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diff --git a/content/know/concept/_index.md b/content/know/concept/_index.md
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+---
+title: "List of concepts"
+date: 2021-02-22T20:38:58+01:00
+draft: false
+layout: "know-list"
+---
+
+This is an alphabetical list of the concepts in this knowledge base.
diff --git a/content/know/concept/blochs-theorem/index.pdc b/content/know/concept/blochs-theorem/index.pdc
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+---
+title: "Bloch's theorem"
+firstLetter: "B"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+
+date: 2021-02-22T20:02:14+01:00
+draft: false
+markup: pandoc
+---
+
+# Bloch's theorem
+In quantum mechanics, **Bloch's theorem** states that,
+given a potential $V(\vec{r})$ which is periodic on a lattice,
+i.e. $V(\vec{r}) = V(\vec{r} + \vec{a})$
+for a primitive lattice vector $\vec{a}$,
+then it follows that the solutions $\psi(\vec{r})$
+to the time-independent Schrödinger equation
+take the following form,
+where the function $u(\vec{r})$ is periodic on the same lattice,
+i.e. $u(\vec{r}) = u(\vec{r} + \vec{a})$:
+
+$$
+\begin{aligned}
+	\boxed{
+		\psi(\vec{r}) = u(\vec{r}) e^{i \vec{k} \cdot \vec{r}}
+	}
+\end{aligned}
+$$
+
+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 $\psi(\vec{r})$ and $\psi(\vec{r} + \vec{a})$
+are eigenstates with the same energy:
+
+$$
+\begin{aligned}
+	\hat{H} \psi(\vec{r}) = E \psi(\vec{r})
+	\qquad
+	\hat{H} \psi(\vec{r} + \vec{a}) = E \psi(\vec{r} + \vec{a})
+\end{aligned}
+$$
+
+Now define the unitary translation operator $\hat{T}(\vec{a})$ such that
+$\psi(\vec{r} + \vec{a}) = \hat{T}(\vec{a}) \psi(\vec{r})$.
+From the previous equation, we then know that:
+
+$$
+\begin{aligned}
+	\hat{H} \hat{T}(\vec{a}) \psi(\vec{r})
+	= E \hat{T}(\vec{a}) \psi(\vec{r})
+	= \hat{T}(\vec{a}) \big(E \psi(\vec{r})\big)
+	= \hat{T}(\vec{a}) \hat{H} \psi(\vec{r})
+\end{aligned}
+$$
+
+In other words, if $\hat{H}$ is lattice-periodic,
+then it will commute with $\hat{T}(\vec{a})$,
+i.e. $[\hat{H}, \hat{T}(\vec{a})] = 0$.
+Consequently, $\hat{H}$ and $\hat{T}(\vec{a})$ must share eigenstates $\psi(\vec{r})$:
+
+$$
+\begin{aligned}
+	\hat{H} \:\psi(\vec{r}) = E \:\psi(\vec{r})
+	\qquad
+	\hat{T}(\vec{a}) \:\psi(\vec{r}) = \tau \:\psi(\vec{r})
+\end{aligned}
+$$
+
+Since $\hat{T}$ is unitary,
+its eigenvalues $\tau$ must have the form $e^{i \theta}$, with $\theta$ real.
+Therefore a translation by $\vec{a}$ causes a phase shift,
+for some vector $\vec{k}$:
+
+$$
+\begin{aligned}
+	\psi(\vec{r} + \vec{a})
+	= \hat{T}(\vec{a}) \:\psi(\vec{r})
+	= e^{i \theta} \:\psi(\vec{r})
+	= e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
+\end{aligned}
+$$
+
+Let us now define the following function,
+keeping our arbitrary choice of $\vec{k}$:
+
+$$
+\begin{aligned}
+	u(\vec{r})
+	= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
+\end{aligned}
+$$
+
+As it turns out, this function is guaranteed to be lattice-periodic for any $\vec{k}$:
+
+$$
+\begin{aligned}
+	u(\vec{r} + \vec{a})
+	&= e^{- i \vec{k} \cdot (\vec{r} + \vec{a})} \:\psi(\vec{r} + \vec{a})
+	\\
+	&= e^{- i \vec{k} \cdot \vec{r}} e^{- i \vec{k} \cdot \vec{a}} e^{i \vec{k} \cdot \vec{a}} \:\psi(\vec{r})
+	\\
+	&= e^{- i \vec{k} \cdot \vec{r}} \:\psi(\vec{r})
+	\\
+	&= u(\vec{r})
+\end{aligned}
+$$
+
+Then Bloch's theorem follows from
+isolating the definition of $u(\vec{r})$ for $\psi(\vec{r})$.
diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc
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+---
+title: "Convolution theorem"
+firstLetter: "C"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+
+date: 2021-02-22T21:35:23+01:00
+draft: false
+markup: pandoc
+---
+
+# Convolution theorem
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            A \cdot (f * g)(x) &= \hat{\mathcal{F}}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
+            B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+To prove this, 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 thing again, this time starting from a product in
+the $x$-domain:
+
+$$\begin{aligned}
+    \hat{\mathcal{F}}\{f(x) \: g(x)\}
+    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
+    \\
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
+    \\
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'}
+    = B \cdot (\tilde{f} * \tilde{g})(k)
+\end{aligned}$$
+
+
+## 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:
+
+$$\begin{aligned}
+    \boxed{(f * g)(t) = \hat{\mathcal{L}}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
+\end{aligned}$$
+
+Because the inverse Laplace transform $\hat{\mathcal{L}}^{-1}$ is quite
+unpleasant, the theorem is often stated using the forward transform
+instead:
+
+$$\begin{aligned}
+    \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
+\end{aligned}$$
+
+We prove this by expanding 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:
+
+$$\begin{aligned}
+    \hat{\mathcal{L}}\{(f * g)(t)\}
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
+    \\
+    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
+    \\
+    &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'}
+    = \tilde{f}(s) \: \tilde{g}(s)
+\end{aligned}$$
diff --git a/content/know/concept/dirac-delta-function/index.pdc b/content/know/concept/dirac-delta-function/index.pdc
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+---
+title: "Dirac delta function"
+firstLetter: "D"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-22T21:35:38+01:00
+draft: false
+markup: pandoc
+---
+
+# Dirac delta function
+
+The **Dirac delta function** $\delta(x)$, often just called the **delta function**,
+is an infinitely narrow discontinuous "spike" at $x = 0$ whose area is
+defined to be 1:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(x) =
+        \begin{cases}
+            +\infty & \mathrm{if}\: x = 0 \\
+            0 & \mathrm{if}\: x \neq 0
+        \end{cases}
+        \quad \mathrm{and} \quad
+        \int_{-\varepsilon}^\varepsilon \delta(x) \dd{x} = 1
+    }
+\end{aligned}$$
+
+It is sometimes also called the **sampling function**, due 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 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/):
+
+$$\begin{aligned}
+    \boxed{
+        \delta(x)
+        %= \lim_{n \to +\infty} \!\Big\{\frac{\sin(n x)}{\pi x}\Big\}
+        = \frac{1}{2\pi} \int_{-\infty}^\infty \exp(i k x) \dd{k}
+        \:\:\propto\:\: \hat{\mathcal{F}}\{1\}
+    }
+\end{aligned}$$
+
+When the argument of $\delta(x)$ is scaled, the delta function is itself scaled:
+
+$$\begin{aligned}
+    \boxed{
+        \delta(s x) = \frac{1}{|s|} \delta(x)
+    }
+\end{aligned}$$
+
+*__Proof.__ Because it is symmetric, $\delta(s x) = \delta(|s| x)$. Then by
+substituting $\sigma = |s| x$:*
+
+$$\begin{aligned}
+    \int \delta(|s| x) \dd{x}
+    &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
+\end{aligned}$$
+
+*__Q.E.D.__*
+
+An even more impressive property is the behaviour of the derivative of
+$\delta(x)$:
+
+$$\begin{aligned}
+    \boxed{
+        \int f(\xi) \: \delta'(x - \xi) \dd{\xi} = f'(x)
+    }
+\end{aligned}$$
+
+*__Proof.__ Note which variable is used for the
+differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:*
+
+$$\begin{aligned}
+    \int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi}
+    &= \dv{x} \int f(\xi) \: \delta(x - \xi) \dd{x}
+    = f'(x)
+\end{aligned}$$
+
+*__Q.E.D.__*
+
+This property also generalizes nicely for the higher-order derivatives:
+
+$$\begin{aligned}
+    \boxed{
+        \int f(\xi) \: \dv[n]{\delta(x - \xi)}{x} \dd{\xi} = \dv[n]{f(x)}{x}
+    }
+\end{aligned}$$
diff --git a/content/know/concept/dirac-notation/index.pdc b/content/know/concept/dirac-notation/index.pdc
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+---
+title: "Dirac notation"
+firstLetter: "D"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:35:46+01:00
+draft: false
+markup: pandoc
+---
+
+# Dirac notation
+
