Improve knowledge baseHEADmaster

13 files changed, 137 insertions, 130 deletions

diff --git a/source/know/concept/bose-einstein-distribution/index.md b/source/know/concept/bose-einstein-distribution/index.md
index 5640e69..ea5ca68 100644
--- a/source/know/concept/bose-einstein-distribution/index.md
+++ b/source/know/concept/bose-einstein-distribution/index.md
@@ -14,19 +14,19 @@ which do not obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-
 distribute themselves across the available states
 in a system at equilibrium.
 
-Consider a single-particle state $$s$$,
+Consider a single-particle state $$\ket{i}$$,
 which can contain any number of bosons.
-Since the occupation number $$N$$ is variable,
+Since the occupation number $$n_i$$ is variable,
 we use the [grand canonical ensemble](/know/concept/grand-canonical-ensemble/),
 whose grand partition function $$\mathcal{Z}$$ is as shown below,
-where $$\varepsilon$$ is the energy per particle,
+where $$\varepsilon_i$$ is the energy per particle,
 and $$\mu$$ is the chemical potential.
 We evaluate the sum in $$\mathcal{Z}$$ as a geometric series:
 
 $$\begin{aligned}
     \mathcal{Z}
-    = \sum_{N = 0}^\infty \Big( e^{-\beta (\varepsilon - \mu)} \Big)^{N}
-    = \frac{1}{1 - e^{-\beta (\varepsilon - \mu)}}
+    = \sum_{m = 0}^\infty \Big( e^{-\beta (\varepsilon_i - \mu)} \Big)^{m}
+    = \frac{1}{1 - e^{-\beta (\varepsilon_i - \mu)}}
 \end{aligned}$$
 
 The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
@@ -35,42 +35,42 @@ is the Landau potential $$\Omega$$, given by:
 $$\begin{aligned}
     \Omega
     = - k T \ln{\mathcal{Z}}
-    = k T \ln\!\big( 1 - e^{-\beta (\varepsilon - \mu)} \big)
+    = k T \ln\!\big( 1 - e^{-\beta (\varepsilon_i - \mu)} \big)
 \end{aligned}$$
 
-The average number of particles $$\expval{N}$$ in $$s$$
+The average number of particles $$\expval{n_i}$$ in $$\ket{i}$$
 is then found by taking a derivative of $$\Omega$$:
 
 $$\begin{aligned}
-    \expval{N}
+    \expval{n_i}
     = - \pdv{\Omega}{\mu}
     = k T \pdv{\ln{\mathcal{Z}}}{\mu}
-    = \frac{e^{-\beta (\varepsilon - \mu)}}{1 - e^{-\beta (\varepsilon - \mu)}}
+    = \frac{e^{-\beta (\varepsilon_i - \mu)}}{1 - e^{-\beta (\varepsilon_i - \mu)}}
 \end{aligned}$$
 
-By multiplying both the numerator and the denominator by $$e^{\beta(\varepsilon \!-\! \mu)}$$,
+By multiplying both the numerator and the denominator by $$e^{\beta(\varepsilon_i - \mu)}$$,
 we arrive at the standard form of the **Bose-Einstein distribution** $$f_B$$:
 
 $$\begin{aligned}
     \boxed{
-        \expval{N}
-        = f_B(\varepsilon)
-        = \frac{1}{e^{\beta (\varepsilon - \mu)} - 1}
+        \expval{n_i}
+        = f_B(\varepsilon_i)
+        = \frac{1}{e^{\beta (\varepsilon_i - \mu)} - 1}
     }
 \end{aligned}$$
 
-This gives the expected occupation number $$\expval{N}$$
-of state $$s$$ with energy $$\varepsilon$$,
+This gives the expected occupation number $$\expval{n_i}$$
+of state $$\ket{i}$$ with energy $$\varepsilon_i$$,
 given a temperature $$T$$ and chemical potential $$\mu$$.
 
