Improve knowledge baseHEADmaster

21 files changed, 450 insertions, 358 deletions

diff --git a/source/know/concept/blochs-theorem/index.md b/source/know/concept/blochs-theorem/index.md
index d7fcf90..c6278f3 100644
--- a/source/know/concept/blochs-theorem/index.md
+++ b/source/know/concept/blochs-theorem/index.md
@@ -12,14 +12,14 @@ given a potential $$V(\vb{r})$$ which is periodic on a lattice,
 i.e. $$V(\vb{r}) = V(\vb{r} + \vb{a})$$
 for a primitive lattice vector $$\vb{a}$$,
 then it follows that the solutions $$\psi(\vb{r})$$
-to the time-independent Schrödinger equation
-take the following form,
+to the time-independent Schrödinger equation take the following form,
 where the function $$u(\vb{r})$$ is periodic on the same lattice,
 i.e. $$u(\vb{r}) = u(\vb{r} + \vb{a})$$:
 
 $$\begin{aligned}
 	\boxed{
-		\psi(\vb{r}) = u(\vb{r}) e^{i \vb{k} \cdot \vb{r}}
+		\psi(\vb{r})
+		= u(\vb{r}) e^{i \vb{k} \cdot \vb{r}}
 	}
 \end{aligned}$$
 
@@ -33,9 +33,11 @@ then both $$\psi(\vb{r})$$ and $$\psi(\vb{r} + \vb{a})$$
 are eigenstates with the same energy:
 
 $$\begin{aligned}
-	\hat{H} \psi(\vb{r}) = E \psi(\vb{r})
-	\qquad
-	\hat{H} \psi(\vb{r} + \vb{a}) = E \psi(\vb{r} + \vb{a})
+	\hat{H} \psi(\vb{r})
+	= E \psi(\vb{r})
+	\qquad \qquad
+	\hat{H} \psi(\vb{r} + \vb{a})
+	= E \psi(\vb{r} + \vb{a})
 \end{aligned}$$
 
 Now define the unitary translation operator $$\hat{T}(\vb{a})$$ such that
@@ -52,18 +54,21 @@ $$\begin{aligned}
 In other words, if $$\hat{H}$$ is lattice-periodic,
 then it will commute with $$\hat{T}(\vb{a})$$,
 i.e. $$[\hat{H}, \hat{T}(\vb{a})] = 0$$.
-Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$ must share eigenstates $$\psi(\vb{r})$$:
+Consequently, $$\hat{H}$$ and $$\hat{T}(\vb{a})$$
+must share eigenstates $$\psi(\vb{r})$$:
 
 $$\begin{aligned}
-	\hat{H} \:\psi(\vb{r}) = E \:\psi(\vb{r})
+	\hat{H} \psi(\vb{r})
+	= E \psi(\vb{r})
 	\qquad \qquad
-	\hat{T}(\vb{a}) \:\psi(\vb{r}) = \tau \:\psi(\vb{r})
+	\hat{T}(\vb{a}) \psi(\vb{r})
+	= \tau \psi(\vb{r})
 \end{aligned}$$
 
 Since $$\hat{T}$$ is unitary,
 its eigenvalues $$\tau$$ must have the form $$e^{i \theta}$$, with $$\theta$$ real.
 Therefore a translation by $$\vb{a}$$ causes a phase shift,
-for some vector $$\vb{k}$$:
+so there exists a vector $$\vb{k}$$ such that:
 
 $$\begin{aligned}
 	\psi(\vb{r} + \vb{a})
diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 5f4add0..3821512 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -67,7 +67,7 @@ but unfortunately also quite difficult to work with.
 In addition, $$f$$ is a 7-dimensional function,
 so the BTE is already hard to solve without collisions!
 We only present the simplest case,
-known as the **Bhatnagar-Gross-Krook approximation**:
+the **Bhatnagar-Gross-Krook approximation**:
 if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known,
 then each collision brings the system closer to $$f_0$$:
 
@@ -90,14 +90,15 @@ $$\begin{aligned}
     n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
 \end{aligned}$$
 
-Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so:
+Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$
+can be averaged like so:
 
 $$\begin{aligned}
-    \Expval{Q}
-    = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+    \Expval{Q}\!(\vb{r}, t)
+    = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{v}) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
 \end{aligned}$$
 
-With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate,
+With that in mind, we multiply the collisionless BTE by $$Q(\vb{v})$$ and integrate,
 assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$:
 
 $$\begin{aligned}
@@ -136,7 +137,8 @@ $$\begin{aligned}
 
 If we set $$Q = m$$, then the mass density $$\rho = n \Expval{Q}$$,
 and we find that the **zeroth moment** of the BTE describes conservation of mass,
-where $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity:
+where $$\vb{V} \equiv \Expval{\vb{v}} = n^{-1} \int \vb{v} f \dd{\vb{v}}$$
+is the fluid velocity:
 
 $$\begin{aligned}
     \boxed{
@@ -231,7 +233,8 @@ $$\begin{aligned}
 
 
 {% include proof/start.html id="proof-moment2" -%}
-We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:
+We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype
+and recognize $$\rho$$ wherever possible:
 
 $$\begin{aligned}
     0
@@ -244,7 +247,8 @@ $$\begin{aligned}
     - \frac{\vb{F}}{2} \cdot \bigg( n \Expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg)
 \end{aligned}$$
 
-We handle these terms one by one. Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:
+We handle these terms one by one.
+Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:
 
 $$\begin{aligned}
     \Expval{|\vb{v}|^2}
diff --git a/source/know/concept/diffie-hellman-key-exchange/index.md b/source/know/concept/diffie-hellman-key-exchange/index.md
index 3525881..a6e0894 100644
--- a/source/know/concept/diffie-hellman-key-exchange/index.md
+++ b/source/know/concept/diffie-hellman-key-exchange/index.md
@@ -32,19 +32,22 @@ there is no efficient algorithm to recover $$n$$.
 
 Suppose that Alice and Bob want to exchange encrypted data in the future,
 so they need to agree on an encryption key to use.
-However, they can only exchange messages with each other over
+However, they can only exchange messages over
 an insecure channel, which is being eavesdropped.
 
 After they publicly agree on the values of $$g$$ and $$p$$,
-Alice and Bob each choose a secret number from $$\{0, ..., p \!-\! 2\}$$, respectively $$a$$ and $$b$$,
+Alice and Bob each choose a secret number from $$\{0, ..., p \!-\! 2\}$$,
+respectively $$a$$ and $$b$$,
 and then privately calculate $$A$$ and $$B$$ as follows:
 
 $$\begin{aligned}
     A
-    \equiv g^a \bmod p
-    \qquad \qquad
+    &\equiv f(a)
+    = g^a \bmod p
+    \\
     B
-    \equiv g^b \bmod p
+    &\equiv f(b)
+    = g^b \bmod p
 \end{aligned}$$
 
 Finally, they transmit these numbers $$A$$ and $$B$$
@@ -69,8 +72,7 @@ but cannot recover $$a$$ or $$b$$.
 This assumption is just that: an assumption.
 So far, nobody has been able to prove or disprove it
 for classical computation.
-However, for quantum computers,
-it has already been *dis*proven!
+However, for quantum computers, it has already been *dis*proven!
 In this case, another method must be used,
 for example the [BB84 protocol](/know/concept/bb84-protocol/).
 
diff --git a/source/know/concept/dyson-equation/index.md b/source/know/concept/dyson-equation/index.md
index ae9eb35..03be06f 100644
--- a/source/know/concept/dyson-equation/index.md
+++ b/source/know/concept/dyson-equation/index.md
@@ -25,8 +25,8 @@ $$\begin{aligned}
     = \delta(\vb{r} - \vb{r}') \: \delta(t - t')
 \end{aligned}$$
 
-From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$
-as follows, so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:
+From this, we define the inverse $$\hat{G}{}_0^{-1}(\vb{r}, t)$$ as follows,
+so that $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:
 
 $$\begin{aligned}
     \hat{G}{}_0^{-1}(\vb{r}, t)
@@ -35,16 +35,15 @@ $$\begin{aligned}
 