+**Dirac notation** is a notation to do calculations in a 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**
+$\braket{V}{W}$, 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:
+
+$$\begin{aligned}
+    \braket{V}{W} = \braket{W}{V}^*
+\end{aligned}$$
+
+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 $\braket{V}{W}$ is then just the familiar dot product $V \cdot W$:
+
+$$\begin{gathered}
+    \braket{V}{W}
+    =
+    \begin{bmatrix}
+        v_1^* & \cdots & v_N^*
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        w_1 \\ \vdots \\ w_N
+    \end{bmatrix}
+    = v_1^* w_1 + ... + v_N^* w_N
+\end{gathered}$$
+
+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:
+
+$$\begin{gathered}
+    \braket{f}{g}
+    = F[g(x)]
+    = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{gathered}$$
+
+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:
+
+$$\begin{aligned}
+    \bra{u} \!\Big(\!\ket{f} \bra{g}\!\Big)\! \ket{w}
+    &= U\big[f(x) \: G[w(x)]\big]
+    = U\Big[ f(x) \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big]
+    \\
+    &= \int_a^b u^*(x) \: f(x) \: \Big(\int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big) \dd{x}
+    \\
+    &= \Big( \int_a^b u^*(x) \: f(x) \dd{x} \Big) \Big( \int_a^b g^*(\xi) \: w(\xi) \dd{\xi} \Big)
+    \\
+    &= \braket{u}{f} \braket{g}{w}
+\end{aligned}$$
diff --git a/content/know/concept/fourier-transform/index.pdc b/content/know/concept/fourier-transform/index.pdc
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+---
+title: "Fourier transform"
+firstLetter: "F"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-22T21:35:54+01:00
+draft: false
+markup: pandoc
+---
+
+# Fourier transform
+
+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):
+
+$$\begin{aligned}
+    \boxed{
+        \tilde{f}(k)
+        = \hat{\mathcal{F}}\{f(x)\}
+        = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x}
+    }
+\end{aligned}$$
+
+The **inverse Fourier transform** (iFT) undoes the forward FT operation:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}
+        = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k}
+    }
+\end{aligned}$$
+
+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 interesting 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:
+
+$$\begin{aligned}
+    \boxed{
+        \hat{\mathcal{F}} \Big\{ \dv[n]{f}{x} \Big\}
+        = (- i s k)^n \tilde{f}(k)
+    }
+\end{aligned}$$
+
+Derivatives in the frequency domain have an analogous property:
+
+$$\begin{aligned}
+    \boxed{
+        \dv[n]{\tilde{f}}{k}
+        = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x}
+        = \hat{\mathcal{F}}\{ (i s x)^n f(x) \}
+    }
+\end{aligned}$$
diff --git a/content/know/concept/gram-schmidt-method/index.pdc b/content/know/concept/gram-schmidt-method/index.pdc
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+---
+title: "Gram-Schmidt method"
+firstLetter: "G"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+
+date: 2021-02-22T21:36:08+01:00
+draft: false
+markup: pandoc
+---
+
+# Gram-Schmidt method
+
+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 $\ket*{n_1}$:
+    
+    $$\begin{aligned}
+        \ket*{n_1} = \frac{\ket*{V_1}}{\sqrt{\braket*{V_1}{V_1}}}
+    \end{aligned}$$
+    
+2.  Begin loop. Take the next non-orthonormal vector $\ket*{V_j}$, and
+    subtract from it its projection onto every already-processed vector:
+    
+    $$\begin{aligned}
+        \ket*{n_j'} = \ket*{V_j} - \ket*{n_1} \braket*{n_1}{V_j} - \ket*{n_2} \braket*{n_2}{V_j} - ... - \ket*{n_{j-1}} \braket*{n_{j-1}}{V_{j-1}}
+    \end{aligned}$$
+    
+    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*:
+    
+    $$\begin{aligned}
+                \ket*{n_j} = \frac{\ket*{n_j'}}{\sqrt{\braket*{n_j'}{n_j'}}}
+    \end{aligned}$$
+    
+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/).
diff --git a/content/know/concept/hilbert-space/index.pdc b/content/know/concept/hilbert-space/index.pdc
new file mode 100644
index 0000000..1faf08a
--- /dev/null
+++ b/content/know/concept/hilbert-space/index.pdc
@@ -0,0 +1,202 @@
+---
+title: "Hilbert space"
+firstLetter: "H"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Quantum mechanics
+
+date: 2021-02-22T21:36:24+01:00
+draft: false
+markup: pandoc
+---
+
+# Hilbert space
+
+A **Hilbert space**, also known as 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 $\braket{U}{V}$
+called the **inner product**, which takes two vectors and returns a
+scalar, and has the following properties:
+
++ **Skew symmetry**: $\braket{U}{V} = (\braket{V}{U})^*$, where ${}^*$ is the complex conjugate.
