We then define the unit tangent vectors $\vu{t}_x$ and $\vu{t}_y$
+to be parallel to the $x$-axis and $y$-axis, respectively:
+
+$$\begin{aligned}
+    \vu{t}_x
+    = \frac{1}{\sqrt{1 + (h_x)^2}}
+    \begin{bmatrix}
+        1 \\ 0 \\ h_x
+    \end{bmatrix}
+    \qquad
+    \vu{t}_y
+    = \frac{1}{\sqrt{1 + (h_y)^2}}
+    \begin{bmatrix}
+        0 \\ 1 \\ h_y
+    \end{bmatrix}
+\end{aligned}$$
+
+Since they were chosen to lie along the axes,
+these vectors are not necessarily orthogonal,
+so we need to normalize the resulting normal vector $\vu{n}$:
+
+$$\begin{aligned}
+    \vu{n}
+    = \vu{t}_x \cross \vu{t}_y
+    = \frac{1}{\sqrt{1 + (h_x)^2 + (h_y)^2}}
+    \begin{bmatrix}
+        - h_x \\ - h_y \\ 1
+    \end{bmatrix}
+\end{aligned}$$
+
+Let us take a look at the divergence of $\vu{n}$,
+or to be precise, its *projection* onto the reference plane
+(although this distinction is not really important for our purposes):
+
+$$\begin{aligned}
+    \nabla \cdot \vu{n}
+    = - \dv{}{x}\bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) - \dv{}{y}\bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg)
+\end{aligned}$$
+
+Compare this with the expression for $\lambda$ we found earlier,
+with the help of variational calculus:
+
+$$\begin{aligned}
+    \lambda
+    &= \dv{}{x}\Big( \pdv{\mathcal{L}}{h_x} \Big) + \dv{}{y}\Big( \pdv{\mathcal{L}}{h_y} \Big)
+    \\
+    &= \dv{}{x}\bigg( \frac{h_x}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg) + \dv{}{y}\bigg( \frac{h_y}{\sqrt{1 + (h_x)^2 + (h_y)^2}} \bigg)
+\end{aligned}$$
+
+The similarity is clearly visible.
+This leads us to the following general expression:
+
+$$\begin{aligned}
+    \boxed{
+        \kappa_1 + \kappa_2
+        = - \nabla \cdot \vu{n}
+    }
+\end{aligned}$$
+
+A useful property is that
+the principal directions of curvature are always orthogonal.
+To show this, consider the most general second-order approximating surface,
+in polar coordinates:
+
+$$\begin{aligned}
+    h(x, y)
+    &= \frac{1}{2} a x^2 + \frac{1}{2} b y^2 + c x y
+    \\
+    &= \frac{1}{2} a r^2 \cos^2\varphi + \frac{1}{2} b r^2 \sin^2\varphi + c r^2 \cos\varphi \sin\varphi
+\end{aligned}$$
+
+Sufficiently close to the extremum, where $h_x$ and $h_y$ are negligible,
+the curvature along a certain direction $\varphi$ is given by
+our earlier formula for a 2D height function:
+
+$$\begin{aligned}
+    \kappa(\varphi)
+    \approx \pdvn{2}{h}{r}
+    = a \cos^2\varphi + b \sin^2\varphi + c \sin(2 \varphi)
+\end{aligned}$$
+
+To find the extremes of $\kappa$,
+we differentiate with respect to $\varphi$ and demand that it is zero:
+
+$$\begin{aligned}
+    0
+    &= - 2 a \cos\varphi \sin\varphi + 2 b \sin\varphi \cos\varphi + 2 c \cos(2 \varphi)
+    \\
+    &= - a \sin(2 \varphi) + b \sin(2 \varphi) + 2 c \cos(2 \varphi)
+\end{aligned}$$
+
+After rearranging this a bit, we arrive at the following transcendental equation:
+
+$$\begin{aligned}
+    \frac{2 c}{a - b}
+    = \frac{\sin(2 \varphi)}{\cos(2 \varphi)}
+    = \tan(2 \varphi)
+\end{aligned}$$
+
+Since the $\tan$ function is $\pi$-periodic,
+this has two solutions, $\varphi_0$ and $\varphi_0 + \pi/2$,
+which are clearly orthogonal,
+hence the principal directions are at an angle of $\pi/2$.
+
+Finally, it is also worth mentioning that
+the principal directions always lie in planes
+containing the normal of the surface.
+
+
+
+## References
+1.  T. Bohr,
+    *Curvature of plane curves and surfaces*,
+    2020, unpublished.
+2.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/curvilinear-coordinates/index.md
@@ -0,0 +1,380 @@
+---
+title: "Curvilinear coordinates"
+date: 2021-03-03
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+In a 3D coordinate system, the isosurface of a coordinate
+(i.e. the surface where that coordinate is constant while the others vary)
+is known as a **coordinate surface**, and the intersections of
+the surfaces of different coordinates are called **coordinate lines**.
+
+A **curvilinear** coordinate system is one where at least one of the coordinate surfaces is curved,
+e.g. in cylindrical coordinates the line between $r$ and $z$ is a circle.
+If the coordinate surfaces are mutually perpendicular,
+it is an **orthogonal** system, which is generally desirable.
+
+A useful attribute of a coordinate system is its **line element** $\dd{\ell}$,
+which represents the differential element of a line in any direction.
+For an orthogonal system, its square $\dd{\ell}^2$ is calculated
+by taking the differential elements of the old Cartesian $(x, y, z)$ system
+and writing them out in the new $(x_1, x_2, x_3)$ system.
+The resulting expression will be of the form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\ell}^2
+        = \dd{x}^2 + \dd{y}^2 + \dd{z}^2
+        = h_1^2 \dd{x_1}^2 + h_2^2 \dd{x_2}^2 + h_3^2 \dd{x_3}^2
+    }
+\end{aligned}$$
+
+Where $h_1$, $h_2$, and $h_3$ are called **scale factors**,
+and need not be constants.
+The equation above only contains quadratic terms
+because the coordinate system is orthogonal by assumption.
+
+Examples of orthogonal curvilinear coordinate systems include
+[spherical coordinates](/know/concept/spherical-coordinates/),
+[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/),
+and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/).
+
+In the following subsections,
+we derive general formulae to convert expressions
+from Cartesian coordinates to the new orthogonal system $(x_1, x_2, x_3)$.
+
+
+## Basis vectors
+
+Consider the the vector form of the line element $\dd{\ell}$,
+denoted by $\dd{\vu{\ell}}$ and expressed as:
+
+$$\begin{aligned}
+    \dd{\vu{\ell}}
+    = \vu{e}_x \dd{x} + \vu{e}_y \dd{y} + \vu{e}_z \dd{z}
+\end{aligned}$$
+
+We can expand the Cartesian differential elements, e.g. $\dd{y}$,
+in the new basis as follows:
+
+$$\begin{aligned}
+    \dd{y}
+    = \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3}
+\end{aligned}$$
+
+If we write this out for $\dd{x}$, $\dd{y}$ and $\dd{z}$,
+and group the terms according to $\dd{x}_1$, $\dd{x}_2$ and $\dd{x}_3$,
+we can compare it the alternative form of $\dd{\vu{\ell}}$:
+
+$$\begin{aligned}
+    \dd{\vu{\ell}}
+    = \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4}
+\end{aligned}$$
+
+From this, we can read off $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$.
+Here we only give $\vu{e}_1$, since $\vu{e}_2$ and $\vu{e}_3$ are analogous:
+
+$$\begin{aligned}
+    \boxed{
+        h_1 \vu{e}_1
+        = \vu{e}_x \pdv{x}{x_1} + \vu{e}_y \pdv{y}{x_1} + \vu{e}_z \pdv{y}{x_1}
+    }
+\end{aligned}$$
+
+
+## Gradient
+
+In an orthogonal coordinate system,
+the gradient $\nabla f$ of a scalar $f$ is as follows,
+where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$:
+
+$$\begin{gathered}
+    \boxed{
+        \nabla f
+        = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1}
+        + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2}
+        + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3}
+    }
+\end{gathered}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-grad"/>
+<label for="proof-grad">Proof</label>
+<div class="hidden">
+<label for="proof-grad">Proof.</label>
+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:
+
+$$\begin{gathered}
+    \nabla f
+    = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1}
+    + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
+    + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
+\end{gathered}$$
+</div>
+</div>
+
+
+## Divergence
+
+The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$
+in an orthogonal system is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{h_1 h_2 h_3}
+        \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-div"/>
+<label for="proof-div">Proof</label>
+<div class="hidden">
+<label for="proof-div">Proof.</label>
+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.
+</div>
+</div>
+
+
+## Laplacian
+
+The Laplacian $\nabla^2 f$ is simply $\nabla \cdot \nabla f$,
+so we can find the general formula
+by combining the two preceding results
+for the gradient and the divergence:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{h_1 h_2 h_3}
+        \bigg(
+            \pdv{}{x_1}\Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{x_1} \!\Big)
+            + \pdv{}{x_2}\Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{x_2} \!\Big)
+            + \pdv{}{x_3}\Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{x_3} \!\Big)
+        \bigg)
+    }
+\end{aligned}$$
+
+
+## Curl
+
+The curl of a vector $\vb{V}$ is as follows
+in a general orthogonal curvilinear system:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big)
+            \\
+            &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big)
+            \\
+            &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-curl"/>
+<label for="proof-curl">Proof</label>
+<div class="hidden">
+<label for="proof-curl">Proof.</label>
+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.
+</div>
+</div>
+
+
+## Differential elements
+
+The point of the scale factors $h_1$, $h_2$ and $h_3$, as can seen from their derivation,
+is to correct for "distortions" of the coordinates compared to the Cartesian system,
+such that the line element $\dd{\ell}$ retains its length.
+This property extends to the surface $\dd{S}$ and volume $\dd{V}$.
+
+When handling a differential volume in curvilinear coordinates,
+e.g. for a volume integral,
+the size of the box $\dd{V}$ must be corrected by the scale factors:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = \dd{x}\dd{y}\dd{z}
+        = h_1 h_2 h_3 \dd{x_1} \dd{x_2} \dd{x_3}
+    }
+\end{aligned}$$
+
+The same is true for the isosurfaces $\dd{S_1}$, $\dd{S_2}$ and $\dd{S_3}$
+where the coordinates $x_1$, $x_2$ and $x_3$ are respectively kept constant:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dd{S_1} &= h_2 h_3 \dd{x_2} \dd{x_3}
+            \\
+            \dd{S_2} &= h_1 h_3 \dd{x_1} \dd{x_3}
+            \\
+            \dd{S_3} &= h_1 h_2 \dd{x_1} \dd{x_2}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Using the same logic, the normal vector element $\dd{\vu{S}}$
+of an arbitrary surface is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_1 h_2 h_3 \dd{x_2} \dd{x_3} + \vu{e}_2 h_1 h_3 \dd{x_1} \dd{x_3} + \vu{e}_3 h_1 h_2 \dd{x_1} \dd{x_2}
+    }
+\end{aligned}$$
+
+Finally, the tangent vector element $\dd{\vu{\ell}}$ takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_1 h_1 \dd{x_1} + \vu{e}_2 h_2 \dd{x_2} + \vu{e}_3 h_3 \dd{x_3}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.
diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/cylindrical-parabolic-coordinates/index.md
new file mode 100644
index 0000000..f9d0475
--- /dev/null
+++ b/source/know/concept/cylindrical-parabolic-coordinates/index.md
@@ -0,0 +1,182 @@
+---
+title: "Cylindrical parabolic coordinates"
+date: 2021-03-04
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**Cylindrical parabolic coordinates** are a coordinate system
+that describes a point in space using three coordinates $(\sigma, \tau, z)$.
+The $z$-axis is unchanged from the Cartesian system,
+hence it is called a *cylindrical* system.
+In the $z$-isoplane, however, confocal parabolas are used.
+These coordinates can be converted to the Cartesian $(x, y, z)$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        x = \frac{1}{2} (\tau^2 - \sigma^2 )
+        \qquad
+        y = \sigma \tau
+        \qquad
+        z = z
+    }
+\end{aligned}$$
+
+Converting the other way is a bit trickier.
+It can be done by solving the following equations,
+and potentially involves some fiddling with signs:
+
+$$\begin{aligned}
+    2 x
+    = \frac{y^2}{\sigma^2} - \sigma^2
+    \qquad \qquad
+    2 x
+    = - \frac{y^2}{\tau^2} + \tau^2
+\end{aligned}$$
+
+Cylindrical parabolic coordinates form an orthogonal
+[curvilinear system](/know/concept/curvilinear-coordinates/),
+so we would like to find its scale factors $h_\sigma$, $h_\tau$ and $h_z$.
+The differentials of the Cartesian coordinates are as follows:
+
+$$\begin{aligned}
+    \dd{x} = - \sigma \dd{\sigma} + \tau \dd{\tau}
+    \qquad
+    \dd{y} = \tau \dd{\sigma} + \sigma \dd{\tau}
+    \qquad
+    \dd{z} = \dd{z}
+\end{aligned}$$
+
+We calculate the line segment $\dd{\ell}^2$,
+skipping many terms thanks to orthogonality:
+
+$$\begin{aligned}
+    \dd{\ell}^2
+    &= (\sigma^2 + \tau^2) \:\dd{\sigma}^2 + (\tau^2 + \sigma^2) \:\dd{\tau}^2 + \dd{z}^2
+\end{aligned}$$
+
+From this, we can directly read off the scale factors $h_\sigma^2$, $h_\tau^2$ and $h_z^2$,
+which turn out to be:
+
+$$\begin{aligned}
+    \boxed{
+        h_\sigma = \sqrt{\sigma^2 + \tau^2}
+        \qquad
+        h_\tau = \sqrt{\sigma^2 + \tau^2}
+        \qquad
+        h_z = 1
+    }
+\end{aligned}$$
+
+With these scale factors, we can use
+the general formulae for orthogonal curvilinear coordinates
+to easily to convert things from the Cartesian system.
+The basis vectors are:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \vu{e}_\sigma
+            &= \frac{- \sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
+            \\
+            \vu{e}_\tau
+            &= \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y
+            \\
+            \vu{e}_z
+            &= \vu{e}_z
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla f
+        = \frac{\vu{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma}
+        + \frac{\vu{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau}
+        + \vu{e}_z \pdv{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{\sigma^2 + \tau^2}
+        \Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\tau} \Big) + \pdv{V_z}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{\sigma^2 + \tau^2} \Big( \pdvn{2}{f}{\sigma} + \pdvn{2}{f}{\tau} \Big) + \pdvn{2}{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \vu{e}_\sigma \Big( \frac{\vu{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big)
+            \\
+            &+ \vu{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big)
+            \\
+            &+ \frac{\vu{e}_z}{\sigma^2 + \tau^2}
+            \Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The differential element of volume $\dd{V}$
+in cylindrical parabolic coordinates is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V} = (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} \dd{z}
+    }
+\end{aligned}$$
+
+The differential elements of the isosurfaces are as follows,
+where $\dd{S_\sigma}$ is the $\sigma$-isosurface, etc.:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dd{S_\sigma} &= \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
+            \\
+            \dd{S_\tau} &= \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
+            \\
+            \dd{S_z} &= (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The normal element $\dd{\vu{S}}$ of a surface and
+the tangent element $\dd{\vu{\ell}}$ of a curve are respectively:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z}
+        + \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z}
+        + \vu{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma}
+        + \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau}
+        + \vu{e}_z \dd{z}
+    }
+\end{aligned}$$
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.
diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/cylindrical-polar-coordinates/index.md
new file mode 100644
index 0000000..a91e53e
--- /dev/null
+++ b/source/know/concept/cylindrical-polar-coordinates/index.md
@@ -0,0 +1,200 @@
+---
+title: "Cylindrical polar coordinates"
+date: 2021-07-26
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+**Cylindrical polar coordinates** are an extension of polar coordinates to 3D,
+which describes the location of a point in space
+using the coordinates $(r, \varphi, z)$.
+The $z$-axis is unchanged from Cartesian coordinates,
+hence it is called a *cylindrical* system.
+
+Cartesian coordinates $(x, y, z)$
+and the cylindrical system $(r, \varphi, z)$ are related by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            x &= r \cos\varphi \\
+            y &= r \sin\varphi \\
+            z &= z
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Conversely, a point given in $(x, y, z)$
+can be converted to $(r, \varphi, z)$
+using these formulae:
+
+$$\begin{aligned}
+    \boxed{
+        r = \sqrt{x^2 + y^2}
+        \qquad
+        \varphi = \mathtt{atan2}(y, x)
+        \qquad
+        z = z
+    }
+\end{aligned}$$
+
+The cylindrical polar coordinates form an orthogonal
+[curvilinear system](/know/concept/curvilinear-coordinates/),
+whose scale factors $h_r$, $h_\varphi$ and $h_z$ we want to find.
+To do so, we calculate the differentials of the Cartesian coordinates:
+
+$$\begin{aligned}
+    \dd{x} = \dd{r} \cos\varphi - \dd{\varphi} r \sin\varphi
+    \qquad
+    \dd{y} = \dd{r} \sin\varphi + \dd{\varphi} r \cos\varphi
+    \qquad
+    \dd{z} = \dd{z}
+\end{aligned}$$
+
+And then we calculate the line element $\dd{\ell}^2$,
+skipping many terms thanks to orthogonality,
+
+$$\begin{aligned}
+    \dd{\ell}^2
+    &= \dd{r}^2 \big( \cos^2(\varphi) + \sin^2(\varphi) \big)
+    + \dd{\varphi}^2 \big( r^2 \sin^2(\varphi) + r^2 \cos^2(\varphi) \big)
+    + \dd{z}^2
+    \\
+    &= \dd{r}^2 + r^2 \: \dd{\varphi}^2 + \dd{z}^2
+\end{aligned}$$
+
+Finally, we can simply read off
+the squares of the desired scale factors
+$h_r^2$, $h_\varphi^2$ and $h_z^2$:
+
+$$\begin{aligned}
+    \boxed{
+        h_r = 1
+        \qquad
+        h_\varphi = r
+        \qquad
+        h_z = 1
+    }
+\end{aligned}$$
+
+With these factors, we can easily convert things from the Cartesian system
+using the standard formulae for orthogonal curvilinear coordinates.
+The basis vectors are:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \vu{e}_r
+            &= \cos\varphi \:\vu{e}_x + \sin\varphi \:\vu{e}_y
+            \\
+            \vu{e}_\varphi
+            &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y
+            \\
+            \vu{e}_z
+            &= \vu{e}_z
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The basic vector operations (gradient, divergence, Laplacian and curl) are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla f
+        = \vu{e}_r \pdv{f}{r}
+        + \vu{e}_\varphi \frac{1}{r} \pdv{f}{\varphi}
+        + \mathbf{e}_z \pdv{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{r} \pdv{(r V_r)}{r}
+        + \frac{1}{r} \pdv{V_\varphi}{\varphi}
+        + \pdv{V_z}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \nabla^2 f
+        = \frac{1}{r} \pdv{}{r}\Big( r \pdv{f}{r} \Big)
+        + \frac{1}{r^2} \pdvn{2}{f}{\varphi}
+        + \pdvn{2}{f}{z}
+    }
+\end{aligned}$$
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \vu{e}_r \Big( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \Big)
+            \\
+            &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big)
+            \\
+            &+ \frac{\vu{e}_z}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The differential element of volume $\dd{V}$
+takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = r \dd{r} \dd{\varphi} \dd{z}
+    }
+\end{aligned}$$
+
+So, for example, an integral over all of space is converted like so:
+
+$$\begin{aligned}
+    \iiint_{-\infty}^\infty f(x, y, z) \dd{V}
+    = \int_{-\infty}^{\infty} \int_0^{2\pi} \int_0^\infty f(r, \varphi, z) \: r \dd{r} \dd{\varphi} \dd{z}
+\end{aligned}$$
+
+The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dd{S}_r = r \dd{\varphi} \dd{z}
+            \qquad
+            \dd{S}_\varphi = \dd{r} \dd{z}
+            \qquad
+            \dd{S}_z = r \dd{r} \dd{\varphi}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_r \: r \dd{\varphi} \dd{z}
+        + \vu{e}_\varphi \dd{r} \dd{z}
+        + \vu{e}_z \: r \dd{r} \dd{\varphi}
+    }
+\end{aligned}$$
+
+And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_r \dd{r}
+        + \vu{e}_\varphi \: r \dd{\varphi}
+        + \vu{e}_z \dd{z}
+    }
+\end{aligned}$$
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.
diff --git a/source/know/concept/debye-length/index.md b/source/know/concept/debye-length/index.md
new file mode 100644
index 0000000..9b87585
--- /dev/null
+++ b/source/know/concept/debye-length/index.md
@@ -0,0 +1,150 @@
+---
+title: "Debye length"
+date: 2021-10-18
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+If a charged object is put in a plasma,
+it repels like charges and attracts opposite charges,
+leading to a **Debye sheath** around the object's surface
+with a net opposite charge.
+This has the effect of **shielding** the object's presence
+from the rest of the plasma.
+
+We start from [Gauss' law](/know/concept/maxwells-equations/)
+for the [electric field](/know/concept/electric-field/) $\vb{E}$,
+expressing $\vb{E}$ as the gradient of a potential $\phi$,
+i.e. $\vb{E} = -\nabla \phi$,
+and splitting the charge density into ions $n_i$ and electrons $n_e$:
+
+$$\begin{aligned}
+    \nabla^2 \phi(\vb{r})
+    = - \frac{1}{\varepsilon_0} \Big( q_i n_i(\vb{r}) + q_e n_e(\vb{r}) + q_t \delta(\vb{r}) \Big)
+\end{aligned}$$
+
+The last term represents a *test particle*,
+which will be shielded.
+This particle is a point charge $q_t$,
+whose density is simply a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(\vb{r})$,
+and is not included in $n_i$ or $n_e$.
+
+For a plasma in thermal equilibrium,
+we have the [Boltzmann relations](/know/concept/boltzmann-relation/)
+for the densities:
+
+$$\begin{aligned}
+    n_i(\vb{r})
+    = n_{i0} \exp\!\bigg( \!-\! \frac{q_i \phi(\vb{r})}{k_B T_i} \bigg)
+    \qquad \quad
+    n_e(\vb{r})
+    = n_{e0} \exp\!\bigg( \!-\! \frac{q_e \phi(\vb{r})}{k_B T_e} \bigg)
+\end{aligned}$$
+
+We assume that electrical interactions are weak compared to thermal effects,
+i.e. $k_B T \gg q \phi$ in both cases.
+Then we Taylor-expand the Boltzmann relations to first order:
+
+$$\begin{aligned}
+    n_i(\vb{r})
+    \approx n_{i0} \bigg( 1 - \frac{q_i \phi(\vb{r})}{k_B T_i} \bigg)
+    \qquad \quad
+    n_e(\vb{r})
+    \approx n_{e0} \bigg( 1 - \frac{q_e \phi(\vb{r})}{k_B T_e} \bigg)
+\end{aligned}$$
+
+Inserting this back into Gauss' law,
+we arrive at the following equation for $\phi(\vb{r})$,
+where we have assumed quasi-neutrality such that $q_i n_{i0} = q_e n_{e0}$:
+
+$$\begin{aligned}
+    \nabla^2 \phi
+    &= - \frac{1}{\varepsilon_0}
+    \bigg( q_i n_{i0} - n_{i0} \frac{q_i^2 \phi}{k_B T_i} + q_e n_{e0} - n_{e0} \frac{q_e^2 \phi}{k_B T_e} + q_t \delta(\vb{r}) \bigg)
+    \\
+    &= \bigg( \frac{n_{i0} q_i^2}{\varepsilon_0 k_B T_i} + \frac{n_{e0} q_e^2}{\varepsilon_0 k_B T_e} \bigg) \phi
+    - \frac{q_t}{\varepsilon_0} \delta(\vb{r})
+\end{aligned}$$
+
+We now define the **ion** and **electron Debye lengths**
+$\lambda_{Di}$ and $\lambda_{De}$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{1}{\lambda_{Di}^2}
+        \equiv \frac{n_{i0} q_i^2}{\varepsilon_0 k_B T_i}
+    }
+    \qquad \quad
+    \boxed{
+        \frac{1}{\lambda_{De}^2}
+        \equiv \frac{n_{e0} q_e^2}{\varepsilon_0 k_B T_e}
+    }
+\end{aligned}$$
+
+And then the **total Debye length** $\lambda_D$ is defined as the sum of their inverses,
+and gives the rough thickness of the Debye sheath:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{1}{\lambda_D^2}
+        \equiv \frac{1}{\lambda_{Di}^2} + \frac{1}{\lambda_{De}^2}
+        = \frac{n_{i0} q_i^2 T_e + n_{e0} q_e^2 T_i}{\varepsilon_0 k_B T_i T_e}
+    }
+\end{aligned}$$
+
+With this, the equation can be put in the form below,
+suggesting exponential decay:
+
+$$\begin{aligned}
+    \nabla^2 \phi(\vb{r})
+    &= \frac{1}{\lambda_D^2} \phi(\vb{r})
+    - \frac{q_t}{\varepsilon_0} \delta(\vb{r})
+\end{aligned}$$
+
+This has the following solution,
+known as the **Yukawa potential**,
+which decays exponentially,
+representing the plasma's **self-shielding**
+over a characteristic distance $\lambda_D$:
+
+$$\begin{aligned}
+    \boxed{
+        \phi(r)
+        = \frac{q_t}{4 \pi \varepsilon_0 r} \exp\!\Big( \!-\!\frac{r}{\lambda_D} \Big)
+    }
+\end{aligned}$$
+
+Note that $r$ is a scalar,
+i.e. the potential depends only on the radial distance to $q_t$.
+This treatment only makes sense
+if the plasma is sufficiently dense,
+such that there is a large number of particles
+in a sphere with radius $\lambda_D$.
+This corresponds to a large [Coulomb logarithm](/know/concept/coulomb-logarithm/) $\ln\!(\Lambda)$:
+
+$$\begin{aligned}
+    1 \ll \frac{4 \pi}{3} n_0 \lambda_D^3 = \frac{2}{9} \Lambda
+\end{aligned}$$
+
+The name *Yukawa potential* originates from particle physics,
+but can in general be used to refer to any potential (electric or energetic)
+of the following form:
+
+$$\begin{aligned}
+    V(r)
+    = \frac{A}{r} \exp(-B r)
+\end{aligned}$$
+
+Where $A$ and $B$ are scaling constants that depend on the problem at hand.
+
+
+
+## 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/density-of-states/index.md b/source/know/concept/density-of-states/index.md
new file mode 100644
index 0000000..5b18537
--- /dev/null
+++ b/source/know/concept/density-of-states/index.md
@@ -0,0 +1,153 @@
+---
+title: "Density of states"
+date: 2021-05-08
+categories:
+- Physics
+- Statistics
+layout: "concept"
+---
+
+The **density of states** $g(E)$ of a physical system is defined such that
+$g(E) \dd{E}$ is the number of states which could be occupied
+with an energy in the interval $[E, E + \dd{E}]$.
+In fact, $E$ need not be an energy;
+it should just be something that effectively identifies the state.
+
+In its simplest form, the density of states is as follows,
+where $\Gamma(E)$ is the number of states with energy
+less than or equal to the argument $E$:
+
+$$\begin{aligned}
+    g(E)
+    = \dv{\Gamma}{E}
+\end{aligned}$$
+
+If the states can be treated as waves,
+which is often the case,
+then we can calculate the density of states $g(k)$ in
+$k$-space, i.e. as a function of the wavenumber $k = |\vb{k}|$.
+Once we have $g(k)$, we use the dispersion relation $E(k)$ to find $g(E)$,
+by demanding that:
+
+$$\begin{aligned}
+    g(k) \dd{k} = g(E) \dd{E}
+    \quad \implies \quad
+    g(E)
+    = g(k) \dv{k}{E}
+\end{aligned}$$
+
+Inverting the dispersion relation $E(k)$ to get $k(E)$ might be difficult,
+in which case the left-hand equation can be satisfied numerically.
+
+
+Define $\Omega_n(k)$ as the number of states with
+a $k$-value less than or equal to the argument,
+or in other words, the volume of a hypersphere with radius $k$.
+Then the $n$-dimensional density of states $g_n(k)$
+has the following general form:
+
+$$\begin{aligned}
+    \boxed{
+        g_n(k)
+        = \frac{D}{2^n k_{\mathrm{min}}^n} \: \dv{\Omega_n}{k}
+    }
+\end{aligned}$$
+
+Where $D$ is each state's degeneracy (e.g. due to spin),
+and $k_{\mathrm{min}}$ is the smallest allowed $k$-value,
+according to the characteristic length $L$ of the system.
+We divide by $2^n$ to limit ourselves to the sector where all axes are positive,
+because we are only considering the magnitude of $k$.
+
+In one dimension $n = 1$, the number of states within a distance $k$ from the
+origin is the distance from $k$ to $-k$
+(we let it run negative, since its meaning does not matter here), given by:
+
+$$\begin{aligned}
+    \Omega_1(k)
+    = 2 k
+\end{aligned}$$
+
+To get $k_{\mathrm{min}}$, we choose to look at a rod of length $L$,
+across which the function is a standing wave, meaning that
+the allowed values of $k$ must be as follows, where $m \in \mathbb{N}$:
+
+$$\begin{aligned}
+    \lambda = \frac{2 L}{m}
+    \quad \implies \quad
+    k = \frac{2 \pi}{\lambda} = \frac{m \pi}{L}
+\end{aligned}$$
+
+Take the smallest option $m = 1$,
+such that $k_{\mathrm{min}} = \pi / L$,
+the 1D density of states $g_1(k)$ is:
+
+$$\begin{aligned}
+    \boxed{
+        g_1(k)
+        = \frac{D L}{2 \pi} \: 2
+        = \frac{D L}{\pi}
+    }
+\end{aligned}$$
+
+In 2D, the number of states within a range $k$ of the
+origin is the area of a circle with radius $k$:
+
+$$\begin{aligned}
+    \Omega_2(k)
+    = \pi k^2
+\end{aligned}$$
+
+Analogously to the 1D case,
+we take the system to be a square of side $L$,
+so $k_{\mathrm{min}} = \pi / L$ again.
+The density of states then becomes:
+
+$$\begin{aligned}
+    \boxed{
+        g_2(k)
+        = \frac{D L^2}{4 \pi^2} \:2 \pi k
+        = \frac{D L^2 k}{2 \pi}
+    }
+\end{aligned}$$
+
+In 3D, the number of states is the volume of a sphere with radius $k$:
+
+$$\begin{aligned}
+    \Omega_3(k)
+    = \frac{4 \pi}{3} k^3
+\end{aligned}$$
+
+For a cube with side $L$, we once again find $k_{\mathrm{min}} = \pi / L$.
+We thus get:
+
+$$\begin{aligned}
+    \boxed{
+        g_3(k)
+        = \frac{D L^3}{8 \pi^3} \:4 \pi k^2
+        = \frac{D L^3 k^2}{2 \pi^2}
+    }
+\end{aligned}$$
+
+All these expressions contain the characteristic length/area/volume $L^n$,
+and therefore give the number of states in that region only.
+Keep in mind that $L$ is free to choose;
+it need not be the physical size of the system.
+In fact, we typically want the density of states
+per unit length/area/volume,
+so we can just set $L = 1$ in our preferred unit of distance.
+
+If the system is infinitely large, or if it has periodic boundaries,
+then $k$ becomes a continuous variable and $k_\mathrm{min} \to 0$.
+But again, $L$ is arbitrary,
+so a finite value can be chosen.
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
+2.  B. Van Zeghbroeck,
+    [Principles of semiconductor devices](https://ecee.colorado.edu/~bart/book/book/chapter2/ch2_4.htm), 2011,
+    University of Colorado.
diff --git a/source/know/concept/density-operator/index.md b/source/know/concept/density-operator/index.md
new file mode 100644
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@@ -0,0 +1,131 @@
+---
+title: "Density operator"
+date: 2021-03-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, the expectation value of an observable
+$\expval{\hat{L}}$ represents the average result from measuring
+$\hat{L}$ on a large number of systems (an **ensemble**)
+prepared in the same state $\Ket{\Psi}$,
+known as a **pure ensemble** or (somewhat confusingly) **pure state**.
+
+But what if the systems of the ensemble are not all in the same state?
+To work with such a **mixed ensemble** or **mixed state**,
+the **density operator** $\hat{\rho}$ or **density matrix** (in a basis) is useful.
+It is defined as follows, where $p_n$ is the probability
+that the system is in state $\Ket{\Psi_n}$,
+i.e. the proportion of systems in the ensemble that are
+in state $\Ket{\Psi_n}$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}
+        = \sum_{n} p_n \Ket{\Psi_n} \Bra{\Psi_n}
+    }
+\end{aligned}$$
+
+Do not let is this form fool you into thinking that $\hat{\rho}$ is diagonal:
+$\Ket{\Psi_n}$ need not be basis vectors.
+Instead, the matrix elements of $\hat{\rho}$ are found as usual,
+where $\Ket{j}$ and $\Ket{k}$ are basis vectors:
+
+$$\begin{aligned}
+    \matrixel{j}{\hat{\rho}}{k}
+    = \sum_{n} p_n \Inprod{j}{\Psi_n} \Inprod{\Psi_n}{k}
+\end{aligned}$$
+
+However, from the special case where $\Ket{\Psi_n}$ are indeed basis vectors,
+we can conclude that $\hat{\rho}$ is positive semidefinite and Hermitian,
+and that its trace (i.e. the total probability) is 100%:
+
+$$\begin{gathered}
+    \boxed{
+        \hat{\rho} \ge 0
+    }
+    \qquad \qquad
+    \boxed{
+        \hat{\rho}^\dagger = \hat{\rho}
+    }
+    \qquad \qquad
+    \boxed{
+        \mathrm{Tr}(\hat{\rho}) = 1
+    }
+\end{gathered}$$
+
+These properties are preserved by all changes of basis.
+If the ensemble is purely $\Ket{\Psi}$,
+then $\hat{\rho}$ is given by a single state vector:
+
+$$\begin{aligned}
+    \hat{\rho} = \Ket{\Psi} \Bra{\Psi}
+\end{aligned}$$
+
+From the special case where $\Ket{\Psi}$ is a basis vector,
+we can conclude that for a pure ensemble,
+$\hat{\rho}$ is idempotent, which means that:
+
+$$\begin{aligned}
+    \hat{\rho}^2 = \hat{\rho}
+\end{aligned}$$
+
+This can be used to find out whether a given $\hat{\rho}$
+represents a pure or mixed ensemble.
+
+Next, we define the ensemble average $\expval{\hat{O}}$
+as the mean of the expectation values of $\hat{O}$ for states in the ensemble.
+We use the same notation as for the pure expectation value,
+since this is only a small extension of the concept to mixed ensembles.
+It is calculated like so:
+
+$$\begin{aligned}
+    \boxed{
+        \expval{\hat{O}}
+        = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O}}{\Psi_n}
+        = \mathrm{Tr}(\hat{\rho} \hat{O})
+    }
+\end{aligned}$$
+
+To prove the latter,
+we write out the trace $\mathrm{Tr}$ as the sum of the diagonal elements, so:
+
+$$\begin{aligned}
+    \mathrm{Tr}(\hat{\rho} \hat{O})
+    &= \sum_{j} \matrixel{j}{\hat{\rho} \hat{O}}{j}
+    = \sum_{j} \sum_{n} p_n \Inprod{j}{\Psi_n} \matrixel{\Psi_n}{\hat{O}}{j}
+    \\
+    &= \sum_{n} \sum_{j} p_n\matrixel{\Psi_n}{\hat{O}}{j} \Inprod{j}{\Psi_n}
+    = \sum_{n} p_n \matrixel{\Psi_n}{\hat{O} \hat{I}}{\Psi_n}
+    = \expval{\hat{O}}
+\end{aligned}$$
+
+In both the pure and mixed cases,
+if the state probabilities $p_n$ are constant with respect to time,
+then the evolution of the ensemble obeys the **Von Neumann equation**:
+
+$$\begin{aligned}
+    \boxed{
+        i \hbar \dv{\hat{\rho}}{t} = \comm{\hat{H}}{\hat{\rho}}
+    }
+\end{aligned}$$
+
+This equivalent to the Schrödinger equation:
+one can be derived from the other.
+We differentiate $\hat{\rho}$ with the product rule,
+and then substitute the opposite side of the Schrödinger equation:
+
+$$\begin{aligned}
+    i \hbar \dv{\hat{\rho}}{t}
+    &= i \hbar \dv{}{t}\sum_n p_n \Ket{\Psi_n} \Bra{\Psi_n}
+    \\
+    &= \sum_n p_n \Big( i \hbar \dv{}{t}\Ket{\Psi_n} \Big) \Bra{\Psi_n} + \sum_n p_n \Ket{\Psi_n} \Big( i \hbar \dv{}{t}\Bra{\Psi_n} \Big)
+    \\
+    &= \sum_n p_n \ket{\hat{H} n} \Bra{n} - \sum_n p_n \Ket{n} \bra{\hat{H} n}
+    = \hat{H} \hat{\rho} - \hat{\rho} \hat{H}
+    = \comm{\hat{H}}{\hat{\rho}}
+\end{aligned}$$
+
+
diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md
new file mode 100644
index 0000000..b7d5386
--- /dev/null
+++ b/source/know/concept/detailed-balance/index.md
@@ -0,0 +1,232 @@
+---
+title: "Detailed balance"
+date: 2021-11-27
+categories:
+- Physics
+- Mathematics
+- Stochastic analysis
+layout: "concept"
+---
+
+Consider a system that can be regarded as a
+[Markov process](/know/concept/markov-process/),
+which means that its components (e.g. particles) are transitioning
+between a known set of states,
+with no history-dependence and no appreciable influence from interactions.
+
+At equilibrium, the principle of **detailed balance** then says that
+for all states, the rate of leaving that state is exactly equal to
+the rate of entering it, for every possible transition.
+In effect, such a system looks "frozen" to an outside observer,
+since all net transition rates are zero.
+
+We will focus on the case where both time and the state space are continuous.
+Given some initial conditions,
+assume that a component's trajectory can be described
+as an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
+with a time-independent drift $f$ and intensity $g$,
+and with a probability density $\phi(t, x)$ governed by the
+[forward Kolmogorov equation](/know/concept/kolmogorov-equations/)
+(in 3D):
+
+$$\begin{aligned}
+    \pdv{\phi}{t}
+    = - \nabla \cdot \big( \vb{u} \phi - D \nabla \phi \big)
+\end{aligned}$$
+
+We start by demanding **stationarity**,
+which is a weaker condition than detailed balance.
+We want the probability $P$ of being in an arbitrary state volume $V$
+to be constant in time:
+
+$$\begin{aligned}
+    0
+    = \pdv{}{t}P(X_t \in V)
+    = \pdv{}{t}\int_V \phi \dd{V}
+    = \int_V \pdv{\phi}{t} \dd{V}
+\end{aligned}$$
+
+We substitute the forward Kolmogorov equation,
+and apply the divergence theorem:
+
+$$\begin{aligned}
+    0
+    = - \int_V \nabla \cdot \big( \vb{u} \phi - D \nabla \phi \big) \dd{V}
+    = - \oint_{\partial V} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
+\end{aligned}$$
+
+In other words, the "flow" of probability *into* the volume $V$
+is equal to the flow *out of* $V$.
+If such a probability density exists,
+it is called a **stationary distribution** $\phi(t, x) = \pi(x)$.
+Because $V$ was arbitrary, $\pi$ can be found by solving:
+
+$$\begin{aligned}
+    0
+    = - \nabla \cdot \big( \vb{u} \pi - D \nabla \pi \big)
+\end{aligned}$$
+
+Therefore, stationarity means that the state transition rates are constant.
+To get detailed balance, however, we demand that
+the transition rates are zero everywhere:
+the probability flux through an arbitrary surface $S$ must vanish
+(compare to closed surface integral above):
+
+$$\begin{aligned}
+    0
+    = - \int_{S} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
+\end{aligned}$$
+
+And since $S$ is arbitrary, this is only satisfied if the flux is trivially zero
+(the above justification can easily be repeated in 1D, 2D, 4D, etc.):
+
+$$\begin{aligned}
+    \boxed{
+        0 = \vb{u} \phi - D \nabla \phi
+    }
+\end{aligned}$$
+
+This is a stronger condition that stationarity,
+but fortunately often satisfied in practice.
+
+The fact that a system in detailed balance appears "frozen"
+implies it is **time-reversible**,
+meaning its statistics are the same for both directions of time.
+Formally, given two arbitrary functions $h(x)$ and $k(x)$,
+we have the property:
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E}\big[ h(X_0) \: k(X_t) \big]
+        = \mathbf{E}\big[ h(X_t) \: k(X_0) \big]
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-reversibility"/>
+<label for="proof-reversibility">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-reversibility">Proof.</label>
+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.
+</div>
+</div>
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/deutsch-jozsa-algorithm/deutsch-circuit.png b/source/know/concept/deutsch-jozsa-algorithm/deutsch-circuit.png
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/deutsch-jozsa-algorithm/deutsch-circuit.png
diff --git a/source/know/concept/deutsch-jozsa-algorithm/deutsch-jozsa-circuit.png b/source/know/concept/deutsch-jozsa-algorithm/deutsch-jozsa-circuit.png
new file mode 100644
index 0000000..335a624
--- /dev/null
+++ b/source/know/concept/deutsch-jozsa-algorithm/deutsch-jozsa-circuit.png
diff --git a/source/know/concept/deutsch-jozsa-algorithm/index.md b/source/know/concept/deutsch-jozsa-algorithm/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/deutsch-jozsa-algorithm/index.md
@@ -0,0 +1,229 @@
+---
+title: "Deutsch-Jozsa algorithm"
+date: 2021-04-08
+categories:
+- Quantum information
+- Algorithms
+layout: "concept"
+---
+
+The **Deutsch algorithm** and its extension, the **Deutsch-Jozsa algorithm**,
+were first to prove that quantum computers can
+solve certain problems more efficiently
+than any classical system.
+
+Given an unknown "black box" binary function $f(x)$ of one or more bits $x$,
+the goal is determine whether $f$ is
+**constant** (i.e. $f(x)$ is the same for all $x$)
+or **balanced** (i.e. exactly 50% of all $x$-values yield $f(x) = 0$,
+and the other 50% yield $f(x) = 1$).
+We can query $f$ as many times as we want with inputs of our choice,
+but we want to solve the problem using as few queries as possible.
+
+The problem is extremely artificial and of no practical use,
+but quantum computers can solve it with a single query,
+while classical computers need up to $2^{N - 1} + 1$ queries
+for an $N$-bit $x$.
+
+
+## Deutsch algorithm
+
+The Deutsch algorithm handles the simplest case,
+where $x$ is only a single bit.
+Only four $f$ exist:
+
++ **Constant**: $(f(0) = f(1) = 0)$ or $(f(0) = f(1) = 1)$.
++ **Balanced**: $(f(0) = 0, f(1) = 1)$, or $(f(0) = 1, f(1) = 0)$.
+
+In other words, we only need to determine if $f(0) = f(1)$ or $f(0) \neq f(1)$.
+To do this, we use the following quantum circuit,
+where $U_f$ is the oracle we query:
+
+<a href="deutsch-circuit.png">
+<img src="deutsch-circuit.png" style="width:48%">
+</a>
+
+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:
+
+<a href="deutsch-jozsa-circuit.png">
+<img src="deutsch-jozsa-circuit.png" style="width:52%">
+</a>
+
+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
new file mode 100644
index 0000000..7bee0cf
--- /dev/null
+++ b/source/know/concept/dielectric-function/index.md
@@ -0,0 +1,138 @@
+---
+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
new file mode 100644
index 0000000..0947e29
--- /dev/null
+++ b/source/know/concept/diffie-hellman-key-exchange/index.md
@@ -0,0 +1,75 @@
+---
+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
new file mode 100644
index 0000000..63edd81
--- /dev/null
+++ b/source/know/concept/dirac-delta-function/index.md
@@ -0,0 +1,119 @@
+---
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(s x) = \frac{1}{|s|} \delta(x)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-scale"/>
+<label for="proof-scale">Proof</label>
+<div class="hidden">
+<label for="proof-scale">Proof.</label>
+Because it is symmetric, $\delta(s x) = \delta(|s| x)$.
+Then by substituting $\sigma = |s| x$:
+
+$$\begin{aligned}
+    \int \delta(|s| x) \dd{x}
+    &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
+\end{aligned}$$
+</div>
+</div>
+
+An even more impressive property is the behaviour of the derivative of $\delta(x)$:
+
+$$\begin{aligned}
+    \boxed{
+        \int f(\xi) \: \delta'(x - \xi) \dd{\xi} = f'(x)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dv1"/>
+<label for="proof-dv1">Proof</label>
+<div class="hidden">
+<label for="proof-dv1">Proof.</label>
+Note which variable is used for the
+differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:
+
+$$\begin{aligned}
+    \int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi}
+    &= \dv{}{x}\int f(\xi) \: \delta(x - \xi) \dd{x}
+    = f'(x)
+\end{aligned}$$
+</div>
+</div>
+
+This property also generalizes nicely for the higher-order derivatives:
+
+$$\begin{aligned}
+    \boxed{
+        \int f(\xi) \: \dvn{n}{\delta(x - \xi)}{x} \dd{\xi} = \dvn{n}{f(x)}{x}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/dirac-notation/index.md b/source/know/concept/dirac-notation/index.md
new file mode 100644
index 0000000..414b903
--- /dev/null
+++ b/source/know/concept/dirac-notation/index.md
@@ -0,0 +1,130 @@
+---
+title: "Dirac notation"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+**Dirac notation** is a notation to do calculations in a [Hilbert space](/know/concept/hilbert-space/)
+without needing to worry about the space's representation. It is
+basically the *lingua franca* of quantum mechanics.
+
+In Dirac notation there are **kets** $\Ket{V}$ from the Hilbert space
+$\mathbb{H}$ and **bras** $\Bra{V}$ from a dual $\mathbb{H}'$ of the
+former. Crucially, the bras and kets are from different Hilbert spaces
+and therefore cannot be added, but every bra has a corresponding ket and
+vice versa.
+
+Bras and kets can be combined in two ways: the **inner product**
+$\Inprod{V}{W}$, which returns a scalar, and the **outer product**
+$\Ket{V} \Bra{W}$, which returns a mapping $\hat{L}$ from kets $\Ket{V}$
+to other kets $\Ket{V'}$, i.e. a linear operator. Recall that the
+Hilbert inner product must satisfy:
+
+$$\begin{aligned}
+    \Inprod{V}{W} = \Inprod{W}{V}^*
+\end{aligned}$$
+
+So far, nothing has been said about the actual representation of bras or
+kets. If we represent kets as $N$-dimensional columns vectors, the
+corresponding bras are given by the kets' adjoints, i.e. their transpose
+conjugates:
+
+$$\begin{aligned}
+    \Ket{V} =
+    \begin{bmatrix}
+        v_1 \\ \vdots \\ v_N
+    \end{bmatrix}
+    \quad \implies \quad
+    \Bra{V} =
+    \begin{bmatrix}
+        v_1^* & \cdots & v_N^*
+    \end{bmatrix}
+\end{aligned}$$
+
+The inner product $\Inprod{V}{W}$ is then just the familiar dot product $V \cdot W$:
+
+$$\begin{gathered}
+    \Inprod{V}{W}
+    =
+    \begin{bmatrix}
+        v_1^* & \cdots & v_N^*
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1 \\ \vdots \\ w_N
+    \end{bmatrix}
+    = v_1^* w_1 + ... + v_N^* w_N
+\end{gathered}$$
+
+Meanwhile, the outer product $\Ket{V} \Bra{W}$ creates an $N \cross N$ matrix:
+
+$$\begin{gathered}
+    \Ket{V} \Bra{W}
+    =
+    \begin{bmatrix}
+        v_1 \\ \vdots \\ v_N
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1^* & \cdots & w_N^*
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        v_1 w_1^* & \cdots & v_1 w_N^* \\
+        \vdots & \ddots & \vdots \\
+        v_N w_1^* & \cdots & v_N w_N^*
+    \end{bmatrix}
+\end{gathered}$$
+
+If the kets are instead represented by functions $f(x)$ of
+$x \in [a, b]$, then the bras represent *functionals* $F[u(x)]$ which
+take an unknown function $u(x)$ as an argument and turn it into a scalar
+using integration:
+
+$$\begin{aligned}
+    \Ket{f} = f(x)
+    \quad \implies \quad
+    \Bra{f}
+    = F[u(x)]
+    = \int_a^b f^*(x) \: u(x) \dd{x}
+\end{aligned}$$
+
+Consequently, the inner product is simply the following familiar integral:
+
+$$\begin{gathered}
+    \Inprod{f}{g}
+    = F[g(x)]
+    = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{gathered}$$
+
+However, the outer product becomes something rather abstract:
+
+$$\begin{gathered}
+    \Ket{f} \Bra{g}
+    = f(x) \: G[u(x)]
+    = f(x) \int_a^b g^*(\xi) \: u(\xi) \dd{\xi}
+\end{gathered}$$
+
+This result makes more sense if we surround it by a bra and a ket:
+
+$$\begin{aligned}
+    \Bra{u} \!\Big(\!\Ket{f} \Bra{g}\!\Big)\! \Ket{w}
+    &= U\big[f(x) \: G[w(x)]\big]
+    = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
+    \\
+    &= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
+    \\
+    &= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
+    \\
+    &= \Inprod{u}{f} \Inprod{g}{w}
+\end{aligned}$$
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/source/know/concept/dispersive-broadening/index.md b/source/know/concept/dispersive-broadening/index.md
new file mode 100644
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@@ -0,0 +1,96 @@
+---
+title: "Dispersive broadening"
+date: 2021-02-27
+categories:
+- Physics
+- Optics
+- Fiber optics
+layout: "concept"
+---
+
+In optical fibers, **dispersive broadening** is a (linear) effect
+where group velocity dispersion (GVD) "smears out" a pulse in the time domain
+due to the different group velocities of its frequencies,
+since pulses always have a non-zero width in the $\omega$-domain.
+No new frequencies are created.
+
+A pulse envelope $A(z, t)$ inside a fiber must obey the nonlinear Schrödinger equation,
+where the parameters $\beta_2$ and $\gamma$ respectively
+control dispersion and nonlinearity:
+
+$$\begin{aligned}
+    0
+    = i \pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma |A|^2 A
+\end{aligned}$$
+
+We set $\gamma = 0$ to ignore all nonlinear effects,
+and consider a Gaussian initial condition:
+
+$$\begin{aligned}
+    A(0, t)
+    = \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big)
+\end{aligned}$$
+
+By [Fourier transforming](/know/concept/fourier-transform/) in $t$,
+the full analytical solution $A(z, t)$ is found to be as follows,
+where it can be seen that the amplitude
+decreases and the width increases with $z$:
+
+$$\begin{aligned}
+    A(z,t) = \sqrt{\frac{P_0}{1 - i \beta_2 z / T_0^2}}
+    \exp\!\bigg(\! -\!\frac{t^2 / (2 T_0^2)}{1 + \beta_2^2 z^2 / T_0^4} \big( 1 + i \beta_2 z / T_0^2 \big) \bigg)
+\end{aligned}$$
+
+To quantify the strength of dispersive effects,
+we define the dispersion length $L_D$
+as the distance over which the half-width at $1/e$ of maximum power
+(initially $T_0$) increases by a factor of $\sqrt{2}$:
+
+$$\begin{aligned}
+    T_0 \sqrt{1 + \beta_2^2 L_D^2 / T_0^4} = T_0 \sqrt{2}
+    \qquad \implies \qquad
+    \boxed{
+        L_D = \frac{T_0^2}{|\beta_2|}
+    }
+\end{aligned}$$
+
+This phenomenon is illustrated below for our example of a Gaussian pulse
+with parameter values $T_0 = 1\:\mathrm{ps}$, $P_0 = 1\:\mathrm{kW}$,
+$\beta_2 = -10 \:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0$:
+
+<a href="pheno-disp.jpg">
+<img src="pheno-disp-small.jpg" style="width:100%">
+</a>
+
+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.
diff --git a/source/know/concept/dispersive-broadening/pheno-disp-small.jpg b/source/know/concept/dispersive-broadening/pheno-disp-small.jpg
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/dispersive-broadening/pheno-disp-small.jpg
diff --git a/source/know/concept/dispersive-broadening/pheno-disp.jpg b/source/know/concept/dispersive-broadening/pheno-disp.jpg
new file mode 100644
index 0000000..a97312b
--- /dev/null
+++ b/source/know/concept/dispersive-broadening/pheno-disp.jpg
diff --git a/source/know/concept/drude-model/index.md b/source/know/concept/drude-model/index.md
new file mode 100644
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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/):
+
+$$\begin{aligned}
+    \boxed{
+        \hat{A}\{h(x)\}
+        = \hat{L}\{h(x)\}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-kolmogorov"/>
+<label for="proof-kolmogorov">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-kolmogorov">Proof.</label>
+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 $t \to 0^+$:
+
+$$\begin{aligned}
+    \hat{L}\{h(X_0)\}
+    \approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0]
+\end{aligned}$$
+</div>
+</div>
+
+The general definition of resembles that of a classical derivative,
+and indeed, the generator $\hat{A}$ can be thought of as a differential operator.
+In that case, we would like an analogue of the classical
+fundamental theorem of calculus to relate it to integration.
+
+Such an analogue is provided by **Dynkin's formula**:
+for a stopping time $\tau$ with a finite expected value $\mathbf{E}[\tau|X_0] < \infty$,
+it states that:
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E}\big[ h(X_\tau) | X_0 \big]
+        = h(X_0) + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dynkin"/>
+<label for="proof-dynkin">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-dynkin">Proof.</label>
+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.
+</div>
+</div>
+
+A common application of Dynkin's formula is predicting
+when the stopping time $\tau$ occurs, and in what state $X_\tau$ this happens.
+Consider an example:
+for a region $\Omega$ of state space with $X_0 \in \Omega$,
+we define the exit time $\tau \equiv \inf\{ t : X_t \notin \Omega \}$,
+provided that $\mathbf{E}[\tau | X_0] < \infty$.
+
+To get information about when and where $X_t$ exits $\Omega$,
+we define the *general reward* $\Gamma$ as follows,
+consisting of a *running reward* $R$ for $X_t$ inside $\Omega$,
+and a *terminal reward* $T$ on the boundary $\partial \Omega$ where we stop at $X_\tau$:
+
+$$\begin{aligned}
+    \Gamma
+    = \int_0^\tau R(X_t) \dd{t} + \: T(X_\tau)
+\end{aligned}$$
+
+For example, for $R = 1$ and $T = 0$, this becomes $\Gamma = \tau$,
+and if $R = 0$, then $T(X_\tau)$ can tell us the exit point.
+Let us now define $h(X_0) = \mathbf{E}[\Gamma | X_0]$,
+and apply Dynkin's formula:
+
+$$\begin{aligned}
+    \mathbf{E}\big[ h(X_\tau) | X_0 \big]
+    &= \mathbf{E}\big[ \Gamma \big| X_0 \big] + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
+    \\
+    &= \mathbf{E}\big[ T(X_\tau) | X_0 \big] + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} + R(X_t) \dd{t} \bigg| X_0 \bigg]
+\end{aligned}$$
+
+The two leftmost terms depend on the exit point $X_\tau$,
+but not directly on $X_t$ for $t < \tau$,
+while the rightmost depends on the whole trajectory $X_t$.
+Therefore, the above formula is fulfilled
+if $h(x)$ satisfies the following equation and boundary conditions:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{cases}
+            \hat{L}\{h(x)\} + R(x) = 0 & \mathrm{for}\; x \in \Omega \\
+            h(x) = T(x) & \mathrm{for}\; x \notin \Omega
+        \end{cases}
+    }
+\end{aligned}$$
+
+In other words, we have just turned a difficult question about a stochastic trajectory $X_t$
+into a classical differential boundary value problem for $h(x)$.
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/dyson-equation/index.md b/source/know/concept/dyson-equation/index.md
new file mode 100644
index 0000000..82020ad
--- /dev/null
+++ b/source/know/concept/dyson-equation/index.md
@@ -0,0 +1,169 @@
+---
+title: "Dyson equation"
+date: 2021-11-01
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+Consider the time-dependent Schrödinger equation,
+describing a wavefunction $\Psi_0(\vb{r}, t)$:
+
+$$\begin{aligned}
+    i \hbar \pdv{}{t}\Psi_0(\vb{r}, t)
+    = \hat{H}_0(\vb{r}) \: \Psi_0(\vb{r}, t)
+\end{aligned}$$
+
+By definition, this equation's
+[fundamental solution](/know/concept/fundamental-solution/)
+$G_0(\vb{r}, t; \vb{r}', t')$ satisfies the following:
+
+$$\begin{aligned}
+    \Big( i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) \Big) G_0(\vb{r}, t; \vb{r}', t')
+    = \delta(\vb{r} - \vb{r}') \: \delta(t - t')
+\end{aligned}$$
+
+From this, we define the inverse $\hat{G}{}_0^{-1}(\vb{r}, t)$
+as follows, so that $\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$:
+
+$$\begin{aligned}
+    \hat{G}{}_0^{-1}(\vb{r}, t)
+    &\equiv i \hbar \pdv{}{t}- \hat{H}_0(\vb{r})
+\end{aligned}$$
+
+Note that $\hat{G}{}_0^{-1}$ is an operator, while $G_0$ is a function.
+For the sake of consistency, we thus define
+the operator $\hat{G}_0(\vb{r}, t)$
+as a multiplication by $G_0$
+and integration over $\vb{r}'$ and $t'$:
+
+$$\begin{aligned}
+    \hat{G}_0(\vb{r}, t) \: f
+    \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
+\end{aligned}$$
+
+For an arbitrary function $f(\vb{r}, t)$,
+so that $\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$.
+Moving on, the Schrödinger equation can be rewritten like so,
+using $\hat{G}{}_0^{-1}$:
+
+$$\begin{aligned}
+    \hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t)
+    = 0
+\end{aligned}$$
+
+Let us assume that $\hat{H}_0$ is simple,
+such that $G_0$ and $\hat{G}{}_0^{-1}$ can be found without issues
+by solving the defining equation above.
+
+Suppose we now add a more complicated and
+possibly time-dependent term $\hat{H}_1(\vb{r}, t)$,
+in which case the corresponding fundamental solution
+$G(\vb{r}, \vb{r}', t, t')$ satisfies:
+
+$$\begin{aligned}
+    \delta(\vb{r} - \vb{r}') \: \delta(t - t')
+    &= \Big( i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) - \hat{H}_1(\vb{r}, t) \Big) G(\vb{r}, t; \vb{r}', t')
+    \\
+    &= \Big( \hat{G}{}_0^{-1}(\vb{r}, t) - \hat{H}_1(\vb{r}, t) \Big) G(\vb{r}, t; \vb{r}', t')
+\end{aligned}$$
+
+This equation is typically too complicated to solve,
+so we would like an easier way to calculate this new $G$.
+The perturbed wavefunction $\Psi(\vb{r}, t)$
+satisfies the Schrödinger equation:
+
+$$\begin{aligned}
+    \Big( \hat{G}{}_0^{-1}(\vb{r}, t) - \hat{H}_1(\vb{r}, t) \Big) \Psi(\vb{r}, t)
+    = 0
+\end{aligned}$$
+
+We know that $\hat{G}{}_0^{-1} \Psi_0 = 0$,
+which we put on the right,
+and then we apply $\hat{G}_0$ in front:
+
+$$\begin{aligned}
+    \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi
+    = \hat{G}_0^{-1} \Psi_0
+    \quad \implies \quad
+    \Psi - \hat{G}_0 \hat{H}_1 \Psi
+    &= \Psi_0
+\end{aligned}$$
+
+This equation is recursive,
+so we iteratively insert it into itself.
+Note that the resulting equations are the same as those from
+[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/):
+
+$$\begin{aligned}
+    \Psi
+    &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi
+    \\
+    &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi
+    \\
+    &= \Psi_0 + \hat{G}_0 \hat{H}_1 \Psi_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi_0
+    + \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \Psi_0 + \: ...
+    \\
+    &= \Psi_0 + \big( \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G}_0 +  \hat{G}_0 \hat{H}_1 \hat{G}_0 \hat{H}_1 \hat{G}_0 + \: ... \big) \hat{H}_1 \Psi_0
+\end{aligned}$$
+
+The parenthesized expression clearly has the same recursive pattern,
+so we denote it by $\hat{G}$ and write the so-called **Dyson equation**:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{G}
+        = \hat{G}_0 + \hat{G}_0 \hat{H}_1 \hat{G}
+    }
+\end{aligned}$$
+
+Such an iterative scheme is excellent for approximating $\hat{G}(\vb{r}, t)$.
+Once a satisfactory accuracy is obtained,
+the perturbed wavefunction $\Psi$ can be calculated from:
+
+$$\begin{aligned}
+    \boxed{
+        \Psi
+        = \Psi_0 + \hat{G} \hat{H}_1 \Psi_0
+    }
+\end{aligned}$$
+
+This relation is equivalent to the Schrödinger equation.
+So now we have the operator $\hat{G}(\vb{r}, t)$,
+but what about the fundamental solution function $G(\vb{r}, t; \vb{r}', t')$?
+Let us take its definition, multiply it by an arbitrary $f(\vb{r}, t)$,
+and integrate over $G$'s second argument pair:
+
+$$\begin{aligned}
+    \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
+    = \iint \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
+    = f
+\end{aligned}$$
+
+Where we have hidden the arguments $(\vb{r}, t)$ for brevity.
+We now apply $\hat{G}_0(\vb{r}, t)$ to this equation
+(which contains an integral over $t''$ independent of $t'$):
+
+$$\begin{aligned}
+    \hat{G}_0 f
+    &= \big( \hat{G}_0 \hat{G}{}_0^{-1} - \hat{G}_0 \hat{H}_1 \big) \iint_{-\infty}^\infty G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
+    \\
+    &= \big( 1 - \hat{G}_0 \hat{H}_1 \big) \iint_{-\infty}^\infty G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
+\end{aligned}$$
+
+Here, the shape of Dyson's equation is clearly recognizable,
+so we conclude that, as expected, the operator $\hat{G}$
+is defined as multiplication by the function $G$ followed by integration:
+
+$$\begin{aligned}
+    \hat{G}(\vb{r}, t) \: f(\vb{r}, t)
+    \equiv \iint_{-\infty}^\infty G(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}}' \dd{t'}
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/ehrenfests-theorem/index.md b/source/know/concept/ehrenfests-theorem/index.md
new file mode 100644
index 0000000..a2a676a
--- /dev/null
+++ b/source/know/concept/ehrenfests-theorem/index.md
@@ -0,0 +1,131 @@
+---
+title: "Ehrenfest's theorem"
+date: 2021-02-24
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the
+time evolution of an observable's expectation value $\expval{\hat{L}}$.
+
+The time-dependent Schrödinger equation is as follows,
+where prime denotes differentiation with respect to time $t$:
+
+$$\begin{aligned}
+    \Ket{\psi'} = \frac{1}{i \hbar} \hat{H} \Ket{\psi}
+    \qquad
+    \Bra{\psi'} = - \frac{1}{i \hbar} \Bra{\psi} \hat{H}
+\end{aligned}$$
+
+Given an observable operator $\hat{L}$ and a state $\Ket{\psi}$,
+the time-derivative of the expectation value $\expval{\hat{L}}$ is as follows
+(due to the product rule of differentiation):
+
+$$\begin{aligned}
+    \dv{\expval{\hat{L}}}{t}
+    &= \matrixel{\psi}{\hat{L}}{\psi'} + \matrixel{\psi'}{\hat{L}}{\psi} + \matrixel{\psi}{\hat{L}'}{\psi}
+    \\
+    &= \frac{1}{i \hbar} \matrixel{\psi}{\hat{L}\hat{H}}{\psi}
+    - \frac{1}{i \hbar} \matrixel{\psi}{\hat{H}\hat{L}}{\psi}
+    + \Expval{\dv{\hat{L}}{t}}
+\end{aligned}$$
+
+The first two terms on the right can be rewritten using a commutator,
+yielding the general form of Ehrenfest's theorem:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{\expval{\hat{L}}}{t}
+        = \frac{1}{i \hbar} \Expval{[\hat{L}, \hat{H}]} + \Expval{\dv{\hat{L}}{t}}
+    }
+\end{aligned}$$
+
+In practice, since most operators are time-independent,
+the last term often vanishes.
+
+As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+this relation proves itself,
+when one simply wraps all terms in $\Bra{\psi}$ and $\Ket{\psi}$.
+
+Two observables of particular interest are the position $\hat{X}$ and momentum $\hat{P}$.
+Applying the above theorem to $\hat{X}$ yields the following,
+which we reduce using the fact that $\hat{X}$ commutes
+with the potential $V(\hat{X})$,
+because one is a function of the other:
+
+$$\begin{aligned}
+    \dv{\expval{\hat{X}}}{t}
+    &= \frac{1}{i \hbar} \Expval{[\hat{X}, \hat{H}]}
+    = \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2] + 2 m [\hat{X}, V(\hat{X})]}
+    = \frac{1}{2 i \hbar m} \Expval{[\hat{X}, \hat{P}^2]}
+    \\
+    &= \frac{1}{2 i \hbar m} \Expval{\hat{P} [\hat{X}, \hat{P}] + [\hat{X}, \hat{P}] \hat{P}}
+    = \frac{2 i \hbar}{2 i \hbar m} \expval{\hat{P}}
+    = \frac{\expval{\hat{P}}}{m}
+\end{aligned}$$
+
+This is the first part of the "original" form of Ehrenfest's theorem,
+which is reminiscent of classical Newtonian mechanics:
+
+$$\begin{gathered}
+    \boxed{
+        \dv{\expval{\hat{X}}}{t} = \frac{\expval{\hat{P}}}{m}
+    }
+\end{gathered}$$
+
+Next, applying the general formula to the expected momentum $\expval{\hat{P}}$
+gives us:
+
+$$\begin{aligned}
+    \dv{\expval{\hat{P}}}{t}
+    &= \frac{1}{i \hbar} \Expval{[\hat{P}, \hat{H}]}
+    = \frac{1}{2 i \hbar m} \Expval{[\hat{P}, \hat{P}^2] + 2 m [\hat{P}, V(\hat{X})]}
+    = \frac{1}{i \hbar} \Expval{[\hat{P}, V(\hat{X})]}
+\end{aligned}$$
+
+To find the commutator, we go to the $\hat{X}$-basis and use a test
+function $f(x)$:
+
+$$\begin{aligned}
+    \Comm{- i \hbar \dv{}{x}}{V(x)} \: f(x)
+    &= - i \hbar \frac{dV}{dx} f(x) - i \hbar V(x) \frac{df}{dx} + i \hbar V(x) \frac{df}{dx}
+    = - i \hbar \frac{dV}{dx} f(x)
+\end{aligned}$$
+
+By inserting this result back into the previous equation, we find the following:
+
+$$\begin{aligned}
+    \dv{\expval{\hat{P}}}{t}
+    &= - \frac{i \hbar}{i \hbar} \Expval{\frac{d V}{d \hat{X}}}
+    = - \Expval{\frac{d V}{d \hat{X}}}
+\end{aligned}$$
+
+This is the second part of Ehrenfest's theorem,
+which is also similar to Newtonian mechanics:
+
+$$\begin{gathered}
+    \boxed{
+        \dv{\expval{\hat{P}}}{t} = - \Expval{\pdv{V}{\hat{X}}}
+    }
+\end{gathered}$$
+
+There is an important consequence of Ehrenfest's original theorems
+for the symbolic derivatives of the Hamiltonian $\hat{H}$
+with respect to $\hat{X}$ and $\hat{P}$:
+
+$$\begin{gathered}
+    \boxed{
+        \Expval{\pdv{\hat{H}}{\hat{P}}}
+        = \dv{\expval{\hat{X}}}{t}
+    }
+    \qquad \quad
+    \boxed{
+        - \Expval{\pdv{\hat{H}}{\hat{X}}}
+        = \dv{\expval{\hat{P}}}{t}
+    }
+\end{gathered}$$
+
+These are easy to prove yourself,
+and are analogous to Hamilton's canonical equations.
diff --git a/source/know/concept/einstein-coefficients/index.md b/source/know/concept/einstein-coefficients/index.md
new file mode 100644
index 0000000..ff44888
--- /dev/null
+++ b/source/know/concept/einstein-coefficients/index.md
@@ -0,0 +1,341 @@
+---
+title: "Einstein coefficients"
+date: 2021-07-11
+categories:
+- Physics
+- Optics
+- Quantum mechanics
+- Two-level system
+- Laser theory
+layout: "concept"
+---
+
+The **Einstein coefficients** quantify
+the emission and absorption of photons by a solid,
+and can be calculated analytically from first principles
+in several useful situations.
+
+
+## Qualitative description
+
+Suppose we have a ground state with energy $E_1$ containing $N_1$ electrons,
+and an excited state with energy $E_2$ containing $N_2$ electrons.
+The resonance $\omega_0 \equiv (E_2 \!-\! E_1)/\hbar$
+is the frequency of the photon emitted
+when an electron falls from $E_2$ to $E_1$.
+
+The first Einstein coefficient is the **spontaneous emission rate** $A_{21}$,
+which gives the probability per unit time
+that an excited electron falls from state 2 to 1,
+so that $N_2(t)$ obeys the following equation,
+which is easily solved:
+
+$$\begin{aligned}
+    \dv{N_2}{t} = - A_{21} N_2
+    \quad \implies \quad
+    N_2(t) = N_2(0) \exp(- t / \tau)
+\end{aligned}$$
+
+Where $\tau = 1 / A_{21}$ is the **natural radiative lifetime** of the excited state,
+which gives the lifetime of an excited electron,
+before it decays to the ground state.
+
+The next coefficient is the **absorption rate** $B_{12}$,
+which is the probability that an incoming photon excites an electron,
+per unit time and per unit spectral energy density
+(i.e. the rate depends on the frequency of the incoming light).
+Then $N_1(t)$ obeys the following equation:
+
+$$\begin{aligned}
+    \dv{N_1}{t} = - B_{12} N_1 u(\omega_0)
+\end{aligned}$$
+
+Where $u(\omega)$ is the spectral energy density of the incoming light,
+put here to express the fact that only photons with frequency $\omega_0$ are absorbed.
+
+There is one more Einstein coefficient: the **stimulated emission rate** $B_{21}$.
+An incoming photon has an associated electromagnetic field,
+which can encourage an excited electron to drop to the ground state,
+such that for $A_{21} = 0$:
+
+$$\begin{aligned}
+    \dv{N_2}{t} = - B_{21} N_2 u(\omega_0)
+\end{aligned}$$
+
+These three coefficients $A_{21}$, $B_{12}$ and $B_{21}$ are related to each other.
+Suppose that the system is in equilibrium,
+i.e. that $N_1$ and $N_2$ are constant.
+We assume that the number of particles in the system is constant,
+implying that $N_1'(t) = - N_2'(t) = 0$, so:
+
+$$\begin{aligned}
+    B_{12} N_1 u(\omega_0) = A_{21} N_2 + B_{21} N_2 u(\omega_0) = 0
+\end{aligned}$$
+
+Isolating this equation for $u(\omega_0)$,
+gives following expression for the radiation:
+
+$$\begin{aligned}
+    u(\omega_0)
+    = \frac{A_{21}}{(N_1 / N_2) B_{12} - B_{21}}
+\end{aligned}$$
+
+We assume that the system is in thermal equilibrium
+with its own black-body radiation, and that there is no external light.
+Then this is a [canonical ensemble](/know/concept/canonical-ensemble/),
+meaning that the relative probability that an electron has $E_2$ compared to $E_1$
+is given by the Boltzmann distribution:
+
+$$\begin{aligned}
+    \frac{\mathrm{Prob}(E_2)}{\mathrm{Prob}(E_1)}
+    = \frac{N_2}{N_1}
+    = \frac{g_2}{g_1} \exp(- \hbar \omega_0 \beta)
+\end{aligned}$$
+
+Where $g_2$ and $g_1$ are the degeneracies of the energy levels.
+Inserting this back into the equation for the spectrum $u(\omega_0)$ yields:
+
+$$\begin{aligned}
+    u(\omega_0)
+    = \frac{A_{21}}{(g_1 / g_2) B_{12} \exp(\hbar \omega_0 \beta) - B_{21}}
+\end{aligned}$$
+
+Since $u(\omega_0)$ represents only black-body radiation,
+our result must agree with [Planck's law](/know/concept/plancks-law/):
+
+$$\begin{aligned}
+    u(\omega_0)
+    = \frac{A_{21}}{B_{21} \big( (g_1 B_{12} / g_2 B_{21}) \exp(\hbar \omega_0 \beta) - 1 \big)}
+    = \frac{\hbar \omega_0^3}{\pi^2 c^3} \frac{1}{\exp(\hbar \omega_0 \beta) - 1}
+\end{aligned}$$
+
+This gives us the following two equations relating the Einstein coefficients:
+
+$$\begin{aligned}
+    \boxed{
+        A_{21} = \frac{\hbar \omega_0^3}{\pi^2 c^3} B_{21}
+        \qquad \quad
+        g_1 B_{12} = g_2 B_{21}
+    }
+\end{aligned}$$
+
+Note that this result holds even if $E_1$ is not the ground state,
+but instead some lower excited state below $E_2$,
+due to the principle of [detailed balance](/know/concept/detailed-balance/).
+Furthermore, it turns out that these relations
+also hold if the system is not in equilibrium.
+
+A notable case is **population inversion**,
+where $B_{21} N_2 > B_{12} N_1$ such that $N_2 > (g_2 / g_1) N_1$.
+This situation is mandatory for lasers, where stimulated emission must dominate,
+such that the light becomes stronger as it travels through the medium.
+
+
+## Coherent light
+
+In fact, we can analytically calculate the Einstein coefficients in some cases,
+by treating incoming light as a perturbation
+to an electron in a two-level system,
+and then finding $B_{12}$ and $B_{21}$ from the resulting transition rate.
+We need to make the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
+in which case the perturbing Hamiltonian $\hat{H}_1(t)$ is given by:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = - q \vec{r} \cdot \vec{E}_0 \cos(\omega t)
+\end{aligned}$$
+
+Where $q = -e$ is the electron charge,
+$\vec{r}$ is the position operator,
+and $\vec{E}_0$ is the amplitude of
+the [electromagnetic wave](/know/concept/electromagnetic-wave-equation/).
+For simplicity, we let the amplitude be along the $z$-axis:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = - q E_0 z \cos(\omega t)
+\end{aligned}$$
+
+This form of $\hat{H}_1$ is a well-known case for
+[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
+which tells us that the transition probability from $\Ket{a}$ to $\Ket{b}$ is:
+
+$$\begin{aligned}
+    P_{ab}
+    = \frac{\big|\!\matrixel{a}{H_1}{b}\!\big|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ba} - \omega) t / 2 \big)}{(\omega_{ba} - \omega)^2}
+\end{aligned}$$
+
+If the nucleus is at $z = 0$,
+then generally $\Ket{1}$ and $\Ket{2}$ will be even or odd functions of $z$,
+meaning that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$
+(see also [Laporte's selection rule](/know/concept/selection-rules/)),
+leading to:
+
+$$\begin{gathered}
+    \matrixel{1}{H_1}{2} = - E_0 d^*
+    \qquad
+    \matrixel{2}{H_1}{1} = - E_0 d
+    \\
+    \matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
+\end{gathered}$$
+
+Where $d \equiv q \matrixel{2}{z}{1}$ is a constant,
+namely the $z$-component of the **transition dipole moment**.
+The chance of an upward jump (i.e. absorption) is:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{E_0^2 |d|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+\end{aligned}$$
+
+Meanwhile, the transition probability for stimulated emission is as follows,
+using the fact that $P_{ab}$ is a sinc-function,
+and is therefore symmetric around $\omega_{ba}$:
+
+$$\begin{aligned}
+    P_{21}
+    = \frac{E_0^2 |d|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+\end{aligned}$$
+
+Surprisingly, the probabilities of absorption and stimulated emission are the same!
+In practice, however, the relative rates of these two processes depends heavily on
+the availability of electrons and holes in both states.
+
+In theory, we could calculate the transition rate $R_{12} = \ipdv{P_{12}}{t}$,
+which would give us Einstein's absorption coefficient $B_{12}$,
+for this specific case of coherent monochromatic light.
+However, the result would not be constant in time $t$,
+so is not really useful.
+
+
+## Polarized light
+
+To solve this "problem", we generalize to (incoherent) polarized polychromatic light.
+To do so, we note that the energy density $u$ of an electric field $E_0$ is given by:
+
+$$\begin{aligned}
+    u = \frac{1}{2} \varepsilon_0 E_0^2
+    \qquad \implies \qquad
+    E_0^2 = \frac{2 u}{\varepsilon_0}
+\end{aligned}$$
+
+Where $\varepsilon_0$ is the vacuum permittivity.
+Putting this in the previous result for $P_{12}$ gives us:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{2 u |d|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+\end{aligned}$$
+
+For a continuous light spectrum,
+this $u$ turns into the spectral energy density $u(\omega)$:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{2 |d|^2}{\varepsilon_0 \hbar^2}
+    \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
+\end{aligned}$$
+
+From here, the derivation is similar to that of
+[Fermi's golden rule](/know/concept/fermis-golden-rule/),
+despite the distinction that we are integrating over frequencies rather than states.
+
+At sufficiently large $t$, the integrand is sharply peaked at $\omega = \omega_0$
+and negligible everywhere else,
+so we take $u(\omega)$ out of the integral and extend the integration limits.
+Then we rewrite and look up the integral,
+which turns out to be $\pi t$:
+
+$$\begin{aligned}
+    P_{12}
+    = \frac{|d|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
+    = \frac{\pi |d|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
+\end{aligned}$$
+
+From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
+is then calculated as follows:
+
+$$\begin{aligned}
+    R_{12}
+    = \pdv{P_{12}}{t}
+    = \frac{\pi |d|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
+\end{aligned}$$
+
+Using the relations from earlier with $g_1 = g_2$,
+the Einstein coefficients are found to be as follows
+for a polarized incoming light spectrum:
+
+$$\begin{aligned}
+    \boxed{
+        B_{21} = B_{12} = \frac{\pi |d|^2}{\varepsilon_0 \hbar^2}
+        \qquad
+        A_{21} = \frac{\omega_0^3 |d|^2}{\pi \varepsilon_0 \hbar c^3}
+    }
+\end{aligned}$$
+
+
+## Unpolarized light
+
+We can generalize the above result even further to unpolarized light.
+Let us return to the matrix elements of the perturbation $\hat{H}_1$,
+and define the polarization unit vector $\vec{n}$:
+
+$$\begin{aligned}
+    \matrixel{2}{\hat{H}_1}{1}
+    = - \vec{d} \cdot \vec{E}_0
+    = - E_0 (\vec{d} \cdot \vec{n})
+\end{aligned}$$
+
+Where $\vec{d} \equiv q \matrixel{2}{\vec{r}}{1}$ is
+the full **transition dipole moment** vector, which is usually complex.
+
+The goal is to calculate the average of $|\vec{d} \cdot \vec{n}|^2$.
+In [spherical coordinates](/know/concept/spherical-coordinates/),
+we integrate over all directions $\vec{n}$ for fixed $\vec{d}$,
+using that $\vec{d} \cdot \vec{n} = |\vec{d}| \cos(\theta)$
+with $|\vec{d}| \equiv |d_x|^2 \!+\! |d_y|^2 \!+\! |d_z|^2$:
+
+$$\begin{aligned}
+    \Expval{|\vec{d} \cdot \vec{n}|^2}
+    = \frac{1}{4 \pi} \int_0^\pi \int_0^{2 \pi} |\vec{d}|^2 \cos^2(\theta) \sin(\theta) \dd{\varphi} \dd{\theta}
+\end{aligned}$$
+
+Where we have divided by $4\pi$ (the surface area of a unit sphere) for normalization,
+and $\theta$ is the polar angle between $\vec{n}$ and $\vec{d}$.
+Evaluating the integrals yields:
+
+$$\begin{aligned}
+    \Expval{|\vec{d} \cdot \vec{n}|^2}
+    = \frac{2 \pi}{4 \pi} |\vec{d}|^2 \int_0^\pi \cos^2(\theta) \sin(\theta) \dd{\theta}
+    = \frac{|\vec{d}|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
+    = \frac{|\vec{d}|^2}{3}
+\end{aligned}$$
+
+With this additional constant factor $1/3$,
+the transition rate $R_{12}$ is modified to:
+
+$$\begin{aligned}
+    R_{12}
+    = \pdv{P_{12}}{t}
+    = \frac{\pi |\vec{d}|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
+\end{aligned}$$
+
+From which it follows that the Einstein coefficients for unpolarized light are given by:
+
+$$\begin{aligned}
+    \boxed{
+        B_{21} = B_{12} = \frac{\pi |\vec{d}|^2}{3 \varepsilon_0 \hbar^2}
+        \qquad
+        A_{21} = \frac{\omega_0^3 |\vec{d}|^2}{3 \pi \varepsilon_0 \hbar c^3}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M. Fox,
+    *Optical properties of solids*, 2nd edition,
+    Oxford.
+2.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/source/know/concept/elastic-collision/index.md b/source/know/concept/elastic-collision/index.md
new file mode 100644
index 0000000..f3c4b7b
--- /dev/null
+++ b/source/know/concept/elastic-collision/index.md
@@ -0,0 +1,154 @@
+---
+title: "Elastic collision"
+date: 2021-10-04
+categories:
+- Physics
+- Classical mechanics
+layout: "concept"
+---
+
+In an **elastic collision**,
+the sum of the colliding objects' kinetic energies
+is the same before and after the collision.
+In contrast, in an **inelastic collision**,
+some of that energy is converted into another form,
+for example heat.
+
+
+## One dimension
+
+In 1D, not only the kinetic energy is conserved, but also the total momentum.
+Let $v_1$ and $v_2$ be the initial velocities of objects 1 and 2,
+and $v_1'$ and $v_2'$ their velocities afterwards:
+
+$$\begin{aligned}
+    \begin{cases}
+        \quad\! m_1 v_1 +\:\:\: m_2 v_2
+        = \quad m_1 v_1' +\:\:\: m_2 v_2'
+        \\
+        \displaystyle\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2
+        = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} m_2 v_2'^2
+    \end{cases}
+\end{aligned}$$
+
+After some rearranging,
+these two equations can be written as follows:
+
+$$\begin{aligned}
+    \begin{cases}
+        m_1 (v_1 - v_1')
+        \qquad\quad\:\;\; = m_2 (v_2' - v_2)
+        \\
+        m_1 (v_1 - v_1') (v_1 + v_1')
+        = m_2 (v_2' - v_2) (v_2 + v_2')
+    \end{cases}
+\end{aligned}$$
+
+Using the first equation to replace $m_1 (v_1 \!-\! v_1')$
+with $m_2 (v_2 \!-\! v_2')$ in the second:
+
+$$\begin{aligned}
+    m_2 (v_1 + v_1') (v_2' - v_2)
+    = m_2 (v_2 + v_2') (v_2' - v_2)
+\end{aligned}$$
+
+Dividing out the common factors
+then leads us to a simplified system of equations:
+
+$$\begin{aligned}
+    \begin{cases}
+        \qquad\;\; v_1 + v_1'
+        = v_2 + v_2'
+        \\
+        m_1 v_1 + m_2 v_2
+        = m_1 v_1' + m_2 v_2'
+    \end{cases}
+\end{aligned}$$
+
+Note that the first relation is equivalent to $v_1 - v_2 = v_2' - v_1'$,
+meaning that the objects' relative velocity
+is reversed by the collision.
+Moving on, we replace $v_1'$ in the second equation:
+
+$$\begin{aligned}
+    m_1 v_1 + m_2 v_2
+    &= m_1 (v_2 + v_2' - v_1) + m_2 v_2'
+    \\
+    (m_1 + m_2) v_2'
+    &= 2 m_1 v_1 + (m_2 - m_1) v_2
+\end{aligned}$$
+
+Dividing by $m_1 + m_2$,
+and going through the same process for $v_1'$,
+we arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            v_1'
+            &= \frac{(m_1 - m_2) v_1 + 2 m_2 v_2}{m_1 + m_2}
+            \\
+            v_2'
+            &= \frac{2 m_1 v_1 + (m_2 - m_1) v_2}{m_1 + m_2}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+To analyze this result,
+for practicality, we simplify it by setting $v_2 = 0$.
+In that case:
+
+$$\begin{aligned}
+    v_1'
+    = \frac{(m_1 - m_2) v_1}{m_1 + m_2}
+    \qquad \quad
+    v_2'
+    = \frac{2 m_1 v_1}{m_1 + m_2}
+\end{aligned}$$
+
+How much of its energy and momentum does object 1 transfer to object 2?
+The following ratios compare $v_1$ and $v_2'$ to quantify the transfer:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    = \frac{2 m_2}{m_1 + m_2}
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    = \frac{4 m_1 m_2}{(m_1 + m_2)^2}
+\end{aligned}$$
+
+If $m_1 = m_2$, both ratios reduce to $1$,
+meaning that all energy and momentum is transferred,
+and object 1 is at rest after the collision.
+Newton's cradle is an example of this.
+
+If $m_1 \ll m_2$, object 1 simply bounces off object 2,
+barely transferring any energy.
+Object 2 ends up with twice object 1's momentum,
+but $v_2'$ is very small and thus negligible:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    \approx 2
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    \approx \frac{4 m_1}{m_2}
+\end{aligned}$$
+
+If $m_1 \gg m_2$, object 1 barely notices the collision,
+so not much is transferred to object 2:
+
+$$\begin{aligned}
+    \frac{m_2 v_2'}{m_1 v_1}
+    \approx \frac{2 m_2}{m_1}
+    \qquad \quad
+    \frac{m_2 v_2'^2}{m_1 v_1^2}
+    \approx \frac{4 m_2}{m_1}
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md
new file mode 100644
index 0000000..c3e6dc0
--- /dev/null
+++ b/source/know/concept/electric-dipole-approximation/index.md
@@ -0,0 +1,165 @@
+---
+title: "Electric dipole approximation"
+date: 2021-09-14
+categories:
+- Physics
+- Quantum mechanics
+- Optics
+- Electromagnetism
+- Perturbation
+layout: "concept"
+---
+
+Suppose that an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+is travelling through an atom, and affecting the electrons.
+The general Hamiltonian of an electron in such a wave is given by:
+
+$$\begin{aligned}
+    \hat{H}
+    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi
+    \\
+    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
+\end{aligned}$$
+
+With charge $q = - e$,
+canonical momentum operator $\vu{P} = - i \hbar \nabla$,
+and magnetic vector potential $\vb{A}(\vb{x}, t)$.