 Note that $$\hat{G}{}_0^{-1}$$ is an operator, while $$G_0$$ is a function.
 For the sake of consistency, we thus define
-the operator $$\hat{G}_0(\vb{r}, t)$$
-as a multiplication by $$G_0$$
-and integration over $$\vb{r}'$$ and $$t'$$:
+its operator version $$\hat{G}_0(\vb{r}, t)$$
+as a multiplication by $$G_0$$ and integration over $$\vb{r}'$$ and $$t'$$:
 
 $$\begin{aligned}
     \hat{G}_0(\vb{r}, t) \: f
-    \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
+    \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}}' \dd{t'}
 \end{aligned}$$
 
-For an arbitrary function $$f(\vb{r}, t)$$,
+Where $$f(\vb{r}, t)$$ is an arbitrary function,
 so that $$\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$$.
 Moving on, the Schrödinger equation can be rewritten like so,
 using $$\hat{G}{}_0^{-1}$$:
@@ -61,7 +60,7 @@ by solving the defining equation above.
 Suppose we now add a more complicated and
 possibly time-dependent term $$\hat{H}_1(\vb{r}, t)$$,
 in which case the corresponding fundamental solution
-$$G(\vb{r}, \vb{r}', t, t')$$ satisfies:
+$$G(\vb{r}, \vb{r}', t, t')$$ (note the lack of a $$0$$ subscript) satisfies:
 
 $$\begin{aligned}
     \delta(\vb{r} - \vb{r}') \: \delta(t - t')
@@ -72,7 +71,7 @@ $$\begin{aligned}
 
 This equation is typically too complicated to solve,
 so we would like an easier way to calculate this new $$G$$.
-The perturbed wavefunction $$\Psi(\vb{r}, t)$$
+Clearly, the perturbed wavefunction $$\Psi(\vb{r}, t)$$
 satisfies the Schrödinger equation:
 
 $$\begin{aligned}
@@ -80,9 +79,8 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-We know that $$\hat{G}{}_0^{-1} \Psi_0 = 0$$,
-which we put on the right,
-and then we apply $$\hat{G}_0$$ in front:
+We know that $$\hat{G}{}_0^{-1} \Psi_0 = 0$$ from earlier,
+which we put on the right, and then apply $$\hat{G}_0$$ to it:
 
 $$\begin{aligned}
     \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi
@@ -110,7 +108,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The parenthesized expression clearly has the same recursive pattern,
-so we denote it by $$\hat{G}$$ and write the so-called **Dyson equation**:
+so we denote it by $$\hat{G}$$ (an operator, not the function $$G$$)
+and write the so-called **Dyson equation**:
 
 $$\begin{aligned}
     \boxed{
@@ -133,8 +132,8 @@ $$\begin{aligned}
 This relation is equivalent to the Schrödinger equation.
 So now we have the operator $$\hat{G}(\vb{r}, t)$$,
 but what about the fundamental solution function $$G(\vb{r}, t; \vb{r}', t')$$?
-Let us take its definition, multiply it by an arbitrary $$f(\vb{r}, t)$$,
-and integrate over $$G$$'s second argument pair:
+Let us take the latter's definition and multiply it by an arbitrary $$f(\vb{r}, t)$$,
+and then integrate over $$G$$'s second argument pair:
 
 $$\begin{aligned}
     \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
@@ -143,8 +142,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where we have hidden the arguments $$(\vb{r}, t)$$ for brevity.
-We now apply $$\hat{G}_0(\vb{r}, t)$$ to this equation
-(which contains an integral over $$t''$$ independent of $$t'$$):
+We apply $$\hat{G}_0(\vb{r}, t)$$ to this equation:
 
 $$\begin{aligned}
     \hat{G}_0 f
@@ -154,8 +152,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Here, the shape of Dyson's equation is clearly recognizable,
-so we conclude that, as expected, the operator $$\hat{G}$$
-is defined as multiplication by the function $$G$$ followed by integration:
+so we conclude that the operator $$\hat{G}$$
+is defined as multiplication by the function $$G$$ followed by integration,
+exactly analogously to $$\hat{G}_0$$ and $$G_0$$,
+which should not be a big surprise:
 
 $$\begin{aligned}
     \hat{G}(\vb{r}, t) \: f(\vb{r}, t)
diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md
index 06f0f45..393b875 100644
--- a/source/know/concept/electric-dipole-approximation/index.md
+++ b/source/know/concept/electric-dipole-approximation/index.md
@@ -138,10 +138,10 @@ $$\begin{aligned}
     \\
     &= - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t)
     \\
-    &\approx - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
+    &\approx - \vu{p} \cdot \vb{E}_0 \exp(- i \omega t)
 \end{aligned}$$
 