 {% comment %}
-The corresponding variance $$\sigma^2 \equiv \expval{N^2} - \expval{N}^2$$ is found to be:
+The corresponding variance $$\sigma^2 \equiv \expval{n_i^2} - \expval{n_i}^2$$ is found to be:
 
 $$\begin{aligned}
     \boxed{
         \sigma^2
-        = k T \pdv{\expval{N}}{\mu}
-        = \expval{N} \big(1 + \expval{N}\!\big)
+        = k T \pdv{\expval{n_i}}{\mu}
+        = \expval{n_i} \big(1 + \expval{n_i}\!\big)
     }
 \end{aligned}$$
 {% endcomment %}
diff --git a/source/know/concept/boussinesq-wave-theory/index.md b/source/know/concept/boussinesq-wave-theory/index.md
index e5fd433..b5f91e1 100644
--- a/source/know/concept/boussinesq-wave-theory/index.md
+++ b/source/know/concept/boussinesq-wave-theory/index.md
@@ -275,7 +275,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The smallest term we will include is $$a h^2 / \lambda^3$$;
-anything smaller (specifically containing $$a^2 / \lambda^2$$) will be discarded.
+anything smaller (i.e. containing a factor of $$a^2 / \lambda^2$$) will be discarded.
 Of course, this decision is arbitrary:
 higher-order approximations exist for deeper water and/or taller waves,
 but we stick with Boussinesq's original choice, leaving:
diff --git a/source/know/concept/coupled-mode-theory/index.md b/source/know/concept/coupled-mode-theory/index.md
index 6a5ec1b..23b6470 100644
--- a/source/know/concept/coupled-mode-theory/index.md
+++ b/source/know/concept/coupled-mode-theory/index.md
@@ -10,8 +10,8 @@ layout: "concept"
 
 Given an optical resonator (e.g. a photonic crystal cavity),
 consider one of its quasinormal modes
-with frequency $$\omega_0$$ and decay rate $$1 / \tau_0$$.
-Its complex amplitude $$A$$ is governed by:
+with frequency $$\omega_0$$ and decay rate $$1 / \tau_0$$ in isolation.
+Its complex amplitude $$A$$ then obeys:
 
 $$\begin{aligned}
     \dv{A}{t}
diff --git a/source/know/concept/debye-length/index.md b/source/know/concept/debye-length/index.md
index 5961c4f..063e308 100644
--- a/source/know/concept/debye-length/index.md
+++ b/source/know/concept/debye-length/index.md
@@ -12,8 +12,7 @@ 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.
+This has the effect of **shielding** the rest of the plasma from the object's presence.
 
 We start from [Gauss' law](/know/concept/maxwells-equations/)
 for the [electric field](/know/concept/electric-field/) $$\vb{E}$$,
@@ -23,12 +22,12 @@ and splitting the charge density into ions $$n_i$$ and electrons $$n_e$$:
 
 $$\begin{aligned}
     \nabla^2 \phi(\vb{r})
-    = - \frac{1}{\varepsilon_0} \Big( q_i n_i(\vb{r}) + q_e n_e(\vb{r}) + q_t \delta(\vb{r}) \Big)
+    = - \frac{1}{\varepsilon_0} \Big( q_i n_i(\vb{r}) + q_e n_e(\vb{r}) + Q \delta(\vb{r}) \Big)
 \end{aligned}$$
 
 The last term represents a *test particle*,
 which will be shielded.
-This particle is a point charge $$q_t$$,
+This particle is a point charge $$Q$$,
 whose density is simply a [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(\vb{r})$$,
 and is not included in $$n_i$$ or $$n_e$$.
 
@@ -63,10 +62,10 @@ where we have assumed quasi-neutrality such that $$q_i n_{i0} = q_e n_{e0}$$:
 $$\begin{aligned}
     \nabla^2 \phi
     &= - \frac{1}{\varepsilon_0}
-    \bigg( q_i n_{i0} - n_{i0} \frac{q_i^2 \phi}{k_B T_i} + q_e n_{e0} - n_{e0} \frac{q_e^2 \phi}{k_B T_e} + q_t \delta(\vb{r}) \bigg)
+    \bigg( q_i n_{i0} - n_{i0} \frac{q_i^2 \phi}{k_B T_i} + q_e n_{e0} - n_{e0} \frac{q_e^2 \phi}{k_B T_e} + Q \delta(\vb{r}) \bigg)
     \\
     &= \bigg( \frac{n_{i0} q_i^2}{\varepsilon_0 k_B T_i} + \frac{n_{e0} q_e^2}{\varepsilon_0 k_B T_e} \bigg) \phi
-    - \frac{q_t}{\varepsilon_0} \delta(\vb{r})
+    - \frac{Q}{\varepsilon_0} \delta(\vb{r})
 \end{aligned}$$
 
 We now define the **ion** and **electron Debye lengths**
@@ -101,24 +100,27 @@ suggesting exponential decay:
 $$\begin{aligned}
     \nabla^2 \phi(\vb{r})
     &= \frac{1}{\lambda_D^2} \phi(\vb{r})
-    - \frac{q_t}{\varepsilon_0} \delta(\vb{r})
+    - \frac{Q}{\varepsilon_0} \delta(\vb{r})
 \end{aligned}$$
 