+We reduce this by fixing the Coulomb gauge $\nabla \cdot \vb{A} = 0$,
+so that $\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$:
+
+$$\begin{aligned}
+    \comm{\vb{A}}{\vu{P}} \psi
+    &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
+    \\
+    &= i \hbar (\nabla \cdot \vb{A}) \psi
+    = 0
+\end{aligned}$$
+
+Where $\psi$ is an arbitrary test function.
+Assuming $\vb{A}$ is so small that $\vb{A}{}^2$ is negligible, we split $\hat{H}$ as follows,
+where $\hat{H}_1$ can be regarded as a perturbation to $\hat{H}_0$:
+
+$$\begin{aligned}
+    \hat{H}
+    = \hat{H}_0 + \hat{H}_1
+    \qquad \quad
+    \hat{H}_0
+    \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi
+    \qquad \quad
+    \hat{H}_1
+    \equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
+\end{aligned}$$
+
+In an electromagnetic wave, $\vb{A}$ is oscillating sinusoidally in time and space:
+
+$$\begin{aligned}
+    \vb{A}(\vb{x}, t) = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t)
+\end{aligned}$$
+
+Mathematically, it is more convenient to represent this with a complex exponential,
+whose real part should be taken at the end of the calculation:
+
+$$\begin{aligned}
+    \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
+\end{aligned}$$
+
+The corresponding perturbative [electric field](/know/concept/electric-field/) $\vb{E}$ is then given by:
+
+$$\begin{aligned}
+    \vb{E}(\vb{x}, t)
+    = - \pdv{\vb{A}}{t}
+    = \vb{E}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
+\end{aligned}$$
+
+Where $\vb{E}_0 = \omega \vb{A}_0$.
+Let us restrict ourselves to visible light,
+whose wavelength $2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$.
+Meanwhile, an atomic orbital is several Bohr $\sim 10^{-10} \:\mathrm{m}$,
+so $\vb{k} \cdot \vb{x}$ is negligible:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{E}(\vb{x}, t)
+        \approx \vb{E}_0 \exp(- i \omega t)
+    }
+\end{aligned}$$
+
+This is the **electric dipole approximation**:
+we ignore all spatial variation of $\vb{E}$,
+and only consider its temporal oscillation.
+Also, since we have not used the word "photon",
+we are implicitly treating the radiation classically,
+and the electron quantum-mechanically.
+
+Next, we want to rewrite $\hat{H}_1$
+to use the electric field $\vb{E}$ instead of the potential $\vb{A}$.
+To do so, we use that $\vu{P} = m \: \idv{\vu{x}}{t}$
+and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \vu{P}
+    = m \idv{\vu{x}}{t}
+    = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}}
+    = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
+\end{aligned}$$
+
+Taking the off-diagonal inner product with
+the two-level system's states $\Ket{1}$ and $\Ket{2}$ gives:
+
+$$\begin{aligned}
+    \matrixel{2}{\vu{P}}{1}
+    = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
+    = m i \omega_0 \matrixel{2}{\vu{x}}{1}
+\end{aligned}$$
+
+Therefore, $\vu{P} / m = i \omega_0 \vu{x}$,
+where $\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$ is the resonance of the energy gap,
+close to which we assume that $\vb{A}$ and $\vb{E}$ are oscillating, i.e. $\omega \approx \omega_0$.
+We thus get:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    &= - \frac{q}{m} \vu{P} \cdot \vb{A}
+    = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t)
+    \\
+    &\approx - q \vu{x} \cdot \vb{E}_0 \exp(- i \omega t)
+    = - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
+\end{aligned}$$
+
+Where $\vu{d} \equiv q \vu{x} = - e \vu{x}$ is
+the **transition dipole moment operator** of the electron,
+hence the name **electric dipole approximation**.
+Finally, we take the real part, yielding:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{H}_1(t)
+        = - \vu{d} \cdot \vb{E}(t)
+        = - q \vu{x} \cdot \vb{E}_0 \cos(\omega t)
+    }
+\end{aligned}$$
+
+If this approximation is too rough,
+$\vb{E}$ can always be Taylor-expanded in $(i \vb{k} \cdot \vb{x})$:
+
+$$\begin{aligned}
+    \vb{E}(\vb{x}, t)
+    = \vb{E}_0 \Big( 1 + (i \vb{k} \cdot \vb{x}) + \frac{1}{2} (i \vb{k} \cdot \vb{x})^2 + \: ... \Big) \exp(- i \omega t)
+\end{aligned}$$
+
+Taking the real part then yields the following series of higher-order correction terms:
+
+$$\begin{aligned}
+    \vb{E}(\vb{x}, t)
+    = \vb{E}_0 \Big( \cos(\omega t) + (\vb{k} \cdot \vb{x}) \sin(\omega t) - \frac{1}{2} (\vb{k} \cdot \vb{x})^2 \cos(\omega t) + \: ... \Big)
+\end{aligned}$$
+
+
+
+## References
+1.  M. Fox,
+    *Optical properties of solids*, 2nd edition,
+    Oxford.
+2.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/source/know/concept/electric-field/index.md b/source/know/concept/electric-field/index.md
new file mode 100644
index 0000000..eef8bd5
--- /dev/null
+++ b/source/know/concept/electric-field/index.md
@@ -0,0 +1,126 @@
+---
+title: "Electric field"
+date: 2021-07-12
+categories:
+- Physics
+- Electromagnetism
+layout: "concept"
+---
+
+The **electric field** $\vb{E}$ is a vector field
+that describes electric effects,
+and is defined as the field that correctly predicts
+the [Lorentz force](/know/concept/lorentz-force/)
+on a particle with electric charge $q$:
+
+$$\begin{aligned}
+    \vb{F}
+    = q \vb{E}
+\end{aligned}$$
+
+This definition implies that the direction of $\vb{E}$
+is from positive to negative charges,
+since opposite charges attracts and like charges repel.
+
+If two opposite point charges with magnitude $q$
+are observed from far away,
+they can be treated as a single object called a **dipole**,
+which has an **electric dipole moment** $\vb{p}$ defined like so,
+where $\vb{d}$ is the vector going from
+the negative to the positive charge (opposite direction of $\vb{E}$):
+
+$$\begin{aligned}
+    \vb{p} = q \vb{d}
+\end{aligned}$$
+
+Alternatively, for consistency with [magnetic fields](/know/concept/magnetic-field/),
+$\vb{p}$ can be defined from the aligning torque $\vb{\tau}$
+experienced by the dipole when placed in an $\vb{E}$-field.
+In other words, $\vb{p}$ satisfies:
+
+$$\begin{aligned}
+    \vb{\tau} = \vb{p} \times \vb{E}
+\end{aligned}$$
+
+Where $\vb{p}$ has units of $\mathrm{C m}$.
+The **polarization density** $\vb{P}$ is defined from $\vb{p}$,
+and roughly speaking represents the moments per unit volume:
+
+$$\begin{aligned}
+    \vb{P} \equiv \dv{\vb{p}}{V}
+    \:\:\iff\:\:
+    \vb{p} = \int_V \vb{P} \dd{V}
+\end{aligned}$$
+
+If $\vb{P}$ has the same magnitude and direction throughout the body,
+then this becomes $\vb{p} = \vb{P} V$, where $V$ is the volume.
+Therefore, $\vb{P}$ has units of $\mathrm{C / m^2}$.
+
+A nonzero $\vb{P}$ complicates things,
+since it contributes to the field and hence modifies $\vb{E}$.
+We thus define
+the "free" **displacement field** $\vb{D}$
+from the "bound" field $\vb{P}$
+and the "net" field $\vb{E}$:
+
+$$\begin{aligned}
+    \vb{D} \equiv \varepsilon_0 \vb{E} + \vb{P}
+    \:\:\iff\:\:
+    \vb{E} = \frac{1}{\varepsilon_0} (\vb{D} - \vb{P})
+\end{aligned}$$
+
+Where the **electric permittivity of free space** $\varepsilon_0$ is a known constant.
+It is important to point out some inconsistencies here:
+$\vb{D}$ and $\vb{P}$ contain a factor of $\varepsilon_0$,
+and therefore measure **flux density**,
+while $\vb{E}$ does not contain $\varepsilon_0$,
+and thus measures **field intensity**.
+Note that this convention is the opposite
+of the magnetic analogues $\vb{B}$, $\vb{H}$ and $\vb{M}$,
+and that $\vb{M}$ has the opposite sign of $\vb{P}$.
+
+The polarization $\vb{P}$ is a function of $\vb{E}$.
+In addition to the inherent polarity
+of the material $\vb{P}_0$ (zero in most cases),
+there is a (possibly nonlinear) response
+to the applied $\vb{E}$-field:
+
+$$\begin{aligned}
+    \vb{P} =
+    \vb{P}_0 + \varepsilon_0 \chi_e^{(1)} \vb{E}
+    + \varepsilon_0 \chi_e^{(2)} |\vb{E}| \: \vb{E}
+    + \varepsilon_0 \chi_e^{(3)} |\vb{E}|^2 \: \vb{E} + ...
+\end{aligned}$$
+
+Where the $\chi_e^{(n)}$ are the **electric susceptibilities** of the medium.
+For simplicity, we often assume that only the $n\!=\!1$ term is nonzero,
+which is the linear response to $\vb{E}$.
+In that case, we define the **absolute permittivity** $\varepsilon$ so that:
+
+$$\begin{aligned}
+    \vb{D}
+    = \varepsilon_0 \vb{E} + \vb{P}
+    = \varepsilon_0 \vb{E} + \varepsilon_0 \chi_e^{(1)} \vb{E}
+    = \varepsilon_0 \varepsilon_r \vb{E}
+    = \varepsilon \vb{E}
+\end{aligned}$$
+
+I.e. $\varepsilon \equiv \varepsilon_r \varepsilon_0$,
+where $\varepsilon_r \equiv 1 + \chi_e^{(1)}$ is
+the [**dielectric function**](/know/concept/dielectric-function/)
+or **relative permittivity**,
+whose calculation is of great interest in physics.
+
+In reality, a material cannot respond instantly to $\vb{E}$,
+meaning that $\chi_e^{(1)}$ is a function of time,
+and that $\vb{P}$ is the convolution of $\chi_e^{(1)}(t)$ and $\vb{E}(t)$:
+
+$$\begin{aligned}
+    \vb{P}(t)
+    = \varepsilon_0 \big(\chi_e^{(1)} * \vb{E}\big)(t)
+    = \varepsilon_0 \int_{-\infty}^\infty \chi_e^{(1)}(t - \tau) \: \vb{E}(\tau) \:d\tau
+\end{aligned}$$
+
+Note that this definition requires $\chi_e^{(1)}(t) = 0$ for $t < 0$
+in order to ensure causality,
+which leads to the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations/).
diff --git a/source/know/concept/electromagnetic-wave-equation/index.md b/source/know/concept/electromagnetic-wave-equation/index.md
new file mode 100644
index 0000000..e6f6677
--- /dev/null
+++ b/source/know/concept/electromagnetic-wave-equation/index.md
@@ -0,0 +1,246 @@
+---
+title: "Electromagnetic wave equation"
+date: 2021-09-09
+categories:
+- Physics
+- Electromagnetism
+- Optics
+layout: "concept"
+---
+
+The electromagnetic wave equation describes
+the propagation of light through various media.
+Since an electromagnetic (light) wave consists of
+an [electric field](/know/concept/electric-field/)
+and a [magnetic field](/know/concept/magnetic-field/),
+we need [Maxwell's equations](/know/concept/maxwells-equations/)
+in order to derive the wave equation.
+
+
+## Uniform medium
+
+We will use all of Maxwell's equations,
+but we start with Ampère's circuital law for the "free" fields $\vb{H}$ and $\vb{D}$,
+in the absence of a free current $\vb{J}_\mathrm{free} = 0$:
+
+$$\begin{aligned}
+    \nabla \cross \vb{H}
+    = \pdv{\vb{D}}{t}
+\end{aligned}$$
+
+We assume that the medium is isotropic, linear,
+and uniform in all of space, such that:
+
+$$\begin{aligned}
+    \vb{D} = \varepsilon_0 \varepsilon_r \vb{E}
+    \qquad \quad
+    \vb{H} = \frac{1}{\mu_0 \mu_r} \vb{B}
+\end{aligned}$$
+
+Which, upon insertion into Ampère's law,
+yields an equation relating $\vb{B}$ and $\vb{E}$.
+This may seem to contradict Ampère's "total" law,
+but keep in mind that $\vb{J}_\mathrm{bound} \neq 0$ here:
+
+$$\begin{aligned}
+    \nabla \cross \vb{B}
+    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{\vb{E}}{t}
+\end{aligned}$$
+
+Now we take the curl, rearrange,
+and substitute $\nabla \cross \vb{E}$ according to Faraday's law:
+
+$$\begin{aligned}
+    \nabla \cross (\nabla \cross \vb{B})
+    = \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdv{}{t}(\nabla \cross \vb{E})
+    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t}
+\end{aligned}$$
+
+Using a vector identity, we rewrite the leftmost expression,
+which can then be reduced thanks to Gauss' law for magnetism $\nabla \cdot \vb{B} = 0$:
+
+$$\begin{aligned}
+    - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{B}}{t}
+    &= \nabla (\nabla \cdot \vb{B}) - \nabla^2 \vb{B}
+    = - \nabla^2 \vb{B}
+\end{aligned}$$
+
+This describes $\vb{B}$.
+Next, we repeat the process for $\vb{E}$:
+taking the curl of Faraday's law yields:
+
+$$\begin{aligned}
+    \nabla \cross (\nabla \cross \vb{E})
+    = - \pdv{}{t}(\nabla \cross \vb{B})
+    = - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+\end{aligned}$$
+
+Which can be rewritten using same vector identity as before,
+and then reduced by assuming that there is no net charge density $\rho = 0$
+in Gauss' law, such that $\nabla \cdot \vb{E} = 0$:
+
+$$\begin{aligned}
+    - \mu_0 \mu_r \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+    &= \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E}
+    = - \nabla^2 \vb{E}
+\end{aligned}$$
+
+We thus arrive at the following two (implicitly coupled)
+wave equations for $\vb{E}$ and $\vb{B}$,
+where we have defined the phase velocity $v \equiv 1 / \sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}$:
+
+$$\begin{aligned}
+    \boxed{
+        \pdvn{2}{\vb{E}}{t} - \frac{1}{v^2} \nabla^2 \vb{E}
+        = 0
+    }
+    \qquad \quad
+    \boxed{
+        \pdvn{2}{\vb{B}}{t} - \frac{1}{v^2} \nabla^2 \vb{B}
+        = 0
+    }
+\end{aligned}$$
+
+Traditionally, it is said that the solutions are as follows,
+where the wavenumber $|\vb{k}| = \omega / v$:
+
+$$\begin{aligned}
+    \vb{E}(\vb{r}, t)
+    &= \vb{E}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t)
+    \\
+    \vb{B}(\vb{r}, t)
+    &= \vb{B}_0 \exp(i \vb{k} \cdot \vb{r} - i \omega t)
+\end{aligned}$$
+
+In fact, thanks to linearity, these **plane waves** can be treated as
+terms in a Fourier series, meaning that virtually
+*any* function $f(\vb{k} \cdot \vb{r} - \omega t)$ is a valid solution.
+
+Keep in mind that in reality $\vb{E}$ and $\vb{B}$ are real,
+so although it is mathematically convenient to use plane waves,
+in the end you will need to take the real part.
+
+
+## Non-uniform medium
+
+A useful generalization is to allow spatial change
+in the relative permittivity $\varepsilon_r(\vb{r})$
+and the relative permeability $\mu_r(\vb{r})$.
+We still assume that the medium is linear and isotropic, so:
+
+$$\begin{aligned}
+    \vb{D}
+    = \varepsilon_0 \varepsilon_r(\vb{r}) \vb{E}
+    \qquad \quad
+    \vb{B}
+    = \mu_0 \mu_r(\vb{r}) \vb{H}
+\end{aligned}$$
+
+Inserting these expressions into Faraday's and Ampère's laws
+respectively yields:
+
+$$\begin{aligned}
+    \nabla \cross \vb{E}
+    = - \mu_0 \mu_r(\vb{r}) \pdv{\vb{H}}{t}
+    \qquad \quad
+    \nabla \cross \vb{H}
+    = \varepsilon_0 \varepsilon_r(\vb{r}) \pdv{\vb{E}}{t}
+\end{aligned}$$
+
+We then divide Ampère's law by $\varepsilon_r(\vb{r})$,
+take the curl, and substitute Faraday's law, giving:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big)
+    = \varepsilon_0 \pdv{}{t}(\nabla \cross \vb{E})
+    = - \mu_0 \mu_r \varepsilon_0 \pdvn{2}{\vb{H}}{t}
+\end{aligned}$$
+
+Next, we exploit linearity by decomposing $\vb{H}$ and $\vb{E}$
+into Fourier series, with terms given by:
+
+$$\begin{aligned}
+    \vb{H}(\vb{r}, t)
+    = \vb{H}(\vb{r}) \exp(- i \omega t)
+    \qquad \quad
+    \vb{E}(\vb{r}, t)
+    = \vb{E}(\vb{r}) \exp(- i \omega t)
+\end{aligned}$$
+
+By inserting this ansatz into the equation,
+we can remove the explicit time dependence:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\varepsilon_r} \nabla \cross \vb{H} \Big) \exp(- i \omega t)
+    = \mu_0 \varepsilon_0 \omega^2 \mu_r \vb{H} \exp(- i \omega t)
+\end{aligned}$$
+
+Dividing out $\exp(- i \omega t)$,
+we arrive at an eigenvalue problem for $\omega^2$,
+with $c = 1 / \sqrt{\mu_0 \varepsilon_0}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \Big( \frac{1}{\varepsilon_r(\vb{r})} \nabla \cross \vb{H}(\vb{r}) \Big)
+        = \Big( \frac{\omega}{c} \Big)^2 \mu_r(\vb{r}) \vb{H}(\vb{r})
+    }
+\end{aligned}$$
+
+Compared to a uniform medium, $\omega$ is often not arbitrary here:
+there are discrete eigenvalues $\omega$,
+corresponding to discrete **modes** $\vb{H}(\vb{r})$.
+
+Next, we go through the same process to find an equation for $\vb{E}$.
+Starting from Faraday's law, we divide by $\mu_r(\vb{r})$,
+take the curl, and insert Ampère's law:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big)
+    = - \mu_0 \pdv{}{t}(\nabla \cross \vb{H})
+    = - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+\end{aligned}$$
+
+Then, by replacing $\vb{E}(\vb{r}, t)$ with our plane-wave ansatz,
+we remove the time dependence:
+
+$$\begin{aligned}
+    \nabla \cross \Big( \frac{1}{\mu_r} \nabla \cross \vb{E} \Big) \exp(- i \omega t)
+    = - \mu_0 \varepsilon_0 \omega^2 \varepsilon_r \vb{E} \exp(- i \omega t)
+\end{aligned}$$
+
+Which, after dividing out $\exp(- i \omega t)$,
+yields an analogous eigenvalue problem with $\vb{E}(r)$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \Big( \frac{1}{\mu_r(\vb{r})} \nabla \cross \vb{E}(\vb{r}) \Big)
+        = \Big( \frac{\omega}{c} \Big)^2 \varepsilon_r(\vb{r}) \vb{E}(\vb{r})
+    }
+\end{aligned}$$
+
+Usually, it is a reasonable approximation
+to say $\mu_r(\vb{r}) = 1$,
+in which case the equation for $\vb{H}(\vb{r})$
+becomes a Hermitian eigenvalue problem,
+and is thus easier to solve than for $\vb{E}(\vb{r})$.
+
+Keep in mind, however, that in any case,
+the solutions $\vb{H}(\vb{r})$ and/or $\vb{E}(\vb{r})$
+must satisfy the two Maxwell's equations that were not explicitly used:
+
+$$\begin{aligned}
+    \nabla \cdot (\varepsilon_r \vb{E}) = 0
+    \qquad \quad
+    \nabla \cdot (\mu_r \vb{H}) = 0
+\end{aligned}$$
+
+This is equivalent to demanding that the resulting waves are *transverse*,
+or in other words,
+the wavevector $\vb{k}$ must be perpendicular to
+the amplitudes $\vb{H}_0$ and $\vb{E}_0$.
+
+
+## References
+1.  J.D. Joannopoulos, S.G. Johnson, J.N. Winn, R.D. Meade,
+    *Photonic crystals: molding the flow of light*,
+    2nd edition, Princeton.
diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/equation-of-motion-theory/index.md
@@ -0,0 +1,197 @@
+---
+title: "Equation-of-motion theory"
+date: 2021-11-08
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In many-body quantum theory, **equation-of-motion theory**
+is a method to calculate the time evolution of a system's properties
+using [Green's functions](/know/concept/greens-functions/).
+
+Starting from the definition of
+the retarded single-particle Green's function $G_{\nu \nu'}^R(t, t')$,
+we simply take the $t$-derivative
+(we could do the same with the advanced function $G_{\nu \nu'}^A$):
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R_{\nu \nu'}(t, t')}{t}
+    &= \pdv{\Theta(t \!-\! t')}{t} \Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+    + \Theta(t \!-\! t') \pdv{}{t}\Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+    \\
+    &= \delta(t \!-\! t') \Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+    + \Theta(t \!-\! t') \Expval{\Comm{\dv{\hat{c}_\nu(t)}{t}}{\hat{c}_{\nu'}^\dagger(t)}_{\mp}}
+\end{aligned}$$
+
+Where we have used that the derivative
+of a [Heaviside step function](/know/concept/heaviside-step-function/) $\Theta$
+is a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$.
+Also, from the [second quantization](/know/concept/second-quantization/),
+$\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$
+for $t = t'$ is zero when $\nu \neq \nu'$.
+
+Since we are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+we know the equation of motion of $\hat{c}_\nu(t)$:
+
+$$\begin{aligned}
+    \dv{\hat{c}_\nu(t)}{t}
+    = \frac{i}{\hbar} \comm{\hat{H}_0(t)}{\hat{c}_\nu(t)} + \frac{i}{\hbar} \comm{\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}
+\end{aligned}$$
+
+Where the single-particle part of the Hamiltonian $\hat{H}_0$
+and the interaction part $\hat{H}_\mathrm{int}$
+are assumed to be time-independent in the Schrödinger picture.
+We thus get:
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R_{\nu \nu'}}{t}
+    &= \delta_{\nu \nu'} \delta(t \!-\! t')+ \frac{i}{\hbar} \Theta(t \!-\! t')
+    \Expval{\Comm{\comm{\hat{H}_0}{\hat{c}_\nu} + \comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
+\end{aligned}$$
+
+The most general form of $\hat{H}_0$, for any basis,
+is as follows, where $u_{\nu' \nu''}$ are constants:
+
+$$\begin{aligned}
+    \hat{H}_0
+    = \sum_{\nu' \nu''} u_{\nu' \nu''} \hat{c}_{\nu'}^\dagger \hat{c}_{\nu''}
+    \quad \implies \quad
+    \comm{\hat{H}_0}{\hat{c}_\nu}
+    = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-commH0"/>
+<label for="proof-commH0">Proof</label>
+<div class="hidden">
+<label for="proof-commH0">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \comm{\hat{H}_0}{\hat{f}_{\!\nu}}
+    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
+    + \comm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
+    \\
+    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
+    - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''}
+    + \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''}
+    - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big)
+    \\
+    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''}
+    - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
+    = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
+\end{aligned}$$
+</div>
+</div>
+
+Substituting this into $G_{\nu \nu'}^R$'s equation of motion,
+we recognize another Green's function $G_{\nu'' \nu'}^R$:
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R_{\nu \nu'}}{t}
+    &= \delta_{\nu \nu'} \delta(t \!-\! t') + \frac{i}{\hbar} \Theta(t \!-\! t')
+    \bigg( \Expval{\comm{\comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
+    - \sum_{\nu''} u_{\nu \nu''} \Expval{\comm{\hat{c}_{\nu''}}{\hat{c}_{\nu'}^\dagger}_{\mp}} \bigg)
+    \\
+    &= \delta_{\nu \nu'} \delta(t \!-\! t')
+    + \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\comm{\comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}}
+    + \sum_{\nu''} u_{\nu \nu''} G_{\nu''\nu'}^R(t, t')
+\end{aligned}$$
+
+Rearranging this as follows yields the main result
+of equation-of-motion theory:
+
+$$\begin{aligned}
+    \boxed{
+        \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t, t')
+        = \delta_{\nu \nu'} \delta(t \!-\! t') + D_{\nu \nu'}^R(t, t')
+    }
+\end{aligned}$$
+
+Where $D_{\nu \nu'}^R$ represents a correction due to interactions $\hat{H}_\mathrm{int}$,
+and also has the form of a retarded Green's function,
+but with $\hat{c}_{\nu}$ replaced by $\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$:
+
+$$\begin{aligned}
+    \boxed{
+        D^R_{\nu'' \nu'}(t, t')
+        \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\comm{\comm{-\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+    }
+\end{aligned}$$
+
+Unfortunately, calculating $D_{\nu \nu'}^R$
+might still not be doable due to $\hat{H}_\mathrm{int}$.
+The key idea of equation-of-motion theory is to either approximate $D_{\nu \nu'}^R$ now,
+or to differentiate it again $i \hbar \idv{D_{\nu \nu'}^R}{t}$,
+and try again for the resulting corrections,
+until a solvable equation is found.
+There is no guarantee that that will ever happen;
+if not, one of the corrections needs to be approximated.
+
+For non-interacting particles $\hat{H}_\mathrm{int} = 0$,
+so clearly $D_{\nu \nu'}^R$ trivially vanishes then.
+Let us assume that $\hat{H}_0$ is also time-independent,
+such that $G_{\nu'' \nu'}^R$ only depends on the difference $t - t'$:
+
+$$\begin{aligned}
+        \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t - t')
+        = \delta_{\nu \nu'} \delta(t - t')
+\end{aligned}$$
+
+We take the [Fourier transform](/know/concept/fourier-transform/)
+$(t \!-\! t') \to (\omega + i \eta)$, where $\eta \to 0^+$ ensures convergence:
+
+$$\begin{aligned}
+        \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
+        = \delta_{\nu \nu'}
+\end{aligned}$$
+
+If we assume a diagonal basis $u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$,
+this reduces to the following:
+
+$$\begin{aligned}
+        \delta_{\nu \nu'}
+        &= \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - \varepsilon_\nu \delta_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
+        \\
+        &= \Big( \hbar (\omega + i \eta) - \varepsilon_\nu \Big) G^R_{\nu \nu'}(\omega)
+\end{aligned}$$
+
+For a non-interacting, time-independent Hamiltonian,
+we therefore arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        G^R_{\nu \nu'}(\omega)
+        = \frac{\delta_{\nu \nu'}}{\hbar (\omega + i \eta) - \varepsilon_\nu}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/euler-bernoulli-law/index.md b/source/know/concept/euler-bernoulli-law/index.md
new file mode 100644
index 0000000..37212d2
--- /dev/null
+++ b/source/know/concept/euler-bernoulli-law/index.md
@@ -0,0 +1,308 @@
+---
+title: "Euler-Bernoulli law"
+date: 2021-06-03
+categories:
+- Physics
+layout: "concept"
+---
+
+**Euler-Bernoulli beam theory** concerns itself with the bending of beams
+(e.g. the metal beams used in large buildings),
+subject to certain simplifying assumptions,
+which are generally valid for beams that are narrow,
+i.e. longitudinally much larger than transversely.
+
+Consider a beam of length $L$, placed upright
+on the $z = 0$ plane, above the origin.
+If we pull the top of this beam in the postive $y$-direction,
+we assume that it bends uniformly,
+i.e. with constant radius of [curvature](/know/concept/curvature/) $R$.
+We also assume that the bending is **shear-free**:
+if we treat the beam as a bundle of elastic strings,
+then there is no friction between them.
+
+The central string has its length unchanged (i.e. still $L$),
+while an arbitrary non-central string is extended or compressed to $L'$.
+The [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) element $u_{zz}$ is then:
+
+$$\begin{aligned}
+    u_{zz}
+    = \frac{L' - L}{L}
+\end{aligned}$$
+
+Because the bending is uniform, the central string
+is an arc with radius $R$ and central angle $\theta$,
+where $L = \theta R$.
+The non-central string has $L' = \theta R'$,
+where $R'$ is geometrically shown to be $R' = R - y$,
+with $y$ being the $y$-coordinate of that string at the beam's base.
+So:
+
+$$\begin{aligned}
+    u_{zz}
+    = \frac{R' - R}{R}
+    = - \frac{y}{R}
+\end{aligned}$$
+
+By assumption, there are no shear stresses
+and no forces acting on the beam's sides,
+so the only nonzero component of the
+[Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
+is $\sigma_{zz}$, given by [Hooke's law](/know/concept/hookes-law/):
+
+$$\begin{aligned}
+    \sigma_{zz} = E u_{zz}
+\end{aligned}$$
+
+Where $E$ is the elastic modulus of the material.
+By Hooke's inverse law,
+the other nonzero strain components are as follows,
+where $\nu$ is Poisson's ratio:
+
+$$\begin{aligned}
+    u_{xx}
+    = u_{yy}
+    = - \frac{\nu}{E} \sigma_{zz}
+    = - \nu u_{zz}
+    = \nu \frac{y}{R}
+\end{aligned}$$
+
+For completeness, we turn the strain tensor $\hat{u}$
+into a full displacement field $\va{u}$:
+
+$$\begin{aligned}
+    \boxed{
+        u_x = \nu \frac{x y}{R}
+        \qquad
+        u_y = \frac{z^2}{2 R} + \nu \frac{y^2 - x^2}{2 R}
+        \qquad
+        u_z = - \frac{y z}{R}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-field"/>
+<label for="proof-field">Proof</label>
+<div class="hidden">
+<label for="proof-field">Proof.</label>
+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.
+</div>
+</div>
+
+In any case, the beam experiences a bending torque with an $x$-component $T_x$ given by:
+
+$$\begin{aligned}
+    T_x
+    = - \int_A y \sigma_{zz} \dd{A}
+    = - \frac{E}{R} \int_A y^2 \dd{A}
+\end{aligned}$$
+
+Where $A$ is the cross-section.
+Th above integral is known as the **area moment**,
+and is typically abbreviated by $I$.
+This brings us to the **Euler-Bernoulli law**:
+
+$$\begin{aligned}
+    \boxed{
+        T_x
+        = - \frac{E I}{R}
+    }
+    \qquad \quad
+    I
+    \equiv \int_A y^2 \dd{A}
+\end{aligned}$$
+
+The product $E I$ is called the **flexural rigidity**,
+i.e. the beam's "stiffness".
+For a small deformation, i.e. a large radius of curvature $R$,
+the law can be approximated by:
+
+$$\begin{aligned}
+    T_x
+    \approx - E I \dvn{2}{y}{z}
+\end{aligned}$$
+
+
+
+## Slender rods
+
+A beam that is very thin in the transverse directions ($x$ and $y$ in this case),
+can be approximated as a single string or rod $y(z)$.
+Each infinitesimal piece $(\dd{y}, \dd{z})$ of the rod
+exerts forces $F_y$ and $F_z$ on the next piece,
+and is feels external forces-per-length $K_y$ and $K_z$, e.g. gravity.
+In order to have equilibrium, the total force must be zero:
+
+$$\begin{aligned}
+    0
+    &= F_y(z + \dd{z}) - F_y(z) + K_y(z) \dd{z}
+    \\
+    0
+    &= F_z(z + \dd{z}) - F_z(z) + K_z(z) \dd{z}
+\end{aligned}$$
+
+Rearranging these relations yields these equations for the internal forces $F_y$ and $F_z$:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{F_y}{z}
+        = - K_y
+    }
+    \qquad \quad
+    \boxed{
+        \dv{F_z}{z}
+        = - K_z
+    }
+\end{aligned}$$
+
+Meanwhile, the rod also feels a torque with $x$-component $T_x$,
+where equilibrium entails:
+
+$$\begin{aligned}
+    0
+    = T_x(z + \dd{z}) - T_x(z) + F_z(z) \dd{y} - F_y(z) \dd{z}
+\end{aligned}$$
+
+This can be rearranged to get a differential equation for $T_x$, namely:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{T_x}{z}
+        = F_y - F_z \dv{y}{z}
+    }
+\end{aligned}$$
+
+If $F_z$ and $\idv{y}{z}$ are small, the last term can be dropped.
+These equations are widely applicable,
+but there is one especially important application,
+so much so that it is usually what is meant by "Euler-Bernoulli law":
+the shape of a laterally loaded rod.
+
+Consider a beam along the $z$-axis, carrying a lateral load $K_y$,
+e.g. its own weight $A g \rho$ or more.
+Assuming there is no other load $K_z = 0$
+and $F_y \ll F_z$, the above equations become:
+
+$$\begin{aligned}
+    T_x
+    = - E I \dvn{2}{y}{z}
+    \qquad
+    \dv{T_x}{z}
+    = F_y
+    \qquad
+    \dv{F_y}{z}
+    = - K_y
+\end{aligned}$$
+
+Which we can simply substitute into each other,
+eventually leading to:
+
+$$\begin{aligned}
+    \boxed{
+        K_y
+        = \dvn{2}{}{z}\Big( E I \dvn{2}{y}{z} \Big)
+    }
+\end{aligned}$$
+
+This is often referred to as the **Euler-Bernoulli law** as well.
+Typically the flexural rigidity $EI$ is a constant in $z$,
+in which case we can reduce the equation to:
+
+$$\begin{aligned}
+    K_y
+    = E I \dvn{4}{y}{z}
+\end{aligned}$$
+
+Which is clearly solved by a fourth-order polynomial,
+given some boundary conditions.
+
+
+
+## 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/euler-equations/index.md b/source/know/concept/euler-equations/index.md
new file mode 100644
index 0000000..f89d577
--- /dev/null
+++ b/source/know/concept/euler-equations/index.md
@@ -0,0 +1,182 @@
+---
+title: "Euler equations"
+date: 2021-03-31
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+The **Euler equations** are a system of partial differential equations
+that govern the movement of **ideal fluids**,
+i.e. fluids without viscosity.
+There exist several forms, depending on
+the surrounding assumptions about the fluid.
+
+
+## Incompressible fluid
+
+In a fluid moving according to the velocity vield $\va{v}(\va{r}, t)$,
+the acceleration felt by a particle is given by
+the **material acceleration field** $\va{w}(\va{r}, t)$,
+which is the [material derivative](/know/concept/material-derivative/) of $\va{v}$:
+
+$$\begin{aligned}
+    \va{w}
+    \equiv \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
+\end{aligned}$$
+
+This infinitesimal particle obeys Newton's second law,
+which can be written as follows:
+
+$$\begin{aligned}
+    \va{w} \dd{m}
+    = \va{w} \rho \dd{V}
+    = \va{f^*} \dd{V}
+\end{aligned}$$
+
+Where $\dd{m}$ and $\dd{V}$ are the particle's mass volume,
+and $\rho$ is the fluid density, which we assume, in this case, to be constant in space and time.
+Then the **effective force density** $\va{f^*}$ represents the net force-per-particle.
+By dividing the law by $\dd{V}$, we find:
+
+$$\begin{aligned}
+    \rho \va{w}
+    = \va{f^*}
+\end{aligned}$$
+
+Next, we want to find another expression for $\va{f^*}$.
+We know that the overall force $\va{F}$ on an arbitrary volume $V$ of the fluid
+is the sum of the gravity body force $\va{F}_g$,
+and the pressure contact force $\va{F}_p$ on the enclosing surface $S$.
+Using the divergence theorem, we then find:
+
+$$\begin{aligned}
+    \va{F}
+    = \va{F}_g + \va{F}_p
+    = \int_V \rho \va{g} \dd{V} - \oint_S p \dd{\va{S}}
+    = \int_V (\rho \va{g} - \nabla p) \dd{V}
+    = \int_V \va{f^*} \dd{V}
+\end{aligned}$$
+
+Where $p(\va{r}, t)$ is the pressure field,
+and $\va{g}(\va{r}, t)$ is the gravitational acceleration field.
+Combining this with Newton's law, we find the following equation for the force density:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \rho \va{w}
+    = \rho \va{g} - \nabla p
+\end{aligned}$$
+
+Dividing this by $\rho$,
+we get the first of the system of Euler equations:
+
+$$\begin{aligned}
+    \boxed{
+        \va{w}
+        = \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho}
+    }
+\end{aligned}$$
+
+The last ingredient is **incompressibility**:
+the same volume must simultaneously
+be flowing in and out of an arbitrary enclosure $S$.
+Then, by the divergence theorem:
+
+$$\begin{aligned}
+    0
+    = \oint_S \va{v} \cdot \dd{\va{S}}
+    = \int_V \nabla \cdot \va{v} \dd{V}
+\end{aligned}$$
+
+Since $S$ and $V$ are arbitrary,
+the integrand must vanish by itself everywhere:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \va{v} = 0
+    }
+\end{aligned}$$
+
+Combining this with the equation for $\va{w}$,
+we get a system of two coupled differential equations:
+these are the Euler equations for an incompressible fluid
+with spatially uniform density $\rho$:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho}
+        \qquad \quad
+        \nabla \cdot \va{v}
+        = 0
+    }
+\end{aligned}$$
+
+The above form is straightforward to generalize to incompressible fluids
+with non-uniform spatial densities $\rho(\va{r}, t)$.
+In other words, these fluids are "lumpy" (variable density),
+but the size of their lumps does not change (incompressibility).
+
+To update the equations, we demand conservation of mass:
+the mass evolution of a volume $V$
+is equal to the mass flow through its boundary $S$.
+Applying the divergence theorem again:
+
+$$\begin{aligned}
+    0
+    = \dv{}{t}\int_V \rho \dd{V} + \oint_S \rho \va{v} \cdot \dd{\va{S}}
+    = \int_V \dv{\rho}{t} + \nabla \cdot (\rho \va{v}) \dd{V}
+\end{aligned}$$
+
+Since $V$ is arbitrary, the integrand must be zero.
+This leads to the following **continuity equation**,
+to which we apply a vector identity:
+
+$$\begin{aligned}
+    0
+    = \dv{\rho}{t} + \nabla \cdot (\rho \va{v})
+    = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho + \rho (\nabla \cdot \va{v})
+\end{aligned}$$
+
+Thanks to incompressibility, the last term disappears,
+leaving us with a material derivative:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \frac{\mathrm{D} \rho}{\mathrm{D} t}
+        = \dv{\rho}{t} + (\va{v} \cdot \nabla) \rho
+    }
+\end{aligned}$$
+
+Putting everything together, Euler's system of equations
+now takes the following form:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho}
+        \qquad
+        \nabla \cdot \va{v}
+        = 0
+        \qquad
+        \frac{\mathrm{D} \rho}{\mathrm{D} t}
+        = 0
+    }
+\end{aligned}$$
+
+Usually, however, when discussing incompressible fluids,
+$\rho$ is assumed to be spatially uniform,
+in which case the latter equation is trivially satisfied.
+
+
+
+## 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/fabry-perot-cavity/cavity.png b/source/know/concept/fabry-perot-cavity/cavity.png
new file mode 100644
index 0000000..f7b0c2a
--- /dev/null
+++ b/source/know/concept/fabry-perot-cavity/cavity.png
diff --git a/source/know/concept/fabry-perot-cavity/index.md b/source/know/concept/fabry-perot-cavity/index.md
new file mode 100644
index 0000000..e2852d4
--- /dev/null
+++ b/source/know/concept/fabry-perot-cavity/index.md
@@ -0,0 +1,233 @@
+---
+title: "Fabry-Pérot cavity"
+date: 2021-09-18
+categories:
+- Physics
+- Optics
+- Laser theory
+layout: "concept"
+---
+
+In its simplest form, a **Fabry-Pérot cavity**
+is a region of light-transmitting medium surrounded by two mirrors,
+which may transmit some of the incoming light.
+Such a setup can be used as e.g. an interferometer or a laser cavity.
+
+Below, we calculate its quasinormal modes in 1D.
+We divide the $x$-axis into three domains: left $L$, center $C$, and right $R$.
+The cavity $C$ has length $\ell$ and is centered on $x = 0$.
+Let $n_L$, $n_C$ and $n_R$ be the respective domains' refractive indices:
+
+<a href="cavity.png">
+<img src="cavity.png" style="width:70%">
+</a>
+
+
+## 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
index 0000000..7fc4f0e
--- /dev/null
+++ b/source/know/concept/fermi-dirac-distribution/index.md
@@ -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
new file mode 100644
index 0000000..f64e9df
--- /dev/null
+++ b/source/know/concept/fermis-golden-rule/index.md
@@ -0,0 +1,86 @@
+---
+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.
diff --git a/source/know/concept/feynman-diagram/conservation.png b/source/know/concept/feynman-diagram/conservation.png
new file mode 100644
index 0000000..1d19fd0
--- /dev/null
+++ b/source/know/concept/feynman-diagram/conservation.png
diff --git a/source/know/concept/feynman-diagram/freegf.png b/source/know/concept/feynman-diagram/freegf.png
new file mode 100644
index 0000000..632ccaf
--- /dev/null
+++ b/source/know/concept/feynman-diagram/freegf.png
diff --git a/source/know/concept/feynman-diagram/fullgf.png b/source/know/concept/feynman-diagram/fullgf.png
new file mode 100644
index 0000000..f376ef2
--- /dev/null
+++ b/source/know/concept/feynman-diagram/fullgf.png
diff --git a/source/know/concept/feynman-diagram/index.md b/source/know/concept/feynman-diagram/index.md
new file mode 100644
index 0000000..15410f7
--- /dev/null
+++ b/source/know/concept/feynman-diagram/index.md
@@ -0,0 +1,332 @@
+---
+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/):
+
+<a href="freegf.png">
+<img src="freegf.png" style="width:60%">
+</a>
+$$\begin{aligned}
+    = i \hbar G_{s_2 s_1}^0(\vb{r}_2, t_2; \vb{r}_1, t_1)
+    = \Expval{\mathcal{T} \Big\{ \hat{\Psi}_{s_2 I}(\vb{r}_2, t_2) \hat{\Psi}_{s_1 I}^\dagger(\vb{r}_1, t_1) \Big\}}
+\end{aligned}$$
+
+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/):
+
+<a href="fullgf.png">
+<img src="fullgf.png" style="width:60%">
+</a>
+$$\begin{aligned}
+    = i \hbar G_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
+    = \Expval{\mathcal{T} \Big\{ \hat{\Psi}_{s_2 H}(\vb{r}_2, t_2) \hat{\Psi}_{s_1 H}^\dagger(\vb{r}_1, t_1) \Big\}}
+\end{aligned}$$
+
+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:
+
+<a href="interaction.png">
+<img src="interaction.png" style="width:60%">
+</a>
+$$\begin{aligned}
+    = \frac{1}{i \hbar} W_{s_2 s_1}(\vb{r}_2, t_2; \vb{r}_1, t_1)
+    = \frac{1}{i \hbar} W(\vb{r}_2, \vb{r}_1; t_1) \: \delta(t_2 - t_1)
+\end{aligned}$$
+
+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:
+
+<a href="perturbation.png">
+<img src="perturbation.png" style="width:35%">
+</a>
+$$\begin{aligned}
+    = \frac{1}{i \hbar} V_s(\vb{r}, t)
+\end{aligned}$$
+
+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 $\tilde{\vb{k}} = (\vb{k}, \omega)$:
+
+<a href="conservation.png">
+<img src="conservation.png" style="width:40%">
+</a>
+$$\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.
diff --git a/source/know/concept/feynman-diagram/interaction.png b/source/know/concept/feynman-diagram/interaction.png
new file mode 100644
index 0000000..3c2d093
--- /dev/null
+++ b/source/know/concept/feynman-diagram/interaction.png
diff --git a/source/know/concept/feynman-diagram/perturbation.png b/source/know/concept/feynman-diagram/perturbation.png
new file mode 100644
index 0000000..2e8fbf6
--- /dev/null
+++ b/source/know/concept/feynman-diagram/perturbation.png
diff --git a/source/know/concept/ficks-laws/index.md b/source/know/concept/ficks-laws/index.md
new file mode 100644
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--- /dev/null
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@@ -0,0 +1,165 @@
+---
+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
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/fourier-transform/index.md
@@ -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:
+
+$$\begin{aligned}
+    \boxed{
+        A B
+        = \bigg( \frac{|s|}{2 \pi} \bigg)^{\!N}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-constants-ND"/>
+<label for="proof-constants-ND">Proof</label>
+<div class="hidden">
+<label for="proof-constants-ND">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{ f(\vb{x}) \}\}
+    &= (2 \pi)^N A B \int f(\vb{x}')
+    \Big( \prod_{n = 1}^N \delta(s(x_n' - x_n)) \Big) \ddn{N}{\vb{x}'}
+    \\
+    &= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
+    = \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
+\end{aligned}$$
+</div>
+</div>
+
+Differentiation is more complicated for $N > 1$,
+but the FT is still useful,
+notably for the Laplacian $\nabla^2 \equiv \idv{ {}^2}{x_1^2} + ... + \idv{ {}^2}{x_N^2}$.
+Let $|\vb{k}|$ be the norm of $\vb{k}$,
+then for a localized $f$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{F}}\{\nabla^2 f(\vb{x})\}
+        = - s^2 |\vb{k}|^2 \tilde{f}(\vb{k})
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-laplacian"/>
+<label for="proof-laplacian">Proof</label>
+<div class="hidden">
+<label for="proof-laplacian">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{\nabla^2 f\}
+    &= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
+    = - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
+\end{aligned}$$
+</div>
+</div>
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/fredholm-alternative/index.md b/source/know/concept/fredholm-alternative/index.md
new file mode 100644
index 0000000..c813fb4
--- /dev/null
+++ b/source/know/concept/fredholm-alternative/index.md
@@ -0,0 +1,61 @@
+---
+title: "Fredholm alternative"
+date: 2021-05-29
+categories:
+- Mathematics
+layout: "concept"
+---
+
+The **Fredholm alternative** is a theorem regarding equations involving
+a linear operator $\hat{L}$ on a [Hilbert space](/know/concept/hilbert-space/),
+and is useful in the context of multiple-scale perturbation theory.
+It is an *alternative* because it gives two mutually exclusive options,
+given here in [Dirac notation](/know/concept/dirac-notation/):
+
+1.  $\hat{L} \Ket{u} = \Ket{f}$ has a unique solution $\Ket{u}$ for every $\Ket{f}$.
+2.  $\hat{L}^\dagger \Ket{w} = 0$ has non-zero solutions.
+    Then regarding $\hat{L} \Ket{u} = \Ket{f}$:
+    1.  If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then it has infinitely many solutions $\Ket{u}$.
+    2.  If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then it has no solutions $\Ket{u}$.
+
+Where $\hat{L}^\dagger$ is the adjoint of $\hat{L}$.
+In other words, $\hat{L} \Ket{u} = \Ket{f}$ has non-trivial solutions if
+and only if for all $\Ket{w}$ (including the trivial case $\Ket{w} = 0$)
+it holds that $\Inprod{w}{f} = 0$.
+
+As a specific example,
+if $\hat{L}$ is a matrix and the kets are vectors,
+this theorem can alternatively be stated as follows using the determinant:
+
+1.  If $\mathrm{det}(\hat{L}) \neq 0$, then $\hat{L} \vec{u} = \vec{f}$
+    has a unique solution $\vec{u}$ for every $\vec{f}$.
+2.  If $\mathrm{det}(\hat{L}) = 0$,
+    then $\hat{L}^\dagger \vec{w} = \vec{0}$ has non-zero solutions.
+    Then regarding $\hat{L} \vec{u} = \vec{f}$:
+    1.  If $\vec{w} \cdot \vec{f} = 0$ for all $\vec{w}$, then it has
+        infinitely many solutions $\vec{u}$.
+    2.  If $\vec{w} \cdot \vec{f} \neq 0$ for any $\vec{w}$, then it has
+        no solutions $\vec{u}$.
+
+Consequently, the Fredholm alternative is also brought up
+in the context of eigenvalue problems.
+Define $\hat{M} = (\hat{L} - \lambda \hat{I})$,
+where $\lambda$ is an eigenvalue of $\hat{L}$
+if and only if $\mathrm{det}(\hat{M}) = 0$.
+Then for the equation $\hat{M} \Ket{u} = \Ket{f}$, we can say that:
+
+1.  If $\lambda$ is *not* an eigenvalue,
+    then there is a unique solution $\Ket{u}$ for each $\Ket{f}$.
+2.  If $\lambda$ is an eigenvalue, then $\hat{M}^\dagger \Ket{w} = 0$
+    has non-zero solutions. Then:
+    1.  If $\Inprod{w}{f} = 0$ for all $\Ket{w}$, then there are
+        infinitely many solutions $\Ket{u}$.
+    2.  If $\Inprod{w}{f} \neq 0$ for any $\Ket{w}$, then there are no
+        solutions $\Ket{u}$.
+
+
+
+## References
+1.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*, 2020,
+    unpublished.
diff --git a/source/know/concept/fundamental-solution/index.md b/source/know/concept/fundamental-solution/index.md
new file mode 100644
index 0000000..f5a51d5
--- /dev/null
+++ b/source/know/concept/fundamental-solution/index.md
@@ -0,0 +1,145 @@
+---
+title: "Fundamental solution"
+date: 2021-11-02
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+Given a linear operator $\hat{L}$ acting on $x \in [a, b]$,
+its **fundamental solution** $G(x, x')$ is defined as the response
+of $\hat{L}$ to a [Dirac delta function](/know/concept/dirac-delta-function/)
+$\delta(x - x')$ for $x \in ]a, b[$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{L}\{ G(x, x') \}
+        = A \delta(x - x')
+    }
+\end{aligned}$$
+
+Where $A$ is a constant, usually $1$.
+Fundamental solutions are often called **Green's functions**,
+but are distinct from the (somewhat related)
+[Green's functions](/know/concept/greens-functions/)
+in many-body quantum theory.
+
+Note that the definition of $G(x, x')$ generalizes that of
+the [impulse response](/know/concept/impulse-response/).
+And likewise, due to the superposition principle,
+once $G$ is known, $\hat{L}$'s response $u(x)$ to
+*any* forcing function $f(x)$ can easily be found as follows:
+
+$$\begin{aligned}
+    \hat{L} \{ u(x) \}
+    = f(x)
+        \quad \implies \quad
+    \boxed{
+        u(x)
+        = \frac{1}{A} \int_a^b f(x') \: G(x, x') \dd{x'}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-solution"/>
+<label for="proof-solution">Proof</label>
+<div class="hidden">
+<label for="proof-solution">Proof.</label>
+$\hat{L}$ 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.
+</div>
+</div>
+
+While the impulse response is typically used for initial value problems,
+the fundamental solution $G$ is used for boundary value problems.
+Suppose those boundary conditions are homogeneous,
+i.e. $u(x)$ or one of its derivatives is zero at the boundaries.
+Then:
+
+$$\begin{aligned}
+    0
+    &= u(a)
+    = \frac{1}{A} \int_a^b f(x') \: G(a, x') \dd{x'}
+    \qquad \implies \quad
+    G(a, x') = 0
+    \\
+    0
+    &= u_x(a)
+    = \frac{1}{A} \int_a^b f(x') \: G_x(a, x') \dd{x'}
+    \quad \implies \quad
+    G_x(a, x') = 0
+\end{aligned}$$
+
+This holds for all $x'$, and analogously for the other boundary $x = b$.
+In other words, the boundary conditions are built into $G$.
+
+What if the boundary conditions are inhomogeneous?
+No problem: thanks to the linearity of $\hat{L}$,
+those conditions can be given to the homogeneous solution $u_h(x)$,
+where $\hat{L}\{ u_h(x) \} = 0$,
+such that the inhomogeneous solution $u_i(x) = u(x) - u_h(x)$
+has homogeneous boundaries again,
+so we can use $G$ as usual to find $u_i(x)$, and then just add $u_h(x)$.
+
+If $\hat{L}$ is self-adjoint
+(see e.g. [Sturm-Liouville theory](/know/concept/sturm-liouville-theory/)),
+then the fundamental solution $G(x, x')$
+has the following **reciprocity** boundary condition:
+
+$$\begin{aligned}
+    \boxed{
+        G(x, x') = G^*(x', x)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-reciprocity"/>
+<label for="proof-reciprocity">Proof</label>
+<div class="hidden">
+<label for="proof-reciprocity">Proof.</label>
+Consider two parameters $x_1'$ and $x_2'$.
+The self-adjointness of $\hat{L}$ means that:
+
+$$\begin{aligned}
+    \int_a^b G^*(x, x_1') \Big( \hat{L} \{ G(x, x_2') \} \Big) \dd{x}
+    &= \int_a^b \Big( \hat{L} \{ G(x, x_1') \} \Big)^* G(x, x_2') \dd{x}
+    \\
+    \int_a^b G^*(x, x_1') \: \delta(x - x_2') \dd{x}
+    &= \int_a^b \delta^*(x - x_1') \: G(x, x_2') \dd{x}
+    \\
+    G^*(x_2', x_1')
+    &= G(x_1', x_2')
+\end{aligned}$$
+</div>
+</div>
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/fundamental-thermodynamic-relation/index.md b/source/know/concept/fundamental-thermodynamic-relation/index.md
new file mode 100644
index 0000000..a251fcd
--- /dev/null
+++ b/source/know/concept/fundamental-thermodynamic-relation/index.md
@@ -0,0 +1,53 @@
+---
+title: "Fundamental thermodynamic relation"
+date: 2021-07-07
+categories:
+- Physics
+- Thermodynamics
+layout: "concept"
+---
+
+The **fundamental thermodynamic relation** combines the first two
+[laws of thermodynamics](/know/concept/laws-of-thermodynamics/),
+and gives the change of the internal energy $U$,
+which is a [thermodynamic potential](/know/concept/thermodynamic-potential/),
+in terms of the change in
+entropy $S$, volume $V$, and the number of particles $N$.
+
+Starting from the first law of thermodynamics,
+we write an infinitesimal change in energy $\dd{U}$ as follows,
+where $T$ is the temperature and $P$ is the pressure:
+
+$$\begin{aligned}
+    \dd{U} &= \dd{Q} + \dd{W} = T \dd{S} - P \dd{V}
+\end{aligned}$$
+
+The term $T \dd{S}$ comes from the second law of thermodynamics,
+and represents the transfer of thermal energy,
+while $P \dd{V}$ represents physical work.
+
+However, we are missing a term, namely matter transfer.
+If particles can enter/leave the system (i.e. the population $N$ is variable),
+then each such particle costs an amount $\mu$ of energy,
+where $\mu$ is known as the **chemical potential**:
+
+$$\begin{aligned}
+    \dd{U} = T \dd{S} - P \dd{V} + \mu \dd{N}
+\end{aligned}$$
+
+To generalize even further, there may be multiple species of particle,
+which each have a chemical potential $\mu_i$.
+In that case, we sum over all species $i$:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{U} = T \dd{S} - P \dd{V} + \sum_{i}^{} \mu_i \dd{N_i}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md
new file mode 100644
index 0000000..5cd2d19
--- /dev/null
+++ b/source/know/concept/ghz-paradox/index.md
@@ -0,0 +1,115 @@
+---
+title: "GHZ paradox"
+date: 2021-03-29
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+layout: "concept"
+---
+
+The **Greenberger-Horne-Zeilinger** or **GHZ paradox**
+is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/)
+that does not use inequalities,
+but the three-particle entangled **GHZ state** $\Ket{\mathrm{GHZ}}$ instead,
+
+$$\begin{aligned}
+    \boxed{
+        \Ket{\mathrm{GHZ}}
+        = \frac{1}{\sqrt{2}} \Big( \Ket{000} + \Ket{111} \Big)
+    }
+\end{aligned}$$
+
+Where $\Ket{0}$ and $\Ket{1}$ are qubit states,
+for example, the eigenvalues of the Pauli matrix $\hat{\sigma}_z$.
+
+If we now apply certain products of the Pauli matrices $\hat{\sigma}_x$ and $\hat{\sigma}_y$
+to the three particles, we find:
+
+
+$$\begin{aligned}
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \Ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0}
+    + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \Big)
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes \Ket{1} \otimes \Ket{1} + \Ket{0} \otimes \Ket{0} \otimes \Ket{0} \Big)
+    = \Ket{\mathrm{GHZ}}
+    \\
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \Ket{\mathrm{GHZ}}
+    &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_y \Ket{0} \otimes \hat{\sigma}_y \Ket{0}
+    + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \Big)
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes i \Ket{1} \otimes i \Ket{1} + \Ket{0} \otimes i \Ket{0} \otimes i \Ket{0} \Big)
+    = - \Ket{\mathrm{GHZ}}
+\end{aligned}$$
+
+In other words, the GHZ state is a simultaneous eigenstate of these composite operators,
+with eigenvalues $+1$ and $-1$, respectively.
+Let us introduce two other product operators,
+such that we have a set of four observables,
+for which $\Ket{\mathrm{GHZ}}$ gives these eigenvalues:
+
+$$\begin{aligned}
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
+    \quad &\implies \quad +1
+    \\
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x
+    \quad &\implies \quad -1
+\end{aligned}$$
+
+According to any local hidden variable (LHV) theory,
+the measurement outcomes of the operators are predetermined,
+and the three particles $A$, $B$ and $C$ can be measured separately,
+or in other words, the eigenvalues can be factorized:
+
+$$\begin{aligned}
+    \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
+    \quad &\implies \quad +1 = m_x^A m_x^B m_x^C
+    \\
+    \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1 = m_x^A m_y^B m_y^C
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y
+    \quad &\implies \quad -1 = m_y^A m_x^B m_y^C
+    \\
+    \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x
+    \quad &\implies \quad -1 = m_y^A m_y^B m_x^C
+\end{aligned}$$
+
+Where $m_x^A = \pm 1$ etc.
+Let us now multiply both sides of these four equations together:
+
+$$\begin{aligned}
+    (+1) (-1) (-1) (-1)
+    &= (m_x^A m_x^B m_x^C) (m_x^A m_y^B m_y^C) (m_y^A m_x^B m_y^C) (m_y^A m_y^B m_x^C)
+    \\
+    -1
+    &= (m_x^A)^2 (m_x^B)^2 (m_x^C)^2 (m_y^A)^2 (m_y^B)^2 (m_y^C)^2
+\end{aligned}$$
+
+This is a contradiction: the left-hand side is $-1$,
+but all six factors on the right are $+1$.
+This means that we must have made an incorrect assumption along the way.
+
+Our only assumption was that we could factorize the eigenvalues,
+so that e.g. particle $A$ could be measured on its own
+without an "action-at-a-distance" effect on $B$ or $C$.
+However, because that leads us to a contradiction,
+we must conclude that action-at-a-distance exists,
+and that therefore all LHV-based theories are invalid.
+
+
+
+## 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/grad-shafranov-equation/index.md b/source/know/concept/grad-shafranov-equation/index.md
new file mode 100644
index 0000000..8bef8af
--- /dev/null
+++ b/source/know/concept/grad-shafranov-equation/index.md
@@ -0,0 +1,226 @@
+---
+title: "Grad-Shafranov equation"
+date: 2022-03-06
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+Nuclear fusion reactors tend to have a torus shape,
+in which the plasma is confined by a **pinch**,
+i.e. by [magnetic fields](/know/concept/magnetic-field/)
+chosen so that the [Lorentz force](/know/concept/lorentz-force/)
+stops particles escaping.
+Effectively, we are taking a cylindrical [screw pinch](/know/concept/screw-pinch/)
+and bending it into a torus.
+
+We would like to find the equilibrium state of the plasma
+in the general case of a reactor with toroidal symmetry.
+Using ideal [magnetohydrodynamics](/know/concept/magnetohydrodynamics/) (MHD),
+we start by assuming that the fluid is stationary,
+and that the confining field $\vb{B}$ is fixed:
+
+$$\begin{aligned}
+    \vb{u}
+    = 0
+    \qquad \qquad
+    \pdv{\vb{u}}{t}
+    = 0
+    \qquad \qquad
+    \pdv{\vb{B}}{t}
+    = 0
+    \qquad \qquad
+    \vb{E}
+    = 0
+\end{aligned}$$
+
+Notice that $\vb{E} = 0$ is a result of the ideal generalized Ohm's law.
+Under these assumptions, the relevant MHD equations to be solved are
+Gauss' law for magnetism, Ampère's law, and the MHD momentum equation, respectively:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \vb{B}
+    \qquad \qquad
+    \mu_0 \vb{J}
+    = \nabla \cross \vb{B}
+    \qquad \qquad
+    \nabla p
+    = \vb{J} \cross \vb{B}
+\end{aligned}$$
+
+The goal is to analyze them in this order,
+exploiting toroidal symmetry along the way,
+to arrive at a general equilibrium condition.
+[Cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/) $(r, \theta, z)$
+are a natural choice, with the $z$-axis running through the middle of the torus.
+
+As preparation, it is a good idea to write $\vb{B}$
+as the curl of a magnetic vector potential $\vb{A}$,
+which looks like this in cylindrical polar coordinates:
+
+$$\begin{aligned}
+    \vb{B}
+    = \nabla \cross \vb{A}
+    = \begin{bmatrix}
+        \displaystyle \frac{1}{r} \pdv{A_z}{\theta} - \pdv{A_\theta}{z} \\
+        \displaystyle \pdv{A_r}{z} - \pdv{A_z}{r} \\
+        \displaystyle \frac{1}{r} \Big( \pdv{(r A_\theta)}{r} - \pdv{A_r}{\theta} \Big)
+    \end{bmatrix}
+    = \begin{bmatrix}
+        \displaystyle - \pdv{A_\theta}{z} \\
+        \displaystyle \pdv{A_r}{z} - \pdv{A_z}{r} \\
+        \displaystyle \frac{1}{r} \pdv{(r A_\theta)}{r}
+    \end{bmatrix}
+\end{aligned}$$
+
+Here, it is convenient to define the so-called **stream function** $\psi$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \psi
+        \equiv r A_\theta
+    }
+\end{aligned}$$
+
+Such that $\vb{B}$ can be written as below,
+where we will regard $B_\theta$ as a given quantity:
+
+$$\begin{aligned}
+    \vb{B}
+    = \begin{bmatrix}
+        \displaystyle -\frac{1}{r} \pdv{\psi}{z} \\
+        B_\theta \\
+        \displaystyle \frac{1}{r} \pdv{\psi}{r}
+    \end{bmatrix}
+    \qquad \mathrm{where} \qquad
+    B_\theta
+    = \pdv{A_r}{z} - \pdv{A_z}{r}
+\end{aligned}$$
+
+
+Inserting this into Gauss' law,
+we see that it is trivially satisfied,
+thanks to circular symmetry guaranteeing that $\ipdv{B_\theta}{\theta} = 0$:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \vb{B}
+    &= - \frac{1}{r} \pdv{}{r}\bigg( \frac{r}{r} \pdv{\psi}{z} \bigg)
+    + \frac{1}{r} \pdv{B_\theta}{\theta}
+    + \pdv{}{z}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+    \\
+    &= - \frac{1}{r} \mpdv{\psi}{r}{z} + \frac{1}{r} \mpdv{\psi}{z}{r}
+    = 0
+\end{aligned}$$
+
+What matters is that we have expressions for the components of $\vb{B}$.
+Moving on, to find the current density $\vb{J}$,
+we use Ampère's law and symmetry to get:
+
+
+$$\begin{aligned}
+    \vb{J}
+    = \frac{1}{\mu_0} \nabla \cross \vb{B}
+    = \frac{1}{\mu_0}
+    \begin{bmatrix}
+        \displaystyle \frac{1}{r} \pdv{B_z}{\theta} - \pdv{B_\theta}{z} \\
+        \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\
+        \displaystyle \frac{1}{r} \Big( \pdv{(r B_\theta)}{r} - \pdv{B_r}{\theta} \Big)
+    \end{bmatrix}
+    = \frac{1}{\mu_0}
+    \begin{bmatrix}
+        \displaystyle 0 \\
+        \displaystyle \pdv{B_r}{z} - \pdv{B_z}{r} \\
+        \displaystyle \frac{1}{r} \pdv{(r B_\theta)}{r}
+    \end{bmatrix}
+\end{aligned}$$
+
+Where we have assumed that $B_\theta$ depends only on $r$, not $z$ or $\theta$.
+Substituting this into the MHD momentum equation
+gives the following pressure gradient $\nabla p$:
+
+$$\begin{aligned}
+    \nabla p
+    &= \vb{J} \cross \vb{B}
+    = \begin{bmatrix}
+        J_\theta B_z - J_z B_\theta \\
+        J_z B_r - J_r B_z \\
+        J_r B_\theta - J_\theta B_r
+    \end{bmatrix}
+    = \begin{bmatrix}
+        J_\theta B_z - J_z B_\theta \\
+        J_z B_r \\
+        - J_\theta B_r
+    \end{bmatrix}
+\end{aligned}$$
+
+Now, the idea is to focus on this $r$-component to get an equation for $\psi$,
+whose solution can then be used to calculate the $\theta$ and $z$-components of $\nabla p$.
+Therefore, we evaluate:
+
+$$\begin{aligned}
+    \pdv{p}{r}
+    &= J_\theta B_z - J_z B_\theta
+    \\
+    &= \frac{1}{\mu_0} \bigg( \pdv{B_r}{z} - \pdv{B_z}{r} \bigg) B_z
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+    \\
+    &= - \frac{1}{\mu_0} \bigg( \pdv{}{z}\Big(\frac{1}{r} \pdv{\psi}{z}\Big)
+    + \pdv{}{r}\Big(\frac{1}{r} \pdv{\psi}{r}\Big) \bigg) \frac{1}{r} \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+    \\
+    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdvn{2}{\psi}{z} + \pdv{}{r}\Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{(r B_\theta)}{r} B_\theta
+\end{aligned}$$
+
+By using the chain rule to rewrite $\ipdv{}{r}= (\ipdv{\psi}{r}) \; \ipdv{}{\psi}$,
+we get $\ipdv{\psi}{r}$ in each term:
+
+$$\begin{aligned}
+    \pdv{\psi}{r} \pdv{p}{\psi}
+    &= - \frac{1}{\mu_0 r} \bigg( \frac{1}{r} \pdvn{2}{\psi}{z} + \pdv{}{r}\Big( \frac{1}{r} \pdv{\psi}{r} \Big) \bigg) \pdv{\psi}{r}
+    - \frac{1}{\mu_0 r} \pdv{\psi}{r} \pdv{(r B_\theta)}{\psi} B_\theta
+\end{aligned}$$
+
+Dividing out $\ipdv{\psi}{r}$ and multiplying by $\mu_0 r^2$
+leads us to the **Grad-Shafranov equation**,
+which gives the equilibrium condition of a plasma in a toroidal reactor:
+
+$$\begin{aligned}
+    \boxed{
+        \pdvn{2}{\psi}{z} + r \pdv{}{r}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+        = - \mu_0 r^2 \pdv{p}{\psi} - r \pdv{(r B_\theta)}{\psi} B_\theta
+    }
+\end{aligned}$$
+
+Weirdly, $\psi$ appears both as an unknown and as a differentiation variable,
+but this equation can still be solved analytically by
+assuming a certain $\psi$-dependence of $p$ and $r B_\theta$.
+
+Suppose that $B_\theta$ is induced by a poloidal electrical current $I_\mathrm{pol}$,
+i.e. a current around the "tube" of the torus,
+then, assuming $I_\mathrm{pol}$ only depends on $r$, we have:
+
+$$\begin{aligned}
+    B_\theta
+    = \frac{\mu_0 I_\mathrm{pol}(r)}{2 \pi r}
+\end{aligned}$$
+
+Inserting this into the Grad-Shafranov equation yields its following alternative form:
+
+$$\begin{aligned}
+    \boxed{
+        \pdvn{2}{\psi}{z} + r \pdv{}{r}\bigg( \frac{1}{r} \pdv{\psi}{r} \bigg)
+        = - \mu_0 r^2 \pdv{p}{\psi} - \frac{\mu_0^2}{8 \pi^2} \pdv{I_\mathrm{pol}^2}{\psi}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
+
diff --git a/source/know/concept/gram-schmidt-method/index.md b/source/know/concept/gram-schmidt-method/index.md
new file mode 100644
index 0000000..483dd71
--- /dev/null
+++ b/source/know/concept/gram-schmidt-method/index.md
@@ -0,0 +1,49 @@
+---
+title: "Gram-Schmidt method"
+date: 2021-02-22
+categories:
+- Mathematics
+- Algorithms
+layout: "concept"
+---
+
+Given a set of linearly independent non-orthonormal vectors
+$\ket{V_1}, \ket{V_2}, ...$ from a [Hilbert space](/know/concept/hilbert-space/),
+the **Gram-Schmidt method**
+turns them into an orthonormal set $\ket{n_1}, \ket{n_2}, ...$ as follows:
+
+1.  Take the first vector $\ket{V_1}$ and normalize it to get $\ket{n_1}$:
+    
+    $$\begin{aligned}
+        \ket{n_1} = \frac{\ket{V_1}}{\sqrt{\inprod{V_1}{V_1}}}
+    \end{aligned}$$
+    
+2.  Begin loop. Take the next non-orthonormal vector $\ket{V_j}$, and
+    subtract from it its projection onto every already-processed vector:
+    
+    $$\begin{aligned}
+        \ket{n_j'} = \ket{V_j} - \ket{n_1} \inprod{n_1}{V_j} - \ket{n_2} \inprod{n_2}{V_j} - ... - \ket{n_{j-1}} \inprod{n_{j-1}}{V_{j-1}}
+    \end{aligned}$$
+    
+    This leaves only the part of $\ket{V_j}$ which is orthogonal to
+    $\ket{n_1}$, $\ket{n_2}$, etc. This why the input vectors must be
+    linearly independent; otherwise $\Ket{n_j'}$ may become zero at some
+    point.
+    
+3.  Normalize the resulting ortho*gonal* vector $\ket{n_j'}$ to make it
+    ortho*normal*:
+    
+    $$\begin{aligned}
+                \ket{n_j} = \frac{\ket{n_j'}}{\sqrt{\inprod{n_j'}{n_j'}}}
+    \end{aligned}$$
+    
+4.  Loop back to step 2, taking the next vector $\ket{V_{j+1}}$.
+
+If you are unfamiliar with this notation, take a look at [Dirac notation](/know/concept/dirac-notation/).
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/source/know/concept/grand-canonical-ensemble/index.md b/source/know/concept/grand-canonical-ensemble/index.md
new file mode 100644
index 0000000..85bf90a
--- /dev/null
+++ b/source/know/concept/grand-canonical-ensemble/index.md
@@ -0,0 +1,74 @@
+---
+title: "Grand canonical ensemble"
+date: 2021-07-11
+categories:
+- Physics
+- Thermodynamics
+- Thermodynamic ensembles
+layout: "concept"
+---
+
+The **grand canonical ensemble** or **μVT ensemble**
+extends the [canonical ensemble](/know/concept/canonical-ensemble/)
+by allowing the exchange of both energy $U$ and particles $N$
+with an external reservoir,
+so that the conserved state functions are
+the temperature $T$, the volume $V$, and the chemical potential $\mu$.
+
+The derivation is practically identical to that of the canonical ensemble.
+We refer to the system of interest as $A$,
+and the reservoir as $B$.
+In total, $A\!+\!B$ has energy $U$ and population $N$.
+
+Let $c_B(U_B)$ be the number of $B$-microstates with energy $U_B$.
+Then the probability that $A$ is in a specific microstate $s_A$ is as follows:
+
+$$\begin{aligned}
+    p(s)
+    = \frac{c_B\big(U - U_A(s_A), N - N_A(s_A)\big)}{\sum_{s_A} c_B\big(U \!-\! U_A(s_A), N \!-\! N_A(s_A)\big)}
+\end{aligned}$$
+
+Then, as for the canonical ensemble,
+we assume $U_B \gg U_A$ and $N_B \gg N_A$,
+and approximate $\ln{p(s_A)}$
+by Taylor-expanding $\ln{c_B}$ around $U_B = U$ and $N_B = N$.
+The resulting probability distribution is known as the **Gibbs distribution**,
+with $\beta \equiv 1/(kT)$:
+
+$$\begin{aligned}
+    \boxed{
+        p(s_A) = \frac{1}{\mathcal{Z}} \exp\!\Big(\!-\! \beta \: \big( U_A(s_A) \!-\! \mu N_A(s_A) \big) \Big)
+    }
+\end{aligned}$$
+
+Where the normalizing **grand partition function** $\mathcal{Z}(\mu, V, T)$ is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \mathcal{Z} \equiv \sum_{s_A}^{} \exp\!\Big(\!-\! \beta \: \big( U_A(s_A) - \mu N_A(s_A) \big) \Big)
+    }
+\end{aligned}$$
+
+In contrast to the canonical ensemble,
+whose [thermodynamic potential](/know/concept/thermodynamic-potential/)
+was the Helmholtz free energy $F$,
+the grand canonical ensemble instead
+minimizes the **grand potential** $\Omega$:
+
+$$\begin{aligned}
+    \boxed{
+        \Omega(T, V, \mu)
+        \equiv - k T \ln{\mathcal{Z}}
+        = \Expval{U_A} - T S_A - \mu \Expval{N_A}
+    }
+\end{aligned}$$
+
+So $\mathcal{Z} = \exp(- \beta \Omega)$.
+This is proven in the same way as for $F$ in the canonical ensemble.
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md
new file mode 100644
index 0000000..48f1e76
--- /dev/null
+++ b/source/know/concept/greens-functions/index.md
@@ -0,0 +1,390 @@
+---
+title: "Green's functions"
+date: 2021-11-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In many-body quantum theory, a **Green's function**
+can be any correlation function between two given operators,
+although it is usually used to refer to the special case
+where the operators are particle creation/annihilation operators
+from the [second quantization](/know/concept/second-quantization/).
+
+They are somewhat related to
+[fundamental solutions](/know/concept/fundamental-solution/),
+which are also called *Green's functions*,
+but in general they are not the same,
+except in a special case, see below.
+
+
+## Single-particle functions
+
+If the two operators are single-particle creation/annihilation operators,
+then we get the **single-particle Green's functions**,
+for which the symbol $G$ is used.
+
+The **time-ordered** or **causal Green's function** $G_{\nu \nu'}$ is as follows,
+where $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/),
+$\nu$ and $\nu'$ are single-particle states,
+and $\hat{c}_\nu$ annihilates a particle from $\nu$, etc.:
+
+$$\begin{aligned}
+    \boxed{
+        G_{\nu \nu'}(t, t')
+        \equiv -\frac{i}{\hbar} \Expval{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t') \Big\}}
+    }
+\end{aligned}$$
+
+The expectation value $\Expval{}$ is
+with respect to thermodynamic equilibrium.
+This is sometimes in the [canonical ensemble](/know/concept/canonical-ensemble/)
+(for some two-particle Green's functions, see below),
+but usually in the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
+since we are adding/removing particles.
+In the latter case, we assume that the chemical potential $\mu$
+is already included in the Hamiltonian $\hat{H}$.
+Explicitly, for a complete set of many-particle states $\Ket{\Psi_n}$, we have:
+
+$$\begin{aligned}
+    G_{\nu \nu'}(t, t')
+    &= -\frac{i}{\hbar Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}} \Big)
+    \\
+    &= -\frac{i}{\hbar Z} \sum_{n}
+    \Matrixel{\Psi_n}{\mathcal{T} \Big\{ \hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')\Big\} \: e^{- \beta \hat{H}}}{\Psi_n}
+\end{aligned}$$
+
+Arguably more prevalent are
+the **retarded Green's function** $G_{\nu \nu'}^R$
+and the **advanced Green's function** $G_{\nu \nu'}^A$
+which are defined like so:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            G_{\nu \nu'}^R(t, t')
+            &\equiv -\frac{i}{\hbar} \Theta(t - t') \Expval{\comm{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+            \\
+            G_{\nu \nu'}^A(t, t')
+            &\equiv \frac{i}{\hbar} \Theta(t' - t) \Expval{\comm{\hat{c}_{\nu}(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where $\Theta$ is a [Heaviside function](/know/concept/heaviside-step-function/),
+and $[,]_{\mp}$ is a commutator for bosons,
+and an anticommutator for fermions.
+Depending on the context,
+we could either be in the [Heisenberg picture](/know/concept/heisenberg-picture/)
+or in the [interaction picture](/know/concept/interaction-picture/),
+hence $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$ are time-dependent.
+
+Furthermore, the **greater Green's function** $G_{\nu \nu'}^>$
+and **lesser Green's function** $G_{\nu \nu'}^<$ are:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            G_{\nu \nu'}^>(t, t')
+            &\equiv -\frac{i}{\hbar} \Expval{\hat{c}_{\nu}(t) \: \hat{c}_{\nu'}^\dagger(t')}
+            \\
+            G_{\nu \nu'}^<(t, t')
+            &\equiv \mp \frac{i}{\hbar} \Expval{\hat{c}_{\nu'}^\dagger(t') \: \hat{c}_{\nu}(t)}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where $-$ is for bosons, and $+$ for fermions.
+With this, the causal, retarded and advanced Green's functions
+can thus be expressed as follows:
+
+$$\begin{aligned}
+    G_{\nu \nu'}(t, t')
+    &= \Theta(t - t') \: G_{\nu \nu'}^>(t, t') + \Theta(t' - t) \: G_{\nu \nu'}^<(t, t')
+    \\
+    G_{\nu \nu'}^R(t, t')
+    &= \Theta(t - t') \big( G_{\nu \nu'}^>(t, t') - G_{\nu \nu'}^<(t, t') \big)
+    \\
+    G_{\nu \nu'}^A(t, t')
+    &= \Theta(t' - t) \big( G_{\nu \nu'}^<(t, t') - G_{\nu \nu'}^>(t, t') \big)
+\end{aligned}$$
+
+If the Hamiltonian involves interactions,
+it might be more natural to use quantum field operators $\hat{\Psi}(\vb{r}, t)$
+instead of choosing a basis of single-particle states $\psi_\nu$.
+In that case, instead of a label $\nu$,
+we use the spin $s$ and position $\vb{r}$, leading to:
+
+$$\begin{aligned}
+    G_{ss'}(\vb{r}, t; \vb{r}', t')
+    &= -\frac{i}{\hbar} \Theta(t - t') \Expval{\mathcal{T}\Big\{ \hat{\Psi}_{s}(\vb{r}, t) \hat{\Psi}_{s'}^\dagger(\vb{r}', t') \Big\}}
+    \\
+    &= \sum_{\nu \nu'} \psi_\nu(\vb{r}) \: \psi^*_{\nu'}(\vb{r}') \: G_{\nu \nu'}(t, t')
+\end{aligned}$$
+
+And analogously for $G_{ss'}^R$, $G_{ss'}^A$, $G_{ss'}^>$ and $G_{ss'}^<$.
+Note that the time-dependence is given to the old $G_{\nu \nu'}$,
+i.e. to $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$,
+because we are in the Heisenberg picture.
+
+If the Hamiltonian is time-independent,
+then it can be shown that all the Green's functions
+only depend on the time-difference $t - t'$:
+
+$$\begin{gathered}
+    G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t - t')
+    \\
+    G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^R(t - t')
+    \qquad \quad
+    G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^A(t - t')
+    \\
+    G_{\nu \nu'}^>(t, t') = G_{\nu \nu'}^>(t - t')
+    \qquad \quad
+    G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
+\end{gathered}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-time-diff"/>
+<label for="proof-time-diff">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-time-diff">Proof.</label>
+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.
+</div>
+</div>
+
+If the Hamiltonian is both time-independent and non-interacting,
+then the time-dependence of $\hat{c}_\nu$
+can simply be factored out as
+$\hat{c}_\nu(t) = \hat{c}_\nu \exp(- i \varepsilon_\nu t / \hbar)$.
+Then the diagonal ($\nu = \nu'$) greater and lesser Green's functions
+can be written in the form below, where $f_\nu$ is either
+the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/)
+or the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/).
+
+$$\begin{aligned}
+    G_{\nu \nu}^>(t, t')
+    &= -\frac{i}{\hbar} \Expval{\hat{c}_{\nu} \hat{c}_{\nu}^\dagger} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+    \\
+    &= -\frac{i}{\hbar} (1 - f_\nu) \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+    \\
+    G_{\nu \nu}^<(t, t')
+    &= \mp \frac{i}{\hbar} \Expval{\hat{c}_{\nu}^\dagger \hat{c}_{\nu}} \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+    \\
+    &= \mp \frac{i}{\hbar} f_\nu \exp\!\big(\!-\! i \varepsilon_\nu (t \!-\! t') / \hbar \big)
+\end{aligned}$$
+
+
+## As fundamental solutions
+
+In the absence of interactions,
+we know from the derivation of
+[equation-of-motion theory](/know/concept/equation-of-motion-theory/)
+that the equation of motion of $G^R(\vb{r}, t; \vb{r}', t')$
+is as follows (neglecting spin):
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R}{t}
+    = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+    + \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\comm{\hat{H}_0}{\hat{\Psi}(\vb{r}, t)}}{\hat{\Psi}^\dagger(\vb{r}', t')}}
+\end{aligned}$$
+
+If $\hat{H}_0$ only contains kinetic energy,
+i.e. there is no external potential,
+it can be shown that:
+
+$$\begin{aligned}
+    \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+    = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-commH0"/>
+<label for="proof-commH0">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-commH0">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \comm{\hat{H}_0}{\hat{\Psi}(\vb{r})}
+    &= \frac{\hbar^2}{2 m} \sum_{\nu \nu'} \hat{c}_{\nu'}
+    \int \delta(\vb{r} - \vb{r}') \: \nabla^2 \psi_{\nu'}(\vb{r}') \dd{\vb{r}'}
+    \\
+    &= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
+    = \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
+\end{aligned}$$
+</div>
+</div>
+
+After substituting this into the equation of motion,
+we recognize $G^R(\vb{r}, t; \vb{r}', t')$ itself:
+
+$$\begin{aligned}
+    i \hbar \pdv{G^R}{t}
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+    + \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r}, t)}{\hat{\Psi}^\dagger(\vb{r}', t')}}
+    \\
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t') - \frac{\hbar^2}{2 m} \nabla_\vb{r}^2
+    \Big( \!-\! \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\hat{\Psi}(\vb{r}, t)}{\hat{\Psi}^\dagger(\vb{r}', t')}} \Big)
+    \\
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+    - \frac{\hbar^2}{2 m} \nabla_\vb{r}^2 G^R(\vb{r}, t; \vb{r}', t')
+\end{aligned}$$
+
+Rearranging this leads to the following,
+which is the definition of a fundamental solution:
+
+$$\begin{aligned}
+    \Big( i \hbar \pdv{}{t}+ \frac{\hbar^2}{2 m} \nabla_\vb{r}^2 \Big) G^R(\vb{r}, t; \vb{r}', t')
+    &= \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')
+\end{aligned}$$
+
+Therefore, the retarded Green's function
+(and, it turns out, the advanced Green's function too)
+is a fundamental solution of the Schrödinger equation
+if there is no potential,
+i.e. the Hamiltonian only contains kinetic energy.
+
+
+## Two-particle functions
+
+We generalize the above to two arbitrary operators $\hat{A}$ and $\hat{B}$,
+giving us the **two-particle Green's functions**,
+or just **correlation functions**.
+The **causal correlation function** $C_{AB}$,
+the **retarded correlation function** $C_{AB}^R$
+and the **advanced correlation function** $C_{AB}^A$ are defined as follows
+(in the Heisenberg picture):
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            C_{AB}(t, t')
+            &\equiv -\frac{i}{\hbar} \Expval{\mathcal{T}\Big\{\hat{A}(t) \hat{B}(t')\Big\}}
+            \\
+            C_{AB}^R(t, t')
+            &\equiv -\frac{i}{\hbar} \Theta(t - t') \Expval{\comm{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
+            \\
+            C_{AB}^A(t, t')
+            &\equiv \frac{i}{\hbar} \Theta(t' - t) \Expval{\comm{\hat{A}(t)}{\hat{B}(t')}_{\mp}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where the expectation value $\Expval{}$ is taken of thermodynamic equilibrium.
+The name *two-particle* comes from the fact that $\hat{A}$ and $\hat{B}$
+will often consist of a sum of products
+of two single-particle creation/annihilation operators.
+
+Like for the single-particle Green's functions,
+if the Hamiltonian is time-independent,
+then it can be shown that the two-particle functions
+only depend on the time-difference $t - t'$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}(t, t') = G_{\nu \nu'}(t \!-\! t')
+    \qquad
+    G_{\nu \nu'}^R(t, t') = G_{\nu \nu'}^>(t \!-\! t')
+    \qquad
+    G_{\nu \nu'}^A(t, t') = G_{\nu \nu'}^<(t \!-\! t')
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md
new file mode 100644
index 0000000..5fd9f38
--- /dev/null
+++ b/source/know/concept/gronwall-bellman-inequality/index.md
@@ -0,0 +1,204 @@
+---
+title: "Grönwall-Bellman inequality"
+date: 2021-11-07
+categories:
+- Mathematics
+layout: "concept"
+---
+
+Suppose we have a first-order ordinary differential equation
+for some function $u(t)$, and that it can be shown from this equation
+that the derivative $u'(t)$ is bounded as follows:
+
+$$\begin{aligned}
+    u'(t)
+    \le \beta(t) \: u(t)
+\end{aligned}$$
+
+Where $\beta(t)$ is known.
+Then **Grönwall's inequality** states that the solution $u(t)$ is bounded:
+
+$$\begin{aligned}
+    \boxed{
+        u(t)
+        \le u(0) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-original"/>
+<label for="proof-original">Proof</label>
+<div class="hidden">
+<label for="proof-original">Proof.</label>
+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.
+</div>
+</div>
+
+Grönwall's inequality can be generalized to non-differentiable functions.
+Suppose we know:
+
+$$\begin{aligned}
+    u(t)
+    \le \alpha(t) + \int_0^t \beta(s) \: u(s) \dd{s}
+\end{aligned}$$
+
+Where $\alpha(t)$ and $\beta(t)$ are known.
+Then the **Grönwall-Bellman inequality** states that:
+
+$$\begin{aligned}
+    \boxed{
+        u(t)
+        \le \alpha(t) + \int_0^t \alpha(s) \: \beta(s) \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \dd{s}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-integral"/>
+<label for="proof-integral">Proof</label>
+<div class="hidden">
+<label for="proof-integral">Proof.</label>
+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.