++ **Positive semidefiniteness**: $\braket{V}{V} \ge 0$, and $\braket{V}{V} = 0$ if $V = \mathbf{0}$.
++ **Linearity in second operand**: $\braket{U}{(a V + b W)} = a \braket{U}{V} + b \braket{U}{W}$.
+
+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
+$|V| = \sqrt{\braket{V}{V}}$ 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
+$\braket{U}{V} = 0$. 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:
+
+$$\begin{gathered}
+    V = \sum_{n = 1}^N v_n \vu{e}_n
+    \quad
+    W = \sum_{n = 1}^N w_n \vu{e}_n
+    \\
+    \quad \implies \quad
+    \braket{V}{W} = \sum_{n = 1}^N \sum_{m = 1}^N v_n^* w_m \braket{\vu{e}_n}{\vu{e}_j}
+\end{gathered}$$
+
+If the basis vectors $\vu{e}_1, ..., \vu{e}_N$ are already
+orthonormal, this reduces to:
+
+$$\begin{aligned}
+    \braket{V}{W} = \sum_{n = 1}^N v_n^* w_n
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \braket{\vu{e}_n}{V} = \sum_{m = 1}^N \delta_{nm} v_m = v_n
+\end{aligned}$$
+
+
+## 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:
+
+$$\begin{aligned}
+    f(x) = \int_a^b \braket{x}{f} \dd{x}
+\end{aligned}$$
+
+Similarly, the inner product $\braket{f}{g}$ must also be redefined as
+follows:
+
+$$\begin{aligned}
+    \braket{f}{g} = \int_a^b f^*(x) \: g(x) \dd{x}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    f(x) = \braket{x}{f} = \matrixel{x}{\hat{I}}{f}
+    = \int_a^b \braket{x}{\xi} \braket{\xi}{f} \dd{\xi}
+    = \int_a^b \braket{x}{\xi} f(\xi) \dd{\xi}
+\end{aligned}$$
+
+Since we want the latter integral to reduce to $f(x)$, it is plain to see that
+$\braket{x}{\xi}$ can only be a [Dirac delta function](/know/concept/dirac-delta-function/),
+i.e $\braket{x}{\xi} = \delta(x - \xi)$:
+
+$$\begin{aligned}
+    \int_a^b \braket{x}{\xi} f(\xi) \dd{\xi}
+    = \int_a^b \delta(x - \xi) f(\xi) \dd{\xi}
+    = f(x)
+\end{aligned}$$
+
+Consequently, $\braket{x}{\xi} = 0$ if $x \neq \xi$ as expected for an
+orthogonal set of basis vectors, but if $x = \xi$ the inner product
+$\braket{x}{\xi}$ 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/content/know/concept/legendre-transform/index.pdc b/content/know/concept/legendre-transform/index.pdc
new file mode 100644
index 0000000..8a0d3e3
--- /dev/null
+++ b/content/know/concept/legendre-transform/index.pdc
@@ -0,0 +1,89 @@
+---
+title: "Legendre transform"
+firstLetter: "L"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-22T21:36:35+01:00
+draft: false
+markup: pandoc
+---
+
+# Legendre transform
+
+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 $-C$ instead) for all $x_0 \in [a, b]$, where $C$ is the
+constant corresponding 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 and Hamiltonian mechanics to each other.
+It is also used to convert between thermodynamic potentials.
diff --git a/content/know/concept/parsevals-theorem/index.pdc b/content/know/concept/parsevals-theorem/index.pdc
new file mode 100644
index 0000000..8f653f8
--- /dev/null
+++ b/content/know/concept/parsevals-theorem/index.pdc
@@ -0,0 +1,76 @@
+---
+title: "Parseval's theorem"
+firstLetter: "P"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-22T21:36:44+01:00
+draft: false
+markup: pandoc
+---
+
+# Parseval's theorem
+
+**Parseval's theorem** relates the inner product of two functions $f(x)$ and $g(x)$ to 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 Fourier transform's definition:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)}
+    }
+    \\
+    \boxed{
+        \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)}
+    }
+\end{aligned}$$
+
+For this reason, physicists like to define their Fourier transform
+with $A = B = 1 / \sqrt{2\pi}$ and $|s| = 1$, because then the FT nicely
+conserves the total probability (quantum mechanics) or the total energy
+(optics).