-Where $$\vu{d} \equiv q \vu{x}$$ is
+Where $$\vu{p} \equiv q \vu{x}$$ is
 the **transition dipole moment operator** of the electron,
 hence the name *electric dipole approximation*.
 Finally, we take the real part, yielding:
@@ -150,7 +150,7 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             \hat{H}_1(t)
-            &= - \vu{d} \cdot \vb{E}(t)
+            &= - \vu{p} \cdot \vb{E}(t)
             \\
             &= - q \vu{x} \cdot \vb{E}_0 \cos(\omega t)
         \end{aligned}
diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md
index c1ed8da..76cee81 100644
--- a/source/know/concept/equation-of-motion-theory/index.md
+++ b/source/know/concept/equation-of-motion-theory/index.md
@@ -100,7 +100,8 @@ $$\begin{aligned}
     \\
     &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''}
     - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
-    = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
+    \\
+    &= - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
 \end{aligned}$$
 {% include proof/end.html id="proof-commutator" %}
 
@@ -160,7 +161,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We take the [Fourier transform](/know/concept/fourier-transform/)
-$$(t \!-\! t') \to (\omega + i \eta)$$, where $$\eta \to 0^+$$ ensures convergence:
+$$(t - t') \to (\omega + i \eta)$$, where $$\eta \to 0^+$$ ensures convergence:
 
 $$\begin{aligned}
         \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega)
@@ -178,7 +179,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 For a non-interacting, time-independent Hamiltonian,
-we therefore arrive at:
+we thus arrive at the famous result:
 
 $$\begin{aligned}
     \boxed{
diff --git a/source/know/concept/fabry-perot-cavity/index.md b/source/know/concept/fabry-perot-cavity/index.md
index d5ea0ea..c648549 100644
--- a/source/know/concept/fabry-perot-cavity/index.md
+++ b/source/know/concept/fabry-perot-cavity/index.md
@@ -10,11 +10,13 @@ layout: "concept"
 ---
 
 In its simplest form, a **Fabry-Pérot cavity**
-is a region of light-transmitting medium surrounded by two mirrors,
-which may transmit some of the incoming light.
-Such a setup can be used as e.g. an interferometer or a laser cavity.
+is a region of light-transmitting medium surrounded by two parallel mirrors,
+which may let some of the light escape.
+Such a setup can be used as e.g. a laser cavity or an interferometer.
+Below, we treat this simple system as an exercise
+for calculating *quasinormal modes* in 1D,
+i.e. modes with complex resonances.
 
-Below, we calculate its quasinormal modes in 1D.
 We divide the $$x$$-axis into three domains: left $$L$$, center $$C$$, and right $$R$$.
 The cavity $$C$$ has length $$\ell$$ and is centered on $$x = 0$$.
 Let $$n_L$$, $$n_C$$ and $$n_R$$ be the respective domains' refractive indices:
@@ -95,8 +97,8 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-We do not want to simply satisfy this equation
-by setting $$A_1$$, $$A_2$$, $$A_3$$ and $$A_4$$,
+We do not want to satisfy this equation
+by simply setting $$A_1$$, $$A_2$$, $$A_3$$ and $$A_4$$,
 so we demand that the system matrix is not invertible,
 i.e. its determinant is zero:
 
@@ -116,7 +118,9 @@ $$\begin{aligned}
     - 2 n_C (n_L + n_R) \cos(k_m n_C \ell)
 \end{aligned}$$
 