+</div>
+</div>
+
+In the special case where $\alpha(t)$ is non-decreasing with $t$,
+the inequality reduces to:
+
+$$\begin{aligned}
+    \boxed{
+        u(t)
+        \le \alpha(t) \exp\!\bigg( \int_0^t \beta(s) \dd{s} \bigg)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-special"/>
+<label for="proof-special">Proof</label>
+<div class="hidden">
+<label for="proof-special">Proof.</label>
+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:
+
+$$\begin{aligned}
+    u(t)
+    &\le \alpha(t) - \alpha(t) \bigg[ \exp\!\bigg( \int_s^t \beta(r) \dd{r} \bigg) \bigg]_{s = 0}^{s = t}
+    \\
+    &\le \alpha(t) - \alpha(t) \exp\!\bigg( \int_t^t \beta(r) \dd{r} \bigg) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
+    \\
+    &\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
+\end{aligned}$$
+</div>
+</div>
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/guiding-center-theory/index.md b/source/know/concept/guiding-center-theory/index.md
new file mode 100644
index 0000000..d429a50
--- /dev/null
+++ b/source/know/concept/guiding-center-theory/index.md
@@ -0,0 +1,516 @@
+---
+title: "Guiding center theory"
+date: 2021-09-21
+categories:
+- Physics
+- Electromagnetism
+- Plasma physics
+layout: "concept"
+---
+
+When discussing the [Lorentz force](/know/concept/lorentz-force/),
+we introduced the concept of *gyration*:
+a particle in a uniform [magnetic field](/know/concept/magnetic-field/) $\vb{B}$
+*gyrates* in a circular orbit around a **guiding center**.
+Here, we will generalize this result
+to more complicated situations,
+for example involving [electric fields](/know/concept/electric-field/).
+
+The particle's equation of motion
+combines the Lorentz force $\vb{F}$
+with Newton's second law:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q \big( \vb{E} + \vb{u} \cross \vb{B} \big)
+\end{aligned}$$
+
+We now allow the fields vary slowly in time and space.
+We thus add deviations $\delta\vb{E}$ and $\delta\vb{B}$:
+
+$$\begin{aligned}
+    \vb{E}
+    \to \vb{E} + \delta\vb{E}(\vb{x}, t)
+    \qquad \quad
+    \vb{B}
+    \to \vb{B} + \delta\vb{B}(\vb{x}, t)
+\end{aligned}$$
+
+Meanwhile, the velocity $\vb{u}$ can be split into
+the guiding center's motion $\vb{u}_{gc}$
+and the *known* Larmor gyration $\vb{u}_L$ around the guiding center,
+such that $\vb{u} = \vb{u}_{gc} + \vb{u}_L$.
+Inserting:
+
+$$\begin{aligned}
+    m \dv{}{t}\big( \vb{u}_{gc} + \vb{u}_L \big)
+    = q \big( \vb{E} + \delta\vb{E} + (\vb{u}_{gc} + \vb{u}_L) \cross (\vb{B} + \delta\vb{B}) \big)
+\end{aligned}$$
+
+We already know that $m \: \idv{\vb{u}_L}{t} = q \vb{u}_L \cross \vb{B}$,
+which we subtract from the total to get:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \delta\vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big)
+\end{aligned}$$
+
+This will be our starting point.
+Before proceeding, we also define
+the average of $\Expval{f}$ of a function $f$ over a single gyroperiod,
+where $\omega_c$ is the cyclotron frequency:
+
+$$\begin{aligned}
+    \Expval{f}
+    \equiv \int_0^{2 \pi / \omega_c} f(t) \dd{t}
+\end{aligned}$$
+
+Assuming that gyration is much faster than the guiding center's motion,
+we can use this average to approximately remove the finer dynamics,
+and focus only on the guiding center.
+
+
+## Uniform electric and magnetic field
+
+Consider the case where $\vb{E}$ and $\vb{B}$ are both uniform,
+such that $\delta\vb{B} = 0$ and $\delta\vb{E} = 0$:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B} \big)
+\end{aligned}$$
+
+Dotting this with the unit vector $\vu{b} \equiv \vb{B} / |\vb{B}|$
+makes all components perpendicular to $\vb{B}$ vanish,
+including the cross product,
+leaving only the (scalar) parallel components
+$u_{gc\parallel}$ and $E_\parallel$:
+
+$$\begin{aligned}
+    m \dv{u_{gc\parallel}}{t}
+    = \frac{q}{m} E_{\parallel}
+\end{aligned}$$
+
+This simply describes a constant acceleration,
+and is easy to integrate.
+Next, the equation for $\vb{u}_{gc\perp}$ is found by
+subtracting $u_{gc\parallel}$'s equation from the original:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc\perp}}{t}
+    = q (\vb{E} + \vb{u}_{gc} \cross \vb{B}) - q E_\parallel \vu{b}
+    = q (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B})
+\end{aligned}$$
+
+Keep in mind that $\vb{u}_{gc\perp}$ explicitly excludes gyration.
+If we try to split $\vb{u}_{gc\perp}$ into a constant and a time-dependent part,
+and choose the most convenient constant,
+we notice that the only way to exclude gyration
+is to demand that $\vb{u}_{gc\perp}$ does not depend on time.
+Therefore:
+
+$$\begin{aligned}
+    0
+    = \vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B}
+\end{aligned}$$
+
+To find $\vb{u}_{gc\perp}$, we take the cross product with $\vb{B}$,
+and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$:
+
+$$\begin{aligned}
+    0
+    = \vb{B} \cross (\vb{E}_\perp + \vb{u}_{gc\perp} \cross \vb{B})
+    = \vb{B} \cross \vb{E} + \vb{u}_{gc\perp} B^2
+\end{aligned}$$
+
+Rearranging this shows that $\vb{u}_{gc\perp}$ is constant.
+The guiding center drifts sideways at this speed,
+hence it is called a **drift velocity** $\vb{v}_E$.
+Curiously, $\vb{v}_E$ is independent of $q$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_E
+        = \frac{\vb{E} \cross \vb{B}}{B^2}
+    }
+\end{aligned}$$
+
+Drift is not specific to an electric field:
+$\vb{E}$ can be replaced by a general force $\vb{F}/q$ without issues.
+In that case, the resulting drift velocity $\vb{v}_F$ does depend on $q$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_F
+        = \frac{\vb{F} \cross \vb{B}}{q B^2}
+    }
+\end{aligned}$$
+
+
+## Non-uniform magnetic field
+
+Next, consider a more general case, where $\vb{B}$ is non-uniform,
+but $\vb{E}$ is still uniform:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross (\vb{B} + \delta\vb{B}) + \vb{u}_L \cross \delta\vb{B} \big)
+\end{aligned}$$
+
+Assuming the gyroradius $r_L$ is small compared to the variation of $\vb{B}$,
+we set $\delta\vb{B}$ to the first-order term
+of a Taylor expansion of $\vb{B}$ around $\vb{x}_{gc}$,
+that is, $\delta\vb{B} = (\vb{x}_L \cdot \nabla) \vb{B}$.
+We thus have:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B}
+    + \vb{u}_{gc} \cross (\vb{x}_L \cdot \nabla) \vb{B}
+    + \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} \big)
+\end{aligned}$$
+
+We approximate this by taking the average over a single gyration,
+as defined earlier:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \big( \vb{E} + \vb{u}_{gc} \cross \vb{B}
+    + \vb{u}_{gc} \cross \Expval{ (\vb{x}_L \cdot \nabla) \vb{B} }
+    + \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} } \big)
+\end{aligned}$$
+
+Where we have used that $\Expval{\vb{u}_{gc}} = \vb{u}_{gc}$.
+The two averaged expressions turn out to be:
+
+$$\begin{aligned}
+    \Expval{ (\vb{x}_L \cdot \nabla) \vb{B} }
+    = 0
+    \qquad \quad
+    \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    \approx - \frac{u_L^2}{2 \omega_c} \nabla B
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-nonuniform-B-averages"/>
+<label for="proof-nonuniform-B-averages">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-nonuniform-B-averages">Proof.</label>
+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 $B \equiv |\vb{B}| \approx B_z$:
+
+$$\begin{aligned}
+    \Expval{ \vb{u}_L \cross (\vb{x}_L \cdot \nabla) \vb{B} }
+    &= - \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \displaystyle \pdv{B_z}{x} \\
+        \displaystyle \pdv{B_z}{y} \\
+        \displaystyle \pdv{B_z}{z}
+    \end{pmatrix}
+    \approx - \frac{u_L^2}{2 \omega_c}
+    \begin{pmatrix}
+        \displaystyle \pdv{B}{x} \\
+        \displaystyle \pdv{B}{y} \\
+        \displaystyle \pdv{B}{z}
+    \end{pmatrix}
+    = - \frac{u_L^2}{2 \omega_c} \nabla B
+\end{aligned}$$
+</div>
+</div>
+
+With this, the guiding center's equation of motion
+is reduced to the following:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc}}{t}
+    = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
+\end{aligned}$$
+
+Let us now split $\vb{u}_{gc}$ into
+components $\vb{u}_{gc\perp}$ and $u_{gc\parallel} \vu{b}$,
+which are respectively perpendicular and parallel
+to the magnetic unit vector $\vu{b}$,
+such that $\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$.
+Consequently:
+
+$$\begin{aligned}
+    \dv{\vb{u}_{gc}}{t}
+    = \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t}
+\end{aligned}$$
+
+Inserting this into the guiding center's equation of motion,
+we now have:
+
+$$\begin{aligned}
+    \dv{\vb{u}_{gc}}{t}
+    = m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t} \bigg)
+    = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
+\end{aligned}$$
+
+The derivative of $\vu{b}$ can be rewritten as follows,
+where $R_c$ is the radius of the field's [curvature](/know/concept/curvature/),
+and $\vb{R}_c$ is the corresponding vector from the center of curvature:
+
+$$\begin{aligned}
+    \dv{\vu{b}}{t}
+    \approx - u_{gc\parallel} \frac{\vb{R}_c}{R_c^2}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-nonuniform-B-curvature"/>
+<label for="proof-nonuniform-B-curvature">Proof</label>
+<div class="hidden">
+<label for="proof-nonuniform-B-curvature">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \dv{\vu{b}}{s}
+    = - \frac{\vu{R}_c \dd{\theta}}{R_c \dd{\theta}}
+    = - \frac{\vu{R}_c}{R_c}
+    = - \frac{\vb{R}_c}{R_c^2}
+\end{aligned}$$
+</div>
+</div>
+
+With this, we arrive at the following equation of motion
+for the guiding center:
+
+$$\begin{aligned}
+    m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} - u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} \bigg)
+    = q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
+\end{aligned}$$
+
+Since both $\vb{R}_c$ and any cross product with $\vb{B}$
+will always be perpendicular to $\vb{B}$,
+we can split this equation into perpendicular and parallel components like so:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc\perp}}{t}
+    &= q \vb{E}_{\perp} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} + q \vb{u}_{gc} \cross \vb{B}
+    \\
+    m \dv{u_{gc\parallel}}{t}
+    &= q E_{\parallel} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\parallel} B
+\end{aligned}$$
+
+The parallel part simply describes an acceleration.
+The perpendicular part is more interesting:
+we rewrite it as follows, defining an effective force $\vb{F}_{\!\perp}$:
+
+$$\begin{aligned}
+    m \dv{\vb{u}_{gc\perp}}{t}
+    = \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B}
+    \qquad \quad
+    \vb{F}_{\!\perp}
+    \equiv q \vb{E}_\perp + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B
+\end{aligned}$$
+
+To solve this, we make a crude approximation now, and improve it later.
+We thus assume that $\vb{u}_{gc\perp}$ is constant in time,
+such that the equation reduces to:
+
+$$\begin{aligned}
+     0
+     \approx \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B}
+     = \vb{F}_{\!\perp} + q \vb{u}_{gc\perp} \cross \vb{B}
+\end{aligned}$$
+
+This is analogous to the previous case of a uniform electric field,
+with $q \vb{E}$ replaced by $\vb{F}_{\!\perp}$,
+so it is also solved by crossing with $\vb{B}$ in front,
+yielding a drift:
+
+$$\begin{aligned}
+    \vb{u}_{gc\perp}
+    \approx \vb{v}_F
+    \equiv \frac{\vb{F}_{\!\perp} \cross \vb{B}}{q B^2}
+\end{aligned}$$
+
+From the definition of $\vb{F}_{\!\perp}$,
+this total $\vb{v}_F$ can be split into three drifts:
+the previously seen electric field drift $\vb{v}_E$,
+the **curvature drift** $\vb{v}_c$,
+and the **grad-$\vb{B}$ drift** $\vb{v}_{\nabla B}$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_c
+        = \frac{m u_{gc\parallel}^2}{q} \frac{\vb{R}_c \cross \vb{B}}{R_c^2 B^2}
+    }
+    \qquad \quad
+    \boxed{
+        \vb{v}_{\nabla B}
+        = \frac{u_L^2}{2 \omega_c} \frac{\vb{B} \cross \nabla B}{B^2}
+    }
+\end{aligned}$$
+
+Such that $\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$.
+We are still missing a correction,
+since we neglected the time dependence of $\vb{u}_{gc\perp}$ earlier.
+This correction is called $\vb{v}_p$,
+where $\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$.
+We revisit the perpendicular equation, which now reads:
+
+$$\begin{aligned}
+    m \dv{}{t}\big( \vb{v}_F + \vb{v}_p \big)
+    = \vb{F}_{\!\perp} + q \big( \vb{v}_F + \vb{v}_p \big) \cross \vb{B}
+\end{aligned}$$
+
+We assume that $\vb{v}_F$ varies much faster than $\vb{v}_p$,
+such that $\idv{}{\vb{v}p}{t}$ is negligible.
+In addition, from the derivation of $\vb{v}_F$,
+we know that $\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$,
+leaving only:
+
+$$\begin{aligned}
+    m \dv{\vb{v}_F}{t}
+    = q \vb{v}_p \cross \vb{B}
+\end{aligned}$$
+
+To isolate this for $\vb{v}_p$,
+we take the cross product with $\vb{B}$ in front,
+like earlier.
+We thus arrive at the following correction,
+known as the **polarization drift** $\vb{v}_p$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{v}_p
+        = - \frac{m}{q B^2} \dv{\vb{v}_F}{t} \cross \vb{B}
+    }
+\end{aligned}$$
+
+In many cases $\vb{v}_E$ dominates $\vb{v}_F$,
+so in some literature $\vb{v}_p$ is approximated as follows:
+
+$$\begin{aligned}
+    \vb{v}_p
+    \approx - \frac{m}{q B^2} \dv{\vb{v}_E}{t} \cross \vb{B}
+    = - \frac{m}{q B^2} \Big( \dv{}{t}(\vb{E}_\perp \cross \vb{B}) \Big) \cross \vb{B}
+    = - \frac{m}{q B^2} \dv{\vb{E}_\perp}{t}
+\end{aligned}$$
+
+The polarization drift stands out from the others:
+it has the opposite sign,
+it is proportional to $m$,
+and it is often only temporary.
+Therefore, it is also called the **inertia drift**.
+
+
+
+## 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/hagen-poiseuille-equation/index.md b/source/know/concept/hagen-poiseuille-equation/index.md
new file mode 100644
index 0000000..fc2975a
--- /dev/null
+++ b/source/know/concept/hagen-poiseuille-equation/index.md
@@ -0,0 +1,197 @@
+---
+title: "Hagen-Poiseuille equation"
+date: 2021-04-13
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+The **Hagen-Poiseuille equation**, or simply the **Poiseuille equation**,
+describes the flow of a fluid with nonzero [viscosity](/know/concept/viscosity/)
+through a cylindrical pipe.
+Due to its viscosity, the fluid clings to the sides,
+limiting the amount that can pass through, for a pipe with radius $R$.
+
+Consider the [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
+of an incompressible fluid with spatially uniform density $\rho$.
+Assuming that the flow is steady $\ipdv{\va{v}}{t} = 0$,
+and that gravity is negligible $\va{g} = 0$, we get:
+
+$$\begin{aligned}
+    (\va{v} \cdot \nabla) \va{v}
+    = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+    \qquad \quad
+    \nabla \cdot \va{v} = 0
+\end{aligned}$$
+
+Into this, we insert the ansatz $\va{v} = \vu{e}_z \: v_z(r)$,
+where $\vu{e}_z$ is the $z$-axis' unit vector.
+In other words, we assume that the flow velocity depends only on $r$;
+not on $\phi$ or $z$.
+Plugging this into the Navier-Stokes equations,
+$\nabla \cdot \va{v}$ is trivially zero,
+and in the other equation we multiply out $\rho$, yielding this,
+where $\eta = \rho \nu$ is the dynamic viscosity:
+
+$$\begin{aligned}
+    \nabla p
+    = \vu{e}_z \: \eta \nabla^2 v_z
+\end{aligned}$$
+
+Because only $\vu{e}_z$ appears on the right-hand side,
+only the $z$-component of $\nabla p$ can be nonzero.
+However, $v_z(r)$ is a function of $r$, not $z$!
+The left thus only depends on $z$, and the right only on $r$,
+meaning that both sides must equal a constant,
+which we call $-G$:
+
+$$\begin{aligned}
+    \dv{p}{z}
+    = -G
+    \qquad \quad
+    \eta \frac{1}{r} \dv{}{r}\Big( r \dv{v_z}{r} \Big)
+    = - G
+\end{aligned}$$
+
+The former equation, for $p(z)$, is easy to solve.
+We get an integration constant $p(0)$:
+
+$$\begin{aligned}
+    p(z)
+    = p(0) - G z
+\end{aligned}$$
+
+This gives meaning to the **pressure gradient** $G$:
+for a pipe of length $L$,
+it describes the pressure difference $\Delta p = p(0) - p(L)$
+that is driving the fluid,
+i.e. $G = \Delta p / L$
+
+As for the latter equation, for $v_z(r)$,
+we start by integrating it once, introducing a constant $A$:
+
+$$\begin{aligned}
+    \dv{}{r}\Big( r \dv{v_z}{r} \Big)
+    = - \frac{G}{\eta} r
+    \quad \implies \quad
+    \dv{v_z}{r}
+    = - \frac{G}{2 \eta} r + \frac{A}{r}
+\end{aligned}$$
+
+Integrating this one more time,
+thereby introducing another constant $B$,
+we arrive at:
+
+$$\begin{aligned}
+    v_z
+    = - \frac{G}{4 \eta} r^2 + A \ln{r} + B
+\end{aligned}$$
+
+The velocity must be finite at $r = 0$, so we set $A = 0$.
+Furthermore, the Navier-Stokes equation's *no-slip* condition
+demands that $v_z = 0$ at the boundary $r = R$,
+so $B = G R^2 / (4 \eta)$.
+This brings us to the **Poiseuille solution** for $v_z(r)$:
+
+$$\begin{aligned}
+    \boxed{
+        v_z(r)
+        = \frac{G}{4 \eta} (R^2 - r^2)
+    }
+\end{aligned}$$
+
+How much fluid can pass through the pipe per unit time?
+This is denoted by the **volumetric flow rate** $Q$,
+which is the integral of $v_z$ over the circular cross-section:
+
+$$\begin{aligned}
+    Q
+    = 2 \pi \int_0^R v_z(r) \: r \dd{r}
+    = \frac{\pi G}{2 \eta} \int_0^R R^2 r - r^3 \dd{r}
+    = \frac{\pi G}{2 \eta} \bigg[ \frac{R^2 r^2}{2} - \frac{r^4}{4} \bigg]_0^R
+\end{aligned}$$
+
+We thus arrive at the main Hagen-Poiseuille equation,
+which predicts $Q$ for a given setup:
+
+$$\begin{aligned}
+    \boxed{
+        Q
+        = \frac{\pi G R^4}{8 \eta}
+    }
+\end{aligned}$$
+
+Consequently, the average flow velocity $\Expval{v_z}$
+is simply $Q$ divided by the cross-sectional area:
+
+$$\begin{aligned}
+    \Expval{v_z}
+    = \frac{Q}{\pi R^2}
+    = \frac{G R^2}{8 \eta}
+\end{aligned}$$
+
+The fluid's viscous stickiness means it exerts a drag force $D$
+on the pipe as it flows. For a pipe of length $L$ and radius $R$,
+we calculate $D$ by multiplying the internal area $2 \pi R L$
+by the [shear stress](/know/concept/cauchy-stress-tensor/)
+$-\sigma_{zr}$ on the wall
+(i.e. the wall applies $\sigma_{zr}$, the fluid responds with $- \sigma_{zr}$):
+
+$$\begin{aligned}
+    D
+    = - 2 \pi R L \: \sigma_{zr} \big|_{r = R}
+    = - 2 \pi R L \eta \dv{v_z}{r}\Big|_{r = R}
+    = 2 \pi R L \eta \frac{G R}{2 \eta}
+    = \pi R^2 L G
+\end{aligned}$$
+
+We would like to get rid of $G$ for being impractical,
+so we substitute $R^2 G = 8 \eta \Expval{v_z}$, yielding:
+
+$$\begin{aligned}
+    \boxed{
+        D
+        = 8 \pi \eta L \Expval{v_z}
+    }
+\end{aligned}$$
+
+Due to this drag, the pressure difference $\Delta p = p(0) - p(L)$
+does work on the fluid, at a rate $P$,
+since power equals force (i.e. pressure times area) times velocity:
+
+$$\begin{aligned}
+    P
+    = 2 \pi \int_0^R \Delta p \: v_z(r) \: r \dd{r}
+\end{aligned}$$
+
+Because $\Delta p$ is independent of $r$,
+we get the same integral we used to calculate $Q$.
+Then, thanks to the fact that $\Delta p = G L$
+and $Q = \pi R^2 \Expval{v_z}$, it follows that:
+
+$$\begin{aligned}
+    P
+    = \Delta p \: Q
+    = G L \pi R^2  \Expval{v_z}
+    = D \Expval{v_z}
+\end{aligned}$$
+
+In conclusion, the power $P$,
+needed to drive a fluid through the pipe at a rate $Q$,
+is given by:
+
+$$\begin{aligned}
+    \boxed{
+        P
+        = 8 \pi \eta L \Expval{v_z}^2
+    }
+\end{aligned}$$
+
+
+
+## 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/hamiltonian-mechanics/index.md b/source/know/concept/hamiltonian-mechanics/index.md
new file mode 100644
index 0000000..610f8bd
--- /dev/null
+++ b/source/know/concept/hamiltonian-mechanics/index.md
@@ -0,0 +1,308 @@
+---
+title: "Hamiltonian mechanics"
+date: 2021-07-03
+categories:
+- Physics
+- Classical mechanics
+layout: "concept"
+---
+
+**Hamiltonian mechanics** is an alternative formulation of classical mechanics,
+which equivalent to Newton's laws,
+but often mathematically advantageous.
+It is built on the shoulders of [Lagrangian mechanics](/know/concept/lagrangian-mechanics/),
+which is in turn built on [variational calculus](/know/concept/calculus-of-variations/).
+
+
+## Definitions
+
+In Lagrangian mechanics, use a Lagrangian $L$,
+which depends on position $q(t)$ and velocity $\dot{q}(t)$,
+to define the momentum $p(t)$ as a derived quantity.
+Hamiltonian mechanics switches the roles of $\dot{q}$ and $p$:
+the **Hamiltonian** $H$ is a function of $q$ and $p$,
+and the velocity $\dot{q}$ is derived from it:
+
+$$\begin{aligned}
+    \pdv{L(q, \dot{q})}{\dot{q}} = p
+    \qquad \quad
+    \pdv{H(q, p)}{p} \equiv \dot{q}
+\end{aligned}$$
+
+Conveniently, this switch turns out to be
+[Legendre transformation](/know/concept/legendre-transform/):
+$H$ is the Legendre transform of $L$,
+with $p = \partial L / \partial \dot{q}$ taken as
+the coordinate to replace $\dot{q}$.
+Therefore:
+
+$$\begin{aligned}
+    \boxed{
+        H(q, p) \equiv \dot{q} \: p - L(q, \dot{q})
+    }
+\end{aligned}$$
+
+This almost always works,
+because $L$ is usually a second-order polynomial of $\dot{q}$,
+and thus convex as required for Legendre transformation.
+In the above expression,
+$\dot{q}$ must be rewritten in terms of $p$ and $q$,
+which is trivial, since $p$ is proportional to $\dot{q}$ by definition.
+
+The Hamiltonian $H$ also has a direct physical meaning:
+for a mass $m$, and for $L = T - V$,
+it is straightforward to show that $H$ represents the total energy $T + V$:
+
+$$\begin{aligned}
+    H
+    = \dot{q} \: p - L
+    = m \dot{q}^2 - L
+    = 2 T - (T - V)
+    = T + V
+\end{aligned}$$
+
+Just as Lagrangian mechanics,
+Hamiltonian mechanics scales well for large systems.
+Its definition is generalized as follows to $N$ objects,
+where $p$ is shorthand for $p_1, ..., p_N$:
+
+$$\begin{aligned}
+    \boxed{
+        H(q, p)
+        \equiv \bigg( \sum_{n = 1}^N \dot{q}_n \: p_n \bigg) - L(q, \dot{q})
+    }
+\end{aligned}$$
+
+The positions and momenta $(q, p)$ form a phase space,
+i.e. they fully describe the state.
+
+An extremely useful concept in Hamiltonian mechanics
+is the **Poisson bracket** (PB),
+which is a binary operation on two quantities $A(q, p)$ and $B(q, p)$,
+denoted by $\{A, B\}$:
+
+$$\begin{aligned}
+    \boxed{
+        \{ A, B \}
+        \equiv \sum_{n = 1}^N \Big( \pdv{A}{q_n} \pdv{B}{p_n} - \pdv{A}{p_n} \pdv{B}{q_n} \Big)
+    }
+\end{aligned}$$
+
+
+## Canonical equations
+
+Lagrangian mechanics has a single Euler-Lagrange equation per object,
+yielding $N$ second-order equations of motion in total.
+In contrast, Hamiltonian mechanics has $2 N$ first-order equations of motion,
+known as **Hamilton's canonical equations**:
+
+$$\begin{aligned}
+    \boxed{
+        - \pdv{H}{q_n} = \dot{p}_n
+        \qquad
+        \pdv{H}{p_n} = \dot{q}_n
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-canoneq"/>
+<label for="proof-canoneq">Proof</label>
+<div class="hidden">
+<label for="proof-canoneq">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \pdv{H}{p_n}
+    &= \pdv{}{p_n}\Big( \sum_{j} \dot{q}_j \: p_j - L \Big)
+    \\
+    &= \sum_{j} \bigg( \Big( \dot{q}_j \pdv{p_j}{p_n} + p_j \pdv{\dot{q}_j}{p_n} \Big)
+    - \Big( \pdv{L}{q_j} \pdv{q_j}{p_n} + \pdv{L}{\dot{q}_j} \pdv{\dot{q}_j}{p_n} \Big) \bigg)
+    \\
+    &= \dot{q}_n + \sum_{j} \Big( p_j \pdv{\dot{q}_j}{p_n}
+    - 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
+    = \dot{q}_n
+\end{aligned}$$
+</div>
+</div>
+
+Just like in Lagrangian mechanics, if $H$ does not explicitly contain $q_n$,
+then $q_n$ is called a **cyclic coordinate**, and leads to the conservation of $p_n$:
+
+$$\begin{aligned}
+    \dot{p}_n = - \pdv{H}{q_n} = 0
+    \quad \implies \quad
+    p_n = \mathrm{conserved}
+\end{aligned}$$
+
+Of course, there may be other conserved quantities.
+Generally speaking, the $t$-derivative of an arbitrary quantity $A(q, p, t)$ is as follows,
+where $\ipdv{}{t}$ is a "soft" derivative
+(only affects explicit occurrences of $t$),
+and $\idv{}{t}$ is a "hard" derivative
+(also affects implicit $t$ inside $q$ and $p$):
+
+$$\begin{aligned}
+    \boxed{
+        \dv{A}{t}
+        = \{ A, H \} + \pdv{A}{t}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-diff-t"/>
+<label for="proof-diff-t">Proof</label>
+<div class="hidden">
+<label for="proof-diff-t">Proof.</label>
+We differentiate via the multivariate chain rule,
+insert the canonical equations,
+and eventually recognize the PB definition:
+
+$$\begin{aligned}
+    \dv{A}{t}
+    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{q_n}{t} + \pdv{A}{p_n} \pdv{p_n}{t} \Big) + \pdv{A}{t}
+    \\
+    &= \sum_{n} \Big( \pdv{A}{q_n} \dot{q}_n + \pdv{A}{p_n} \dot{p}_n \Big) + \pdv{A}{t}
+    \\
+    &= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
+\end{aligned}$$
+</div>
+</div>
+
+Assuming that $H$ does not explicitly depend on $t$,
+the above property naturally leads us to an alternative
+way of writing Hamilton's canonical equations:
+
+$$\begin{aligned}
+    \dot{q}_n = \{ q_n, H \}
+    \qquad \quad
+    \dot{p}_n = \{ p_n, H \}
+\end{aligned}$$
+
+
+
+## Canonical coordinates
+
+So far, we have assumed that the phase space coordinates $(q, p)$
+are the *positions* and *canonical momenta*, respectively,
+and that led us to Hamilton's canonical equations.
+
+In theory, we could make a transformation of the following general form:
+
+$$\begin{aligned}
+    q \to Q(q, p)
+    \qquad \quad
+    p \to P(q, p)
+\end{aligned}$$
+
+However, most choices of $(Q, P)$ would not preserve Hamilton's equations.
+Any $(Q, P)$ that do keep this form
+are known as **canonical coordinates**,
+and the corresponding transformation is a **canonical transformation**.
+That is, any $(Q, P)$ that satisfy:
+
+$$\begin{aligned}
+    - \pdv{H}{Q_n} = \dot{P}_n
+    \qquad \quad
+    \pdv{H}{P_n} = \dot{Q}_n
+\end{aligned}$$
+
+Then we might as well write $H(q, p)$ as $H(Q, P)$.
+So, which $(Q, P)$ fulfill this?
+It turns out that the following must be satisfied for all $n, j$,
+where $\delta_{nj}$ is the Kronecker delta:
+
+$$\begin{aligned}
+    \boxed{
+        \{ Q_n, Q_j \} = \{ P_n, P_j \} = 0
+        \qquad
+        \{ Q_n, P_j \} = \delta_{nj}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-cantrans"/>
+<label for="proof-cantrans">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-cantrans">Proof.</label>
+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. $\{A, B\} = - \{B, A\}$.
+</div>
+</div>
+
+If you have experience with quantum mechanics,
+the latter equation should look suspiciously similar
+to the *canonical commutation relation* $[\hat{Q}, \hat{P}] = i \hbar$.
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/source/know/concept/harmonic-oscillator/index.md b/source/know/concept/harmonic-oscillator/index.md
new file mode 100644
index 0000000..f17203b
--- /dev/null
+++ b/source/know/concept/harmonic-oscillator/index.md
@@ -0,0 +1,287 @@
+---
+title: "Harmonic oscillator"
+date: 2021-06-02
+categories:
+- Physics
+- Mathematics
+layout: "concept"
+---
+
+A **harmonic oscillator** obeys
+the simple 1D version of [Hooke's law](/know/concept/hookes-law/):
+to displace the system away from its equilibrium,
+the needed force $F_d(x)$ scales linearly with the displacement $x(t)$:
+
+$$\begin{aligned}
+    F_d(x) = k x
+\end{aligned}$$
+
+Where $k$ is a system-specific proportionality constant,
+called the **spring constant**,
+since a spring is a good example of a harmonic oscillator,
+at least for small displacements.
+Hooke's law is also often stated for
+the restoring force $F_r(x)$ instead:
+
+$$\begin{aligned}
+    F_r(x) = - k x
+\end{aligned}$$
+
+Let a mass $m$ be attached to the end of the spring.
+After displacing it, we let it go $F_d = 0$,
+so Newton's second law for the restoring force $F_r$ demands that:
+
+$$\begin{aligned}
+    F_r = m x''
+\end{aligned}$$
+
+But $F_r = - k x$,
+meaning $m x'' = - k x$,
+leading to the following equation for $x(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        x'' + \omega_0^2 x = 0
+    }
+\end{aligned}$$
+
+Where $\omega_0 \equiv \sqrt{k / m}$ is the **natural frequency** of the system.
+This differential equation has the following general solution:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = C_1 \sin(\omega_0 t) + C_2 \cos(\omega_0 t)
+    }
+\end{aligned}$$
+
+Where $C_1$ and $C_2$ are constants determined by the initial conditions.
+For example, for $x(0) = 1$ and $x'(0) = 0$, the solution becomes:
+
+$$\begin{aligned}
+    x(t) = \cos(\omega_0 t)
+\end{aligned}$$
+
+When using [Lagrangian](/know/concept/lagrangian-mechanics/)
+or Hamiltonian mechanics,
+we need to know the potential energy $V(x)$
+added to the system by a displacement to $x$.
+This equals the work done by the displacement,
+and is therefore given by:
+
+$$\begin{aligned}
+    V(x) = \int_0^x F_d(x) \:dx = \frac{1}{2} k x^2 = \frac{1}{2} m \omega_0^2 x^2
+\end{aligned}$$
+
+
+## Damped oscillation
+
+If there is a **friction force** $F_f$ affecting the system,
+then the oscillation amplitude will decrease,
+or it might not oscillate at all.
+We define $F_f$ using a **viscous damping coefficient** $c$:
+
+$$\begin{aligned}
+    F_f = - c x'
+\end{aligned}$$
+
+Both $F_r$ and $F_f$ are acting on the system,
+so Newton's second law states that:
+
+$$\begin{aligned}
+    m x'' = - c x' - k x
+\end{aligned}$$
+
+This can be rewritten in the following conventional form
+by defining the **damping coefficient** $\zeta \equiv c / (2 \sqrt{m k})$,
+which determines the expected behaviour of the system:
+
+$$\begin{aligned}
+    \boxed{
+        x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = 0
+    }
+\end{aligned}$$
+
+The general solution is found from the roots $u$ of the auxiliary quadratic equation:
+
+$$\begin{aligned}
+    u^2 + 2 \zeta \omega_0 u + \omega_0^2 = 0
+\end{aligned}$$
+
+The discriminant $D = 4 \zeta^2 \omega_0^2 - 4 \omega_0^2$
+tells us that the behaviour changes substantially
+depending on the damping coefficient $\zeta$,
+with three possibilities: $\zeta < 1$ or $\zeta = 1$ or $\zeta > 1$.
+
+If $\zeta < 1$, there is **underdamping**:
+the system oscillates with exponentially decaying
+amplitude and reduced frequency $\omega_1 \equiv \omega_0 \sqrt{1 - \zeta^2}$.
+The general solution is:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \big( C_1 \sin(\omega_1 t) + C_2 \cos(\omega_1 t) \big) \exp(- \zeta \omega_0 t)
+    }
+\end{aligned}$$
+
+If $\zeta = 1$, there is **critical damping**:
+the system returns to its equilibrium point in minimum time.
+The general solution is given by:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \big( C_1 + C_2 t \big) \exp(- \zeta \omega_0 t)
+    }
+\end{aligned}$$
+
+If $\zeta > 1$, there is **overdamping**:
+the system returns to equilibrium slowly.
+The general solution is as follows,
+where $\omega_1 \equiv \omega_0 \sqrt{\zeta^2 - 1}$:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \big( C_1 \exp(\omega_1 t) + C_2 \exp(- \omega_1 t) \big) \exp(- \zeta \omega_0 t)
+    }
+\end{aligned}$$
+
+
+## Forced oscillation
+
+In the differential equations given above,
+the right-hand side has always been zero,
+meaning that the oscillator is not affected by any external forces.
+What if we put a function there?
+
+$$\begin{aligned}
+    x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = f(t)
+\end{aligned}$$
+
+Obviously, there exist infinitely many $f(t)$ to choose from,
+and each needs a separate analysis.
+However, there is one type of $f(t)$ that deserves special mention,
+namely sinusoids:
+
+$$\begin{aligned}
+    \boxed{
+        x'' + 2 \zeta \omega_0 x' + \omega_0^2 x = \frac{F}{m} \cos(\omega t + \chi)
+    }
+\end{aligned}$$
+
+Where $F$ is a constant force, $\chi$ is an arbitrary phase,
+and the frequency $\omega$ is not necessarily $\omega_0$.
+We solve this case for $x(t)$ in detail.
+Consider the complex version of the equation:
+
+$$\begin{aligned}
+    X'' + 2 \zeta \omega_0 X' + \omega_0^2 X = \frac{F}{m} \exp\!\big(i (\omega t + \chi)\big)
+\end{aligned}$$
+
+Then $x(t) = \Real\{X(t)\}$.
+Inserting the ansatz $X(t) = C \exp(i \omega t)$,
+for some constant $C$:
+
+$$\begin{aligned}
+    - C \omega^2 + C 2 i \zeta \omega_0 \omega + C \omega_0^2 = \frac{F}{m} \exp(i \chi)
+\end{aligned}$$
+
+Where $\exp(i \omega t)$ has already been divided out.
+We isolate this equation for $C$:
+
+$$\begin{aligned}
+    C
+    = \frac{F}{m \big((\omega_0^2 - \omega^2) + 2 i \zeta \omega_0 \omega\big)} \exp(i \chi)
+    = \frac{F \big((\omega_0^2 - \omega^2) - 2 i \zeta \omega_0 \omega\big)}
+    {m \big((\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2\big)}
+    \exp(i \chi)
+\end{aligned}$$
+
+We would like to rewrite this in polar form $C = r \exp(i \theta)$,
+which turns out to be as follows:
+
+$$\begin{aligned}
+    C
+    &= \frac{F}{m \sqrt{(\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2}}
+    \exp\!\bigg(i \chi - i \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)\bigg)
+\end{aligned}$$
+
+For brevity, let us define the **impedance** $Z$
+and the **phase shift** $\phi$
+in the following way:
+
+$$\begin{aligned}
+    Z
+    \equiv \sqrt{(\omega_0^2 - \omega^2)^2 / \omega^2 + 4 \zeta^2 \omega_0^2}
+    \qquad \quad
+    \phi
+    \equiv \arctan\!\Big(\frac{2 \zeta \omega_0 \omega}{\omega_0^2 - \omega^2}\Big)
+\end{aligned}$$
+
+Returning to the original ansatz $X(t) = C \exp(i \omega t)$,
+we take its real part to find $x(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        x(t)
+        = \frac{F}{m \omega Z} \sin(\omega t + \chi - \phi)
+    }
+\end{aligned}$$
+
+Two things are noteworthy here.
+Firstly, $f(t)$ and $x(t)$ are out of phase by $\phi$; there is some lag.
+This is caused by damping, because if $\zeta = 0$, it disappears $\phi = 0$.
+
+Secondly, the amplitude of $x(t)$ depends on $\omega$ and $\omega_0$.
+This brings us to **resonance**,
+where the amplitude can become extremely large.
+Actually, resonance has two subtly different definitions,
+depending on which one of $\omega$ and $\omega_0$ is a free parameter,
+and which one is fixed.
+
+If the natural $\omega_0$ is fixed and the driving $\omega$ is variable,
+we find for which $\omega$ resonance occurs by minimizing the amplitude denominator $\omega Z$.
+We thus find:
+
+$$\begin{aligned}
+    0
+    = \dv{(\omega Z)}{\omega}
+    = \frac{- 4 \omega_0^2 \omega + 4 \omega^3 + 8 \zeta^2 \omega_0^2 \omega}{2 \sqrt{(\omega_0^2 - \omega^2)^2 + 4 \zeta^2 \omega_0^2 \omega^2}}
+    \quad \implies \quad
+    \boxed{
+        \omega = \omega_0 \sqrt{1 - 2 \zeta^2}
+    }
+\end{aligned}$$
+
+Meaning the resonant $\omega$ is lower than $\omega_0$,
+and resonance can only occur if $\zeta < 1 / \sqrt{2}$.
+
+However, if the driving $\omega$ is fixed and the natural is $\omega_0$ is variable,
+the problem is bit more subtle:
+the damping coefficient $\zeta = c / (2 m \omega_0)$
+depends on $\omega_0$.
+This leads us to:
+
+$$\begin{aligned}
+    0
+    = \dv{(\omega Z)}{\omega_0}
+    = \frac{4 \omega_0^3 - 4 \omega^2 \omega_0}{2 \sqrt{(\omega_0^2 - \omega^2)^2 + c^2 \omega^2 / m^2}}
+    \quad \implies \quad
+    \boxed{
+        \omega_0 = \omega
+    }
+\end{aligned}$$
+
+Surprisingly, the damping does not affect $\omega_0$, if $\omega$ is given.
+However, in both cases, the damping *does* matter for the eventual amplitude:
+$c \to 0$ leads to $x \to \infty$,
+and resonance disappears or becomes negligible for $c \to \infty$.
+
+
+
+## References
+1.  M.L. Boas,
+    *Mathematical methods in the physical sciences*, 2nd edition,
+    Wiley.
diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
new file mode 100644
index 0000000..30b5f5d
--- /dev/null
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -0,0 +1,96 @@
+---
+title: "Heaviside step function"
+date: 2021-02-25
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+The **Heaviside step function** $\Theta(t)$,
+is a discontinuous function used for enforcing causality
+or for representing a signal switched on at $t = 0$.
+It is defined as:
+
+$$\begin{aligned}
+    \boxed{
+        \Theta(t) =
+        \begin{cases}
+            0 & \mathrm{if}\: t < 0 \\
+            1 & \mathrm{if}\: t > 1
+        \end{cases}
+    }
+\end{aligned}$$
+
+The value of $\Theta(t \!=\! 0)$ varies between definitions;
+common choices are $0$, $1$ and $1/2$.
+In practice, this rarely matters, and some authors even
+change their definition on the fly for convenience.
+For physicists, $\Theta(0) = 1$ is generally best, such that:
+
+$$\begin{aligned}
+    \boxed{
+        \forall n \in \mathbb{R}: \Theta^n(t) = \Theta(t)
+    }
+\end{aligned}$$
+
+Unsurprisingly, the first-order derivative of $\Theta(t)$ is
+the [Dirac delta function](/know/concept/dirac-delta-function/):
+
+$$\begin{aligned}
+    \boxed{
+        \Theta'(t) = \delta(t)
+    }
+\end{aligned}$$
+
+The [Fourier transform](/know/concept/fourier-transform/)
+of $\Theta(t)$ is as follows,
+where $\pv{}$ is the Cauchy principal value,
+$A$ and $s$ are constants from the FT's definition,
+and $\mathrm{sgn}$ is the signum function:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{\Theta}(\omega)
+        = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-fourier"/>
+<label for="proof-fourier">Proof</label>
+<div class="hidden">
+<label for="proof-fourier">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \tilde{\Theta}(\omega)
+    &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
+    = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
+\end{aligned}$$
+</div>
+</div>
+
+The use of $\pv{}$ without an integral is an abuse of notation,
+and means that this result only makes sense when wrapped in an integral.
+Formally, $\pv{\{1 / \omega\}}$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
+
diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
new file mode 100644
index 0000000..b6c49d7
--- /dev/null
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -0,0 +1,115 @@
+---
+title: "Heisenberg picture"
+date: 2021-02-24
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+The **Heisenberg picture** is an alternative formulation of quantum
+mechanics, and is equivalent to the traditionally-taught Schrödinger equation.
+
+In the Schrödinger picture, the operators (observables) are fixed
+(as long as they do not depend on time), while the state
+$\Ket{\psi_S(t)}$ changes according to the Schrödinger equation,
+which can be written using the generator of translations $\hat{U}(t)$ like so,
+for a time-independent $\hat{H}_S$:
+
+$$\begin{aligned}
+    \Ket{\psi_S(t)} = \hat{U}(t) \Ket{\psi_S(0)}
+    \qquad \quad
+    \boxed{
+        \hat{U}(t) \equiv \exp\!\bigg(\!-\! i \frac{\hat{H}_S t}{\hbar} \bigg)
+    }
+\end{aligned}$$
+
+In contrast, the Heisenberg picture reverses the roles:
+the states $\Ket{\psi_H}$ are invariant,
+and instead the operators vary with time.
+An advantage of this is that the basis states remain the same.
+
+Given a Schrödinger-picture state $\Ket{\psi_S(t)}$, and operator
+$\hat{L}_S(t)$ which may or may not depend on time, they can be
+converted to the Heisenberg picture by the following change of basis:
+
+$$\begin{aligned}
+    \boxed{
+        \Ket{\psi_H} \equiv \Ket{\psi_S(0)}
+        \qquad
+        \hat{L}_H(t) \equiv \hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)
+    }
+\end{aligned}$$
+
+Since $\hat{U}(t)$ is unitary, the expectation value of a given operator is unchanged:
+
+$$\begin{aligned}
+    \expval{\hat{L}_H}
+    &= \matrixel{\psi_H}{\hat{L}_H(t)}{\psi_H}
+    = \matrixel{\psi_S(0)}{\hat{U}^\dagger(t) \: \hat{L}_S(t) \: \hat{U}(t)}{\psi_S(0)}
+    \\
+    &= \matrixel{\hat{U}(t) \psi_S(0)}{\hat{L}_S(t)}{\hat{U}(t) \psi_S(0)}
+    = \matrixel{\psi_S(t)}{\hat{L}_S}{\psi_S(t)}
+    = \expval{\hat{L}_S}
+\end{aligned}$$
+
+The Schrödinger and Heisenberg pictures therefore respectively
+correspond to active and passive transformations by $\hat{U}(t)$
+in [Hilbert space](/know/concept/hilbert-space/).
+The two formulations are thus entirely equivalent,
+and can be derived from one another,
+as will be shown shortly.
+
+In the Heisenberg picture, the states are constant,
+so the time-dependent Schrödinger equation is not directly useful.
+Instead, we will use it derive a new equation for $\hat{L}_H(t)$.
+The key is that the generator $\hat{U}(t)$ is defined from the Schrödinger equation:
+
+$$\begin{aligned}
+    \dv{}{t}\hat{U}(t) = - \frac{i}{\hbar} \hat{H}_S(t) \: \hat{U}(t)
+\end{aligned}$$
+
+Where $\hat{H}_S(t)$ may depend on time. We differentiate the definition of
+$\hat{L}_H(t)$ and insert the other side of the Schrödinger equation
+when necessary:
+
+$$\begin{aligned}
+    \dv{}{\hat{L}H}{t}
+    &= \dv{\hat{U}^\dagger}{t} \hat{L}_S \hat{U}
+    + \hat{U}^\dagger \hat{L}_S \dv{\hat{U}}{t}
+    + \hat{U}^\dagger \dv{\hat{L}_S}{t} \hat{U}
+    \\
+    &= \frac{i}{\hbar} \hat{U}^\dagger \hat{H}_S (\hat{U} \hat{U}^\dagger) \hat{L}_S \hat{U}
+    - \frac{i}{\hbar} \hat{U}^\dagger \hat{L}_S (\hat{U} \hat{U}^\dagger) \hat{H}_S \hat{U}
+    + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    \\
+    &= \frac{i}{\hbar} \hat{H}_H \hat{L}_H
+    - \frac{i}{\hbar} \hat{L}_H \hat{H}_H
+    + \Big( \dv{\hat{L}_S}{t} \Big)_H
+    = \frac{i}{\hbar} \comm{\hat{H}_H}{\hat{L}_H} + \Big( \dv{\hat{L}_S}{t} \Big)_H
+\end{aligned}$$
+
+We thus get the equation of motion for operators in the Heisenberg picture:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{}{t}\hat{L}_H(t) = \frac{i}{\hbar} \comm{\hat{H}_H(t)}{\hat{L}_H(t)} + \Big( \dv{}{t}\hat{L}_S(t) \Big)_H
+    }
+\end{aligned}$$
+
+This equation is closer to classical mechanics than the Schrödinger picture:
+inserting the position $\hat{X}$ and momentum $\hat{P} = - i \hbar \: \idv{}{\hat{X}}$
+gives the following Newton-style equations:
+
+$$\begin{aligned}
+    \dv{\hat{X}}{t}
+    &= \frac{i}{\hbar} \comm{\hat{H}}{\hat{X}}
+    = \frac{\hat{P}}{m}
+    \\
+    \dv{\hat{P}}{t}
+    &= \frac{i}{\hbar} \comm{\hat{H}}{\hat{P}}
+    = - \dv{V(\hat{X})}{\hat{X}}
+\end{aligned}$$
+
+For a proof, see [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/),
+which is closely related to the Heisenberg picture.
diff --git a/source/know/concept/hellmann-feynman-theorem/index.md b/source/know/concept/hellmann-feynman-theorem/index.md
new file mode 100644
index 0000000..6b458db
--- /dev/null
+++ b/source/know/concept/hellmann-feynman-theorem/index.md
@@ -0,0 +1,91 @@
+---
+title: "Hellmann-Feynman theorem"
+date: 2021-11-29
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+Consider the time-independent Schrödinger equation,
+where the Hamiltonian $\hat{H}$ depends on a general parameter $\lambda$,
+whose meaning or type we will not specify:
+
+$$\begin{aligned}
+    \hat{H}(\lambda) \Ket{\psi_n(\lambda)}
+    = E_n(\lambda) \Ket{\psi_n(\lambda)}
+\end{aligned}$$
+
+Assuming all eigenstates $\Ket{\psi_n}$ are normalized,
+this gives us the following basic relation:
+
+$$\begin{aligned}
+    \matrixel{\psi_m}{\hat{H}}{\psi_n}
+    = E_n \Inprod{\psi_m}{\psi_n}
+    = \delta_{mn} E_n
+\end{aligned}$$
+
+We differentiate this with respect to $\lambda$,
+which could be a scalar or a vector.
+This yields:
+
+$$\begin{aligned}
+    \delta_{mn} \nabla_\lambda E_n
+    &= \nabla_\lambda \matrixel{\psi_m}{\hat{H}}{\psi_n}
+    \\
+    &= \matrixel{\nabla_\lambda \psi_m}{\hat{H}}{\psi_n}
+    + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    + \matrixel{\psi_m}{\hat{H}}{\nabla_\lambda \psi_n}
+    \\
+    &= E_m \Inprod{\psi_m}{\nabla_\lambda \psi_n} + E_n \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+\end{aligned}$$
+
+In order to simplify this,
+we differentiate the orthogonality relation
+$\Inprod{\psi_m}{\psi_n} = \delta_{mn}$,
+which ends up telling us that
+$\Inprod{\nabla_\lambda \psi_m}{\psi_n} = - \Inprod{\psi_m}{\nabla_\lambda \psi_n}$:
+
+$$\begin{aligned}
+    0
+    = \nabla_\lambda \delta_{mn}
+    = \nabla_\lambda \Inprod{\psi_m}{\psi_n}
+    = \Inprod{\nabla_\lambda \psi_m}{\psi_n} + \Inprod{\psi_m}{\nabla_\lambda \psi_n}
+\end{aligned}$$
+
+Using this result to replace $\Inprod{\nabla_\lambda \psi_m}{\psi_n}$
+in the previous equation leads to:
+
+$$\begin{aligned}
+    \delta_{mn} \nabla_\lambda E_n
+    &= (E_m - E_n) \Inprod{\psi_m}{\nabla_\lambda \psi_n} + \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+\end{aligned}$$
+
+For $m = n$, we therefore arrive at the **Hellmann-Feynman theorem**,
+which is useful when doing numerical calculations
+to minimize energies with respect to $\lambda$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla_\lambda E_n
+        = \matrixel{\psi_n}{\nabla_\lambda \hat{H}}{\psi_n}
+    }
+\end{aligned}$$
+
+While for $m \neq n$, we get the **Epstein generalization**
+of the Hellmann-Feynman theorem, which is for example relevant for
+the [Berry phase](/know/concept/berry-phase/):
+
+$$\begin{aligned}
+    \boxed{
+        (E_n - E_m) \Inprod{\psi_m}{\nabla_\lambda \psi_n}
+        = \matrixel{\psi_m}{\nabla_\lambda \hat{H}}{\psi_n}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  G. Grosso, G.P. Parravicini,
+    *Solid state physics*,
+    2nd edition, Elsevier.
diff --git a/source/know/concept/hermite-polynomials/index.md b/source/know/concept/hermite-polynomials/index.md
new file mode 100644
index 0000000..17e61df
--- /dev/null
+++ b/source/know/concept/hermite-polynomials/index.md
@@ -0,0 +1,94 @@
+---
+title: "Hermite polynomials"
+date: 2021-09-08
+categories:
+- Mathematics
+- Statistics
+layout: "concept"
+---
+
+The **Hermite polynomials** are a set of functions
+that appear in physics and statistics,
+although slightly different definitions are used in those fields.
+
+
+## Physicists' definition
+
+The **Hermite equation** is an eigenvalue problem for $n$,
+and the Hermite polynomials $H_n(x)$ are its eigenfunctions $u(x)$,
+subject to the boundary condition that $u$ grows at most polynomially,
+in which case the eigenvalues $n$ are non-negative integers:
+
+$$\begin{aligned}
+    \boxed{
+        u'' - 2 x u' + 2 n u = 0
+    }
+\end{aligned}$$
+
+The $n$th-order Hermite polynomial $H_n(x)$
+is therefore as follows, according to physicists:
+
+$$\begin{aligned}
+    H_n(x)
+    &= (-1)^n \exp(x^2) \dvn{n}{}{x}\exp(- x^2)
+    \\
+    &= \Big( 2 x - \dv{}{x}\Big)^n 1
+\end{aligned}$$
+
+This form is known as a *Rodrigues' formula*.
+The first handful of Hermite polynomials are:
+
+$$\begin{gathered}
+    H_0(x) = 1
+    \qquad \quad
+    H_1(x) = 2 x
+    \qquad \quad
+    H_2(x) = 4 x^2 - 2
+    \\
+    H_3(x) = 8 x^3 - 12 x
+    \qquad \quad
+    H_4(x) = 16 x^4 - 48 x^2 + 12
+\end{gathered}$$
+
+And then more $H_n$ can be computed quickly
+using the following recurrence relation:
+
+$$\begin{aligned}
+    \boxed{
+        H_{n + 1}(x) = 2 x H_n(x) - 2n H_{n-1}(x)
+    }
+\end{aligned}$$
+
+They (almost) form an *Appell sequence*,
+meaning their derivatives are like so:
+
+$$\begin{aligned}
+    \boxed{
+        \dvn{k}{}{x}H_n(x)
+        = 2^k \frac{n!}{(n - k)!} H_{n - k}(x)
+    }
+\end{aligned}$$
+
+Importantly, all $H_n$ are orthogonal with respect to the weight function $w(x) \equiv \exp(- x^2)$:
+
+$$\begin{aligned}
+    \boxed{
+        \Inprod{H_n}{w H_m}
+        \equiv \int_{-\infty}^\infty H_n(x) \: H_m(x) \: w(x) \dd{x}
+        = \sqrt{\pi} 2^n n! \: \delta_{nm}
+    }
+\end{aligned}$$
+
+Where $\delta_{nm}$ is the Kronecker delta.
+Finally, they form a basis in the [Hilbert space](/know/concept/hilbert-space/)
+of all functions $f(x)$ for which $\Inprod{f}{w f}$ is finite.
+This means that every such $f$ can be expanded in $H_n$:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \sum_{n = 0}^\infty a_n H_n(x)
+        = \sum_{n = 0}^\infty \frac{\Inprod{H_n}{w f}}{\Inprod{H_n}{w H_n}} H_n(x)
+    }
+\end{aligned}$$
+
diff --git a/source/know/concept/hilbert-space/index.md b/source/know/concept/hilbert-space/index.md
new file mode 100644
index 0000000..d2b9770
--- /dev/null
+++ b/source/know/concept/hilbert-space/index.md
@@ -0,0 +1,196 @@
+---
+title: "Hilbert space"
+date: 2021-02-22
+categories:
+- Mathematics
+- Quantum mechanics
+layout: "concept"
+---
+
+A **Hilbert space**, also called an **inner product space**, is an
+abstract **vector space** with a notion of length and angle.
+
+
+## Vector space
+
+An abstract **vector space** $\mathbb{V}$ is a generalization of the
+traditional concept of vectors as "arrows". It consists of a set of
+objects called **vectors** which support the following (familiar)
+operations:
+
++ **Vector addition**: the sum of two vectors $V$ and $W$, denoted $V + W$.
++ **Scalar multiplication**: product of a vector $V$ with a scalar $a$, denoted $a V$.
+
+In addition, for a given $\mathbb{V}$ to qualify as a proper vector
+space, these operations must obey the following axioms:
+
++ **Addition is associative**: $U + (V + W) = (U + V) + W$
++ **Addition is commutative**: $U + V = V + U$
++ **Addition has an identity**: there exists a $\mathbf{0}$ such that $V + 0 = V$
++ **Addition has an inverse**: for every $V$ there exists $-V$ so that $V + (-V) = 0$
++ **Multiplication is associative**: $a (b V) = (a b) V$
++ **Multiplication has an identity**: There exists a $1$ such that $1 V = V$
++ **Multiplication is distributive over scalars**: $(a + b)V = aV + bV$
++ **Multiplication is distributive over vectors**: $a (U + V) = a U + a V$
+
+A set of $N$ vectors $V_1, V_2, ..., V_N$ is **linearly independent** if
+the only way to satisfy the following relation is to set all the scalar coefficients $a_n = 0$:
+
+$$\begin{aligned}
+    \mathbf{0} = \sum_{n = 1}^N a_n V_n
+\end{aligned}$$
+
+In other words, these vectors cannot be expressed in terms of each
+other. Otherwise, they would be **linearly dependent**.
+
+A vector space $\mathbb{V}$ has **dimension** $N$ if only up to $N$ of
+its vectors can be linearly indepedent. All other vectors in
+$\mathbb{V}$ can then be written as a **linear combination** of these $N$ **basis vectors**.
+
+Let $\vu{e}_1, ..., \vu{e}_N$ be the basis vectors, then any
+vector $V$ in the same space can be **expanded** in the basis according to
+the unique weights $v_n$, known as the **components** of $V$
+in that basis:
+
+$$\begin{aligned}
+    V = \sum_{n = 1}^N v_n \vu{e}_n
+\end{aligned}$$
+
+Using these, the vector space operations can then be implemented as follows:
+
+$$\begin{gathered}
+    V = \sum_{n = 1} v_n \vu{e}_n
+    \quad
+    W = \sum_{n = 1} w_n \vu{e}_n
+    \\
+    \quad \implies \quad
+    V + W = \sum_{n = 1}^N (v_n + w_n) \vu{e}_n
+    \qquad
+    a V = \sum_{n = 1}^N a v_n \vu{e}_n
+\end{gathered}$$
+
+
+## Inner product
+
+A given vector space $\mathbb{V}$ can be promoted to a **Hilbert space**
+or **inner product space** if it supports an operation $\Inprod{U}{V}$
+called the **inner product**, which takes two vectors and returns a
+scalar, and has the following properties:
+
++ **Skew symmetry**: $\Inprod{U}{V} = (\Inprod{V}{U})^*$, where ${}^*$ is the complex conjugate.
++ **Positive semidefiniteness**: $\Inprod{V}{V} \ge 0$, and $\Inprod{V}{V} = 0$ if $V = \mathbf{0}$.
++ **Linearity in second operand**: $\Inprod{U}{(a V + b W)} = a \Inprod{U}{V} + b \Inprod{U}{W}$.
+
+The inner product describes the lengths and angles of vectors, and in
+Euclidean space it is implemented by the dot product.
+
+The **magnitude** or **norm** $|V|$ of a vector $V$ is given by
+$|V| = \sqrt{\Inprod{V}{V}}$ and represents the real positive length of $V$.
+A **unit vector** has a norm of 1.
+
+Two vectors $U$ and $V$ are **orthogonal** if their inner product
+$\Inprod{U}{V} = 0$. If in addition to being orthogonal, $|U| = 1$ and
+$|V| = 1$, then $U$ and $V$ are known as **orthonormal** vectors.
+
+Orthonormality is desirable for basis vectors, so if they are
+not already like that, it is common to manually turn them into a new
+orthonormal basis using e.g. the [Gram-Schmidt method](/know/concept/gram-schmidt-method).
+
+As for the implementation of the inner product, it is given by:
+
+$$\begin{gathered}
+    V = \sum_{n = 1}^N v_n \vu{e}_n
+    \quad
+    W = \sum_{n = 1}^N w_n \vu{e}_n
+    \\
+    \quad \implies \quad
+    \Inprod{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \Inprod{\vu{e}_n}{\vu{e}_j}
+\end{gathered}$$
+
+If the basis vectors $\vu{e}_1, ..., \vu{e}_N$ are already
+orthonormal, this reduces to:
+
+$$\begin{aligned}
+    \Inprod{V}{W} = \sum_{n = 1}^N v_n^* w_n
+\end{aligned}$$
+
+As it turns out, the components $v_n$ are given by the inner product
+with $\vu{e}_n$, where $\delta_{nm}$ is the Kronecker delta:
+
+$$\begin{aligned}
+    \Inprod{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n
+\end{aligned}$$
+
+
+## Infinite dimensions
+
+As the dimensionality $N$ tends to infinity, things may or may not
+change significantly, depending on whether $N$ is **countably** or
+**uncountably** infinite.
+
+In the former case, not much changes: the infinitely many **discrete**
+basis vectors $\vu{e}_n$ can all still be made orthonormal as usual,
+and as before:
+
+$$\begin{aligned}
+    V = \sum_{n = 1}^\infty v_n \vu{e}_n
+\end{aligned}$$
+
+A good example of such a countably-infinitely-dimensional basis are the
+solution eigenfunctions of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
+
+However, if the dimensionality is uncountably infinite, the basis
+vectors are **continuous** and cannot be labeled by $n$. For example, all
+complex functions $f(x)$ defined for $x \in [a, b]$ which
+satisfy $f(a) = f(b) = 0$ form such a vector space.
+In this case $f(x)$ is expanded as follows, where $x$ is a basis vector:
+
+$$\begin{aligned}
+    f(x) = \int_a^b \Inprod{x}{f} \dd{x}
+\end{aligned}$$
+
+Similarly, the inner product $\Inprod{f}{g}$ must also be redefined as
+follows:
+
+$$\begin{aligned}
+    \Inprod{f}{g} = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{aligned}$$
+
+The concept of orthonormality must be also weakened. A finite function
+$f(x)$ can be normalized as usual, but the basis vectors $x$ themselves
+cannot, since each represents an infinitesimal section of the real line.
+
+The rationale in this case is that action of the identity operator $\hat{I}$ must
+be preserved, which is given here in [Dirac notation](/know/concept/dirac-notation/):
+
+$$\begin{aligned}
+    \hat{I} = \int_a^b \Ket{\xi} \Bra{\xi} \dd{\xi}
+\end{aligned}$$
+
+Applying the identity operator to $f(x)$ should just give $f(x)$ again:
+
+$$\begin{aligned}
+    f(x) = \Inprod{x}{f} = \matrixel{x}{\hat{I}}{f}
+    = \int_a^b \Inprod{x}{\xi} \Inprod{\xi}{f} \dd{\xi}
+    = \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
+\end{aligned}$$
+
+Since we want the latter integral to reduce to $f(x)$, it is plain to see that
+$\Inprod{x}{\xi}$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
+i.e $\Inprod{x}{\xi} = \delta(x - \xi)$:
+
+$$\begin{aligned}
+    \int_a^b \Inprod{x}{\xi} f(\xi) \dd{\xi}
+    = \int_a^b \delta(x - \xi) f(\xi) \dd{\xi}
+    = f(x)
+\end{aligned}$$
+
+Consequently, $\Inprod{x}{\xi} = 0$ if $x \neq \xi$ as expected for an
+orthogonal set of vectors, but if $x = \xi$ the inner product
+$\Inprod{x}{\xi}$ is infinite, unlike earlier.
+
+Technically, because the basis vectors $x$ cannot be normalized, they
+are not members of a Hilbert space, but rather of a superset called a
+**rigged Hilbert space**. Such vectors have no finite inner product with
+themselves, but do have one with all vectors from the actual Hilbert
+space.
diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md
new file mode 100644
index 0000000..17bd5a6
--- /dev/null
+++ b/source/know/concept/holomorphic-function/index.md
@@ -0,0 +1,189 @@
+---
+title: "Holomorphic function"
+date: 2021-02-25
+categories:
+- Mathematics
+- Complex analysis
+layout: "concept"
+---
+
+In complex analysis, a complex function $f(z)$ of a complex variable $z$
+is called **holomorphic** or **analytic** if it is complex differentiable in the
+neighbourhood of every point of its domain.
+This is a very strong condition.
+
+As a result, holomorphic functions are infinitely differentiable and
+equal their Taylor expansion at every point. In physicists' terms,
+they are extremely "well-behaved" throughout their domain.
+
+More formally, a given function $f(z)$ is holomorphic in a certain region
+if the following limit exists for all $z$ in that region,
+and for all directions of $\Delta z$:
+
+$$\begin{aligned}
+    \boxed{
+        f'(z) = \lim_{\Delta z \to 0} \frac{f(z + \Delta z) - f(z)}{\Delta z}
+    }
+\end{aligned}$$
+
+We decompose $f$ into the real functions $u$ and $v$ of real variables $x$ and $y$:
+
+$$\begin{aligned}
+    f(z) = f(x + i y) = u(x, y) + i v(x, y)
+\end{aligned}$$
+
+Since we are free to choose the direction of $\Delta z$, we choose $\Delta x$ and $\Delta y$:
+
+$$\begin{aligned}
+    f'(z)
+    &= \lim_{\Delta x \to 0} \frac{f(z + \Delta x) - f(z)}{\Delta x}
+    = \pdv{u}{x} + i \pdv{v}{x}
+    \\
+    &= \lim_{\Delta y \to 0} \frac{f(z + i \Delta y) - f(z)}{i \Delta y}
+    = \pdv{v}{y} - i \pdv{u}{y}
+\end{aligned}$$
+
+For $f(z)$ to be holomorphic, these two results must be equivalent.
+Because $u$ and $v$ are real by definition,
+we thus arrive at the **Cauchy-Riemann equations**:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{u}{x} = \pdv{v}{y}
+        \qquad
+        \pdv{v}{x} = - \pdv{u}{y}
+    }
+\end{aligned}$$
+
+Therefore, a given function $f(z)$ is holomorphic if and only if its real
+and imaginary parts satisfy these equations. This gives an idea of how
+strict the criteria are to qualify as holomorphic.
+
+
+## Integration formulas
+
+Holomorphic functions satisfy **Cauchy's integral theorem**, which states
+that the integral of $f(z)$ over any closed curve $C$ in the complex plane is zero,
+provided that $f(z)$ is holomorphic for all $z$ in the area enclosed by $C$:
+
+$$\begin{aligned}
+    \boxed{
+        \oint_C f(z) \dd{z} = 0
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-int-theorem"/>
+<label for="proof-int-theorem">Proof</label>
+<div class="hidden">
+<label for="proof-int-theorem">Proof.</label>
+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.
+</div>
+</div>
+
+An interesting consequence is **Cauchy's integral formula**, which
+states that the value of $f(z)$ at an arbitrary point $z_0$ is
+determined by its values on an arbitrary contour $C$ around $z_0$:
+
+$$\begin{aligned}
+    \boxed{
+        f(z_0) = \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-int-formula"/>
+<label for="proof-int-formula">Proof</label>
+<div class="hidden">
+<label for="proof-int-formula">Proof.</label>
+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 $z_0$:
+
+$$\begin{aligned}
+    \lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
+    &= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
+    = f(z_0)
+\end{aligned}$$
+</div>
+</div>
+
+Similarly, **Cauchy's differentiation formula**,
+or **Cauchy's integral formula for derivatives**
+gives all derivatives of a holomorphic function as follows,
+and also guarantees their existence:
+
+$$\begin{aligned}
+    \boxed{
+        f^{(n)}(z_0)
+        = \frac{n!}{2 \pi i} \oint_C \frac{f(z)}{(z - z_0)^{n + 1}} \dd{z}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-diff-formula"/>
+<label for="proof-diff-formula">Proof</label>
+<div class="hidden">
+<label for="proof-diff-formula">Proof.</label>
+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 $n$.
+</div>
+</div>
+
diff --git a/source/know/concept/hookes-law/index.md b/source/know/concept/hookes-law/index.md
new file mode 100644
index 0000000..57ab27f
--- /dev/null
+++ b/source/know/concept/hookes-law/index.md
@@ -0,0 +1,239 @@
+---
+title: "Hooke's law"
+date: 2021-04-02
+categories:
+- Physics
+- Continuum physics
+layout: "concept"
+---
+
+In its simplest form, **Hooke's law** dictates that
+changing the length of an elastic object requires
+a force that is proportional the desired length difference.
+In its most general form, it gives a linear relationship
+between the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
+to the [Cauchy strain tensor](/know/concept/cauchy-strain-tensor/) $\hat{u}$.
+
+Importantly, all forms of Hooke's law are only valid for small deformations,
+since the stress-strain relationship becomes nonlinear otherwise.
+
+
+## Simple form
+
+The simple form of the law is traditionally quoted for springs,
+since they have a spring constant $k$ giving the ratio
+between the force $F$ and extension $x$:
+
+$$\begin{aligned}
+    \boxed{
+        F
+        = k x
+    }
+\end{aligned}$$
+
+In general, all solids are elastic for small extensions,
+and therefore also obey Hooke's law.
+In light of this fact, we replace the traditional spring
+with a rod of length $L$ and cross-section $A$.
+
+The constant $k$ depends on, among several things,
+the spring's length $L$ and cross-section $A$,
+so for our generalization, we want a new parameter
+to describe the proportionality independently of the rod's dimensions.
+To achieve this, we realize that the force $F$ is spread across $A$,
+and that the extension $x$ should be take relative to $L$.
+
+$$\begin{aligned}
+    \frac{F}{A}
+    = \Big( k \frac{L}{A} \Big) \frac{x}{L}
+\end{aligned}$$
+
+The force-per-area $F/A$ on a solid is the definition of **stress**,
+and the relative elongation $x/L$ is the defintion of **strain**.
+If $F$ acts along the $x$-axis, we can then write:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{xx}
+        = E u_{xx}
+    }
+\end{aligned}$$
+
+Where the proportionality constant $E$,
+known as the **elastic modulus** or **Young's modulus**,
+is the general material parameter that we wanted:
+
+$$\begin{aligned}
+    E
+    = k \frac{L}{A}
+\end{aligned}$$
+
+Due to the microscopic structure of some (usually crystalline) materials,
+$E$ might be dependent on the direction of the force $F$.
+For simplicity, we only consider **isotropic** materials,
+which have the same properties measured from any direction.
+
+However, we are still missing something.
+When a spring is pulled,
+it becomes narrower as its coils move apart,
+and this effect is also seen when stretching solids in general:
+if we pull our rod along the $x$-axis, we expect it to deform in $y$ and $z$ as well.
+This is described by **Poisson's ratio** $\nu$:
+
+$$\begin{aligned}
+    \boxed{
+        \nu
+        \equiv - \frac{u_{yy}}{u_{xx}}
+    }
+\end{aligned}$$
+
+Note that $u_{yy} = u_{zz}$ because the material is assumed to be isotropic.
+Intuitively, you may expect that the volume of the object is conserved,
+but for most materials that is not accurate.
+
+In summary, for our example case with a force $F = T A$ pulling at the rod
+along the $x$-axis, the full stress and strain tensors are given by:
+
+$$\begin{aligned}
+    \hat{\sigma} =
+    \begin{bmatrix}
+        T & 0 & 0 \\
+        0 & 0 & 0 \\
+        0 & 0 & 0
+    \end{bmatrix}
+    \qquad
+    \hat{u} =
+    \begin{bmatrix}
+        T/E & 0 & 0 \\
+        0 & -\nu T/E & 0 \\
+        0 & 0 & -\nu T/E
+    \end{bmatrix}
+\end{aligned}$$
+
+
+## General isotropic form
+
+The general form of Hooke's law is a linear relationship
+between the stress and strain tensors:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\sigma}
+        = 2 \mu \: \hat{u} + \lambda \Tr(\hat{u}) \: \hat{1}
+    }
+\end{aligned}$$
+
+Where $\Tr{}$ is the trace.
+This is often written in index notation,
+with the Kronecker delta $\delta_{ij}$:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{ij}
+        = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk}
+    }
+\end{aligned}$$
+
+The constants $\mu$ and $\lambda$ are called the **Lamé coefficients**,
+and are related to $E$ and $\nu$ in a way we can derive
+by returning to the example with a tension $T = F/A$ along $x$.
+For $\sigma_{xx}$, we have:
+
+$$\begin{aligned}
+    T
+    = \sigma_{xx}
+    &= 2 \mu u_{xx} + \lambda (u_{xx} + u_{yy} + u_{zz})
+    \\
+    &= \frac{2 \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T)
+    \\
+    &= \frac{T}{E} \Big( 2 \mu + \lambda (1 - 2 \nu) \Big)
+\end{aligned}$$
+
+Meanwhile, the other diagonal stresses $\sigma_{yy} = \sigma_{zz}$
+are expressed in terms of the strain like so:
+
+$$\begin{aligned}
+    0
+    = \sigma_{yy}
+    &= 2 \mu u_{yy} + \lambda (u_{xx} + u_{yy} + u_{zz})
+    \\
+    &= - \frac{2 \nu \mu}{E} T + \frac{\lambda}{E} T - \frac{\nu \lambda}{E} (T + T)
+    \\
+    &= \frac{E}{T} \Big( \!-\! 2 \nu \mu + \lambda (1 - 2 \nu) \Big)
+\end{aligned}$$
+
+After dividing out superfluous factors from the two preceding equations,
+we arrive at:
+
+$$\begin{aligned}
+    E
+    = 2 \mu + \lambda (1 - 2 \nu)
+    \qquad \quad
+    2 \nu \mu
+    = \lambda (1 - 2 \nu)
+\end{aligned}$$
+
+Solving this system of equations for the Lamé coefficients
+yields the following result:
+
+$$\begin{aligned}
+    \boxed{
+        \lambda
+        = \frac{E \nu}{(1 - 2 \nu)(1 + \nu)}
+        \qquad \quad
+        \mu
+        = \frac{E}{2 (1 + \nu)}
+    }
+\end{aligned}$$
+
+Which can straightforwardly be inverted
+to express $E$ and $\nu$ as a function of $\mu$ and $\lambda$:
+
+$$\begin{aligned}
+    \boxed{
+        E
+        = \mu \frac{3 \lambda + 2 \mu}{\lambda + \mu}
+        \qquad \quad
+        \nu
+        = \frac{\lambda}{2 (\lambda + \mu)}
+    }
+\end{aligned}$$
+
+Hooke's law itself can also be inverted,
+i.e. we can express the strain as a function of stress.
+First, observe that the trace of the stress tensor satisfies:
+
+$$\begin{aligned}
+    \Tr(\hat{\sigma})
+    = \sum_{i} \sigma_{ii}
+    = 2 \mu \sum_{i} u_{ii} + \lambda \sum_{i} \sum_{k} u_{kk}
+    = (2 \mu + 3 \lambda) \sum_{i} u_{ii}
+\end{aligned}$$
+
+Inserting this into Hooke's law
+yields an equation that only contains one strain component $u_{ij}$:
+
+$$\begin{aligned}
+    \sigma_{ij}
+    = 2 \mu u_{ij} + \frac{\lambda}{2 \mu + 3 \lambda} \delta_{ij} \sum_{k} \sigma_{kk}
+\end{aligned}$$
+
+Which is therefore trivial to isolate for $u_{ij}$,
+leading us to Hooke's inverted law:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            u_{ij}
+            &= \frac{\sigma_{ij}}{2 \mu} - \frac{\lambda}{2 \mu (3 \lambda + 2 \mu)} \delta_{ij} \sum_{k} \sigma_{kk}
+            \\
+            &= \frac{1 + \nu}{E} \sigma_{ij} - \frac{\nu}{E} \delta_{ij} \sum_{k} \sigma_{kk}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+
+## 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/hydrostatic-pressure/index.md b/source/know/concept/hydrostatic-pressure/index.md
new file mode 100644
index 0000000..4add09e
--- /dev/null
+++ b/source/know/concept/hydrostatic-pressure/index.md
@@ -0,0 +1,211 @@
+---
+title: "Hydrostatic pressure"
+date: 2021-03-12
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+layout: "concept"
+---
+
+The pressure $p$ inside a fluid at rest,
+the so-called **hydrostatic pressure**,
+is an important quantity.
+Here we will properly define it,
+and derive the equilibrium condition for the fluid to be at rest,
+both with and without an arbitrary gravity field.
+
+
+## Without gravity
+
+Inside the fluid, we can imagine small arbitrary partition surfaces,
+with normal vector $\vu{n}$ and area $\dd{S}$,
+yielding the following vector element $\dd{\va{S}}$:
+
+$$\begin{aligned}
+    \dd{\va{S}}
+    = \vu{n} \dd{S}
+\end{aligned}$$
+
+The orientation of these surfaces does not matter.
+The **pressure** $p(\va{r})$ is defined as the force-per-area
+of these tiny surface elements:
+
+$$\begin{aligned}
+    \dd{\va{F}}
+    = - p(\va{r}) \dd{\va{S}}
+\end{aligned}$$
+
+The negative sign is there because a positive pressure is conventionally defined
+to push from the positive (normal) side of $\dd{\va{S}}$ to the negative side.
+The total force $\va{F}$ on a larger surface inside the fluid is
+then given by the surface integral over many adjacent $\dd{\va{S}}$:
+
+$$\begin{aligned}
+    \va{F}
+    = - \int_S p(\va{r}) \dd{\va{S}}
+\end{aligned}$$
+
+If we now consider a *closed* surface,
+which encloses a "blob" of the fluid,
+then we can use the divergence theorem to get a volume integral:
+
+$$\begin{aligned}
+    \va{F}
+    = - \oint_S p \dd{\va{S}}
+    = - \int_V \nabla p \dd{V}
+\end{aligned}$$
+
+Since the total force on the blob is simply the sum of the forces $\dd{\va{F}}$
+on all its constituent volume elements $\dd{V}$,
+we arrive at the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\va{F}}
+        = - \nabla p \dd{V}
+    }
+\end{aligned}$$
+
+If the fluid is at rest, then all forces on the blob cancel out
+(otherwise it would move).
+Since we are currently neglecting all forces other than pressure,
+this is equivalent to demanding that $\dd{\va{F}} = 0$,
+which implies that $\nabla p = 0$, i.e. the pressure is constant.
+
+$$\begin{aligned}
+    \boxed{
+        \nabla p = 0
+    }
+\end{aligned}$$
+
+
+## With gravity
+
+If we include gravity, then,
+in addition to the pressure's *contact force* $\va{F}_p$ from earlier,
+there is also a *body force* $\va{F}_g$ acting on
+the arbitrary blob $V$ of fluid enclosed by $S$:
+
+$$\begin{aligned}
+    \va{F}_g
+    = \int_V \rho \va{g} \dd{V}
+\end{aligned}$$
+
+Where $\rho$ is the fluid's density (which need not be constant)
+and $\va{g}$ is the gravity field given in units of force-per-mass.
+For a fluid at rest, these forces must cancel out:
+
+$$\begin{aligned}
+    \va{F}
+    = \va{F}_g + \va{F}_p
+    = \int_V \rho \va{g} - \nabla p \dd{V}
+    = 0
+\end{aligned}$$
+
+Since this a single integral over an arbitrary volume,
+it implies that every point of the fluid must
+locally satisfy the following equilibrium condition:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla p
+        = \rho \va{g}
+    }
+\end{aligned}$$
+
+On Earth (or another body with strong gravity),
+it is reasonable to treat $\va{g}$ as only pointing in the downward $z$-direction,
+in which case the above condition turns into:
+
+$$\begin{aligned}
+    p
+    = \rho g_0 z
+\end{aligned}$$
+
+Where $g_0$ is the magnitude of the $z$-component of $\va{g}$.
+We can generalize the equilibrium condition by treating
+the gravity field as the gradient of the gravitational potential $\Phi$:
+
+$$\begin{aligned}
+    \va{g}(\va{r})
+    = - \nabla \Phi(\va{r})
+\end{aligned}$$
+
+With this, the equilibrium condition is turned into the following equation:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \Phi + \frac{\nabla p}{\rho}
+        = 0
+    }
+\end{aligned}$$
+
+In practice, the density $\rho$ of the fluid
+may be a function of the pressure $p$ (compressibility)
+and/or temperature $T$ (thermal expansion).
+We will tackle the first complication, but neglect the second,
+i.e. we assume that the temperature is equal across the fluid.
+
+We then define the **pressure potential** $w(p)$ as
+the indefinite integral of the density:
+
+$$\begin{aligned}
+    w(p)
+    \equiv \int \frac{1}{\rho(p)} \dd{p}
+\end{aligned}$$
+
+Using this, we can rewrite the equilibrium condition as a single gradient like so:
+
+$$\begin{aligned}
+    0
+    = \nabla \Phi + \frac{\nabla p}{\rho}
+    = \nabla \Phi + \dv{w}{p} \nabla p
+    = \nabla \Big( \Phi + w(p) \Big)
+\end{aligned}$$
+
+From this, let us now define the
+**effective gravitational potential** $\Phi^*$ as follows:
+
+$$\begin{aligned}
+    \Phi^* \equiv \Phi + w(p)
+\end{aligned}$$
+
+This results in the cleanest form yet of the equilibrium condition, namely:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \Phi^*
+        = 0
+    }
+\end{aligned}$$
+
+At every point in the fluid, despite $p$ being variable,
+the force that is applied by the pressure must have the same magnitude in all directions at that point.
+This statement is known as **Pascal's law**,
+and is due to the fact that all forces must cancel out
+for an arbitrary blob:
+
+$$\begin{aligned}
+    \va{F}
+    = \va{F}_g + \va{F}_p
+    = 0
+\end{aligned}$$
+
+Let the blob be a cube with side $a$.
+Now, $\va{F}_p$ is a contact force,
+meaning it acts on the surface, and is thus proportional to $a^2$,
+however, $\va{F}_g$ is a body force,
+meaning it acts on the volume, and is thus proportional to $a^3$.
+Since we are considering a *point* in the fluid,
+$a$ is infinitesimally small,
+so that $\va{F}_p$ dominates $\va{F}_g$.
+Consequently, at equilibrium, $\va{F}_p$ must cancel out by itself,
+which means that the pressure is the same in all directions.
+
+
+
+## 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/imaginary-time/index.md b/source/know/concept/imaginary-time/index.md
new file mode 100644
index 0000000..5dc9264
--- /dev/null
+++ b/source/know/concept/imaginary-time/index.md
@@ -0,0 +1,173 @@
+---
+title: "Imaginary time"
+date: 2021-11-11
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+Let $\hat{A}_S$ and $\hat{B}_S$ be time-independent in the Schrödinger picture.
+Then, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+consider the following expectation value
+with respect to thermodynamic equilibium
+(as found in [Green's functions](/know/concept/greens-functions/) for example):
+
+$$\begin{aligned}
+    \expval{\hat{A}_H(t) \hat{B}_H(t')}
+    &= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_{0,S}(t)) \: \hat{A}_H(t) \: \hat{B}_H(t') \Big)
+\end{aligned}$$
+
+Where the "simple" Hamiltonian $\hat{H}_{0,S}$ is time-independent.
+Suppose a (maybe time-dependent) "difficult" $\hat{H}_{1,S}$ is added,
+so that the total Hamiltonian is $\hat{H}_S = \hat{H}_{0,S} + \hat{H}_{1,S}$.
+Then it is easier to consider the expectation value
+in the [interaction picture](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \expval{\hat{A}_H(t) \hat{B}_H(t')}
+    &= \frac{1}{Z} \Tr\!\Big( \exp(-\beta \hat{H}_S(t)) \: \hat{K}_I(0, t) \hat{A}_I(t) \hat{K}_I(t, t') \hat{B}_I(t') \hat{K}_I(t', 0) \Big)
+\end{aligned}$$
+
+Where $\hat{K}_I(t, t_0)$ is the time evolution operator of $\hat{H}_{1,S}$.
+In front, we have $\exp(-\beta \hat{H}_S(t))$,
+while $\hat{K}_I$ is an exponential of an integral of $\hat{H}_{1,I}$, so we are stuck.
+Keep in mind that exponentials of operators
+cannot just be factorized, i.e. in general
+$\exp(\hat{A} \!+\! \hat{B}) \neq \exp(\hat{A}) \exp(\hat{B})$
+
+To get around this, a useful mathematical trick is
+to use an **imaginary time** variable $\tau$ instead of the real time $t$.
+Fixing a $t$, we "redefine" the interaction picture along the imaginary axis:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{A}_I(\tau)
+        \equiv \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \: \hat{A}_S \: \exp\!\bigg( \!-\! \frac{\tau \hat{H}_{0,S}}{\hbar}\bigg)
+    }
+\end{aligned}$$
+
+Ironically, $\tau$ is real; the point is that this formula
+comes from the real-time definition by replacing $t \to -i \tau$.
+The Heisenberg and Schrödinger pictures can be redefined in the same way.
+
+In fact, by substituting $t \to -i \tau$,
+all the key results of the interaction picture can be updated,
+for example the Schrödinger equation for $\Ket{\psi_S(\tau)}$ becomes:
+
+$$\begin{aligned}
+    \hbar \dv{}{t}\Ket{\psi_S(\tau)}
+    = - \hat{H}_S \Ket{\psi_S(\tau)}
+    \quad \implies \quad
+    \Ket{\psi_S(\tau)}
+    = \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar} \bigg) \Ket{\psi_H}
+\end{aligned}$$
+
+And the interaction picture's time evolution operator $\hat{K}_I$
+turns out to be given by:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{K}_I(\tau, \tau_0)
+        = \mathcal{T} \bigg\{ \exp\!\bigg( \!-\! \frac{1}{\hbar} \int_{\tau_0}^\tau \hat{H}_{1,I}(\tau') \dd{\tau'} \bigg) \bigg\}
+    }
+\end{aligned}$$
+
+Where $\mathcal{T}$ is the
+[time-ordered product](/know/concept/time-ordered-product/)
+with respect to $\tau$.