+
+To prove this, we insert the inverse FT into the inner product
+definition:
+
+$$\begin{aligned}
+    \braket{f}{g}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\}\big)^* \: \hat{\mathcal{F}}^{-1}\{\tilde{g}(k)\} \dd{x}
+    \\
+    &= B^2 \int
+    \Big( \int \tilde{f}^*(k_1) \exp(i s k_1 x) \dd{k_1} \Big)
+    \Big( \int \tilde{g}(k) \exp(- i s k x) \dd{k} \Big)
+    \dd{x}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \tilde{g}(k) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s x (k_1 - k)) \dd{x} \Big) \dd{k_1} \dd{k}
+    \\
+    &= 2 \pi B^2 \iint \tilde{f}^*(k_1) \: \tilde{g}(k) \: \delta(s (k_1 - k)) \dd{k_1} \dd{k}
+    \\
+    &= \frac{2 \pi B^2}{|s|} \int_{-\infty}^\infty \tilde{f}^*(k) \: \tilde{g}(k) \dd{k}
+    = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}}{\tilde{g}}
+\end{aligned}$$
+
+Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+Note that we can just as well do it in the opposite direction,
+which yields an equivalent result:
+
+$$\begin{aligned}
+    \braket*{\tilde{f}}{\tilde{g}}
+    &= \int_{-\infty}^\infty \big( \hat{\mathcal{F}}\{f(x)\}\big)^* \: \hat{\mathcal{F}}\{g(x)\} \dd{k}
+    \\
+    &= A^2 \int
+    \Big( \int f^*(x_1) \exp(- i s k x_1) \dd{x_1} \Big)
+    \Big( \int g(x) \exp(i s k x) \dd{x} \Big)
+    \dd{k}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) g(x) \Big( \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i s k (x_1 - x)) \dd{k} \Big) \dd{x_1} \dd{x}
+    \\
+    &= 2 \pi A^2 \iint f^*(x_1) \: g(x) \: \delta(s (x_1 - x)) \dd{x_1} \dd{x}
+    \\
+    &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
+    = \frac{2 \pi A^2}{|s|} \braket{f}{g}
+\end{aligned}$$
diff --git a/content/know/concept/partial-fraction-decomposition/index.pdc b/content/know/concept/partial-fraction-decomposition/index.pdc
new file mode 100644
index 0000000..1f4207f
--- /dev/null
+++ b/content/know/concept/partial-fraction-decomposition/index.pdc
@@ -0,0 +1,60 @@
+---
+title: "Partial fraction decomposition"
+firstLetter: "P"
+publishDate: 2021-02-22
+categories:
+- Mathematics
+
+date: 2021-02-22T21:36:56+01:00
+draft: false
+markup: pandoc
+---
+
+# Partial fraction decomposition
+
+**Partial fraction decomposition** or **expansion** is a method to rewrite a
+quotient of two polynomials $g(x)$ and $h(x)$, where the numerator
+$g(x)$ is of lower order than $h(x)$, as a sum 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 $c_{1,1}, ..., c_{1,m}$.
diff --git a/content/know/concept/pauli-exclusion-principle/index.pdc b/content/know/concept/pauli-exclusion-principle/index.pdc
new file mode 100644
index 0000000..aa9609b
--- /dev/null
+++ b/content/know/concept/pauli-exclusion-principle/index.pdc
@@ -0,0 +1,125 @@
+---
+title: "Pauli exclusion principle"
+firstLetter: "P"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:37:14+01:00
+draft: false
+markup: pandoc
+---
+
+# Pauli exclusion principle
+
+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/content/know/concept/probability-current/index.pdc b/content/know/concept/probability-current/index.pdc
new file mode 100644
index 0000000..c67956a
--- /dev/null
+++ b/content/know/concept/probability-current/index.pdc
@@ -0,0 +1,98 @@
+---
+title: "Probability current"
+firstLetter: "P"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:37:26+01:00
+draft: false
+markup: pandoc
+---
+
+# Probability current
+