-This has the following solution,
-known as the **Yukawa potential**,
-which decays exponentially,
-representing the plasma's **self-shielding**
-over a characteristic distance $$\lambda_D$$:
+This has the solution below, known as the **Yukawa potential**,
+which looks like Coulomb's law but with an extra exponential factor,
+representing the plasma's **self-shielding**:
 
 $$\begin{aligned}
     \boxed{
         \phi(r)
-        = \frac{q_t}{4 \pi \varepsilon_0 r} \exp\!\Big( \!-\!\frac{r}{\lambda_D} \Big)
+        = \frac{Q}{4 \pi \varepsilon_0 r} \exp\!\Big( \!-\!\frac{r}{\lambda_D} \Big)
     }
 \end{aligned}$$
 
+We call it *self*-shielding because in reality
+$$Q$$ is simply an electron or ion of the plasma.
+This explains why plasmas are macroscopically neutral,
+despite consisting of charged particles.
+
 Note that $$r$$ is a scalar,
-i.e. the potential depends only on the radial distance to $$q_t$$.
+i.e. the potential depends only on the radial distance to $$Q$$.
 This treatment only makes sense
 if the plasma is sufficiently dense,
 such that there is a large number of particles
diff --git a/source/know/concept/fermi-dirac-distribution/index.md b/source/know/concept/fermi-dirac-distribution/index.md
index 2a38eb3..7554e5a 100644
--- a/source/know/concept/fermi-dirac-distribution/index.md
+++ b/source/know/concept/fermi-dirac-distribution/index.md
@@ -13,18 +13,18 @@ layout: "concept"
 which obey the [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/),
 distribute themselves across the available states in a system at equilibrium.
 
-Consider one single-particle state $$s$$,
+Consider one single-particle state $$\ket{i}$$,
 which can contain $$0$$ or $$1$$ fermions.
-Because the occupation number $$N$$ is variable,
+Because the occupation number $$n_i$$ 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 of $$s$$
+where $$\varepsilon_i$$ is the energy of $$\ket{i}$$
 and $$\mu$$ is the chemical potential:
 
 $$\begin{aligned}
     \mathcal{Z}
-    = \sum_{N = 0}^1 \Big( e^{-\beta (\varepsilon - \mu)} \Big)^N
-    = 1 + e^{-\beta (\varepsilon - \mu)}
+    = \sum_{m = 0}^1 \Big( e^{-\beta (\varepsilon_i - \mu)} \Big)^m
+    = 1 + e^{-\beta (\varepsilon_i - \mu)}
 \end{aligned}$$
 
 The corresponding [thermodynamic potential](/know/concept/thermodynamic-potential/)
@@ -33,43 +33,43 @@ is the Landau potential $$\Omega$$, given by:
 $$\begin{aligned}
     \Omega
     = - k T \ln{\mathcal{Z}}
-    = - k T \ln\!\Big( 1 + e^{-\beta (\varepsilon - \mu)} \Big)
+    = - k T \ln\!\Big( 1 + e^{-\beta (\varepsilon_i - \mu)} \Big)
 \end{aligned}$$
 
-The average number of particles $$\expval{N}$$
-in $$s$$ is then found by taking a derivative of $$\Omega$$:
+The average number of particles $$\expval{n_i}$$
+in $$\ket{i}$$ is then found by taking a derivative of $$\Omega$$:
 
 $$\begin{aligned}
-    \expval{N}
+    \expval{n_i}
     = - \pdv{\Omega}{\mu}
     = k T \pdv{\ln{\mathcal{Z}}}{\mu}
-    = \frac{e^{-\beta (\varepsilon - \mu)}}{1 + e^{-\beta (\varepsilon - \mu)}}
+    = \frac{e^{-\beta (\varepsilon_i - \mu)}}{1 + e^{-\beta (\varepsilon_i - \mu)}}
 \end{aligned}$$
 
-By multiplying both the numerator and the denominator by $$e^{\beta (\varepsilon \!-\! \mu)}$$,
+By multiplying both the numerator and the denominator by $$e^{\beta (\varepsilon_i - \mu)}$$,
 we arrive at the standard form of
 the **Fermi-Dirac distribution** or **Fermi function** $$f_F$$:
 
 $$\begin{aligned}
     \boxed{
-        \expval{N}
-        = f_F(\varepsilon)
-        = \frac{1}{e^{\beta (\varepsilon - \mu)} + 1}
+        \expval{n_i}
+        = f_F(\varepsilon_i)
+        = \frac{1}{e^{\beta (\varepsilon_i - \mu)} + 1}
     }
 \end{aligned}$$
 
-This gives the expected occupation number $$\expval{N}$$
-of state $$s$$ with energy $$\varepsilon$$,
+This gives the expected occupation number $$\expval{n_i}$$
+of state $$\ket{i}$$ with energy $$\varepsilon_i$$,
 given a temperature $$T$$ and chemical potential $$\mu$$.
 