+This operator works as expected:
+
+$$\begin{aligned}
+    \Ket{\psi_I(\tau)}
+    = \hat{K}_I(\tau, \tau_0) \Ket{\psi_I(\tau_0)}
+\end{aligned}$$
+
+Where $\Ket{\psi_I(\tau)}$ is related to
+the Schrödinger and Heisenberg pictures as follows:
+
+$$\begin{aligned}
+    \Ket{\psi_I(\tau)}
+    \equiv \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \Ket{\psi_S(\tau)}
+    = \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar}\bigg) \Ket{\psi_H}
+\end{aligned}$$
+
+It is interesting to combine this definition
+with the action of time evolution $\hat{K}_I(\tau, \tau_0)$:
+
+$$\begin{aligned}
+    \Ket{\psi_I(\tau)}
+    &= \hat{K}_I(\tau, \tau_0) \Ket{\psi_I(\tau_0)}
+    \\
+    \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau \hat{H}_S}{\hbar}\bigg) \Ket{\psi_H}
+    &= \hat{K}_I(\tau, \tau_0) \exp\!\bigg(\frac{\tau_0 \hat{H}_{0,S}}{\hbar}\bigg) \exp\!\bigg( \!-\! \frac{\tau_0 \hat{H}_S}{\hbar}\bigg) \Ket{\psi_H}
+\end{aligned}$$
+
+Rearranging this leads to the following useful
+alternative expression for $\hat{K}_I(\tau, \tau_0)$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{K}_I(\tau, \tau_0)
+        = \exp\!\bigg(\frac{\tau \hat{H}_{0,S}}{\hbar}\bigg)
+        \exp\!\bigg(\!-\! \frac{(\tau \!-\! \tau_0) \hat{H}_{S}}{\hbar}\bigg)
+        \exp\!\bigg(\!-\! \frac{\tau_0 \hat{H}_{0,S}}{\hbar}\bigg)
+    }
+\end{aligned}$$
+
+Returning to our initial example,
+we can set $\tau = \hbar \beta$ and $\tau_0 = 0$,
+so $\hat{K}_I(\tau, \tau_0)$ becomes:
+
+$$\begin{aligned}
+    \hat{K}_I(\hbar \beta, 0)
+    &= \exp\!\big(\beta \hat{H}_{0,S}\big) \exp\!\big(\!-\! \beta \hat{H}_{S}\big)
+    \\
+    \implies \quad
+    \exp\!\big(\!-\! \beta \hat{H}_{S}\big)
+    &= \exp\!\big(\!-\! \beta \hat{H}_{0,S}\big) \hat{K}_I(\hbar \beta, 0)
+\end{aligned}$$
+
+Using the easily-shown fact that
+$\hat{K}_I(\hbar \beta, 0) \hat{K}_I(0, \tau) = \hat{K}_I(\hbar \beta, \tau)$,
+we can therefore rewrite the thermodynamic expectation value like so:
+
+$$\begin{aligned}
+    \expval{\hat{A}_H(\tau) \hat{B}_H(\tau')}
+    &= \frac{1}{Z} \Tr\!\Big(\! \exp(-\beta \hat{H}_{0,S}) \hat{K}_I(\hbar \beta, \tau)
+    \hat{A}_I(\tau) \hat{K}_I(\tau, \tau') \hat{B}_I(\tau') \hat{K}_I(\tau', 0) \!\Big)
+\end{aligned}$$
+
+We now introduce a time-ordering $\mathcal{T}$,
+letting us reorder the (bosonic) $\hat{K}_I$-operators inside,
+and thereby reduce the expression considerably:
+
+$$\begin{aligned}
+    \Expval{\mathcal{T}\Big\{\hat{A}_H \hat{B}_H\Big\}}
+    &= \frac{1}{Z} \Tr\!\Big( \mathcal{T} \Big\{ \hat{K}_I(\hbar \beta, \tau) \hat{K}_I(\tau, \tau') \hat{K}_I(\tau', 0)
+    \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp(-\beta \hat{H}_{0,S}) \Big)
+    \\
+    &= \frac{1}{Z} \Tr\!\Big( \mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\} \exp(-\beta \hat{H}_{0,S}) \Big)
+\end{aligned}$$
+
+Where $Z = \Tr\!\big(\exp(-\beta \hat{H}_S)\big) = \Tr\!\big(\hat{K}_I(\hbar \beta, 0) \exp(-\beta \hat{H}_{0,S})\big)$.
+If we now define $\Expval{}_0$ as the expectation value with respect
+to the unperturbed equilibrium involving only $\hat{H}_{0,S}$,
+we arrive at the following way of writing this time-ordered expectation:
+
+$$\begin{aligned}
+    \boxed{
+        \Expval{\mathcal{T}\Big\{\hat{A}_H \hat{B}_H\Big\}}
+        = \frac{\Expval{\mathcal{T}\Big\{ \hat{K}_I(\hbar \beta, 0) \hat{A}_I(\tau) \hat{B}_I(\tau') \Big\}}_0}{\Expval{\hat{K}_I(\hbar \beta, 0)}_0}
+    }
+\end{aligned}$$
+
+For another application of imaginary time,
+see e.g. the [Matsubara Green's function](/know/concept/matsubara-greens-function/).
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/impulse-response/index.md b/source/know/concept/impulse-response/index.md
new file mode 100644
index 0000000..65849aa
--- /dev/null
+++ b/source/know/concept/impulse-response/index.md
@@ -0,0 +1,81 @@
+---
+title: "Impulse response"
+date: 2021-03-09
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+The **impulse response** $u_p(t)$ of a system whose behaviour is described
+by a linear operator $\hat{L}$, is defined as the reponse of the system
+when forced by the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta(t)$:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{L} \{ u_p(t) \} = \delta(t)
+    }
+\end{aligned}$$
+
+This can be used to find the response $u(t)$ of $\hat{L}$ to
+*any* forcing function $f(t)$, i.e. not only $\delta(t)$,
+by simply taking the convolution with $u_p(t)$:
+
+$$\begin{aligned}
+    \hat{L} \{ u(t) \} = f(t)
+    \quad \implies \quad
+    \boxed{
+        u(t) = (f * u_p)(t)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-main"/>
+<label for="proof-main">Proof</label>
+<div class="hidden">
+<label for="proof-main">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
+    &= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
+\end{aligned}$$
+</div>
+</div>
+
+This is useful for solving initial value problems,
+because any initial condition can be satisfied
+due to the linearity of $\hat{L}$,
+by choosing the initial values of the homogeneous solution $\hat{L}\{ u_h(t) \} = 0$
+such that the total solution $(f * u_p)(t) + u_h(t)$
+has the desired values.
+
+Meanwhile, for boundary value problems,
+the related [fundamental solution](/know/concept/fundamental-solution/)
+is preferable.
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/index.md b/source/know/concept/index.md
new file mode 100644
index 0000000..69b8ca9
--- /dev/null
+++ b/source/know/concept/index.md
@@ -0,0 +1,33 @@
+---
+title: "List of concepts"
+date: 2021-02-22
+layout: "default"
+---
+
+# List of concepts
+
+{% assign by_letter = site.pages
+  | where_exp: "item", "item.layout == 'concept'"
+  | group_by_exp: "item", "item.title | truncate: 1, ''"
+  | sort: "name"
+%}
+
+This is an alphabetical list of the concepts in this knowledge base.
+
+<div style="text-align: center;">
+&hyphen; 
+{% for letter in by_letter %}
+<a href="#{{ letter.name }}">{{ letter.name }}</a>
+&hyphen; 
+{% endfor %}
+</div>
+
+{% for letter in by_letter %}
+<h2 id="{{ letter.name }}">{{ letter.name }}</h2>
+<ul>
+  {% assign items = letter.items | sort: "title" %}
+  {% for item in items %}
+  <li><a href="{{ item.url }}">{{ item.title }}</a></li>
+  {% endfor %}
+</ul>
+{% 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):
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E} \bigg( \int_a^b G_t \dd{B_t} \bigg)^2
+        = \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-isometry"/>
+<label for="proof-isometry">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-isometry">Proof.</label>
+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 $h$:
+
+$$\begin{aligned}
+    \mathbf{E} \bigg[ \sum_{t = a}^{t = b} G_t (B_{t + h} \!-\! B_t) \bigg]^2
+    &= \sum_{t = a}^{t = b} \mathbf{E} \big[ G_t^2 \big] h
+    \longrightarrow
+    \int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
+\end{aligned}$$
+</div>
+</div>
+
+Furthermore, Itō integrals are [martingales](/know/concept/martingale/),
+meaning that the average noise contribution is zero,
+which makes intuitive sense,
+since true white noise cannot be biased.
+
+<div class="accordion">
+<input type="checkbox" id="proof-martingale"/>
+<label for="proof-martingale">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-martingale">Proof.</label>
+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.
+</div>
+</div>
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/ito-process/index.md b/source/know/concept/ito-process/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/ito-process/index.md
@@ -0,0 +1,361 @@
+---
+title: "Itō process"
+date: 2021-11-06
+categories:
+- Mathematics
+- Stochastic analysis
+layout: "concept"
+---
+
+Given two [stochastic processes](/know/concept/stochastic-process/)
+$F_t$ and $G_t$, consider the following random variable $X_t$,
+where $B_t$ is the [Wiener process](/know/concept/wiener-process/),
+i.e. Brownian motion:
+
+$$\begin{aligned}
+    X_t
+    = X_0 + \int_0^t F_s \dd{s} + \int_0^t G_s \dd{B_s}
+\end{aligned}$$
+
+Where the latter is an [Itō integral](/know/concept/ito-integral/),
+assuming $G_t$ is Itō-integrable.
+We call $X_t$ an **Itō process** if $F_t$ is locally integrable,
+and the initial condition $X_0$ is known,
+i.e. $X_0$ is $\mathcal{F}_0$-measurable,
+where $\mathcal{F}_t$ is the filtration
+to which $F_t$, $G_t$ and $B_t$ are adapted.
+The above definition of $X_t$ is often abbreviated as follows,
+where $X_0$ is implicit:
+
+$$\begin{aligned}
+    \dd{X_t}
+    = F_t \dd{t} + G_t \dd{B_t}
+\end{aligned}$$
+
+Typically, $F_t$ is referred to as the **drift** of $X_t$,
+and $G_t$ as its **intensity**.
+Because the Itō integral of $G_t$ is a
+[martingale](/know/concept/martingale/),
+it does not contribute to the mean of $X_t$:
+
+$$\begin{aligned}
+    \mathbf{E}[X_t]
+    = \int_0^t \mathbf{E}[F_s] \dd{s}
+\end{aligned}$$
+
+Now, consider the following **Itō stochastic differential equation** (SDE),
+where $\xi_t = \idv{B_t}{t}$ is white noise,
+informally treated as the $t$-derivative of $B_t$:
+
+$$\begin{aligned}
+    \dv{X_t}{t}
+    = f(X_t, t) + g(X_t, t) \: \xi_t
+\end{aligned}$$
+
+An Itō process $X_t$ is said to satisfy this equation
+if $f(X_t, t) = F_t$ and $g(X_t, t) = G_t$,
+in which case $X_t$ is also called an **Itō diffusion**.
+All Itō diffusions are [Markov processes](/know/concept/markov-process/),
+since only the current value of $X_t$ determines the future,
+and $B_t$ is also a Markov process.
+
+
+## Itō's lemma
+
+Classically, given $y \equiv h(x(t), t)$,
+the chain rule of differentiation states that:
+
+$$\begin{aligned}
+    \dd{y}
+    = \pdv{h}{t} \dd{t} + \pdv{h}{x} \dd{x}
+\end{aligned}$$
+
+However, for a stochastic process $Y_t \equiv h(X_t, t)$,
+where $X_t$ is an Itō process,
+the chain rule is modified to the following,
+known as **Itō's lemma**:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{Y_t}
+        = \bigg( \pdv{h}{t} + \pdv{h}{x} F_t + \frac{1}{2} \pdvn{2}{h}{x} G_t^2 \bigg) \dd{t} + \pdv{h}{x} G_t \dd{B_t}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-lemma"/>
+<label for="proof-lemma">Proof</label>
+<div class="hidden">
+<label for="proof-lemma">Proof.</label>
+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 $\dd{t}$.
+</div>
+</div>
+
+The most important application of Itō's lemma
+is to perform coordinate transformations,
+to make the solution of a given Itō SDE easier.
+
+
+## Coordinate transformations
+
+The simplest coordinate transformation is a scaling of the time axis.
+Defining $s \equiv \alpha t$, the goal is to keep the Itō process.
+We know how to scale $B_t$, be setting $W_s \equiv \sqrt{\alpha} B_{s / \alpha}$.
+Let $Y_s \equiv X_t$ be the new variable on the rescaled axis, then:
+
+$$\begin{aligned}
+    \dd{Y_s}
+    = \dd{X_t}
+    &= f(X_t) \dd{t} + g(X_t) \dd{B_t}
+    \\
+    &= \frac{1}{\alpha} f(Y_s) \dd{s} + \frac{1}{\sqrt{\alpha}} g(Y_s) \dd{W_s}
+\end{aligned}$$
+
+$W_s$ is a valid Wiener process,
+and the other changes are small,
+so this is still an Itō process.
+
+To solve SDEs analytically, it is usually best
+to have additive noise, i.e. $g = 1$.
+This can be achieved using the **Lamperti transform**:
+define $Y_t \equiv h(X_t)$, where $h$ is given by:
+
+$$\begin{aligned}
+    \boxed{
+        h(x)
+        = \int_{x_0}^x \frac{1}{g(y)} \dd{y}
+    }
+\end{aligned}$$
+
+Then, using Itō's lemma, it is straightforward
+to show that the intensity becomes $1$.
+Note that the lower integration limit $x_0$ does not enter:
+
+$$\begin{aligned}
+    \dd{Y_t}
+    &= \bigg( f(X_t) \: h'(X_t) + \frac{1}{2} g^2(X_t) \: h''(X_t) \bigg) \dd{t} + g(X_t) \: h'(X_t) \dd{B_t}
+    \\
+    &= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g^2(X_t) \frac{g'(X_t)}{g^2(X_t)} \bigg) \dd{t} + \frac{g(X_t)}{g(X_t)} \dd{B_t}
+    \\
+    &= \bigg( \frac{f(X_t)}{g(X_t)} - \frac{1}{2} g'(X_t) \bigg) \dd{t} + \dd{B_t}
+\end{aligned}$$
+
+Similarly, we can eliminate the drift $f = 0$,
+thereby making the Itō process a martingale.
+This is done by defining $Y_t \equiv h(X_t)$, with $h(x)$ given by:
+
+$$\begin{aligned}
+    \boxed{
+        h(x)
+        = \int_{x_0}^x \exp\!\bigg( \!-\!\! \int_{x_1}^y \frac{2 f(z)}{g^2(z)} \dd{z} \bigg) \dd{y}
+    }
+\end{aligned}$$
+
+The goal is to make the parenthesized first term (see above)
+of Itō's lemma disappear, which this $h(x)$ does indeed do.
+Note that $x_0$ and $x_1$ do not enter:
+
+$$\begin{aligned}
+    0
+    &= f(x) \: h'(x) + \frac{1}{2} g^2(x) \: h''(x)
+    \\
+    &= \Big( f(x) - \frac{1}{2} g^2(x) \frac{2 f(x)}{g^2(x)} \Big) \exp\!\bigg( \!-\!\! \int_{x_1}^x \frac{2 f(y)}{g^2(y)} \dd{y} \bigg)
+\end{aligned}$$
+
+
+## Existence and uniqueness
+
+It is worth knowing under what condition a solution to a given SDE exists,
+in the sense that it is finite on the entire time axis.
+Suppose the drift $f$ and intensity $g$ satisfy these inequalities,
+for some known constant $K$ and for all $x$:
+
+$$\begin{aligned}
+    x f(x) \le K (1 + x^2)
+    \qquad \quad
+    g^2(x) \le K (1 + x^2)
+\end{aligned}$$
+
+When this is satisfied, we can find the following upper bound
+on an Itō process $X_t$,
+which clearly implies that $X_t$ is finite for all $t$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E}[X_t^2]
+        \le \big(X_0^2 + 3 K t\big) \exp\!\big(3 K t\big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-existence"/>
+<label for="proof-existence">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-existence">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \mathbf{E}[Y_t]
+    &\le (Y_0 + 3 K t) \exp\!\bigg( \int_0^t 3 K \dd{s} \bigg)
+    \\
+    &\le (Y_0 + 3 K t) \exp\!\big(3 K t\big)
+\end{aligned}$$
+</div>
+</div>
+
+If a solution exists, it is also worth knowing whether it is unique.
+Suppose that $f$ and $g$ satisfy the following inequalities,
+for some constant $K$ and for all $x$ and $y$:
+
+$$\begin{aligned}
+    \big| f(x) - f(y) \big| \le K \big| x - y \big|
+    \qquad \quad
+    \big| g(x) - g(y) \big| \le K \big| x - y \big|
+\end{aligned}$$
+
+Let $X_t$ and $Y_t$ both be solutions to a given SDE,
+but the initial conditions need not be the same,
+such that the difference is initially $X_0 \!-\! Y_0$.
+Then the difference $X_t \!-\! Y_t$ is bounded by:
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E}\big[ (X_t - Y_t)^2 \big]
+        \le (X_0 - Y_0)^2 \exp\!\Big( \big(2 K \!+\!  K^2 \big) t \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-uniqueness"/>
+<label for="proof-uniqueness">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-uniqueness">Proof.</label>
+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):
+
+$$\begin{aligned}
+    \mathbf{E}[Z_t]
+    &\le Z_0 \exp\!\bigg( \int_0^t 2 K \!+\! K^2 \dd{s} \bigg)
+    \\
+    &\le Z_0 \exp\!\Big( \big( 2 K \!+\! K^2 \big) t \Big)
+\end{aligned}$$
+</div>
+</div>
+
+Using these properties, it can then be shown
+that if all of the above conditions are satisfied,
+then the SDE has a unique solution,
+which is $\mathcal{F}_t$-adapted, continuous, and exists for all times.
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/jellium/index.md b/source/know/concept/jellium/index.md
new file mode 100644
index 0000000..6e395b4
--- /dev/null
+++ b/source/know/concept/jellium/index.md
@@ -0,0 +1,416 @@
+---
+title: "Jellium"
+date: 2021-11-23
+categories:
+- Physics
+- Quantum mechanics
+- Perturbation
+layout: "concept"
+---
+
+**Jellium**, also called the **uniform** or **homogeneous electron gas**,
+is a theoretical material where all electrons are free,
+and the ions' positive charge is smeared into a uniform background "jelly".
+This simple model lets us study electron interactions easily.
+
+
+## Without interactions
+
+Let us start by neglecting electron-electron interactions.
+This is clearly a dubious assumption, but we will stick with it for now.
+For an infinitely large sample of jellium,
+the single-electron states are simply plane waves.
+We consider an arbitrary cube of volume $V$,
+and impose periodic boundary conditions on it,
+such that the single-particle orbitals are (suppressing spin):
+
+$$\begin{aligned}
+    \Inprod{\vb{r}}{\psi_{\vb{k}}}
+    = \psi_{\vb{k}}(\vb{r})
+    = \frac{1}{\sqrt{V}} \exp(i \vb{k} \cdot \vb{r})
+    \qquad \quad
+    \vb{k} = \frac{2 \pi}{V^{1/3}} (n_x, n_y, n_z)
+\end{aligned}$$
+
+Where $n_x, n_y, n_z \in \mathbb{Z}$.
+This is a discrete (but infinite) set of independent orbitals,
+so it is natural to use the
+[second quantization](/know/concept/second-quantization/)
+to write the non-interacting Hamiltonian $\hat{H}_0$,
+where $\hbar^2 |\vb{k}|^2 / (2 m)$ is the kinetic energy
+of the orbital with wavevector $\vb{k}$, and $s$ is the spin:
+
+$$\begin{aligned}
+    \hat{H}_0
+    = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}
+\end{aligned}$$
+
+Assuming that the temperature $T = 0$,
+the $N$-electron ground state of this Hamiltonian
+is known as the **Fermi sea** or **Fermi sphere** $\Ket{\mathrm{FS}}$,
+and is constructed by filling up the single-electron states
+starting from the lowest energy:
+
+$$\begin{aligned}
+    \Ket{\mathrm{FS}}
+    = \prod_{s} \prod_{j = 1}^{N/2} \hat{c}_{s,\vb{k}_j}^\dagger \Ket{0}
+\end{aligned}$$
+
+Because $T = 0$, all the electrons stay in their assigned state.
+The energy and wavenumber $|\vb{k}|$ of the highest filled orbital
+are called the **Fermi energy** $\epsilon_F$ and **Fermi wavenumber** $k_F$,
+and obey the expected kinetic energy relation:
+
+$$\begin{aligned}
+    \boxed{
+        \epsilon_F
+        = \frac{\hbar^2}{2 m} k_F^2
+    }
+\end{aligned}$$
+
+The Fermi sea can be visualized in $\vb{k}$-space as a sphere with radius $k_F$.
+Because $\vb{k}$ is discrete, the sphere's surface is not smooth,
+but in the limit $V \to \infty$ it becomes perfect.
+
+Now, we would like a relation between the system's parameters,
+e.g. $N$ and $V$, and the resulting values of $\epsilon_F$ or $k_F$.
+The total population $N$ must be given by:
+
+$$\begin{aligned}
+    N
+    = \sum_{s} \sum_{\vb{k}} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
+    = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
+\end{aligned}$$
+
+Where we have turned the sum over $\vb{k}$ into an integral with a constant factor,
+by using that each orbital exclusively occupies a volume $(2 \pi)^3 / V$ in $\vb{k}$-space.
+
+At zero temperature, this inner product can only be $0$ or $1$,
+depending on whether $\vb{k}$ is outside or inside the Fermi sphere.
+We can therefore rewrite using a
+[Heaviside step function](/know/concept/heaviside-step-function/):
+
+$$\begin{aligned}
+    N
+    = \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
+    = 2 \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
+\end{aligned}$$
+
+Where we realized that spin does not matter,
+and replaced the sum over $s$ by a factor $2$.
+In order to evaluate this 3D integral,
+we go to [spherical coordinates](/know/concept/spherical-coordinates/)
+$(|\vb{k}|, \theta, \varphi)$:
+
+$$\begin{aligned}
+    N
+    &= \frac{V}{4 \pi^3} \int_0^{2 \pi} \int_0^\pi \int_0^\infty \Theta(k_F - |\vb{k}|) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
+    \\
+    &= \frac{V}{4 \pi^3} 4 \pi \int_0^{k_F} |\vb{k}|^2 \dd{|\vb{k}|}
+    = \frac{V}{\pi^2} \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_0^{k_F}
+    = \frac{V}{3 \pi^2} k_F^3
+\end{aligned}$$
+
+Using that the electron density $n = N/V$,
+we thus arrive at the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        k_F^3
+        = 3 \pi^2 n
+    }
+\end{aligned}$$
+
+This result also justifies our assumption that $T = 0$:
+we can accurately calculate the density $n$ for many conducting materials,
+and this relation then gives $k_F$ and $\epsilon_F$.
+It turns out that $\epsilon_F$ is usually very large
+compared to the thermal energy $k_B T$ at reasonable temperatures,
+so we can conclude that thermal fluctuations are negligible.
+
+Now, $\epsilon_F$ is the highest single-electron energy,
+but about the total $N$-particle energy $E^{(0)}$?
+
+$$\begin{aligned}
+    E^{(0)}
+    = \matrixel{\mathrm{FS}}{\hat{H}_0}{\mathrm{FS}}
+    = \sum_{s} \sum_{\vb{k}} \frac{\hbar^2 |\vb{k}|^2}{2 m} \matrixel{\mathrm{FS}}{\hat{c}_{s,\vb{k}}^\dagger \hat{c}_{s,\vb{k}}}{\mathrm{FS}}
+\end{aligned}$$
+
+Once again, we turn the sum over $\vb{k}$ into an integral,
+and recognize the spin's irrelevance:
+
+$$\begin{aligned}
+    E^{(0)}
+    &= \sum_{s} \frac{V}{(2 \pi)^3} \int_{-\infty}^\infty \frac{\hbar^2 |\vb{k}|^2}{2 m}
+    \matrixel{\mathrm{FS}}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}}}{\mathrm{FS}} \dd{\vb{k}}
+    \\
+    &= \frac{\hbar^2 V}{8 \pi^3 m} \int_{-\infty}^\infty |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \dd{\vb{k}}
+\end{aligned}$$
+
+In spherical coordinates,
+we evaluate the integral and find that $E^{(0)}$ is proportional to $k_F^5$:
+
+$$\begin{aligned}
+    E^{(0)}
+    &= \frac{\hbar^2 V}{8 \pi^3 m} \int_0^{2 \pi}
+    \int_0^\pi \int_0^\infty \Big( |\vb{k}|^2 \: \Theta(k_F - |\vb{k}|) \Big) |\vb{k}|^2 \sin(\theta) \dd{|\vb{k}|} \dd{\theta} \dd{\varphi}
+    \\
+    &= \frac{\hbar^2 V}{8 \pi^3 m} 4 \pi \int_0^{k_F} |\vb{k}|^4 \dd{|\vb{k}|}
+    = \frac{\hbar^2 V}{2 \pi^2 m} \bigg[ \frac{|\vb{k}|^5}{5} \bigg]_0^{k_F}
+    = \frac{\hbar^2 V}{10 \pi^2 m} k_F^5
+\end{aligned}$$
+
+In general, it is more useful to consider
+the average kinetic energy per electron $E^{(0)} / N$,
+which we find to be as follows, using that $k_F^3 = 3 \pi^2 n$:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{E^{(0)}}{N}
+        = \frac{3 \hbar^2}{10 m} k_F^2
+        = \frac{3}{5} \epsilon_F
+    }
+    \:\sim\: n^{2/3}
+\end{aligned}$$
+
+Traditionally, this is expressed using a dimensionless parameter $r_s$,
+defined as the radius of a sphere containing a single electron,
+measured in Bohr radii $a_0 \equiv 4 \pi \varepsilon_0 \hbar^2 / (e^2 m)$:
+
+$$\begin{aligned}
+        \frac{4 \pi}{3} (a_0 r_s)^3
+        = \frac{1}{n}
+        = \frac{3 \pi^2}{k_F^3}
+        \quad \implies \quad
+        r_s
+        = \Big( \frac{3}{4 \pi a_0^3 n} \Big)^{1/3}
+        = \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 k_F}
+\end{aligned}$$
+
+Such that the ground state energy can be rewritten in Rydberg units of energy like so:
+
+$$\begin{aligned}
+    \frac{E^{(0)}}{N}
+    = \frac{3 \hbar^2}{10 m} \frac{4 \pi \varepsilon_0 e^2}{4 \pi \varepsilon_0 e^2} \frac{a_0^2 k_F^2}{a_0^2}
+    = \frac{3 e^2}{40 \pi \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{2/3} \frac{1}{a_0 r_s^2}
+    \approx \frac{2.21}{r_s^2} \; \mathrm{Ry}
+\end{aligned}$$
+
+
+## With interactions
+
+To include Coulomb interactions, let us try
+[time-independent pertubation theory](/know/concept/time-independent-perturbation-theory/).
+Clearly, this will give better results when the interaction is relatively weak, if ever.
+
+The Coulomb potential is proportional to the inverse distance,
+and the average electron spacing is roughly $n^{-1/3}$,
+so the interaction energy $E_\mathrm{int}$ should scale as $n^{1/3}$.
+We already know that the kinetic energy $E_\mathrm{kin} = E^{(0)}$ scales as $n^{2/3}$,
+meaning perturbation theory should be reasonable
+if $1 \gg E_\mathrm{int} / E_\mathrm{kin} \sim n^{-1/3}$,
+so in the limit of high density $n \to \infty$.
+
+The two-body Coulomb interaction operator $\hat{W}$
+is as follows in second-quantized form:
+
+$$\begin{aligned}
+    \hat{W}
+    = \frac{1}{2 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{e^2}{\varepsilon_0 |\vb{q}|^2}
+    \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
+\end{aligned}$$
+
+The first-order correction $E^{(1)}$ to the ground state (i.e. Fermi sea) energy
+is then given by:
+
+$$\begin{aligned}
+    E^{(1)}
+    = \matrixel{\mathrm{FS}}{\hat{W}}{\mathrm{FS}}
+    = \frac{e^2}{2 \varepsilon_0 V} \sum_{s_1 s_2} \sum_{\vb{k}_1 \vb{k}_2} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    \matrixel{\mathrm{FS}}{
+        \hat{c}_{s_1, \vb{k}_1 + \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2 - \vb{q}}^\dagger \hat{c}_{s_2, \vb{k}_2} \hat{c}_{s_1, \vb{k}_1}
+    }{\mathrm{FS}}
+\end{aligned}$$
+
+This inner product can only be nonzero
+if the two creation operators $\hat{c}^\dagger$
+are for the same orbitals as the two annihilation operators $\hat{c}$.
+Since $\vb{q} \neq 0$, this means that $s_1 = s_2$,
+and that momentum is conserved: $\vb{k}_2 = \vb{k}_1 \!+\! \vb{q}$.
+And of course both $\vb{k}_1$ and $\vb{k}_1 \!+\! \vb{q}$
+must be inside the Fermi sphere,
+to avoid annihilating an empty orbital.
+Let $s = s_1$ and $\vb{k} = \vb{k}_1$:
+
+$$\begin{aligned}
+    E^{(1)}
+    &= \frac{e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    \matrixel{\mathrm{FS}}{
+        \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}} \hat{c}_{s, \vb{k}}
+    }{\mathrm{FS}}
+    \\
+    &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    \matrixel{\mathrm{FS}}{
+        \big( \hat{c}_{s, \vb{k} + \vb{q}}^\dagger \hat{c}_{s, \vb{k} + \vb{q}}\big) \big(\hat{c}_{s, \vb{k}}^\dagger \hat{c}_{s, \vb{k}}\big)
+    }{\mathrm{FS}}
+    \\
+    &= \frac{- e^2}{2 \varepsilon_0 V} \sum_{s} \sum_{\vb{k}} \sum_{\vb{q} \neq 0} \frac{1}{|\vb{q}|^2}
+    \Theta(k_F - |\vb{k}|) \:\Theta(k_F - |\vb{k} \!+\! \vb{q}|)
+\end{aligned}$$
+
+Next, we convert the sum over $\vb{q}$ into an integral in spherical coordinates.
+Clearly, $\vb{q}$ is the "jump" made by an electron from one orbital to another,
+so the largest possible jump
+goes from a point on the Fermi surface to the opposite point,
+and thus has length $2 k_F$.
+This yields the integration limit, and therefore leads to:
+
+$$\begin{aligned}
+    E^{(1)}
+    &= \frac{- e^2}{(2 \pi)^3 \varepsilon_0} \sum_{\vb{k}}
+    \int_0^{2 \pi} \!\!\int_0^\pi \!\!\int_0^\infty \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \frac{|\vb{q}|^2}{|\vb{q}|^2}
+    \sin(\theta_q) \dd{|\vb{q}|} \dd{\theta_q} \dd{\varphi_q}
+    \\
+    &= \frac{- e^2}{2 \pi^2 \varepsilon_0} \sum_{\vb{k}}
+    \int_0^{2 k_F} \Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|) \dd{|\vb{q}|}
+\end{aligned}$$
+
+Where we have used that the direction of $\vb{q}$,
+i.e. $(\theta_q,\varphi_q)$, is irrelevant,
+as long as we define $\theta_k$ as
+the angle between $\vb{q}$ and $\vb{k} \!+\! \vb{q}$
+when we go to spherical coordinates $(|\vb{k}|, \theta_k, \varphi_k)$ for $\vb{k}$:
+
+$$\begin{aligned}
+    E^{(1)}
+    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^\infty
+    \!\Theta(k_F \!-\! |\vb{k}|) \: \Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|)
+    \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} \int_0^{2 k_F} \!\!\!\!\int_0^{2 \pi} \!\!\!\int_0^\pi \!\!\!\int_0^{k_F}
+    \!\Theta(k_F \!-\! |\vb{k} \!+\! \vb{q}|)
+    \: |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
+\end{aligned}$$
+
+Unfortunately, this last step function is less easy to translate into integration limits.
+In effect, we are trying to calculate the intersection volume of two spheres,
+both with radius $k_F$, one centered on the origin (for $\vb{k}$),
+and the other centered on $\vb{q}$ (for $\vb{k} \!+\! \vb{q}$).
+Imagine a triangle with side lengths $|\vb{k}|$, $|\vb{q}|$ and $|\vb{k} \!+\! \vb{q}|^2$,
+where $\theta_k$ is the angle between $|\vb{k}|$ and $|\vb{k} \!+\! \vb{q}|$.
+The *law of cosines* then gives the following relation:
+
+$$\begin{aligned}
+    |\vb{k}|^2
+    = |\vb{q}|^2 + |\vb{k} \!+\! \vb{q}|^2 - 2 |\vb{q}| |\vb{k} \!+\! \vb{q}| \cos(\theta_k)
+\end{aligned}$$
+
+We already know that $|\vb{k}| < k_F$ and $0 < |\vb{q}| < 2 k_F$,
+so by isolating for $\cos(\theta_k)$,
+we can obtain bounds on $\theta_k$ and $|\vb{k}|$.
+Let $|\vb{k}| \to k_F$ in both cases, then:
+
+$$\begin{aligned}
+    \cos(\theta_k)
+    = \frac{|\vb{k} \!+\! \vb{q}|^2 + |\vb{q}|^2 - |\vb{k}|^2}{2 |\vb{k} \!+\! \vb{q}| |\vb{q}|}
+    &\:\:\underset{|\vb{q}| \to 0}{>}\:\:\: \frac{k_F^2 + |\vb{q}|^2 - k_F^2}{2 k_F |\vb{q}|}
+    = \frac{|\vb{q}|}{2 k_F}
+    \\
+    &\underset{|\vb{q}| \to 2 k_F}{<}\:\: \frac{k_F^2 + 4 k_F ^2 - k_F^2}{2 k_F 2 k_F}
+    = 1
+\end{aligned}$$
+
+Meaning that $0 < \theta_k < \arccos{|\vb{q}| / (2 k_F)}$.
+To get a lower limit for $|\vb{k}|$, we "cheat" by artificially demanding
+that $\vb{k}$ does not cross the halfway point between the spheres,
+with the result that $|\vb{k}| \cos(\theta_k) > |\vb{q}|/2$.
+Then, thanks to symmetry (both spheres have the same radius),
+we just multiply the integral by $2$,
+for $\vb{k}$ on the other side of the halfway point.
+
+Armed with these integration limits, we return to calculating $E^{(1)}$,
+substituting $\xi \equiv \cos(\theta_k)$:
+
+$$\begin{aligned}
+    E^{(1)}
+    &= \frac{- e^2 V}{16 \pi^5 \varepsilon_0} 2 \int_0^{2 k_F} \!\!\!\int_0^{2 \pi} \!\!\int_0^{\arccos{|\vb{q}| / (2 k_F)}}
+    \!\!\int_{|\vb{q}|/(2 \cos{\theta_k})}^{k_F} |\vb{k}|^2 \sin(\theta_k) \dd{|\vb{k}|} \dd{\theta_k} \dd{\varphi_k} \dd{|\vb{q}|}
+    \\
+    &= \frac{e^2 V}{8 \pi^5 \varepsilon_0} 2 \pi \int_0^{2 k_F} \!\!\!\int_1^{|\vb{q}| / (2 k_F)}
+    \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \frac{\sin(\theta_k)}{\sin(\theta_k)} \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    \!\!\int_{|\vb{q}|/(2 \xi)}^{k_F} |\vb{k}|^2 \dd{|\vb{k}|} \dd{\xi} \dd{|\vb{q}|}
+\end{aligned}$$
+
+Where we have used that $\varphi_k$ does not appear in the integrand.
+Evaluating these integrals:
+
+$$\begin{aligned}
+    E^{(1)}
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    \bigg[ \frac{|\vb{k}|^3}{3} \bigg]_{|\vb{q}|/(2 \xi)}^{k_F} \dd{\xi} \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F} \!\!\!\int_{|\vb{q}| / (2 k_F)}^1
+    \bigg( \frac{k_F^3}{3} - \frac{|\vb{q}|^3}{24 \xi^3} \bigg) \dd{\xi} \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    \bigg[ \frac{k_F^3}{3} x + \frac{|\vb{q}|^3}{48 \xi^2} \bigg]_{|\vb{q}| / (2 k_F)}^1 \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \int_0^{2 k_F}
+    \bigg( \frac{k_F^3}{3} + \frac{|\vb{q}|^3}{48} - \frac{k_F^2 |\vb{q}|}{4} \bigg) \dd{|\vb{q}|}
+    \\
+    &= \frac{- e^2 V}{4 \pi^4 \varepsilon_0} \bigg[ \frac{k_F^3 |\vb{q}|}{3} + \frac{|\vb{q}|^4}{192} - \frac{k_F^2 |\vb{q}|^2}{8} \bigg]_0^{2 k_F}
+    \\
+    &= \frac{- e^2 V}{16 \pi^4 \varepsilon_0} k_F^4
+    = \frac{- e^2 N}{16 \pi^4 \varepsilon_0 n} k_F^4
+    = -\frac{3 e^2 N}{16 \pi^2 \varepsilon_0} k_F
+\end{aligned}$$
+
+Per particle, the first-order energy correction $E^{(1)}$
+is therefore found to be as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{E^{(1)}}{N}
+        = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} k_F
+    }
+\end{aligned}$$
+
+This can also be written using the parameter $r_s$ introduced above, leading to:
+
+$$\begin{aligned}
+    \frac{E^{(1)}}{N}
+    = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \frac{a_0 k_F}{a_0}
+    = -\frac{3 e^2}{16 \pi^2 \varepsilon_0} \Big( \frac{9 \pi}{4} \Big)^{1/3} \frac{1}{a_0 r_s}
+\end{aligned}$$
+
+Consequently, for sufficiently high densities $n$,
+the total energy $E$ per particle is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{E}{N}
+        \approx \bigg( \frac{2.21}{r_s^2} - \frac{0.92}{r_s} \bigg) \; \mathrm{Ry}
+    }
+\end{aligned}$$
+
+Unfortunately, this is as far as we can go.
+In theory, the second-order energy correction $E^{(2)}$ is as shown below,
+but it turns out that it (and all higher orders) diverge:
+
+$$\begin{aligned}
+    E^{(2)}
+    = \sum_{\Psi_n \neq \mathrm{FS}} \frac{\big| \matrixel{\mathrm{FS}}{\hat{W}}{\Psi_n} \big|^2}{E^{(0)} - E_n}
+\end{aligned}$$
+
+The only cure for this is to go to infinite order,
+where all the infinities add up to a finite result.
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/kolmogorov-equations/index.md b/source/know/concept/kolmogorov-equations/index.md
new file mode 100644
index 0000000..a6579b2
--- /dev/null
+++ b/source/know/concept/kolmogorov-equations/index.md
@@ -0,0 +1,246 @@
+---
+title: "Kolmogorov equations"
+date: 2021-11-14
+categories:
+- Mathematics
+- Statistics
+- Stochastic analysis
+layout: "concept"
+---
+
+Consider the following general [Itō diffusion](/know/concept/ito-calculus/)
+$X_t \in \mathbb{R}$, which is assumed to satisfy
+the conditions for unique existence on the entire time axis:
+
+$$\begin{aligned}
+    \dd{X}_t
+    = f(X_t, t) \dd{t} + g(X_t, t) \dd{B_t}
+\end{aligned}$$
+
+Let $\mathcal{F}_t$ be the filtration to which $X_t$ is adapted,
+then we define $Y_s$ as shown below,
+namely as the [conditional expectation](/know/concept/conditional-expectation/)
+of $h(X_t)$, for an arbitrary bounded function $h(x)$,
+given the information $\mathcal{F}_s$ available at time $s \le t$.
+Because $X_t$ is a [Markov process](/know/concept/markov-process/),
+$Y_s$ must be $X_s$-measurable,
+so it is a function $k$ of $X_s$ and $s$:
+
+$$\begin{aligned}
+    Y_s
+    \equiv \mathbf{E}[h(X_t) | \mathcal{F}_s]
+    = \mathbf{E}[h(X_t) | X_s]
+    = k(X_s, s)
+\end{aligned}$$
+
+Consequently, we can apply Itō's lemma to find $\dd{Y_s}$
+in terms of $k$, $f$ and $g$:
+
+$$\begin{aligned}
+    \dd{Y_s}
+    &= \bigg( \pdv{k}{s} + \pdv{k}{x} f + \frac{1}{2} \pdvn{2}{k}{x} g^2 \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s}
+    \\
+    &= \bigg( \pdv{k}{s} + \hat{L} k \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s}
+\end{aligned}$$
+
+Where we have defined the linear operator $\hat{L}$
+to have the following action on $k$:
+
+$$\begin{aligned}
+    \hat{L} k
+    \equiv \pdv{k}{x} f + \frac{1}{2} \pdvn{2}{k}{x} g^2
+\end{aligned}$$
+
+At this point, we need to realize that $Y_s$ is
+a [martingale](/know/concept/martingale/) with respect to $\mathcal{F}_s$,
+since $Y_s$ is $\mathcal{F}_s$-adapted and finite,
+and it satisfies the martingale property,
+for $r \le s \le t$:
+
+$$\begin{aligned}
+    \mathbf{E}[Y_s | \mathcal{F}_r]
+    = \mathbf{E}\Big[ \mathbf{E}[h(X_t) | \mathcal{F}_s] \Big| \mathcal{F}_r \Big]
+    = \mathbf{E}\big[ h(X_t) \big| \mathcal{F}_r \big]
+    = Y_r
+\end{aligned}$$
+
+Where we used the tower property of conditional expectations,
+because $\mathcal{F}_r \subset \mathcal{F}_s$.
+However, an Itō diffusion can only be a martingale
+if its drift term (the one containing $\dd{s}$) vanishes,
+so, looking at $\dd{Y_s}$, we must demand that:
+
+$$\begin{aligned}
+    \pdv{k}{s} + \hat{L} k
+    = 0
+\end{aligned}$$
+
+Because $k(X_s, s)$ is a Markov process,
+we can write it with a transition density $p(s, X_s; t, X_t)$,
+where in this case $s$ and $X_s$ are given initial conditions,
+$t$ is a parameter, and the terminal state $X_t$ is a random variable.
+We thus have:
+
+$$\begin{aligned}
+    k(x, s)
+    = \int_{-\infty}^\infty p(s, x; t, y) \: h(y) \dd{y}
+\end{aligned}$$
+
+We insert this into the equation that we just derived for $k$, yielding:
+
+$$\begin{aligned}
+    0
+    = \int_{-\infty}^\infty \!\! \Big( \pdv{}{s}p(s, x; t, y) + \hat{L} p(s, x; t, y) \Big) h(y) \dd{y}
+\end{aligned}$$
+
+Because $h$ is arbitrary, and this must be satisfied for all $h$,
+the transition density $p$ fulfills:
+
+$$\begin{aligned}
+    0
+    = \pdv{}{s}p(s, x; t, y) + \hat{L} p(s, x; t, y)
+\end{aligned}$$
+
+Here, $t$ is a known parameter and $y$ is a "known" integration variable,
+leaving only $s$ and $x$ as free variables for us to choose.
+We therefore define the **likelihood function** $\psi(s, x)$,
+which gives the likelihood of an initial condition $(s, x)$
+given that the terminal condition is $(t, y)$:
+
+$$\begin{aligned}
+    \boxed{
+        \psi(s, x)
+        \equiv p(s, x; t, y)
+    }
+\end{aligned}$$
+
+And from the above derivation,
+we conclude that $\psi$ satisfies the following PDE,
+known as the **backward Kolmogorov equation**:
+
+$$\begin{aligned}
+    \boxed{
+        - \pdv{\psi}{s}
+        = \hat{L} \psi
+        = f \pdv{\psi}{x} + \frac{1}{2} g^2 \pdvn{2}{\psi}{x}
+    }
+\end{aligned}$$
+
+Moving on, we can define the traditional
+**probability density function** $\phi(t, y)$ from the transition density $p$,
+by fixing the initial $(s, x)$
+and leaving the terminal $(t, y)$ free:
+
+$$\begin{aligned}
+    \boxed{
+        \phi(t, y)
+        \equiv p(s, x; t, y)
+    }
+\end{aligned}$$
+
+With this in mind, for $(s, x) = (0, X_0)$,
+the unconditional expectation $\mathbf{E}[Y_t]$
+(i.e. the conditional expectation without information)
+will be constant in time, because $Y_t$ is a martingale:
+
+$$\begin{aligned}
+    \mathbf{E}[Y_t]
+    = \mathbf{E}[k(X_t, t)]
+    = \int_{-\infty}^\infty k(y, t) \: \phi(t, y) \dd{y}
+    = \Inprod{k}{\phi}
+    = \mathrm{const}
+\end{aligned}$$
+
+This integral has the form of an inner product,
+so we switch to [Dirac notation](/know/concept/dirac-notation/).
+We differentiate with respect to $t$,
+and use the backward equation $\ipdv{k}{t} + \hat{L} k = 0$:
+
+$$\begin{aligned}
+    0
+    = \pdv{}{t}\Inprod{k}{\phi}
+    = \Inprod{k}{\pdv{\phi}{t}} + \Inprod{\pdv{k}{t}}{\phi}
+    = \Inprod{k}{\pdv{\phi}{t}} - \Inprod{\hat{L} k}{\phi}
+    = \Inprod{k}{\pdv{\phi}{t} - \hat{L}{}^\dagger \phi}
+\end{aligned}$$
+
+Where $\hat{L}{}^\dagger$ is by definition the adjoint operator of $\hat{L}$,
+which we calculate using partial integration,
+where all boundary terms vanish thanks to the *existence* of $X_t$;
+in other words, $X_t$ cannot reach infinity at any finite $t$,
+so the integrand must decay to zero for $|y| \to \infty$:
+
+$$\begin{aligned}
+    \Inprod{\hat{L} k}{\phi}
+    &= \int_{-\infty}^\infty \pdv{k}{y} f \phi + \frac{1}{2} \pdvn{2}{k}{y} g^2 \phi \dd{y}
+    \\
+    &= \bigg[ k f \phi + \frac{1}{2} \pdv{k}{y} g^2 \phi \bigg]_{-\infty}^\infty
+    - \int_{-\infty}^\infty k \pdv{}{y}(f \phi) + \frac{1}{2} \pdv{k}{y} \pdv{}{y}(g^2 \phi) \dd{y}
+    \\
+    &= \bigg[ -\frac{1}{2} k g^2 \phi \bigg]_{-\infty}^\infty
+    + \int_{-\infty}^\infty - k \pdv{}{y}(f \phi) + \frac{1}{2} k \pdvn{2}{}{y}(g^2 \phi) \dd{y}
+    \\
+    &= \int_{-\infty}^\infty k \: \big( \hat{L}{}^\dagger \phi \big) \dd{y}
+    = \Inprod{k}{\hat{L}{}^\dagger \phi}
+\end{aligned}$$
+
+Since $k$ is arbitrary, and $\ipdv{\Inprod{k}{\phi}}{t} = 0$ for all $k$,
+we thus arrive at the **forward Kolmogorov equation**,
+describing the evolution of the probability density $\phi(t, y)$:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\phi}{t}
+        = \hat{L}{}^\dagger \phi
+        = - \pdv{}{y}(f \phi) + \frac{1}{2} \pdvn{2}{}{y}(g^2 \phi)
+    }
+\end{aligned}$$
+
+This can be rewritten in a way
+that highlights the connection between Itō diffusions and physical diffusion,
+if we define the **diffusivity** $D$, **advection** $u$, and **probability flux** $J$:
+
+$$\begin{aligned}
+    D
+    \equiv \frac{1}{2} g^2
+    \qquad \quad
+    u
+    = f - \pdv{D}{x}
+    \qquad \quad
+    J
+    \equiv u \phi - D \pdv{\phi}{x}
+\end{aligned}$$
+
+Such that the forward Kolmogorov equation takes the following **conservative form**,
+so called because it looks like a physical continuity equation:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\phi}{t}
+        = - \pdv{J}{x}
+        = - \pdv{}{x}\Big( u \phi - D \pdv{\phi}{x} \Big)
+    }
+\end{aligned}$$
+
+Note that if $u = 0$, then this reduces to
+[Fick's second law](/know/concept/ficks-laws/).
+The backward Kolmogorov equation can also be rewritten analogously,
+although it is less noteworthy:
+
+$$\begin{aligned}
+    \boxed{
+        - \pdv{\psi}{t}
+        = u \pdv{\psi}{x} + \pdv{}{x}\Big( D \pdv{\psi}{x} \Big)
+    }
+\end{aligned}$$
+
+Notice that the diffusivity term looks the same
+in both the forward and backward equations;
+we say that diffusion is self-adjoint.
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/kramers-kronig-relations/index.md b/source/know/concept/kramers-kronig-relations/index.md
new file mode 100644
index 0000000..f66ab0a
--- /dev/null
+++ b/source/know/concept/kramers-kronig-relations/index.md
@@ -0,0 +1,135 @@
+---
+title: "Kramers-Kronig relations"
+date: 2021-02-25
+categories:
+- Mathematics
+- Complex analysis
+- Physics
+- Optics
+layout: "concept"
+---
+
+Let $\chi(t)$ be a complex function describing
+the response of a system to an impulse $f(t)$ starting at $t = 0$.
+The **Kramers-Kronig relations** connect the real and imaginary parts of $\chi(t)$,
+such that one can be reconstructed from the other.
+Suppose we can only measure $\chi_r(t)$ or $\chi_i(t)$:
+
+$$\begin{aligned}
+    \chi(t) = \chi_r(t) + i \chi_i(t)
+\end{aligned}$$
+
+Assuming that the system was at rest until $t = 0$,
+the response $\chi(t)$ cannot depend on anything from $t < 0$,
+since the known impulse $f(t)$ had not started yet,
+This principle is called **causality**, and to enforce it,
+we use the [Heaviside step function](/know/concept/heaviside-step-function/)
+$\Theta(t)$ to create a **causality test** for $\chi(t)$:
+
+$$\begin{aligned}
+    \chi(t) = \chi(t) \: \Theta(t)
+\end{aligned}$$
+
+If we [Fourier transform](/know/concept/fourier-transform/) this equation,
+then it will become a convolution in the frequency domain
+thanks to the [convolution theorem](/know/concept/convolution-theorem/),
+where $A$, $B$ and $s$ are constants from the FT definition:
+
+$$\begin{aligned}
+    \tilde{\chi}(\omega)
+    = (\tilde{\chi} * \tilde{\Theta})(\omega)
+    = B \int_{-\infty}^\infty \tilde{\chi}(\omega') \: \tilde{\Theta}(\omega - \omega') \dd{\omega'}
+\end{aligned}$$
+
+We look up the FT of the step function $\tilde{\Theta}(\omega)$,
+which involves the signum function $\mathrm{sgn}(t)$,
+the [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$,
+and the Cauchy principal value $\pv{}$.
+We arrive at:
+
+$$\begin{aligned}
+    \tilde{\chi}(\omega)
+    &= \frac{A B}{|s|} \pv{\int_{-\infty}^\infty \tilde{\chi}(\omega')
+    \Big( \pi \delta(\omega - \omega') + i \:\mathrm{sgn} \frac{1}{\omega - \omega'} \Big) \dd{\omega'}}
+    \\
+    &= \Big( \frac{1}{2} \frac{2 \pi A B}{|s|} \Big) \tilde{\chi}(\omega)
+    + i \Big( \frac{\mathrm{sgn}(s)}{2 \pi} \frac{2 \pi A B}{|s|} \Big)
+    \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}}
+\end{aligned}$$
+
+From the definition of the Fourier transform we know that $2 \pi A B / |s| = 1$:
+
+$$\begin{aligned}
+    \tilde{\chi}(\omega)
+    &= \frac{1}{2} \tilde{\chi}(\omega)
+    + \mathrm{sgn}(s) \frac{i}{2 \pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}}
+\end{aligned}$$
+
+We isolate this equation for $\tilde{\chi}(\omega)$
+to get the final version of the causality test:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{\chi}(\omega)
+        = - \mathrm{sgn}(s) \frac{i}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}(\omega')}{\omega - \omega'} \dd{\omega'}}
+    }
+\end{aligned}$$
+
+By inserting $\tilde{\chi}(\omega) = \tilde{\chi}_r(\omega) + i \tilde{\chi}_i(\omega)$
+and splitting the equation into real and imaginary parts,
+we get the Kramers-Kronig relations:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \tilde{\chi}_r(\omega)
+            &= \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{\omega' - \omega} \dd{\omega'}}
+            \\
+            \tilde{\chi}_i(\omega)
+            &= - \mathrm{sgn}(s) \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{\omega' - \omega} \dd{\omega'}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+If the time-domain response function $\chi(t)$ is real
+(so far we have assumed it to be complex),
+then we can take advantage of the fact that
+the FT of a real function satisfies
+$\tilde{\chi}(-\omega) = \tilde{\chi}^*(\omega)$, i.e. $\tilde{\chi}_r(\omega)$
+is even and $\tilde{\chi}_i(\omega)$ is odd. We multiply the fractions by
+$(\omega' + \omega)$ above and below:
+
+$$\begin{aligned}
+    \tilde{\chi}_r(\omega)
+    &= \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}}
+    + \frac{\omega}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_i(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} \bigg)
+    \\
+    \tilde{\chi}_i(\omega)
+    &= - \mathrm{sgn}(s) \bigg( \frac{1}{\pi} \pv{\int_{-\infty}^\infty \frac{\omega' \tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}}
+    + \frac{\omega}{\pi} \pv{\int_{-\infty}^\infty \frac{\tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}} \bigg)
+\end{aligned}$$
+
+For $\tilde{\chi}_r(\omega)$, the second integrand is odd, so we can drop it.
+Similarly, for $\tilde{\chi}_i(\omega)$, the first integrand is odd.
+We therefore find the following variant of the Kramers-Kronig relations:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \tilde{\chi}_r(\omega)
+            &= \mathrm{sgn}(s) \frac{2}{\pi} \pv{\int_0^\infty \frac{\omega' \tilde{\chi}_i(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}}
+            \\
+            \tilde{\chi}_i(\omega)
+            &= - \mathrm{sgn}(s) \frac{2 \omega}{\pi} \pv{\int_0^\infty \frac{\tilde{\chi}_r(\omega')}{ {\omega'}^2 - \omega^2} \dd{\omega'}}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+To reiterate: this version is only valid if $\chi(t)$ is real in the time domain.
+
+
+
+## References
+1.  M. Wubs,
+    *Optical properties of solids: Kramers-Kronig relations*, 2013,
+    unpublished.
diff --git a/source/know/concept/kubo-formula/index.md b/source/know/concept/kubo-formula/index.md
new file mode 100644
index 0000000..40d90e1
--- /dev/null
+++ b/source/know/concept/kubo-formula/index.md
@@ -0,0 +1,170 @@
+---
+title: "Kubo formula"
+date: 2021-09-23
+categories:
+- Physics
+- Quantum mechanics
+- Perturbation
+layout: "concept"
+---
+
+Consider the following quantum Hamiltonian,
+split into a main time-independent term $\hat{H}_{0,S}$
+and a small time-dependent perturbation $\hat{H}_{1,S}$,
+which is turned on at $t = t_0$:
+
+$$\begin{aligned}
+    \hat{H}_S(t)
+    = \hat{H}_{0,S} + \hat{H}_{1,S}(t)
+\end{aligned}$$
+
+And let $\Ket{\psi_S(t)}$ be the corresponding solutions to the Schrödinger equation.
+Then, given a time-independent observable $\hat{A}$,
+its expectation value $\expval{\hat{A}}$ evolves like so,
+where the subscripts $S$ and $I$
+respectively refer to the Schrödinger
+and [interaction pictures](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \expval{\hat{A}}(t)
+    = \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)}
+    &= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)}
+    \\
+    &= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)}
+\end{aligned}$$
+
+Where the time evolution operator $\hat{K}_I(t, t_0)$ is as follows,
+which we Taylor-expand:
+
+$$\begin{aligned}
+    \hat{K}_I(t, t_0)
+    = \mathcal{T} \bigg\{ \exp\!\bigg( \frac{1}{i \hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \bigg\}
+    \approx 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'}
+\end{aligned}$$
+
+With this, the following product of operators (as encountered earlier) can be written as:
+
+$$\begin{aligned}
+    \hat{K}_I^\dagger \hat{A}_I \hat{K}_I
+    &\approx \bigg( 1 + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg) \hat{A}_I(t)
+    \bigg( 1 - \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \dd{t'} \bigg)
+    \\
+    &\approx \hat{A}_I(t)
+    - \frac{i}{\hbar} \int_{t_0}^t \hat{A}_I(t) \hat{H}_{1,I}(t') \dd{t'}
+    + \frac{i}{\hbar} \int_{t_0}^t \hat{H}_{1,I}(t') \hat{A}_I(t) \dd{t'}
+\end{aligned}$$
+
+Where we have dropped the last term,
+because $\hat{H}_{1}$ is assumed to be so small
+that it only matters to first order.
+Here, we notice a commutator, so we can rewrite:
+
+$$\begin{aligned}
+    \hat{K}_I^\dagger \hat{A}_I \hat{K}_I
+    &= \hat{A}_I(t) - \frac{i}{\hbar} \int_{t_0}^t \Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')} \dd{t'}
+\end{aligned}$$
+
+Returning to $\expval{\hat{A}}$,
+we have the following formula,
+where $\Expval{}$ is the expectation value for $\Ket{\psi(t)}$,
+and $\Expval{}_0$ is the expectation value for $\Ket{\psi_I(t_0)}$:
+
+$$\begin{aligned}
+    \expval{\hat{A}}(t)
+    = \expval{\hat{K}_I^\dagger \hat{A}_I \hat{K}_I}_0
+    = \expval{\hat{A}_I(t)}_0 - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+\end{aligned}$$
+
+Now we define $\delta\!\expval{\hat{A}}\!(t)$
+as the change of $\expval{\hat{A}}$ due to the perturbation $\hat{H}_1$,
+and insert $\expval{\hat{A}}(t)$:
+
+$$\begin{aligned}
+    \delta\!\expval{\hat{A}}\!(t)
+    \equiv \expval{\hat{A}}(t) - \expval{\hat{A}_I}_0
+    = - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+\end{aligned}$$
+
+Finally, we introduce
+a [Heaviside step function](/know/concept/heaviside-step-function) $\Theta$
+and change the integration limit accordingly,
+leading to the **Kubo formula**
+describing the response of $\expval{\hat{A}}$ to first order in $\hat{H}_1$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta\!\expval{\hat{A}}\!(t)
+        = \int_{t_0}^\infty C^R_{A H_1}(t, t') \dd{t'}
+    }
+\end{aligned}$$
+
+Where we have defined the **retarded correlation function** $C^R_{A H_1}(t, t')$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        C^R_{A H_1}(t, t')
+        \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0
+    }
+\end{aligned}$$
+
+Note that observables are bosonic,
+because in the [second quantization](/know/concept/second-quantization/)
+they consist of products of even numbers
+of particle creation/annihiliation operators.
+Therefore, this correlation function
+is a two-particle [Green's function](/know/concept/greens-functions/).
+
+A common situation is that $\hat{H}_1$ consists of
+a time-independent operator $\hat{B}$
+and a time-dependent function $f(t)$,
+allowing us to split $C^R_{A H_1}$ as follows:
+
+$$\begin{aligned}
+    \hat{H}_{1,S}(t)
+    = \hat{B}_S \: f(t)
+    \quad \implies \quad
+    C^R_{A H_1}(t, t')
+    = C^R_{A B}(t, t') f(t')
+\end{aligned}$$
+
+Since $C_{AB}^R$ is a Green's function,
+we know that it only depends on the difference $t - t'$,
+as long as the system was initially in thermodynamic equilibrium,
+and $\hat{H}_{0,S}$ is time-independent:
+
+$$\begin{aligned}
+    C^R_{A B}(t, t')
+    = C^R_{A B}(t - t')
+\end{aligned}$$
+
+With this, the Kubo formula can be written as follows,
+where we have set $t_0 = - \infty$:
+
+$$\begin{aligned}
+    \delta\!\expval{A}\!(t)
+    = \int_{-\infty}^\infty C^R_{A B}(t - t') f(t') \dd{t'}
+    = (C^R_{A B} * f)(t)
+\end{aligned}$$
+
+This is a convolution,
+so the [convolution theorem](/know/concept/convolution-theorem/)
+states that the [Fourier transform](/know/concept/fourier-transform/)
+of $\delta\!\expval{\hat{A}}\!(t)$ is simply the product
+of the transforms of $C^R_{AB}$ and $f$:
+
+$$\begin{aligned}
+    \boxed{
+        \delta\!\expval{\hat{A}}\!(\omega)
+        = \tilde{C}{}^R_{A B}(\omega) \: \tilde{f}(\omega)
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
+2.  K.S. Thygesen,
+    *Advanced solid state physics: linear response theory*,
+    2013, unpublished.
diff --git a/source/know/concept/lagrange-multiplier/index.md b/source/know/concept/lagrange-multiplier/index.md
new file mode 100644
index 0000000..4c01aed
--- /dev/null
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@@ -0,0 +1,121 @@
+---
+title: "Lagrange multiplier"
+date: 2021-03-02
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+The method of **Lagrange multipliers** or **undetermined multipliers**
+is a technique for optimizing (i.e. finding the extrema of)
+a function $f(x, y, z)$,
+subject to a given constraint $\phi(x, y, z) = C$,
+where $C$ is a constant.
+
+If we ignore the constraint $\phi$,
+optimizing $f$ simply comes down to finding stationary points:
+
+$$\begin{aligned}
+    0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z}
+\end{aligned}$$
+
+This problem is easy:
+$\dd{x}$, $\dd{y}$, and $\dd{z}$ are independent and arbitrary,
+so all we need to do is find the roots of
+the partial derivatives $f_x$, $f_y$ and $f_z$,
+which we respectively call $x_0$, $y_0$ and $z_0$,
+and then the extremum is simply $(x_0, y_0, z_0)$.
+
+But the constraint $\phi$, over which we have no control,
+adds a relation between $\dd{x}$, $\dd{y}$, and $\dd{z}$,
+so if two are known, the third is given by $\phi = C$.
+The problem is then a system of equations:
+
+$$\begin{aligned}
+    0 &= \dd{f} = f_x \dd{x} + f_y \dd{y} + f_z \dd{z}
+    \\
+    0 &= \dd{\phi} = \phi_x \dd{x} + \phi_y \dd{y} + \phi_z \dd{z}
+\end{aligned}$$
+
+Solving this directly would be a delicate balancing act
+of all the partial derivatives.
+
+To help us solve this, we introduce a "dummy" parameter $\lambda$,
+the so-called **Lagrange multiplier**,
+and contruct a new function $L$ given by:
+
+$$\begin{aligned}
+    L(x, y, z) = f(x, y, z) + \lambda \phi(x, y, z)
+\end{aligned}$$
+
+At the extremum, $\dd{L} = \dd{f} + \lambda \dd{\phi} = 0$,
+so now the problem is a "single" equation again:
+
+$$\begin{aligned}
+    0 = \dd{L}
+    = (f_x + \lambda \phi_x) \dd{x} + (f_y + \lambda \phi_y) \dd{y} + (f_z + \lambda \phi_z) \dd{z}
+\end{aligned}$$
+
+Assuming $\phi_z \neq 0$, we now choose $\lambda$ such that $f_z + \lambda \phi_z = 0$.
+This choice represents satisfying the constraint,
+so now the remaining $\dd{x}$ and $\dd{y}$ are independent again,
+and we simply have to find the roots of $f_x + \lambda \phi_x$ and $f_y + \lambda \phi_y$.
+
+In effect, after introducing $\lambda$,
+we have four unknowns $(x, y, z, \lambda)$,
+but also four equations:
+
+$$\begin{aligned}
+    L_x = L_y = L_z = 0
+    \qquad \quad
+    \phi = C
+\end{aligned}$$
+
+We are only really interested in the first three unknowns $(x, y, z)$,
+so $\lambda$ is sometimes called the **undetermined multiplier**,
+since it is just an algebraic helper whose value is irrelevant.
+
+This method generalizes nicely to multiple constraints or more variables:
+suppose that we want to find the extrema of $f(x_1, ..., x_N)$
+subject to $M < N$ conditions:
+
+$$\begin{aligned}
+    \phi_1(x_1, ..., x_N) = C_1 \qquad \cdots \qquad \phi_M(x_1, ..., x_N) = C_M
+\end{aligned}$$
+
+This once again turns into a delicate system of $M+1$ equations to solve:
+
+$$\begin{aligned}
+    0 &= \dd{f} = f_{x_1} \dd{x_1} + ... + f_{x_N} \dd{x_N}
+    \\
+    0 &= \dd{\phi_1} = \phi_{1, x_1} \dd{x_1} + ... + \phi_{1, x_N} \dd{x_N}
+    \\
+    &\vdots
+    \\
+    0 &= \dd{\phi_M} = \phi_{M, x_1} \dd{x_1} + ... + \phi_{M, x_N} \dd{x_N}
+\end{aligned}$$
+
+Then we introduce $M$ Lagrange multipliers $\lambda_1, ..., \lambda_M$
+and define $L(x_1, ..., x_N)$:
+
+$$\begin{aligned}
+    L = f + \sum_{m = 1}^M \lambda_m \phi_m
+\end{aligned}$$
+
+As before, we set $\dd{L} = 0$ and choose the multipliers $\lambda_1, ..., \lambda_M$
+to eliminate $M$ of its $N$ terms:
+
+$$\begin{aligned}
+    0 = \dd{L}
+    = \sum_{n = 1}^N \Big( f_{x_n} + \sum_{m = 1}^M \lambda_m \phi_{x_n} \Big) \dd{x_n}
+\end{aligned}$$
+
+
+## References
+1.  G.B. Arfken, H.J. Weber,
+    *Mathematical methods for physicists*, 6th edition, 2005,
+    Elsevier.
+2.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/lagrangian-mechanics/index.md b/source/know/concept/lagrangian-mechanics/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/lagrangian-mechanics/index.md
@@ -0,0 +1,129 @@
+---
+title: "Lagrangian mechanics"
+date: 2021-07-01
+categories:
+- Physics
+- Classical mechanics
+layout: "concept"
+---
+
+**Lagrangian mechanics** is a formulation of classical mechanics,
+which is equivalent to Newton's laws,
+but offers some advantages.
+Its mathematical backbone is the
+[calculus of variations](/know/concept/calculus-of-variations/),
+and hence it is built on the **principle of least action**,
+which states that the path taken by a system
+will be a minimum of the **action** (i.e. energy cost) of that path.
+
+For a moving object with position $x(t)$ and velocity $\dot{x}(t)$,
+we define the Lagrangian $L$ as the difference
+between its kinetic and potential energies:
+
+$$\begin{aligned}
+    \boxed{
+        L(x, \dot{x}, t) \equiv T - V = \frac{1}{2} m \dot{x}^2 - V(x)
+    }
+\end{aligned}$$
+
+From variational calculus we then get the Euler-Lagrange equation,
+which in this case turns out to just be Newton's second law:
+
+$$\begin{aligned}
+    \dv{}{t}\Big( \pdv{L}{\dot{x}} \Big) = \pdv{L}{x}
+    \qquad \implies \qquad
+    m \ddot{x} = - \pdv{V}{x} = F
+\end{aligned}$$
+
+But compared to Newtonian mechanics,
+Lagrangian mechanics scales better for large systems.
+For example, to describe the dynamics of $N$ objects $x_1(t), ..., x_N(t)$,
+we only need a single $L$
+from which the equations of motion can easily be derived.
+Getting these equations directly from Newton's laws could get messy.
+
+At no point have we assumed Cartesian coordinates:
+the Euler-Lagrange equations keep their form
+for any independent coordinates $q_1(t), ..., q_N(t)$:
+
+$$\begin{aligned}
+    \dv{}{t}\Big( \pdv{L}{\dot{q}_n} \Big) = \pdv{L}{q_n}
+\end{aligned}$$
+
+We define the **canonical momentum conjugate** $p_n(t)$
+and the **generalized force conjugate** $F_n(t)$ as follows,
+such that we can always get Newton's second law:
+
+$$\begin{aligned}
+    \boxed{
+        p_n \equiv \pdv{L}{\dot{q}_n} \qquad F_n \equiv \pdv{L}{q_n}
+    }
+    \qquad \implies \qquad
+    \dv{p_n}{t} = F_n
+\end{aligned}$$
+
+But this is actually a bit misleading,
+since $p_n$ need not be a momentum, nor $F_n$ a force,
+although often they are.
+For example, $p_n$ could be angular momentum, and $F_n$ torque.
+
+Another advantage of Lagrangian mechanics is that
+the conserved quantities can be extracted from $L$ using Noether's theorem.
+In the simplest case, if $L$ does not depend on $q_n$
+(then known as a **cyclic coordinate**),
+then we know that the "momentum" $p_n$ is a conserved quantity:
+
+$$\begin{aligned}
+    F_n = \pdv{L}{q_n} = 0
+    \qquad \implies \qquad
+    \dv{p_n}{t} = 0
+\end{aligned}$$
+
+Now, as the number of particles $N$ increases to infinity,
+variational calculus will give infinitely many coupled equations,
+which is obviously impractical.
+
+Such a system can be regarded as continuous, so the $N$ functions $q_n$
+can be replaced by a single density function $u(x,t)$.
+This approach can also be used for continuous fields,
+in which case the complex conjugate $u^*$ is often included.
+The Lagrangian $L$ then becomes:
+
+$$\begin{aligned}
+    L(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t)
+    = \int_{-\infty}^\infty \mathcal{L}(u, u^*, u_x, u_x^*, u_t, u_t^*, x, t) \dd{x}
+\end{aligned}$$
+
+Where $\mathcal{L}$ is known as the **Lagrangian density**.
+By inserting this into the functional $J$
+used for the derivation of the Euler-Lagrange equations, we get:
+
+$$\begin{aligned}
+    J[u]
+    = \int_{t_0}^{t_1} L \dd{t}
+    = \int_{t_0}^{t_1} \! \int_{-\infty}^\infty \mathcal{L} \dd{x} \dd{t}
+\end{aligned}$$
+
+This is simply 2D variational problem,
+so the Euler-Lagrange equations will be two PDEs:
+
+$$\begin{aligned}
+    0 &= \pdv{\mathcal{L}}{u} - \pdv{}{x}\Big( \pdv{\mathcal{L}}{u_x} \Big) - \pdv{}{t}\Big( \pdv{\mathcal{L}}{u_t} \Big)
+    \\
+    0 &= \pdv{\mathcal{L}}{u^*} - \pdv{}{x}\Big( \pdv{\mathcal{L}}{u_x^*} \Big) - \pdv{}{t}\Big( \pdv{\mathcal{L}}{u_t^*} \Big)
+\end{aligned}$$
+
+If $\mathcal{L}$ is real,
+then these two Euler-Lagrange equations will in fact be identical.
+
+Finally, note that for abstract fields,
+the Lagrangian density $\mathcal{L}$ rarely has
+a physical interpretation, and is not unique.
+Instead, it must be reverse-engineered from a relevant equation.
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/source/know/concept/laguerre-polynomials/index.md b/source/know/concept/laguerre-polynomials/index.md
new file mode 100644
index 0000000..130dff2
--- /dev/null
+++ b/source/know/concept/laguerre-polynomials/index.md
@@ -0,0 +1,125 @@
+---
+title: "Laguerre polynomials"
+date: 2021-09-08
+categories:
+- Mathematics
+layout: "concept"
+---
+
+The **Laguerre polynomials** are a set of useful functions that arise in physics.
+They are the non-singular eigenfunctions $u(x)$ of **Laguerre's equation**,
+with the corresponding eigenvalues $n$ being non-negative integers:
+
+$$\begin{aligned}
+    \boxed{
+        x u'' + (1 - x) u' + n u = 0
+    }
+\end{aligned}$$
+
+The $n$th-order Laguerre polynomial $L_n(x)$
+is given in the form of a *Rodrigues' formula* by:
+
+$$\begin{aligned}
+    L_n(x)
+    &= \frac{1}{n!} \exp(x) \dvn{n}{}{x}\big(x^n \exp(-x)\big)
+    \\
+    &= \frac{1}{n!} \Big( \dv{}{x}- 1 \Big)^n x^n
+\end{aligned}$$
+
+The first couple of Laguerre polynomials $L_n(x)$ are therefore as follows:
+
+$$\begin{gathered}
+    L_0(x) = 1
+    \qquad \quad
+    L_1(x) = 1 - x
+    \qquad \quad
+    L_2(x) = \frac{1}{2} (x^2 - 4 x + 2)
+\end{gathered}$$
+
+Based on Laguerre's equation,
+**Laguerre's generalized equation** is as follows,
+with an arbitrary real (but usually integer) parameter $\alpha$,
+and $n$ still a non-negative integer:
+
+$$\begin{aligned}
+    \boxed{
+        x u'' + (\alpha + 1 - x) u' + n u = 0
+    }
+\end{aligned}$$
+
+Its solutions, denoted by $L_n^\alpha(x)$,
+are the **generalized** or **associated Laguerre polynomials**,
+which also have a Rodrigues' formula.
+Note that if $\alpha = 0$ then $L_n^\alpha = L_n$:
+
+$$\begin{aligned}
+    L_n^\alpha(x)
+    &= \frac{1}{n!} x^{-\alpha} \exp(x) \dvn{n}{}{x}\big( x^{n + \alpha} \exp(-x) \big)
+    \\
+    &= \frac{x^{-\alpha}}{n!} \Big( \dv{}{x}- 1 \Big)^n x^{n + \alpha}
+\end{aligned}$$
+
+The first couple of associated Laguerre polynomials $L_n^\alpha(x)$ are therefore as follows:
+
+$$\begin{aligned}
+    L_0^\alpha(x) = 1
+    \qquad
+    L_1^\alpha(x) = \alpha + 1 - x
+    \qquad
+    L_2^\alpha(x) = \frac{1}{2} (x^2 - 2 \alpha x - 4 x + \alpha^2 + 3 \alpha + 2)
+\end{aligned}$$
+
+And then more $L_n^\alpha$ can be computed quickly
+using the following recurrence relation:
+
+$$\begin{aligned}
+    \boxed{
+        L_{n + 1}^\alpha(x)
+        = \frac{(\alpha + 2 n + 1 - x) L_n^\alpha(x) - (\alpha + n) L_{n - 1}^\alpha(x)}{n + 1}
+    }
+\end{aligned}$$
+
+The derivatives are also straightforward to calculate
+using the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        \dvn{k}{}{x}L_n^\alpha(x)
+        = (-1)^k L_{n - k}^{\alpha + k}(x)
+    }
+\end{aligned}$$
+
+Noteworthy is that these polynomials (both normal and associated)
+are all mutually orthogonal for $x \in [0, \infty[$,
+with respect to the weight function $w(x) \equiv x^\alpha \exp(-x)$:
+
+$$\begin{aligned}
+    \boxed{
+        \Inprod{L_m^\alpha}{w L_n^\alpha}
+        = \int_0^\infty L_m^\alpha(x) \: L_n^\alpha(x) \: w(x) \dd{x}
+        = \frac{\Gamma(n + \alpha + 1)}{n!} \delta_{nm}
+    }
+\end{aligned}$$
+
+Where $\delta_{nm}$ is the Kronecker delta.
+Moreover, they form a basis in
+the [Hilbert space](/know/concept/hilbert-space/)
+of all functions $f(x)$ for which $\Inprod{f}{w f}$ is finite.
+Any such $f$ can thus be expanded as follows:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \sum_{n = 0}^\infty a_n L_n^\alpha(x)
+        = \sum_{n = 0}^\infty \frac{\Inprod{L_n}{w f}}{\Inprod{L_n}{w L_n}} L_n^\alpha(x)
+    }
+\end{aligned}$$
+
+Finally, the $L_n^\alpha(x)$ are related to
+the [Hermite polynomials](/know/concept/hermite-polynomials/) $H_n(x)$ like so:
+
+$$\begin{aligned}
+    H_{2n(x)} &= (-1)^n 2^{2n} n! \: L_n^{-1/2}(x^2)
+    \\
+    H_{2n + 1(x)} &= (-1)^n 2^{2n + 1} n! \: L_n^{1/2}(x^2)
+\end{aligned}$$
diff --git a/source/know/concept/landau-quantization/index.md b/source/know/concept/landau-quantization/index.md
new file mode 100644
index 0000000..e28b396
--- /dev/null
+++ b/source/know/concept/landau-quantization/index.md
@@ -0,0 +1,122 @@
+---
+title: "Landau quantization"
+date: 2021-07-01
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+When a particle with charge $q$ is moving in a homogeneous
+[magnetic field](/know/concept/magnetic-field/),
+quantum mechanics decrees that its allowed energies split
+into degenerate discrete **Landau levels**,
+a phenomenon known as **Landau quantization**.
+
+Starting from the Hamiltonian $\hat{H}$ for a particle with mass $m$
+in a vector potential $\vec{A}(\hat{Q})$:
+
+$$\begin{aligned}
+    \hat{H}
+    &= \frac{1}{2 m} \big( \hat{p} - q \vec{A} \big)^2
+\end{aligned}$$
+
+We choose $\vec{A} = (- \hat{y} B, 0, 0)$,
+yielding a magnetic field $\vec{B} = \nabla \times \vec{A}$
+pointing in the $z$-direction with strength $B$.
+The Hamiltonian becomes:
+
+$$\begin{aligned}
+    \hat{H}
+    &= \frac{\big( \hat{p}_x - q B \hat{y} \big)^2}{2 m}  + \frac{\hat{p}_y^2}{2 m} + \frac{\hat{p}_z^2}{2 m}
+\end{aligned}$$
+
+The only position operator occurring in $\hat{H}$ is $\hat{y}$,
+so $[\hat{H}, \hat{p}_x] = [\hat{H}, \hat{p}_z] = 0$.
+Because $\hat{p}_z$ appears in an unmodified kinetic energy term,
+and the corresponding $\hat{z}$ does not occur at all,
+the particle has completely free motion in the $z$-direction.
+Likewise, because $\hat{x}$ does not occur in $\hat{H}$,
+we can replace $\hat{p}_x$ by its eigenvalue $\hbar k_x$,
+although the motion is not free, due to $q B \hat{y}$.
+
+Based on the absence of $\hat{x}$ and $\hat{z}$,
+we make the following ansatz for the wavefunction $\Psi$:
+a plane wave in the $x$ and $z$ directions, multiplied by an unknown $\phi(y)$:
+
+$$\begin{aligned}
+    \Psi(x, y, z)
+    = \phi(y) \exp(i k_x x + i k_z z)
+\end{aligned}$$
+
+Inserting this into the time-independent Schrödinger equation gives,
+after dividing out the plane wave exponential $\exp(i k_x x + i k_z z)$:
+
+$$\begin{aligned}
+    E \phi
+    &= \frac{1}{2 m} \Big( (\hbar k_x - q B y)^2 + \hat{p}_y^2 + \hbar^2 k_z^2 \Big) \phi
+\end{aligned}$$
+
+By defining the cyclotron frequency $\omega_c \equiv q B / m$ and rearranging,
+we can turn this into a 1D quantum harmonic oscillator in $y$,
+with a couple of extra terms:
+
+$$\begin{aligned}
+    \Big( E - \frac{\hbar^2 k_z^2}{2 m} \Big) \phi
+    &= \bigg( \frac{1}{2} m \omega_c^2 \Big( y - \frac{\hbar k_x}{m \omega_c} \Big)^2 + \frac{\hat{p}_y^2}{2 m} \bigg) \phi
+\end{aligned}$$
+
+The potential minimum is shifted by $y_0 = \hbar k_x / (m \omega_c)$,
+and a plane wave in $z$ contributes to the energy $E$.
+In any case, the energy levels of this type of system are well-known:
+
+$$\begin{aligned}
+    \boxed{
+        E_n = \hbar \omega_c \Big(n + \frac{1}{2}\Big) + \frac{\hbar^2 k_z^2}{2 m}
+    }
+\end{aligned}$$
+
+And $\Psi_n$ is then as follows,
+where $\phi$ is the known quantum harmonic oscillator solution:
+
+$$\begin{aligned}
+    \Psi_n(x, y, z)
+    = \phi_n(y - y_0) \exp(i k_x x + i k_z z)
+\end{aligned}$$
+
+Note that this wave function contains $k_x$ (also inside $y_0$),
+but $k_x$ is absent from the energy $E_n$.
+This implies degeneracy:
+assuming periodic boundary conditions $\Psi(x\!+\!L_x) = \Psi(x)$,
+then $k_x$ can take values of the form $2 \pi n / L_x$, for $n \in \mathbb{Z}$.
+
+However, $k_x$ also occurs in the definition of $y_0$, so the degeneracy
+is finite, since $y_0$ must still lie inside the system,
+or, more formally, $y_0 \in [0, L_y]$:
+
+$$\begin{aligned}
+    0 \le y_0 = \frac{\hbar k_x}{m \omega_c} = \frac{\hbar 2 \pi n}{q B L_x} \le L_y
+\end{aligned}$$
+
+Isolating this for $n$, we find the following upper bound of the degeneracy:
+
+$$\begin{aligned}
+    \boxed{
+        n \le
+        \frac{q B L_x L_y}{2 \pi \hbar} = \frac{q B A}{h}
+    }
+\end{aligned}$$
+
+Where $A \equiv L_x L_y$ is the area of the confinement in the $(x,y)$-plane.
+Evidently, the degeneracy of each level increases with larger $B$,
+but since $\omega_c = q B / m$, the energy gap between each level increases too.
+In other words: the [density of states](/know/concept/density-of-states/)
+is a constant with respect to the energy,
+but the states get distributed across the $E_n$ differently depending on $B$.
+
+
+
+## References
+1.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.
diff --git a/source/know/concept/langmuir-waves/index.md b/source/know/concept/langmuir-waves/index.md
new file mode 100644
index 0000000..b5f3fd4
--- /dev/null
+++ b/source/know/concept/langmuir-waves/index.md
@@ -0,0 +1,256 @@
+---
+title: "Langmuir waves"
+date: 2021-10-30
+categories:
+- Physics
+- Plasma physics
+- Plasma waves
+- Perturbation
+layout: "concept"
+---
+
+In plasma physics, **Langmuir waves** are oscillations in the electron density,
+which may or may not propagate, depending on the temperature.
+
+Assuming no [magnetic field](/know/concept/magnetic-field/) $\vb{B} = 0$,
+no ion motion $\vb{u}_i = 0$ (since $m_i \gg m_e$),
+and therefore no ion-electron momentum transfer,
+the [two-fluid equations](/know/concept/two-fluid-equations/)
+tell us that:
+
+$$\begin{aligned}
+    m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
+    = q_e n_e \vb{E} - \nabla p_e
+    \qquad \quad
+    \pdv{n_e}{t} + \nabla \cdot (n_e \vb{u}_e) = 0
+\end{aligned}$$
+
+These are the electron momentum and continuity equations.
+We also need [Gauss' law](/know/concept/maxwells-equations/):
+
+$$\begin{aligned}
+    \varepsilon_0 \nabla \cdot \vb{E}
+    = q_e (n_e - n_i)
+\end{aligned}$$
+
+We split $n_e$, $\vb{u}_e$ and $\vb{E}$ into a base component
+(subscript $0$) and a perturbation (subscript $1$):
+
+$$\begin{aligned}
+    n_e
+    = n_{e0} + n_{e1}
+    \qquad \quad
+    \vb{u}_e
+    = \vb{u}_{e0} + \vb{u}_{e1}
+    \qquad \quad
+    \vb{E}
+    = \vb{E}_0 + \vb{E}_1
+\end{aligned}$$
+
+Where the perturbations $n_{e1}$, $\vb{u}_{e1}$ and $\vb{E}_1$ are very small,
+and the equilibrium components $n_{e0}$, $\vb{u}_{e0}$ and $\vb{E}_0$
+by definition satisfy:
+
+$$\begin{aligned}
+    \pdv{n_{e0}}{t} = 0
+    \qquad
+    \pdv{\vb{u}_{e0}}{t} = 0
+    \qquad
+    \nabla n_{e0} = 0
+    \qquad
+    \vb{u}_{e0} = 0
+    \qquad
+    \vb{E}_0 = 0
+\end{aligned}$$
+
+We insert this decomposistion into the electron continuity equation,
+arguing that $n_{e1} \vb{u}_{e1}$ is small enough to neglect, leading to:
+
+$$\begin{aligned}
+    0
+    &= \pdv{(n_{e0}\!+\! n_{e1})}{t} + \nabla \cdot \Big( (n_{e0} \!+\! n_{e1}) \: (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \Big)
+    \\
+    &= \pdv{n_{e1}}{t} + \nabla \cdot \Big( n_{e0} \vb{u}_{e1} + n_{e1} \vb{u}_{e1} \Big)
+    \\
+    &\approx \pdv{n_{e1}}{t} + \nabla \cdot (n_{e0} \vb{u}_{e1})
+    = \pdv{n_{e1}}{t} + n_{e0} \nabla \cdot \vb{u}_{e1}
+\end{aligned}$$
+
+Likewise, we insert it into Gauss' law,
+and use the plasma's quasi-neutrality $n_i = n_{e0}$ to get:
+
+$$\begin{aligned}
+    \varepsilon_0 \nabla \cdot \big( \vb{E}_0 \!+\! \vb{E}_1 \big)
+    = q_e (n_{e0} + n_{e1} - n_i)
+    \quad \implies \quad
+    \varepsilon_0 \nabla \cdot \vb{E}_1
+    = q_e n_{e1}
+\end{aligned}$$
+
+Since we are looking for linear waves,
+we make the following ansatz for the perturbations:
+
+$$\begin{aligned}
+    n_{e1}(\vb{r}, t)
+    &= n_{e1} \exp(i \vb{k} \cdot \vb{r} - i \omega t)
+    \\
+    \vb{u}_{e1}(\vb{r}, t)
+    &= \vb{u}_{e1} \exp(i \vb{k} \cdot \vb{r}  - i \omega t)
+    \\
+    \vb{E}_1(\vb{r}, t)
+    &= \vb{E}_1 \:\exp(i \vb{k} \cdot \vb{r} - i \omega t)
+\end{aligned}$$
+
+Inserting this into the continuity equation and Gauss' law yields, respectively:
+
+$$\begin{aligned}
+    - i \omega n_{e1} = - i n_{e0} \vb{k} \cdot \vb{u}_{e1}
+    \qquad \quad
+    -\! i \varepsilon_0 \vb{k} \cdot \vb{E}_1 = q_e n_{e1}
+\end{aligned}$$
+
+However, there are three unknowns $n_{e1}$, $\vb{u}_{e1}$ and $\vb{E}_1$,
+so one more equation is needed.
+
+
+## Cold Langmuir waves
+
+We therefore turn to the electron momentum equation.