+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:
+
+$$\begin{aligned}
+    P = \int_{V} | \psi |^2 \dd[3]{\vec{r}}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \pdv{P}{t}
+    &= \int_{V} \psi \pdv{\psi^*}{t} + \psi^* \pdv{\psi}{t} \dd[3]{\vec{r}}
+    = \frac{i}{\hbar} \int_{V} \psi (\hat{H} \psi^*) - \psi^* (\hat{H} \psi) \dd[3]{\vec{r}}
+    \\
+    &= \frac{i}{\hbar} \int_{V} \psi \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi^* + V(\vec{r}) \psi^* \Big)
+    - \psi^* \Big( \!-\! \frac{\hbar^2}{2 m} \nabla^2 \psi + V(\vec{r}) \psi \Big) \dd[3]{\vec{r}}
+    \\
+    &= \frac{i \hbar}{2 m} \int_{V} - \psi \nabla^2 \psi^* + \psi^* \nabla^2 \psi \dd[3]{\vec{r}}
+    = - \int_{V} \nabla \cdot \vec{J} \dd[3]{\vec{r}}
+\end{aligned}$$
+
+Where we have defined the probability current $\vec{J}$ as follows in
+the $\vec{r}$-basis:
+
+$$\begin{aligned}
+    \vec{J}
+    = \frac{i \hbar}{2 m} (\psi \nabla \psi^* - \psi^* \nabla \psi)
+    = \mathrm{Re} \Big\{ \psi \frac{i \hbar}{m} \psi^* \Big\}
+\end{aligned}$$
+
+Let us rewrite this using the momentum operator
+$\hat{p} = -i \hbar \nabla$ as follows, noting that $\hat{p} / m$ is
+simply the velocity operator $\hat{v}$:
+
+$$\begin{aligned}
+    \boxed{
+        \vec{J}
+        = \frac{1}{2 m} ( \psi^* \hat{p} \psi - \psi \hat{p} \psi^*)
+        = \mathrm{Re} \Big\{ \psi^* \frac{\hat{p}}{m} \psi \Big\}
+        = \mathrm{Re} \{ \psi^* \hat{v} \psi \}
+    }
+\end{aligned}$$
+
+Returning to the derivation of $\vec{J}$, we now have the following
+equation:
+
+$$\begin{aligned}
+    \pdv{P}{t}
+    = \int_{V} \pdv{|\psi|^2}{t} \dd[3]{\vec{r}}
+    = - \int_{V} \nabla \cdot \vec{J} \dd[3]{\vec{r}}
+\end{aligned}$$
+
+By removing the integrals, we thus arrive at the **continuity equation**
+for $\vec{J}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vec{J}
+        = - \pdv{|\psi|^2}{t}
+    }
+\end{aligned}$$
+
+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
+$\vec{J}$ 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 $\vec{A}$, 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:
+
+$$\begin{aligned}
+    \boxed{
+        \vec{J}
+        = \mathrm{Re} \Big\{ \psi^* \frac{\hat{p} - q \vec{A}}{m} \psi \Big\}
+    }
+\end{aligned}$$
diff --git a/content/know/concept/slater-determinant/index.pdc b/content/know/concept/slater-determinant/index.pdc
new file mode 100644
index 0000000..8bc4291
--- /dev/null
+++ b/content/know/concept/slater-determinant/index.pdc
@@ -0,0 +1,54 @@
+---
+title: "Slater determinant"
+firstLetter: "S"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:38:03+01:00
+draft: false
+markup: pandoc
+---
+
+# Slater determinant
+
+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/content/know/concept/sturm-liouville-theory/index.pdc b/content/know/concept/sturm-liouville-theory/index.pdc
new file mode 100644
index 0000000..7ccd625
--- /dev/null
+++ b/content/know/concept/sturm-liouville-theory/index.pdc
@@ -0,0 +1,346 @@
+---
+title: "Sturm-Liouville theory"
+firstLetter: "S"
+publishDate: 2021-02-23
+categories:
+- Mathematics
+- Physics
+
+date: 2021-02-23T08:52:28+01:00
+draft: false
+markup: pandoc
+---
+
+# Sturm-Liouville theory
+
+**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:
+
+$$\begin{aligned}
+    \braket*{f}{\hat{L} g}
+    = \braket*{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+What is $\hat{L}^\dagger$, given the above definition of $\hat{L}$?