 {% comment %}
-The corresponding variance $$\sigma^2 \equiv \expval{N^2} - \expval{N}^2$$ is found to be:
+The corresponding variance $$\sigma^2 \equiv \expval{n_i^2} - \expval{n_i}^2$$ is found to be:
 
 $$\begin{aligned}
     \boxed{
         \sigma^2
-        = k T \pdv{\expval{N}}{\mu}
-        = \expval{N} \big(1 - \expval{N}\big)
+        = k T \pdv{\expval{n_i}}{\mu}
+        = \expval{n_i} \big(1 - \expval{n_i}\big)
     }
 \end{aligned}$$
 {% endcomment %}
diff --git a/source/know/concept/korteweg-de-vries-equation/index.md b/source/know/concept/korteweg-de-vries-equation/index.md
index e8035d1..13b1ee2 100644
--- a/source/know/concept/korteweg-de-vries-equation/index.md
+++ b/source/know/concept/korteweg-de-vries-equation/index.md
@@ -152,7 +152,7 @@ rather than transform the coordinate system,
 the velocity is incorporated into his ansatz for $$f$$;
 in other words, he assumed that the entire liquid is moving at $$q_0$$.
 For a wave going in the positive $$x$$-direction,
-the linearized problem then predicts a profile $$\eta(x \!-\! (\sqrt{g h} \!+\! q_0))$$,
+the linearized problem then predicts a profile $$\eta(x \!-\! (\sqrt{g h} \!+\! q_0) t)$$,
 so de Vries chose $$q_0 = -\sqrt{g h}$$ to make it stationary.
 Analogously, $$q_0 = \sqrt{g h}$$ for a backward-moving wave.
 With this in mind, the ansatz is:
diff --git a/source/know/concept/kubo-formula/index.md b/source/know/concept/kubo-formula/index.md
index 4cb39ac..41fda3c 100644
--- a/source/know/concept/kubo-formula/index.md
+++ b/source/know/concept/kubo-formula/index.md
@@ -27,14 +27,15 @@ respectively refer to the Schrödinger
 and [interaction pictures](/know/concept/interaction-picture/):
 
 $$\begin{aligned}
-    \expval{\hat{A}}(t)
+    \expval{\hat{A}(t)}
     = \matrixel{\psi_S(t)}{\hat{A}_S}{\psi_S(t)}
     &= \matrixel{\psi_I(t)}{\hat{A}_I(t)}{\psi_I(t)}
     \\
     &= \matrixel{\psi_I(t_0)\,}{\,\hat{K}_I^\dagger(t, t_0) \hat{A}_I(t) \hat{K}_I(t, t_0)\,}{\,\psi_I(t_0)}
 \end{aligned}$$
 
-Where the time evolution operator $$\hat{K}_I(t, t_0)$$ is as follows,
+Where the [time evolution operator](/know/concept/time-evolution-operator/)
+$$\hat{K}_I(t, t_0)$$ is as follows,
 which we Taylor-expand:
 
 $$\begin{aligned}
@@ -71,18 +72,18 @@ where $$\Expval{}$$ is the expectation value for $$\Ket{\psi(t)}$$,
 and $$\Expval{}_0$$ is the expectation value for $$\Ket{\psi_I(t_0)}$$:
 
 $$\begin{aligned}
-    \expval{\hat{A}}(t)
+    \expval{\hat{A}(t)}
     = \expval{\hat{K}_I^\dagger \hat{A}_I \hat{K}_I}_0
     = \expval{\hat{A}_I(t)}_0 - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
-Now we define $$\delta\!\expval{\hat{A}}\!(t)$$
+Now we define $$\delta\!\expval{\hat{A}(t)}$$
 as the change of $$\expval{\hat{A}}$$ due to the perturbation $$\hat{H}_1$$,
-and insert $$\expval{\hat{A}}(t)$$:
+and insert $$\expval{\hat{A}(t)}$$:
 