+For now, let us assume that the electrons have no thermal motion,
+i.e. the electron temperature $T_e = 0$, so that $p_e = 0$, leaving:
+
+$$\begin{aligned}
+    m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
+    = q_e n_e \vb{E}
+\end{aligned}$$
+
+Inserting the decomposition then gives the following,
+where we neglect $(\vb{u}_{e1} \cdot \nabla) \vb{u}_{e1}$
+because $\vb{u}_{e1}$ is so small by assumption:
+
+$$\begin{gathered}
+    m_e (n_{e0} \!+\! n_{e1}) \Big( \pdv{(\vb{u}_{e0} \!+\! \vb{u}_{e1})}{t}
+    + \big( (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \cdot \nabla \big) (\vb{u}_{e0} \!+\! \vb{u}_{e1}) \Big)
+    = q_e \big( n_{e0} \!+\! n_{e1} \big) \big( \vb{E}_0 \!+\! \vb{E}_1 \big)
+    \\
+    \implies \qquad
+    q_e \vb{E}_1
+    = m_e \Big( \pdv{\vb{u}_{e1}}{t} + \big(\vb{u}_{e1} \cdot \nabla \big) \vb{u}_{e1} \Big)
+    \approx m_e \pdv{\vb{u}_{e1}}{t}
+\end{gathered}$$
+
+And then inserting our plane-wave ansatz yields
+the third equation we were looking for:
+
+$$\begin{aligned}
+    -i \omega m_e \vb{u}_{e1} = q_e \vb{E}_1
+\end{aligned}$$
+
+Solving this system of three equations for $\omega^2$
+gives the following dispersion relation:
+
+$$\begin{aligned}
+    \omega^2
+    = \frac{\omega n_{e0}}{n_{e1}} \vb{k} \cdot \vb{u}_{e1}
+    = \frac{i \omega n_{e0} q_e}{\omega m_e n_{e1}} \vb{k} \cdot \vb{E}_1
+    = \frac{i n_{e0} n_{e1} q_e^2}{i \varepsilon_0 m_e n_{e1}}
+    = \frac{n_{e0} q_e^2}{\varepsilon_0 m_e}
+\end{aligned}$$
+
+This result is known as the **plasma frequency** $\omega_p$,
+and describes the frequency of **cold Langmuir waves**,
+otherwise known as **plasma oscillations**:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_p
+        = \sqrt{\frac{n_{0e} q_e^2}{\varepsilon_0 m_e}}
+    }
+\end{aligned}$$
+
+Note that this is a dispersion relation $\omega(k) = \omega_p$,
+but that $\omega_p$ does not contain $k$.
+This means that cold Langmuir waves do not propagate:
+the oscillation is "stationary".
+
+
+## Warm Langmuir waves
+
+Next, we generalize this result to nonzero $T_e$,
+in which case the pressure $p_e$ is involved:
+
+$$\begin{aligned}
+    m_e n_{e0} \pdv{}{\vb{u}{e1}}{t}
+    = q_e n_{e0} \vb{E}_1 - \nabla p_e
+\end{aligned}$$
+
+From the two-fluid thermodynamic equation of state,
+we know that $\nabla p_e$ can be written as:
+
+$$\begin{aligned}
+    \nabla p_e
+    = \gamma k_B T_e \nabla n_e
+    = \gamma k_B T_e \nabla (n_{e0} + n_{e1})
+    = \gamma k_B T_e \nabla n_{e1}
+\end{aligned}$$
+
+With this, insertion of our plane-wave ansatz
+into the electron equation results in:
+
+$$\begin{aligned}
+    -i \omega m_e n_{e0} \vb{u}_{e1} = q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k}
+\end{aligned}$$
+
+Which once again closes the system of three equations.
+Solving for $\omega^2$ then gives:
+
+$$\begin{aligned}
+    \omega^2
+    = \frac{\omega n_{e0}}{n_{e1}} \vb{k} \cdot \vb{u}_{e1}
+    &= \frac{i \omega n_{e0}}{\omega n_{e0} m_e n_{e1}} \vb{k} \cdot \Big( q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k} \Big)
+    \\
+    &= \frac{n_{e0} q_e^2}{\varepsilon_0 m_e} - \frac{i \omega}{\omega m_e n_{e1}} i \gamma k_B T_e n_{e1} \big(\vb{k} \cdot \vb{k}\big)
+\end{aligned}$$
+
+Recognizing the first term as the plasma frequency $\omega_p^2$,
+we therefore arrive at the **Bohm-Gross dispersion relation** $\omega(\vb{k})$
+for **warm Langmuir waves**:
+
+$$\begin{aligned}
+    \boxed{
+        \omega^2
+        = \omega_p^2 + \frac{\gamma k_B T_e}{m_e} |\vb{k}|^2
+    }
+\end{aligned}$$
+
+This expression is typically quoted for 1D oscillations,
+in which case $\gamma = 3$ and $k = |\vb{k}|$:
+
+$$\begin{aligned}
+    \omega^2
+    = \omega_p^2 + \frac{3 k_B T_e}{m_e} k^2
+\end{aligned}$$
+
+Unlike for $T_e = 0$, these "warm" waves do propagate,
+carrying information at group velocity $v_g$,
+which, in the limit of large $k$, is given by:
+
+$$\begin{aligned}
+    v_g
+    = \pdv{\omega}{k}
+    \to \sqrt{\frac{3 k_B T_e}{m_e}}
+\end{aligned}$$
+
+This is the root-mean-square velocity of the
+[Maxwell-Boltzmann speed distribution](/know/concept/maxwell-boltzmann-distribution/),
+meaning that information travels at the thermal velocity for large $k$.
+
+
+
+## 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/laplace-transform/index.md b/source/know/concept/laplace-transform/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/laplace-transform/index.md
@@ -0,0 +1,125 @@
+---
+title: "Laplace transform"
+date: 2021-07-02
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+The **Laplace transform** is an integral transform
+that losslessly converts a function $f(t)$ of a real variable $t$,
+into a function $\tilde{f}(s)$ of a complex variable $s$,
+where $s$ is sometimes called the **complex frequency**,
+analogously to the [Fourier transform](/know/concept/fourier-transform/).
+The transform is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}(s)
+        \equiv \hat{\mathcal{L}}\{f(t)\}
+        \equiv \int_0^\infty f(t) \exp(- s t) \dd{t}
+    }
+\end{aligned}$$
+
+Depending on $f(t)$, this integral may diverge.
+This is solved by restricting the domain of $\tilde{f}(s)$
+to $s$ where $\mathrm{Re}\{s\} > s_0$,
+for an $s_0$ large enough to compensate for the growth of $f(t)$.
+
+The **inverse Laplace transform** $\hat{\mathcal{L}}{}^{-1}$ involves complex integration,
+and is therefore a lot more difficult to calculate.
+Fortunately, it is usually avoidable by rewriting a given $s$-space expression
+using [partial fraction decomposition](/know/concept/partial-fraction-decomposition/),
+and then looking up the individual terms.
+
+
+## Derivatives
+
+The derivative of a transformed function is the transform
+of the original mutliplied by its variable.
+This is especially useful for transforming ODEs with variable coefficients:
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}{}'(s) = - \hat{\mathcal{L}}\{t f(t)\}
+    }
+\end{aligned}$$
+
+This property generalizes nicely to higher-order derivatives of $s$, so:
+
+$$\begin{aligned}
+    \boxed{
+        \dvn{n}{\tilde{f}}{s} = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dv-s"/>
+<label for="proof-dv-s">Proof</label>
+<div class="hidden">
+<label for="proof-dv-s">Proof.</label>
+The exponential $\exp(- s t)$ is the only thing that depends on $s$ here:
+
+$$\begin{aligned}
+    \dvn{n}{\tilde{f}}{s}
+    &= \dvn{n}{}{s}\int_0^\infty f(t) \exp(- s t) \dd{t}
+    \\
+    &= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t}
+    = (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
+\end{aligned}$$
+</div>
+</div>
+
+The Laplace transform of a derivative introduces the initial conditions into the result.
+Notice that $f(0)$ is the initial value in the original $t$-domain:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{L}}\{ f'(t) \} = - f(0) + s \tilde{f}(s)
+    }
+\end{aligned}$$
+
+This property generalizes to higher-order derivatives,
+although it gets messy quickly.
+Once again, the initial values of the lower derivatives appear:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{L}} \big\{ f^{(n)}(t) \big\}
+        = - \sum_{j = 0}^{n - 1} s^j f^{(n - 1 - j)}(0) + s^n \tilde{f}(s)
+    }
+\end{aligned}$$
+
+Where $f^{(n)}(t)$ is shorthand for the $n$th derivative of $f(t)$,
+and $f^{(0)}(t) = f(t)$.
+As an example, $\hat{\mathcal{L}}\{f'''(t)\}$ becomes
+$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$.
+
+<div class="accordion">
+<input type="checkbox" id="proof-dv-t"/>
+<label for="proof-dv-t">Proof</label>
+<div class="hidden">
+<label for="proof-dv-t">Proof.</label>
+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.
+</div>
+</div>
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/larmor-precession/index.md b/source/know/concept/larmor-precession/index.md
new file mode 100644
index 0000000..ff80619
--- /dev/null
+++ b/source/know/concept/larmor-precession/index.md
@@ -0,0 +1,102 @@
+---
+title: "Larmor precession"
+date: 2021-07-02
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+Consider a stationary spin-1/2 particle,
+placed in a [magnetic field](/know/concept/magnetic-field/)
+with magnitude $B$ pointing in the $z$-direction.
+In that case, its Hamiltonian $\hat{H}$ is given by:
+
+$$\begin{aligned}
+    \hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z}
+\end{aligned}$$
+
+Where $\gamma = - q / m$ is the gyromagnetic ratio,
+and $\hat{\sigma}_z$ is the Pauli spin matrix for the $z$-direction.
+Since $\hat{H}$ is proportional to $\hat{\sigma}_z$,
+they share eigenstates $\Ket{\downarrow}$ and $\Ket{\uparrow}$.
+The respective eigenenergies $E_{\downarrow}$ and $E_{\uparrow}$ are as follows:
+
+$$\begin{aligned}
+    E_{\downarrow} = \frac{\hbar}{2} \gamma B
+    \qquad
+    E_{\uparrow} = - \frac{\hbar}{2} \gamma B
+\end{aligned}$$
+
+Because $\hat{H}$ is time-independent,
+the general time-dependent solution $\Ket{\chi(t)}$ is of the following form,
+where $a$ and $b$ are constants,
+and the exponentials are "twiddle factors":
+
+$$\begin{aligned}
+    \Ket{\chi(t)}
+    = a \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow}
+    \:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow}
+\end{aligned}$$
+
+For our purposes, we can safely assume that $a$ and $b$ are real,
+and then say that there exists an angle $\theta$
+satisfying $a = \sin(\theta / 2)$ and $b = \cos(\theta / 2)$, such that:
+
+$$\begin{aligned}
+    \Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow}
+    \:+\: \cos(\theta / 2) \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow}
+\end{aligned}$$
+
+Now, we find the expectation values of the spin operators
+$\expval{\hat{S}_x}$, $\expval{\hat{S}_y}$, and $\expval{\hat{S}_z}$.
+The first is:
+
+$$\begin{aligned}
+    \matrixel{\chi}{\hat{S}_x}{\chi}
+    &= \frac{\hbar}{2}
+    \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}}
+    \cdot
+    \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
+    \cdot
+    \begin{bmatrix} a \exp(- i E_{\downarrow} t / \hbar) \\ b \exp(- i E_{\uparrow} t / \hbar) \end{bmatrix}
+    \\
+    &= \frac{\hbar}{2}
+    \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}}
+    \cdot
+    \begin{bmatrix} b \exp(- i E_{\uparrow} t / \hbar) \\ a \exp(- i E_{\downarrow} t / \hbar) \end{bmatrix}
+    \\
+    &= \frac{\hbar}{2} \Big( a b \exp(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar)
+    + b a \exp(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big)
+    \\
+    &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( \exp(i \gamma B t) + \exp(- i \gamma B t) \Big)
+    \\
+    &= \frac{\hbar}{2} \cos(\gamma B t) \Big( \cos(\theta/2) \sin(\theta/2) + \cos(\theta/2) \sin(\theta/2) \Big)
+    \\
+    &= \frac{\hbar}{2} \sin(\theta) \cos(\gamma B t)
+\end{aligned}$$
+
+The other two are calculated in the same way,
+with the following results:
+
+$$\begin{aligned}
+    \matrixel{\chi}{\hat{S}_y}{\chi} = - \frac{\hbar}{2} \sin(\theta) \sin(\gamma B t)
+    \qquad
+    \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos(\theta)
+\end{aligned}$$
+
+The result is that the spin axis is off by $\theta$ from the $z$-direction,
+and is rotating (or **precessing**) around the $z$-axis at the **Larmor frequency** $\omega$:
+
+$$\begin{aligned}
+    \boxed{
+        \omega = \gamma B
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/source/know/concept/laser-rate-equations/index.md b/source/know/concept/laser-rate-equations/index.md
new file mode 100644
index 0000000..dec7e4b
--- /dev/null
+++ b/source/know/concept/laser-rate-equations/index.md
@@ -0,0 +1,324 @@
+---
+title: "Laser rate equations"
+date: 2022-03-16
+categories:
+- Physics
+- Optics
+- Laser theory
+layout: "concept"
+---
+
+The [Maxwell-Bloch equations](/know/concept/maxwell-bloch-equations/) (MBEs)
+give a fundamental description of light-matter interaction
+for a two-level quantum system for the purposes of laser theory.
+They govern the [electric field](/know/concept/electric-field/) $\vb{E}^{+}$,
+the induced polarization $\vb{P}^{+}$,
+and the total population inversion $D$:
+
+$$\begin{aligned}
+    - \mu_0 \pdvn{2}{\vb{P}^{+}}{t}
+    &= \nabla \cross \nabla \cross \vb{E}^{+} + \frac{n^2}{c^2} \pdvn{2}{\vb{E}^{+}}{t}
+    \\
+    \pdv{\vb{P}^{+}}{t}
+    &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+}
+    - \frac{i |g|^2}{\hbar} \vb{E}^{+} D
+    \\
+    \pdv{D}{t}
+    &= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
+\end{aligned}$$
+
+Where $n$ is the background medium's refractive index,
+$\omega_0$ the two-level system's gap resonance frequency,
+$|g| \equiv |\matrixel{e}{\vu{x}}{g}|$ the transition dipole moment,
+$\gamma_\perp$ and $\gamma_\parallel$ empirical decay rates,
+and $D_0$ the equilibrium inversion.
+Note that $\vb{E}^{-} = (\vb{E}^{+})^*$.
+
+Let us make the following ansatz,
+where $\vb{E}_0^{+}$ and $\vb{P}_0^{+}$ are slowly-varying envelopes
+of a plane wave with angular frequency $\omega \approx \omega_0$:
+
+$$\begin{aligned}
+    \vb{E}^{+}(\vb{r}, t)
+    = \frac{1}{2} \vb{E}_0^{+}(\vb{r}, t) \: e^{-i \omega t}
+    \qquad \qquad
+    \vb{P}^{+}(\vb{r}, t)
+    = \frac{1}{2} \vb{P}_0^{+}(\vb{r}, t) \: e^{-i \omega t}
+\end{aligned}$$
+
+We insert this into the first MBE,
+and assume that $\vb{E}_0^{+}$ and $\vb{P}_0^{+}$
+vary so slowly that their second-order derivatives are negligible,
+i.e. $\ipdvn{2}{\vb{E}_0^{+}\!}{t} \approx 0$ and $\ipdvn{2}{\vb{P}_0^{+}\!}{t} \approx 0$,
+giving:
+
+$$\begin{aligned}
+    \mu_0 \bigg( i 2 \omega \pdv{\vb{P}_0^{+}}{t} + \omega^2 \vb{P}_0^{+} \bigg)
+    = \nabla \cross \nabla \cross \vb{E}_0^{+}
+    - \frac{n^2}{c^2} \bigg( i 2 \omega \pdv{\vb{E}_0^{+}}{t} + \omega^2 \vb{E}_0^{+} \bigg)
+\end{aligned}$$
+
+To get rid of the double curl,
+consider the time-independent
+[electromagnetic wave equation](/know/concept/electromagnetic-wave-equation/),
+where $\Omega$ is an eigenfrequency of the optical cavity
+in which lasing will occur:
+
+$$\begin{aligned}
+    \nabla \cross \nabla \cross \vb{E}_0^{+}
+    = \frac{n^2}{c^2} \Omega^2 \vb{E}_0^{+}
+\end{aligned}$$
+
+For simplicity, we restrict ourselves to a single-mode laser,
+where there is only one $\Omega$ and $\vb{E}_0^{+}$ to care about.
+Substituting the above equation into the first MBE yields:
+
+$$\begin{aligned}
+    i 2 \omega \pdv{\vb{P}_0^{+}}{t} + \omega^2 \vb{P}_0^{+}
+    = \varepsilon_0 n^2 \bigg( (\Omega^2 - \omega^2) \vb{E}_0^{+} - i 2 \omega \pdv{\vb{E}_0^{+}}{t} \bigg)
+\end{aligned}$$
+
+Where we used $1 / c^2 = \mu_0 \varepsilon_0$.
+Assuming the light is more or less on-resonance $\omega \approx \Omega$,
+we can approximate $\Omega^2 \!-\! \omega^2 \approx 2 \omega (\Omega \!-\! \omega)$, so:
+
+$$\begin{aligned}
+    i 2 \pdv{\vb{P}_0^{+}}{t} + \omega \vb{P}_0^{+}
+    = \varepsilon_0 n^2 \bigg( 2 (\Omega - \omega) \vb{E}_0^{+} - i 2 \pdv{\vb{E}_0^{+}}{t} \bigg)
+\end{aligned}$$
+
+Moving on to the second MBE,
+inserting the ansatz $\vb{P}^{+} = \vb{P}_0^{+} e^{-i \omega t} / 2$ leads to:
+
+$$\begin{aligned}
+    \pdv{\vb{P}_0^{+}}{t}
+    = - \Big( \gamma_\perp + i (\omega_0 - \omega) \Big) \vb{P}_0^{+} - \frac{i |g|^2}{\hbar} \vb{E}_0^{+} D
+\end{aligned}$$
+
+Typically, $\gamma_\perp$ is much larger than the rate of any other decay process,
+in which case $\ipdv{}{\vb{P}0^{+}\!}{t}$ is negligible compared to $\gamma_\perp \vb{P}_0^{+}$.
+Effectively, this means that the polarization $\vb{P}_0^{+}$
+near-instantly follows the electric field $\vb{E}^{+}\!$.
+Setting $\ipdv{}{\vb{P}0^{+}\!}{t} \approx 0$, the second MBE becomes:
+
+$$\begin{aligned}
+    \vb{P}^{+}
+    = -\frac{i |g|^2}{\hbar (\gamma_\perp + i (\omega_0 \!-\! \omega))} \vb{E}^{+} D
+    = \frac{|g|^2 \gamma(\omega)}{\hbar \gamma_\perp} \vb{E}^{+} D
+\end{aligned}$$
+
+Where the Lorentzian gain curve $\gamma(\omega)$
+(which also appears in the [SALT equation](/know/concept/salt-equation/))
+represents a laser's preferred spectrum for amplification,
+and is defined like so:
+
+$$\begin{aligned}
+    \gamma(\omega)
+    \equiv \frac{\gamma_\perp}{(\omega - \omega_0) + i \gamma_\perp}
+\end{aligned}$$
+
+Note that $\gamma(\omega)$ satisfies the following relation,
+which will be useful to us later:
+
+$$\begin{aligned}
+    \gamma^*(\omega) - \gamma(\omega)
+    = \frac{\gamma_\perp (i \gamma_\perp + i \gamma_\perp)}{(\omega - \omega_0)^2 + \gamma_\perp^2}
+    = i 2 |\gamma(\omega)|^2
+\end{aligned}$$
+
+Returning to the first MBE with $\ipdv{\vb{P}_0^{+}}{t} \approx 0$,
+we substitute the above expression for $\vb{P}_0^{+}$:
+
+$$\begin{aligned}
+    \pdv{\vb{E}_0^{+}}{t}
+    &= i (\omega - \Omega) \vb{E}_0^{+} + i \frac{\omega}{2 \varepsilon_0 n^2} \vb{P}_0^{+}
+    \\
+    &= i (\omega - \Omega) \vb{E}_0^{+} + i \frac{|g|^2 \omega \gamma(\omega)}{2 \hbar \varepsilon_0 \gamma_\perp n^2} \vb{E}_0^{+} D
+\end{aligned}$$
+
+Next, we insert our ansatz for $\vb{E}^{+}\!$ and $\vb{P}^{+}\!$
+into the third MBE, and rewrite $\vb{P}_0^{+}$ as above.
+Using our identity for $\gamma(\omega)$,
+and the fact that $\vb{E}_0^{+} \cdot \vb{E}_0^{-} = |\vb{E}|^2$, we find:
+
+$$\begin{aligned}
+    \pdv{D}{t}
+    &= \gamma_\parallel (D_0 - D) + \frac{i}{2 \hbar}
+    \Big( \frac{|g|^2 \gamma^*(\omega)}{\hbar \gamma_\perp} \vb{E}_0^{-} D \cdot \vb{E}_0^{+}
+    - \frac{|g|^2 \gamma(\omega)}{\hbar \gamma_\perp} \vb{E}_0^{+} D \cdot \vb{E}_0^{-} \Big)
+    \\
+    &= \gamma_\parallel (D_0 - D) + \frac{i |g|^2}{2 \hbar^2 \gamma_\perp} \Big( \gamma^*(\omega) - \gamma(\omega) \Big) |\vb{E}|^2 D
+    \\
+    &= \gamma_\parallel (D_0 - D) - \frac{|g|^2}{\hbar^2 \gamma_\perp} |\gamma(\omega)|^2 |\vb{E}|^2 D
+\end{aligned}$$
+
+This is the prototype of the first laser rate equation.
+However, in order to have a practical set,
+we need an equation for $|\vb{E}|^2$,
+which we can obtain using the first MBE:
+
+$$\begin{aligned}
+    \pdv{|\vb{E}|^2}{t}
+    &= \vb{E}_0^{+} \pdv{\vb{E}_0^{-}}{t} + \vb{E}_0^{-} \pdv{\vb{E}_0^{+}}{t}
+    \\
+    &= -i (\omega - \Omega^*) |\vb{E}|^2 - i \frac{|g|^2 \omega \gamma^*(\omega)}{2 \hbar \varepsilon_0 \gamma_\perp n^2} |\vb{E}|^2 D
+    + i (\omega - \Omega) |\vb{E}|^2 + i \frac{|g|^2 \omega \gamma(\omega)}{2 \hbar \varepsilon_0 \gamma_\perp n^2} |\vb{E}|^2 D
+    \\
+    &= i (\Omega^* - \Omega) |\vb{E}|^2
+    + i \frac{|g|^2 \omega}{2 \hbar \varepsilon_0 \gamma_\perp n^2} \Big(\gamma(\omega) - \gamma^*(\omega)\Big) |\vb{E}|^2 D
+    \\
+    &= 2 \Imag(\Omega) |\vb{E}|^2 + \frac{|g|^2 \omega}{\hbar \varepsilon_0 \gamma_\perp n^2} |\gamma(\omega)|^2 |\vb{E}|^2 D
+\end{aligned}$$
+
+Where $\Imag(\Omega) < 0$ represents the fact that the laser cavity is leaky.
+We now have the **laser rate equations**,
+although they are still in an unidiomatic form:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \pdv{|\vb{E}|^2}{t}
+            &= 2 \Imag(\Omega) |\vb{E}|^2 + \frac{|g|^2 \omega}{\hbar \varepsilon_0 \gamma_\perp n^2} |\gamma(\omega)|^2 |\vb{E}|^2 D
+            \\
+            \pdv{D}{t}
+            &= \gamma_\parallel (D_0 - D) - \frac{|g|^2}{\hbar^2 \gamma_\perp} |\gamma(\omega)|^2 |\vb{E}|^2 D
+        \end{aligned}
+    }
+\end{aligned}$$
+
+To rewrite this, we replace $|\vb{E}|^2$ with the photon number $N_p$ as follows,
+with $U = \varepsilon_0 n^2 |\vb{E}|^2 / 2$ being the energy density of the light:
+
+$$\begin{aligned}
+    N_{p}
+    = \frac{U}{\hbar \omega}
+    = \frac{\varepsilon_0 n^2}{2 \hbar \omega} |\vb{E}|^2
+\end{aligned}$$
+
+Furthermore, consider the definition of the inversion $D$:
+because a photon emission annihilates an electron-hole pair,
+it reduces $D$ by $2$.
+Since lasing is only possible for $D > 0$,
+we can replace $D$ with the conduction band's electron population $N_e$,
+which is reduced by $1$ whenever a photon is emitted.
+The laser rate equations then take the following standard form:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \pdv{N_p}{t}
+            &= - \gamma_p N_p + G N_p N_e
+            \\
+            \pdv{N_e}{t}
+            &= R_\mathrm{pump} - \gamma_e N_e - G N_p N_e
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where $\gamma_e$ is a redefinition of $\gamma_\parallel$
+depending on the electron decay processes,
+and the photon loss rate $\gamma_p$, the gain $G$,
+and the carrier supply rate $R_\mathrm{pump}$
+are defined like so:
+
+$$\begin{aligned}
+    \gamma_p
+    = - 2 \Imag(\Omega)
+    = \frac{Q}{\Real(\Omega)}
+    \qquad \quad
+    G
+    \equiv \frac{|g|^2 \omega}{\hbar \varepsilon_0 \gamma_\perp n^2} |\gamma(\omega)|^2
+    \qquad \quad
+    R_\mathrm{pump}
+    \equiv \gamma_\parallel D_0
+\end{aligned}$$
+
+With $Q$ being the cavity mode's quality factor.
+The nonlinear coupling term $G N_p N_e$ represents
+[stimulated emission](/know/concept/einstein-coefficients/),
+which is the key to lasing.
+
+To understand the behaviour of a laser,
+consider these equations in a steady state,
+i.e. where $N_p$ and $N_e$ are constant in $t$:
+
+$$\begin{aligned}
+    0
+    &= - \gamma_p N_p + G N_p N_e
+    \\
+    0
+    &= R_\mathrm{pump} - \gamma_e N_e - G N_p N_e
+\end{aligned}$$
+
+In addition to the trivial solution $N_p = 0$,
+we can also have $N_p > 0$.
+Isolating $N_p$'s equation for $N_e$ and inserting that into $N_e$'s equation, we find:
+
+$$\begin{aligned}
+     N_e
+    = \frac{\gamma_p}{G}
+    \qquad \implies \qquad
+    \boxed{
+        N_p
+        = \frac{1}{\gamma_p} \bigg( R_\mathrm{pump} - \frac{\gamma_e \gamma_p}{G} \bigg)
+    }
+\end{aligned}$$
+
+The quantity $R_\mathrm{thr} \equiv \gamma_e \gamma_p / G$ is called the **lasing threshold**:
+if $R_\mathrm{pump} \ge R_\mathrm{thr}$, the laser is active,
+meaning that $N_p$ is big enough to cause
+a "chain reaction" of stimulated emission
+that consumes all surplus carriers to maintain a steady state.
+
+The point is that $N_e$ is independent of the electron supply $R_\mathrm{pump}$,
+because all additional electrons are almost immediately
+annihilated by stimulated emission.
+Consequently $N_p$ increases linearly as $R_\mathrm{pump}$ is raised,
+at a much steeper slope than would be possible below threshold.
+The output of the cavity is proportional to $N_p$,
+so the brightness is also linear.
+
+Unfortunately, by deriving the laser rate equations from the MBEs,
+we lost some interesting and important effects,
+most notably spontaneous emission,
+which is needed for $N_p$ to grow if $R_\mathrm{pump}$ is below threshold.
+
+For this reason, the laser rate equations are typically presented
+in a more empirical form, which "bookkeeps" the processes affecting $N_p$ and $N_e$.
+Consider the following example:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \pdv{N_p}{t}
+            &= - (\gamma_\mathrm{out} + \gamma_\mathrm{abs} + \gamma_\mathrm{loss}) N_p + \gamma_\mathrm{spon} N_e + G_\mathrm{stim} N_p N_e
+            \\
+            \pdv{N_e}{t}
+            &= R_\mathrm{pump} + \gamma_\mathrm{abs} N_p
+            - (\gamma_\mathrm{spon} + \gamma_\mathrm{n.r.} + \gamma_\mathrm{leak}) N_e - G_\mathrm{stim} N_p N_e
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where $\gamma_\mathrm{out}$ represents the cavity's usable output,
+$\gamma_\mathrm{abs}$ the medium's absorption,
+$\gamma_\mathrm{loss}$ scattering losses,
+$\gamma_\mathrm{spon}$ spontaneous emission,
+$\gamma_\mathrm{n.r.}$ non-radiative electron-hole recombination,
+and $\gamma_\mathrm{leak}$ the fact that
+some carriers leak away before they can be used for emission.
+
+Unsurprisingly, this form is much harder to analyze,
+but more accurately describes the dynamics inside a laser.
+To make matters even worse, many of these decay rates depend on $N_p$ or $N_e$,
+so solutions can only be obtained numerically.
+
+
+
+## References
+1.  D. Meschede,
+    *Optics, light and lasers*,
+    Wiley.
+2.  L.A. Coldren, S.W. Corzine, M.L. Mašanović,
+    *Diode lasers and photonic integrated circuits*, 2nd edition,
+    Wiley.
diff --git a/source/know/concept/laws-of-thermodynamics/index.md b/source/know/concept/laws-of-thermodynamics/index.md
new file mode 100644
index 0000000..2d7af9e
--- /dev/null
+++ b/source/know/concept/laws-of-thermodynamics/index.md
@@ -0,0 +1,103 @@
+---
+title: "Laws of thermodynamics"
+date: 2021-07-07
+categories:
+- Physics
+- Thermodynamics
+layout: "concept"
+---
+
+The **laws of thermodynamics** are of great importance
+to physics, chemistry and engineering,
+since they restrict what a device or process can physically achieve.
+For example, the impossibility of *perpetual motion*
+is a consequence of these laws.
+
+
+## First law
+
+The **first law of thermodynamics** states that energy is conserved.
+When a system goes from one equilibrium to another,
+the change $\Delta U$ of its energy $U$ is equal to
+the work $\Delta W$ done by external forces,
+plus the energy transferred by heating ($\Delta Q > 0$) or cooling ($\Delta Q < 0$):
+
+$$\begin{aligned}
+    \boxed{
+        \Delta U = \Delta W + \Delta Q
+    }
+\end{aligned}$$
+
+The internal energy $U$ is a state variable,
+so is independent of the path taken between equilibria.
+However, the work $\Delta W$ and heating $\Delta Q$ do depend on the path,
+so the first law means that
+the act of transferring energy is path-dependent,
+but the result has no "memory" of that path.
+
+
+## Second law
+
+The **second law of thermodynamics** states that
+the total entropy never decreases.
+An important consequence is that
+no machine can convert energy into work with 100% efficiency.
+
+It is possible for the local entropy $S_{\mathrm{loc}}$
+of a system to decrease, but doing so requires work,
+and therefore the entropy of the surroundings $S_{\mathrm{sur}}$
+must increase accordingly, such that:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta S_{\mathrm{tot}} = \Delta S_{\mathrm{loc}} + \Delta S_{\mathrm{sur}} \ge 0
+    }
+\end{aligned}$$
+
+Since the total entropy never decreases,
+the equilibrium state of a system must be a maximum
+of its entropy $S$, and therefore $S$ can be used as
+a [thermodynamic "potential"](/know/concept/thermodynamic-potential/).
+
+The only situation where $\Delta S = 0$ is a reversible process,
+since then it must be possible to return to
+the previous equilibrium state by doing the same work in the opposite direction.
+
+According to the first law,
+if a process is reversible, or if it is only heating/cooling,
+then (after one reversible cycle) the energy change
+is simply the heat transfer $\dd{U} = \dd{Q}$.
+An entropy change $\dd{S}$ is then expressed as follows
+(since $\ipdv{S}{U} = 1 / T$ by definition):
+
+$$\begin{aligned}
+    \boxed{
+        \dd{S}
+        = \Big( \pdv{S}{U} \Big)_{V, N} \dd{U}
+        = \frac{\dd{Q}}{T}
+    }
+\end{aligned}$$
+
+Confusingly, this equation is sometimes also called the second law of thermodynamics.
+
+
+## Third law
+
+The **third law of thermodynamics** states that
+the entropy $S$ of a system goes to zero when the temperature reaches absolute zero:
+
+$$\begin{aligned}
+    \boxed{
+        \lim_{T \to 0} S = 0
+    }
+\end{aligned}$$
+
+From this, the absolute quantity of $S$ is defined, otherwise we would
+only be able to speak of entropy differences $\Delta S$.
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/lawson-criterion/index.md b/source/know/concept/lawson-criterion/index.md
new file mode 100644
index 0000000..c1848dc
--- /dev/null
+++ b/source/know/concept/lawson-criterion/index.md
@@ -0,0 +1,127 @@
+---
+title: "Lawson criterion"
+date: 2021-10-06
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+For sustained nuclear fusion to be possible,
+the **Lawson criterion** must be met,
+from which some required properties
+of the plasma and the reactor chamber can be deduced.
+
+Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion,
+but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way.
+If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction,
+then the following inequality must be satisfied
+in order to have harvestable energy:
+
+$$\begin{aligned}
+    P_\mathrm{loss}
+    \le P_\mathrm{fus} + P_\mathrm{aux}
+\end{aligned}$$
+
+We can rewrite $P_\mathrm{aux}$ using the definition
+of the **energy gain factor** $Q$,
+which is the ratio of the output and input powers of the fusion reaction:
+
+$$\begin{aligned}
+    Q
+    \equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}}
+    \quad \implies \quad
+    P_\mathrm{aux}
+    = \frac{P_\mathrm{fus}}{Q}
+\end{aligned}$$
+
+Returning to the inequality, we can thus rearrange its right-hand side as follows:
+
+$$\begin{aligned}
+    P_\mathrm{loss}
+    \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q}
+    = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big)
+    = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
+\end{aligned}$$
+
+We assume that the plasma has equal species densities $n_i = n_e$,
+so its total density $n = 2 n_i$.
+Then $P_\mathrm{fus}$ is as follows,
+where $f_{ii}$ is the frequency
+with which a given ion collides with other ions,
+and $E_\mathrm{fus}$ is the energy released by a single fusion reaction:
+
+$$\begin{aligned}
+    P_\mathrm{fus}
+    = f_{ii} n_i E_\mathrm{fus}
+    = \big( n_i \Expval{\sigma v} \big) n_i E_\mathrm{fus}
+    = \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus}
+\end{aligned}$$
+
+Where $\Expval{\sigma v}$ is the mean product
+of the velocity $v$ and the collision cross-section $\sigma$.
+
+Furthermore, assuming that both species have the same temperature $T_i = T_e = T$,
+the total energy density $W$ of the plasma is given by:
+
+$$\begin{aligned}
+    W
+    = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e
+    = 3 k_B T n
+\end{aligned}$$
+
+Where $k_B$ is Boltzmann's constant.
+From this, we can define the **confinement time** $\tau_E$
+as the characteristic lifetime of energy in the reactor, before leakage.
+Therefore:
+
+$$\begin{aligned}
+    \tau_E
+    \equiv \frac{W}{P_\mathrm{loss}}
+    \quad \implies \quad
+    P_\mathrm{loss}
+    = \frac{3 n k_B T}{\tau_E}
+\end{aligned}$$
+
+Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$
+into the inequality, we arrive at:
+
+$$\begin{aligned}
+    \frac{3 n k_B T}{\tau_E}
+    \le \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
+\end{aligned}$$
+
+This can be rearranged to the form below,
+which is the original Lawson criterion:
+
+$$\begin{aligned}
+    n \tau_E
+    \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\Expval{\sigma v} E_\mathrm{fus}}
+\end{aligned}$$
+
+However, it turns out that the highest fusion power density
+is reached when $T$ is at the minimum of $T^2 / \Expval{\sigma v}$.
+Therefore, we multiply by $T$ to get the Lawson triple product:
+
+$$\begin{aligned}
+    \boxed{
+        n T \tau_E
+        \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}}
+    }
+\end{aligned}$$
+
+For some reason,
+it is often assumed that the fusion is infinitely profitable $Q \to \infty$,
+in which case the criterion reduces to:
+
+$$\begin{aligned}
+    n T \tau_E
+    \ge \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}}
+\end{aligned}$$
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/source/know/concept/legendre-polynomials/index.md b/source/know/concept/legendre-polynomials/index.md
new file mode 100644
index 0000000..338b23f
--- /dev/null
+++ b/source/know/concept/legendre-polynomials/index.md
@@ -0,0 +1,119 @@
+---
+title: "Legendre polynomials"
+date: 2021-09-08
+categories:
+- Mathematics
+layout: "concept"
+---
+
+The **Legendre polynomials** are a set of functions that sometimes arise in physics.
+They are the eigenfunctions $u(x)$ of **Legendre's differential equation**,
+which is a ([Sturm-Liouville](/know/concept/sturm-liouville-theory/))
+eigenvalue problem for $\ell (\ell + 1)$,
+where $\ell$ turns out to be a non-negative integer:
+
+$$\begin{aligned}
+    \boxed{
+        (1 - x^2) u'' - 2 x u' + \ell (\ell + 1) u = 0
+    }
+\end{aligned}$$
+
+The $\ell$th-degree Legendre polynomial $P_\ell(x)$
+is given in the form of a *Rodrigues' formula* by:
+
+$$\begin{aligned}
+    P_\ell(x)
+    &= \frac{1}{2^\ell \ell!} \dvn{\ell}{}{x}(x^2 - 1)^\ell
+\end{aligned}$$
+
+The first handful of Legendre polynomials $P_\ell(x)$ are therefore as follows:
+
+$$\begin{gathered}
+    P_0(x) = 1
+    \qquad \quad
+    P_1(x) = x
+    \qquad \quad
+    P_2(x) = \frac{1}{2} (3 x^2 - 1)
+    \\
+    P_3(x) = \frac{1}{2} (5 x^3 - 3 x)
+    \qquad \quad
+    P_4(x) = \frac{1}{8} (35 x^4 - 30 x^2 + 3)
+\end{gathered}$$
+
+And then more $P_\ell$ can be computed quickly
+using **Bonnet's recursion formula**:
+
+$$\begin{aligned}
+    \boxed{
+        (\ell + 1) P_{\ell + 1}(x) = (2 \ell + 1) x P_\ell(x) - \ell P_{\ell - 1}(x)
+    }
+\end{aligned}$$
+
+The derivative of a given $P_\ell$ can be calculated recursively
+using the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{}{x}P_{\ell + 1}
+        = (\ell + 1) P_\ell(x) + x \dv{}{x}P_\ell(x)
+    }
+\end{aligned}$$
+
+Noteworthy is that the Legendre polynomials
+are mutually orthogonal for $x \in [-1, 1]$:
+
+$$\begin{aligned}
+    \boxed{
+        \Inprod{P_m}{P_n}
+        = \int_{-1}^{1} P_m(x) \: P_n(x) \dd{x}
+        = \frac{2}{2 n + 1} \delta_{nm}
+    }
+\end{aligned}$$
+
+As was to be expected from Sturm-Liouville theory.
+Likewise, they form a complete basis in the
+[Hilbert space](/know/concept/hilbert-space/)
+of piecewise continuous functions $f(x)$ on $x \in [-1, 1]$,
+meaning:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \sum_{\ell = 0}^\infty a_\ell P_\ell(x)
+        = \sum_{\ell = 0}^\infty \frac{\Inprod{P_\ell}{f}}{\Inprod{P_\ell}{P_\ell}} P_\ell(x)
+    }
+\end{aligned}$$
+
+Each Legendre polynomial $P_\ell$ comes with
+a set of **associated Legendre polynomials** $P_\ell^m(x)$
+of order $m$ and degree $\ell$.
+These are the non-singular solutions of the **general Legendre equation**,
+where $m$ and $\ell$ are integers satisfying $-\ell \le m \le \ell$:
+
+$$\begin{aligned}
+    \boxed{
+        (1 - x^2) u'' - 2 x u' + \Big( \ell (\ell + 1) - \frac{m^2}{1 - x^2} \Big) u = 0
+    }
+\end{aligned}$$
+
+The $\ell$th-degree $m$th-order associated Legendre polynomial $P_\ell^m$
+is as follows for $m \ge 0$:
+
+$$\begin{aligned}
+    P_\ell^m(x)
+    = (-1)^m (1 - x^2)^{m/2} \dvn{m}{}{x}P_\ell(x)
+\end{aligned}$$
+
+Here, the $(-1)^m$ in front is called the **Condon-Shortley phase**,
+and is omitted by some authors.
+For negative orders $m$,
+an additional constant factor is necessary:
+
+$$\begin{aligned}
+    P_\ell^{-m}(x) = (-1)^m \frac{(\ell - m)!}{(\ell + m)!} P_\ell^m(x)
+\end{aligned}$$
+
+Beware, the name is misleading:
+if $m$ is odd, then $P_\ell^m$ is actually not a polynomial.
+Moreover, not all $P_\ell^m$ are mutually orthogonal
+(but some are).
diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
new file mode 100644
index 0000000..fb46c9d
--- /dev/null
+++ b/source/know/concept/legendre-transform/index.md
@@ -0,0 +1,91 @@
+---
+title: "Legendre transform"
+date: 2021-02-22
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+The **Legendre transform** of a function $f(x)$ is a new function $L(f')$,
+which depends only on the derivative $f'(x)$ of $f(x)$, and from which
+the original function $f(x)$ can be reconstructed. The point is,
+analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)),
+that $L(f')$ contains the same information as $f(x)$, just in a different form.
+
+Let us choose an arbitrary point $x_0 \in [a, b]$ in the domain of
+$f(x)$. Consider a line $y(x)$ tangent to $f(x)$ at $x = x_0$, which has
+a slope $f'(x_0)$ and intersects the $y$-axis at $-C$:
+
+$$\begin{aligned}
+    y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
+\end{aligned}$$
+
+The Legendre transform $L(f')$ is defined such that $L(f'(x_0)) = C$
+(or sometimes $-C$) for all $x_0 \in [a, b]$,
+where $C$ corresponds to the tangent line at $x = x_0$. This yields:
+
+$$\begin{aligned}
+    L(f'(x)) = f'(x) \: x - f(x)
+\end{aligned}$$
+
+We want this function to depend only on the derivative $f'$, but
+currently $x$ still appears here as a variable. We fix that problem in
+the easiest possible way: by assuming that $f'(x)$ is invertible for all
+$x \in [a, b]$. If $x(f')$ is the inverse of $f'(x)$, then $L(f')$ is
+given by:
+
+$$\begin{aligned}
+    \boxed{
+        L(f') = f' \: x(f') - f(x(f'))
+    }
+\end{aligned}$$
+
+The only requirement for the existence of the Legendre transform is thus
+the invertibility of $f'(x)$ in the target interval $[a,b]$, which can
+only be true if $f(x)$ is either convex or concave, i.e. its derivative
+$f'(x)$ is monotonic.
+
+Crucially, the derivative of $L(f')$ with respect to $f'$ is simply
+$x(f')$. In other words, the roles of $f'$ and $x$ are switched by the
+transformation: the coordinate becomes the derivative and vice versa.
+This is demonstrated here:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f')
+    }
+\end{aligned}$$
+
+Furthermore, Legendre transformation is an *involution*, meaning it is
+its own inverse. Let $g(L')$ be the Legendre transform of $L(f')$:
+
+$$\begin{aligned}
+    g(L') = L' \: f'(L') - L(f'(L'))
+    = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x)
+\end{aligned}$$
+
+Moreover, the inverse of a (forward) transform always exists, because
+the Legendre transform of a convex function is itself convex. Convexity
+of $f(x)$ means that $f''(x) > 0$ for all $x \in [a, b]$, which yields
+the following proof:
+
+$$\begin{aligned}
+    L''(f')
+    = \dv{x(f')}{f'}
+    = \dv{x}{f'(x)}
+    = \frac{1}{f''(x)}
+    > 0
+\end{aligned}$$
+
+Legendre transformation is important in physics,
+since it connects [Lagrangian](/know/concept/lagrangian-mechanics/)
+and [Hamiltonian](/know/concept/hamiltonian-mechanics/) mechanics to each other.
+It is also used to convert between [thermodynamic potentials](/know/concept/thermodynamic-potential/).
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/lehmann-representation/index.md b/source/know/concept/lehmann-representation/index.md
new file mode 100644
index 0000000..dd8c112
--- /dev/null
+++ b/source/know/concept/lehmann-representation/index.md
@@ -0,0 +1,228 @@
+---
+title: "Lehmann representation"
+date: 2021-11-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In many-body quantum theory, the **Lehmann representation**
+is an alternative way to write the [Green's functions](/know/concept/greens-functions/),
+obtained by expanding in the many-particle eigenstates
+under the assumption of a time-independent Hamiltonian $\hat{H}$.
+
+First, we write out the greater Green's function $G_{\nu \nu'}^>(t, t')$,
+and then expand its expected value $\Expval{}$ (at thermodynamic equilibrium)
+into a sum of many-particle basis states $\Ket{n}$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^>(t, t')
+    = - \frac{i}{\hbar} \Expval{\hat{c}_\nu(t) \hat{c}_{\nu'}^\dagger(t')}
+    &= - \frac{i}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{c}_\nu(t) \hat{c}_{\nu'}^\dagger(t') e^{-\beta \hat{H}}}{n}
+\end{aligned}$$
+
+Where $\beta = 1 / (k_B T)$, and $Z$ is the grand partition function
+(see [grand canonical ensemble](/know/concept/grand-canonical-ensemble/));
+the operator $e^{\beta \hat{H}}$ gives the weight of each term at equilibrium.
+Since $\Ket{n}$ is an eigenstate of $\hat{H}$ with energy $E_n$,
+this gives us a factor of $e^{\beta E_n}$.
+Furthermore, we are in the [Heisenberg picture](/know/concept/heisenberg-picture/),
+so we write out the time-dependence of $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^>(t, t')
+    &= - \frac{i}{\hbar Z} \sum_{n} e^{-\beta E_n} \Matrixel{n}{e^{i \hat{H} t / \hbar} \hat{c}_\nu e^{- i \hat{H} t / \hbar}
+    e^{i \hat{H} t' / \hbar} \hat{c}_{\nu'}^\dagger e^{- i \hat{H} t' / \hbar}}{n}
+    \\
+    &= - \frac{i}{\hbar Z} \sum_{n} e^{-\beta E_n}
+    \Matrixel{n}{e^{i \hat{H} (t - t') / \hbar} \hat{c}_\nu e^{- i \hat{H} (t - t') / \hbar} \hat{c}_{\nu'}^\dagger}{n}
+\end{aligned}$$
+
+Where we used that the trace $\Tr\!(x) = \sum_{n} \matrixel{n}{x}{n}$
+is invariant under cyclic permutations of $x$.
+The $\Ket{n}$ form a basis of eigenstates of $\hat{H}$,
+so we insert an identity operator $\sum_{n'} \Ket{n'} \Bra{n'}$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^>(t - t')
+    &= - \frac{i}{\hbar Z} \sum_{n n'} e^{- \beta E_n}
+    \Matrixel{n}{e^{i \hat{H} (t - t') / \hbar} \hat{c}_\nu e^{- i \hat{H} (t - t') / \hbar}}{n'} \Matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
+    \\
+    &= - \frac{i}{\hbar Z} \sum_{n n'} e^{-\beta E_n}
+    \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} e^{i (E_n - E_{n'}) (t - t') / \hbar}
+\end{aligned}$$
+
+Note that $G_{\nu \nu'}^>$ now only depends on the time difference $t - t'$,
+because $\hat{H}$ is time-independent.
+Next, we take the [Fourier transform](/know/concept/fourier-transform/)
+$t \to \omega$ (with $t' = 0$):
+
+$$\begin{aligned}
+    G_{\nu \nu'}^>(\omega)
+    &= - \frac{i}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
+    \int_{-\infty}^\infty e^{i (E_n - E_{n'}) t / \hbar} \: e^{i \omega t} \dd{t}
+\end{aligned}$$
+
+Here, we recognize the integral
+as a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$,
+thereby introducing a factor of $2 \pi$,
+and arriving at the Lehmann representation of $G_{\nu \nu'}^>$:
+
+$$\begin{aligned}
+    \boxed{
+        G_{\nu \nu'}^>(\omega)
+        = - \frac{2 \pi i}{Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
+        \: \delta(E_n - E_{n'} + \hbar \omega)
+    }
+\end{aligned}$$
+
+We now go through the same process for the lesser Green's function $G_{\nu \nu'}^<(t, t')$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^<(t - t')
+    &= \mp \frac{i}{\hbar Z} \sum_{n} \matrixel{n}{\hat{c}_{\nu'}^\dagger(t') \hat{c}_\nu(t) e^{-\beta \hat{H}}}{n}
+    \\
+    &= \mp \frac{i}{\hbar Z} e^{-\beta E_n} \sum_{n n'} \matrixel{n}{\hat{c}_{\nu'}^\dagger}{n'} \matrixel{n'}{\hat{c}_\nu}{n}
+   e^{i (E_{n'} - E_n) (t - t') / \hbar}
+\end{aligned}$$
+
+Where $-$ is for bosons, and $+$ for fermions.
+Fourier transforming yields the following:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^<(\omega)
+    &= \mp \frac{2 \pi i}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_{\nu'}^\dagger}{n'} \matrixel{n'}{\hat{c}_\nu}{n}
+    \: \delta(E_{n'} - E_n + \hbar \omega)
+\end{aligned}$$
+
+We swap $n$ and $n'$, leading to the following
+Lehmann representation of $G_{\nu \nu'}^<$:
+
+$$\begin{aligned}
+    \boxed{
+        G_{\nu \nu'}^<(\omega)
+        = \mp \frac{2 \pi i}{Z} \sum_{n n'} e^{-\beta E_{n'}} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
+        \: \delta(E_n - E_{n'} + \hbar \omega)
+    }
+\end{aligned}$$
+
+Due to the delta function $\delta$,
+each term is only nonzero for $E_n' = E_n + \hbar \omega$,
+so we write:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^<(\omega)
+    = \mp \frac{2 \pi i}{\hbar Z} \sum_{n n'} e^{-\beta (E_n + \hbar \omega)}
+    \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \: \delta(E_n - E_{n'} + \hbar \omega)
+\end{aligned}$$
+
+Therefore, we arrive at the following useful relation
+between $G_{\nu \nu'}^<$ and $G_{\nu \nu'}^>$:
+
+$$\begin{aligned}
+    \boxed{
+        G_{\nu \nu'}^<(\omega)
+        = \pm e^{-\beta \hbar \omega} G_{\nu \nu'}^>(\omega)
+    }
+\end{aligned}$$
+
+Moving on, let us do the same for
+the retarded Green's function $G_{\nu \nu'}^R(t, t')$, given by:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^R(t \!-\! t')
+    &= \Theta(t \!-\! t') \Big( G_{\nu \nu'}^>(t - t') - G_{\nu \nu'}^<(t - t') \Big)
+    \\
+    &= - \frac{i}{\hbar Z} \Theta(t \!-\! t') \sum_{n n'}
+    \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
+     \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
+\end{aligned}$$
+
+We take the Fourier transform, but to ensure convergence,
+we must introduce an infinitesimal positive $\eta \to 0^+$ to the exponent
+(and eventually take the limit):
+
+$$\begin{aligned}
+    G_{\nu \nu'}^R(\omega)
+    &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^\infty \Theta(t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t}
+    \\
+    &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_0^\infty e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t}
+    \\
+    &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big)
+    \bigg[ \frac{\hbar e^{i (\hbar \omega + E_n - E_{n'}) t / \hbar} e^{- \eta t}}{i (\hbar \omega + E_n - E_{n'}) - \hbar \eta} \bigg]_0^\infty
+\end{aligned}$$
+
+Leading us to the following Lehmann representation
+of the retarded Green's function $G_{\nu \nu'}^R$:
+
+$$\begin{aligned}
+    \boxed{
+        G_{\nu \nu'}^R(\omega)
+        = \frac{1}{Z} \sum_{n n'}
+        \frac{\matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}}{\hbar (\omega + i \eta) + E_n - E_{n'}}
+        \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    }
+\end{aligned}$$
+
+Finally, we go through the same steps for the advanced Green's function $G_{\nu \nu'}^A(t, t')$:
+
+$$\begin{aligned}
+    G_{\nu \nu'}^A(t \!-\! t')
+    &= \Theta(t' \!-\! t) \Big( G_{\nu \nu'}^<(t - t') - G_{\nu \nu'}^>(t - t') \Big)
+    \\
+    &= \frac{i}{\hbar Z} \Theta(t' \!-\! t) \sum_{n n'}
+    \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}
+     \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar}
+\end{aligned}$$
+
+For the Fourier transform, we must again introduce $\eta \to 0^+$
+(although note the sign):
+
+$$\begin{aligned}
+    G_{\nu \nu'}^A(\omega)
+    &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^\infty \Theta(-t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t}
+    \\
+    &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^0 e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t}
+    \\
+    &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big)
+    \bigg[ \frac{\hbar e^{i (\hbar \omega + E_n - E_{n'}) t / \hbar} e^{\eta t}}{i (\hbar \omega + E_n - E_{n'}) + \hbar \eta} \bigg]_{-\infty}^0
+\end{aligned}$$
+
+Therefore, the Lehmann representation of
+the advanced Green's function $G_{\nu \nu'}^A$ is as follows:
+
+$$\begin{aligned}
+    \boxed{
+        G_{\nu \nu'}^A(\omega)
+        = \frac{1}{Z} \sum_{n n'}
+        \frac{\matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}}{\hbar (\omega - i \eta) + E_n - E_{n'}}
+        \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    }
+\end{aligned}$$
+
+As a final note, let us take the complex conjugate of this expression:
+
+$$\begin{aligned}
+    \big( G_{\nu \nu'}^A(\omega) \big)^*
+    = \frac{1}{Z} \sum_{n n'}
+    \frac{\matrixel{n}{\hat{c}_{\nu'}}{n'} \matrixel{n'}{\hat{c}_\nu^\dagger}{n}}{\hbar (\omega + i \eta) + E_n - E_{n'}}
+    \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+\end{aligned}$$
+
+Note the subscripts $\nu$ and $\nu'$.
+Comparing this to $G_{\nu \nu'}^R$ gives us another useful relation:
+
+$$\begin{aligned}
+    \boxed{
+        G^R_{\nu \nu'}(\omega)
+        = \big( G^A_{\nu' \nu}(\omega) \big)^*
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
new file mode 100644
index 0000000..e3df901
--- /dev/null
+++ b/source/know/concept/lindhard-function/index.md
@@ -0,0 +1,400 @@
+---
+title: "Lindhard function"
+date: 2022-01-24 # Originally 2021-10-12, major rewrite
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+The **Lindhard function** describes the response of
+[jellium](/know/concept/jellium) (i.e. a free electron gas)
+to an external perturbation, and is a quantum-mechanical
+alternative to the [Drude model](/know/concept/drude-model/).
+
+We start from the [Kubo formula](/know/concept/kubo-formula/)
+for the electron density operator $\hat{n}$,
+which describes the change in $\Expval{\hat{n}}$
+due to a time-dependent perturbation $\hat{H}_1$:
+
+$$\begin{aligned}
+    \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
+    = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
+\end{aligned}$$
+
+Where the subscript $I$ refers to the [interaction picture](/know/concept/interaction-picture/),
+and the expectation $\Expval{}_0$ is for
+a thermal equilibrium before the perturbation was applied.
+Now consider a harmonic $\hat{H}_1$:
+
+$$\begin{aligned}
+    \hat{H}_{1,S}(t)
+    = e^{i (\omega + i \eta) t} \int_{-\infty}^\infty U(\vb{r}) \: \hat{n}_S(\vb{r}) \dd{\vb{r}}
+\end{aligned}$$
+
+Where $S$ is the Schrödinger picture,
+$\eta$ is a positive infinitesimal to ensure convergence later,
+and $U(\vb{r})$ is an arbitrary potential function.
+The Kubo formula becomes:
+
+$$\begin{aligned}
+    \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
+    = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
+\end{aligned}$$
+
+Here, $\chi$ is the density-density correlation function,
+i.e. a two-particle [Green's function](/know/concept/greens-functions/):
+
+$$\begin{aligned}
+    \chi(\vb{r}, \vb{r}'; t, t')
+    \equiv - \frac{i}{\hbar} \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{n}_I(\vb{r}', t')}}_0
+\end{aligned}$$
+
+Let us assume that the unperturbed system (i.e. without $U$) is spatially uniform,
+so that $\chi$ only depends on the difference $\vb{r} - \vb{r}'$.
+We then take its [Fourier transform](/know/concept/fourier-transform/)
+$\vb{r}\!-\!\vb{r}' \to \vb{q}$:
+
+$$\begin{aligned}
+    \chi(\vb{q}; t, t')
+    &= \int_{-\infty}^\infty \chi(\vb{r} - \vb{r}'; t, t') \: e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{r}}
+    \\
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint
+    \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i \vb{q}_1 \cdot \vb{r}} e^{i \vb{q}_2 \cdot \vb{r}'} e^{- i \vb{q} \cdot (\vb{r} - \vb{r}')} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
+\end{aligned}$$
+
+Where both $\hat{n}_I$ have been written as inverse Fourier transforms,
+giving a factor $(2 \pi)^{-2 D}$, with $D$ being the number of spatial dimensions.
+We rearrange to get a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$:
+
+$$\begin{aligned}
+    \chi(\vb{q}; t, t')
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^{2D}} \iiint
+    \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i (\vb{q}_1 - \vb{q}) \cdot \vb{r}} e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2} \dd{\vb{r}}
+    \\
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \iint
+    \Expval{\Comm{\hat{n}_I(\vb{q}_1, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: \delta(\vb{q}_1 \!-\! \vb{q}) \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_1} \dd{\vb{q}_2}
+    \\
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \int
+    \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2}
+\end{aligned}$$
+
+On the left, $\vb{r}'$ does not appear, so it must also disappear on the right.
+If we choose an arbitrary (hyper)cube of volume $V$ in real space,
+then clearly $\int_V \dd{\vb{r}'} = V$. Therefore:
+
+$$\begin{aligned}
+    \chi(\vb{q}; t, t')
+    &= -\frac{i}{\hbar} \frac{\Theta(t \!-\! t')}{(2 \pi)^D} \frac{1}{V} \int_V \int_{-\infty}^\infty
+    \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0
+    \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
+\end{aligned}$$
+
+For $V \to \infty$ we get a Dirac delta function,
+but in fact the conclusion holds for finite $V$ too:
+
+$$\begin{aligned}
+    \chi(\vb{q}; t, t')
+    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \int_{-\infty}^\infty
+    \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(\vb{q}_2, t')}}_0 \: \delta(\vb{q}_2 \!+\! \vb{q}) \dd{\vb{q}_2}
+    \\
+    &= -\frac{i}{\hbar} \Theta(t \!-\! t') \frac{1}{V} \Expval{\Comm{\hat{n}_I(\vb{q}, t)}{\hat{n}_I(-\vb{q}, t')}}_0
+\end{aligned}$$
+
+Similarly, if the unperturbed Hamiltonian $\hat{H}_0$ is time-independent,
+$\chi$ only depends on the time difference $t - t'$.
+Note that $\delta{\Expval{\hat{n}}}$ already has the form of a Fourier transform,
+which gives us an opportunity to rewrite $\chi$
+in the [Lehmann representation](/know/concept/lehmann-representation/):
+
+$$\begin{aligned}
+    \chi(\vb{q}, \omega)
+    = \frac{1}{Z V} \sum_{\nu \nu'}
+    \frac{\matrixel{\nu}{\hat{n}_S(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}_S(-\vb{q})}{\nu}}{\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
+    \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big)
+\end{aligned}$$
+
+Where $\Ket{\nu}$ and $\Ket{\nu'}$ are many-electron eigenstates of $\hat{H}_0$,
+and $Z$ is the [grand partition function](/know/concept/grand-canonical-ensemble/).
+According to the [convolution theorem](/know/concept/convolution-theorem/)
+$\delta{\Expval{\hat{n}}}(\vb{q}, \omega) = \chi(\vb{q}, \omega) \: U(\vb{q})$.
+In anticipation, we swap $\nu$ and $\nu''$ in the second term,
+so the general response function is written as:
+
+$$\begin{aligned}
+    \chi(\vb{q}, \omega)
+    = \frac{1}{Z V} \sum_{\nu \nu'} \bigg(
+    \frac{\matrixel{\nu}{\hat{n}(\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(-\vb{q})}{\nu}}
+    {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
+    - \frac{\matrixel{\nu}{\hat{n}(-\vb{q})}{\nu'} \matrixel{\nu'}{\hat{n}(\vb{q})}{\nu}}
+    {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
+\end{aligned}$$
+
+All operators are in the Schrödinger picture from now on, hence we dropped the subscript $S$.
+
+To proceed, we need to rewrite $\hat{n}(\vb{q})$ somehow.
+If we neglect electron-electron interactions,
+the single-particle states are simply plane waves, in which case:
+
+$$\begin{aligned}
+    \hat{n}(\vb{q})
+    = \sum_{\sigma \vb{k}} \hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k} + \vb{q}}
+    \qquad \qquad
+    \hat{n}(-\vb{q})
+    = \hat{n}^\dagger(\vb{q})
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-density"/>
+<label for="proof-density">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-density">Proof.</label>
+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.
+</div>
+</div>
+
+When neglecting interactions, it is tradition to rename $\chi$ to $\chi_0$.
+We insert $\hat{n}$, suppressing spin:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{1}{Z V} \sum_{\vb{k} \vb{k}'} \sum_{\nu \nu'} \bigg(
+    \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}
+    \matrixel{\nu'}{\hat{c}_{\vb{k}' + \vb{q}}^\dagger \hat{c}_{\vb{k}'}}{\nu}}
+    {\hbar (\omega + i \eta) + E_\nu - E_{\nu'}}
+    - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'}
+    \matrixel{\nu'}{\hat{c}_{\vb{k}'}^\dagger \hat{c}_{\vb{k}' + \vb{q}}}{\nu}}
+    {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
+\end{aligned}$$
+
+Here, $\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}$
+is only nonzero if $\Ket{\nu'}$ is contructed from $\Ket{\nu}$
+by moving an electron from $\vb{k}$ to $\vb{k} \!+\! \vb{q}$,
+and analogously for the other inner products.
+As a result, $\vb{k} = \vb{k}'$ (and $\sigma = \sigma'$).
+
+For the same reason, the energy difference $E_\nu \!-\! E_{\nu'}$
+can simply be replaced by the cost of the single-particle excitation
+$\xi_{\vb{k}} \!-\! \xi_{\vb{k} + \vb{q}}$,
+where $\xi_{\vb{k}}$ is the energy of a $\vb{k}$-orbital.
+Therefore:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu \nu'} \bigg(
+    \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu'}
+    \matrixel{\nu'}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu'}
+    \matrixel{\nu'}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu}
+\end{aligned}$$
+
+Notice that we have eliminated all dependence on $\Ket{\nu'}$,
+so we remove it by $\sum_{\nu} \Ket{\nu} \Bra{\nu} = 1$:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu} \bigg(
+    \frac{\matrixel{\nu}{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}} \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}{\nu}}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    - \frac{\matrixel{\nu}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}} \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\nu}}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \bigg) e^{-\beta E_\nu}
+    \\
+    &= \frac{1}{Z V} \sum_{\vb{k}} \sum_{\nu}
+    \frac{\matrixel{\nu}{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}
+    {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0}}{\nu}}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+\end{aligned}$$
+
+Where we recognized the commutator,
+and eliminated $E_\nu$ using $\hat{H}_0 \Ket{n} = E_\nu \Ket{\nu}$.
+The resulting expression has the form of a matrix trace $\Tr$
+and a thermal expectation $\Expval{}_0$:
+
+$$\begin{aligned}
+    \chi_0
+    &= \frac{1}{Z V} \sum_{\vb{k}} \frac{\Tr\!\big(\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}
+    {\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}} \: e^{- \beta \hat{H}_0} \big)}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    = \frac{1}{V} \sum_{\vb{k}}
+    \frac{\expval{\comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}}_0}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+\end{aligned}$$
+
+This commutator can be evaluated,
+and in this particular case it turns out to be:
+
+$$\begin{aligned}
+    \comm{\hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k} + \vb{q}}}{\hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k}}}
+    = \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-commutator"/>
+<label for="proof-commutator">Proof</label>
+<div class="hidden">
+<label for="proof-commutator">Proof.</label>
+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.
+</div>
+</div>
+
+We substitute this result into $\chi_0$,
+and reintroduce the spin index $\sigma$ associated with $\vb{k}$:
+
+$$\begin{aligned}
+    \chi_0(\vb{q}, \omega)
+    = \frac{1}{V} \sum_{\sigma \vb{k}}
+    \frac{\expval{\hat{c}_{\sigma,\vb{k}}^\dagger \hat{c}_{\sigma,\vb{k}} - \hat{c}_{\sigma,\vb{k}+\vb{q}}^\dagger \hat{c}_{\sigma,\vb{k}+\vb{q}}}_0}
+    {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+\end{aligned}$$
+
+The operator $\hat{c}_{\sigma.\vb{k}}^\dagger \hat{c}_{\sigma.\vb{k}}$
+simply counts the number of electrons in state $(\sigma, \vb{k})$,
+which is given by the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) $n_F$.
+This gives us the **Lindhard response function**:
+
+$$\begin{aligned}
+    \boxed{
+        \chi_0(\vb{q}, \omega)
+        = \frac{1}{V} \sum_{\sigma \vb{k}}
+        \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
+        {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    }
+\end{aligned}$$
+
+From this, we would like to get the
+[dielectric function](/know/concept/dielectric-function/) $\varepsilon_r$.
+Recall its definition, where $U_\mathrm{tot}$, $U_\mathrm{ext}$, and $U_\mathrm{ind}$
+are the total, external and induced potentials, respectively:
+
+$$\begin{aligned}
+    U_\mathrm{tot}
+    = U_\mathrm{ext} + U_\mathrm{ind}
+    = \frac{U_\mathrm{ext}}{\varepsilon_r}
+\end{aligned}$$
+
+Note that these are all *energy* potentials:
+this choice is justified because all energy potentials
+are caused by electric fields in this case.
+The *electric* potential is recoverable as
+$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$,
+where $q_e < 0$ is the charge of an electron.
+
+From the Lindhard response function $\chi_0$,
+we get the induced particle density offset $\delta{\Expval{\hat{n}}}$
+caused by a potential $U$.
+The density $\delta{\Expval{\hat{n}}}$ should be self-consistent,
+implying $U = U_\mathrm{tot}$.
+In other words, we have a linear relation
+$\delta{\Expval{\hat{n}}} = \chi_0 U_\mathrm{tot}$,
+so the standard formula for $\varepsilon_r$ gives:
+
+$$\begin{aligned}
+    \boxed{
+        \varepsilon_r(\vb{q}, \omega)
+        = 1 - \frac{U_{ee}(\vb{q})}{V}
+        \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+    }
+\end{aligned}$$
+
+Where $U_{ee}(\vb{q}) = q_e^2 / (\varepsilon_0 |\vb{q}|^2)$
+is Coulomb repulsion.
+This is the **Lindhard dielectric function** of a free
+non-interacting electron gas,
+at any temperature and for any dimensionality.
+
+
+
+## References
+1.  K.S. Thygesen,
+    *Advanced solid state physics: linear response theory*,
+    2013, unpublished.
+2.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
+3.  G. Grosso, G.P. Parravicini,
+    *Solid state physics*,
+    2nd edition, Elsevier.
diff --git a/source/know/concept/lorentz-force/index.md b/source/know/concept/lorentz-force/index.md
new file mode 100644
index 0000000..293cdbc
--- /dev/null
+++ b/source/know/concept/lorentz-force/index.md
@@ -0,0 +1,190 @@
+---
+title: "Lorentz force"
+date: 2021-09-08
+categories:
+- Physics
+- Electromagnetism
+- Plasma physics
+layout: "concept"
+---
+
+The **Lorentz force** is an empirical force used to define
+the [electric field](/know/concept/electric-field/) $\vb{E}$
+and [magnetic field](/know/concept/magnetic-field/) $\vb{B}$.
+For a particle with charge $q$ moving with velocity $\vb{u}$,
+the Lorentz force $\vb{F}$ is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{F}
+        = q (\vb{E} + \vb{u} \cross \vb{B})
+    }
+\end{aligned}$$
+
+
+## Uniform electric field
+
+Consider the simple case of an electric field $\vb{E}$
+that is uniform in all of space.
+In the absence of a magnetic field $\vb{B} = 0$
+and any other forces,
+Newton's second law states:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q \vb{E}
+\end{aligned}$$
+
+This is straightforward to integrate in time,
+for a given initial velocity vector $\vb{u}_0$:
+
+$$\begin{aligned}
+    \vb{u}(t)
+    = \frac{q}{m} \vb{E} t + \vb{u}_0
+\end{aligned}$$
+
+And then the particle's position $\vb{x}(t)$
+is found be integrating once more,
+with $\vb{x}(0) = \vb{x}_0$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{x}(t)
+        = \frac{q}{2 m} \vb{E} t^2 + \vb{u}_0 t + \vb{x}_0
+    }
+\end{aligned}$$
+
+In summary, unsurprisingly, a uniform electric field $\vb{E}$
+accelerates the particle with a constant force $\vb{F} = q \vb{E}$.
+Note that the direction depends on the sign of $q$.
+
+
+## Uniform magnetic field
+
+Consider the simple case of a uniform magnetic field
+$\vb{B} = (0, 0, B)$ in the $z$-direction,
+without an electric field $\vb{E} = 0$.
+If there are no other forces,
+Newton's second law states:
+
+$$\begin{aligned}
+    \vb{F}
+    = m \dv{\vb{u}}{t}
+    = q \vb{u} \cross \vb{B}
+\end{aligned}$$
+
+Evaluating the cross product yields
+three coupled equations for the components of $\vb{u}$:
+
+$$\begin{aligned}
+    \dv{u_x}{t}
+    = \frac{q B}{m} u_y
+    \qquad \quad
+    \dv{u_y}{t}
+    = - \frac{q B}{m} u_x
+    \qquad \quad
+    \dv{u_z}{t}
+    = 0
+\end{aligned}$$
+
+Differentiating the first equation with respect to $t$,
+and substituting $\idv{u_y}{t}$ from the second,
+we arrive at the following harmonic oscillator:
+
+$$\begin{aligned}
+    \dvn{2}{u_x}{t} = - \omega_c^2 u_x
+\end{aligned}$$
+
+Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
+which may be negative:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_c
+        \equiv \frac{q B}{m}
+    }
+\end{aligned}$$
+
+Suppose we choose our initial conditions so that
+the solution for $u_x(t)$ is given by:
+
+$$\begin{aligned}
+    u_x(t)
+    = u_\perp \cos(\omega_c t)
+\end{aligned}$$
+
+Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity.
+Then $u_y(t)$ is found to be:
+
+$$\begin{aligned}
+    u_y(t)
+    = \frac{m}{q B} \dv{u_x}{t}
+    = - \frac{m \omega_c}{q B} u_\perp \sin(\omega_c t)
+    = - u_\perp \sin(\omega_c t)
+\end{aligned}$$
+
+This means that the particle moves in a circle,
+in a direction determined by the sign of $\omega_c$.
+
+Integrating the velocity yields the position,
+where we refer to the integration constants $x_{gc}$ and $y_{gc}$
+as the **guiding center**, around which the particle orbits or **gyrates**:
+
+$$\begin{aligned}
+    x(t)
+    = \frac{u_\perp}{\omega_c} \sin(\omega_c t) + x_{gc}
+    \qquad \quad
+    y(t)
+    = \frac{u_\perp}{\omega_c} \cos(\omega_c t) + y_{gc}
+\end{aligned}$$
+
+The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:
+
+$$\begin{aligned}
+    \boxed{
+        r_L
+        \equiv \frac{u_\perp}{|\omega_c|}
+        = \frac{m u_\perp}{|q| B}
+    }
+\end{aligned}$$
+
+Finally, it is easy to integrate the equation
+for the $z$-axis velocity $u_z$, which is conserved:
+
+$$\begin{aligned}
+    z(t)
+    = z_{gc}
+    = u_z t + z_0
+\end{aligned}$$
+
+In conclusion, the particle's motion parallel to $\vb{B}$
+is not affected by the magnetic field,
+while its motion perpendicular to $\vb{B}$
+is circular around an imaginary guiding center.
+The end result is that particles follow a helical path
+when moving through a uniform magnetic field:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{x}(t)
+        = \frac{u_\perp}{\omega_c}
+        \begin{pmatrix}
+            \sin(\omega_c t) \\ \cos(\omega_c t) \\ 0
+        \end{pmatrix}
+        + \vb{x}_{gc}(t)
+    }
+\end{aligned}$$
+
+Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$
+is the position of the guiding center.
+For a detailed look at how $\vb{B}$ and $\vb{E}$
+can affect the guiding center's motion,
+see [guiding center theory](/know/concept/guiding-center-theory/).
+
+
+
+## References
+1.  F.F. Chen,
+    *Introduction to plasma physics and controlled fusion*,
+    3rd edition, Springer.
diff --git a/source/know/concept/lubrication-theory/index.md b/source/know/concept/lubrication-theory/index.md
new file mode 100644
index 0000000..04acd30
--- /dev/null
+++ b/source/know/concept/lubrication-theory/index.md
@@ -0,0 +1,215 @@
+---
+title: "Lubrication theory"
+date: 2021-06-03
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+**Lubricants** are widely used
+to reduce friction between two moving surfaces.
+In fluid mechanics, **lubrication theory**
+is the study of fluids that are tightly constrained in one dimension,
+especially those in small gaps between moving surfaces.
+
+For simplicity, we limit ourselves to 2D
+by assuming that everything is constant along the $z$-axis.
+Consider a gap of width $d$ (along $y$) and length $L$ (along $x$),
+where $d \ll L$, containing the fluid.
+Outside the gap, the lubricant has a
+[Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \approx U L / \nu$.
+
+Inside the gap, the Reynolds number $\mathrm{Re}_\mathrm{gap}$ is different.
+This is because advection will dominate along the $x$-axis (gap length),
+and viscosity along the $y$-axis (gap width).
+Therefore:
+
+$$\begin{aligned}
+    \mathrm{Re}_\mathrm{gap}
+    \approx \frac{|(\va{v} \cdot \nabla) \va{v}|}{|\nu \nabla^2 \va{v}|}
+    \approx \frac{U^2 / L}{\nu U / d^2}
+    \approx \frac{d^2}{L^2} \mathrm{Re}
+\end{aligned}$$
+
+If $d$ is small enough compared to $L$,
+then $\mathrm{Re}_\mathrm{gap} \ll 1$.
+More formally, we need $d \ll L / \sqrt{\mathrm{Re}}$,
+so we are inside the boundary layer,
+in the realm of the [Prandtl equations](/know/concept/prandtl-equations/).
+
+Let $\mathrm{Re}_\mathrm{gap} \ll 1$.
+We are thus dealing with *Stokes flow*, in which case
+the [Navier-Stokes equations](/know/concept/navier-stokes/equations/)
+can be reduced to the following *Stokes equations*:
+
+$$\begin{aligned}
+    \pdv{p}{x}
+    = \eta \: \Big( \pdvn{2}{v_x}{x} + \pdvn{2}{v_x}{y} \Big)
+    \qquad \quad
+    \pdv{p}{y}
+    = \eta \: \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big)
+\end{aligned}$$
+
+Let the $y = 0$ plane be an infinite flat surface,
+sliding in the positive $x$-direction at a constant velocity $U$.
+On the other side of the gap,
+an arbitrary surface is described by $h(x)$.
+
+Since the gap is so narrow,
+and the surfaces' movements cause large shear stresses inside,
+$v_y$ is negligible compared to $v_x$.
+Furthermore, because the gap is so long,
+we assume that $\ipdv{v_x}{x}$ is negligible compared to $\ipdv{v_x}{y}$.
+This reduces the Stokes equations to:
+
+$$\begin{aligned}
+    \pdv{p}{x}
+    = \eta \pdvn{2}{v_x}{y}
+    \qquad \quad
+    \pdv{p}{y}
+    = 0
+\end{aligned}$$
+
+This result could also be derived from the Prandtl equations.
+In any case, it tells us that $p$ only depends on $x$,
+allowing us to integrate the former equation:
+
+$$\begin{aligned}
+    v_x
+    = \frac{p'}{2 \eta} y^2 + C_1 y + C_2
+\end{aligned}$$
+
+Where $C_1$ and $C_2$ are integration constants.
+At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$.
+Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to:
+
+$$\begin{aligned}
+    v_x
+    = \frac{p'}{2 \eta} y^2 - \Big( \frac{p'}{2 \eta} h + \frac{U}{h} \Big) y + U
+\end{aligned}$$
+
+The moving bottom surface drags fluid in the $x$-direction
+at a volumetric rate $Q$, given by:
+
+$$\begin{aligned}
+    Q
+    = \int_0^{h(x)} v_x(x, y) \dd{y}
+    = \bigg[ \frac{p'}{6 \eta} y^3 - \frac{p'}{4 \eta} h y^2 - \frac{U}{2 h} y^2 + U y \bigg]_0^{h}
+    = - \frac{p'}{12 \eta} h^3 + \frac{U}{2} h
+\end{aligned}$$
+
+Assuming that the lubricant is incompressible,
+meaning that the same volume of fluid must be leaving a point as is entering it.
+In other words, $Q$ is independent of $x$,
+which allows us to write $p'(x)$ in terms of
+measurable constants and the known function $h(x)$:
+
+$$\begin{aligned}
+    \boxed{
+        p'
+        = 6 \eta \: \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big)
+    }
+\end{aligned}$$
+
+Then we insert this into our earlier expression for $v_x$, yielding:
+
+$$\begin{aligned}
+    v_x
+    &= 3 y (y - h) \Big( \frac{U}{h^2} - \frac{2 Q}{h^3} \Big) - \frac{U h}{h^2} y + \frac{U h^2}{h^2}
+\end{aligned}$$
+
+Which, after some rearranging, can be written in the following form:
+
+$$\begin{aligned}
+    \boxed{
+        v_x
+        = U \frac{(3 y - h) (y - h)}{h^2} - Q \frac{6 y (y - h)}{h^3}
+    }
+\end{aligned}$$
+
+With this, we can find $v_y$ by exploiting incompressibility,
+i.e. the continuity equation states:
+
+$$\begin{aligned}
+    \pdv{v_y}{y}
+    = - \pdv{v_x}{x}
+    = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big)
+\end{aligned}$$
+
+Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$:
+
+$$\begin{aligned}
+    \boxed{
+        v_y
+        = - 2 h' \frac{U h - 3 Q}{h^4} y^2 (h - y)
+    }
+\end{aligned}$$
+
+Typically, the lubricant is not in a preexisting pressure differential,
+i.e it is not getting pumped through the system.
+Although the pressure gradient $p'$ need not be zero,
+we therefore expect that its integral vanishes:
+
+$$\begin{aligned}
+    0
+    = \int_L p'(x) \dd{x}
+    = 6 \eta U \int_L \frac{1}{h(x)^2} \dd{x} - 12 \eta Q \int_L \frac{1}{h(x)^3} \dd{x}
+\end{aligned}$$
+
+Isolating this for $Q$, and defining $q$ as below, yields a simple equation:
+
+$$\begin{aligned}
+    Q
+    = \frac{1}{2} U q
+    \qquad \quad
+    q
+    \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}}
+\end{aligned}$$
+
+We substitute this into $v_x$ and rearrange to get an interesting expression:
+
+$$\begin{aligned}
+    v_x
+    &= U \frac{3 y^2 - h y - 3 h y + h^2}{h^2} - U q \frac{3 y^2 - 3 h y}{h^3}
+    \\
+    &= U \Big( 1 - \frac{y}{h} \Big) \Big( 1 - \frac{3 y (h - q)}{h^2} \Big)
+\end{aligned}$$
+
+The first factor is always positive,
+but the second can be negative,
+if for some $y$-values:
+
+$$\begin{aligned}
+    h^2 < 3 y (h - q)
+    \quad \implies \quad
+    y > \frac{h^2}{3 (h - q)}
+\end{aligned}$$
+
+Since $h > y$, such $y$-values will only exist
+if $h$ is larger than some threshold:
+
+$$\begin{aligned}
+    3 (h - q) > h
+    \quad \implies \quad
+    h > \frac{3}{2} q
+\end{aligned}$$
+
+If this condition is satisfied,
+there will be some flow reversal:
+rather than just getting dragged by the shearing motion,
+the lubricant instead "rolls" inside the gap.
+This is confirmed by $v_y$:
+
+$$\begin{aligned}
+    v_y
+    = - U h' \frac{2 h - 3 q}{h^4} y^2 (h - y)
+\end{aligned}$$
+
+
+
+## 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/magnetic-field/index.md b/source/know/concept/magnetic-field/index.md
new file mode 100644
index 0000000..3b4c10a
--- /dev/null
+++ b/source/know/concept/magnetic-field/index.md
@@ -0,0 +1,104 @@
+---
+title: "Magnetic field"
+date: 2021-07-12
+categories:
+- Physics
+- Electromagnetism
+layout: "concept"
+---
+
+The **magnetic field** $\vb{B}$ is a vector field
+that describes magnetic effects,
+and is defined as the field that correctly predicts
+the [Lorentz force](/know/concept/lorentz-force/)
+on a particle with electric charge $q$:
+
+$$\begin{aligned}
+    \vb{F}
+    = q \vb{v} \cross \vb{B}
+\end{aligned}$$
+
+If an object is placed in a magnetic field $\vb{B}$,
+and wants to rotate to align itself with the field,
+then its **magnetic dipole moment** $\vb{m}$
+is defined from the aligning torque $\vb{\tau}$:
+
+$$\begin{aligned}
+    \vb{\tau} = \vb{m} \times \vb{B}
+\end{aligned}$$
+
+Where $\vb{m}$ has units of $\mathrm{J / T}$.
+From this, the **magnetization** $\vb{M}$ is defined as follows,
+and roughly represents the moments per unit volume:
+
+$$\begin{aligned}
+    \vb{M} \equiv \dv{\vb{m}}{V}
+    \:\:\iff\:\:
+    \vb{m} = \int_V \vb{M} \dd{V}
+\end{aligned}$$
+
+If $\vb{M}$ has the same magnitude and orientation throughout the body,
+then $\vb{m} = \vb{M} V$, where $V$ is the volume.
+Therefore, $\vb{M}$ has units of $\mathrm{A / m}$.
+
+A nonzero $\vb{M}$ complicates things,
+since it contributes to the field
+and hence modifies $\vb{B}$.
+We thus define
+the "free" **auxiliary field** $\vb{H}$
+from the "bound" field $\vb{M}$
+and the "net" field $\vb{B}$:
+
+$$\begin{aligned}
+    \vb{H} \equiv \frac{1}{\mu_0} \vb{B} - \vb{M}
+    \:\:\iff\:\:
+    \vb{B} = \mu_0 (\vb{H} + \vb{M})
+\end{aligned}$$
+
+Where the **magnetic permeability of free space** $\mu_0$ is a known constant.
+It is important to point out some inconsistencies here:
+$\vb{B}$ contains a factor of $\mu_0$, and thus measures **flux density**,
+while $\vb{H}$ and $\vb{M}$ do not contain $\mu_0$,
+and therefore measure **field intensity**.
+Note that this convention is the opposite of the analogous
+[electric fields](/know/concept/electric-field/)
+$\vb{E}$, $\vb{D}$ and $\vb{P}$.
+Also note that $\vb{P}$ has the opposite sign convention of $\vb{M}$.
+
+Some objects, called **ferromagnets** or **permanent magnets**,
+have an inherently nonzero $\vb{M}$.
+Others objects, when placed in a $\vb{B}$-field,
+may instead gain an induced $\vb{M}$.
+
+When $\vb{M}$ is induced,
+its magnitude is usually proportional
+to the applied field strength $\vb{H}$:
+
+$$\begin{aligned}
+    \vb{B}
+    = \mu_0(\vb{H} + \vb{M})
+    = \mu_0 (\vb{H} + \chi_m \vb{H})
+    = \mu_0 \mu_r \vb{H}
+    = \mu \vb{H}
+\end{aligned}$$
+
+Where $\chi_m$ is the **volume magnetic susceptibility**,
+and $\mu_r \equiv 1 + \chi_m$ and $\mu \equiv \mu_r \mu_0$ are
+the **relative permeability** and **absolute permeability**
+of the medium, respectively.
+Materials with intrinsic magnetization, i.e. ferromagnets,
+do not have a well-defined $\chi_m$.
+
+If $\chi_m > 0$, the medium is **paramagnetic**,
+meaning it strengthens the net field $\vb{B}$.
+Otherwise, if $\chi_m < 0$, the medium is **diamagnetic**,
+meaning it counteracts the applied field $\vb{H}$.
+
+For $|\chi_m| \ll 1$, as is often the case,
+the magnetization $\vb{M}$ can be approximated by:
+
+$$\begin{aligned}
+    \vb{M}
+    = \chi_m \vb{H}
+    \approx \chi_m \vb{B} / \mu_0
+\end{aligned}$$
diff --git a/source/know/concept/magnetohydrodynamics/index.md b/source/know/concept/magnetohydrodynamics/index.md
new file mode 100644
index 0000000..1d8bc12
--- /dev/null
+++ b/source/know/concept/magnetohydrodynamics/index.md
@@ -0,0 +1,398 @@
+---
+title: "Magnetohydrodynamics"
+date: 2021-10-21
+categories:
+- Physics
+- Plasma physics
+- Electromagnetism
+layout: "concept"
+---
+
+**Magnetohydrodynamics** (MHD) describes the dynamics
+of fluids that are electrically conductive.
+Notably, it is often suitable to describe plasmas,
+and can be regarded as a special case of the
+[two-fluid model](/know/concept/two-fluid-equations/);
+we will derive it as such,
+but the results are not specific to plasmas.
+
+In the two-fluid model, we described the plasma as two separate fluids,
+but in MHD we treat it as a single conductive fluid.
+The macroscopic pressure $p$
+and electric current density $\vb{J}$ are:
+
+$$\begin{aligned}
+    p
+    = p_i + p_e
+    \qquad \quad
+    \vb{J}
+    = q_i n_i \vb{u}_i + q_e n_e \vb{u}_e
+\end{aligned}$$
+
+Meanwhile, the macroscopic mass density $\rho$
+and center-of-mass flow velocity $\vb{u}$
+are as follows, although the ions dominate due to their large mass:
+
+$$\begin{aligned}
+    \rho
+    = m_i n_i + m_e n_e
+    \approx m_i n_i
+    \qquad \quad
+    \vb{u}
+    = \frac{1}{\rho} \Big( m_i n_i \vb{u}_i + m_e n_e \vb{u}_e \Big)
+    \approx \vb{u}_i
+\end{aligned}$$
+
+With these quantities in mind,
+we add up the two-fluid continuity equations,
+multiplied by their respective particles' masses:
+
+$$\begin{aligned}
+    0
+    &= m_i \pdv{n_i}{t} + m_e \pdv{n_e}{t} + m_i \nabla \cdot (n_i \vb{u}_i) + m_e \nabla \cdot (n_e \vb{u}_e)
+\end{aligned}$$
+
+After some straightforward rearranging,
+we arrive at the single-fluid continuity relation:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u})
+        = 0
+    }
+\end{aligned}$$
+
+Next, consider the two-fluid momentum equations
+for the ions and electrons, respectively:
+
+$$\begin{aligned}
+    m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+    &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
+    \\
+    m_e n_e \frac{\mathrm{D} \vb{u}_e}{\mathrm{D} t}
+    &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
+\end{aligned}$$
+
+We will assume that electrons' inertia
+is negligible compared to the [Lorentz force](/know/concept/lorentz-force/).