+We start from the inner product $\braket*{f}{\hat{L} g}$:
+
+$$\begin{aligned}
+    \braket*{f}{\hat{L} g}
+    &= \int_a^b f^*(x) \hat{L}\{g(x)\} \dd{x}
+    = \int_a^b (f^* p_0) g'' + (f^* p_1) g' + (f^* p_2) g \dd{x}
+    \\
+    &= \big[ (f^* p_0) g' + (f^* p_1) g \big]_a^b - \int_a^b (f^* p_0)' g' + (f^* p_1)' g - (f^* p_2) g \dd{x}
+    \\
+    &= \big[ f^* \big( p_0 g' \!+\! p_1 g \big) \!-\! (f^* p_0)' g \big]_a^b + \int_a^b \! \big( (f p_0)'' - (f p_1)' + (f p_2) \big)^* g \dd{x}
+    \\
+    &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \int_a^b \big( \hat{L}^\dagger\{f\} \big)^* g \dd{x}
+\end{aligned}$$
+
+We now have an expression for $\hat{L}^\dagger$, but are left with an
+annoying boundary term:
+
+$$\begin{aligned}
+    \braket*{f}{\hat{L} g}
+    &= \big[ f^* \big( p_0 g' + (p_1 - p_0') g \big) - (f^*)' p_0 g \big]_a^b + \braket*{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \braket*{f}{\hat{L} g}
+    &= \big[ p_0 \big( f^* g' - (f^*)' g \big) \big]_a^b + \braket{\hat{L}^\dagger f}{g}
+    = \braket*{\hat{L}^\dagger f}{g}
+\end{aligned}$$
+
+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 $\braket{f}{w g}$ 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:
+
+$$\begin{aligned}
+    \int_a^b u_m^* \hat{L}_{SL}\{u_n\} - u_n \hat{L}_{SL}\{u_m^*\} \:dx
+    &= (\lambda_m^* - \lambda_n) \int_a^b u_n u_m^* w \:dx
+    \\
+    \braket*{u_m}{\hat{L}_{SL} u_n} - \braket*{\hat{L}_{SL} u_m}{u_n}
+    &= (\lambda_m^* - \lambda_n) \braket{u_m}{w u_n}
+\end{aligned}$$
+
+The operator $\hat{L}_{SL}$ is self-adjoint by definition,
+so the left-hand side vanishes, leaving us with:
+
+$$\begin{aligned}
+    0
+    &= (\lambda_m^* - \lambda_n) \braket{u_m}{w u_n}
+\end{aligned}$$
+
+When $m = n$, the inner product $\braket{u_n}{w u_n}$ 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 $\braket{u_m}{w u_n} = 0$, 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**:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{u_m(x)}{w(x) u_n(x)}
+        = \braket{u_n}{w u_n} \delta_{nm}
+        = A_n \delta_{nm}
+    }
+\end{aligned}$$
+
+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're 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:
+
+$$\begin{aligned}
+    \int_a^b f(x) w(x) u_m^*(x) \dd{x}
+    &= \int_a^b \Big(\sum_{n = 0}^\infty a_n u_n(x) w(x) u_m^*(x)\Big) \dd{x}
+    \\
+    \braket{u_m}{w f}
+    &= \sum_{n = 0}^\infty a_n \braket{u_m}{w u_n}
+\end{aligned}$$
+
+Because the eigenfunctions of an SLP are mutually orthogonal,
+the summation disappears:
+
+$$\begin{aligned}
+    \braket{u_m}{w f}
+    &= \sum_{n = 0}^\infty a_n \braket{u_m}{w u_n}
+    = \sum_{n = 0}^\infty a_n A_n \delta_{nm}
+    = a_m A_m
+\end{aligned}$$
+
+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 $u_n(x) / A_n$:
+
+$$\begin{aligned}
+    \boxed{
+        a_n
+        = \frac{\braket{u_n}{w f}}{A_n}
+        = \frac{\braket{u_n}{w f}}{\braket{u_n}{w u_n}}
+    }
+\end{aligned}$$
+
+As a final remark, we can see something interesting
+by rearranging the generalized Fourier series
+after inserting the expression for $a_n$:
+
+$$\begin{aligned}
+    f(x)
+    &= \sum_{n = 0}^\infty \frac{1}{A_n} \braket{u_n}{w f} u_n(x)
+    = \int_a^b \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) f(\xi) u_n(x) \Big) \dd{\xi}
+    \\
+    &= \int_a^b f(\xi) \Big(\sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) \Big) \dd{\xi}
+    %= \int_a^b f(\xi) \delta(x - \xi) \dd{\xi}
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \boxed{
+        \sum_{n = 0}^\infty \frac{1}{A_n} u_n^*(\xi) w(\xi) u_n(x) = \delta(x - \xi)
+    }
+\end{aligned}$$
+
diff --git a/content/know/concept/time-independent-perturbation-theory/index.pdc b/content/know/concept/time-independent-perturbation-theory/index.pdc
new file mode 100644
index 0000000..4f30ae8
--- /dev/null
+++ b/content/know/concept/time-independent-perturbation-theory/index.pdc