 $$\begin{aligned}
-    \delta\!\expval{\hat{A}}\!(t)
-    \equiv \expval{\hat{A}}(t) - \expval{\hat{A}_I}_0
+    \delta\!\expval{\hat{A}(t)}
+    \equiv \expval{\hat{A}(t)} - \expval{\hat{A}_I(t)}_0
     = - \frac{i}{\hbar} \int_{t_0}^t \Expval{\Comm{\hat{A}_I(t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
@@ -94,7 +95,7 @@ describing the response of $$\expval{\hat{A}}$$ to first order in $$\hat{H}_1$$:
 
 $$\begin{aligned}
     \boxed{
-        \delta\!\expval{\hat{A}}\!(t)
+        \delta\!\expval{\hat{A}(t)}
         = \int_{t_0}^\infty C^R_{A H_1}(t, t') \dd{t'}
     }
 \end{aligned}$$
@@ -142,7 +143,7 @@ With this, the Kubo formula can be written as follows,
 where we have set $$t_0 = - \infty$$:
 
 $$\begin{aligned}
-    \delta\!\expval{A}\!(t)
+    \delta\!\expval{\hat{A}(t)}
     = \int_{-\infty}^\infty C^R_{A B}(t - t') f(t') \dd{t'}
     = (C^R_{A B} * f)(t)
 \end{aligned}$$
@@ -150,12 +151,12 @@ $$\begin{aligned}
 This is a convolution,
 so the [convolution theorem](/know/concept/convolution-theorem/)
 states that the [Fourier transform](/know/concept/fourier-transform/)
-of $$\delta\!\expval{\hat{A}}\!(t)$$ is simply the product
+of $$\delta\!\expval{\hat{A}(t)}$$ is simply the product
 of the transforms of $$C^R_{AB}$$ and $$f$$:
 
 $$\begin{aligned}
     \boxed{
-        \delta\!\expval{\hat{A}}\!(\omega)
+        \delta\!\expval{\hat{A}(\omega)}
         = \tilde{C}{}^R_{A B}(\omega) \: \tilde{f}(\omega)
     }
 \end{aligned}$$
diff --git a/source/know/concept/larmor-precession/index.md b/source/know/concept/larmor-precession/index.md
index 774af7b..601dae7 100644
--- a/source/know/concept/larmor-precession/index.md
+++ b/source/know/concept/larmor-precession/index.md
@@ -36,8 +36,8 @@ and the exponentials are "twiddle factors":
 
 $$\begin{aligned}
     \Ket{\chi(t)}
-    = a \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow}
-    \:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow}
+    = a e^{- i E_{\downarrow} t / \hbar} \Ket{\downarrow}
+    \:+\: b e^{- i E_{\uparrow} t / \hbar} \Ket{\uparrow}
 \end{aligned}$$
 
 For our purposes, we can safely assume that $$a$$ and $$b$$ are real,
@@ -45,8 +45,8 @@ and then say that there exists an angle $$\theta$$
 satisfying $$a = \sin(\theta / 2)$$ and $$b = \cos(\theta / 2)$$, such that:
 
 $$\begin{aligned}
-    \Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow}
-    \:+\: \cos(\theta / 2) \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow}
+    \Ket{\chi(t)} = \sin(\theta / 2) \: e^{- i E_{\downarrow} t / \hbar} \Ket{\downarrow}
+    \:+\: \cos(\theta / 2) \: e^{- i E_{\uparrow} t / \hbar} \Ket{\uparrow}
 \end{aligned}$$
 