+Let $\tau_\mathrm{char}$ be the characteristic timescale of the plasma's dynamics,
+i.e. nothing noticable happens in times shorter than $\tau_\mathrm{char}$,
+then this assumption can be written as:
+
+$$\begin{aligned}
+    1
+    \gg \frac{\big| m_e n_e \mathrm{D} \vb{u}_e / \mathrm{D} t \big|}{\big| q_e n_e \vb{u}_e \cross \vb{B} \big|}
+    \sim \frac{m_e n_e |\vb{u}_e| / \tau_\mathrm{char}}{q_e n_e |\vb{u}_e| |\vb{B}|}
+    = \frac{m_e}{q_e |\vb{B}| \tau_\mathrm{char}}
+    = \frac{1}{\omega_{ce} \tau_\mathrm{char}}
+    \ll 1
+\end{aligned}$$
+
+Where we have recognized the cyclotron frequency $\omega_c$ (see Lorentz force article).
+In other words, our assumption is equivalent to
+the electron gyration period $2 \pi / \omega_{ce}$
+being small compared to the macroscopic dynamics' timescale $\tau_\mathrm{char}$.
+By construction, we can thus ignore the left-hand side
+of the electron momentum equation, leaving:
+
+$$\begin{aligned}
+    m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+    &= q_i n_i (\vb{E} + \vb{u}_i \cross \vb{B}) - \nabla p_i - f_{ie} m_i n_i (\vb{u}_i - \vb{u}_e)
+    \\
+    0
+    &= q_e n_e (\vb{E} + \vb{u}_e \cross \vb{B}) - \nabla p_e - f_{ei} m_e n_e (\vb{u}_e - \vb{u}_i)
+\end{aligned}$$
+
+We add up these momentum equations,
+recognizing the pressure $p$ and current $\vb{J}$:
+
+$$\begin{aligned}
+    m_i n_i \frac{\mathrm{D} \vb{u}_i}{\mathrm{D} t}
+    &= (q_i n_i + q_e n_e) \vb{E} + \vb{J} \cross \vb{B} - \nabla p
+    - f_{ie} m_i n_i (\vb{u}_i \!-\! \vb{u}_e) - f_{ei} m_e n_e (\vb{u}_e \!-\! \vb{u}_i)
+    \\
+    &= (q_i n_i + q_e n_e) \vb{E} + \vb{J} \cross \vb{B} - \nabla p
+\end{aligned}$$
+
+Where we have used $f_{ie} m_i n_i = f_{ei} m_e n_e$
+because momentum is conserved by the underlying
+[Rutherford scattering](/know/concept/rutherford-scattering/) process,
+which is [elastic](/know/concept/elastic-collision/).
+In other words, the momentum given by ions to electrons
+is equal to the momentum received by electrons from ions.
+
+Since the two-fluid model assumes that
+the [Debye length](/know/concept/debye-length/) $\lambda_D$
+is small compared to a "blob" $\dd{V}$ of the fluid,
+we can invoke quasi-neutrality $q_i n_i + q_e n_e = 0$.
+Using that $\rho \approx m_i n_i$ and $\vb{u} \approx \vb{u}_i$,
+we thus arrive at the **momentum equation**:
+
+$$\begin{aligned}
+    \boxed{
+        \rho \frac{\mathrm{D} \vb{u}}{\mathrm{D} t}
+        = \vb{J} \cross \vb{B} - \nabla p
+    }
+\end{aligned}$$
+
+However, we found this by combining two equations into one,
+so some information was implicitly lost;
+we need a second momentum equation.
+Therefore, we return to the electrons' momentum equation,
+after a bit of rearranging:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u}_e \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
+    = \frac{f_{ei} m_e}{q_e} (\vb{u}_e - \vb{u}_i)
+\end{aligned}$$
+
+Again using quasi-neutrality $q_i n_i = - q_e n_e$,
+the current density $\vb{J} = q_e n_e (\vb{u}_e \!-\! \vb{u}_i)$,
+so:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u}_e \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
+    = \eta \vb{J}
+    \qquad \quad
+    \eta
+    \equiv \frac{f_{ei} m_e}{n_e q_e^2}
+\end{aligned}$$
+
+Where $\eta$ is the electrical resistivity of the plasma,
+see [Spitzer resistivity](/know/concept/spitzer-resistivity/)
+for more information, and a rough estimate of this quantity for a plasma.
+
+Now, using that $\vb{u} \approx \vb{u}_i$,
+we add $(\vb{u} \!-\! \vb{u}_i) \cross \vb{B} \approx 0$ to the equation,
+and insert $\vb{J}$ again:
+
+$$\begin{aligned}
+    \eta \vb{J}
+    &= \vb{E} + \vb{u} \cross \vb{B} + (\vb{u}_e - \vb{u}_i) \cross \vb{B} - \frac{\nabla p_e}{q_e n_e}
+    \\
+    &= \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e} - \frac{\nabla p_e}{q_e n_e}
+\end{aligned}$$
+
+Next, we want to get rid of the pressure term.
+To do so, we take the curl of the equation:
+
+$$\begin{aligned}
+    \nabla \cross (\eta \vb{J})
+    = - \pdv{\vb{B}}{t} + \nabla \cross (\vb{u} \cross \vb{B}) + \nabla \cross \frac{\vb{J} \cross \vb{B}}{q_e n_e}
+    - \nabla \cross \frac{\nabla p_e}{q_e n_e}
+\end{aligned}$$
+
+Where we have used Faraday's law.
+This is the **induction equation**,
+and is used to compute $\vb{B}$.
+The pressure term can be rewritten using the ideal gas law $p_e = k_B T_e n_e$:
+
+$$\begin{aligned}
+    \nabla \cross \frac{\nabla p_e}{q_e n_e}
+    = \frac{k_B}{q_e} \nabla \cross \frac{\nabla (n_e T_e)}{n_e}
+    = \frac{k_B}{q_e} \nabla \cross \Big( \nabla T_e + T_e \frac{\nabla n_e}{n_e} \Big)
+\end{aligned}$$
+
+The curl of a gradient is always zero,
+and we notice that $\nabla n_e / n_e = \nabla\! \ln(n_e)$.
+Then we use the vector identity $\nabla \cross (f \nabla g) = \nabla f \cross \nabla g$,
+leading to:
+
+$$\begin{aligned}
+    \nabla \cross \frac{\nabla p_e}{q_e n_e}
+    = \frac{k_B}{q_e} \nabla \cross \big( T_e \: \nabla\! \ln(n_e) \big)
+    = \frac{k_B}{q_e} \big( \nabla T_e \cross \nabla\! \ln(n_e) \big)
+    = \frac{k_B}{q_e n_e} \big( \nabla T_e \cross \nabla n_e \big)
+\end{aligned}$$
+
+It is reasonable to assume that $\nabla T_e$ and $\nabla n_e$
+point in roughly the same direction,
+in which case the pressure term can be neglected.
+Consequently, $p_e$ has no effect on the dynamics of $\vb{B}$,
+so we argue that it can be dropped from the original (non-curled) equation too, leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e}
+        = \eta \vb{J}
+    }
+\end{aligned}$$
+
+This is known as the **generalized Ohm's law**,
+since it contains the relation $\vb{E} = \eta \vb{J}$.
+
+Next, consider [Ampère's law](/know/concept/maxwells-equations/),
+where we would like to neglect the last term:
+
+$$\begin{aligned}
+    \nabla \cross \vb{B}
+    = \mu_0 \vb{J} + \frac{1}{c^2} \pdv{\vb{E}}{t}
+\end{aligned}$$
+
+From Faraday's law, we can obtain a scale estimate for $\vb{E}$.
+Recall that $\tau_\mathrm{char}$ is the characteristic timescale of the plasma,
+and let $\lambda_\mathrm{char} \gg \lambda_D$ be its characteristic lengthscale:
+
+$$\begin{aligned}
+    \nabla \cross \vb{E}
+    = - \pdv{\vb{B}}{t}
+    \quad \implies \quad
+    |\vb{E}|
+    \sim \frac{\lambda_\mathrm{char}}{\tau_\mathrm{char}} |\vb{B}|
+\end{aligned}$$
+
+From this, we find when we can neglect
+the last term in Ampère's law:
+the characteristic velocity $v_\mathrm{char}$
+must be tiny compared to $c$,
+i.e. the plasma must be non-relativistic:
+
+$$\begin{aligned}
+    1
+    \gg \frac{\big| (\ipdv{\vb{E}}{t}) / c^2 \big|}{\big| \nabla \cross \vb{B} \big|}
+    \sim \frac{|\vb{E}| / \tau_\mathrm{char}}{|\vb{B}| c^2 / \lambda_\mathrm{char}}
+    \sim \frac{|\vb{B}| \lambda_\mathrm{char}^2 / \tau_\mathrm{char}^2}{|\vb{B}| c^2}
+    = \frac{v_\mathrm{char}^2}{c^2}
+    \ll 1
+\end{aligned}$$
+
+We thus have the following reduced form of Ampère's law,
+in addition to Faraday's law:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cross \vb{B}
+        = \mu_0 \vb{J}
+    }
+    \qquad \quad
+    \boxed{
+        \nabla \cross \vb{E}
+        = - \pdv{\vb{B}}{t}
+    }
+\end{aligned}$$
+
+Finally, we revisit the thermodynamic equation of state,
+for a single fluid this time.
+Using the product rule of differentiation yields:
+
+$$\begin{aligned}
+    0
+    &= \frac{\mathrm{D}}{\mathrm{D} t} \Big( \frac{p}{\rho^\gamma} \Big)
+    = \frac{\mathrm{D} p}{\mathrm{D} t} \rho^{-\gamma} - p \gamma \rho^{-\gamma - 1} \frac{\mathrm{D} \rho}{\mathrm{D} t}
+\end{aligned}$$
+
+The continuity equation allows us to rewrite
+the [material derivative](/know/concept/material-derivative/)
+$\mathrm{D} \rho / \mathrm{D} t$ as follows:
+
+$$\begin{aligned}
+    \pdv{\rho}{t} + \nabla \cdot (\rho \vb{u})
+    = \pdv{\rho}{t} + \rho \nabla \cdot \vb{u} + \vb{u} \cdot \nabla \rho
+    = \rho \nabla \cdot \vb{u} + \frac{\mathrm{D} \rho}{\mathrm{D} t}
+    = 0
+\end{aligned}$$
+
+Inserting this into the equation of state
+leads us to a differential equation for $p$:
+
+$$\begin{aligned}
+    0
+    = \frac{\mathrm{D} p}{\mathrm{D} t} + p \gamma \frac{1}{\rho} \rho \nabla \cdot \vb{u}
+    \quad \implies \quad
+    \boxed{
+        \frac{\mathrm{D} p}{\mathrm{D} t} = - p \gamma \nabla \cdot \vb{u}
+    }
+\end{aligned}$$
+
+This closes the set of 14 MHD equations for 14 unknowns.
+Originally, the two-fluid model had 16 of each,
+but we have merged $n_i$ and $n_e$ into $\rho$,
+and $p_i$ and $p_i$ into $p$.
+
+
+## Ohm's law variants
+
+It is worth discussing the generalized Ohm's law in more detail.
+Its full form was:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u} \cross \vb{B} + \frac{\vb{J} \cross \vb{B}}{q_e n_e}
+    = \eta \vb{J}
+\end{aligned}$$
+
+However, most authors neglect some of its terms:
+this form is used for **Hall MHD**,
+where $\vb{J} \cross \vb{B}$ is called the *Hall term*.
+This term can be dropped in any of the following cases:
+
+$$\begin{gathered}
+    1
+    \gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \vb{u} \cross \vb{B} \big|}
+    \sim \frac{\rho v_\mathrm{char} / \tau_\mathrm{char}}{v_\mathrm{char} |\vb{B}| q_i n_i}
+    \approx \frac{m_i n_i}{|\vb{B}| q_i n_i \tau_\mathrm{char}}
+    = \frac{1}{\omega_{ci} \tau_\mathrm{char}}
+    \ll 1
+    \\
+    1
+    \gg \frac{\big| \vb{J} \cross \vb{B} / q_e n_e \big|}{\big| \eta \vb{J} \big|}
+    \sim \frac{|\vb{J}| |\vb{B}| q_e^2 n_e}{f_{ei} m_e |\vb{J}| q_e n_e}
+    = \frac{|\vb{B}| q_e}{f_{ei} m_e}
+    = \frac{\omega_{ce}}{f_{ei}}
+    \ll 1
+\end{gathered}$$
+
+Where we have used the MHD momentum equation with $\nabla p \approx 0$
+to obtain the scale estimate $\vb{J} \cross \vb{B} \sim \rho v_\mathrm{char} / \tau_\mathrm{char}$.
+In other words, if the ion gyration period is short $\tau_\mathrm{char} \gg \omega_{ci}$,
+and/or if the electron gyration period is long
+compared to the electron-ion collision period $\omega_{ce} \ll f_{ei}$,
+then we are left with this form of Ohm's law, used in **resistive MHD**:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u} \cross \vb{B}
+    = \eta \vb{J}
+\end{aligned}$$
+
+Finally, we can neglect the resisitive term $\eta \vb{J}$
+if the Lorentz force is much larger.
+We formalize this condition as follows,
+where we have used Ampère's law to find $\vb{J} \sim \vb{B} / \mu_0 \lambda_\mathrm{char}$:
+
+$$\begin{aligned}
+    1
+    \ll \frac{\big| \vb{u} \cross \vb{B} \big|}{\big| \eta \vb{J} \big|}
+    \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta \vb{J}}
+    \sim \frac{v_\mathrm{char} |\vb{B}|}{\eta |\vb{B}| / \mu_0 \lambda_\mathrm{char}}
+    = \mathrm{R_m}
+    \gg 1
+\end{aligned}$$
+
+Where we have defined the **magnetic Reynolds number** $\mathrm{R_m}$ as follows,
+which is analogous to the fluid [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re}$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{R_m}
+        \equiv \frac{v_\mathrm{char} \lambda_\mathrm{char}}{\eta / \mu_0}
+    }
+\end{aligned}$$
+
+If $\mathrm{R_m} \ll 1$, the plasma is "electrically viscous",
+such that resistivity needs to be accounted for,
+whereas if $\mathrm{R_m} \gg 1$, the resistivity is negligible,
+in which case we have **ideal MHD**:
+
+$$\begin{aligned}
+    \vb{E} + \vb{u} \cross \vb{B}
+    = 0
+\end{aligned}$$
+
+
+
+## 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/markov-process/index.md b/source/know/concept/markov-process/index.md
new file mode 100644
index 0000000..4a681e3
--- /dev/null
+++ b/source/know/concept/markov-process/index.md
@@ -0,0 +1,61 @@
+---
+title: "Markov process"
+date: 2021-11-14
+categories:
+- Mathematics
+- Stochastic analysis
+layout: "concept"
+---
+
+Given a [stochastic process](/know/concept/stochastic-process/)
+$\{X_t : t \ge 0\}$ on a filtered probability space
+$(\Omega, \mathcal{F}, \{\mathcal{F}_t\}, P)$,
+it is said to be a **Markov process**
+if it satisfies the following requirements:
+
+1.  $X_t$ is $\mathcal{F}_t$-adapted,
+    meaning that the current and all past values of $X_t$
+    can be reconstructed from the filtration $\mathcal{F}_t$.
+2.  For some function $h(x)$,
+    the [conditional expectation](/know/concept/conditional-expectation/)
+    $\mathbf{E}[h(X_t) | \mathcal{F}_s] = \mathbf{E}[h(X_t) | X_s]$,
+    i.e. at time $s \le t$, the expectation of $h(X_t)$ depends only on the current $X_s$.
+    Note that $h$ must be bounded and *Borel-measurable*,
+    meaning $\sigma(h(X_t)) \subseteq \mathcal{F}_t$.
+
+This last condition is called the **Markov property**,
+and demands that the future of $X_t$ does not depend on the past,
+but only on the present $X_s$.
+
+If both $t$ and $X_t$ are taken to be discrete,
+then $X_t$ is known as a **Markov chain**.
+This brings us to the concept of the **transition probability**
+$P(X_t \in A | X_s = x)$, which describes the probability that
+$X_t$ will be in a given set $A$, if we know that currently $X_s = x$.
+
+If $t$ and $X_t$ are continuous, we can often (but not always) express $P$
+using a **transition density** $p(s, x; t, y)$,
+which gives the probability density that the initial condition $X_s = x$
+will evolve into the terminal condition $X_t = y$.
+Then the transition probability $P$ can be calculated like so,
+where $B$ is a given Borel set (see [$\sigma$-algebra](/know/concept/sigma-algebra/)):
+
+$$\begin{aligned}
+    P(X_t \in B | X_s = x)
+    = \int_B p(s, x; t, y) \dd{y}
+\end{aligned}$$
+
+A prime examples of a continuous Markov process is
+the [Wiener process](/know/concept/wiener-process/).
+Note that this is also a [martingale](/know/concept/martingale/):
+often, a Markov process happens to be a martingale, or vice versa.
+However, those concepts are not to be confused:
+the Markov property does not specify *what* the expected future must be,
+and the martingale property says nothing about the history-dependence.
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/martingale/index.md b/source/know/concept/martingale/index.md
new file mode 100644
index 0000000..f64c22e
--- /dev/null
+++ b/source/know/concept/martingale/index.md
@@ -0,0 +1,62 @@
+---
+title: "Martingale"
+date: 2021-10-31
+categories:
+- Mathematics
+- Stochastic analysis
+layout: "concept"
+---
+
+A **martingale** is a type of
+[stochastic process](/know/concept/stochastic-process/)
+with important and useful properties,
+especially for stochastic calculus.
+
+For a stochastic process $\{ M_t : t \ge 0 \}$
+on a probability filtered space $(\Omega, \mathcal{F}, \{ \mathcal{F}_t \}, P)$,
+then $M_t$ is a martingale if it satisfies all of the following:
+
+1.  $M_t$ is $\mathcal{F}_t$-adapted, meaning
+    the filtration $\mathcal{F}_t$ contains enough information
+    to reconstruct the current and all past values of $M_t$.
+2.  For all times $t \ge 0$, the expectation value exists $\mathbf{E}(M_t) < \infty$.
+3.  For all $s, t$ satisfying $0 \le s \le t$,
+    the [conditional expectation](/know/concept/conditional-expectation/)
+    $\mathbf{E}(M_t | \mathcal{F}_s) = M_s$,
+    meaning the increment $M_t \!-\! M_s$ is always expected
+    to be zero $\mathbf{E}(M_t \!-\! M_s | \mathcal{F}_s) = 0$.
+
+The last condition is called the **martingale property**,
+and basically means that a martingale is an unbiased random walk.
+Accordingly, the [Wiener process](/know/concept/wiener-process/) $B_t$
+(Brownian motion) is an example of a martingale,
+since each of its increments $B_t \!-\! B_s$ has mean $0$ by definition.
+
+Martingales are easily confused with
+[Markov processes](/know/concept/markov-process/),
+because stochastic processes will often be both,
+e.g. the Wiener process.
+However, these are distinct concepts:
+the martingale property says nothing about history-dependence,
+and the Markov property does not say *what* the future expectation should be.
+
+Modifying property (3) leads to two common generalizations.
+The stochastic process $M_t$ above is a **submartingale**
+if the current value is a lower bound for the expectation:
+
+3.  For $0 \le s \le t$, the conditional expectation $\mathbf{E}(M_t | \mathcal{F}_s) \ge M_s$.
+
+Analogouly, $M_t$ is a **supermartingale**
+if the current value is an upper bound instead:
+
+3.  For $0 \le s \le t$, the conditional expectation $\mathbf{E}(M_t | \mathcal{F}_s) \le M_s$.
+
+Clearly, submartingales and supermartingales are *biased* random walks,
+since they will tend to increase and decrease with time, respectively.
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
diff --git a/source/know/concept/material-derivative/index.md b/source/know/concept/material-derivative/index.md
new file mode 100644
index 0000000..6a02a22
--- /dev/null
+++ b/source/know/concept/material-derivative/index.md
@@ -0,0 +1,115 @@
+---
+title: "Material derivative"
+date: 2021-03-30
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+- Continuum physics
+layout: "concept"
+---
+
+Inside a fluid (or any other continuum), we might be interested in
+the time evolution of a certain intensive quantity $f$,
+e.g. the temperature or pressure,
+represented by a scalar field $f(\va{r}, t)$.
+
+If the fluid is static, the evolution of $f$ is simply $\ipdv{f}{t}$,
+since each point of the fluid is motionless.
+However, if the fluid is moving, we have a problem:
+the fluid molecules at position $\va{r} = \va{r}_0$ are not necessarily
+the same ones at time $t = t_0$ and $t = t_1$.
+Those molecules take $f$ with them as they move,
+so we need to account for this transport somehow.
+
+To do so, we choose an infinitesimal "blob" or **parcel** of the fluid,
+which always contains the same specific molecules,
+and track its position $\va{r}(t)$ through time as it moves and deforms.
+The value of $f$ for this parcel is then given by:
+
+$$\begin{aligned}
+    f(\va{r}, t)
+    = f(\va{r}(t), t)
+    = f\big(x(t), y(t), z(t), t\big)
+\end{aligned}$$
+
+In effect, we have simply made the coordinate $\va{r}$ dependent on time,
+and have specifically chosen the time-dependence to track the parcel.
+The net evolution of $f$ is then its "true" (i.e. non-partial) derivative with respect to $t$,
+allowing us to apply the chain rule:
+
+$$\begin{aligned}
+    \dv{}{t}f\big(x(t), y(t), z(t), t\big)
+    &= \pdv{f}{t} + \pdv{f}{x} \dv{x}{t} + \pdv{f}{y} \dv{y}{t} + \pdv{f}{z} \dv{z}{t}
+    \\
+    &= \pdv{f}{t} + v_x \pdv{f}{x} + v_y \pdv{f}{y} + v_z \pdv{f}{z}
+\end{aligned}$$
+
+Where $v_x$, $v_y$ and $v_z$ are the parcel's velocity components.
+Let $\va{v} = (v_x, v_y, v_z)$ be the velocity vector field,
+then we can rewrite this expression like so:
+
+$$\begin{aligned}
+    \dv{}{t}f\big(x(t), y(t), z(t), t\big)
+    &= \pdv{f}{t} + (\va{v} \cdot \nabla) f
+\end{aligned}$$
+
+Note that $\va{v} = \va{v}(\va{r}, t)$,
+that is, the velocity can change with time ($t$-dependence),
+and depends on which parcel we track ($\va{r}$-dependence).
+
+Of course, the parcel is in our imagination:
+$\va{r}$ does not really depend on $t$;
+after all, we are dealing with a continuum.
+Nevertheless, the right-hand side of the equation is very useful,
+and is known as the **material derivative** or **comoving derivative**:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D}f}{\mathrm{D}t}
+        \equiv \pdv{f}{t} + (\va{v} \cdot \nabla) f
+    }
+\end{aligned}$$
+
+The first term is called the **local rate of change**,
+and the second is the **advective rate of change**.
+In effect, the latter moves the frame of reference along with the material,
+so that we can find the evolution of $f$
+without needing to worry about the continuum's motion.
+
+That was for a scalar field $f(\va{r}, t)$,
+but in fact the definition also works for vector fields $\va{U}(\va{r}, t)$:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{U}}{\mathrm{D}t}
+        \equiv \pdv{\va{U}}{t} + (\va{v} \cdot \nabla) \va{U}
+    }
+\end{aligned}$$
+
+Where the advective term is to be evaluated in the following way in Cartesian coordinates:
+
+$$\begin{aligned}
+    (\va{v} \cdot \nabla) \va{U}
+    =
+    \begin{bmatrix} v_x \\ v_y \\ v_z \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        \displaystyle\pdv{U_x}{x} & \displaystyle\pdv{U_x}{y} & \displaystyle\pdv{U_x}{z} \\
+        \displaystyle\pdv{U_y}{x} & \displaystyle\pdv{U_y}{y} & \displaystyle\pdv{U_y}{z} \\
+        \displaystyle\pdv{U_z}{x} & \displaystyle\pdv{U_z}{y} & \displaystyle\pdv{U_z}{z}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        v_x \displaystyle\pdv{U_x}{x} & v_y \displaystyle\pdv{U_x}{y} & v_z \displaystyle\pdv{U_x}{z} \\
+        v_x \displaystyle\pdv{U_y}{x} & v_y \displaystyle\pdv{U_y}{y} & v_z \displaystyle\pdv{U_y}{z} \\
+        v_x \displaystyle\pdv{U_z}{x} & v_y \displaystyle\pdv{U_z}{y} & v_z \displaystyle\pdv{U_z}{z}
+    \end{bmatrix}
+\end{aligned}$$
+
+
+
+## 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/matsubara-greens-function/index.md b/source/know/concept/matsubara-greens-function/index.md
new file mode 100644
index 0000000..d54bbf2
--- /dev/null
+++ b/source/know/concept/matsubara-greens-function/index.md
@@ -0,0 +1,390 @@
+---
+title: "Matsubara Green's function"
+date: 2021-11-12
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+The **Matsubara Green's function** is an
+[imaginary-time](/know/concept/imaginary-time/) version
+of the real-time [Green's functions](/know/concept/greens-functions/).
+We define as follows in the imaginary-time
+[Heisenberg picture](/know/concept/heisenberg-picture/):
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(\tau, \tau')
+        \equiv -\frac{1}{\hbar} \Expval{\mathcal{T} \big\{ \hat{A}(\tau) \hat{B}(\tau') \big\}}
+    }
+\end{aligned}$$
+
+Where the expectation value $\Expval{}$ is with respect to thermodynamic equilibrium,
+and $\mathcal{T}$ is the [time-ordered product](/know/concept/time-ordered-product/) pseudo-operator.
+Because the Hamiltonian $\hat{H}$ cannot depend on the imaginary time,
+$C_{AB}$ is a function of the difference $\tau \!-\! \tau'$ only:
+
+$$\begin{aligned}
+    C_{AB}(\tau, \tau')
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar}
+    e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big)
+    \\
+    &= - \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} \Big)
+\end{aligned}$$
+
+For $\tau > \tau'$, we see by expanding in the many-particle eigenstates $\Ket{n}$
+that we need to demand $\hbar \beta > \tau \!-\! \tau'$ to prevent
+$C_{AB}$ from diverging for increasing temperatures:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau')
+    &= - \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{e^{-\beta \hat{H}} e^{(\tau - \tau') \hat{H} / \hbar}
+    \hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n}
+    \\
+    &= - \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{A} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B}}{n} e^{-\beta E_n} e^{(\tau - \tau') E_n / \hbar}
+\end{aligned}$$
+
+And likewise, for $\tau < \tau'$,
+we must demand that $\tau \!-\! \tau' > -\hbar \beta$
+for the same reason:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau')
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{B}(\tau') \hat{A}(\tau) \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{B} e^{(\tau - \tau') \hat{H} / \hbar} \hat{A}}{n} e^{-\beta E_n} e^{- (\tau - \tau') E_n / \hbar}
+\end{aligned}$$
+
+With $-$ for bosons, and $+$ for fermions,
+due to the time-ordered product for $\tau > \tau'$.
+
+On this domain $[-\hbar \beta, \hbar \beta]$,
+the Matsubara Green's function $C_{AB}$
+obeys a useful shift relation:
+it is $\hbar \beta$-periodic for bosons,
+and $\hbar \beta$-antiperiodic for fermions:
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(\tau \!-\! \tau') =
+        \begin{cases}
+            \pm C_{AB}(\tau \!-\! \tau' \!+\! \hbar \beta)
+            & \mathrm{if\;} \tau \!-\! \tau' < 0
+            \\
+            \pm C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta)
+            & \mathrm{if\;} \tau \!-\! \tau' > 0
+        \end{cases}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-period"/>
+<label for="proof-period">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-period">Proof.</label>
+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:
+
+$$\begin{aligned}
+    C_{AB}(\tau \!-\! \tau' \!-\! \hbar \beta)
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{-(\tau - \tau' - \hbar \beta) \hat{H} / \hbar}
+    \hat{B} e^{(\tau - \tau' - \hbar \beta) \hat{H} / \hbar} \hat{A} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-(\tau - \tau') \hat{H} / \hbar} \hat{B} e^{(\tau - \tau') \hat{H} / \hbar} e^{-\beta \hat{H}} \hat{A} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} e^{\tau \hat{H} / \hbar} \hat{A} e^{-\tau \hat{H} / \hbar}
+    e^{\tau' \hat{H} / \hbar} \hat{B} e^{-\tau' \hat{H} / \hbar} \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \hat{A}(\tau) \hat{B}(\tau') \Big)
+    \\
+    &= \mp \frac{1}{\hbar Z} \Tr\!\Big( e^{-\beta \hat{H}} \mathcal{T}\big\{ \hat{A}(\tau) \hat{B}(\tau') \big\} \Big)
+    \\
+    &= \pm C_{AB}(\tau \!-\! \tau')
+\end{aligned}$$
+</div>
+</div>
+
+Due to this limited domain $\tau \in [-\hbar \beta, \hbar \beta]$,
+the [Fourier transform](/know/concept/fourier-transform/)
+of $C_{AB}(\tau)$ consists of discrete frequencies
+$k_n \equiv n \pi / (\hbar \beta)$.
+The forward and inverse Fourier transforms
+are therefore defined as given below (with $\tau' = 0$).
+It is convention to write $C_{AB}(i k_n)$ instead of $C_{AB}(k_n)$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            C_{AB}(i k_n)
+            &\equiv \frac{1}{2} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+            \\
+            C_{AB}(\tau)
+            &= \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty C_{AB}(i k_n) e^{-i k_n \tau}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-FT-def"/>
+<label for="proof-FT-def">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-FT-def">Proof.</label>
+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:
+
+$$\begin{aligned}
+    C_{AB}(\tau)
+    &= \frac{\pi}{\hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta\big( (\tau' \!-\! \tau) \pi / \hbar \beta \big) \dd{\tau'}
+    \\
+    &= \frac{\pi \hbar \beta}{\pi \hbar \beta} \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'}
+    \\
+    &= \int_{-\hbar \beta}^{\hbar \beta} C_{AB}(\tau') \: \delta(\tau' \!-\! \tau) \dd{\tau'}
+    \\
+    &= C_{AB}(\tau)
+\end{aligned}$$
+</div>
+</div>
+
+Let us now define the **Matsubara frequencies** $\omega_n$
+as a species-dependent subset of $k_n$:
+
+$$\begin{aligned}
+    \boxed{
+        \omega_n \equiv
+        \begin{cases}
+            \displaystyle\frac{2 n \pi}{\hbar \beta}
+            & \mathrm{bosons}
+            \\
+            \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta}
+            & \mathrm{fermions}
+        \end{cases}
+    }
+\end{aligned}$$
+
+With this, we can rewrite the definition of the forward Fourier transform as follows:
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(i \omega_n)
+        = \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i \omega_n \tau} \dd{\tau}
+        = \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i \omega_n \tau} \dd{\tau}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-FT-alt"/>
+<label for="proof-FT-alt">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-FT-alt">Proof.</label>
+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':
+
+$$\begin{aligned}
+    C_{AB}(i k_n)
+    &= \frac{1}{2} \int_{-\hbar \beta}^0 C_{AB}(\tau \!+\! \hbar \beta) \: e^{i k_n (\tau + \hbar \beta)} \dd{\tau}
+    + \frac{1}{2} \int_0^{\hbar \beta} C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+    \\
+    &= \frac{1}{2} \int_{-\hbar \beta}^0 \Big( C_{AB}(\tau) \pm C_{AB}(\tau) \: e^{i k_n \hbar \beta} \Big) \: e^{i k_n \tau} \dd{\tau}
+    \\
+    &= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
+\end{aligned}$$
+</div>
+</div>
+
+If we actually evaluate this,
+we obtain the following form of $C_{AB}$,
+which is almost identical to the
+[Lehmann representation](/know/concept/lehmann-representation/)
+of the "ordinary" retarded and advanced Green's functions:
+
+$$\begin{aligned}
+    \boxed{
+        C_{AB}(i \omega_m)
+        = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{i \hbar \omega_m + E_n - E_{n'}}
+        \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-Lehmann"/>
+<label for="proof-Lehmann">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-Lehmann">Proof.</label>
+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.
+</div>
+</div>
+
+This gives us the primary use of the Matsubara Green's function $C_{AB}$:
+calculating the retarded $C_{AB}^R$ and advanced $C_{AB}^A$.
+Once we have an expression for Matsubara's $C_{AB}$,
+we can recover $C_{AB}^R$ and $C_{AB}^A$ by substituting
+$i \omega_m \to \omega \!+\! i \eta$ and $i \omega_m \to \omega \!-\! i \eta$ respectively.
+
+In general, we can define the **canonical Green's function** $C_{AB}(z)$
+on the complex plane:
+
+$$\begin{aligned}
+    C_{AB}(z)
+    = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{A}}{n'} \matrixel{n'}{\hat{B}}{n}}{z + E_n - E_{n'}}
+    \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big)
+\end{aligned}$$
+
+This is a [holomorphic function](/know/concept/holomorphic-function/),
+except for poles on the real axis.
+It turns out that $C_{AB}(z)$ must have these properties
+for the substitution $i \omega_n \to \omega \!\pm\! i \eta$ to be valid.
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md
new file mode 100644
index 0000000..b61bc9a
--- /dev/null
+++ b/source/know/concept/matsubara-sum/index.md
@@ -0,0 +1,142 @@
+---
+title: "Matsubara sum"
+date: 2021-11-13
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+A **Matsubara sum** is a summation of the following form,
+which notably appears as the inverse
+[Fourier transform](/know/concept/fourier-transform/) of the
+[Matsubara Green's function](/know/concept/matsubara-greens-function/):
+
+$$\begin{aligned}
+    S_{B,F}
+    = \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
+\end{aligned}$$
+
+Where $i \omega_n$ are the Matsubara frequencies
+for bosons ($B$) or fermions ($F$),
+and $g(z)$ is a function on the complex plane
+that is [holomorphic](/know/concept/holomorphic-function/)
+except for a known set of simple poles,
+and $\tau$ is a real parameter
+(e.g. the [imaginary time](/know/concept/imaginary-time/))
+satisfying $-\hbar \beta < \tau < \hbar \beta$.
+
+Now, consider the following integral
+over a (for now) unspecified counter-clockwise contour $C$,
+with a (for now) unspecified weighting function $h(z)$:
+
+$$\begin{aligned}
+    \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
+    = \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big)
+\end{aligned}$$
+
+Where we have applied the residue theorem
+to get a sum over all simple poles $z_p$
+of either $g$ or $h$ (but not both) enclosed by $C$.
+Clearly, we could make this look like a Matsubara sum,
+if we choose $h$ such that it has poles at $i \omega_n$.
+
+Therefore, we choose the weighting function $h(z)$ as follows,
+where $n_B(z)$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
+and $n_F(z)$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/):
+
+$$\begin{aligned}
+    h(z)
+    =
+    \begin{cases}
+        n_{B,F}(z) & \mathrm{if}\; \tau \ge 0
+        \\
+        -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0
+    \end{cases}
+    \qquad \qquad
+    n_{B,F}(z)
+    = \frac{1}{e^{\hbar \beta z} \mp 1}
+\end{aligned}$$
+
+The distinction between the signs of $\tau$ is needed
+to ensure that the integrand $h(z) e^{z \tau}$ decays for $|z| \to \infty$,
+both for $\Real(z) > 0$ and $\Real(z) < 0$.
+This choice of $h$ indeed has poles at the respective
+Matsubara frequencies $i \omega_n$ of bosons and fermions,
+and the residues are:
+
+$$\begin{aligned}
+    \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_B(z) \big)
+    &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} - 1} \bigg)
+    = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} - 1} \bigg)
+    \\
+    &= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} - 1} \bigg)
+    = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{1 + \hbar \beta \eta - 1} \bigg)
+    = \frac{1}{\hbar \beta}
+    \\
+    \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_F(z) \big)
+    &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} + 1} \bigg)
+    = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} + 1} \bigg)
+    \\
+    &= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} + 1} \bigg)
+    = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{- 1 - \hbar \beta \eta + 1} \bigg)
+    = - \frac{1}{\hbar \beta}
+\end{aligned}$$
+
+In the definition of $h$, the sign flip for $\tau \le 0$
+is introduced because negating the argument also negates the residues,
+i.e. $\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$.
+With this $h$, our contour integral can be rewritten as follows:
+
+$$\begin{aligned}
+    \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
+    &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
+    + \sum_{i \omega_n} e^{i \omega_n \tau} g(i \omega_n) \: \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_{B,F}(z) \big)
+    \\
+    &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
+    \pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
+\end{aligned}$$
+
+Where $+$ is for bosons, and $-$ for fermions.
+Here, we recognize the last term as the Matsubara sum $S_{F,B}$,
+for which we isolate, yielding:
+
+$$\begin{aligned}
+    S_{B,F}
+    = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
+    \pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
+\end{aligned}$$
+
+Now we must choose $C$. Assuming $g(z)$ does not interfere,
+we know that $h(z) e^{z \tau}$ decays to zero
+for $|z| \to \infty$, so a useful choice would be a circle of radius $R$.
+If we then let $R \to \infty$, the contour encloses
+the whole complex plane, including all of the integrand's poles.
+However, thanks to the integrand's decay,
+the resulting contour integral must vanish:
+
+$$\begin{aligned}
+    C
+    = R e^{i \theta}
+    \quad \implies \quad
+    \lim_{R \to \infty}
+    \oint_C g(z) \: h(z) \: e^{z \tau} \dd{z}
+    = 0
+\end{aligned}$$
+
+We thus arrive at the following results
+for bosonic and fermionic Matsubara sums $S_{B,F}$:
+
+$$\begin{aligned}
+    \boxed{
+        S_{B,F}
+        = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{ {z \to z_p}}{\mathrm{Res}}\big(g(z)\big)
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Bruus, K. Flensberg,
+    *Many-body quantum theory in condensed matter physics*,
+    2016, Oxford.
diff --git a/source/know/concept/maxwell-bloch-equations/index.md b/source/know/concept/maxwell-bloch-equations/index.md
new file mode 100644
index 0000000..86d7063
--- /dev/null
+++ b/source/know/concept/maxwell-bloch-equations/index.md
@@ -0,0 +1,447 @@
+---
+title: "Maxwell-Bloch equations"
+date: 2021-10-02
+categories:
+- Physics
+- Quantum mechanics
+- Two-level system
+- Electromagnetism
+- Laser theory
+layout: "concept"
+---
+
+For an electron in a two-level system with time-independent states
+$\Ket{g}$ (ground) and $\Ket{e}$ (excited),
+consider the following general solution
+to the full Schrödinger equation:
+
+$$\begin{aligned}
+    \Ket{\Psi}
+    &= c_g \: \Ket{g} \exp(-i E_g t / \hbar) + c_e \: \Ket{e} \exp(-i E_e t / \hbar)
+\end{aligned}$$
+
+Perturbing this system with
+an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+introduces a time-dependent sinusoidal term $\hat{H}_1$ to the Hamiltonian.
+In the [electric dipole approximation](/know/concept/electric-dipole-approximation/),
+$\hat{H}_1$ is given by:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = - \hat{\vb{p}} \cdot \vb{E}(t)
+    \qquad \qquad
+    \vu{p}
+    \equiv q \vu{x}
+    \qquad \qquad
+    \vb{E}(t)
+    = \vb{E}_0 \cos(\omega t)
+\end{aligned}$$
+
+Where $\vb{E}$ is an [electric field](/know/concept/electric-field/),
+and $\hat{\vb{p}}$ is the dipole moment operator.
+From [Rabi oscillation](/know/concept/rabi-oscillation/),
+we know that the time-varying coefficients $c_g$ and $c_e$
+can then be described by:
+
+$$\begin{aligned}
+    \dv{c_g}{t}
+    &= i \frac{q \matrixel{g}{\vu{x}}{e} \cdot \vb{E}_0}{2 \hbar} \exp\!\big( i \omega t \!-\! i \omega_0 t \big) \: c_e
+    \\
+    \dv{c_e}{t}
+    &= i \frac{q \matrixel{e}{\vu{x}}{g} \cdot \vb{E}_0}{2 \hbar} \exp\!\big(\!-\! i \omega t \!+\! i \omega_0 t \big) \: c_g
+\end{aligned}$$
+
+We want to rearrange these equations a bit.
+Therefore, we split the electric field $\vb{E}$ like so,
+where the amplitudes $\vb{E}_0^{-}$ and $\vb{E}_0^{+}$ may be slowly varying:
+
+$$\begin{aligned}
+    \vb{E}(t)
+    = \vb{E}^{-}(t) + \vb{E}^{+}(t)
+    = \vb{E}_0^{-} \exp(i \omega t) + \vb{E}_0^{+} \exp(-i \omega t)
+\end{aligned}$$
+
+Since $\vb{E}$ is real, $\vb{E}_0^{+} = (\vb{E}_0^{-})^*$.
+Similarly, we define the transition dipole moment $\vb{p}_0^{-}$:
+
+$$\begin{aligned}
+    \vb{p}_0^{-}
+    \equiv q \matrixel{e}{\vu{x}}{g}
+    \qquad \qquad
+    \vb{p}_0^{+}
+    \equiv (\vb{p}_0^{-})^*
+    = q \matrixel{g}{\vu{x}}{e}
+\end{aligned}$$
+
+With these, the equations for $c_g$ and $c_e$ can be rewritten as shown below.
+Note that $\vb{E}^{-}$ and $\vb{E}^{+}$ include the driving plane wave, and the
+[rotating wave approximation](/know/concept/rotating-wave-approximation/) is still made:
+
+$$\begin{aligned}
+    \dv{c_g}{t}
+    &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp(- i \omega_0 t) \: c_e
+    \\
+    \dv{c_e}{t}
+    &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp(i \omega_0 t) \: c_g
+\end{aligned}$$
+
+
+## Optical Bloch equations
+
+For $\Ket{\Psi}$ as defined above,
+the corresponding pure [density operator](/know/concept/density-operator/)
+$\hat{\rho}$ is as follows:
+
+$$\begin{aligned}
+    \hat{\rho}
+    = \Ket{\Psi} \Bra{\Psi}
+    =
+    \begin{bmatrix}
+        c_e c_e^* & c_e c_g^* \exp(-i \omega_0 t) \\
+        c_g c_e^* \exp(i \omega_0 t) & c_g c_g^*
+    \end{bmatrix}
+    \equiv
+    \begin{bmatrix}
+        \rho_{ee} & \rho_{eg} \\
+        \rho_{ge} & \rho_{gg}
+    \end{bmatrix}
+\end{aligned}$$
+
+Where $\omega_0 \equiv (E_e \!-\! E_g) / \hbar$ is the resonance frequency.
+We take the $t$-derivative of the matrix elements,
+and insert the equations for $c_g$ and $c_e$:
+
+$$\begin{aligned}
+    \dv{\rho_{gg}}{t}
+    &= \dv{c_g}{t} c_g^* + c_g \dv{c_g^*}{t}
+    \\
+    &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp(- i \omega_0 t) \: c_e c_g^*
+    - \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp(i \omega_0 t) \: c_g c_e^*
+    \\
+    \dv{\rho_{ee}}{t}
+    &= \dv{c_e}{t} c_e^* + c_e \dv{c_e^*}{t}
+    \\
+    &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \exp(i \omega_0 t) \: c_g c_e^*
+    - \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \exp(- i \omega_0 t) \: c_e c_g^*
+    \\
+    \dv{\rho_{ge}}{t}
+    &= \dv{c_g}{t} c_e^* \exp(i \omega_0 t) + c_g \dv{c_e^*}{t} \exp(i \omega_0 t) + i \omega_0 c_g c_e^* \exp(i \omega_0 t)
+    \\
+    &= \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \: c_e c_e^*
+    - \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \: c_g c_g^*
+    + i \omega_0 c_g c_e^* \exp(i \omega_0 t)
+    \\
+    \dv{\rho_{eg}}{t}
+    &= \dv{c_e}{t} c_g^* \exp(-i \omega_0 t) + c_e \dv{c_g^*}{t} \exp(-i \omega_0 t) - i \omega_0 c_e c_g^* \exp(- i \omega_0 t)
+    \\
+    &= \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \: c_g c_g^*
+    - \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \: c_e c_e^*
+    - i \omega_0 c_e c_g^* \: \exp(- i \omega_0 t)
+\end{aligned}$$
+
+Recognizing the density matrix elements allows us
+to reduce these equations to:
+
+$$\begin{aligned}
+    \dv{\rho_{gg}}{t}
+    &= \frac{i}{\hbar} \Big( \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} - \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} \Big)
+    \\
+    \dv{\rho_{ee}}{t}
+    &= \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} - \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} \Big)
+    \\
+    \dv{\rho_{ge}}{t}
+    &= i \omega_0 \rho_{ge} + \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \big( \rho_{ee} - \rho_{gg} \big)
+    \\
+    \dv{\rho_{eg}}{t}
+    &= - i \omega_0 \rho_{eg} + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \big( \rho_{gg} - \rho_{ee} \big)
+\end{aligned}$$
+
+These equations are correct if nothing else is affecting $\hat{\rho}$.
+But in practice, these quantities decay due to various processes,
+e.g. spontaneous emission (see [Einstein coefficients](/know/concept/einstein-coefficients/)).
+
+Let $\rho_{ee}$ decays with rate $\gamma_e$.
+Since the total probability $\rho_{ee} + \rho_{gg} = 1$,
+we thus have:
+
+$$\begin{aligned}
+    \Big( \dv{\rho_{ee}}{t} \Big)_{e}
+    = - \gamma_e \rho_{ee}
+    \quad \implies \quad
+    \Big( \dv{\rho_{gg}}{t} \Big)_{e}
+    = \gamma_e \rho_{ee}
+\end{aligned}$$
+
+Meanwhile, for whatever reason,
+let $\rho_{gg}$ decay into $\rho_{ee}$ with rate $\gamma_g$:
+
+$$\begin{aligned}
+    \Big( \dv{\rho_{gg}}{t} \Big)_{g}
+    = - \gamma_g \rho_{gg}
+    \quad \implies \quad
+    \Big( \dv{\rho_{gg}}{t} \Big)_{g}
+    = \gamma_g \rho_{gg}
+\end{aligned}$$
+
+And finally, let the diagonal (perpendicular) matrix elements
+both decay with rate $\gamma_\perp$:
+
+$$\begin{aligned}
+    \Big( \dv{\rho_{eg}}{t} \Big)_{\perp}
+    = - \gamma_\perp \rho_{eg}
+    \qquad \qquad
+    \Big( \dv{\rho_{ge}}{t} \Big)_{\perp}
+    = - \gamma_\perp \rho_{ge}
+\end{aligned}$$
+
+Putting everything together,
+we arrive at the **optical Bloch equations** governing $\hat{\rho}$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \dv{\rho_{gg}}{t}
+            &= \gamma_e \rho_{ee} - \gamma_g \rho_{gg}
+            + \frac{i}{\hbar} \Big( \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} - \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} \Big)
+            \\
+            \dv{\rho_{ee}}{t}
+            &=  \gamma_g \rho_{gg} - \gamma_e \rho_{ee}
+            + \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \rho_{ge} - \vb{p}_0^{+} \cdot \vb{E}^{-} \rho_{eg} \Big)
+            \\
+            \dv{\rho_{ge}}{t}
+            &= - \Big( \gamma_\perp - i \omega_0 \Big) \rho_{ge}
+            + \frac{i}{\hbar} \vb{p}_0^{+} \cdot \vb{E}^{-} \Big( \rho_{ee} - \rho_{gg} \Big)
+            \\
+            \dv{\rho_{eg}}{t}
+            &= - \Big( \gamma_\perp + i \omega_0 \Big) \rho_{eg}
+            + \frac{i}{\hbar} \vb{p}_0^{-} \cdot \vb{E}^{+} \Big( \rho_{gg} - \rho_{ee} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Many authors simplify these equations a bit by choosing
+$\gamma_g = 0$ and $\gamma_\perp = \gamma_e / 2$.
+
+
+## Including Maxwell's equations
+
+This two-level system has a dipole moment $\vb{p}$ as follows,
+where we use [Laporte's selection rule](/know/concept/selection-rules/)
+to remove diagonal terms, by assuming that
+the electron's orbitals are odd or even:
+
+$$\begin{aligned}
+    \vb{p}
+    &= \matrixel{\Psi}{\hat{\vb{p}}}{\Psi}
+    \\
+    &= q \Big( c_g c_g^* \matrixel{g}{\vu{x}}{g} + c_e c_e^* \matrixel{e}{\vu{x}}{e}
+    + c_g c_e^* \matrixel{e}{\vu{x}}{g} \exp(i \omega_0 t) + c_e c_g^* \matrixel{g}{\vu{x}}{e} \exp(-i \omega_0 t) \Big)
+    \\
+    &= q \Big( \rho_{ge} \matrixel{e}{\vu{x}}{g} + \rho_{eg} \matrixel{g}{\vu{x}}{e} \Big)
+    = \vb{p}_0^{-} \rho_{ge}(t) + \vb{p}_0^{+} \rho_{eg}(t)
+    \equiv \vb{p}^{-}(t) + \vb{p}^{+}(t)
+\end{aligned}$$
+
+Where we have split $\vb{p}$ analogously to $\vb{E}$
+by defining $\vb{p}^{+} \equiv \vb{p}_0^{+} \rho_{eg}$.
+Its equation of motion can then be found from the optical Bloch equations:
+
+$$\begin{aligned}
+    \dv{\vb{p}^{+}}{t}
+    = \vb{p}_0^{+} \dv{\rho_{eg}}{t}
+    = - \vb{p}_0^{+} \Big( \gamma_\perp + i \omega_0 \Big) \rho_{eg}
+    + \frac{i}{\hbar} \vb{p}_0^{+} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \Big( \rho_{gg} - \rho_{ee} \Big)
+\end{aligned}$$
+
+Some authors do not bother multiplying $\rho_{ge}$ by $\vb{p}_0^{+}$.
+In any case, we arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{\vb{p}^{+}}{t}
+        = - \Big( \gamma_\perp + i \omega_0 \Big) \vb{p}^{+}
+        - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} d
+    }
+\end{aligned}$$
+
+Where we have defined the **population inversion** $d \in [-1, 1]$ as follows,
+which quantifies the electron's excitedness:
+
+$$\begin{aligned}
+    d
+    \equiv \rho_{ee} - \rho_{gg}
+\end{aligned}$$
+
+From the optical Bloch equations,
+we find its equation of motion to be:
+
+$$\begin{aligned}
+    \dv{d}{t}
+    &= \dv{\rho_{ee}}{t} - \dv{\rho_{gg}}{t}
+    = 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+    + \frac{i 2}{\hbar} \Big( \vb{p}^{-} \cdot \vb{E}^{+} - \vb{p}^{+} \cdot \vb{E}^{-} \Big)
+\end{aligned}$$
+
+We can rewrite the first two terms in the following intuitive form,
+which describes a decay with
+rate $\gamma_\parallel \equiv \gamma_g + \gamma_e$
+towards an equilbrium $d_0$:
+
+$$\begin{aligned}
+    2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+    = \gamma_\parallel (d_0 - d)
+    \qquad \qquad
+    d_0
+    \equiv \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-inversion-decay"/>
+<label for="proof-inversion-decay">Proof</label>
+<div class="hidden">
+<label for="proof-inversion-decay">Proof.</label>
+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:
+
+$$\begin{aligned}
+    2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+    &= \gamma_g - \gamma_e - (\gamma_g + \gamma_e) d
+    \\
+    &= (\gamma_g + \gamma_e) \Big( \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e} - d \Big)
+    \\
+    &= \gamma_\parallel ( d_0 - d )
+\end{aligned}$$
+</div>
+</div>
+
+With this, the equation for the population inversion $d$
+takes the following final form:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{d}{t}
+        = \gamma_\parallel (d_0 - d) + \frac{i 2}{\hbar} \Big( \vb{p}^{-} \cdot \vb{E}^{+} - \vb{p}^{+} \cdot \vb{E}^{-} \Big)
+    }
+\end{aligned}$$
+
+Finally, we would like a relation between the polarization
+and the electric field $\vb{E}$,
+for which we turn to [Maxwell's equations](/know/concept/maxwells-equations/).
+We start from Faraday's law,
+and split $\vb{B} = \mu_0 (\vb{H} + \vb{M})$:
+
+$$\begin{aligned}
+    \nabla \cross \vb{E}
+    = - \pdv{\vb{B}}{t}
+    = - \mu_0 \pdv{\vb{H}}{t} - \mu_0 \pdv{\vb{M}}{t}
+\end{aligned}$$
+
+We assume that there is no magnetization $\vb{M} = 0$.
+Then we we take the curl of both sides,
+and replace $\nabla \cross \vb{H}$ with Ampère's circuital law:
+
+$$\begin{aligned}
+    \nabla \cross \big( \nabla \cross \vb{E} \big)
+    = - \mu_0 \pdv{}{t} \big( \nabla \cross \vb{H} \big)
+    = - \mu_0 \pdv{}{t} \Big( \vb{J}_\mathrm{free} + \pdv{\vb{D}}{t} \Big)
+\end{aligned}$$
+
+Inserting the definition $\vb{D} = \varepsilon_0 \vb{E} + \vb{P}$
+together with Ohm's law $\vb{J}_\mathrm{free} = \sigma \vb{E}$ yields:
+
+$$\begin{aligned}
+    \nabla \cross \big( \nabla \cross \vb{E} \big)
+    = - \mu_0 \sigma \pdv{\vb{E}}{t} - \mu_0 \varepsilon_0 \pdvn{2}{\vb{E}}{t} - \mu_0 \pdvn{2}{\vb{P}}{t}
+\end{aligned}$$
+
+Where $\sigma$ is the active material's conductivity, if any;
+almost all authors assume $\sigma = 0$.
+
+Recall that we are describing the dynamics of a two-level system.
+In reality, such a system (e.g. a quantum dot)
+is suspended in a passive background medium,
+which reacts with a polarization $\vb{P}_\mathrm{med}$
+to the electric field $\vb{E}$.
+If the medium is linear, i.e. $\vb{P}_\mathrm{med} = \varepsilon_0 \chi \vb{E}$,
+then:
+
+$$\begin{aligned}
+    \mu_0 \pdvn{2}{\vb{P}}{t}
+    &= - \nabla \cross \big( \nabla \cross \vb{E} \big) - \mu_0 \sigma \pdv{\vb{E}}{t}
+    - \mu_0 \varepsilon_0 \pdvn{2}{\vb{E}}{t} - \mu_0 \pdvn{2}{\vb{P}_\mathrm{med}}{t}
+    \\
+    &= - \nabla \cross \big( \nabla \cross \vb{E} \big) - \mu_0 \sigma \pdv{\vb{E}}{t}
+    - \mu_0 \pdvn{2}{}{t}\Big( \varepsilon_0 \vb{E} + \varepsilon_0 \chi \vb{E} \Big)
+    \\
+    &= - \nabla \cross \big( \nabla \cross \vb{E} \big) - \mu_0 \sigma \pdv{\vb{E}}{t}
+    - \mu_0 \varepsilon_0 \varepsilon_r \pdvn{2}{\vb{E}}{t}
+\end{aligned}$$
+
+Where $\varepsilon_r \equiv 1 + \chi_e$ is the medium's relative permittivity.
+The speed of light $c^2 = 1 / (\mu_0 \varepsilon_0)$,
+and the refractive index $n^2 = \mu_r \varepsilon_r$,
+where $\mu_r = 1$ due to our assumption that $\vb{M} = 0$, so:
+
+$$\begin{aligned}
+    \boxed{
+        \mu_0 \pdvn{2}{\vb{P}}{t}
+        = - \nabla \cross \big( \nabla \cross \vb{E} \big) - \mu_0 \sigma \pdv{\vb{E}}{t} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}}{t}
+    }
+\end{aligned}$$
+
+$\vb{E}$ and $\vb{P}$ can trivially be replaced by $\vb{E}^{+}$ and $\vb{P}^{+}$.
+It is also simple to convert $\vb{p}^{+}$ and $d$
+into the macroscopic $\vb{P}^{+}$ and total $D$
+by summing over all two-level systems in the medium:
+
+$$\begin{aligned}
+    \vb{P}^{+}(\vb{x}, t)
+    &= \sum_{\nu} \vb{p}^{+}_\nu \: \delta(\vb{x} - \vb{x}_\nu)
+    \\
+    D(\vb{x}, t)
+    &= \sum_{\nu} d_\nu \: \delta(\vb{x} - \vb{x}_\nu)
+\end{aligned}$$
+
+We thus arrive at the **Maxwell-Bloch equations**,
+which are the foundation of laser theory:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \mu_0 \pdvn{2}{\vb{P}^{+}}{t}
+            &= - \nabla \cross \nabla \cross \vb{E}^{+} - \mu_0 \sigma \pdv{\vb{E}^{+}}{t} - \frac{n^2}{c^2} \pdvn{2}{\vb{E}^{+}}{t}
+            \\
+            \pdv{\vb{P}^{+}}{t}
+            &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+}
+            - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} D
+            \\
+            \pdv{D}{t}
+            &= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  F. Kärtner,
+    [Ultrafast optics: lecture notes](https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-977-ultrafast-optics-spring-2005/lecture-notes/),
+    2005, MIT.
+2.  H. Haken,
+    *Light: volume 2: laser light dynamics*,
+    1985, North-Holland.
+3.  H.J. Metcalf, P. van der Straten,
+    *Laser cooling and trapping*,
+    1999, Springer.
diff --git a/source/know/concept/maxwell-boltzmann-distribution/index.md b/source/know/concept/maxwell-boltzmann-distribution/index.md
new file mode 100644
index 0000000..5595514
--- /dev/null
+++ b/source/know/concept/maxwell-boltzmann-distribution/index.md
@@ -0,0 +1,214 @@
+---
+title: "Maxwell-Boltzmann distribution"
+date: 2021-05-08
+categories:
+- Physics
+- Statistics
+- Thermodynamics
+layout: "concept"
+---
+
+The **Maxwell-Boltzmann distributions** are a set of closely related
+probability distributions with applications in classical statistical physics.
+
+
+## Velocity vector distribution
+
+In the [canonical ensemble](/know/concept/canonical-ensemble/)
+(where a fixed-size system can exchange energy with its environment),
+the probability of a microstate with energy $E$ is given by the Boltzmann distribution:
+
+$$\begin{aligned}
+    f(E)
+    \:\propto\: \exp\!\big(\!-\! \beta E\big)
+\end{aligned}$$
+
+Where $\beta = 1 / k_B T$.
+We split $E = K + U$,
+with $K$ and $U$ the total kinetic and potential energy contributions.
+If there are $N$ particles in the system,
+with positions $\tilde{r} = (\vec{r}_1, ..., \vec{r}_N)$
+and momenta $\tilde{p} = (\vec{p}_1, ..., \vec{p}_N)$,
+then $K$ only depends on $\tilde{p}$,
+and $U$ only depends on $\tilde{r}$,
+so the probability of a specific microstate
+$(\tilde{r}, \tilde{p})$ is as follows:
+
+$$\begin{aligned}
+    f(\tilde{r}, \tilde{p})
+    \:\propto\: \exp\!\Big(\!-\! \beta \big( K(\tilde{p}) + U(\tilde{r}) \big) \Big)
+\end{aligned}$$
+
+Since this is classical physics,
+we can split the exponential.
+In quantum mechanics,
+the canonical commutation relation would prevent that.
+Anyway, splitting yields:
+
+$$\begin{aligned}
+    f(\tilde{r}, \tilde{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big) \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+\end{aligned}$$
+
+Classically, the probability
+distributions of the momenta and positions are independent:
+
+$$\begin{aligned}
+    f_K(\tilde{p})
+    \:\propto\: \exp\!\big(\!-\! \beta K(\tilde{p}) \big)
+    \qquad \qquad
+    f_U(\tilde{r})
+    \:\propto\: \exp\!\big(\!-\! \beta U(\tilde{r}) \big)
+\end{aligned}$$
+
+We cannot evaluate $f_U(\tilde{r})$ further without knowing $U(\tilde{r})$ for a system.
+We thus turn to $f_K(\tilde{p})$, and see that the total kinetic
+energy $K(\tilde{p})$ is simply the sum of the particles' individual
+kinetic energies $K_n(\vec{p}_n)$, which are well-known:
+
+$$\begin{aligned}
+    K(\tilde{p})
+    = \sum_{n = 1}^N K_n(\vec{p}_n)
+    \qquad \mathrm{where} \qquad
+    K_n(\vec{p}_n)
+    = \frac{|\vec{p}_n|^2}{2 m}
+\end{aligned}$$
+
+Consequently, the probability distribution $f(p_x, p_y, p_z)$ for the
+momentum vector of a single particle is as follows,
+after normalization:
+
+$$\begin{aligned}
+    f(p_x, p_y, p_z)
+    = \Big( \frac{1}{2 \pi m k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{(p_x^2 + p_y^2 + p_z^2)}{2 m k_B T} \Big)
+\end{aligned}$$
+
+We now rewrite this using the velocities $v_x = p_x / m$,
+and update the normalization, giving:
+
+$$\begin{aligned}
+    \boxed{
+        f(v_x, v_y, v_z)
+        = \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} \exp\!\Big( \!-\!\frac{m (v_x^2 + v_y^2 + v_z^2)}{2 k_B T} \Big)
+    }
+\end{aligned}$$
+
+This is the **Maxwell-Boltzmann velocity vector distribution**.
+Clearly, this is a product of three exponentials,
+so the velocity in each direction is independent of the others:
+
+$$\begin{aligned}
+    f(v_x)
+    = \sqrt{\frac{m}{2 \pi k_B T}} \exp\!\Big( \!-\!\frac{m v_x^2}{2 k_B T} \Big)
+\end{aligned}$$
+
+The distribution is thus an isotropic gaussian with standard deviations given by:
+
+$$\begin{aligned}
+    \sigma_x = \sigma_y = \sigma_z
+    = \sqrt{\frac{k_B T}{m}}
+\end{aligned}$$
+
+
+## Speed distribution
+
+We know the distribution of the velocities along each axis,
+but what about the speed $v = |\vec{v}|$?
+Because we do not care about the direction of $\vec{v}$, only its magnitude,
+the [density of states](/know/concept/density-of-states/) $g(v)$ is not constant:
+it is the rate-of-change of the volume of a sphere of radius $v$:
+
+$$\begin{aligned}
+    g(v)
+    = \dv{}{v}\Big( \frac{4 \pi}{3} v^3 \Big)
+    = 4 \pi v^2
+\end{aligned}$$
+
+Multiplying the velocity vector distribution by $g(v)$
+and substituting $v^2 = v_x^2 + v_y^2 + v_z^2$
+then gives us the **Maxwell-Boltzmann speed distribution**:
+
+$$\begin{aligned}
+    \boxed{
+        f(v)
+        = 4 \pi \Big( \frac{m}{2 \pi k_B T} \Big)^{3/2} v^2 \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
+    }
+\end{aligned}$$
+
+Some notable points on this distribution are
+the most probable speed $v_{\mathrm{mode}}$,
+the mean average speed $v_{\mathrm{mean}}$,
+and the root-mean-square speed $v_{\mathrm{rms}}$:
+
+$$\begin{aligned}
+    f'(v_\mathrm{mode})
+    = 0
+    \qquad
+    v_\mathrm{mean}
+    = \int_0^\infty v \: f(v) \dd{v}
+    \qquad
+    v_\mathrm{rms}
+    = \bigg( \int_0^\infty v^2 \: f(v) \dd{v} \bigg)^{1/2}
+\end{aligned}$$
+
+Which can be calculated to have the following exact expressions:
+
+$$\begin{aligned}
+    \boxed{
+        v_{\mathrm{mode}}
+        = \sqrt{\frac{2 k_B T}{m}}
+    }
+    \qquad
+    \boxed{
+        v_{\mathrm{mean}}
+        = \sqrt{\frac{8 k_B T}{\pi m}}
+    }
+    \qquad
+    \boxed{
+        v_{\mathrm{rms}}
+        = \sqrt{\frac{3 k_B T}{m}}
+    }
+\end{aligned}$$
+
+
+## Kinetic energy distribution
+
+Using the speed distribution,
+we can work out the kinetic energy distribution.
+Because $K$ is not proportional to $v$,
+we must do this by demanding that:
+
+$$\begin{aligned}
+    f(K) \dd{K}
+    = f(v) \dd{v}
+    \quad \implies \quad
+    f(K)
+    = f(v) \dv{v}{K}
+\end{aligned}$$
+
+We know that $K = m v^2 / 2$,
+meaning $\dd{K} = m v \dd{v}$
+so the energy distribution $f(K)$ is:
+
+$$\begin{aligned}
+    f(K)
+    = \frac{f(v)}{m v}
+    = \sqrt{\frac{2 m}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} v \exp\!\Big( \!-\!\frac{m v^2}{2 k_B T} \Big)
+\end{aligned}$$
+
+Substituting $v = \sqrt{2 K/m}$ leads to
+the **Maxwell-Boltzmann kinetic energy distribution**:
+
+$$\begin{aligned}
+    \boxed{
+        f(K)
+        = 2 \sqrt{\frac{K}{\pi}} \: \bigg( \frac{1}{k_B T} \bigg)^{3/2} \exp\!\Big( \!-\!\frac{K}{k_B T} \Big)
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/maxwell-relations/index.md b/source/know/concept/maxwell-relations/index.md
new file mode 100644
index 0000000..b546ef3
--- /dev/null
+++ b/source/know/concept/maxwell-relations/index.md
@@ -0,0 +1,290 @@
+---
+title: "Maxwell relations"
+date: 2021-07-08
+categories:
+- Physics
+- Thermodynamics
+layout: "concept"
+---
+
+The **Maxwell relations** are a useful set of relations in thermodynamics.
+They arise from the fact that the order of differentiation is irrelevant
+for well-behaved functions (sometimes known as the *Schwarz theorem*),
+applied to the [thermodynamic potentials](/know/concept/thermodynamic-potential/).
+
+We start by proving the general "recipe".
+Given that the differential element of some $z$ is defined in terms of
+two constant quantities $A$ and $B$ and two independent variables $x$ and $y$:
+
+$$\begin{aligned}
+    \dd{z} \equiv A \dd{x} + B \dd{y}
+\end{aligned}$$
+
+Then the quantities $A$ and $B$ can be extracted
+by dividing by $\dd{x}$ and $\dd{y}$ respectively:
+
+$$\begin{aligned}
+    A = \Big( \pdv{z}{x} \Big)_y
+    \qquad
+    B = \Big( \pdv{z}{y} \Big)_x
+\end{aligned}$$
+
+By differentiating $A$ and $B$,
+and using that the order of differentiation is irrelevant, we find:
+
+$$\begin{aligned}
+    \mpdv{z}{y}{x} =
+    \boxed{
+        \Big( \pdv{A}{y} \Big)_x
+        = \Big( \pdv{B}{x} \Big)_y
+    }
+    = \mpdv{z}{x}{y}
+\end{aligned}$$
+
+Using this, all Maxwell relations are derived.
+Each relation also has a reciprocal form:
+
+$$\begin{aligned}
+    \Big( \pdv{A}{y} \Big)_x^{-1} =
+    \boxed{
+        \Big( \pdv{y}{A} \Big)_x
+        = \Big( \pdv{x}{B} \Big)_y
+    }
+    = \Big( \pdv{B}{x} \Big)_y^{-1}
+\end{aligned}$$
+
+The following quantities are useful to rewrite some of the Maxwell relations:
+the iso-$P$ thermal expansion coefficient $\alpha$,
+the iso-$T$ combressibility $\kappa_T$,
+the iso-$S$ combressibility $\kappa_S$,
+the iso-$V$ heat capacity $C_V$,
+and the iso-$P$ heat capacity $C_P$:
+
+$$\begin{gathered}
+    \alpha \equiv \frac{1}{V} \Big( \pdv{V}{T} \Big)_{P,N}
+    \\
+    \kappa_T \equiv - \frac{1}{V} \Big( \pdv{V}{P} \Big)_{T,N}
+    \qquad \quad
+    \kappa_S \equiv - \frac{1}{V} \Big( \pdv{V}{P} \Big)_{S,N}
+    \\
+    C_V \equiv T \Big( \pdv{S}{T} \Big)_{V,N}
+    \qquad \qquad
+    C_P \equiv T \Big( \pdv{S}{T} \Big)_{P,N}
+\end{gathered}$$
+
+
+## Internal energy
+
+The following Maxwell relations can be derived
+from the internal energy $U(S, V, N)$:
+
+$$\begin{gathered}
+    \mpdv{U}{V}{S} =
+    \boxed{
+        \Big( \pdv{T}{V} \Big)_S = - \Big( \pdv{P}{S} \Big)_V
+    }
+    = \mpdv{U}{S}{V}
+    \\
+    \mpdv{U}{V}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{V} \Big)_N = - \Big( \pdv{P}{N} \Big)_V
+    }
+    = \mpdv{U}{N}{V}
+    \\
+    \mpdv{U}{S}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{S} \Big)_N = \Big( \pdv{T}{N} \Big)_S
+    }
+    = \mpdv{U}{N}{S}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{V}{T} \Big)_S = - \Big( \pdv{S}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{V}{\mu} \Big)_N = - \Big( \pdv{N}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{S}{\mu} \Big)_N = \Big( \pdv{N}{T} \Big)_S
+    }
+\end{gathered}$$
+
+
+## Enthalpy
+
+The following Maxwell relations can be derived
+from the enthalpy $H(S, P, N)$:
+
+$$\begin{gathered}
+    \mpdv{H}{P}{S} =
+    \boxed{
+        \Big( \pdv{T}{P} \Big)_S = \Big( \pdv{V}{S} \Big)_P
+    }
+    = \mpdv{H}{S}{P}
+    \\
+    \mpdv{H}{P}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{P} \Big)_N = \Big( \pdv{V}{N} \Big)_P
+    }
+    = \mpdv{H}{N}{P}
+    \\
+    \mpdv{H}{N}{S} =
+    \boxed{
+        \Big( \pdv{T}{N} \Big)_S = \Big( \pdv{\mu}{S} \Big)_N
+    }
+    = \mpdv{H}{S}{N}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{P}{T} \Big)_S = \Big( \pdv{S}{V} \Big)_P
+    }
+    \\
+    \boxed{
+        \Big( \pdv{P}{\mu} \Big)_N = \Big( \pdv{N}{V} \Big)_P
+    }
+    \\
+    \boxed{
+        \Big( \pdv{N}{T} \Big)_S = \Big( \pdv{S}{\mu} \Big)_N
+    }
+\end{gathered}$$
+
+
+## Helmholtz free energy
+
+The following Maxwell relations can be derived
+from the Helmholtz free energy $F(T, V, N)$:
+
+$$\begin{gathered}
+    - \mpdv{F}{V}{T} =
+    \boxed{
+        \Big( \pdv{S}{V} \Big)_T = \Big( \pdv{P}{T} \Big)_V
+    }
+    = - \mpdv{F}{T}{V}
+    \\
+    \mpdv{F}{V}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{V} \Big)_N = - \Big( \pdv{P}{N} \Big)_V
+    }
+    = \mpdv{F}{N}{V}
+    \\
+    \mpdv{F}{T}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{T} \Big)_N = - \Big( \pdv{S}{N} \Big)_T
+    }
+    = \mpdv{F}{N}{T}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{V}{S} \Big)_T = \Big( \pdv{T}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{V}{\mu} \Big)_N = - \Big( \pdv{N}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{T}{\mu} \Big)_N = - \Big( \pdv{N}{S} \Big)_T
+    }
+\end{gathered}$$
+
+
+## Gibbs free energy
+
+The following Maxwell relations can be derived
+from the Gibbs free energy $G(T, P, N)$:
+
+$$\begin{gathered}
+    \mpdv{G}{T}{P} =
+    \boxed{
+        \Big( \pdv{V}{T} \Big)_P = - \Big( \pdv{S}{P} \Big)_T
+    }
+    = \mpdv{G}{P}{T}
+    \\
+    \mpdv{G}{N}{P} =
+    \boxed{
+        \Big( \pdv{V}{N} \Big)_P = \Big( \pdv{\mu}{P} \Big)_N
+    }
+    = \mpdv{G}{P}{N}
+    \\
+    \mpdv{G}{T}{N} =
+    \boxed{
+        \Big( \pdv{\mu}{T} \Big)_N = - \Big( \pdv{S}{N} \Big)_T
+    }
+    = \mpdv{G}{N}{T}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{T}{V} \Big)_P = - \Big( \pdv{P}{S} \Big)_T
+    }
+    \\
+    \boxed{
+        \Big( \pdv{N}{V} \Big)_P = \Big( \pdv{P}{\mu} \Big)_N
+    }
+    \\
+    \boxed{
+        \Big( \pdv{T}{\mu} \Big)_N = - \Big( \pdv{N}{S} \Big)_T
+    }
+\end{gathered}$$
+
+
+## Landau potential
+
+The following Maxwell relations can be derived
+from the Gibbs free energy $\Omega(T, V, \mu)$:
+
+$$\begin{gathered}
+    - \mpdv{\Omega}{V}{T} =
+    \boxed{
+        \Big( \pdv{S}{V} \Big)_T = \Big( \pdv{P}{T} \Big)_V
+    }
+    = - \mpdv{\Omega}{T}{V}
+    \\
+    - \mpdv{\Omega}{\mu}{V} =
+    \boxed{
+        \Big( \pdv{P}{\mu} \Big)_V = \Big( \pdv{N}{V} \Big)_\mu
+    }
+    = - \mpdv{\Omega}{V}{\mu}
+    \\
+    - \mpdv{\Omega}{T}{\mu} =
+    \boxed{
+        \Big( \pdv{N}{T} \Big)_\mu = \Big( \pdv{S}{\mu} \Big)_T
+    }
+    = - \mpdv{\Omega}{\mu}{T}
+\end{gathered}$$
+
+And the corresponding reciprocal relations are then given by:
+
+$$\begin{gathered}
+    \boxed{
+        \Big( \pdv{V}{S} \Big)_T = \Big( \pdv{T}{P} \Big)_V
+    }
+    \\
+    \boxed{
+        \Big( \pdv{\mu}{P} \Big)_V = \Big( \pdv{V}{N} \Big)_\mu
+    }
+    \\
+    \boxed{
+        \Big( \pdv{T}{N} \Big)_\mu = \Big( \pdv{\mu}{S} \Big)_T
+    }
+\end{gathered}$$
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/maxwells-equations/index.md b/source/know/concept/maxwells-equations/index.md
new file mode 100644
index 0000000..033ebf7
--- /dev/null
+++ b/source/know/concept/maxwells-equations/index.md
@@ -0,0 +1,259 @@
+---
+title: "Maxwell's equations"
+date: 2021-09-09
+categories:
+- Physics
+- Electromagnetism
+layout: "concept"
+---
+
+In physics, **Maxwell's equations** govern
+all macroscopic electromagnetism,
+and notably lead to the
+[electromagnetic wave equation](/know/concept/electromagnetic-wave-equation/),
+which describes the existence of light.
+
+
+## Gauss' law
+
+**Gauss' law** states that the electric flux $\Phi_E$ through
+a closed surface $S(V)$ is equal to the total charge $Q$
+contained in the enclosed volume $V$,
+divided by the vacuum permittivity $\varepsilon_0$:
+
+$$\begin{aligned}
+    \Phi_E
+    = \oint_{S(V)} \vb{E} \cdot \dd{\vb{A}}
+    = \frac{1}{\varepsilon_0} \int_{V} \rho \dd{V}
+    = \frac{Q}{\varepsilon_0}
+\end{aligned}$$
+
+Where $\vb{E}$ is the [electric field](/know/concept/electric-field/),
+and $\rho$ is the charge density in $V$.
+Gauss' law is usually more useful when written in its vector form,
+which can be found by applying the divergence theorem
+to the surface integral above.
+It states that the divergence of $\vb{E}$ is proportional to $\rho$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{E} = \frac{\rho}{\varepsilon_0}
+    }
+\end{aligned}$$
+
+This law can just as well be expressed for
+the displacement field $\vb{D}$
+and polarization density $\vb{P}$.
+We insert $\vb{E} = (\vb{D} - \vb{P}) / \varepsilon_0$
+into Gauss' law for $\vb{E}$, multiplied by $\varepsilon_0$:
+
+$$\begin{aligned}
+    \rho
+    = \nabla \cdot \big( \vb{D} - \vb{P} \big)
+    = \nabla \cdot \vb{D} - \nabla \cdot \vb{P}
+\end{aligned}$$
+
+To proceed, we split the net charge density $\rho$
+into a "free" part $\rho_\mathrm{free}$
+and a "bound" part $\rho_\mathrm{bound}$,
+respectively corresponding to $\vb{D}$ and $\vb{P}$,
+such that $\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}$.
+This yields:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{D} = \rho_{\mathrm{free}}
+    }
+    \qquad \quad
+    \boxed{
+        \nabla \cdot \vb{P} = - \rho_{\mathrm{bound}}
+    }
+\end{aligned}$$
+
+By integrating over an arbitrary volume $V$
+we can get integral forms of these equations:
+
+$$\begin{aligned}
+    \Phi_D
+    &= \oint_{S(V)} \vb{D} \cdot \dd{\vb{A}}
+    = \int_{V} \rho_{\mathrm{free}} \dd{V}
+    = Q_{\mathrm{free}}
+    \\
+    \Phi_P
+    &= \oint_{S(V)} \vb{P} \cdot \dd{\vb{A}}
+    = - \int_{V} \rho_{\mathrm{bound}} \dd{V}
+    = - Q_{\mathrm{bound}}
+\end{aligned}$$
+
+
+## Gauss' law for magnetism
+
+**Gauss' law for magnetism** states that magnetic flux $\Phi_B$
+through a closed surface $S(V)$ is zero.
+In other words, all magnetic field lines entering
+the volume $V$ must leave it too:
+
+$$\begin{aligned}
+    \Phi_B
+    = \oint_{S(V)} \vb{B} \cdot \dd{\vb{A}}
+    = 0
+\end{aligned}$$
+
+Where $\vb{B}$ is the [magnetic field](/know/concept/magnetic-field/).
+Thanks to the divergence theorem,
+this can equivalently be stated in vector form as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{B} = 0
+    }
+\end{aligned}$$
+
+A consequence of this law is the fact that magnetic monopoles cannot exist,
+i.e. there is no such thing as "magnetic charge",
+in contrast to electric charge.
+
+
+## Faraday's law of induction
+
+**Faraday's law of induction** states that a magnetic field $\vb{B}$
+that changes with time will induce an electric field $E$.
+Specifically, the change in magnetic flux through a non-closed surface $S$
+creates an electromotive force around the contour $C(S)$.
+This is written as:
+
+$$\begin{aligned}
+    \oint_{C(S)} \vb{E} \cdot \dd{\vb{l}}
+    = - \dv{}{t}\int_{S} \vb{B} \cdot \dd{\vb{A}}
+\end{aligned}$$
+
+By using Stokes' theorem on the contour integral,
+the vector form of this law is found to be:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \times \vb{E} = - \pdv{\vb{B}}{t}
+    }
+\end{aligned}$$
+
+
+## Ampère's circuital law
+
+**Ampère's circuital law**, with Maxwell's correction,
+states that a magnetic field $\vb{B}$
+can be induced along a contour $C(S)$ by two things:
+a current density $\vb{J}$ through the enclosed surface $S$,
+and a change of the electric field flux $\Phi_E$ through $S$:
+
+$$\begin{aligned}
+    \oint_{C(S)} \vb{B} \cdot d\vb{l}
+    = \mu_0 \Big( \int_S \vb{J} \cdot d\vb{A} + \varepsilon_0 \dv{}{t}\int_S \vb{E} \cdot d\vb{A} \Big)
+\end{aligned}$$
+$$\begin{aligned}
+    \boxed{
+        \nabla \times \vb{B} = \mu_0 \Big( \vb{J} + \varepsilon_0 \pdv{\vb{E}}{t} \Big)
+    }
+\end{aligned}$$
+
+Where $\mu_0$ is the vacuum permeability.
+This relation also exists for the "bound" fields $\vb{H}$ and $\vb{D}$,
+and for $\vb{M}$ and $\vb{P}$.
+We insert $\vb{B} = \mu_0 (\vb{H} + \vb{M})$
+and $\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0$
+into Ampère's law, after dividing it by $\mu_0$ for simplicity:
+
+$$\begin{aligned}
+    \nabla \cross \big( \vb{H} + \vb{M} \big)
+    &= \vb{J} + \pdv{}{t}\big( \vb{D} - \vb{P} \big)
+\end{aligned}$$
+
+To proceed, we split the net current density $\vb{J}$
+into a "free" part $\vb{J}_\mathrm{free}$
+and a "bound" part $\vb{J}_\mathrm{bound}$,
+such that $\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}$.
+This leads us to:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \times \vb{H} = \vb{J}_{\mathrm{free}} + \pdv{\vb{D}}{t}
+    }
+    \qquad \quad
+    \boxed{
+        \nabla \times \vb{M} = \vb{J}_{\mathrm{bound}} - \pdv{\vb{P}}{t}
+    }
+\end{aligned}$$
+
+By integrating over an arbitrary surface $S$
+we can get integral forms of these equations:
+
+$$\begin{aligned}
+    \oint_{C(S)} \vb{H} \cdot d\vb{l}
+    &= \int_S \vb{J}_{\mathrm{free}} \cdot \dd{\vb{A}} + \dv{}{t}\int_S \vb{D} \cdot \dd{\vb{A}}
+    \\
+    \oint_{C(S)} \vb{M} \cdot d\vb{l}
+    &= \int_S \vb{J}_{\mathrm{bound}} \cdot \dd{\vb{A}} - \dv{}{t}\int_S \vb{P} \cdot \dd{\vb{A}}
+\end{aligned}$$
+
+Note that $\vb{J}_\mathrm{bound}$ can be split into
+the **magnetization current density** $\vb{J}_M = \nabla \cross \vb{M}$
+and the **polarization current density** $\vb{J}_P = \ipdv{\vb{P}}{t}$:
+
+$$\begin{aligned}
+    \vb{J}_\mathrm{bound}
+    = \vb{J}_M + \vb{J}_P
+    = \nabla \cross \vb{M} + \pdv{\vb{P}}{t}
+\end{aligned}$$
+
+
+## Redundancy of Gauss' laws
+
+In fact, both of Gauss' laws are redundant,
+because they are already implied by Faraday's and Ampère's laws.
+Suppose we take the divergence of Faraday's law:
+
+$$\begin{aligned}
+    0
+    = \nabla \cdot \nabla \cross \vb{E}
+    = - \nabla \cdot \pdv{\vb{B}}{t}
+    = - \pdv{}{t}(\nabla \cdot \vb{B})
+\end{aligned}$$
+
+Since the divergence of a curl is always zero,
+the right-hand side must vanish.
+We know that $\vb{B}$ can vary in time,
+so our only option to satisfy this is to demand that $\nabla \cdot \vb{B} = 0$.
+We thus arrive arrive at Gauss' law for magnetism from Faraday's law.
+
+The same technique works for Ampère's law.
+Taking its divergence gives us:
+
+$$\begin{aligned}
+    0
+    = \frac{1}{\mu_0} \nabla \cdot \nabla \cross \vb{B}
+    = \nabla \cdot \vb{J} + \varepsilon_0 \pdv{}{t}(\nabla \cdot \vb{E})
+\end{aligned}$$
+
+We integrate this over an arbitrary volume $V$,
+and apply the divergence theorem:
+
+$$\begin{aligned}
+    0
+    &= \int_V \nabla \cdot \vb{J} \dd{V} + \pdv{}{t}\int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V}
+    \\
+    &= \oint_S \vb{J} \cdot \dd{S} + \pdv{}{t}\int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V}
+\end{aligned}$$
+
+The first integral represents the current (charge flux)
+through the surface of $V$.
+Electric charge is not created or destroyed,
+so the second integral *must* be the total charge in $V$:
+
+$$\begin{aligned}
+    Q
+    = \int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V}
+    \quad \implies \quad
+    \nabla \cdot \vb{E}
+    = \frac{\rho}{\varepsilon_0}
+\end{aligned}$$
+
+And we thus arrive at Gauss' law from Ampère's law and charge conservation.
diff --git a/source/know/concept/meniscus/index.md b/source/know/concept/meniscus/index.md
new file mode 100644
index 0000000..1373d63
--- /dev/null
+++ b/source/know/concept/meniscus/index.md
@@ -0,0 +1,189 @@
+---
+title: "Meniscus"
+date: 2021-03-11
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+- Surface tension
+layout: "concept"
+---
+
+When a fluid interface, e.g. the surface of a liquid,
+touches a flat solid wall, it will curve to meet it.
+This small rise or fall is called a **meniscus**,
+and is caused by surface tension and gravity.
+
+In 2D, let the vertical $y$-axis be a flat wall,
+and the fluid tend to $y = 0$ when $x \to \infty$.
+Close to the wall, i.e. for small $x$, the liquid curves up or down
+to touch the wall at a height $y = d$.
+
+Three forces are at work here:
+the first two are the surface tension $\alpha$ of the fluid surface,
+and the counter-pull $\alpha \sin\phi$ of the wall against the tension,
+where $\phi$ is the contact angle.
+The third is the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) gradient
+inside the small portion of the fluid above/below the ambient level,
+which exerts a total force on the wall given by
+(for $\phi < \pi/2$ so that $d > 0$):
+
+$$\begin{aligned}
+    \int_0^d \rho g y \dd{y}
+    = \frac{1}{2} \rho g d^2
+\end{aligned}$$
+
+If you were wondering about the units,
+keep in mind that there is an implicit $z$-direction here too.
+This results in the following balance equation for the forces at the wall:
+
+$$\begin{aligned}
+    \alpha
+    = \alpha \sin\phi + \frac{1}{2} \rho g d^2
+\end{aligned}$$
+
+We isolate this relation for $d$
+and use some trigonometric magic to rewrite it:
+
+$$\begin{aligned}
+    d
+    = \sqrt{\frac{\alpha}{\rho g}} \sqrt{2 (1 - \sin\phi)}
+    = \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)}
+\end{aligned}$$
+
+Here, we recognize the definition of the capillary length $L_c = \sqrt{\alpha / (\rho g)}$,
+yielding an expression for $d$
+that is valid both for $\phi < \pi/2$ (where $d > 0$)
+and $\phi > \pi/2$ (where $d < 0$):
+
+$$\begin{aligned}
+    \boxed{
+        d
+        = 2 L_c \sin\!\Big(\frac{\pi/2 - \phi}{2}\Big)
+    }
+\end{aligned}$$
+
+Next, we would like to know the exact shape of the meniscus.