-Finally, some further rearranging gives a convenient transcendental equation:
+Finally, some further rearranging gives a convenient transcendental equation,
+keeping in mind that $$k_m$$ and the indices $$n_L$$, $$n_C$$ and $$n_R$$
+are generally complex numbers:
 
 $$\begin{aligned}
     \boxed{
@@ -223,9 +227,9 @@ $$\begin{aligned}
     &= (1 - r_R) A_3 e^{i k_m (n_C - n_R) \ell/2}
 \end{aligned}$$
 
-Note that we have not demanded continuity of the electric field.
-This is because the mirrors are infinitely thin "magic" planes;
-had we instead included the full microscopic mirror structure,
+Note that we have not demanded continuity of the electric field,
+because the mirrors are infinitely thin "magic" planes in this case.
+If we had instead included the full microscopic mirror structure,
 then we would have demanded continuity as before.
 
 
diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
index 9f5d4ec..6412914 100644
--- a/source/know/concept/heaviside-step-function/index.md
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -45,15 +45,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The [Fourier transform](/know/concept/fourier-transform/)
-of $$\Theta(t)$$ is as follows,
-where $$\pv{}$$ is the Cauchy principal value,
+of $$\Theta(t)$$ is as follows, where $$\mathcal{P}$$
+is the [Cauchy principal value](/know/concept/cauchy-principal-value/),
 $$A$$ and $$s$$ are constants from the FT's definition,
 and $$\mathrm{sgn}$$ is the signum function:
 
 $$\begin{aligned}
     \boxed{
         \tilde{\Theta}(\omega)
-        = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big)
+        = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \:\mathrm{sgn}(s) \:\mathcal{P} \frac{1}{\omega} \Big)
     }
 \end{aligned}$$
 
@@ -77,18 +77,18 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The first term is proportional to the Dirac delta function.
-The second integral is problematic, so we take the Cauchy principal value $$\pv{}$$
-and look up the integral:
+The second integral is problematic, so we take
+the Cauchy principal value $$\mathcal{P}$$ and look up the integral:
 
 $$\begin{aligned}
     \tilde{\Theta}(\omega)
-    &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
-    = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
+    &= A \pi \delta(s \omega) + \frac{A}{2} \:\mathcal{P}\! \int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}
+    = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \:\mathcal{P} \frac{1}{\omega}
 \end{aligned}$$
 {% include proof/end.html id="proof-fourier" %}
 
 
-The use of $$\pv{}$$ without an integral is an abuse of notation,
+The use of $$\mathcal{P}$$ without an integral is an abuse of notation,
 and means that this result only makes sense when wrapped in an integral.
-Formally, $$\pv{\{1 / \omega\}}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
+Formally, $$\mathcal{P}\{1 / \omega\}$$ is a [Schwartz distribution](/know/concept/schwartz-distribution/).
 
diff --git a/source/know/concept/heisenberg-picture/index.md b/source/know/concept/heisenberg-picture/index.md
index 54bf397..3ffe29a 100644
--- a/source/know/concept/heisenberg-picture/index.md
+++ b/source/know/concept/heisenberg-picture/index.md
@@ -118,7 +118,7 @@ This result is arguably more intuitive than the Schrödinger picture,
 because it allows us to think about observables (i.e. operators) in a more classical way.
 For example, inserting the position $$\hat{X}$$
 and momentum $$\hat{P} = - i \hbar \: \idv{}{\hat{X}}$$
-gives the following Newton-style relations:
+gives the following Newton-style relations (details omitted):
 
 $$\begin{aligned}
     \dv{\hat{X}}{t}
@@ -130,7 +130,6 @@ $$\begin{aligned}
     = - \dv{V(\hat{X})}{\hat{X}}
 \end{aligned}$$
 