 Now, we find the expectation values of the spin operators
@@ -56,23 +56,23 @@ The first is:
 $$\begin{aligned}
     \matrixel{\chi}{\hat{S}_x}{\chi}
     &= \frac{\hbar}{2}
-    \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}}
+    \begin{bmatrix} a e^{i E_{\downarrow} t / \hbar} \\ b e^{i E_{\uparrow} t / \hbar} \end{bmatrix}^{\mathrm{T}}
     \cdot
     \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
     \cdot
-    \begin{bmatrix} a \exp(- i E_{\downarrow} t / \hbar) \\ b \exp(- i E_{\uparrow} t / \hbar) \end{bmatrix}
+    \begin{bmatrix} a e^{- i E_{\downarrow} t / \hbar} \\ b e^{- i E_{\uparrow} t / \hbar} \end{bmatrix}
     \\
     &= \frac{\hbar}{2}
-    \begin{bmatrix} a \exp(i E_{\downarrow} t / \hbar) \\ b \exp(i E_{\uparrow} t / \hbar) \end{bmatrix}^{\mathrm{T}}
+    \begin{bmatrix} a e^{i E_{\downarrow} t / \hbar} \\ b e^{i E_{\uparrow} t / \hbar} \end{bmatrix}^{\mathrm{T}}
     \cdot
-    \begin{bmatrix} b \exp(- i E_{\uparrow} t / \hbar) \\ a \exp(- i E_{\downarrow} t / \hbar) \end{bmatrix}
+    \begin{bmatrix} b e^{- i E_{\uparrow} t / \hbar} \\ a e^{- i E_{\downarrow} t / \hbar} \end{bmatrix}
     \\
-    &= \frac{\hbar}{2} \Big( a b \exp(i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar)
-    + b a \exp(i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar) \Big)
+    &= \frac{\hbar}{2} \Big( a b e^{i (E_{\downarrow} \!-\! E_{\uparrow}) t / \hbar}
+    + b a e^{i (E_{\uparrow} \!-\! E_{\downarrow}) t / \hbar} \Big)
     \\
-    &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( \exp(i \gamma B t) + \exp(- i \gamma B t) \Big)
+    &= \frac{\hbar}{2} \cos(\theta/2) \sin(\theta/2) \Big( e^{i \gamma B t} + e^{- i \gamma B t} \Big)
     \\
-    &= \frac{\hbar}{2} \cos(\gamma B t) \Big( \cos(\theta/2) \sin(\theta/2) + \cos(\theta/2) \sin(\theta/2) \Big)
+    &= \frac{\hbar}{2} \cos(\gamma B t) \cdot 2 \cos(\theta/2) \sin(\theta/2)
     \\
     &= \frac{\hbar}{2} \sin(\theta) \cos(\gamma B t)
 \end{aligned}$$
diff --git a/source/know/concept/lindhard-function/index.md b/source/know/concept/lindhard-function/index.md
index fd620df..5f11d36 100644
--- a/source/know/concept/lindhard-function/index.md
+++ b/source/know/concept/lindhard-function/index.md
@@ -19,7 +19,7 @@ which describes the change in $$\Expval{\hat{n}}$$
 due to a time-dependent perturbation $$\hat{H}_1$$:
 
 $$\begin{aligned}
-    \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
+    \delta\!\Expval{\hat{n}(\vb{r}, t)}
     = -\frac{i}{\hbar} \int_{-\infty}^\infty \Theta(t - t') \Expval{\Comm{\hat{n}_I(\vb{r}, t)}{\hat{H}_{1,I}(t')}}_0 \dd{t'}
 \end{aligned}$$
 
@@ -39,7 +39,7 @@ and $$U(\vb{r})$$ is an arbitrary potential function.
 The Kubo formula becomes:
 
 $$\begin{aligned}
-    \delta\!\Expval{ {\hat{n}}}\!(\vb{r}, t)
+    \delta\!\Expval{\hat{n}(\vb{r}, t)}
     = \iint_{-\infty}^\infty \chi(\vb{r}, \vb{r}'; t, t') \: U(\vb{r}') \: e^{i (\omega + i \eta) t'} \dd{t'} \dd{\vb{r}'}
 \end{aligned}$$
 
@@ -95,8 +95,9 @@ $$\begin{aligned}
     \: e^{i (\vb{q}_2 + \vb{q}) \cdot \vb{r}'} \dd{\vb{q}_2} \dd{\vb{r}'}
 \end{aligned}$$
 
-For $$V \to \infty$$ we get a Dirac delta function,
-but in fact the conclusion holds for finite $$V$$ too:
+This gives a Dirac delta function for $$V \to \infty$$
+(a limit that we will take properly later,
+but beware that some authors set $$V = 1$$ until then):
 
 $$\begin{aligned}
     \chi(\vb{q}; t, t')
@@ -107,8 +108,9 @@ $$\begin{aligned}
 \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,
+$$\chi$$ only depends on the time difference $$t\!-\!t'$$.
+Note that $$\delta{\Expval{\hat{n}}}$$ already has the form of a Fourier transform
+$$t\!-\!t' \to \omega\!+\!i \eta$$,
 which gives us an opportunity to rewrite $$\chi$$
 in the [Lehmann representation](/know/concept/lehmann-representation/):
 
@@ -119,12 +121,12 @@ $$\begin{aligned}
     \Big( e^{-\beta E_\nu} - e^{- \beta E_{\nu'}} \Big)
 \end{aligned}$$
 
-Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-electron eigenstates of $$\hat{H}_0$$,
+Where $$\Ket{\nu}$$ and $$\Ket{\nu'}$$ are many-particle 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:
+To get ready for the calculations ahead,
+we swap $$\nu$$ and $$\nu'$$ in the second term,
+so the response function is as shown below.
+All operators are in the Schrödinger picture from now on:
 
 $$\begin{aligned}
     \chi(\vb{q}, \omega)
@@ -135,7 +137,6 @@ $$\begin{aligned}
     {\hbar (\omega + i \eta) + E_{\nu'} - E_\nu} \bigg) e^{-\beta E_\nu}
 \end{aligned}$$
 
-All operators are in the Schrödinger picture from now on, hence we dropped the subscript $$S$$.
 