+To do this, we need to describe the liquid surface differently,
+using the elevation angle $\theta$ relative to the $y = 0$ plane.
+The curve $\theta(s)$ is a function of the arc length $s$,
+where $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$,
+and is governed by:
+
+$$\begin{aligned}
+    \dv{x}{s}
+    = \cos\theta
+    \qquad
+    \dv{y}{s}
+    = \sin\theta
+    \qquad
+    \dv{\theta}{s}
+    = \frac{1}{R}
+\end{aligned}$$
+
+The last equation describes the curvature radius $R$
+of the surface along the $x$-axis.
+Since we are considering a flat wall,
+there is no curvature in the orthogonal principal direction.
+
+Just below the liquid surface in the meniscus,
+we expect the hydrostatic pressure
+and the [Young-Laplace law](/know/concept/young-laplace-law/)
+to agree about the pressure $p$,
+where $p_0$ is the external air pressure:
+
+$$\begin{aligned}
+    p_0 - \rho g y
+    = p_0 - \frac{\alpha}{R}
+\end{aligned}$$
+
+Rearranging this yields that $R = L_c^2 / y$.
+Inserting this into the curvature equation gives us:
+
+$$\begin{aligned}
+    \dv{\theta}{s}
+    = \frac{y}{L_c^2}
+\end{aligned}$$
+
+By differentiating this equation with respect to $s$
+and using $\idv{y}{s} = \sin\theta$, we arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        L_c^2 \dvn{2}{\theta}{s} = \sin\theta
+    }
+\end{aligned}$$
+
+To solve this equation, we multiply it by $\idv{\theta}{s}$,
+which is nonzero close to the wall:
+
+$$\begin{aligned}
+    L_c^2 \dvn{2}{\theta}{s} \dv{\theta}{s}
+    = \dv{\theta}{s} \sin\theta
+\end{aligned}$$
+
+We integrate both sides with respect to $s$
+and set the integration constant to $1$,
+such that we get zero when $\theta \to 0$ away from the wall:
+
+$$\begin{aligned}
+    \frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2
+    = 1 - \cos\theta
+\end{aligned}$$
+
+Isolating this for $\idv{\theta}{s}$ and using a trigonometric identity then yields:
+
+$$\begin{aligned}
+    \dv{\theta}{s}
+    = \frac{1}{L_c} \sqrt{2 (1 - \cos\theta)}
+    = \frac{1}{L_c} \sqrt{4 \sin^2\!\Big( \frac{\theta}{2} \Big)}
+    = - \frac{2}{L_c} \sin\!\Big( \frac{\theta}{2} \Big)
+\end{aligned}$$
+
+We use trigonometric relations on the equations
+for $\idv{x}{s}$ and $\idv{y}{s}$ to get $\theta$-derivatives:
+
+$$\begin{aligned}
+    \dv{x}{\theta}
+    &= \dv{x}{s} \dv{s}{\theta}
+    = \bigg( 1 - 2 \sin^2\!\Big( \frac{\theta}{2} \Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1}
+    = L_c \sin\!\Big( \frac{\theta}{2} \Big) - \frac{L_c}{2 \sin(\theta/2)}
+    \\
+    \dv{y}{\theta}
+    &= \dv{y}{s} \dv{s}{\theta}
+    = \bigg( 2 \sin\!\Big(\frac{\theta}{2}\Big) \cos\!\Big(\frac{\theta}{2}\Big) \bigg) \bigg( \dv{\theta}{s} \bigg)^{-1}
+    = - L_c \cos\!\Big( \frac{\theta}{2} \Big)
+\end{aligned}$$
+
+Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall.
+Then, by integrating the above equations, we get the following solutions:
+
+$$\begin{gathered}
+    \boxed{
+        \frac{x}{L_c}
+        = 2 \cos\!\Big(\frac{\theta_0}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta_0}{4}\Big) \bigg|
+        - 2 \cos\!\Big(\frac{\theta}{2}\Big) + \log\!\bigg| \tan\!\Big(\frac{\theta}{4}\Big) \bigg|
+    }
+    \\
+    \boxed{
+        \frac{y}{L_c}
+        = - 2 \sin\!\Big(\frac{\theta}{2}\Big)
+    }
+\end{gathered}$$
+
+Where the integration constant has been chosen such that $y \to 0$ for $\theta \to 0$ away from the wall,
+and $x = 0$ for $\theta = \theta_0$.
+This result is consistent with our earlier expression for $d$:
+
+$$\begin{aligned}
+    d
+    = y(\theta_0)
+    = - 2 L_c \sin\!\Big(\frac{\theta_0}{2}\Big)
+    = 2 L_c \sin\!\Big( \frac{\pi/2 - \phi}{2} \Big)
+\end{aligned}$$
+
+
+## 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/metacentric-height/index.md b/source/know/concept/metacentric-height/index.md
new file mode 100644
index 0000000..7588196
--- /dev/null
+++ b/source/know/concept/metacentric-height/index.md
@@ -0,0 +1,179 @@
+---
+title: "Metacentric height"
+date: 2022-03-11
+categories:
+- Physics
+- Fluid mechanics
+layout: "concept"
+---
+
+Consider an object with center of mass $G$,
+floating in a large body of liquid whose surface is flat at $z = 0$.
+For our purposes, it is easiest to use a coordinate system
+whose origin is at the area centroid
+of the object's cross-section through the liquid's surface, namely:
+
+$$\begin{aligned}
+    (x_0, y_0)
+    \equiv \frac{1}{A_{wl}} \iint_{wl} (x, y) \dd{A}
+\end{aligned}$$
+
+Where $A_{wl}$ is the cross-sectional area
+enclosed by the "waterline" around the "boat".
+Note that the boat's center of mass $G$
+does not coincide with the origin in general,
+as is illustrated in the following sketch
+of our choice of coordinate system:
+
+<a href="sketch.png">
+<img src="sketch.png" style="width:67%">
+</a>
+
+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.
diff --git a/source/know/concept/metacentric-height/sketch.png b/source/know/concept/metacentric-height/sketch.png
new file mode 100644
index 0000000..1ee1d04
--- /dev/null
+++ b/source/know/concept/metacentric-height/sketch.png
diff --git a/source/know/concept/microcanonical-ensemble/index.md b/source/know/concept/microcanonical-ensemble/index.md
new file mode 100644
index 0000000..8490767
--- /dev/null
+++ b/source/know/concept/microcanonical-ensemble/index.md
@@ -0,0 +1,120 @@
+---
+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:
+
+$$\begin{aligned}
+    A(0, t)
+    = \sqrt{P_0} \sech\!\Big(\frac{t}{T_0}\Big)
+\end{aligned}$$
+
+<a href="pheno-mi.jpg">
+<img src="pheno-mi-small.jpg" style="width:100%">
+</a>
+
+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.
diff --git a/source/know/concept/modulational-instability/pheno-mi-small.jpg b/source/know/concept/modulational-instability/pheno-mi-small.jpg
new file mode 100644
index 0000000..995ec81
--- /dev/null
+++ b/source/know/concept/modulational-instability/pheno-mi-small.jpg
diff --git a/source/know/concept/modulational-instability/pheno-mi.jpg b/source/know/concept/modulational-instability/pheno-mi.jpg
new file mode 100644
index 0000000..e45f074
--- /dev/null
+++ b/source/know/concept/modulational-instability/pheno-mi.jpg
diff --git a/source/know/concept/multi-photon-absorption/index.md b/source/know/concept/multi-photon-absorption/index.md
new file mode 100644
index 0000000..d5ed2d3
--- /dev/null
+++ b/source/know/concept/multi-photon-absorption/index.md
@@ -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/) $\delta$:
+
+$$\begin{aligned}
+    \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
+    = 2 \pi \: \delta(x) \: t
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-sinc"/>
+<label for="proof-sinc">Proof</label>
+<div class="hidden">
+<label for="proof-sinc">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \lim_{t \to \infty} \bigg| \frac{e^{i x t} - 1}{x} \bigg|^2
+    = 4 \pi^2 \delta^2(x)
+    = 2 \pi \: \delta(x) \: t
+\end{aligned}$$
+</div>
+</div>
+
+
+## One-photon absorption
+
+To warm up, we start at first-order perturbation theory.
+Thanks to our choice of initial condition,
+nothing at all happens to any of the even-numbered states $\Ket{e}$:
+
+$$\begin{aligned}
+    c_e^{(1)}(t)
+    &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(0)} \: e^{i \omega_{eu} \tau} \dd{\tau}
+    = 0
+\end{aligned}$$
+
+While the odd-numbered states $\Ket{u}$ have a nonzero correction $c_u^{(1)}$,
+where $\vb{p}_{u0} = \matrixel{u}{\vu{p}}{0}$:
+
+$$\begin{aligned}
+    c_u^{(1)}(t)
+    &= -\frac{i}{\hbar} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{0} \: c_0^{(0)} \: e^{i \omega_{u0} \tau} \dd{\tau}
+    \\
+    &= i \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \int_0^t e^{i (\omega_{u0} - \omega) \tau} \dd{\tau}
+    \\
+    &= i \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \bigg[ \frac{e^{i (\omega_{u0} - \omega) \tau}}{i (\omega_{u0} - \omega)} \bigg]_0^t
+\end{aligned}$$
+
+Consequently, the first-order correction
+(in the rotating wave approximation) is given by:
+
+$$\begin{aligned}
+    \boxed{
+        c_u^{(1)}(t)
+        \approx \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \frac{e^{i (\omega_{u0} - \omega) t} - 1}{\omega_{u0} - \omega}
+    }
+\end{aligned}$$
+
+Since $\big| c_u^{(1)}(t) \big|^2$ is the probability
+of finding the electron in $\Ket{u}$,
+its transition rate $R_u^{(1)}(t)$ is as follows,
+averaged since the beginning $t = 0$:
+
+$$\begin{aligned}
+    R_u^{(1)}(t)
+    = \frac{\big| c_u^{(1)}(t) \big|^2}{t}
+    = \frac{1}{t} \bigg| \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \bigg|^2
+    \cdot \bigg| \frac{e^{i (\omega_{u0} - \omega) t} - 1}{\omega_{u0} - \omega} \bigg|^2
+\end{aligned}$$
+
+For large $t \to \infty$, we can use the formula we proved earlier
+to get [Fermi's golden rule](/know/concept/fermis-golden-rule/):
+
+$$\begin{aligned}
+    \boxed{
+        R_u^{(1)}
+        = 2 \pi \bigg| \frac{\vb{p}_{u0} \cdot \vb{E}}{\hbar} \bigg|^2 \delta(\omega_{u0} - \omega)
+    }
+\end{aligned}$$
+
+This well-known formula represents **one-photon absorption**:
+it peaks at $\omega_{u0} = \omega$, i.e. when one photon $\hbar \omega$
+has the exact energy of the transition $\hbar \omega_{u0}$.
+Note that this transition is only possible when $\matrixel{u}{\vu{p}}{0} \neq 0$,
+i.e. for any odd-numbered final state $\Ket{u}$.
+
+
+## Two-photon absorption
+
+Next, we go to second-order perturbation theory.
+Based on the previous result, this time
+all odd-numbered states $\Ket{u}$ are unaffected:
+
+$$\begin{aligned}
+    c_u^{(2)}(t)
+    &= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(1)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau}
+    = 0
+\end{aligned}$$
+
+While the even-numbered states $\Ket{e}$ have the following correction,
+using $\omega_{eu} \!+\! \omega_{u0} = \omega_{e0}$:
+
+$$\begin{aligned}
+    c_e^{(2)}(t)
+    &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(1)}(\tau) \: e^{i \omega_{eu} \tau} \dd{\tau}
+    \\
+    &= i \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)}
+    \int_0^t e^{i (\omega_{eu} + \omega_{u0} - 2 \omega) \tau} - e^{i (\omega_{eu} - \omega) \tau} \dd{\tau}
+    \\
+    &= i \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)}
+    \bigg[ \frac{e^{i (\omega_{e0} - 2 \omega) \tau}}{i (\omega_{e0} - 2 \omega)}
+    - \frac{e^{i (\omega_{eu} - \omega) \tau}}{i (\omega_{eu} - \omega)} \bigg]_0^t
+\end{aligned}$$
+
+The second term represents one-photon absorption between $\Ket{u}$ and $\Ket{e}$.
+We do not care about that, so we drop it, leaving only the first term:
+
+$$\begin{aligned}
+    \boxed{
+        c_e^{(2)}(t)
+        \approx \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)}
+        \frac{e^{i (\omega_{e0} - 2 \omega) t} - 1}{\omega_{e0} - 2 \omega}
+    }
+\end{aligned}$$
+
+As before, we can define a rate $R_e^{(2)}(t)$
+for all transitions represented by this term:
+
+$$\begin{aligned}
+    R_e^{(2)}(t)
+    = \frac{\big| c_e^{(2)}(t) \big|^2}{t}
+    = \frac{1}{t} \bigg| \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \bigg|^2
+    \cdot \bigg| \frac{e^{i (\omega_{e0} - 2 \omega) t} - 1}{\omega_{e0} - 2 \omega} \bigg|^2
+\end{aligned}$$
+
+Which for $t \to \infty$ takes a similar form to Fermi's golden rule,
+using the formula we proved:
+
+$$\begin{aligned}
+    \boxed{
+        R_e^{(2)}
+        = 2 \pi \bigg| \sum_{u}^{\mathrm{odd}} \frac{(\vb{p}_{eu} \cdot \vb{E}) (\vb{p}_{u0} \cdot \vb{E})}{\hbar^2 (\omega_{u0} - \omega)} \bigg|^2
+        \delta(\omega_{e0} - 2 \omega)
+    }
+\end{aligned}$$
+
+This represents **two-photon absorption**, since it peaks at $\omega_{e0} = 2 \omega$:
+two identical photons $\hbar \omega$ are absorbed simultaneously
+to bridge the energy gap $\hbar \omega_{e0}$.
+Surprisingly, such a transition can only occur when $\matrixel{e}{\vu{p}}{0} = 0$,
+i.e. for any even-numbered final state $\Ket{e}$.
+Notice that the rate is proportional to $|\vb{E}|^4$,
+so this effect is only noticeable at high light intensities.
+
+
+## Three-photon absorption
+
+For third-order perturbation theory,
+all even-numbered states $\Ket{e}$ are unchanged:
+
+$$\begin{aligned}
+    c_e^{(3)}(t)
+    &= -\frac{i}{\hbar} \sum_{u}^{\mathrm{odd}} \int_0^t \matrixel{e}{\hat{H}_1(\tau)}{u} \: c_u^{(2)}(\tau) \: e^{i \omega_{eu} \tau} \dd{\tau}
+    = 0
+\end{aligned}$$
+
+And the odd-numbered states $\Ket{u}$ get the following third-order corrections:
+
+$$\begin{aligned}
+    c_u^{(3)}(t)
+    &= -\frac{i}{\hbar} \sum_{e}^{\mathrm{even}} \int_0^t \matrixel{u}{\hat{H}_1(\tau)}{e} \: c_e^{(2)}(\tau) \: e^{i \omega_{ue} \tau} \dd{\tau}
+    \\
+    &= i \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}}
+    \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}{\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)}
+    \int_0^t e^{i (\omega_{ue} + \omega_{e0} - 3 \omega) \tau} - e^{i (\omega_{ue} - \omega) \tau} \dd{\tau}
+    \\
+    &= i \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}}
+    \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}{\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)}
+    \bigg[ \frac{e^{i (\omega_{u0} - 3 \omega) \tau}}{i (\omega_{u0} - 3 \omega)}
+    - \frac{e^{i (\omega_{ue} - \omega) \tau}}{i (\omega_{ue} - \omega)} \bigg]_0^t
+\end{aligned}$$
+
+Once again, the second term is uninteresting,
+so we drop it and look at the first term only:
+
+$$\begin{aligned}
+    \boxed{
+        c_u^{(3)}(t)
+        \approx \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}}
+        \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}
+        {\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)}
+        \frac{e^{i (\omega_{u0} - 3 \omega) t} - 1}{\omega_{u0} - 3 \omega}
+    }
+\end{aligned}$$
+
+The resulting transition rate $R_u^{(3)}(t)$
+is found to have the following familiar form:
+
+$$\begin{aligned}
+    R_u^{(3)}(t)
+    = \frac{\big| c_u^{(3)}(t) \big|^2}{t}
+    = \frac{1}{t} \bigg| \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}}
+    \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}
+    {\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \bigg|^2
+    \cdot \bigg| \frac{e^{i (\omega_{u0} - 3 \omega) t} - 1}{\omega_{u0} - 3 \omega} \bigg|^2
+\end{aligned}$$
+
+Applying our formula to this yields the following analogue of Fermi's golden rule:
+
+$$\begin{aligned}
+    \boxed{
+        R_u^{(3)}
+        = 2 \pi \bigg| \sum_{e}^{\mathrm{even}} \sum_{u'}^{\mathrm{odd}}
+        \frac{(\vb{p}_{ue} \cdot \vb{E}) (\vb{p}_{eu'} \cdot \vb{E}) (\vb{p}_{u'0} \cdot \vb{E})}
+        {\hbar^3 (\omega_{u'0} - \omega) (\omega_{e0} - 2 \omega)} \bigg|^2 \delta(\omega_{u0} - 3 \omega)
+    }
+\end{aligned}$$
+
+This represents **three-photon absorption**, since it peaks at $\omega_{u0} = 3 \omega$:
+three identical photons $\hbar \omega$ are absorbed simultaneously
+to bridge the energy gap $\hbar \omega_{u0}$.
+This process is similar to one-photon absorption,
+in the sense that it can only occur if $\matrixel{u}{\vu{p}}{0} \neq 0$.
+The rate is proportional to $|\vb{E}|^6$,
+so this effect only appears at extremely high light intensities.
+
+
+## N-photon absorption
+
+A pattern has appeared in these calculations:
+in $N$th-order perturbation theory,
+we get a term representing $N$-photon absorption,
+with a transition rate proportional to $|\vb{E}|^{2N}$.
+Indeed, we can derive infinitely many formulas in this way,
+although the results become increasingly unrealistic
+due to the dependence on $\vb{E}$.
+
+If $N$ is odd, only odd-numbered destinations $\Ket{u}$ are allowed
+(assuming the electron starts in the ground state $\Ket{0}$),
+and if $N$ is even, only even-numbered destinations $\Ket{e}$.
+Note that nothing has been said about the energies of these states
+(other than $\Ket{0}$ being the minimum);
+everything is determined by the matrix elements $\matrixel{f}{\vu{p}}{i}$.
+
+
+
+## References
+1.  R.W. Boyd,
+    *Nonlinear optics*, 4th edition,
+    Academic Press.
+2.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
diff --git a/source/know/concept/navier-cauchy-equation/index.md b/source/know/concept/navier-cauchy-equation/index.md
new file mode 100644
index 0000000..b9db548
--- /dev/null
+++ b/source/know/concept/navier-cauchy-equation/index.md
@@ -0,0 +1,108 @@
+---
+title: "Navier-Cauchy equation"
+date: 2021-04-02
+categories:
+- Physics
+- Continuum physics
+layout: "concept"
+---
+
+The **Navier-Cauchy equation** describes **elastodynamics**:
+the movements inside an elastic solid
+in response to external forces and/or internal stresses.
+
+For a particle of the solid, whose position is given by the displacement field $\va{u}$,
+Newton's second law is as follows,
+where $\dd{m}$ and $\dd{V}$ are the particle's mass and volume, respectively:
+
+$$\begin{aligned}
+    \va{f^*} \dd{V}
+    = \pdvn{2}{\va{u}}{t} \dd{m}
+    = \rho \pdvn{2}{\va{u}}{t} \dd{V}
+\end{aligned}$$
+
+Where $\rho$ is the mass density,
+and $\va{f^*}$ is the effective force density,
+defined from the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$
+like so, with $\va{f}$ being an external body force, e.g. from gravity:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \va{f} + \nabla \cdot \hat{\sigma}^\top
+\end{aligned}$$
+
+We can therefore write Newton's second law as follows,
+while switching to index notation,
+where $\nabla_j = \ipdv{}{x_j}$ is the partial derivative
+with respect to the $j$th coordinate:
+
+$$\begin{aligned}
+    f_i + \sum_{j} \nabla_j \sigma_{ij}
+    = \rho \pdvn{2}{u_i}{t}
+\end{aligned}$$
+
+The components $\sigma_{ij}$ of the Cauchy stress tensor
+are given by [Hooke's law](/know/concept/hookes-law/),
+where $\mu$ and $\lambda$ are the Lamé coefficients,
+which describe the material:
+
+$$\begin{aligned}
+    \sigma_{ij}
+    = 2 \mu u_{ij} + \lambda \delta_{ij} \sum_{k} u_{kk}
+\end{aligned}$$
+
+In turn, the components $u_{ij}$ of the
+[Cauchy strain tensor](/know/concept/cauchy-strain-tensor/)
+are defined as follows,
+where $u_i$ are once again the components of the displacement vector $\va{u}$:
+
+$$\begin{aligned}
+    u_{ij}
+    = \frac{1}{2} \big( \nabla_i u_j + \nabla_j u_i \big)
+\end{aligned}$$
+
+To derive the Navier-Cauchy equation,
+we start by inserting Hooke's law into Newton's law:
+
+$$\begin{aligned}
+    \rho \pdvn{2}{u_i}{t}
+    &= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \sum_{j} \nabla_j \bigg( \delta_{ij} \sum_{k} u_{kk} \bigg)
+    \\
+    &= f_i + 2 \mu \sum_{j} \nabla_j u_{ij} + \lambda \nabla_i \sum_{j} u_{jj}
+\end{aligned}$$
+
+And then into this we insert the definition of the strain components $u_{ij}$, yielding:
+
+$$\begin{aligned}
+    \rho \pdvn{2}{u_i}{t}
+    &= f_i + \mu \sum_{j} \nabla_j \big( \nabla_i u_j + \nabla_j u_i \big) + \lambda \nabla_i \sum_{j} \nabla_j u_{j}
+\end{aligned}$$
+
+Rearranging this a bit leads us to the Navier-Cauchy equation written in index notation:
+
+$$\begin{aligned}
+    \boxed{
+        \rho \pdvn{2}{u_i}{t}
+        = f_i + \mu \sum_{j} \nabla_j^2 u_i + (\mu + \lambda) \nabla_i \sum_{j} \nabla_j u_j
+    }
+\end{aligned}$$
+
+Traditionally, it is written in vector notation instead,
+in which case it looks like this:
+
+$$\begin{aligned}
+    \boxed{
+        \rho \pdvn{2}{\va{u}}{t}
+        = \va{f} + \mu \nabla^2 \va{u} + (\mu + \lambda) \nabla (\nabla \cdot \va{u})
+    }
+\end{aligned}$$
+
+A special case is the **Navier-Cauchy equilibrium equation**,
+where the left-hand side is just zero.
+That version describes **elastostatics**: the deformation of a solid at rest.
+
+
+## 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/navier-stokes-equations/index.md b/source/know/concept/navier-stokes-equations/index.md
new file mode 100644
index 0000000..cdffebf
--- /dev/null
+++ b/source/know/concept/navier-stokes-equations/index.md
@@ -0,0 +1,128 @@
+---
+title: "Navier-Stokes equations"
+date: 2021-04-12
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+While the [Euler equations](/know/concept/euler-equations/) govern *ideal* "dry" fluids,
+the **Navier-Stokes equations** govern *nonideal* "wet" fluids,
+i.e. fluids with nonzero [viscosity](/know/concept/viscosity/).
+
+
+## Incompressible fluid
+
+First of all, we can reuse the incompressibility condition for ideal fluids, without modifications:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \va{v} = 0
+    }
+\end{aligned}$$
+
+Furthermore, from the derivation of the Euler equations,
+we know that Newton's second law can be written as follows,
+for an infinitesimal particle of the fluid:
+
+$$\begin{aligned}
+    \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \va{f^*}
+\end{aligned}$$
+
+$\mathrm{D}/\mathrm{D}t$ is the [material derivative](/know/concept/material-derivative/),
+$\rho$ is the density, and $\va{f^*}$ is the effective force density,
+expressed in terms of an external body force $\va{f}$ (e.g. gravity)
+and the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/) $\hat{\sigma}$:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \va{f} + \nabla \cdot \hat{\sigma}^\top
+\end{aligned}$$
+
+From the definition of viscosity,
+the stress tensor's elements are like so for a Newtonian fluid:
+
+$$\begin{aligned}
+    \sigma_{ij}
+    = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
+\end{aligned}$$
+
+Where $\eta$ is the dynamic viscosity.
+Inserting this, we calculate $\nabla \cdot \hat{\sigma}^\top$ in index notation:
+
+$$\begin{aligned}
+    \big( \nabla \cdot \hat{\sigma}^\top \big)_i
+    = \sum_{j} \nabla_j \sigma_{ij}
+    &= \sum_{j} \Big( \!-\! \delta_{ij} \nabla_j p + \eta (\nabla_i \nabla_j v_j + \nabla_j^2 v_i) \Big)
+    \\
+    &= - \nabla_i p + \eta \nabla_i \sum_{j} \nabla_j v_j + \eta \sum_{j} \nabla_j^2 v_i
+\end{aligned}$$
+
+Thanks to incompressibility $\nabla \cdot \va{v} = 0$,
+the middle term vanishes, leaving us with:
+
+$$\begin{aligned}
+    \va{f^*}
+    = \va{f} - \nabla p + \eta \nabla^2 \va{v}
+\end{aligned}$$
+
+We assume that the only body force is gravity $\va{f} = \rho \va{g}$.
+Newton's second law then becomes:
+
+$$\begin{aligned}
+    \rho \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+    = \rho \va{g} - \nabla p + \eta \nabla^2 \va{v}
+\end{aligned}$$
+
+Dividing by $\rho$, and replacing $\eta$
+with the kinematic viscosity $\nu = \eta/\rho$,
+yields the main equation:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+    }
+\end{aligned}$$
+
+Finally, we can optionally allow incompressible fluids
+with an inhomogeneous "lumpy" density $\rho$,
+by demanding conservation of mass,
+just like for the Euler equations:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \rho}{\mathrm{D} t}
+        = 0
+    }
+\end{aligned}$$
+
+Putting it all together, the Navier-Stokes equations for an incompressible fluid are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{\mathrm{D} \va{v}}{\mathrm{D} t}
+        = \va{g} - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+        \qquad
+        \nabla \cdot \va{v} = 0
+        \qquad
+        \frac{\mathrm{D} \rho}{\mathrm{D} t}
+        = 0
+    }
+\end{aligned}$$
+
+Due to the definition of viscosity $\nu$ as the molecular "stickiness",
+we have boundary conditions for the velocity field $\va{v}$:
+at any interface, $\va{v}$ must be continuous.
+Likewise, Newton's third law demands that the normal component
+of stress $\hat{\sigma} \cdot \vu{n}$ is continuous there.
+
+
+
+## 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/newtons-bucket/index.md b/source/know/concept/newtons-bucket/index.md
new file mode 100644
index 0000000..30fe079
--- /dev/null
+++ b/source/know/concept/newtons-bucket/index.md
@@ -0,0 +1,91 @@
+---
+title: "Newton's bucket"
+date: 2021-05-13
+categories:
+- Physics
+- Fluid mechanics
+- Fluid statics
+layout: "concept"
+---
+
+**Newton's bucket** is a cylindrical bucket
+that rotates at angular velocity $\omega$.
+Due to [viscosity](/know/concept/viscosity/),
+any liquid in the bucket is affected by the rotation,
+eventually achieving the exact same $\omega$.
+
+However, once in equilibrium, the liquid's surface is not flat,
+but curved upwards from the center.
+This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$:
+
+$$\begin{aligned}
+    \va{f}
+    = \omega^2 \va{r}
+\end{aligned}$$
+
+Where $\va{r}$ is the molecule's position relative to the axis of rotation.
+This (fictitious) force can be written as the gradient
+of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$:
+
+$$\begin{aligned}
+    \Phi_\mathrm{f}
+    = - \frac{\omega^2}{2} r^2
+    = - \frac{\omega^2}{2} (x^2 + y^2)
+\end{aligned}$$
+
+In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$,
+where $\va{g} = - \nabla \Phi_\mathrm{g}$:
+
+$$\begin{aligned}
+    \Phi_\mathrm{g}
+    = \mathrm{g} z
+\end{aligned}$$
+
+Overall, the molecule therefore feels an "effective" force
+with a potential $\Phi$ given by:
+
+$$\begin{aligned}
+    \Phi
+    = \Phi_\mathrm{g} + \Phi_\mathrm{f}
+    = \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2)
+\end{aligned}$$
+
+At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$
+in the liquid is the one that satisfies:
+
+$$\begin{aligned}
+    \frac{\nabla p}{\rho}
+    = - \nabla \Phi
+\end{aligned}$$
+
+Removing the gradients gives integration constants $p_0$ and $\Phi_0$,
+so the equilibrium equation is:
+
+$$\begin{aligned}
+    p - p_0
+    = - \rho (\Phi - \Phi_0)
+\end{aligned}$$
+
+We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$,
+where $z_0$ is the liquid height at the center:
+
+$$\begin{aligned}
+    p
+    = p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2)
+\end{aligned}$$
+
+At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure.
+The $z$-coordinate at which this is satisfied is as follows,
+telling us that the surface is parabolic:
+
+$$\begin{aligned}
+    z
+    = z_0 + \frac{\omega^2}{2 \mathrm{g}} (x^2 + y^2)
+\end{aligned}$$
+
+
+
+## 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/no-cloning-theorem/index.md b/source/know/concept/no-cloning-theorem/index.md
new file mode 100644
index 0000000..e5e5be4
--- /dev/null
+++ b/source/know/concept/no-cloning-theorem/index.md
@@ -0,0 +1,70 @@
+---
+title: "No-cloning theorem"
+date: 2021-03-06
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+layout: "concept"
+---
+
+In quantum mechanics, the **no-cloning theorem** states
+there is no general way to make copies of an arbitrary quantum state $\ket{\psi}$.
+This has profound implications for quantum information.
+
+To prove this theorem, let us pretend that a machine exists
+that can do just that: copy arbitrary quantum states.
+Given an input $\ket{\psi}$ and a blank $\ket{?}$,
+this machines turns $\ket{?}$ into $\ket{\psi}$:
+
+$$\begin{aligned}
+    \ket{\psi} \ket{?}
+    \:\:\longrightarrow\:\:
+    \ket{\psi} \ket{\psi}
+\end{aligned}$$
+
+We can use this device to make copies of the basis vectors $\ket{0}$ and $\ket{1}$:
+
+$$\begin{aligned}
+    \ket{0} \ket{?}
+    \:\:\longrightarrow\:\:
+    \ket{0} \ket{0}
+    \qquad \quad
+    \ket{1} \ket{?}
+    \:\:\longrightarrow\:\:
+    \ket{1} \ket{1}
+\end{aligned}$$
+
+If we feed this machine a superposition $\ket{\psi} = \alpha \ket{0} + \beta \ket{1}$,
+we *want* the following behaviour:
+
+$$\begin{aligned}
+    \Big( \alpha \ket{0} + \beta \ket{1} \Big) \ket{?}
+    \:\:\longrightarrow\:\:
+    &\Big( \alpha \ket{0} + \beta \ket{1} \Big) \Big( \alpha \ket{0} + \beta \ket{1} \Big)
+    \\
+    &= \Big( \alpha^2 \ket{0} \ket{0} + \alpha \beta \ket{0} \ket{1} + \alpha \beta \ket{1} \ket{0} + \beta^2 \ket{1} \ket{1} \Big)
+\end{aligned}$$
+
+Note the appearance of the cross terms with a factor of $\alpha \beta$.
+The problem is that the fundamental linearity of quantum mechanics
+dictates different behaviour:
+
+$$\begin{aligned}
+    \Big( \alpha \ket{0} + \beta \ket{1} \Big) \ket{?}
+    = \alpha \ket{0} \ket{?} + \beta \ket{1} \ket{?}
+    \:\:\longrightarrow\:\:
+    \alpha \ket{0} \ket{0} + \beta \ket{1} \ket{1}
+\end{aligned}$$
+
+This is clearly not the same as before: we have a contradiction,
+which implies that such a general cloning machine cannot ever exist.
+
+
+## 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/optical-wave-breaking/index.md b/source/know/concept/optical-wave-breaking/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/optical-wave-breaking/index.md
@@ -0,0 +1,229 @@
+---
+title: "Optical wave breaking"
+date: 2021-02-27
+categories:
+- Physics
+- Optics
+- Fiber optics
+- Nonlinear optics
+layout: "concept"
+---
+
+In fiber optics, **optical wave breaking** (OWB) is a nonlinear effect
+caused by interaction between
+[group velocity dispersion](/know/concept/dispersive-broadening/) (GVD) and
+[self-phase modulation](/know/concept/self-phase-modulation/) (SPM).
+It only happens in the normal dispersion regime ($\beta_2 > 0$)
+for pulses meeting a certain criterium, as we will see.
+
+SPM creates low frequencies at the front of the pulse, and high ones at the back,
+and if $\beta_2 > 0$, GVD lets low frequencies travel faster than high ones.
+When those effects interact, the pulse gets temporally stretched
+in a surprisingly sophisticated way.
+
+To illustrate this, the instantaneous frequency $\omega_i(z, t) = -\ipdv{\phi}{t}$
+has been plotted below for a theoretical Gaussian input pulse experiencing OWB,
+with settings $T_0 = 100\:\mathrm{fs}$, $P_0 = 5\:\mathrm{kW}$,
+$\beta_2 = 2\:\mathrm{ps}^2/\mathrm{m}$ and $\gamma = 0.1/\mathrm{W}/\mathrm{m}$.
+
+In the left panel, we see the typical S-shape caused by SPM,
+and the arrows indicate the direction that GVD is pushing the curve in.
+This leads to steepening at the edges, i.e. the S gradually turns into a Z.
+Shortly before the slope would become infinite,
+small waves start "falling off" the edge of the pulse,
+hence the name *wave breaking*:
+
+<a href="pheno-break-inst.jpg">
+<img src="pheno-break-inst-small.jpg" style="width:100%">
+</a>
+
+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:
+
+<a href="pheno-break-sgram.jpg">
+<img src="pheno-break-sgram-small.jpg" style="width:80%">
+</a>
+
+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:
+
+<a href="pheno-break.jpg">
+<img src="pheno-break-small.jpg" style="width:100%">
+</a>
+
+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.
+
diff --git a/source/know/concept/optical-wave-breaking/pheno-break-inst-small.jpg b/source/know/concept/optical-wave-breaking/pheno-break-inst-small.jpg
new file mode 100644
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+++ b/source/know/concept/optical-wave-breaking/pheno-break-inst-small.jpg
diff --git a/source/know/concept/optical-wave-breaking/pheno-break-inst.jpg b/source/know/concept/optical-wave-breaking/pheno-break-inst.jpg
new file mode 100644
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+++ b/source/know/concept/optical-wave-breaking/pheno-break-inst.jpg
diff --git a/source/know/concept/optical-wave-breaking/pheno-break-sgram-small.jpg b/source/know/concept/optical-wave-breaking/pheno-break-sgram-small.jpg
new file mode 100644
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+++ b/source/know/concept/optical-wave-breaking/pheno-break-sgram-small.jpg
diff --git a/source/know/concept/optical-wave-breaking/pheno-break-sgram.jpg b/source/know/concept/optical-wave-breaking/pheno-break-sgram.jpg
new file mode 100644
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+++ b/source/know/concept/optical-wave-breaking/pheno-break-sgram.jpg
diff --git a/source/know/concept/optical-wave-breaking/pheno-break-small.jpg b/source/know/concept/optical-wave-breaking/pheno-break-small.jpg
new file mode 100644
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+++ b/source/know/concept/optical-wave-breaking/pheno-break-small.jpg
diff --git a/source/know/concept/optical-wave-breaking/pheno-break.jpg b/source/know/concept/optical-wave-breaking/pheno-break.jpg
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+++ b/source/know/concept/optical-wave-breaking/pheno-break.jpg
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -0,0 +1,82 @@
+---
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \Inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+            \\
+            \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-fourier"/>
+<label for="proof-fourier">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-fourier">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \inprod{\tilde{f}}{\tilde{g}}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
+    \\
+    &= A^2 \int
+    \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
+    \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
+    \dd{k}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
+    \\
+    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
+    = \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
+\end{aligned}$$
+</div>
+</div>
+
+For this reason, physicists like to define the Fourier transform
+with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely
+conserves the functions' normalization.
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/partial-fraction-decomposition/index.md b/source/know/concept/partial-fraction-decomposition/index.md
new file mode 100644
index 0000000..86e1e27
--- /dev/null
+++ b/source/know/concept/partial-fraction-decomposition/index.md
@@ -0,0 +1,61 @@
+---
+title: "Partial fraction decomposition"
+date: 2021-02-22
+categories:
+- Mathematics
+layout: "concept"
+---
+
+**Partial fraction decomposition** or **partial fraction expansion**
+is a method to rewrite quotients of two polynomials $g(x)$ and $h(x)$,
+where the numerator $g(x)$ is of lower order than $h(x)$,
+as sums of fractions with $x$ in the denominator:
+
+$$\begin{aligned}
+    f(x) = \frac{g(x)}{h(x)} = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ...
+\end{aligned}$$
+
+Where $h_n$ etc. are the roots of the denominator $h(x)$. If all $N$ of
+these roots are distinct, then it is sufficient to simply posit:
+
+$$\begin{aligned}
+    \boxed{
+        f(x) = \frac{c_1}{x - h_1} + \frac{c_2}{x - h_2} + ... + \frac{c_N}{x - h_N}
+    }
+\end{aligned}$$
+
+The constants $c_n$ can either be found the hard way,
+by multiplying the denominators around and solving a system of $N$
+equations, or the easy way by using this trick:
+
+$$\begin{aligned}
+    \boxed{
+        c_n = \lim_{x \to h_n} \big( f(x) (x - h_n) \big)
+    }
+\end{aligned}$$
+
+If $h_1$ is a root with multiplicity $m > 1$, then the sum takes the form of:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2} + ...
+    }
+\end{aligned}$$
+
+Where $c_{1,j}$ are found by putting the terms on a common denominator, e.g.
+
+$$\begin{aligned}
+    \frac{c_{1,1}}{x - h_1} + \frac{c_{1,2}}{(x - h_1)^2}
+    = \frac{c_{1,1} (x - h_1) + c_{1,2}}{(x - h_1)^2}
+\end{aligned}$$
+
+And then, using the linear independence of $x^0, x^1, x^2, ...$, solving
+a system of $m$ equations to find all $c_{1,1}, ..., c_{1,m}$.
+
+
+
+## References
+1.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/path-integral-formulation/index.md b/source/know/concept/path-integral-formulation/index.md
new file mode 100644
index 0000000..9c2e301
--- /dev/null
+++ b/source/know/concept/path-integral-formulation/index.md
@@ -0,0 +1,182 @@
+---
+title: "Path integral formulation"
+date: 2021-07-03
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, the **path integral formulation**
+is an alternative description of quantum mechanics,
+which is equivalent to the "traditional" Schrödinger equation.
+Whereas the latter is based on [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
+the former comes from [Lagrangian mechanics](/know/concept/lagrangian-mechanics/).
+
+It expresses the [propagator](/know/concept/propagator/) $K$
+using the following sum over all possible paths $x(t)$,
+which all go from the initial position $x_0$ at time $t_0$
+to the destination $x_N$ at time $t_N$:
+
+$$\begin{aligned}
+    \boxed{
+        K(x_N, t_N; x_0, t_0)
+        = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar)
+    }
+\end{aligned}$$
+
+Where $A$ normalizes.
+$S[x]$ is the classical action of the path $x$, whose minimization yields
+the Euler-Lagrange equation from Lagrangian mechanics.
+Note that each path is given an equal weight,
+even unrealistic paths that make big detours.
+
+This apparent problem solves itself,
+thanks to the fact that paths close to the classical optimum $x_c(t)$
+have an action close to $S_c = S[x_c]$,
+while the paths far away have very different actions.
+Since $S[x]$ is inside a complex exponential,
+this means that paths close to $x_c$ add contructively,
+and the others add destructively and cancel out.
+
+An interesting way too look at it is by varying $\hbar$:
+as its value decreases, minor action differences yield big phase differences,
+which make the quantum wave function stay closer to $x_c$.
+In the limit $\hbar \to 0$, quantum mechanics thus turns into classical mechanics.
+
+## Time-slicing derivation
+
+The most popular way to derive the path integral formulation proceeds as follows:
+starting from the definition of the propagator $K$,
+we divide the time interval $t_N - t_0$ into $N$ "slices"
+of equal width $\Delta t = (t_N - t_0) / N$,
+where $N$ is large:
+
+$$\begin{aligned}
+    K(x_N, t_N; x_0, t_0)
+    &= \matrixel{x_N}{e^{- i \hat{H} (t_N - t_0) / \hbar}}{x_0}
+    = \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar} \cdots e^{- i \hat{H} \Delta t / \hbar}}{x_0}
+\end{aligned}$$
+
+Between the exponentials we insert $N\!-\!1$ identity operators
+$\hat{I} = \int \Ket{x} \Bra{x} \dd{x}$,
+and define $x_j = x(t_j)$ for an arbitrary path $x(t)$:
+
+$$\begin{aligned}
+    K
+    &= \int\cdots\int \matrixel{x_N}{e^{- i \hat{H} \Delta t / \hbar}}{x_{N-1}} \cdots \matrixel{x_1}{e^{- i \hat{H} \Delta t / \hbar}}{x_0}
+    \dd{x_1} \cdots \dd{x_{N - 1}}
+\end{aligned}$$
+
+For sufficiently small time steps $\Delta t$ (i.e. large $N$
+we make the following approximation
+(which would be exact, were it not for the fact that
+$\hat{T}$ and $\hat{V}$ are operators):
+
+$$\begin{aligned}
+    e^{- i \hat{H} \Delta t / \hbar}
+    = e^{- i (\hat{T} + \hat{V}) \Delta t / \hbar}
+    \approx e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}
+\end{aligned}$$
+
+Since $\hat{V} = V(x_j)$,
+we can take it out of the inner product as a constant factor:
+
+$$\begin{aligned}
+    \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar} e^{- i \hat{V} \Delta t / \hbar}}{x_j}
+    = e^{- i V(x_j) \Delta t / \hbar} \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
+\end{aligned}$$
+
+Here we insert the identity operator
+expanded in the momentum basis $\hat{I} = \int \Ket{p} \Bra{p} \dd{p}$,
+and commute it with the kinetic energy $\hat{T} = \hat{p}^2 / (2m)$ to get:
+
+$$\begin{aligned}
+    \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
+    = \int_{-\infty}^\infty \Inprod{x_{j+1}}{p} \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \Inprod{p}{x_j} \dd{p}
+\end{aligned}$$
+
+In the momentum basis $\Ket{p}$,
+the position basis vectors
+are represented by plane waves:
+
+$$\begin{aligned}
+    \Inprod{p}{x_j}
+    = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( \!-\! i \frac{x_j p}{\hbar} \Big)
+    \qquad
+    \Inprod{x_{j+1}}{p}
+    = \frac{1}{\sqrt{2 \pi \hbar}} \exp\!\Big( i \frac{x_{j+1} p}{\hbar} \Big)
+\end{aligned}$$
+
+With this, we return to the inner product and further evaluate the integral:
+
+$$\begin{aligned}
+    \matrixel{x_{j+1}}{e^{- i \hat{T} \Delta t / \hbar}}{x_j}
+    &= \frac{1}{2 \pi \hbar} \int_{-\infty}^\infty
+    \exp\!\Big(\!-\! i \frac{p^2 \Delta t}{2 m \hbar}\Big) \exp\!\Big(i \frac{(x_{j+1} - x_j) p}{\hbar}\Big) \:dp
+    \\
+    &= \frac{1}{2 \pi \hbar} \sqrt{\frac{2 \pi m \hbar}{i \Delta t}} \exp\!\Big( i \frac{m (x_{j+1} - x_j)^2}{2 \hbar \Delta t} \Big)
+\end{aligned}$$
+
+Inserting this back into the definition of the propagator $K(x_N, t_N; x_0, t_0)$ yields:
+
+$$\begin{aligned}
+    K
+    = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2}
+    \int\cdots\int
+    \exp\!\bigg(\! \sum_{j = 0}^{N - 1} i \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \hbar \Delta t} - \frac{V(x_j) \Delta t}{\hbar} \Big) \!\bigg)
+    \dd{x_1} \cdots \dd{x_{N-1}}
+\end{aligned}$$
+
+For large $N$ and small $\Delta t$, the sum in the exponent becomes an integral:
+
+$$\begin{aligned}
+    \frac{i}{\hbar} \sum_{j = 0}^{N - 1} \Big( \frac{m (x_{j+1} \!-\! x_j)^2}{2 \Delta t^2} - V(x_j) \Big) \Delta t
+    \quad \to \quad
+    \frac{i}{\hbar} \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
+\end{aligned}$$
+
+Upon closer inspection, this integral turns out to be the classical action $S[x]$,
+with the integrand being the Lagrangian $L$:
+
+$$\begin{aligned}
+    S[x(t)]
+    = \int_{t_0}^{t_N} L(x, \dot{x}, \tau) \dd{\tau}
+    = \int_{t_0}^{t_N} \Big( \frac{1}{2} m \dot{x}^2 - V(x) \Big) \dd{\tau}
+\end{aligned}$$
+
+The definition of the propagator $K$ is then further reduced to the following:
+
+$$\begin{aligned}
+    K
+    = \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2}
+    \int\cdots\int \exp(i S[x] / \hbar) \dd{x_1} \cdots \dd{x_{N-1}}
+\end{aligned}$$
+
+Finally, for the purpose of normalization,
+we define the integral over all paths $x(t)$ as follows,
+where we write $D[x]$ instead of $\dd{x}$:
+
+$$\begin{aligned}
+    \int D[x]
+    \equiv \lim_{N \to \infty} \Big( \frac{- i m}{2 \pi \hbar \Delta t} \Big)^{\!N / 2} \int\cdots\int \dd{x_1} \cdots \dd{x_{N-1}}
+\end{aligned}$$
+
+We thus arrive at **Feynman's path integral**,
+which sums over all possible paths $x(t)$:
+
+$$\begin{aligned}
+    K
+    = \int \exp(i S[x] / \hbar) \:D[x]
+    = A \sum_{\mathrm{all}\:x(t)} \exp(i S[x] / \hbar)
+\end{aligned}$$
+
+
+
+## References
+1.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.
+2.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.
diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
new file mode 100644
index 0000000..56af39b
--- /dev/null
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -0,0 +1,119 @@
+---
+title: "Pauli exclusion principle"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In quantum mechanics, the **Pauli exclusion principle** is a theorem with
+profound consequences for how the world works.
+
+Suppose we have a composite state
+$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$, where the two
+identical particles $x_1$ and $x_2$ each can occupy the same two allowed
+states $a$ and $b$. We then define the permutation operator $\hat{P}$ as
+follows:
+
+$$\begin{aligned}
+    \hat{P} \Ket{a}\Ket{b} = \Ket{b}\Ket{a}
+\end{aligned}$$
+
+That is, it swaps the states of the particles. Obviously, swapping the
+states twice simply gives the original configuration again, so:
+
+$$\begin{aligned}
+    \hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b}
+\end{aligned}$$
+
+Therefore, $\Ket{a}\Ket{b}$ is an eigenvector of $\hat{P}^2$ with
+eigenvalue $1$. Since $[\hat{P}, \hat{P}^2] = 0$, $\Ket{a}\Ket{b}$
+must also be an eigenket of $\hat{P}$ with eigenvalue $\lambda$,
+satisfying $\lambda^2 = 1$, so we know that $\lambda = 1$ or $\lambda = -1$:
+
+$$\begin{aligned}
+    \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
+\end{aligned}$$
+
+As it turns out, in nature, each class of particle has a single
+associated permutation eigenvalue $\lambda$, or in other words: whether
+$\lambda$ is $-1$ or $1$ depends on the type of particle that $x_1$
+and $x_2$ are. Particles with $\lambda = -1$ are called
+**fermions**, and those with $\lambda = 1$ are known as **bosons**. We
+define $\hat{P}_f$ with $\lambda = -1$ and $\hat{P}_b$ with
+$\lambda = 1$, such that:
+
+$$\begin{aligned}
+    \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
+    \qquad
+    \hat{P}_b \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = \Ket{a}\Ket{b}
+\end{aligned}$$
+
+Another fundamental fact of nature is that identical particles cannot be
+distinguished by any observation. Therefore it is impossible to tell
+apart $\Ket{a}\Ket{b}$ and the permuted state $\Ket{b}\Ket{a}$,
+regardless of the eigenvalue $\lambda$. There is no physical difference!
+
+But this does not mean that $\hat{P}$ is useless: despite not having any
+observable effect, the resulting difference between fermions and bosons
+is absolutely fundamental. Consider the following superposition state,
+where $\alpha$ and $\beta$ are unknown:
+
+$$\begin{aligned}
+    \Ket{\Psi(a, b)}
+    = \alpha \Ket{a}\Ket{b} + \beta \Ket{b}\Ket{a}
+\end{aligned}$$
+
+When we apply $\hat{P}$, we can "choose" between two "intepretations" of
+its action, both shown below. Obviously, since the left-hand sides are
+equal, the right-hand sides must be equal too:
+
+$$\begin{aligned}
+    \hat{P} \Ket{\Psi(a, b)}
+    &= \lambda \alpha \Ket{a}\Ket{b} + \lambda \beta \Ket{b}\Ket{a}
+    \\
+    \hat{P} \Ket{\Psi(a, b)}
+    &= \alpha \Ket{b}\Ket{a} + \beta \Ket{a}\Ket{b}
+\end{aligned}$$
+
+This gives us the equations $\lambda \alpha = \beta$ and
+$\lambda \beta = \alpha$. In fact, just from this we could have deduced
+that $\lambda$ can be either $-1$ or $1$. In any case, for bosons
+($\lambda = 1$), we thus find that $\alpha = \beta$:
+
+$$\begin{aligned}
+    \Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
+\end{aligned}$$
+
+Where $C$ is a normalization constant. As expected, this state is
+**symmetric**: switching $a$ and $b$ gives the same result. Meanwhile, for
+fermions ($\lambda = -1$), we find that $\alpha = -\beta$:
+
+$$\begin{aligned}
+    \Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
+\end{aligned}$$
+
+This state is called **antisymmetric** under exchange: switching $a$ and $b$
+causes a sign change, as we would expect for fermions.
+
+Now, what if the particles $x_1$ and $x_2$ are in the same state $a$?
+For bosons, we just need to update the normalization constant $C$:
+
+$$\begin{aligned}
+    \Ket{\Psi(a, a)}_b
+    = C \Ket{a}\Ket{a}
+\end{aligned}$$
+
+However, for fermions, the state is unnormalizable and thus unphysical:
+
+$$\begin{aligned}
+    \Ket{\Psi(a, a)}_f
+    = C \big( \Ket{a}\Ket{a} - \Ket{a}\Ket{a} \big)
+    = 0
+\end{aligned}$$
+
+And this is the Pauli exclusion principle: **fermions may never
+occupy the same quantum state**. One of the many notable consequences of
+this is that the shells of atoms only fit a limited number of
+electrons (which are fermions), since each must have a different quantum number.
diff --git a/source/know/concept/plancks-law/index.md b/source/know/concept/plancks-law/index.md
new file mode 100644
index 0000000..7ea06f5
--- /dev/null
+++ b/source/know/concept/plancks-law/index.md
@@ -0,0 +1,140 @@
+---
+title: "Planck's law"
+date: 2021-09-09
+categories:
+- Physics
+layout: "concept"
+---
+
+**Planck's law** describes the radiation spectrum of a **black body**:
+a theoretical object in thermal equilibrium,
+which absorbs photons,
+re-radiates them, and then re-absorbs them.
+
+Since the photon population varies with time,
+this is a [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
+and photons are bosons
+(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
+this system must obey the
+[Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
+with a chemical potential $\mu = 0$ (due to the freely varying population):
+
+$$\begin{aligned}
+    f_B(E)
+    = \frac{1}{\exp(\beta E) - 1}
+\end{aligned}$$
+
+Each photon has an energy $E = \hbar \omega = \hbar c k$,
+so the [density of states](/know/concept/density-of-states/)
+is as follows in 3D:
+
+$$\begin{aligned}
+    g(E)
+    = 2 \frac{g(k)}{E'(k)}
+    = \frac{V k^2}{\pi^2 \hbar c}
+    = \frac{V E^2}{\pi^2 \hbar^3 c^3}
+    = \frac{8 \pi V E^2}{h^3 c^3}
+\end{aligned}$$
+
+Where the factor of $2$ accounts for the photon's polarization degeneracy.
+We thus expect that the number of photons $N(E)$
+with an energy between $E$ and $E + \dd{E}$ is given by:
+
+$$\begin{aligned}
+    N(E) \dd{E}
+    = f_B(E) \: g(E) \dd{E}
+    = \frac{8 \pi V}{h^3 c^3} \frac{E^2}{\exp(\beta E) - 1} \dd{E}
+\end{aligned}$$
+
+By substituting $E = h \nu$, we find that the number of photons $N(\nu)$
+with a frequency between $\nu$ and $\nu + \dd{\nu}$ must be as follows:
+
+$$\begin{aligned}
+    N(\nu) \dd{\nu}
+    = \frac{8 \pi V}{c^3} \frac{\nu^2}{\exp(\beta h \nu) - 1} \dd{\nu}
+\end{aligned}$$
+
+Multiplying by the energy $h \nu$ yields the distribution of the radiated energy,
+which we divide by the volume $V$ to get Planck's law,
+also called the **Plank distribution**,
+describing a black body's radiated spectral energy density per unit volume:
+
+$$\begin{aligned}
+    \boxed{
+        u(\nu)
+        = \frac{8 \pi h}{c^3} \frac{\nu^3}{\exp(\beta h \nu) - 1}
+    }
+\end{aligned}$$
+
+
+## Wien's displacement law
+
+The Planck distribution peaks at a particular frequency $\nu_{\mathrm{max}}$,
+which can be found by solving the following equation for $\nu$:
+
+$$\begin{aligned}
+    0
+    = u'(\nu)
+    \quad \implies \quad
+    0
+    = 3 \nu^2 (\exp(\beta h \nu) - 1) - \nu^3 \beta h \exp(\beta h \nu)
+\end{aligned}$$
+
+By defining $x \equiv \beta h \nu_{\mathrm{max}}$,
+this turns into the following transcendental equation:
+
+$$\begin{aligned}
+    3
+    = (3 - x) \exp(x)
+\end{aligned}$$
+
+Whose numerical solution leads to **Wien's displacement law**, given by:
+
+$$\begin{aligned}
+    \boxed{
+        \frac{h \nu_{\mathrm{max}}}{k_B T}
+        \approx 2.822
+    }
+\end{aligned}$$
+
+Which states that the peak frequency $\nu_{\mathrm{max}}$
+is proportional to the temperature $T$.
+
+
+## Stefan-Boltzmann law
+
+Because $u(\nu)$ represents the radiated spectral energy density,
+we can find the total radiated energy $U$ per unit volume by integrating over $\nu$:
+
+$$\begin{aligned}
+    U
+    &= \int_0^\infty u(\nu) \dd{\nu}
+    = \frac{8 \pi h}{c^3} \int_0^\infty \frac{\nu^3}{\exp(\beta h \nu) - 1} \dd{\nu}
+    \\
+    &= \frac{8 \pi h}{\beta^3 h^3 c^3} \int_0^\infty \frac{(\beta h \nu)^3}{\exp(\beta h \nu) - 1} \dd{\nu}
+    = \frac{8 \pi}{\beta^4 h^3 c^3} \int_0^\infty \frac{x^3}{\exp(x) - 1} \dd{x}
+\end{aligned}$$
+
+This definite integral turns out to be $\pi^4/15$,
+leading us to the **Stefan-Boltzmann law**,
+which states that the radiated energy is proportional to $T^4$:
+
+$$\begin{aligned}
+    \boxed{
+        U = \frac{4 \sigma}{c} T^4
+    }
+\end{aligned}$$
+
+Where $\sigma$ is the **Stefan-Boltzmann constant**, which is defined as follows:
+
+$$\begin{aligned}
+    \sigma
+    \equiv \frac{2 \pi^5 k_B^4}{15 c^2 h^3}
+\end{aligned}$$
+
+
+
+## References
+1.  H. Gould, J. Tobochnik,
+    *Statistical and thermal physics*, 2nd edition,
+    Princeton.
diff --git a/source/know/concept/prandtl-equations/index.md b/source/know/concept/prandtl-equations/index.md
new file mode 100644
index 0000000..9e9e745
--- /dev/null
+++ b/source/know/concept/prandtl-equations/index.md
@@ -0,0 +1,205 @@
+---
+title: "Prandtl equations"
+date: 2021-05-29
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+In fluid dynamics, the **Prandtl equations** or **boundary layer equations**
+describe the movement of a [viscous](/know/concept/viscosity/) fluid
+with a large [Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \gg 1$
+close to a solid surface.
+
+Fluids with a large Reynolds number
+are often approximated as having zero viscosity,
+since the simpler [Euler equations](/know/concept/euler-equations)
+can then be used instead of the [Navier-Stokes equations](/know/concept/navier-stokes-equations/).
+
+However, in reality, a viscous fluid obeys the *no-slip* boundary condition:
+at every solid surface the local velocity must be zero.
+This implies the existence of a **boundary layer**:
+a thin layer of fluid "stuck" to solid objects in the flow,
+where viscosity plays an important role.
+This is in contrast to the ideal flow far away from the surface.
+
+We consider a simple theoretical case in 2D:
+a large flat surface located at $y = 0$ for all $x \in \mathbb{R}$,
+with a fluid *trying* to flow parallel to it at $U$.
+The 2D treatment can be justified by assuming that everything is constant in the $z$-direction.
+We will not solve this case,
+but instead derive general equations
+to describe the flow close to a flat surface.
+
+At the wall, there is a very thin boundary layer of thickness $\delta$,
+where the fluid is assumed to be completely stationary $\va{v} = 0$.
+We are mainly interested in the region $\delta < y \ll L$,
+where $L$ is the distance at which the fluid becomes practically ideal.
+This the so-called **slip-flow** region,
+in which the fluid is not stationary,
+but still viscosity-dominated.
+
+In 2D, the steady Navier-Stokes equations are as follows,
+where the flow $\va{v} = (v_x, v_y)$:
+
+$$\begin{aligned}
+    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    &= - \frac{1}{\rho} \pdv{p}{x} + \nu \Big( \pdvn{2}{v_x}{x} + \pdvn{2}{v_x}{y} \Big)
+    \\
+    v_x \pdv{v_y}{x} + v_y \pdv{v_y}{y}
+    &= - \frac{1}{\rho} \pdv{p}{y} + \nu \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big)
+    \\
+    \pdv{v_x}{x} + \pdv{v_y}{y}
+    &= 0
+\end{aligned}$$
+
+The latter represents the fluid's incompressibility.
+We non-dimensionalize these equations,
+and assume that changes along the $y$-axis
+happen on a short scale (say, $\delta$),
+and along the $x$-axis on a longer scale (say, $L$).
+Let $\tilde{x}$ and $\tilde{y}$ be dimenionless variables of order $1$:
+
+$$\begin{aligned}
+    x
+    = L \tilde{x}
+    \qquad \quad
+    y
+    = \delta \tilde{x}
+    \qquad \quad
+    \pdv{}{x}
+    = \frac{1}{L} \pdv{}{\tilde{x}}
+    \qquad \quad
+    \pdv{}{y}
+    = \frac{1}{\delta} \pdv{}{\tilde{y}}
+\end{aligned}$$
+
+Furthermore, we choose velocity scales
+to be consistent with the incompressibility condition,
+and a pressure scale inspired
+by [Bernoulli's theorem](/know/concept/bernoullis-theorem/):
+
+$$\begin{aligned}
+    v_x
+    = U \tilde{v}_x
+    \qquad \quad
+    v_y
+    = \frac{U \delta}{L} \tilde{v}_y
+    \qquad \quad
+    p
+    = \rho U^2 \tilde{p}
+\end{aligned}$$
+
+We insert these scalings into the Navier-Stokes equations, yielding:
+
+$$\begin{aligned}
+    \frac{U^2}{L} \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \frac{U^2}{L} \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \frac{U^2}{L} \pdv{\tilde{p}}{\tilde{x}}
+    + \nu \Big( \frac{U}{L^2} \pdvn{2}{\tilde{v}_x}{\tilde{x}} + \frac{U}{\delta^2} \pdvn{2}{\tilde{v}_x}{\tilde{y}} \Big)
+    \\
+    \frac{U^2 \delta}{L^2} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{U^2 \delta}{L^2} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \frac{U^2}{\delta} \pdv{\tilde{p}}{\tilde{y}}
+    + \nu \Big( \frac{U \delta}{L^3} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{U}{L \delta} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
+\end{aligned}$$
+
+For future convenience,
+we multiply the former equation by $L / U^2$, and the latter by $\delta / U^2$:
+
+$$\begin{aligned}
+    \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{x}}
+    + \nu \Big( \frac{1}{U L} \pdvn{2}{\tilde{v}_x}{\tilde{x}} + \frac{L}{U \delta^2} \pdvn{2}{\tilde{v}_x}{\tilde{y}} \Big)
+    \\
+    \frac{\delta^2}{L^2} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{\delta^2}{L^2} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{y}}
+    + \nu \Big( \frac{\delta^2}{U L^3} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
+\end{aligned}$$
+
+We would like to estimate $\delta$.
+Intuitively, we expect that higher viscosities $\nu$ give thicker layers,
+and that faster velocities $U$ give thinner layers.
+Furthermore, we expect *downstream thickening*:
+with distance $x$, viscous stresses slow down the slip-flow,
+leading to a gradual increase of $\delta(x)$.
+Some dimensional analysis thus yields the following estimate:
+
+$$\begin{aligned}
+    \delta
+    \approx \sqrt{\frac{\nu x}{U}}
+    \sim \sqrt{\frac{\nu L}{U}}
+\end{aligned}$$
+
+We thus insert $\delta = \sqrt{\nu L / U}$ into the Navier-Stokes equations, giving us:
+
+$$\begin{aligned}
+    \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{x}}
+    + \nu \Big( \frac{1}{U L} \pdvn{2}{\tilde{v}_x}{\tilde{x}} + \frac{1}{\nu} \pdvn{2}{\tilde{v}_x}{\tilde{y}} \Big)
+    \\
+    \frac{\nu}{U L} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{\nu}{U L} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{y}}
+    + \nu \Big( \frac{\nu}{U^2 L^2} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{U L} \pdvn{2}{\tilde{v}_y}{\tilde{y}} \Big)
+\end{aligned}$$
+
+Here, we recognize the definition of the Reynolds number $\mathrm{Re} = U L / \nu$:
+
+$$\begin{aligned}
+    \tilde{v}_x \pdv{\tilde{v}_x}{\tilde{x}} + \tilde{v}_y \pdv{\tilde{v}_x}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{x}}
+    + \frac{1}{\mathrm{Re}} \pdvn{2}{\tilde{v}_x}{\tilde{x}} + \pdvn{2}{\tilde{v}_x}{\tilde{y}}
+    \\
+    \frac{1}{\mathrm{Re}} \tilde{v}_x \pdv{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \tilde{v}_y \pdv{\tilde{v}_y}{\tilde{y}}
+    &= - \pdv{\tilde{p}}{\tilde{y}}
+    + \frac{1}{\mathrm{Re}^2} \pdvn{2}{\tilde{v}_y}{\tilde{x}} + \frac{1}{\mathrm{Re}} \pdvn{2}{\tilde{v}_y}{\tilde{y}}
+\end{aligned}$$
+
+Recall that we are only considering large Reynolds numbers $\mathrm{Re} \gg 1$,
+in which case $\mathrm{Re}^{-1} \ll 1$,
+so we can drop many terms, leaving us with these redimensionalized equations:
+
+$$\begin{aligned}
+    v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+    = - \frac{1}{\rho} \pdv{p}{x} + \nu \pdvn{2}{v_x}{y}
+    \qquad \quad
+    \pdv{p}{y}
+    = 0
+\end{aligned}$$
+
+The second one tells us that for a given $x$-value,
+the pressure is the same at the surface
+as in the main flow $y > L$, where the fluid is ideal.
+In the latter regime, we apply Bernoulli's theorem to rewrite $p$,
+using the *Bernoulli head* $H$ and the mainstream velocity $U(x)$:
+
+$$\begin{aligned}
+    p
+    = \rho H - \frac{1}{2} \rho U^2
+    = p_0 - \frac{1}{2} \rho U^2
+\end{aligned}$$
+
+Inserting this into the reduced Navier-Stokes equations,
+we arrive at the Prandtl equations:
+
+$$\begin{aligned}
+    \boxed{
+        v_x \pdv{v_x}{x} + v_y \pdv{v_x}{y}
+        = U \dv{U}{x} + \nu \pdvn{2}{v_x}{y}
+        \qquad \quad
+        \pdv{v_x}{x} + \pdv{v_y}{y}
+        = 0
+    }
+\end{aligned}$$
+
+A notable application of these equations is
+the [Blasius boundary layer](/know/concept/blasius-boundary-layer/),
+where the surface in question
+is a semi-infinite plane.
+
+
+
+## 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/probability-current/index.md b/source/know/concept/probability-current/index.md
new file mode 100644
index 0000000..6f85149
--- /dev/null
+++ b/source/know/concept/probability-current/index.md
@@ -0,0 +1,99 @@
+---
+title: "Probability current"
+date: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+layout: "concept"
+---
+
+In quantum mechanics, the **probability current** describes the movement
+of the probability of finding a particle at given point in space.
+In other words, it treats the particle as a heterogeneous fluid with density $|\psi|^2$.
+Now, the probability of finding the particle within a volume $V$ is:
+
+$$\begin{aligned}
+    P = \int_{V} | \psi |^2 \ddn{3}{\vb{r}}
+\end{aligned}$$
+
+As the system evolves in time, this probability may change, so we take
+its derivative with respect to time $t$, and when necessary substitute
+in the other side of the Schrödinger equation to get:
+
+$$\begin{aligned}
+    \pdv{P}{t}
+    &= \int_{V} \psi \pdv{\psi^*}{t} + \psi^* \pdv{\psi}{t} \ddn{3}{\vb{r}}
+    = \frac{i}{\hbar} \int_{V} \psi (\hat{H} \psi^*) - \psi^* (\hat{H} \psi) \ddn{3}{\vb{r}}
+    \\
+    &= \frac{i}{\hbar} \int_{V} \psi \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi^* + V(\vb{r}) \psi^* \Big)
+    - \psi^* \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi + V(\vb{r}) \psi \Big) \ddn{3}{\vb{r}}
+    \\
+    &= \frac{i \hbar}{2 m} \int_{V} - \psi \nabla^2 \psi^* + \psi^* \nabla^2 \psi \ddn{3}{\vb{r}}
+    = - \int_{V} \nabla \cdot \vb{J} \ddn{3}{\vb{r}}
+\end{aligned}$$
+
+Where we have defined the probability current $\vb{J}$ as follows in
+the $\vb{r}$-basis:
+
+$$\begin{aligned}
+    \vb{J}
+    = \frac{i \hbar}{2 m} (\psi \nabla \psi^* - \psi^* \nabla \psi)
+    = \mathrm{Re} \Big\{ \psi \frac{i \hbar}{m} \psi^* \Big\}
+\end{aligned}$$
+
+Let us rewrite this using the momentum operator
+$\vu{p} = -i \hbar \nabla$ as follows, noting that $\vu{p} / m$ is
+simply the velocity operator $\vu{v}$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{J}
+        = \frac{1}{2 m} ( \psi^* \vu{p} \psi - \psi \vu{p} \psi^*)
+        = \mathrm{Re} \Big\{ \psi^* \frac{\vu{p}}{m} \psi \Big\}
+        = \mathrm{Re} \{ \psi^* \vu{v} \psi \}
+    }
+\end{aligned}$$
+
+Returning to the derivation of $\vb{J}$, we now have the following
+equation:
+
+$$\begin{aligned}
+    \pdv{P}{t}
+    = \int_{V} \pdv{|\psi|^2}{t} \ddn{3}{\vb{r}}
+    = - \int_{V} \nabla \cdot \vb{J} \ddn{3}{\vb{r}}
+\end{aligned}$$
+
+By removing the integrals, we thus arrive at the **continuity equation**
+for $\vb{J}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{J}
+        = - \pdv{|\psi|^2}{t}
+    }
+\end{aligned}$$
+
+This states that the total probability is conserved, and is reminiscent of charge
+conservation in electromagnetism. In other words, the probability at a
+point can only change by letting it "flow" towards or away from it. Thus
+$\vb{J}$ represents the flow of probability, which is analogous to the
+motion of a particle.
+
+As a bonus, this still holds for a particle in an electromagnetic vector
+potential $\vb{A}$, thanks to the gauge invariance of the Schrödinger
+equation. We can thus extend the definition to a particle with charge
+$q$ in an SI-unit field, neglecting spin:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{J}
+        = \mathrm{Re} \Big\{ \psi^* \frac{\vu{p} - q \vb{A}}{m} \psi \Big\}
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  L.E. Ballentine,
+    *Quantum mechanics: a modern development*, 2nd edition,
+    World Scientific.
diff --git a/source/know/concept/propagator/index.md b/source/know/concept/propagator/index.md
new file mode 100644
index 0000000..3ed7fb7
--- /dev/null
+++ b/source/know/concept/propagator/index.md
@@ -0,0 +1,69 @@
+---
+title: "Propagator"
+date: 2021-07-04
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+In quantum mechanics, the **propagator** $K(x_f, t_f; x_i, t_i)$
+gives the probability amplitude that a particle
+starting at $x_i$ at $t_i$ ends up at position $x_f$ at $t_f$.
+It is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        K(x_f, t_f; x_i, t_i)
+        \equiv \matrixel{x_f}{\hat{U}(t_f, t_i)}{x_i}
+    }
+\end{aligned}$$
+
+Where $\hat{U} \equiv \exp(- i t \hat{H} / \hbar)$ is the time-evolution operator.
+The probability that a particle travels
+from $(x_i, t_i)$ to $(x_f, t_f)$ is then given by:
+
+$$\begin{aligned}
+    P
+    &= \big| K(x_f, t_f; x_i, t_i) \big|^2
+\end{aligned}$$
+
+Given a general (i.e. non-collapsed) initial state $\psi_i(x) \equiv \psi(x, t_i)$,
+we must integrate over $x_i$:
+
+$$\begin{aligned}
+    P
+    &= \bigg| \int_{-\infty}^\infty K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \bigg|^2
+\end{aligned}$$
+
+And if the final state $\psi_f(x) \equiv \psi(x, t_f)$
+is not a basis vector either, then we integrate twice:
+
+$$\begin{aligned}
+    P
+    &= \bigg| \iint_{-\infty}^\infty \psi_f^*(x_f) \: K(x_f, t_f; x_i, t_i) \: \psi_i(x_i) \dd{x_i} \dd{x_f} \bigg|^2
+\end{aligned}$$
+
+Given a $\psi_i(x)$, the propagator can also be used
+to find the full final wave function:
+
+$$\begin{aligned}
+    \boxed{
+        \psi(x_f, t_f)
+        = \int_{-\infty}^\infty \psi_i(x_i) K(x_f, t_f; x_i, t_i) \:dx_i
+    }
+\end{aligned}$$
+
+Sometimes the name "propagator" is also used to refer to
+the [fundamental solution](/know/concept/fundamental-solution/) $G$
+of the time-dependent Schrödinger equation,
+which is related to $K$ by:
+
+$$\begin{aligned}
+    \boxed{
+        G(x_f, t_f; x_i, t_i)
+        = - \frac{i}{\hbar} \: \Theta(t_f - t_i) \: K(x_f, t_f; x_i, t_i)
+    }
+\end{aligned}$$
+
+Where $\Theta(t)$ is the [Heaviside step function](/know/concept/heaviside-step-function/).
diff --git a/source/know/concept/pulay-mixing/index.md b/source/know/concept/pulay-mixing/index.md
new file mode 100644
index 0000000..030ad46
--- /dev/null
+++ b/source/know/concept/pulay-mixing/index.md
@@ -0,0 +1,160 @@
+---
+title: "Pulay mixing"
+date: 2021-03-02
+categories:
+- Numerical methods
+layout: "concept"
+---
+
+Some numerical problems are most easily solved *iteratively*,
+by generating a series $\rho_1$, $\rho_2$, etc.
+converging towards the desired solution $\rho_*$.
+**Pulay mixing**, also often called
+**direct inversion in the iterative subspace** (DIIS),
+can speed up the convergence for some types of problems,
+and also helps to avoid periodic divergences.
+
+The key concept it relies on is the **residual vector** $R_n$
+of the $n$th iteration, which in some way measures the error of the current $\rho_n$.
+Its exact definition varies,
+but is generally along the lines of the difference between
+the input of the iteration and the raw resulting output:
+
+$$\begin{aligned}
+	R_n
+	= R[\rho_n]
+	= \rho_n^\mathrm{new}[\rho_n] - \rho_n
+\end{aligned}$$
+
+It is not always clear what to do with $\rho_n^\mathrm{new}$.
+Directly using it as the next input ($\rho_{n+1} = \rho_n^\mathrm{new}$)
+often leads to oscillation,
+and linear mixing ($\rho_{n+1} = (1\!-\!f) \rho_n + f \rho_n^\mathrm{new}$)
+can take a very long time to converge properly.
+Pulay mixing offers an improvement.
+
+The idea is to construct the next iteration's input $\rho_{n+1}$
+as a linear combination of the previous inputs $\rho_1$, $\rho_2$, ..., $\rho_n$,
+such that it is as close as possible to the optimal $\rho_*$:
+
+$$\begin{aligned}
+	\boxed{
+		\rho_{n+1}
+		= \sum_{m = 1}^n \alpha_m \rho_m
+	}
+\end{aligned}$$
+
+To do so, we make two assumptions.
+Firstly, the current $\rho_n$ is already close to $\rho_*$,
+so that such a linear combination makes sense.
+Secondly, the iteration is linear,
+such that the raw output $\rho_{n+1}^\mathrm{new}$
+is also a linear combination with the *same coefficients*:
+
+$$\begin{aligned}
+	\rho_{n+1}^\mathrm{new}
+	= \sum_{m = 1}^n \alpha_m \rho_m^\mathrm{new}
+\end{aligned}$$
+
+We will return to these assumptions later.
+The point is that $R_{n+1}$ is also a linear combination:
+
+$$\begin{aligned}
+	R_{n+1}
+	= \rho_{n+1}^\mathrm{new} - \rho_{n+1}
+	= \sum_{m = 1}^n \alpha_m \rho_m^\mathrm{new} - \sum_{m = 1}^n \alpha_m \rho_m
+	= \sum_{m = 1}^n \alpha_m R_m
+\end{aligned}$$
+
+The goal is to choose the coefficients $\alpha_m$ such that
+the norm of the error $|R_{n+1}| \approx 0$,
+subject to the following constraint to preserve the normalization of $\rho_{n+1}$:
+
+$$\begin{aligned}
+	\sum_{m=1}^n \alpha_m = 1
+\end{aligned}$$
+
+We thus want to minimize the following quantity,
+where $\lambda$ is a [Lagrange multiplier](/know/concept/lagrange-multiplier/):
+
+$$\begin{aligned}
+	\Inprod{R_{n+1}}{R_{n+1}} + \lambda \sum_{m = 1}^n \alpha_m^*
+	= \sum_{m=1}^n \alpha_m^* \Big( \sum_{k=1}^n \alpha_k \Inprod{R_m}{R_k} + \lambda \Big)
+\end{aligned}$$
+
+By differentiating the right-hand side with respect to $\alpha_m^*$
+and demanding that the result is zero,
+we get a system of equations that we can write in matrix form,
+which is cheap to solve:
+
+$$\begin{aligned}
+	\begin{bmatrix}
+		\Inprod{R_1}{R_1} & \cdots & \Inprod{R_1}{R_n} & 1 \\
+		\vdots & \ddots & \vdots & \vdots \\
+		\Inprod{R_n}{R_1} & \cdots & \Inprod{R_n}{R_n} & 1 \\
+		1 & \cdots & 1 & 0
+	\end{bmatrix}
+	\cdot
+	\begin{bmatrix}
+		\alpha_1 \\ \vdots \\ \alpha_n \\ \lambda
+	\end{bmatrix}
+	=
+	\begin{bmatrix}
+		0 \\ \vdots \\ 0 \\ 1
+	\end{bmatrix}
+\end{aligned}$$
+
+From this, we can also see that the Lagrange multiplier
+$\lambda = - \Inprod{R_{n+1}}{R_{n+1}}$,
+where $R_{n+1}$ is the *predicted* residual of the next iteration,
+subject to the two assumptions.
+
+However, in practice, the earlier inputs $\rho_1$, $\rho_2$, etc.
+are much further from $\rho_*$ than $\rho_n$,
+so usually only the most recent $N\!+\!1$ inputs $\rho_{n - N}$, ..., $\rho_n$ are used:
+
+$$\begin{aligned}
+	\rho_{n+1}
+	= \sum_{m = n-N}^n \alpha_m \rho_m
+\end{aligned}$$
+
+You might be confused by the absence of any $\rho_m^\mathrm{new}$
+in the creation of $\rho_{n+1}$, as if the iteration's outputs are being ignored.
+This is due to the first assumption,
+which states that $\rho_n^\mathrm{new}$ and $\rho_n$ are already similar,
+such that they are basically interchangeable.
+
+Speaking of which, about those assumptions:
+while they will clearly become more accurate as $\rho_n$ approaches $\rho_*$,
+they might be very dubious in the beginning.
+A consequence of this is that the early iterations might get "trapped"
+in a suboptimal subspace spanned by $\rho_1$, $\rho_2$, etc.
+To say it another way, we would be varying $n$ coefficients $\alpha_m$
+to try to optimize a $D$-dimensional $\rho_{n+1}$,
+where in general $D \gg n$, at least in the beginning.
+
+There is an easy fix to this problem:
+add a small amount of the raw residual $R_m$
+to "nudge" $\rho_{n+1}$ towards the right subspace,
+where $\beta \in [0,1]$ is a tunable parameter:
+
+$$\begin{aligned}
+	\boxed{
+		\rho_{n+1}
+		= \sum_{m = N}^n \alpha_m (\rho_m + \beta R_m)
+	}
+\end{aligned}$$
+
+In other words, we end up introducing a small amount of the raw outputs $\rho_m^\mathrm{new}$,
+while still giving more weight to iterations with smaller residuals.
+
+Pulay mixing is very effective for certain types of problems,
+e.g. density functional theory,
+where it can accelerate convergence by up to two orders of magnitude!
+
+
+
+## References
+1.  P. Pulay,
+    [Convergence acceleration of iterative sequences. The case of SCF iteration](https://doi.org/10.1016/0009-2614(80)80396-4),
+    1980, Elsevier.
diff --git a/source/know/concept/quantum-entanglement/index.md b/source/know/concept/quantum-entanglement/index.md
new file mode 100644
index 0000000..72aa91b
--- /dev/null
+++ b/source/know/concept/quantum-entanglement/index.md
@@ -0,0 +1,151 @@
+---
+title: "Quantum entanglement"
+date: 2021-03-07
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+layout: "concept"
+---
+
+Consider a composite quantum system which consists of two subsystems $A$ and $B$,
+respectively with basis states $\Ket{a_n}$ and $\Ket{b_n}$.
+All accessible states of the sytem $\Ket{\Psi}$ lie in
+the tensor product of the subsystems'
+[Hilbert spaces](/know/concept/hilbert-space/) $\mathbb{H}_A$ and $\mathbb{H}_B$:
+
+$$\begin{aligned}
+    \Ket{\Psi} \in \mathbb{H}_A \otimes \mathbb{H}_B
+\end{aligned}$$
+
+A subset of these states can be written as the tensor product (i.e. Kronecker product in a basis)
+of a state $\Ket{\alpha}$ in $A$ and a state $\Ket{\beta}$ in $B$,
+often abbreviated as $\Ket{\alpha} \Ket{\beta}$:
+
+$$\begin{aligned}
+    \Ket{\Psi}
+    = \Ket{\alpha} \Ket{\beta}
+    = \Ket{\alpha} \otimes \Ket{\beta}
+\end{aligned}$$
+
+The states that can be written in this way are called **separable**,
+and states that cannot are called **entangled**.
+Therefore, we are dealing with **quantum entanglement**
+if the state of subsystem $A$ cannot be fully described
+independently of the state of subsystem $B$, and vice versa.
+
+To detect and quantify entanglement,
+we can use the [density operator](/know/concept/density-operator/) $\hat{\rho}$.
+For a pure ensemble in a given (possibly entangled) state $\Ket{\Psi}$,
+$\hat{\rho}$ is given by:
+
+$$\begin{aligned}
+    \hat{\rho} = \Ket{\Psi} \Bra{\Psi}
+\end{aligned}$$
+
+From this, we would like to extract the corresponding state of subsystem $A$.
+For that purpose, we define the **reduced density operator** $\hat{\rho}_A$ of subsystem $A$ as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}_A
+        = \Tr_B(\hat{\rho})
+        = \sum_m \Bra{b_m} \Big( \hat{\rho} \Big) \Ket{b_m}
+    }
+\end{aligned}$$
+
+Where $\Tr_B(\hat{\rho})$ is called the **partial trace** of $\hat{\rho}$,
+which basically eliminates subsystem $B$ from $\hat{\rho}$.
+For a pure composite state $\Ket{\Psi}$,
+the resulting $\hat{\rho}_A$ describes a pure state in $A$ if $\Ket{\Psi}$ is separable,
+else, if $\Ket{\Psi}$ is entangled, it describes a mixed state in $A$.
+In the former case we simply find:
+
+$$\begin{aligned}
+    \boxed{
+        \Ket{\Psi} = \Ket{\alpha} \otimes \Ket{\beta}
+        \quad \implies \quad
+        \hat{\rho}_A = \Ket{\alpha} \Bra{\alpha}
+    }
+\end{aligned}$$
+
+We call $\Ket{\Psi}$ **maximally entangled**
+if its reduced density operators are **maximally mixed**,
+where $N$ is the dimension of $\mathbb{H}_A$ and $\hat{I}$ is the identity matrix:
+
+$$\begin{aligned}
+    \hat{\rho}_A
+    = \frac{1}{N} \hat{I}
+\end{aligned}$$
+
+Suppose that we are given an entangled pure state
+$\Ket{\Psi} \neq \Ket{\alpha} \otimes \Ket{\beta}$.
+Then the partial traces $\hat{\rho}_A$ and $\hat{\rho}_B$
+of $\hat{\rho} = \Ket{\Psi} \Bra{\Psi}$ are mixed states with the same probabilities $p_n$
+(assuming $\mathbb{H}_A$ and $\mathbb{H}_B$ have the same dimensions,
+which is usually the case):
+
+$$\begin{aligned}
+    \hat{\rho}_A
+    = \Tr_B(\hat{\rho})
+    = \sum_n p_n \Ket{a_n} \Bra{a_n}
+    \qquad \quad
+    \hat{\rho}_B
+    = \Tr_A(\hat{\rho})
+    = \sum_n p_n \Ket{b_n} \Bra{b_n}
+\end{aligned}$$
+
+There exists an orthonormal choice
+of the subsystem basis states $\Ket{a_n}$ and $\Ket{b_n}$,
+such that $\Ket{\Psi}$ can be written as follows,
+where $p_n$ are the probabilities in the reduced density operators:
+
+$$\begin{aligned}
+    \Ket{\Psi}
+    = \sum_n \sqrt{p_n} \Big( \Ket{a_n} \otimes \Ket{b_n} \Big)
+\end{aligned}$$
+
+This is the **Schmidt decomposition**,
+and the **Schmidt number** is the number of nonzero terms in the summation,
+which can be used to determine if the state $\Ket{\Psi}$
+is entangled (greater than one) or separable (equal to one).
+
+By looking at the Schmidt decomposition, we can notice that,
+if $\hat{O}_A$ and $\hat{O}_B$ are the subsystem observables
+with basis eigenstates $\Ket{a_n}$ and $\Ket{b_n}$,
+then measurement results of these operators
+will be perfectly correlated across $A$ and $B$.
+This is a general property of entangled systems,
+but beware: correlation does not imply entanglement!
+
+But what if the composite system is in a mixed state $\hat{\rho}$?
+The state is separable if and only if:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\rho}
+        = \sum_m p_m \Big( \hat{\rho}_A \otimes \hat{\rho}_B \Big)
+    }
+\end{aligned}$$
+
+Where $p_m$ are probabilities,
+and $\hat{\rho}_A$ and $\hat{\rho}_B$ can be any subsystem states.
+In reality, it is very hard to determine, using this criterium,
+whether an arbitrary given $\hat{\rho}$ is separable or not.
+
+As a final side note, the expectation value
+of an obervable $\hat{O}_A$ acting only on $A$ is given by:
+
+$$\begin{aligned}
+    \expval{\hat{O}_A}
+    = \Tr\!\big(\hat{\rho} \hat{O}_A\big)
+    = \Tr_A\!\big(\Tr_B(\hat{\rho} \hat{O}_A)\big)
+    = \Tr_A\!\big(\Tr_B(\hat{\rho}) \hat{O}_A)\big)
+    = \Tr_A\!\big(\hat{\rho}_A \hat{O}_A\big)
+\end{aligned}$$
+
+
+## References
+1.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.
diff --git a/source/know/concept/quantum-fourier-transform/index.md b/source/know/concept/quantum-fourier-transform/index.md
new file mode 100644
index 0000000..1ebf1d1
--- /dev/null
+++ b/source/know/concept/quantum-fourier-transform/index.md
@@ -0,0 +1,198 @@
+---
+title: "Quantum Fourier transform"
+date: 2021-05-01
+categories:
+- Algorithms
+- Quantum information
+layout: "concept"
+---
+
+The **quantum Fourier transform (QFT)** is a quantum counterpart
+of the classical discrete Fourier transform.
+It is defined like so, where $\Ket{x}$
+is an $n$-qubit computational basis state $\Ket{x_1} \cdots \Ket{x_n}$,
+and $\omega_N$ is an $N$th complex root of unity $\omega_N^N = 1$ with $N = 2^n$:
+
+$$\begin{aligned}
+    \boxed{
+        \Ket{x}
+        \:\to\:
+        \frac{1}{\sqrt{N}} \sum_{k = 0}^{N - 1} \omega_N^{xk} \Ket{k}
+    }
+\end{aligned}$$
+
+Note that $\Ket{x}$ and $\Ket{k}$ refer to the same basis set;
+we use these names to clarify which "space" we are considering.