-Where the commutators have been treated as known.
 These equations would not be valid in the Schrödinger picture,
 unless we took their expectation value
 to get [Ehrenfest's theorem](/know/concept/ehrenfests-theorem/).
diff --git a/source/know/concept/hellmann-feynman-theorem/index.md b/source/know/concept/hellmann-feynman-theorem/index.md
index c6bf720..d02b285 100644
--- a/source/know/concept/hellmann-feynman-theorem/index.md
+++ b/source/know/concept/hellmann-feynman-theorem/index.md
@@ -72,7 +72,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-While for $$m \neq n$$, we get the **Epstein generalization**
+While for $$m \neq n$$, we get the (unfortunately named) **Epstein generalization**
 of the Hellmann-Feynman theorem, which is for example relevant for
 the [Berry phase](/know/concept/berry-phase/):
 
diff --git a/source/know/concept/langmuir-waves/index.md b/source/know/concept/langmuir-waves/index.md
index 2dbce8f..736ef71 100644
--- a/source/know/concept/langmuir-waves/index.md
+++ b/source/know/concept/langmuir-waves/index.md
@@ -54,13 +54,13 @@ are assumed to satisfy:
 
 $$\begin{aligned}
     \pdv{n_{e0}}{t} = 0
-    \qquad
+    \qquad \quad
     \pdv{\vb{u}_{e0}}{t} = 0
-    \qquad
+    \qquad \quad
     \nabla n_{e0} = 0
-    \qquad
+    \qquad \quad
     \vb{u}_{e0} = 0
-    \qquad
+    \qquad \quad
     \vb{E}_0 = 0
 \end{aligned}$$
 
@@ -73,8 +73,7 @@ $$\begin{aligned}
     \\
     &= \pdv{n_{e1}}{t} + \nabla \cdot \Big( n_{e0} \vb{u}_{e1} + n_{e1} \vb{u}_{e1} \Big)
     \\
-    &\approx \pdv{n_{e1}}{t} + \nabla \cdot (n_{e0} \vb{u}_{e1})
-    = \pdv{n_{e1}}{t} + n_{e0} \nabla \cdot \vb{u}_{e1}
+    &\approx \pdv{n_{e1}}{t} + n_{e0} \nabla \cdot \vb{u}_{e1}
 \end{aligned}$$
 
 Likewise, we insert it into Gauss' law,
@@ -83,7 +82,7 @@ and use the plasma's quasi-neutrality $$n_i = n_{e0}$$ to get:
 $$\begin{aligned}
     \varepsilon_0 \nabla \cdot \big( \vb{E}_0 \!+\! \vb{E}_1 \big)
     = q_e (n_{e0} + n_{e1} - n_i)
-    \quad \implies \quad
+    \qquad \implies \qquad
     \varepsilon_0 \nabla \cdot \vb{E}_1
     = q_e n_{e1}
 \end{aligned}$$
@@ -106,12 +105,13 @@ Inserting this into the continuity equation and Gauss' law yields, respectively:
 
 $$\begin{aligned}
     - i \omega n_{e1} = - i n_{e0} \vb{k} \cdot \vb{u}_{e1}
-    \qquad \quad
+    \qquad \qquad
     -\! i \varepsilon_0 \vb{k} \cdot \vb{E}_1 = q_e n_{e1}
 \end{aligned}$$
 
+These form a system of equations to be solved.
 However, there are three unknowns $$n_{e1}$$, $$\vb{u}_{e1}$$ and $$\vb{E}_1$$,
-so one more equation is needed.
+so one more equation is needed before we can do so.
 
 
 
@@ -180,7 +180,8 @@ the oscillation is stationary.
 ## Warm Langmuir waves
 
 Next, we generalize this result to nonzero $$T_e$$,
-in which case the pressure $$p_e$$ is involved:
+in which case the pressure $$p_e$$ is involved,
+so the electron momentum equation is given by:
 
 $$\begin{aligned}
     m_e n_{e0} \pdv{\vb{u}_{e1}}{t}
@@ -198,10 +199,11 @@ $$\begin{aligned}
 \end{aligned}$$
 
 With this, insertion of our plane-wave ansatz
-into the electron equation results in:
+into the momentum equation results in:
 
 $$\begin{aligned}
-    -i \omega m_e n_{e0} \vb{u}_{e1} = q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k}
+    -i \omega m_e n_{e0} \vb{u}_{e1}
+    = q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k}
 \end{aligned}$$
 