 To proceed, we need to rewrite $$\hat{n}(\vb{q})$$ somehow.
 If we neglect electron-electron interactions,
@@ -180,9 +181,8 @@ 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.
+This is exact in the limit $$V \to \infty$$,
+in which case $$\vb{k}$$ also becomes a continuous variable again.
 
 We apply this standard trick from condensed matter physics to $$\hat{n}$$,
 and $$V$$ cancels out:
@@ -341,45 +341,49 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-From this, we would like to get the
-[dielectric function](/know/concept/dielectric-function/) $$\varepsilon_r$$.
-Recall its definition, where $$U_\mathrm{tot}$$, $$U_\mathrm{ext}$$, and $$U_\mathrm{ind}$$
-are the total, external and induced potentials, respectively:
+This is its most general form, but for practical calculations
+we need to formally take the limit $$V \to \infty$$
+and then use $$\sum_{\vb{k}} = V / (2 \pi)^{D} \int_{-\infty}^{\infty} \dd{\vb{k}}$$.
+Furthermore, electrons are spin-1/2 particles,
+so each orbital contains two, meaning
+$$\sum_{\sigma}$$ simply gives a constant factor:
 
 $$\begin{aligned}
-    U_\mathrm{tot}
-    = U_\mathrm{ext} + U_\mathrm{ind}
-    = \frac{U_\mathrm{ext}}{\varepsilon_r}
+    \boxed{
+        \chi_0(\vb{q}, \omega)
+        = \frac{2}{(2 \pi)^{D}} \int_{-\infty}^{\infty}
+        \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}
+        {\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
+    }
 \end{aligned}$$
 
-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
-$$\Phi_\mathrm{tot} = q_e U_\mathrm{tot}$$,
+From this, we would like to get the
+[dielectric function](/know/concept/dielectric-function/) $$\varepsilon_r$$.
+When an external [electric field](/know/concept/electric-field/) is applied,
+the electrons respond and thereby modify the net field inside the material.
+We include this effect in our *energy* potential $$U$$,
+such that the net *electric* potential
+$$\Phi_\mathrm{tot} = U / q_e$$,
 where $$q_e < 0$$ is the charge of an electron.
+This is not the same as including direct electron-electron interactions!
 
-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:
+We thus have a linear relation for the induced *particle* density
+$$\delta\!\Expval{\hat{n}(\vb{q}, \omega)} = \chi_0(\vb{q}, \omega) \: U(\vb{q})$$
+thanks to the [convolution theorem](/know/concept/convolution-theorem/).
+The corresponding induced *charge* density is given by
+$$\rho_\mathrm{ind} = q_e^2 \chi_0 \Phi_\mathrm{tot}$$,
+so the standard formula for $$\varepsilon_r$$ yields:
 
 $$\begin{aligned}
     \boxed{
         \varepsilon_r(\vb{q}, \omega)
-        = 1 - \frac{U_{ee}(\vb{q})}{V}
-        \sum_{\sigma \vb{k}} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}}
+        = 1 - U_{ee}(\vb{q}) \frac{2}{(2 \pi)^{D}}
+        \int_{-\infty}^{\infty} \frac{n_F(\xi_{\vb{k}}) - n_F(\xi_{\vb{k} + \vb{q}})}{\hbar (\omega + i \eta) + \xi_{\vb{k}} - \xi_{\vb{k} + \vb{q}}} \dd{\vb{k}}
     }
 \end{aligned}$$
 
-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,
+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.
 
 
diff --git a/source/know/concept/matsubara-greens-function/index.md b/source/know/concept/matsubara-greens-function/index.md
index 5e753db..6f60edf 100644
--- a/source/know/concept/matsubara-greens-function/index.md
+++ b/source/know/concept/matsubara-greens-function/index.md
@@ -64,7 +64,7 @@ $$\begin{aligned}
 With $$-$$ for bosons, and $$+$$ for fermions,
 due to the time-ordered product for $$\tau > \tau'$$.
 