+Furthermore, note the sign of the exponent of $\omega_N$,
+which is the opposite of the classical DFT convention.
+In other words, the *forward* QFT corresponds to the *inverse* DFT.
+
+The **inverse quantum Fourier transform (iQFT)** has a different sign in the exponent:
+
+$$\begin{aligned}
+    \boxed{
+        \Ket{k}
+        \:\to\:
+        \frac{1}{\sqrt{N}} \sum_{x = 0}^{N - 1} \omega_N^{-xk} \Ket{x}
+    }
+\end{aligned}$$
+
+The above definitions of the QFT and iQFT describe
+the effect on a single basis vector $\Ket{x}$,
+so the effect on an abitrary superposition follows from linearity,
+e.g. for the QFT:
+
+$$\begin{aligned}
+    \sum_{x = 0}^{N - 1} c_x \Ket{x}
+    \:\to\:
+    \sum_{x = 0}^{N - 1} c_x \bigg( \frac{1}{\sqrt{N}} \sum_{k = 0}^{N - 1} \omega_N^{xk} \Ket{k} \bigg)
+\end{aligned}$$
+
+Classically, such a double sum takes
+$\mathcal{O}(N^2) = \mathcal{O}(2^{2n})$ time to evaluate naively,
+with a potential improvement to $\mathcal{O}(N \log{N}) = \mathcal{O}(n 2^n)$
+for smarter algorithms.
+Quantum computers can do a QFT in $\mathcal{O}(\log^2(N)) = \mathcal{O}(n^2)$ time,
+and approximate it in $\mathcal{O}(n \log{n})$ time.
+
+To find out how, we look at the forward QFT,
+which maps a basis state $\Ket{x}$ to another state $\Ket{\tilde{x}}$:
+
+$$\begin{aligned}
+    \Ket{x}
+    \:\to\:
+    \Ket{\tilde{x}}
+    = \frac{1}{\sqrt{N}} \sum_{k = 0}^{N - 1} \omega_N^{xk} \Ket{k}
+    = \frac{1}{\sqrt{N}} \sum_{k = 0}^{N - 1} \exp\!\bigg( \frac{i 2 \pi x k}{N} \bigg) \Ket{k}
+\end{aligned}$$
+
+We can decompose $k$ into its binary representation
+$k_1 2^{n-1} + k_2 2^{n-2} + ... + k_n 2^{0}$:
+
+$$\begin{aligned}
+    \Ket{\tilde{x}}
+    &= \frac{1}{\sqrt{2^n}} \sum_{k = 0}^{2^n - 1} \exp\!\bigg( \frac{i 2 \pi x}{2^n} \sum_{j = 1}^{n} k_j 2^{n-j} \bigg) \Ket{k}
+    \\
+    &= \frac{1}{\sqrt{2^n}} \sum_{k = 0}^{2^n - 1} \exp\!\bigg( i 2 \pi x \sum_{j = 1}^{n} \frac{k_j}{2^j} \bigg) \Ket{k}
+\end{aligned}$$
+
+Expanding the exponential-of-a-sum into a product-of-exponentials then yields:
+
+$$\begin{aligned}
+    \Ket{\tilde{x}}
+    &= \frac{1}{\sqrt{2^n}} \sum_{k = 0}^{2^n - 1} \prod_{j = 1}^{n} \exp\!\bigg( i 2 \pi x \frac{k_j}{2^j} \bigg) \Ket{k}
+\end{aligned}$$
+
+If $k_j = 0$, the exponential is zero.
+We can thus separate this state into its individual qubits:
+
+$$\begin{aligned}
+    \Ket{\tilde{x}}
+    &= \bigotimes_{j = 1}^{n} \frac{1}{\sqrt{2}} \bigg( \Ket{0} + \exp\!\Big( \frac{i 2 \pi x}{2^j} \Big) \Ket{1} \bigg)
+\end{aligned}$$
+
+Next, we use the trick from before:
+decompose $x$ as $x_1 2^{n-1} + x_2 2^{n-2} + ... + x_n 2^{0}$:
+
+$$\begin{aligned}
+    \Ket{\tilde{x}}
+    &= \bigotimes_{j = 1}^{n} \frac{1}{\sqrt{2}} \bigg( \Ket{0} + \exp\!\Big( i 2 \pi \sum_{r = 1}^{n} x_r 2^{n-r-j} \Big) \Ket{1} \bigg)
+\end{aligned}$$
+
+The factor $2^{n-r-j}$ may be smaller or bigger than $1$,
+so it is convenient for us to use the following notation
+for non-integer binary numbers.
+Note the decimal point in the middle:
+
+$$\begin{aligned}
+    \left[ a_1 \cdots a_{d} \:.\: a_{d+1} \cdots a_{n} \right]
+    = \sum_{r = 1}^{n} a_r 2^{d - r}
+\end{aligned}$$
+
+In the above QFT state, the position of the decimal point is $d = n - j$,
+so we write it as:
+
+$$\begin{aligned}
+    \Ket{\tilde{x}}
+    &= \bigotimes_{j = 1}^{n} \frac{1}{\sqrt{2}} \bigg( \Ket{0}
+    + \exp\!\Big( i 2 \pi \big[ x_1 \cdots x_{n-j} \:.\: x_{n-j+1} \cdots x_n \big] \Big) \Ket{1} \bigg)
+\end{aligned}$$
+
+Because $\exp(i 2 \pi m) = 1$ for all integers $m$,
+we can discard bits before the decimal point:
+
+$$\begin{aligned}
+    \Ket{\tilde{x}}
+    &= \frac{1}{\sqrt{2^n}} \bigotimes_{j = 1}^{n} \bigg( \Ket{0} + \exp\!\Big( i 2 \pi \big[ 0\:.\: x_{n-j+1} \cdots x_n \big] \Big) \Ket{1} \bigg)
+    \\
+    &= \frac{1}{\sqrt{2^n}} \bigg( \Ket{0} + \exp\!\Big( i 2 \pi \big[ 0. x_n \big] \Big) \Ket{1} \bigg)
+    \otimes \bigg( \Ket{0} + \exp\!\Big( i 2 \pi \big[ 0. x_{n-1} x_n \big] \Big) \Ket{1} \bigg)
+    \otimes \cdots
+\end{aligned}$$
+
+Furthermore, each exponential can be factorized,
+with every factor containing one bit of $x$:
+
+$$\begin{aligned}
+    \exp\!\Big( i 2 \pi \big[ 0\:.\: x_{n-j+1} \cdots x_n \big] \Big)
+    = \exp\!\Big( i 2 \pi \frac{x_{n-j+1}}{2} \Big) \cdots \exp\!\Big( i 2 \pi \frac{x_{n}}{2^{j}} \Big)
+\end{aligned}$$
+
+This suggests a way to implement the QFT using
+[quantum gates](/know/concept/quantum-gate/) in a circuit.
+If the $j$th qubit is in $\Ket{+} = (\Ket{0} + \Ket{1})/\sqrt{2}$,
+then for each bit $x_{n-j+r}$ where $r \in \{1, ..., j\}$:
+
++ If $x_{n-j+r} = 0$, do nothing.
++ If $x_{n-j+r} = 1$, add a relative phase $2 \pi / 2^{r}$.
+
+The full QFT algorithm therefore proceeds as follows,
+for the $(n\!-\!j\!+\!1)$'th input $\Ket{x_{n-j+1}}$:
+
+1.  Apply the Hadamard gate $H$.
+    If $x_{n-j+1} = 0$, this puts the qubit in $\Ket{+}$.
+    If $x_{n-j+1} = 1$, this puts the qubit in $\Ket{+}$,
+    and then adds a phase $\pi$, yielding $\Ket{-}$.
+2.  Apply the phase shift gate $R_{\phi}$ controlled by $x_{n-j+2}$,
+    with angle $\phi = 2 \pi / 2^{2}$.
+3.  Apply $R_{\phi}$ controlled by $x_{n-j+3}$,
+    with angle $\phi = 2 \pi / 2^{3}$...
+4.  And so on, until the $n$th bit $x_n$ is reached,
+    and used to control $R_\phi$ with $\phi = 2 \pi / 2^{j}$.
+
+And so on, for each $j \in \{1, ..., n\}$,
+and we reach the above expression for $\Ket{\tilde{x}}$.
+We started from the $(n\!-\!j\!+\!1)$'th input qubit,
+i.e. we read the input in reverse order.
+Therefore all the qubits need to be swapped back to front,
+either before or after the above algorithm is run.
+
+The quantum circuit to execute the mentioned steps is illustrated below,
+excluding the swapping part to get the right order.
+Here, $R_m$ means $R_\phi$ with $\phi = 2 \pi / 2^m$:
+
+<a href="qft-circuit-noswap.png">
+<img src="qft-circuit-noswap.png" style="width:100%">
+</a>
+
+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:
+
+<a href="qft-circuit-swap.png">
+<img src="qft-circuit-swap.png" style="width:85%">
+</a>
+
+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.
+
diff --git a/source/know/concept/quantum-fourier-transform/qft-circuit-noswap.png b/source/know/concept/quantum-fourier-transform/qft-circuit-noswap.png
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/quantum-fourier-transform/qft-circuit-noswap.png
diff --git a/source/know/concept/quantum-fourier-transform/qft-circuit-swap.png b/source/know/concept/quantum-fourier-transform/qft-circuit-swap.png
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/quantum-fourier-transform/qft-circuit-swap.png
diff --git a/source/know/concept/quantum-gate/cnot.png b/source/know/concept/quantum-gate/cnot.png
new file mode 100644
index 0000000..5389d11
--- /dev/null
+++ b/source/know/concept/quantum-gate/cnot.png
diff --git a/source/know/concept/quantum-gate/cu.png b/source/know/concept/quantum-gate/cu.png
new file mode 100644
index 0000000..a75b9e7
--- /dev/null
+++ b/source/know/concept/quantum-gate/cu.png
diff --git a/source/know/concept/quantum-gate/index.md b/source/know/concept/quantum-gate/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/quantum-gate/index.md
@@ -0,0 +1,296 @@
+---
+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 $\Ket{\psi_2}$:
+
+<a href="swap.png">
+<img src="swap.png" style="width:22%">
+</a>
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{SWAP} =
+        \begin{bmatrix}
+            1 & 0 & 0 & 0 \\
+            0 & 0 & 1 & 0 \\
+            0 & 1 & 0 & 0 \\
+            0 & 0 & 0 & 1
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+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:
+
+<a href="cnot.png">
+<img src="cnot.png" style="width:22%">
+</a>
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{CNOT} =
+        \begin{bmatrix}
+            1 & 0 & 0 & 0 \\
+            0 & 1 & 0 & 0 \\
+            0 & 0 & 0 & 1 \\
+            0 & 0 & 1 & 0
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+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:
+
+<a href="cu.png">
+<img src="cu.png" style="width:22%">
+</a>
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{CU} =
+        \begin{bmatrix}
+            1 & 0 & 0 & 0 \\
+            0 & 1 & 0 & 0 \\
+            0 & 0 & u_{00} & u_{01} \\
+            0 & 0 & u_{10} & u_{11}
+        \end{bmatrix}
+    }
+\end{aligned}$$
+
+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.
diff --git a/source/know/concept/quantum-gate/swap.png b/source/know/concept/quantum-gate/swap.png
new file mode 100644
index 0000000..2262c53
--- /dev/null
+++ b/source/know/concept/quantum-gate/swap.png
diff --git a/source/know/concept/quantum-teleportation/index.md b/source/know/concept/quantum-teleportation/index.md
new file mode 100644
index 0000000..24df7ed
--- /dev/null
+++ b/source/know/concept/quantum-teleportation/index.md
@@ -0,0 +1,145 @@
+---
+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
new file mode 100644
index 0000000..8f2b918
--- /dev/null
+++ b/source/know/concept/rabi-oscillation/index.md
@@ -0,0 +1,215 @@
+---
+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
new file mode 100644
index 0000000..fb465be
--- /dev/null
+++ b/source/know/concept/random-phase-approximation/dyson.png
diff --git a/source/know/concept/random-phase-approximation/index.md b/source/know/concept/random-phase-approximation/index.md
new file mode 100644
index 0000000..ab43a0e
--- /dev/null
+++ b/source/know/concept/random-phase-approximation/index.md
@@ -0,0 +1,179 @@
+---
+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:
+
+<a href="rpasigma.png">
+<img src="rpasigma.png" style="width:92%">
+</a>
+
+Where we have defined the **screened interaction** $W^\mathrm{RPA}$,
+denoted by a double wavy line:
+
+<a href="screened.png">
+<img src="screened.png" style="width:95%">
+</a>
+
+Rearranging the above sequence of diagrams quickly leads to the following
+[Dyson equation](/know/concept/dyson-equation/):
+
+<a href="dyson.png">
+<img src="dyson.png" style="width:55%">
+</a>
+
+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 $\tilde{\vb{q}} \equiv (\vb{q}, i \omega_n^F)$:
+
+<a href="pairbubble.png">
+<img src="pairbubble.png" style="width:45%">
+</a>
+
+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.
diff --git a/source/know/concept/random-phase-approximation/pairbubble.png b/source/know/concept/random-phase-approximation/pairbubble.png
new file mode 100644
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+++ b/source/know/concept/random-phase-approximation/pairbubble.png
diff --git a/source/know/concept/random-phase-approximation/rpasigma.png b/source/know/concept/random-phase-approximation/rpasigma.png
new file mode 100644
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+++ b/source/know/concept/random-phase-approximation/rpasigma.png
diff --git a/source/know/concept/random-phase-approximation/screened.png b/source/know/concept/random-phase-approximation/screened.png
new file mode 100644
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diff --git a/source/know/concept/random-variable/index.md b/source/know/concept/random-variable/index.md
new file mode 100644
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@@ -0,0 +1,202 @@
+---
+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
new file mode 100644
index 0000000..fc4cc69
--- /dev/null
+++ b/source/know/concept/renyi-entropy/index.md
@@ -0,0 +1,108 @@
+---
+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.
diff --git a/source/know/concept/repetition-code/bit-flip-detect.png b/source/know/concept/repetition-code/bit-flip-detect.png
new file mode 100644
index 0000000..4928dc1
--- /dev/null
+++ b/source/know/concept/repetition-code/bit-flip-detect.png
diff --git a/source/know/concept/repetition-code/bit-flip-encode.png b/source/know/concept/repetition-code/bit-flip-encode.png
new file mode 100644
index 0000000..e1afe96
--- /dev/null
+++ b/source/know/concept/repetition-code/bit-flip-encode.png
diff --git a/source/know/concept/repetition-code/index.md b/source/know/concept/repetition-code/index.md
new file mode 100644
index 0000000..0e992c4
--- /dev/null
+++ b/source/know/concept/repetition-code/index.md
@@ -0,0 +1,343 @@
+---
+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/):
+
+<a href="bit-flip-encode.png">
+<img src="bit-flip-encode.png" style="width:32%">
+</a>
+
+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:
+
+<table style="width:30%;margin:auto;text-align:center;">
+    <tr>
+        <th>Error</th>
+        <th>$ZZI$</th>
+        <th>$IZZ$</th>
+    </tr>
+    <tr>
+        <td>$I$</td>
+        <td>$+1$</td>
+        <td>$+1$</td>
+    </tr>
+    <tr>
+        <td>$X_1$</td>
+        <td>$-1$</td>
+        <td>$+1$</td>
+    </tr>
+    <tr>
+        <td>$X_2$</td>
+        <td>$-1$</td>
+        <td>$-1$</td>
+    </tr>
+    <tr>
+        <td>$X_1$</td>
+        <td>$+1$</td>
+        <td>$-1$</td>
+    </tr>
+</table>
+
+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:
+
+<a href="bit-flip-detect.png">
+<img src="bit-flip-detect.png" style="width:62%">
+</a>
+
+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 href="phase-flip-encode.png">
+<img src="phase-flip-encode.png" style="width:40%">
+</a>
+
+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:
+
+<a href="phase-flip-detect.png">
+<img src="phase-flip-detect.png" style="width:70%">
+</a>
+
+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:
+
+<a href="shor-code-encode.png">
+<img src="shor-code-encode.png" style="width:55%">
+</a>
+
+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.
+
diff --git a/source/know/concept/repetition-code/phase-flip-detect.png b/source/know/concept/repetition-code/phase-flip-detect.png
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diff --git a/source/know/concept/repetition-code/shor-code-encode.png b/source/know/concept/repetition-code/shor-code-encode.png
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diff --git a/source/know/concept/residue-theorem/index.md b/source/know/concept/residue-theorem/index.md
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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 $C$:
+
+$$\begin{aligned}
+    \boxed{
+        \oint_C f(z) \dd{z} = i 2 \pi \sum_{z_p} R_p
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-res-theorem"/>
+<label for="proof-res-theorem">Proof</label>
+<div class="hidden">
+<label for="proof-res-theorem">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \oint_C f(z) \dd{z}
+    &= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
+    = \sum_{p} R_p \: 2 \pi i
+\end{aligned}$$
+</div>
+</div>
+
+This theorem might not seem very useful,
+but in fact, by cleverly choosing the contour $C$,
+it lets us evaluate many integrals along the real axis,
+most notably [Fourier transforms](/know/concept/fourier-transform/).
+It can also be used to derive the [Kramers-Kronig relations](/know/concept/kramers-kronig-relations).
diff --git a/source/know/concept/reynolds-number/index.md b/source/know/concept/reynolds-number/index.md
new file mode 100644
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@@ -0,0 +1,158 @@
+---
+title: "Reynolds number"
+date: 2021-05-04
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+layout: "concept"
+---
+
+The [Navier-Stokes equations](/know/concept/navier-stokes-equations/)
+are infamously tricky to solve,
+so we would like a way to qualitatively predict
+the behaviour of a fluid without needing the flow $\va{v}$.
+Consider the main equation:
+
+$$\begin{aligned}
+    \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
+    = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+\end{aligned}$$
+
+In this case, the gravity term $\va{g}$
+has been absorbed into the pressure term:
+$p \to p\!+\!\rho \Phi$,
+where $\Phi$ is the gravitational scalar potential,
+i.e. $\va{g} = - \nabla \Phi$.
+
+Let us introduce the dimensionless variables $\va{v}'$, $\va{r}'$, $t'$ and $p'$,
+where $U$ and $L$ are respectively a characteristic velocity and length
+of the system at hand:
+
+$$\begin{aligned}
+    \va{v} = U \va{v}'
+    \qquad
+    \va{r} = L \va{r}'
+    \qquad
+    t = \frac{L}{U} t'
+    \qquad
+    p = \rho U^2 p'
+\end{aligned}$$
+
+In this non-dimenionsalization, the differential operators are scaled as follows:
+
+$$\begin{aligned}
+    \pdv{}{t}
+    = \frac{U}{L} \pdv{}{t'}
+    \qquad \quad
+    \nabla
+    = \frac{1}{L} \nabla'
+\end{aligned}$$
+
+Putting everything into the main Navier-Stokes equation then yields:
+
+$$\begin{aligned}
+    \frac{U^2}{L} \pdv{\va{v}}{t'} + \frac{U^2}{L} (\va{v}' \cdot \nabla') \va{v}'
+    = - \frac{U^2}{L} \nabla' p' + \frac{U \nu}{L^2} \nabla'^2 \va{v}'
+\end{aligned}$$
+
+After dividing out $U^2/L$,
+we arrive at the form of the original equation again:
+
+$$\begin{aligned}
+    \pdv{\va{v}}{t'} + (\va{v}' \cdot \nabla') \va{v}'
+    = - \nabla' p' + \frac{\nu}{U L} \nabla'^2 \va{v}'
+\end{aligned}$$
+
+The constant factor of the last term
+leads to the definition of the **Reynolds number** $\mathrm{Re}$:
+
+$$\begin{aligned}
+    \boxed{
+        \mathrm{Re}
+        \equiv \frac{U L}{\nu}
+    }
+\end{aligned}$$
+
+If we choose $U$ and $L$ appropriately for a given system,
+the Reynolds number allows us to predict the general trends.
+It can be regarded as the inverse of an "effective viscosity":
+when $\mathrm{Re}$ is large, viscosity only has a minor role,
+but when $\mathrm{Re}$ is small, it dominates the dynamics.
+
+Another way is thus to see the Reynolds number
+as the characteristic ratio between the advective term
+(see [material derivative](/know/concept/material-derivative/))
+to the [viscosity](/know/concept/viscosity/) term,
+since $\va{v} \sim U$:
+
+$$\begin{aligned}
+     \mathrm{Re}
+     \approx \frac{\big| (\va{v} \cdot \nabla) \va{v} \big|}{\big| \nu \nabla^2 \va{v} \big|}
+     \approx \frac{U^2 / L}{\nu U / L^2}
+     = \frac{U L}{\nu}
+\end{aligned}$$
+
+In other words, $\mathrm{Re}$
+describes the relative strength of intertial and viscous forces.
+Returning to the dimensionless Navier-Stokes equation:
+
+$$\begin{aligned}
+    \pdv{\va{v}}{t'} + (\va{v}' \cdot \nabla') \va{v}'
+    = - \nabla' p' + \frac{1}{\mathrm{Re}} \nabla'^2 \va{v}'
+\end{aligned}$$
+
+For large $\mathrm{Re} \gg 1$,
+we can neglect the latter term,
+such that redimensionalizing yields:
+
+$$\begin{aligned}
+    \pdv{\va{v}}{t} + (\va{v} \cdot \nabla) \va{v}
+    = - \frac{\nabla p}{\rho}
+\end{aligned}$$
+
+Which is simply the main [Euler equation](/know/concept/euler-equations/)
+for an ideal fluid, i.e. a fluid without viscosity.
+
+
+
+## Stokes flow
+
+A notable case is so-called **Stokes flow** or **creeping flow**,
+meaning flow at $\mathrm{Re} \ll 1$.
+In this limit, the Navier-Stokes equations can be linearized:
+since $\mathrm{Re}$ is the advective-to-viscous ratio,
+$\mathrm{Re} \ll 1$ implies that we can ignore the advective term, leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\va{v}}{t}
+        = - \frac{\nabla p}{\rho} + \nu \nabla^2 \va{v}
+    }
+\end{aligned}$$
+
+This equation is called the **unsteady Stokes equation**.
+Usually, however, such flows are assumed to be steady
+(i.e. time-invariant), leading to the **steady Stokes equation**,
+with $\eta = \rho \nu$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla p
+        = \eta \nabla^2 \va{v}
+    }
+\end{aligned}$$
+
+This equation is much easier to solve than the full Navier-Stokes equation
+thanks to being linear,
+and has some interesting properties, such as time-reversibility.
+
+
+
+## References
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
+2.  R. Fitzpatrick,
+    [Dimensionless numbers in incompressible flow](https://farside.ph.utexas.edu/teaching/336L/Fluid/node17.html),
+    University of Texas.
diff --git a/source/know/concept/ritz-method/index.md b/source/know/concept/ritz-method/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/ritz-method/index.md
@@ -0,0 +1,369 @@
+---
+title: "Ritz method"
+date: 2022-09-18
+categories:
+- Physics
+- Mathematics
+- Perturbation
+- Quantum mechanics
+- Numerical methods
+layout: "concept"
+---
+
+In various branches of physics,
+the **Ritz method** is a technique to approximately find the lowest solutions to an eigenvalue problem.
+Some call it the **Rayleigh-Ritz method**, the **Ritz-Galerkin method**,
+or simply the **variational method**.
+
+
+
+## Background
+
+In the context of [variational calculus](/know/concept/calculus-of-variations/),
+consider the following functional to be optimized:
+
+$$\begin{aligned}
+    R[u]
+    = \frac{1}{S} \int_a^b p(x) \big|u_x(x)\big|^2 - q(x) \big|u(x)\big|^2 \dd{x}
+\end{aligned}$$
+
+Where $u(x) \in \mathbb{C}$ is the unknown function,
+and $p(x), q(x) \in \mathbb{R}$ are given.
+In addition, $S$ is the norm of $u$, which we demand be constant
+with respect to a weight function $w(x) \in \mathbb{R}$:
+
+$$\begin{aligned}
+    S
+    = \int_a^b w(x) \big|u(x)\big|^2 \dd{x}
+\end{aligned}$$
+
+To handle this normalization requirement,
+we introduce a [Lagrange multiplier](/know/concept/lagrange-multiplier/) $\lambda$,
+and define the Lagrangian $\Lambda$ for the full constrained optimization problem as:
+
+$$\begin{aligned}
+    \Lambda
+    \equiv \frac{1}{S} \bigg( \big( p |u_x|^2 - q |u|^2 \big) - \lambda \big( w |u|^2 \big) \bigg)
+\end{aligned}$$
+
+The resulting Euler-Lagrange equation is then calculated in the standard way, yielding:
+
+$$\begin{aligned}
+    0
+    &= \pdv{\Lambda}{u^*} - \dv{}{x}\Big( \pdv{\Lambda}{u_x^*} \Big)
+    \\
+    &= - \frac{1}{S} \bigg( q u + \lambda w u + \dv{}{x}\big( p u_x \big) \bigg)
+\end{aligned}$$
+
+Which is clearly satisfied if and only if the following equation is fulfilled:
+
+$$\begin{aligned}
+    \dv{}{x}\big( p u_x \big) + q u
+    = - \lambda w u
+\end{aligned}$$
+
+This has the familiar form of a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/) (SLP),
+with $\lambda$ representing an eigenvalue.
+SLPs have useful properties, but before we can take advantage of those,
+we need to handle an important detail: the boundary conditions (BCs) on $u$.
+The above equation is only a valid SLP for certain BCs,
+as seen in the derivation of Sturm-Liouville theory.
+
+Let us return to the definition of $R[u]$,
+and integrate it by parts:
+
+$$\begin{aligned}
+    R[u]
+    &= \frac{1}{S} \int_a^b p u_x u_x^* - q u u^* \dd{x}
+    \\
+    &= \frac{1}{S} \Big[ p u_x u^* \Big]_a^b - \frac{1}{S} \int_a^b \dv{}{x}\Big(p u_x\Big) u^* + q u u^* \dd{x}
+\end{aligned}$$
+
+The boundary term vanishes for a subset of the BCs that make a valid SLP,
+including Dirichlet BCs $u(a) = u(b) = 0$, Neumann BCs $u_x(a) = u_x(b) = 0$, and periodic BCs.
+Therefore, we assume that this term does indeed vanish,
+such that we can use Sturm-Liouville theory later:
+
+$$\begin{aligned}
+    R[u]
+    &= - \frac{1}{S} \int_a^b \bigg( \dv{}{x}\Big(p u_x\Big) + q u \bigg) u^* \dd{x}
+    \equiv - \frac{1}{S} \int_a^b u^* \hat{H} u \dd{x}
+\end{aligned}$$
+
+Where $\hat{H}$ is the self-adjoint Sturm-Liouville operator.
+Because the constrained Euler-Lagrange equation is now an SLP,
+we know that it has an infinite number of real discrete eigenvalues $\lambda_n$ with a lower bound,
+corresponding to mutually orthogonal eigenfunctions $u_n(x)$.
+
+To understand the significance of this result,
+suppose we have solved the SLP,
+and now insert one of the eigenfunctions $u_n$ into $R$:
+
+$$\begin{aligned}
+    R[u_n]
+    &= - \frac{1}{S_n} \int_a^b u_n^* \hat{H} u_n \dd{x}
+    = \frac{1}{S_n} \int_a^b u_n^* \lambda_n w u_n \dd{x}
+    \\
+    &= \frac{1}{S_n} \lambda_n \int_a^b w |u_n|^2 \dd{x}
+    = \frac{S_n}{S_n} \lambda_n
+\end{aligned}$$
+
+Where $S_n$ is the normalization of $u_n$.
+In other words, when given $u_n$,
+the functional $R$ yields the corresponding eigenvalue $\lambda_n$:
+
+$$\begin{aligned}
+    \boxed{
+        R[u_n]
+        = \lambda_n
+    }
+\end{aligned}$$
+
+This powerful result was not at all clear from $R$'s initial definition.
+
+
+
+## Justification
+
+But what if we do not know the eigenfunctions? Is $R$ still useful?
+Yes, as we shall see. Suppose we make an educated guess $u(x)$
+for the ground state (i.e. lowest-eigenvalue) solution $u_0(x)$:
+
+$$\begin{aligned}
+    u(x)
+    = u_0(x) + \sum_{n = 1}^\infty c_n u_n(x)
+\end{aligned}$$
+
+Here, we are using the fact that the eigenfunctions of an SLP form a complete set,
+so our (known) guess $u$ can be expanded in the true (unknown) eigenfunctions $u_n$.
+We are assuming that $u$ is already quite close to its target $u_0$,
+such that the (unknown) expansion coefficients $c_n$ are small;
+specifically $|c_n|^2 \ll 1$.
+Let us start from what we know:
+
+$$\begin{aligned}
+    \boxed{
+        R[u]
+        = - \frac{\displaystyle\int u^* \hat{H} u \dd{x}}{\displaystyle\int u^* w u \dd{x}}
+    }
+\end{aligned}$$
+
+This quantity is known as the **Rayleigh quotient**.
+Inserting our ansatz $u$,
+and using that the true $u_n$ have corresponding eigenvalues $\lambda_n$:
+
+$$\begin{aligned}
+    R[u]
+    &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \: \hat{H} \Big\{ u_0 + \sum_n c_n u_n \Big\} \dd{x}}
+    {\displaystyle\int w \Big( u_0 + \sum_n c_n u_n \Big) \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \dd{x}}
+    \\
+    &= - \frac{\displaystyle\int \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( \!-\! \lambda_0 w u_0 - \sum_n c_n \lambda_n w u_n \Big) \dd{x}}
+    {\displaystyle\int w  \Big( u_0^* + \sum_n c_n^* u_n^* \Big) \Big( u_0 + \sum_n c_n u_n \Big) \dd{x}}
+\end{aligned}$$
+
+For convenience, we switch to [Dirac notation](/know/concept/dirac-notation/)
+before evaluating further.
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \Big( \Bra{u_0} + \sum_n c_n^* \Bra{u_n} \Big) \cdot \Big( \lambda_0 \Ket{w u_0} + \sum_n c_n \lambda_n \Ket{w u_n} \Big)}
+    {\displaystyle \Big( \Bra{u_0} + \sum_n c_n^* \Bra{u_n} \Big) \cdot \Big( \Ket{w u_0} + \sum_n c_n \Ket{w u_n} \Big)}
+    \\
+    &= \frac{\displaystyle \lambda_0 \Inprod{u_0}{w u_0} + \lambda_0 \sum_{n = 1}^\infty c_n^* \Inprod{u_n}{w u_0}
+    + \sum_{n = 1}^\infty c_n \lambda_n \Inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \lambda_n \Inprod{u_m}{w u_n}}
+    {\displaystyle \Inprod{u_0}{w u_0} + \sum_{n = 1}^\infty c_n^* \Inprod{u_n}{w u_0}
+    + \sum_{n = 1}^\infty c_n \Inprod{u_0}{w u_n} + \sum_{m n} c_n c_m^* \Inprod{u_m}{w u_n}}
+\end{aligned}$$
+
+Using orthogonality $\Inprod{u_m}{w u_n} = S_n \delta_{mn}$,
+and the fact that $n \neq 0$ by definition, we find:
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \lambda_0 S_0 + \lambda_0 \sum_n c_n^* S_n \delta_{n0}
+    + \sum_n c_n \lambda_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* \lambda_n S_n \delta_{mn}}
+    {\displaystyle S_0 + \sum_n c_n^* S_n \delta_{n0} + \sum_n c_n S_n \delta_{n0} + \sum_{m n} c_n c_m^* S_n \delta_{mn}}
+    \\
+    &= \frac{\displaystyle \lambda_0 S_0 + 0 + 0 + \sum_{n} c_n c_n^* \lambda_n S_n}
+    {\displaystyle S_0 + 0 + 0 + \sum_{n} c_n c_n^* S_n}
+    = \frac{\displaystyle \lambda_0 S_0 + \sum_{n} |c_n|^2 \lambda_n S_n}
+    {\displaystyle S_0 + \sum_{n} |c_n|^2 S_n}
+\end{aligned}$$
+
+It is always possible to choose our normalizations such that $S_n = S$ for all $u_n$, leaving:
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \lambda_0 S + \sum_{n} |c_n|^2 \lambda_n S}
+    {\displaystyle S + \sum_{n} |c_n|^2 S}
+    = \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_n}
+    {\displaystyle 1 + \sum_{n} |c_n|^2}
+\end{aligned}$$
+
+And finally, after rearranging the numerator, we arrive at the following relation:
+
+$$\begin{aligned}
+    R
+    &= \frac{\displaystyle \lambda_0 + \sum_{n} |c_n|^2 \lambda_0 + \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)}
+    {\displaystyle 1 + \sum_{n} |c_n|^2}
+    = \lambda_0 + \frac{\displaystyle \sum_{n} |c_n|^2 (\lambda_n - \lambda_0)}
+    {\displaystyle 1 + \sum_{n} |c_n|^2}
+\end{aligned}$$
+
+Thus, if we improve our guess $u$,
+then $R[u]$ approaches the true eigenvalue $\lambda_0$.
+For numerically finding $u_0$ and $\lambda_0$, this gives us a clear goal: minimize $R$, because:
+
+$$\begin{aligned}
+    \boxed{
+        R[u]
+        = \lambda_0 + \frac{\displaystyle \sum_{n = 1}^\infty |c_n|^2 (\lambda_n - \lambda_0)}
+        {\displaystyle 1 + \sum_{n = 1}^\infty |c_n|^2}
+        \ge \lambda_0
+    }
+\end{aligned}$$
+
+In the context of quantum mechanics, this is not surprising,
+since any superposition of multiple states
+is guaranteed to have a higher energy than the ground state.
+
+Note that the convergence to $\lambda_0$ goes as $|c_n|^2$,
+while $u$ converges to $u_0$ as $|c_n|$ by definition,
+so even a fairly bad guess $u$ will give a decent estimate for $\lambda_0$.
+
+
+
+## The method
+
+In the following, we stick to Dirac notation,
+since the results hold for both continuous functions $u(x)$ and discrete vectors $\vb{u}$,
+as long as the operator $\hat{H}$ is self-adjoint.
+Suppose we express our guess $\Ket{u}$ as a linear combination
+of *known* basis vectors $\Ket{f_n}$ with weights $a_n \in \mathbb{C}$:
+
+$$\begin{aligned}
+    \Ket{u}
+    = \sum_{n = 0}^\infty c_n \Ket{u_n}
+    = \sum_{n = 0}^\infty a_n \Ket{f_n}
+    \approx \sum_{n = 0}^{N - 1} a_n \Ket{f_n}
+\end{aligned}$$
+
+For numerical tractability, we truncate the sum at $N$ terms,
+and for generality, we allow $\Ket{f_n}$ to be non-orthogonal,
+as described by an *overlap matrix* with elements $S_{mn}$:
+
+$$\begin{aligned}
+    \Inprod{f_m}{w f_n} = S_{m n}
+\end{aligned}$$
+
+From the discussion above,
+we know that the ground-state eigenvalue $\lambda_0$ is estimated by:
+
+$$\begin{aligned}
+    \lambda_0
+    \approx \lambda
+    = R[u]
+    = \frac{\inprod{u}{\hat{H} u}}{\Inprod{u}{w u}}
+    = \frac{\displaystyle \sum_{m n} a_m^* a_n \inprod{f_m}{\hat{H} f_n}}{\displaystyle \sum_{m n} a_m^* a_n \Inprod{f_m}{w f_n}}
+    \equiv \frac{\displaystyle \sum_{m n} a_m^* a_n H_{m n}}{\displaystyle \sum_{m n} a_m^* a_n S_{mn}}
+\end{aligned}$$
+
+And we also know that our goal is to minimize $R[u]$,
+so we vary $a_k^*$ to find its extremum:
+
+$$\begin{aligned}
+    0
+    = \pdv{R}{a_k^*}
+    &= \frac{\displaystyle \Big( \sum_{n} a_n H_{k n} \Big) \Big( \sum_{m n} a_n a_m^* S_{mn} \Big)
+    - \Big( \sum_{n} a_n S_{k n} \Big) \Big( \sum_{m n} a_n a_m^* H_{mn} \Big)}
+    {\Big( \displaystyle \sum_{m n} a_n a_m^* S_{mn} \Big)^2}
+    \\
+    &= \frac{\displaystyle \Big( \sum_{n} a_n H_{k n} \Big) - R[u] \Big( \sum_{n} a_n S_{k n}\Big)}{\Inprod{u}{w u}}
+    = \frac{\displaystyle \sum_{n} a_n \big(H_{k n} - \lambda S_{k n}\big)}{\Inprod{u}{w u}}
+\end{aligned}$$
+
+Clearly, this is only satisfied if the following holds for all $k = 0, 1, ..., N\!-\!1$:
+
+$$\begin{aligned}
+    0
+    = \sum_{n = 0}^{N - 1} a_n \big(H_{k n} - \lambda S_{k n}\big)
+\end{aligned}$$
+
+For illustrative purposes,
+we can write this as a matrix equation
+with $M_{k n} \equiv H_{k n} - \lambda S_{k n}$:
+
+$$\begin{aligned}
+    \begin{bmatrix}
+        M_{0,0} & M_{0,1} & \cdots & M_{0,N-1} \\
+        M_{1,0} & \ddots & & \vdots \\
+        \vdots & & \ddots & \vdots \\
+        M_{N-1,0} & \cdots & \cdots & M_{N-1,N-1}
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        a_0 \\ a_1 \\ \vdots \\ a_{N-1}
+    \end{bmatrix}
+    =
+    \begin{bmatrix}
+        0 \\ 0 \\ \vdots \\ 0
+    \end{bmatrix}
+\end{aligned}$$
+
+Note that this looks like an eigenvalue problem for $\lambda$.
+Indeed, demanding that $\overline{M}$ cannot simply be inverted
+(i.e. the solution is non-trivial)
+yields a characteristic polynomial for $\lambda$:
+
+$$\begin{aligned}
+    0
+    = \det\!\Big[ \overline{M} \Big]
+    = \det\!\Big[ \overline{H} - \lambda \overline{S} \Big]
+\end{aligned}$$
+
+This gives a set of $\lambda$,
+which are the exact eigenvalues of $\overline{H}$,
+and the estimated eigenvalues of $\hat{H}$
+(recall that $\overline{H}$ is $\hat{H}$ expressed in a truncated basis).
+The eigenvector $\big[ a_0, a_1, ..., a_{N-1} \big]$ of the lowest $\lambda$
+gives the optimal weights to approximate $\Ket{u_0}$ in the basis $\{\Ket{f_n}\}$.
+Likewise, the higher $\lambda$'s eigenvectors approximate
+excited (i.e. non-ground) eigenstates of $\hat{H}$,
+although in practice the results are less accurate the higher we go.
+
+The overall accuracy is determined by how good our truncated basis is,
+i.e. how large a subspace it spans
+of the [Hilbert space](/know/concept/hilbert-space/) in which the true $\Ket{u_0}$ resides.
+Clearly, adding more basis vectors will improve the results,
+at the cost of computation.
+For example, if $\hat{H}$ represents a helium atom,
+a good choice for $\{\Ket{f_n}\}$ would be hydrogen orbitals,
+since those are qualitatively similar.
+
+You may find this result unsurprising;
+it makes some intuitive sense that approximating $\hat{H}$
+in a limited basis would yield a matrix $\overline{H}$ giving rough eigenvalues.
+The point of this discussion is to rigorously show
+the validity of this approach.
+
+If we only care about the ground state,
+then we already know $\lambda$ from $R[u]$,
+so all we need to do is solve the above matrix equation for $a_n$.
+Keep in mind that $\overline{M}$ is singular,
+and $a_n$ are only defined up to a constant factor.
+
+Nowadays, there exist many other methods to calculate eigenvalues
+of complicated operators $\hat{H}$,
+but an attractive feature of the Ritz method is that it is single-step,
+whereas its competitors tend to be iterative.
+That said, the Ritz method cannot recover from a poorly chosen basis.
+
+
+
+## References
+1.  G.B. Arfken, H.J. Weber,
+    *Mathematical methods for physicists*, 6th edition, 2005,
+    Elsevier.
+2.  O. Bang,
+    *Applied mathematics for physicists: lecture notes*, 2019,
+    unpublished.
diff --git a/source/know/concept/rotating-wave-approximation/index.md b/source/know/concept/rotating-wave-approximation/index.md
new file mode 100644
index 0000000..efb9739
--- /dev/null
+++ b/source/know/concept/rotating-wave-approximation/index.md
@@ -0,0 +1,120 @@
+---
+title: "Rotating wave approximation"
+date: 2022-02-01
+categories:
+- Physics
+- Quantum mechanics
+- Two-level system
+- Optics
+layout: "concept"
+---
+
+Consider the following periodic perturbation $\hat{H}_1$ to a quantum system,
+which represents e.g. an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
+in the [electric dipole approximation](/know/concept/electric-dipole-approximation/):
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = \hat{V} \cos(\omega t)
+    = \frac{\hat{V}}{2} \Big( e^{i \omega t} + e^{-i \omega t} \Big)
+\end{aligned}$$
+
+Where $\hat{V}$ is some operator, and we assume that $\omega$
+is fairly close to a resonance frequency $\omega_0$
+of the system that is getting perturbed by $\hat{H}_1$.
+
+As an example, consider a two-level system
+consisting of states $\Ket{g}$ and $\Ket{e}$,
+with a resonance frequency $\omega_0 = (E_e \!-\! E_g) / \hbar$.
+From the derivation of
+[time-dependent perturbation theory](/know/concept/time-dependent-perturbation-theory/),
+we know that the state $\Ket{\Psi} = c_g \Ket{g} + c_e \Ket{e}$ evolves as:
+
+$$\begin{aligned}
+    i \hbar \dv{c_g}{t}
+    &= \matrixel{g}{\hat{H}_1(t)}{g} \: c_g(t) + \matrixel{g}{\hat{H}_1(t)}{e} \: c_e(t) \: e^{- i \omega_0 t}
+    \\
+    i \hbar \dv{c_e}{t}
+    &= \matrixel{e}{\hat{H}_1(t)}{g} \: c_g(t) \: e^{i \omega_0 t} + \matrixel{e}{\hat{H}_1(t)}{e} \: c_e(t)
+\end{aligned}$$
+
+Typically, $\hat{V}$ has odd spatial parity, in which case
+[Laporte's selection rule](/know/concept/selection-rules/)
+reduces this to:
+
+$$\begin{aligned}
+    \dv{c_g}{t}
+    &= \frac{1}{i \hbar} \matrixel{g}{\hat{H}_1}{e} \: c_e \: e^{- i \omega_0 t}
+    \\
+    \dv{c_e}{t}
+    &= \frac{1}{i \hbar} \matrixel{e}{\hat{H}_1}{g} \: c_g \: e^{i \omega_0 t}
+\end{aligned}$$
+
+We now insert the general $\hat{H}_1$ defined above,
+and define $V_{eg} \equiv \matrixel{e}{\hat{V}}{g}$ to get:
+
+$$\begin{aligned}
+    \dv{c_g}{t}
+    &= \frac{V_{eg}^*}{i 2 \hbar}
+    \Big( e^{i (\omega - \omega_0) t} + e^{- i (\omega + \omega_0) t} \Big) \: c_e
+    \\
+    \dv{c_e}{t}
+    &= \frac{V_{eg}}{i 2 \hbar}
+    \Big( e^{i (\omega + \omega_0) t} + e^{- i (\omega - \omega_0) t} \Big) \: c_g
+\end{aligned}$$
+
+At last, here we make the **rotating wave approximation**:
+since $\omega$ is assumed to be close to $\omega_0$,
+we argue that $\omega \!+\! \omega_0$ is so much larger than $\omega \!-\! \omega_0$
+that those oscillations turn out negligible
+if the system is observed over a reasonable time interval.
+
+Specifically, since both exponentials have the same weight,
+the fast ($\omega \!+\! \omega_0$) oscillations
+have a tiny amplitude compared to the slow ($\omega \!-\! \omega_0$) ones.
+Furthermore, since they average out to zero over most realistic time intervals,
+the fast terms can be dropped, leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            e^{i (\omega - \omega_0) t} + e^{- i (\omega + \omega_0) t}
+            &\approx e^{i (\omega - \omega_0) t}
+            \\
+            e^{i (\omega + \omega_0) t} + e^{- i (\omega - \omega_0) t}
+            &\approx e^{- i (\omega - \omega_0) t}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Such that our example set of equations can be approximated as shown below,
+and its analysis can continue;
+see [Rabi oscillation](/know/concept/rabi-oscillation/) for more:
+
+$$\begin{aligned}
+    \dv{c_g}{t}
+    &= \frac{V_{eg}^*}{i 2 \hbar} c_e \: e^{i (\omega - \omega_0) t}
+    \\
+    \dv{c_e}{t}
+    &= \frac{V_{eg}}{i 2 \hbar} c_g \: e^{- i (\omega - \omega_0) t}
+\end{aligned}$$
+
+This approximation's name is a bit confusing:
+the idea is that going from the Schrödinger to
+the [interaction picture](/know/concept/interaction-picture/)
+has the effect of removing the exponentials of $\omega_0$ from the above equations,
+i.e. multiplying them by $e^{i \omega_0 t}$ and $e^{- i \omega_0 t}$
+respectively, which can be regarded as a rotation.
+
+Relative to this rotation, when we split the wave $\cos(\omega t)$
+into two exponentials, one co-rotates, and the other counter-rotates.
+We keep only the co-rotating waves, hence the name.
+
+The rotating wave approximation is usually used in the context
+of the two-level quantum system for light-matter interactions,
+as in the above example.
+However, it is not specific to that case,
+and it more generally refers to any approximation
+where fast-oscillating terms are neglected.
+
+
diff --git a/source/know/concept/runge-kutta-method/index.md b/source/know/concept/runge-kutta-method/index.md
new file mode 100644
index 0000000..1e8e4a2
--- /dev/null
+++ b/source/know/concept/runge-kutta-method/index.md
@@ -0,0 +1,261 @@
+---
+title: "Runge-Kutta method"
+date: 2022-03-10
+categories:
+- Mathematics
+- Numerical methods
+layout: "concept"
+---
+
+A **Runge-Kutta method** (RKM) is a popular approach
+to numerically solving systems of ordinary differential equations.
+Let $\vb{x}(t)$ be the vector we want to find,
+governed by $\vb{f}(t, \vb{x})$:
+
+$$\begin{aligned}
+    \vb{x}'(t)
+    = \vb{f}\big(t, \vb{x}(t)\big)
+\end{aligned}$$
+
+Like in all numerical methods, the $t$-axis is split into discrete steps.
+If a step has size $h$, then as long as $h$ is small enough,
+we can make the following approximation:
+
+$$\begin{aligned}
+    \vb{x}'(t) + a h \vb{x}''(t)
+    &\approx \vb{x}'(t \!+\! a h)
+    \\
+    &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t \!+\! a h)\big)
+    \\
+    &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t) \!+\! a h \vb{x}'(t) \big)
+\end{aligned}$$
+
+For sufficiently small $h$,
+higher-order derivates can also be included,
+albeit still at $t \!+\! a h$:
+
+$$\begin{aligned}
+    \vb{x}'(t) + a h \vb{x}''(t) + b h^2 \vb{x}'''(t)
+    &\approx \vb{f}\big(t \!+\! a h,\, \vb{x}(t) \!+\! a h \vb{x}'(t) \!+\! b h^2 \vb{x}''(t) \big)
+\end{aligned}$$
+
+Although these approximations might seem innocent,
+they actually make it quite complicated to determine the error order of a given RKM.
+
+Now, consider a Taylor expansion around the current $t$,
+truncated at a chosen order $n$:
+
+$$\begin{aligned}
+    \vb{x}(t \!+\! h)
+    &= \vb{x}(t) + h \vb{x}'(t) + \frac{h^2}{2} \vb{x}''(t) + \frac{h^3}{6} \vb{x}'''(t) + \:...\, + \frac{h^n}{n!} \vb{x}^{(n)}(t)
+    \\
+    &= \vb{x}(t) + h \bigg[ \vb{x}'(t) + \frac{h}{2} \vb{x}''(t) + \frac{h^2}{6} \vb{x}'''(t) + \:...\, + \frac{h^{n-1}}{n!} \vb{x}^{(n)}(t) \bigg]
+\end{aligned}$$
+
+We are free to split the terms as follows,
+choosing real factors $\omega_{mj}$ subject to $\sum_{j} \omega_{mj} = 1$:
+
+$$\begin{aligned}
+    \vb{x}(t \!+\! h)
+    &= \vb{x} + h \bigg[ \sum_{j = 1}^{N_1} \omega_{1j} \, \vb{x}'
+    + \frac{h}{2} \sum_{j = 1}^{N_2} \omega_{2j} \, \vb{x}''
+    + \:...\, + \frac{h^{n-1}}{n!} \sum_{j = 1}^{N_n} \omega_{nj} \, \vb{x}^{(n)} \bigg]
+\end{aligned}$$
+
+Where the integers $N_1,...,N_n$ are also free to choose,
+but for reasons that will become clear later,
+the most general choice for an RKM is $N_1 = n$, $N_n = 1$, and:
+
+$$\begin{aligned}
+    N_{n-1}
+    = N_n \!+\! 2
+    ,\quad
+    \cdots
+    ,\quad
+    N_{n-m}
+    = N_{n-m+1} \!+\! m \!+\! 1
+    ,\quad
+    \cdots
+    ,\quad
+    N_{2}
+    = N_3 \!+\! n \!-\! 1
+\end{aligned}$$
+
+In other words, $N_{n-m}$ is the $m$th triangular number.
+This is not so important,
+since this is not a practical way to describe RKMs,
+but it is helpful to understand how they work.
+
+
+## Example derivation
+
+For example, let us truncate at $n = 3$,
+such that $N_1 = 3$, $N_2 = 3$ and $N_3 = 1$.
+The following derivation is very general,
+except it requires all $\alpha_j \neq 0$.
+Renaming $\omega_{mj}$, we start from:
+
+$$\begin{aligned}
+    \vb{x}(t \!+\! h)
+    &= \vb{x} + h \bigg[ (\alpha_1 + \alpha_2 + \alpha_3) \, \vb{x}'
+    + \frac{h}{2} (\beta_2 + \beta_{31} + \beta_{32}) \, \vb{x}''
+    + \frac{h^2}{6} \gamma_3 \, \vb{x}''' \bigg]
+    \\
+    &= \vb{x} + h \bigg[ \alpha_1 \vb{x}'
+    + \Big( \alpha_2 \vb{x}' + \frac{h}{2} \beta_2 \vb{x}'' \Big)
+    + \Big( \alpha_3 \vb{x}' + \frac{h}{2} (\beta_{31} + \beta_{32}) \vb{x}'' + \frac{h^2}{6} \gamma_3 \vb{x}''' \Big) \bigg]
+\end{aligned}$$
+
+As discussed earlier, the parenthesized expressions
+can be approximately rewritten with $\vb{f}$:
+
+$$\begin{aligned}
+    \vb{x}(t \!+\! h)
+    = \vb{x} + h &\bigg[ \alpha_1 \vb{f}(t, \vb{x})
+    + \alpha_2 \vb{f}\Big( t \!+\! \frac{h \beta_2}{2 \alpha_2}, \;
+    \vb{x} \!+\! \frac{h \beta_2}{2 \alpha_2} \vb{x}' \Big)
+    \\
+    & + \alpha_3 \vb{f}\Big( t \!+\! \frac{h (\beta_{31} \!\!+\!\! \beta_{32})}{2 \alpha_3}, \;
+    \vb{x} \!+\! \frac{h \beta_{31}}{2 \alpha_3} \vb{x}' \!+\! \frac{h \beta_{32}}{2 \alpha_3} \vb{x}'
+    \!+\! \frac{h^2 \gamma_3}{6 \alpha_3} \vb{x}'' \Big) \bigg]
+    \\
+    = \vb{x} + h &\bigg[ \alpha_1 \vb{k}_1
+    + \alpha_2 \vb{f}\Big( t \!+\! \frac{h \beta_2}{2 \alpha_2}, \;
+    \vb{x} \!+\! \frac{h \beta_2}{2 \alpha_2} \vb{k}_1 \!\Big)
+    \\
+    & + \alpha_3 \vb{f}\Big( t \!+\! \frac{h (\beta_{31} \!\!+\!\! \beta_{32})}{2 \alpha_3}, \;
+    \vb{x} \!+\! \frac{h \beta_{31}}{2 \alpha_3} \vb{k}_1 \!+\! \frac{h \beta_{32}}{2 \alpha_3}
+    \vb{f}\Big( t \!+\! \frac{h \gamma_3}{3 \beta_{32}}, \;
+    \vb{x} \!+\! \frac{h \gamma_3}{3 \beta_{32}} \vb{k}_1 \!\Big) \!\Big) \bigg]
+\end{aligned}$$
+
+Here, we can see an opportunity to save some computational time
+by reusing an evaluation of $\vb{f}$.
+Technically, this is optional, but it would be madness not to,
+so we choose:
+
+$$\begin{aligned}
+    \frac{\beta_2}{2 \alpha_2}
+    = \frac{\gamma_3}{3 \beta_{32}}
+\end{aligned}$$
+
+Such that the next step of $\vb{x}$'s numerical solution is as follows,
+recalling that $\sum_{j} \alpha_j = 1$:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{x}(t \!+\! h)
+        = \vb{x}(t) + h \Big( \alpha_1 \vb{k}_1 + \alpha_2 \vb{k}_2 + \alpha_3 \vb{k}_3 \Big)
+    }
+\end{aligned}$$
+
+Where $\vb{k}_1$, $\vb{k}_2$ and $\vb{k}_3$ are different estimates
+of the average slope $\vb{x}'$ between $t$ and $t \!+\! h$,
+whose weighted average is used to make the $t$-step.
+They are given by:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \vb{k}_1
+            &\equiv \vb{f}(t, \vb{x})
+            \\
+            \vb{k}_2
+            &\equiv \vb{f}\bigg( t + \frac{h \beta_2}{2 \alpha_2}, \;
+            \vb{x} + \frac{h \beta_2}{2 \alpha_2} \vb{k}_1 \bigg)
+            \\
+            \vb{k}_3
+            &\equiv \vb{f}\bigg( t + \frac{h (\beta_{31} \!\!+\!\! \beta_{32})}{2 \alpha_3}, \;
+            \vb{x} + \frac{h \beta_{31}}{2 \alpha_3} \vb{k}_1 + \frac{h \beta_{32}}{2 \alpha_3} \vb{k}_2 \bigg)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Despite the contraints on $\alpha_j$ and $\beta_j$,
+there is an enormous freedom of choice here,
+all leading to valid RKMs, although not necessarily good ones.
+
+
+## General form
+
+A more practical description goes as follows:
+in an $s$-stage RKM, a weighted average is taken
+of up to $s$ slope estimates $\vb{k}_j$ with weights $b_j$.
+Let $\sum_{j} b_j = 1$, then:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{x}(t \!+\! h)
+        = \vb{x}(t) + h \sum_{j = 1}^{s} b_j \vb{k}_j
+    }
+\end{aligned}$$
+
+Where the estimates $\vb{k}_1, ..., \vb{k}_s$
+depend on each other, and are calculated one by one as:
+
+$$\begin{aligned}
+    \boxed{
+        \vb{k}_m
+        = \vb{f}\bigg( t + h c_m,\; \vb{x} + h \sum_{j = 1}^{m - 1} a_{mj} \vb{k}_j \bigg)
+    }
+\end{aligned}$$
+
+With $c_1 = 1$ and $\sum_{j = 1} a_{mj} = c_m$.
+Writing this out for the first few $m$, the pattern is clear:
+
+$$\begin{aligned}
+    \vb{k}_1
+    &= \vb{f}(t, \vb{x})
+    \\
+    \vb{k}_2
+    &= \vb{f}\big( t + h c_2,\; \vb{x} + h a_{21} \vb{k}_1 \big)
+    \\
+    \vb{k}_3
+    &= \vb{f}\big( t + h c_3,\; \vb{x} + h (a_{31} \vb{k}_1 + a_{32} \vb{k}_2) \big)
+    \\
+    \vb{k}_4
+    &= \:...
+\end{aligned}$$
+
+The coefficients of a given RKM are usually
+compactly represented in a **Butcher tableau**:
+
+$$\begin{aligned}
+    \begin{array}{c|ccc}
+        0 \\
+        c_2 & a_{21} \\
+        c_3 & a_{31} & a_{32} \\
+        \vdots & \vdots & \vdots & \ddots \\
+        c_s & a_{s1} & a_{s2} & \cdots & a_{s,s-1} \\
+        \hline
+        & b_1 & b_2 & \cdots & b_{s-1} & b_s
+    \end{array}
+\end{aligned}$$
+
+Each RKM has an **order** $p$,
+such that the global truncation error is $\mathcal{O}(h^p)$,
+i.e. the accumulated difference between the numerical
+and the exact solutions is proportional to $h^p$.
+
+The surprise is that $p$ need not be equal to the Taylor expansion order $n$,
+nor the stage count $s$.
+Typically, $s = n$ for computational efficiency, but $s \ge n$ is possible in theory.
+
+The order $p$ of a given RKM is determined by
+a complicated set of equations on the coefficients,
+and the lowest possible $s$ for a desired $p$
+is in fact only partially known.
+For $p \le 4$ the bound is $s \ge p$,
+whereas for $p \ge 5$ the only proven bound is $s \ge p \!+\! 1$,
+but for $p \ge 7$ no such efficient methods have been found so far.
+
+If you need an RKM with a certain order, look it up.
+There exist many efficient methods for $p \le 4$ where $s = p$,
+and although less popular, higher $p$ are also available.
+
+
+
+## References
+1.  J.C. Butcher,
+    *Numerical methods for ordinary differential equations*, 3rd edition,
+    Wiley.
diff --git a/source/know/concept/rutherford-scattering/index.md b/source/know/concept/rutherford-scattering/index.md
new file mode 100644
index 0000000..aec42be
--- /dev/null
+++ b/source/know/concept/rutherford-scattering/index.md
@@ -0,0 +1,242 @@
+---
+title: "Rutherford scattering"
+date: 2021-10-02
+categories:
+- Physics
+- Plasma physics
+layout: "concept"
+---
+
+**Rutherford scattering** or **Coulomb scattering**
+is an [elastic pseudo-collision](/know/concept/elastic-collision/)
+of two electrically charged particles.
+It is not a true collision, and is caused by Coulomb repulsion.
+
+The general idea is illustrated below.
+Consider two particles 1 and 2, with the same charge sign.
+Let 2 be initially at rest, and 1 approach it with velocity $\vb{v}_1$.
+Coulomb repulsion causes 1 to deflect by an angle $\theta$,
+and pushes 2 away in the process:
+
+<a href="two-body.png">
+<img src="two-body.png" style="width:50%">
+</a>
+
+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:
+
+<a href="one-body.png">
+<img src="one-body.png" style="width:60%">
+</a>
+
+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.
diff --git a/source/know/concept/rutherford-scattering/one-body.png b/source/know/concept/rutherford-scattering/one-body.png
new file mode 100644
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+++ b/source/know/concept/rutherford-scattering/one-body.png
diff --git a/source/know/concept/rutherford-scattering/two-body.png b/source/know/concept/rutherford-scattering/two-body.png
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/rutherford-scattering/two-body.png
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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@@ -0,0 +1,281 @@
+---
+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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@@ -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:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta m = 0 \:\:\mathrm{or}\: \pm 1
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dipole-m"/>
+<label for="proof-dipole-m">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-dipole-m">Proof.</label>
+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.
+</div>
+</div>
+
+Meanwhile, for the total angular momentum $\ell$ we have the following:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell = \pm 1
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dipole-l"/>
+<label for="proof-dipole-l">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-dipole-l">Proof.</label>
+We start from the following relation
+(which is already quite a chore to prove):
+
+$$\begin{aligned}
+    \Comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}
+    = 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r})
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-dipole-l-comm"/>
+<label for="proof-dipole-l-comm">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-dipole-l-comm">Proof.</label>
+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 $\vu{r} = (\hat{x}, \hat{y}, \hat{z})$.
+</div>
+</div>
+
+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:
+
+$$\begin{aligned}
+    (\ell_f - \ell_i)^2
+    = 1
+\end{aligned}$$
+</div>
+</div>
+
+
+## Rotational rules
+
+Given a general (pseudo)scalar operator $\hat{s}$,
+which, by nature, must satisfy the
+following relations with the angular momentum operators:
+
+$$\begin{aligned}
+    \comm{\hat{L}^2}{\hat{s}} = 0
+    \qquad
+    \comm{\hat{L}_z}{\hat{s}} = 0
+    \qquad
+    \comm{\hat{L}_{\pm}}{\hat{s}} = 0
+\end{aligned}$$
+
+Where $\hat{L}_\pm \equiv \hat{L}_x \pm i \hat{L}_y$.
+The inner product of any such $\hat{s}$ must obey these selection rules:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell = 0
+    }
+    \qquad \quad
+    \boxed{
+        \Delta m = 0
+    }
+\end{aligned}$$
+
+It is common to write this in the following more complete way, where
+$\matrixel{\ell_f}{|\hat{s}|}{\ell_i}$ is the **reduced matrix element**,
+which is identical to $\matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}$, but
+with a different notation to say that it does not depend on $m_f$ or $m_i$:
+
+$$\begin{aligned}
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{s}}{\ell_i m_i}
+        = \delta_{\ell_f \ell_i} \delta_{m_f m_i} \matrixel{\ell_f}{|\hat{s}|}{\ell_i}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-rot-scalar"/>
+<label for="proof-rot-scalar">Proof</label>
+<div class="hidden" markdown="1">
+<label for="proof-rot-scalar">Proof.</label>
+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.
+</div>
+</div>
+
+Similarly, given a general (pseudo)vector operator $\vu{V}$,
+which, by nature, must satisfy the following commutation relations,
+where $\hat{V}_\pm \equiv \hat{V}_x \pm i \hat{V}_y$:
+
+$$\begin{gathered}
+    \comm{\hat{L}_z}{\hat{V}_z} = 0
+    \qquad
+    \comm{\hat{L}_z}{\hat{V}_{\pm}} = \pm \hbar \hat{V}_{\pm}
+    \qquad
+    \comm{\hat{L}_{\pm}}{\hat{V}_z} = \mp \hbar \hat{V}_{\pm}
+    \\
+    \comm{\hat{L}_{\pm}}{\hat{V}_{\pm}} = 0
+    \qquad
+    \comm{\hat{L}_{\pm}}{\hat{V}_{\mp}} = \pm 2 \hbar \hat{V}_z
+\end{gathered}$$
+
+The inner product of any such $\vu{V}$ must obey the following selection rules:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta \ell
+        = 0 \:\:\mathrm{or}\: \pm 1
+    }
+    \qquad
+    \boxed{
+        \Delta m
+        = 0 \:\:\mathrm{or}\: \pm 1
+    }
+\end{aligned}$$
+
+In fact, the complete result involves the Clebsch-Gordan coefficients (from spin addition):
+
+$$\begin{gathered}
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{z}}{\ell_i m_i}
+        = C^{\ell_i \: 1 \: \ell_f}_{m_i \: 0 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i}
+    }
+    \\
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{+}}{\ell_i m_i}
+        = - \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: 1 \:m_f} \matrixel{\ell_f}{|\hat{V}|}{\ell_i}
+    }
+    \\
+    \boxed{
+        \matrixel{\ell_f m_f}{\hat{V}_{-}}{\ell_i m_i}
+        = \sqrt{2} C^{\ell_i \: 1 \: \ell_f}_{m_i \: -1 \:m_f} \matrixel{\ell_f}{|\hat{V}}{|\ell_i}
+    }
+\end{gathered}$$
+
+
+## Superselection rule
+
+Selection rules are not always about atomic electron transitions, or angular momenta even.
+
+According to the **principle of indistinguishability**,
+permuting identical particles never leads to an observable difference.
+In other words, the particles are fundamentally indistinguishable,
+so for any observable $\hat{O}$ and multi-particle state $\Ket{\Psi}$, we can say:
+
+$$\begin{aligned}
+    \matrixel{\Psi}{\hat{O}}{\Psi}
+    = \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi}
+\end{aligned}$$
+
+Where $\hat{P}$ is an arbitrary permutation operator.
+Indistinguishability implies that $\comm{\hat{P}}{\hat{O}} = 0$
+for all $\hat{O}$ and $\hat{P}$,
+which lets us prove the above equation, using that $\hat{P}$ is unitary:
+
+$$\begin{aligned}
+    \matrixel{\hat{P} \Psi}{\hat{O}}{\hat{P} \Psi}
+    = \matrixel{\Psi}{\hat{P}^{-1} \hat{O} \hat{P}}{\Psi}
+    = \matrixel{\Psi}{\hat{P}^{-1} \hat{P} \hat{O}}{\Psi}
+    = \matrixel{\Psi}{\hat{O}}{\Psi}
+\end{aligned}$$
+
+Consider a symmetric state $\Ket{s}$ and an antisymmetric state $\Ket{a}$
+(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
+which obey the following for a permutation $\hat{P}$:
+
+$$\begin{aligned}
+    \hat{P} \Ket{s}
+    = \Ket{s}
+    \qquad
+    \hat{P} \Ket{a}
+    = - \Ket{a}
+\end{aligned}$$
+
+Any obervable $\hat{O}$ then satisfies the equation below,
+again thanks to the fact that $\hat{P} = \hat{P}^{-1}$:
+
+$$\begin{aligned}
+    \matrixel{s}{\hat{O}}{a}
+    = \matrixel{\hat{P} s}{\hat{O}}{a}
+    = \matrixel{s}{\hat{P}^{-1} \hat{O}}{a}
+    = \matrixel{s}{\hat{O} \hat{P}}{a}
+    = \matrixel{s}{\hat{O}}{\hat{P} a}
+    = - \matrixel{s}{\hat{O}}{a}
+\end{aligned}$$
+
+This leads us to the **superselection rule**,
+which states that there can never be any interference
+between states of different permutation symmetry:
+
+$$\begin{aligned}
+    \boxed{
+        \matrixel{s}{\hat{O}}{a}
+        = 0
+    }
+\end{aligned}$$
+
+
+
+## References
+1.  D.J. Griffiths, D.F. Schroeter,
+    *Introduction to quantum mechanics*, 3rd edition,
+    Cambridge.
diff --git a/source/know/concept/self-energy/dyson.png b/source/know/concept/self-energy/dyson.png
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@@ -0,0 +1,307 @@
+---
+title: "Self-energy"
+date: 2021-11-21
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+Suppose we have a time-independent Hamiltonian $\hat{H} = \hat{H}_0 + \hat{W}$,
+consisting of a simple $\hat{H}_0$ and a difficult interaction $\hat{W}$,
+for example describing Coulomb repulsion between electrons.
+
+The concept of [imaginary time](/know/concept/imaginary-time/)
+exists to handle such difficult time-independent Hamiltonians
+at nonzero temperatures. Therefore, we know that the
+[Matsubara Green's function](/know/concept/matsubara-greens-function/)
+$G$ can be written as follows, where $\mathcal{T}$ is the
+[time-ordered product](/know/concept/time-ordered-product/),
+and $\beta = 1 / (k_B T)$:
+
+$$\begin{aligned}
+    G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)
+    = \frac{\Expval{\mathcal{T}\Big\{ \hat{K}(\hbar \beta, 0) \hat{\Psi}_{s_b}(\vb{r}_b, \tau_b) \hat{\Psi}_{s_a}^\dagger(\vb{r}_a, \tau_a) \Big\}}}
+    {\hbar \Expval{\hat{K}(\hbar \beta, 0)}}
+\end{aligned}$$
+
+Where we know that the time evolution operator $\hat{K}$
+is as follows in the [interaction picture](/know/concept/interaction-picture/):
+
+$$\begin{aligned}
+    \hat{K}(\tau_2, \tau_1)
+    &= \mathcal{T}\bigg\{ \exp\!\bigg( \!-\!\frac{1}{\hbar} \int_{\tau_1}^{\tau_2} \hat{W}(\tau) \dd{\tau} \bigg) \bigg\}
+    \\
+    &= \sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n
+    \mathcal{T}\bigg\{ \bigg( \int_{\tau_1}^{\tau_2} \hat{W}(\tau) \dd{\tau} \bigg)^n \bigg\}
+\end{aligned}$$
+
+Where $\hat{W}$ is the two-body operator in the interaction picture.
+We insert this into the full Green's function above,
+and abbreviate
+$G_{ba} \equiv G_{s_b s_a}(\vb{r}_b, \tau_b; \vb{r}_a, \tau_a)$
+and $\hat{\Psi}_a \equiv \hat{\Psi}_{s_a}(\vb{r}_a, \tau_a)$:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \int\cdots\int_0^{\hbar \beta}
+    \Expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\!\frac{1}{\hbar} \Big)^n \int\cdots\int_0^{\hbar \beta}
+    \Expval{\mathcal{T}\Big\{ \hat{W}(\tau_1) \cdots \hat{W}(\tau_n) \Big\}} \dd{\tau_1} \cdots \dd{\tau_n}}
+\end{aligned}$$
+
+Next, we write out the interaction operator $\hat{W}$
+in the [second quantization](/know/concept/second-quantization/),
+assuming there is no spin-flipping,
+and that $W(\vb{r}_1, \vb{r}_2) = W(\vb{r}_2, \vb{r}_1)$
+(hence $1/2$ to avoid double-counting):
+
+$$\begin{aligned}
+    \hat{W}(\tau_1)
+    &= \frac{1}{2} \sum_{s_1 s_2} \iint_{-\infty}^\infty \hat{\Psi}_{s_1}^\dagger(\vb{r}_1, \tau_1) \hat{\Psi}_{s_2}^\dagger(\vb{r}_2, \tau_1)
+    W(\vb{r}_1, \vb{r}_2) \hat{\Psi}_{s_2}(\vb{r}_2, \tau_1) \hat{\Psi}_{s_1}(\vb{r}_1, \tau_1) \dd{\vb{r}_1} \dd{\vb{r}_2}
+\end{aligned}$$
+
+We integrate this over $\tau_1$ and over a dummy $\tau_2$.
+Defining $W_{j'j} \equiv W(\vb{r}_j', \vb{r}_j) \: \delta(\tau_1 \!-\! \tau_2)$ we get:
+
+$$\begin{aligned}
+    \int_0^{\hbar \beta} \hat{W}(\tau_1) \dd{\tau_1}
+    &= \frac{1}{2} \iint \hat{\Psi}_{s_1}^\dagger(\vb{r}_1, \tau_1) \hat{\Psi}_{s_2}^\dagger(\vb{r}_2, \tau_2)
+    \: W_{1,2} \: \hat{\Psi}_{s_2}(\vb{r}_2, \tau_2) \hat{\Psi}_{s_1}(\vb{r}_1, \tau_1) \dd{\tau_2} \dd{\vb{r}_1} \dd{\vb{r}_2}
+    \\
+    &= \frac{1}{2} \iint \hat{\Psi}_1^\dagger \hat{\Psi}_2^\dagger  W_{1,2} \hat{\Psi}_2 \hat{\Psi}_1 \dd{1} \dd{2}
+\end{aligned}$$
+
+Where we have further abbreviated $\int \dd{j} \equiv \sum_{s_j} \int \dd{\vb{r}_j} \int \dd{\tau_j}$.
+The full $G_{ba}$ thus becomes:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n+1}
+    \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{num} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{1}{2 \hbar} \Big)^n (-\hbar)^{2n}
+    \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( G^0_\mathrm{den} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+\end{aligned}$$
+
+Where we have realized that both the numerator and denominator
+contain many-particle non-interacting Green's functions, defined as:
+
+$$\begin{aligned}
+    G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)
+    &= \Big( \!-\!\frac{1}{\hbar} \Big)^{2 n + 1}
+    \Expval{\mathcal{T}\Big\{ \hat{\Psi}_{1'}^\dagger \hat{\Psi}_{1}^\dagger \hat{\Psi}_{1} \hat{\Psi}_{1'} \cdots
+    \hat{\Psi}_{n'}^\dagger \hat{\Psi}_{n}^\dagger \hat{\Psi}_{n} \hat{\Psi}_{n'} \hat{\Psi}_b \hat{\Psi}_a^\dagger \Big\}}
+    \\
+    G^0_\mathrm{den}(1'1 \cdots n'n; 1'1 \cdots n'n)
+    &= \Big( \!-\!\frac{1}{\hbar} \Big)^{2 n}
+    \Expval{\mathcal{T}\Big\{ \hat{\Psi}_{1'}^\dagger \hat{\Psi}_{1}^\dagger \hat{\Psi}_{1} \hat{\Psi}_{1'} \cdots
+    \hat{\Psi}_{n'}^\dagger \hat{\Psi}_{n}^\dagger \hat{\Psi}_{n} \hat{\Psi}_{n'} \Big\}}
+\end{aligned}$$
+
+By applying [Wick's theorem](/know/concept/wicks-theorem/),
+we can rewrite these as a sum of products of single-particle Green's functions,
+so for instance $G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)$ becomes:
+
+$$\begin{aligned}
+    G^0_\mathrm{num}(b1'1 \cdots n'n; a1'1 \cdots n'n)
+    = \mathrm{det} \begin{bmatrix}
+        G^0_{ba} & G^0_{b1'} & G^0_{b1} & G^0_{b2'} & \cdots & G^0_{bn'} & G^0_{bn} \\
+        G^0_{1'a} & G^0_{1'1'} & G^0_{1'1} & G^0_{1'2'} & \cdots & G^0_{1'n'} & G^0_{1'n} \\
+        \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
+        G^0_{n'a} & G^0_{n'1'} & G^0_{n'1} & G^0_{n'2'} & \cdots & G^0_{n'n'} & G^0_{n'n} \\
+        G^0_{na} & G^0_{n1'} & G^0_{n1} & G^0_{n2'} & \cdots & G^0_{nn'} & G^0_{nn}
+    \end{bmatrix}
+\end{aligned}$$
+
+And analogously for $G^0_\mathrm{den}$.
+If we are studying bosons instead of fermions,
+the above determinant would need to be replaced by a *permanent*.
+We assume fermions from now on.
+
+We thus have sums over all permutations $p$
+of products of single-particle Green's function,
+times $(-1)^p$ to account for swaps of fermionic operators:
+
+$$\begin{aligned}
+    G_{ba}
+    &= -\frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n + 1} G^0_{(p,m)} \Big) \dd{1}' \dd{1} \cdots \dd{n'} \dd{n}}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{n!} \Big( \!-\! \frac{\hbar}{2} \Big)^n
+    \int\cdots\int W_{1'1} \cdots W_{n'n} \: \Big( \sum_{p} (-1)^p \prod_{m = 1}^{2 n} G^0_{(p,m)} \Big) \dd{1'} \dd{1} \cdots \dd{n'} \dd{n}}
+\end{aligned}$$
+
+These integrals over products of interactions and Green's functions
+are the perfect place to apply [Feynman diagrams](/know/concept/feynman-diagram/).
+Conveniently, it turns out that the factor $(-1)^p$
+is equivalent to the rule that each diagram must be multiplied by $(-1)^F$,
+with $F$ the number of fermion loops.
+Keep in mind that fermion lines absorb a factor $-\hbar$ each (see above),
+and interactions $-1/\hbar$.
+
+The denominator turns into a sum of all possible diagrams
+(including equivalent ones) for each total order $n$
+(the order is the number of interaction lines).
+The endpoints $a$ and $b$ do not appear here,
+so we conclude that all those diagrams only have internal vertices;
+we will therefore refer to them as **internal diagrams**.
+
+And in the numerator, we sum over all diagrams of total order $n$
+containing the external vertices $a$ and $b$.
+Some of them are **connected**,
+so all vertices (including $a$ and $b$) are in the same graph,
+but most are **disconnected**.
+Because disconnected diagrams have no shared lines or vertices to integrate over,
+they can simply be factored into separate diagrams.
+
+If it contains $a$ and $b$, we call it an **external diagram**,
+and then clearly all disconnected parts must be internal diagrams
+($a$ and $b$ are always connected,
+since they are the only vertices with just one fermion line;
+all internal vertices must have two).
+We thus find:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!}
+    \bigg[ \sum_{m = 0}^{n} \frac{n!}{m! (n \!-\! m)!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+\end{aligned}$$
+
+Where the total order is the sum of the orders of all considered diagrams,
+and the new factor is needed for all the possible choices
+of vertices to put in the external part.
+Note that the external diagram does not directly depend on $n$,
+so we reorganize:
+
+$$\begin{aligned}
+    G_{ba}
+    &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{2^{n-m} (n \!-\! m)!}
+    \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; (n \!-\! m)}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\hbar \displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+\end{aligned}$$
+
+Since both $n$ and $m$ start at zero,
+and the sums include all possible diagrams,
+we see that the second sum in the numerator does not actually depend on $m$:
+
+$$\begin{aligned}
+    \hbar G_{ba}
+    &= \frac{\displaystyle\sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+    \bigg[ \sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}} \bigg]}
+    {\displaystyle\sum_{n = 0}^\infty \frac{1}{2^n n!} \binom{\mathrm{0\;or\;more\;internal}}{\mathrm{total\;order} \; n}_{\!\Sigma\mathrm{all}}}
+    \\
+    &= \sum_{m = 0}^{\infty} \frac{1}{2^m m!} \binom{1 \; \mathrm{external}}{\mathrm{order} \; m}_{\!\Sigma\mathrm{all}}
+\end{aligned}$$
+
+In other words, all the disconnected diagrams simply cancel out,
+and we are left with a sum over all possible fully connected diagrams
+that contain $a$ and $b$. Furthermore, it can be shown using combinatorics
+that exactly $2^m m!$ diagrams at each order are topologically equivalent,
+so we are left with non-equivalent diagrams only.
+Let $G(b,a) = G_{ba}$:
+
+<a href="fullgf.png">
+<img src="fullgf.png" style="width:90%">
+</a>
+
+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 $b$:
+
+<a href="selfenergy.png">
+<img src="selfenergy.png" style="width:90%">
+</a>
+
+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 $\Sigma(y, x)$:
+
+<a href="dyson.png">
+<img src="dyson.png" style="width:95%">
+</a>
+
+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.
diff --git a/source/know/concept/self-energy/selfenergy.png b/source/know/concept/self-energy/selfenergy.png
new file mode 100644
index 0000000..59b8542
--- /dev/null
+++ b/source/know/concept/self-energy/selfenergy.png
diff --git a/source/know/concept/self-phase-modulation/index.md b/source/know/concept/self-phase-modulation/index.md
new file mode 100644
index 0000000..b2a4b58
--- /dev/null
+++ b/source/know/concept/self-phase-modulation/index.md
@@ -0,0 +1,98 @@
+---
+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:
+
+$$\begin{aligned}
+    A(z, t) = \sqrt{P_0} \exp\!\Big(\!-\!\frac{t^2}{2 T_0^2}\Big) \exp\!\bigg( i \gamma z P_0 \exp\!\Big(\!-\!\frac{t^2}{T_0^2}\Big) \bigg)
+\end{aligned}$$
+
+<a href="pheno-spm.jpg">
+<img src="pheno-spm-small.jpg" style="width:100%">
+</a>
+
+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.
diff --git a/source/know/concept/self-phase-modulation/pheno-spm-small.jpg b/source/know/concept/self-phase-modulation/pheno-spm-small.jpg
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/self-phase-modulation/pheno-spm-small.jpg
diff --git a/source/know/concept/self-phase-modulation/pheno-spm.jpg b/source/know/concept/self-phase-modulation/pheno-spm.jpg
new file mode 100644
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+++ b/source/know/concept/self-phase-modulation/pheno-spm.jpg
diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
new file mode 100644
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@@ -0,0 +1,140 @@
+---
+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:
+
+<a href="pheno-steep.jpg">
+<img src="pheno-steep-small.jpg" style="width:100%">
+</a>
+
+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.
diff --git a/source/know/concept/self-steepening/pheno-steep-small.jpg b/source/know/concept/self-steepening/pheno-steep-small.jpg
new file mode 100644
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+++ b/source/know/concept/self-steepening/pheno-steep-small.jpg
diff --git a/source/know/concept/self-steepening/pheno-steep.jpg b/source/know/concept/self-steepening/pheno-steep.jpg
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/self-steepening/pheno-steep.jpg
diff --git a/source/know/concept/shors-algorithm/index.md b/source/know/concept/shors-algorithm/index.md
new file mode 100644
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+++ b/source/know/concept/shors-algorithm/index.md
@@ -0,0 +1,301 @@
+---
+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 $N^2 \le 2^q < 2 N^2$:
+
+<a href="shors-circuit.png">
+<img src="shors-circuit.png" style="width:70%">
+</a>
+
+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.
+
diff --git a/source/know/concept/shors-algorithm/shors-circuit.png b/source/know/concept/shors-algorithm/shors-circuit.png
new file mode 100644
index 0000000..d373199
--- /dev/null
+++ b/source/know/concept/shors-algorithm/shors-circuit.png
diff --git a/source/know/concept/sigma-algebra/index.md b/source/know/concept/sigma-algebra/index.md
new file mode 100644
index 0000000..30be914
--- /dev/null
+++ b/source/know/concept/sigma-algebra/index.md
@@ -0,0 +1,54 @@
+---
+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
index 0000000..2391e42
--- /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:
+
+<a href="simons-circuit.png">
+<img src="simons-circuit.png" style="width:52%">
+</a>
+
+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
--- /dev/null
+++ b/source/know/concept/simons-algorithm/simons-circuit.png
diff --git a/source/know/concept/slater-determinant/index.md b/source/know/concept/slater-determinant/index.md
new file mode 100644
index 0000000..eca8048
--- /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
new file mode 100644
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--- /dev/null
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@@ -0,0 +1,103 @@
+---
+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
new file mode 100644
index 0000000..464a1e7
--- /dev/null
+++ b/source/know/concept/step-index-fiber/bessel.jpg
diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
new file mode 100644
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--- /dev/null
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@@ -0,0 +1,421 @@
+---
+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 $\mu$:
+
+$$\begin{aligned}
+    \begin{cases}
+        \displaystyle
+        0 = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R
+        & \mathrm{for}\; \mu \neq 0
+        \\
+        \displaystyle
+        0 = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R
+        & \mathrm{for}\; \mu = 0
+    \end{cases}
+\end{aligned}$$
+
+<a href="bessel.jpg">
+<img src="bessel.jpg" style="width:100%">
+</a>
+
+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:
+
+<a href="modes.jpg">
+<img src="modes.jpg" style="width:100%">
+</a>
+
+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
new file mode 100644
index 0000000..85682d7
--- /dev/null
+++ b/source/know/concept/step-index-fiber/modes.jpg
diff --git a/source/know/concept/stochastic-process/index.md b/source/know/concept/stochastic-process/index.md
new file mode 100644
index 0000000..512110a
--- /dev/null
+++ b/source/know/concept/stochastic-process/index.md
@@ -0,0 +1,58 @@
+---
+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
index 0000000..d52b0b4
--- /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:
+
+<table style="width:70%;margin:auto;text-align:center;">
+    <tr>
+        <th>$(a_1, a_2)$</th>
+        <th>Operator</th>
+        <th>Result</th>
+    </tr>
+    <tr>
+        <td>$00$</td>
+        <td>$\hat{I}$</td>
+        <td>$\ket{\Phi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B + \Ket{1}_A \Ket{1}_B \Big)$</td>
+    </tr>
+    <tr>
+        <td>$01$</td>
+        <td>$\hat{\sigma}_z$</td>
+        <td>$\ket{\Phi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{0}_B - \Ket{1}_A \Ket{1}_B \Big)$</td>
+    </tr>
+    <tr>
+        <td>$10$</td>
+        <td>$\hat{\sigma}_x$</td>
+        <td>$\ket{\Psi^{+}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B + \Ket{1}_A \Ket{0}_B \Big)$</td>
+    </tr>
+    <tr>
+        <td>$11$</td>
+        <td>$\hat{\sigma}_x \hat{\sigma}_z$</td>
+        <td>$\ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big(\Ket{0}_A \Ket{1}_B - \Ket{1}_A \Ket{0}_B \Big)$</td>
+    </tr>
+</table>
+
+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
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--- /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.
diff --git a/source/know/concept/toffoli-gate/and.png b/source/know/concept/toffoli-gate/and.png
new file mode 100644
index 0000000..2a25dc8
--- /dev/null
+++ b/source/know/concept/toffoli-gate/and.png
diff --git a/source/know/concept/toffoli-gate/index.md b/source/know/concept/toffoli-gate/index.md
new file mode 100644
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--- /dev/null
+++ b/source/know/concept/toffoli-gate/index.md
@@ -0,0 +1,94 @@
+---
+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:
+
+<a href="toffoli.png">
+<img src="toffoli.png" style="width:19%">
+</a>
+
+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 href="nand.png">
+<img src="nand.png" style="width:38%">
+</a>
+
+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:
+
+<a href="not.png">
+<img src="not.png" style="width:32%">
+</a>
+
+<a href="and.png">
+<img src="and.png" style="width:35%">
+</a>
+
+<a href="xor.png">
+<img src="xor.png" style="width:35%">
+</a>
+
+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 $B$:
+
+<a href="or.png">
+<img src="or.png" style="width:50%">
+</a>
+
+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.
diff --git a/source/know/concept/toffoli-gate/nand.png b/source/know/concept/toffoli-gate/nand.png
new file mode 100644
index 0000000..e9a10ca
--- /dev/null
+++ b/source/know/concept/toffoli-gate/nand.png
diff --git a/source/know/concept/toffoli-gate/not.png b/source/know/concept/toffoli-gate/not.png
new file mode 100644
index 0000000..5e33f3f
--- /dev/null
+++ b/source/know/concept/toffoli-gate/not.png
diff --git a/source/know/concept/toffoli-gate/or.png b/source/know/concept/toffoli-gate/or.png
new file mode 100644
index 0000000..147fc00
--- /dev/null
+++ b/source/know/concept/toffoli-gate/or.png
diff --git a/source/know/concept/toffoli-gate/toffoli.png b/source/know/concept/toffoli-gate/toffoli.png
new file mode 100644
index 0000000..c9ba923
--- /dev/null
+++ b/source/know/concept/toffoli-gate/toffoli.png
diff --git a/source/know/concept/toffoli-gate/xor.png b/source/know/concept/toffoli-gate/xor.png
new file mode 100644
index 0000000..8295466
--- /dev/null
+++ b/source/know/concept/toffoli-gate/xor.png
diff --git a/source/know/concept/two-fluid-equations/index.md b/source/know/concept/two-fluid-equations/index.md
new file mode 100644
index 0000000..b9f1e94
--- /dev/null
+++ b/source/know/concept/two-fluid-equations/index.md
@@ -0,0 +1,273 @@
+---
+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.