 Which once again closes the system of three equations.
@@ -209,7 +211,8 @@ Solving for $$\omega^2$$ then gives:
 
 $$\begin{aligned}
     \omega^2
-    = \frac{\omega n_{e0}}{n_{e1}} \vb{k} \cdot \vb{u}_{e1}
+    &= \frac{\omega n_{e0}}{n_{e1}} \vb{k} \cdot \vb{u}_{e1}
+    \\
     &= \frac{i \omega n_{e0}}{\omega n_{e0} m_e n_{e1}} \vb{k} \cdot \Big( q_e n_{e0} \vb{E}_1 - i \gamma k_B T_e n_{e1} \vb{k} \Big)
     \\
     &= \frac{n_{e0} q_e^2}{\varepsilon_0 m_e} - \frac{i \omega}{\omega m_e n_{e1}} i \gamma k_B T_e n_{e1} \big(\vb{k} \cdot \vb{k}\big)
@@ -235,13 +238,13 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Unlike for $$T_e = 0$$, these "warm" waves do propagate,
-carrying information at group velocity $$v_g$$,
-which, in the limit of large $$k$$, is given by:
+because $$k$$ appears in the dispersion relation.
+They carry information at group velocity $$v_g = \ipdv{w}{k}$$,
+which in the limit of large $$k$$ becomes:
 
 $$\begin{aligned}
-    v_g
-    = \pdv{\omega}{k}
-    \to \sqrt{\frac{3 k_B T_e}{m_e}}
+    \lim_{k \to \infty} v_g
+    = \sqrt{\frac{3 k_B T_e}{m_e}}
 \end{aligned}$$
 
 This is the root-mean-square velocity of the
diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
index d09613f..0d168aa 100644
--- a/source/know/concept/legendre-transform/index.md
+++ b/source/know/concept/legendre-transform/index.md
@@ -11,9 +11,8 @@ layout: "concept"
 The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$,
 which depends only on the derivative $$f'(x)$$ of $$f(x)$$,
 and from which the original $$f(x)$$ can be reconstructed.
-The point is that $$L(f')$$ contains the same information as $$f(x)$$,
-just in a different form,
-analogously to e.g. the [Fourier transform](/know/concept/fourier-transform/).
+The point is that $$L(f')$$ contains the same information as $$f(x)$$
+in a different form, like e.g. a [Fourier transform](/know/concept/fourier-transform/).
 
 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 $$x = x_0$$,
@@ -23,18 +22,17 @@ $$\begin{aligned}
     y(x)
     &= f'(x_0) (x - x_0) + f(x_0)
     \\
-    &= f'(x_0) \: x - C
+    &= f'(x_0) \: x - C(x_0)
 \end{aligned}$$
 
-Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$.
+Where $$C(x) \equiv f'(x) \: x - f(x)$$.
 We now define the *Legendre transform* $$L(f')$$,
-such that for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$
-(some authors use $$-C$$ instead).
-Renaming $$x_0$$ to $$x$$:
+such that for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C(x_0)$$
+(some authors use $$-C$$ instead):
 
 $$\begin{aligned}
     L(f'(x))
-    &= f'(x) \: x - f(x)
+    &\equiv f'(x) \: x - f(x)
 \end{aligned}$$
 
 We want this function to depend only on the derivative $$f'$$,
diff --git a/source/know/concept/maxwell-relations/index.md b/source/know/concept/maxwell-relations/index.md
index 892ced1..f51acea 100644
--- a/source/know/concept/maxwell-relations/index.md
+++ b/source/know/concept/maxwell-relations/index.md
@@ -9,7 +9,7 @@ layout: "concept"
 ---
 
 The **Maxwell relations** are a useful set of relations in thermodynamics.
-They arise from the fact that the order of differentiation is irrelevant
+They arise from the fact that the ordering of differentiation is irrelevant
 for well-behaved functions (sometimes known as the *Schwarz theorem*),
 applied to the [thermodynamic potentials](/know/concept/thermodynamic-potential/).
 