-On this domain $$[-\hbar \beta, \hbar \beta]$$,
+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,
@@ -133,7 +133,7 @@ $$\begin{aligned}
 {% include proof/end.html id="proof-period" %}
 
 
-Due to this limited domain $$\tau \in [-\hbar \beta, \hbar \beta]$$,
+Due to this limited domain $$\tau \in \:]\!-\!\hbar \beta, \hbar \beta[$$,
 the [Fourier transform](/know/concept/fourier-transform/)
 of $$C_{AB}(\tau)$$ consists of discrete frequencies
 $$k_n \equiv n \pi / (\hbar \beta)$$.
@@ -288,7 +288,7 @@ $$\begin{aligned}
     \matrixel{n'}{\hat{B}}{n} e^{(E_n - E_{n'})(\tau - \tau') / \hbar}
 \end{aligned}$$
 
-We take the Fourier transform by integrating over $$[0, \hbar \beta]$$:
+We take the Fourier transform by integrating over $$]0, \hbar \beta[$$:
 
 $$\begin{aligned}
     C_{AB}(i \omega_m)
@@ -324,7 +324,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Since $$\tau \!-\! \tau' < 0$$ this time,
-we take the Fourier transform over $$[-\hbar \beta, 0]$$:
+we take the Fourier transform over $$]\!-\!\hbar \beta, 0[$$:
 
 $$\begin{aligned}
     C_{AB}(i \omega_m)
@@ -341,7 +341,7 @@ $$\begin{aligned}
     \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)
+    \Big( e^{- \beta E_n} \mp 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)
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md
index 0e04455..3347476 100644
--- a/source/know/concept/matsubara-sum/index.md
+++ b/source/know/concept/matsubara-sum/index.md
@@ -23,7 +23,7 @@ $$\begin{aligned}
 $$g(z)$$ is a *meromorphic* function on the complex frequency plane,
 i.e. it is [holomorphic](/know/concept/holomorphic-function/)
 except for a known set of simple poles,
-and $$\tau \in [-\hbar \beta, \hbar \beta]$$ is a real parameter.
+and $$\tau \in \:]\!-\!\hbar \beta, \hbar \beta[$$ is a real parameter.
 The Matsubara frequencies $$i \omega_n$$ are defined as follows
 for bosons (subscript $$B$$) or fermions (subscript $$F$$):
 
@@ -77,9 +77,9 @@ $$\begin{aligned}
     h(z)
     \equiv
     \begin{cases}
-        n_{B,F}(z) & \mathrm{if}\; \tau \ge 0
+        n_{B,F}(z) & \mathrm{if}\; 0 \le \tau < \hbar \beta
         \\
-        -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0
+        -n_{B,F}(-z) & \mathrm{if}\; \!-\!\hbar \beta < \tau \le 0
     \end{cases}
 \end{aligned}$$
 
@@ -107,7 +107,7 @@ $$\begin{aligned}
     &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} + 1} \bigg)
     = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} + 1} \bigg)
     \\
-    &= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{e^{\hbar \beta \eta} + 1} \bigg)
+    &= \lim_{\eta \to 0}\!\bigg( \frac{\eta}{-e^{\hbar \beta \eta} + 1} \bigg)
     = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{- 1 - \hbar \beta \eta + 1} \bigg)
     = - \frac{1}{\hbar \beta}
 \end{aligned}$$
diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
index 9821718..5b83b69 100644
--- a/source/know/concept/pauli-exclusion-principle/index.md
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -106,7 +106,7 @@ $$\begin{aligned}
     = C \Ket{a}\Ket{a}
 \end{aligned}$$
 
-However, for fermions, the state is unnormalizable and thus unphysical:
+However, for fermions, the state is unnormalizable and therefore unphysical:
 
 $$\begin{aligned}
     \Ket{\Psi(a, a)}_f
diff --git a/source/know/concept/salt-equation/index.md b/source/know/concept/salt-equation/index.md
index d47383f..e6ed5e5 100644
--- a/source/know/concept/salt-equation/index.md
+++ b/source/know/concept/salt-equation/index.md
@@ -80,7 +80,7 @@ $$\begin{aligned}
     + \frac{i}{\hbar} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n \: D
 \end{aligned}$$
 
-With being $$\vb{p}_0^{+} \vb{p}_0^{-}$$ a dyadic product.
+With $$\vb{p}_0^{+} \vb{p}_0^{-}$$ denoting a dyadic product.
 Isolating the latter equation for $$\vb{p}_n$$ gives us:
 
 $$\begin{aligned}