@@ -0,0 +1,329 @@
+---
+title: "Time-independent perturbation theory"
+firstLetter: "T"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:38:18+01:00
+draft: false
+markup: pandoc
+---
+
+# Time-independent perturbation theory
+
+**Time-independent perturbation theory**, sometimes also called
+**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:
+
+$$\begin{aligned}
+    \matrixel{k}{\hat{H}_1}{n} - E_n^{(1)} \braket{k}{n} + \sum_{m \neq n} c_m (\varepsilon_m - \varepsilon_n) \braket{k}{m} = 0
+\end{aligned}$$
+
+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 correction to the energy $E_n^{(2)}$, we
+take the corresponding equation and put $\bra{n}$ in front of it:
+
+$$\begin{aligned}
+    \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} + \matrixel{n}{\hat{H}_0}{\psi_n^{(2)}}
+    &= E_n^{(2)} \braket{n}{n} + E_n^{(1)} \braket{n}{\psi_n^{(1)}} + \varepsilon_n \braket{n}{\psi_n^{(2)}}
+\end{aligned}$$
+
+Because $\hat{H}_0$ is Hermitian, we know that
+$\matrixel{n}{\hat{H}_0}{\psi_n^{(2)}} = \varepsilon_n \braket{n}{\psi_n^{(2)}}$,
+i.e. we apply it to the bra, which lets us eliminate two terms. Also,
+since $\ket{n}$ is normalized, we find:
+
+$$\begin{aligned}
+    E_n^{(2)}
+    = \matrixel{n}{\hat{H}_1}{\psi_n^{(1)}} - E_n^{(1)} \braket{n}{\psi_n^{(1)}}
+\end{aligned}$$
+
+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 $\bra{n, \delta}$:
+
+$$\begin{aligned}
+    \matrixel{n, \delta}{\hat{H}_1}{n} + \matrixel{n, \delta}{\hat{H}_0}{\psi_n^{(1)}}
+    &= E_n^{(1)} \braket{n, \delta}{n} + \varepsilon_n \braket{n, \delta}{\psi_n^{(1)}}
+\end{aligned}$$
+
+Since $\hat{H}_0$ is Hermitian, we use the same trick as before to
+reduce the problem to:
+
+$$\begin{aligned}
+    \matrixel{n, \delta}{\hat{H}_1}{n}
+    &= E_n^{(1)} \braket{n, \delta}{n}
+\end{aligned}$$
+
+We express $\ket{n}$ as a linear combination of the eigenbasis vectors
+$\ket{n, d}$ to get:
+
+$$\begin{aligned}
+    \sum_{d = 1}^{D} c_d \matrixel{n, \delta}{\hat{H}_1}{n, d}
+    = E_n^{(1)} \sum_{d = 1}^{D} c_d \braket{n, \delta}{n, d}
+    = c_{\delta} E_n^{(1)}
+\end{aligned}$$
+
+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 $d_1 \neq d_2$:
+
+$$\begin{aligned}
+    \hat{L} \ket{n, b_1} = \ell_1 \ket{n, b_1}
+    \qquad
+    \hat{L} \ket{n, b_2} = \ell_2 \ket{n, b_2}
+\end{aligned}$$
+
+When this 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 $\hat{L}$.
diff --git a/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
new file mode 100644
index 0000000..482650e
--- /dev/null
+++ b/content/know/concept/wentzel-kramers-brillouin-approximation/index.pdc
@@ -0,0 +1,198 @@
+---
+title: "Wentzel-Kramers-Brillouin approximation"
+firstLetter: "W"
+publishDate: 2021-02-22
+categories:
+- Quantum mechanics
+- Physics
+
+date: 2021-02-22T21:38:35+01:00
+draft: false
+markup: pandoc
+---
+
+# Wentzel-Kramers-Brillouin approximation
+
+In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
+approximation** is a method 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 rearrangement 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 the
+case in classical mechanics, but not necessarily so in quantum
+mechanics, but we stick with it for now. We rewrite the Schrödinger
+equation:
+
+$$\begin{aligned}
+    0
+    = \dv[2]{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
+    = \dv[2]{\psi}{x} + \frac{p^2}{\hbar^2} \psi
+\end{aligned}$$
+
+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 not allowed; a ball
+cannot simply go through a potential bump without the necessary energy.
+However, in quantum mechanics, particles can **tunnel** through barriers.
+
+Conveniently, all 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-mechanical region ($E < V$) it is exponential. Note that for
+$E \approx V$ the approximation breaks down, due to the appearance of
+$p(x)$ in the denominator.