@@ -54,7 +54,7 @@ $$\begin{aligned}
     = \Big( \pdv{B}{x} \Big)_y^{-1}
 \end{aligned}$$
 
-The following quantities are useful to rewrite some of the Maxwell relations:
+The following quantities can be useful to rewrite some of the Maxwell relations:
 the iso-$$P$$ thermal expansion coefficient $$\alpha$$,
 the iso-$$T$$ combressibility $$\kappa_T$$,
 the iso-$$S$$ combressibility $$\kappa_S$$,
@@ -73,6 +73,10 @@ $$\begin{gathered}
     C_P \equiv T \Big( \pdv{S}{T} \Big)_{P,N}
 \end{gathered}$$
 
+But for simplicity and brevity,
+we will not do any such rewriting in this article.
+
+
 
 ## Internal energy
 
@@ -116,6 +120,7 @@ $$\begin{gathered}
 \end{gathered}$$
 
 
+
 ## Enthalpy
 
 The following Maxwell relations can be derived
@@ -158,6 +163,7 @@ $$\begin{gathered}
 \end{gathered}$$
 
 
+
 ## Helmholtz free energy
 
 The following Maxwell relations can be derived
@@ -200,6 +206,7 @@ $$\begin{gathered}
 \end{gathered}$$
 
 
+
 ## Gibbs free energy
 
 The following Maxwell relations can be derived
@@ -242,10 +249,11 @@ $$\begin{gathered}
 \end{gathered}$$
 
 
+
 ## Landau potential
 
 The following Maxwell relations can be derived
-from the Gibbs free energy $$\Omega(T, V, \mu)$$:
+from the Landau potential $$\Omega(T, V, \mu)$$:
 
 $$\begin{gathered}
     - \mpdv{\Omega}{V}{T} =
diff --git a/source/know/concept/multi-photon-absorption/index.md b/source/know/concept/multi-photon-absorption/index.md
index 80dbc9b..481c19d 100644
--- a/source/know/concept/multi-photon-absorption/index.md
+++ b/source/know/concept/multi-photon-absorption/index.md
@@ -30,7 +30,6 @@ Here, we have made the
 to neglect the $$e^{i \omega t}$$ term,
 because it turns out to be irrelevant in this discussion.
 
-
 We call the ground state $$\Ket{0}$$,
 but other than that, the other states need *not* be sorted by energy.
 However, we demand that the following holds
@@ -187,7 +186,7 @@ i.e. for any odd-numbered final state $$\Ket{u}$$.
 ## Two-photon absorption
 
 Next, we go to second-order perturbation theory.
-Based on the previous result, this time
+Thanks to the previous result $$c_e^{(1)}(t) = 0$$, this time
 all odd-numbered states $$\Ket{u}$$ are unaffected:
 
 $$\begin{aligned}
@@ -248,7 +247,7 @@ two identical photons $$\hbar \omega$$ are absorbed simultaneously
 to bridge the energy gap $$\hbar \omega_{e0}$$.
 Surprisingly, such a transition can only occur when $$\matrixel{e}{\vu{p}}{0} = 0$$,
 i.e. for any even-numbered final state $$\Ket{e}$$.
-Notice that the rate is proportional to $$|\vb{E}|^4$$,
+The rate is proportional to $$|\vb{E}|^4$$,
 so this effect is only noticeable at high light intensities.
 
 
@@ -339,7 +338,7 @@ due to the dependence on $$\vb{E}$$.
 If $$N$$ is odd, only odd-numbered destinations $$\Ket{u}$$ are allowed
 (assuming the electron starts in the ground state $$\Ket{0}$$),
 and if $$N$$ is even, only even-numbered destinations $$\Ket{e}$$.
-Note that nothing has been said about the energies of these states
+Nothing has been said about the energies of these states
 (other than $$\Ket{0}$$ being the minimum);
 everything is determined by the matrix elements $$\matrixel{f}{\vu{p}}{i}$$.
 
diff --git a/source/know/concept/pauli-exclusion-principle/index.md b/source/know/concept/pauli-exclusion-principle/index.md
index 5b83b69..15130d9 100644
--- a/source/know/concept/pauli-exclusion-principle/index.md
+++ b/source/know/concept/pauli-exclusion-principle/index.md
@@ -8,57 +8,69 @@ categories:
 layout: "concept"
 ---
 
-In quantum mechanics, the **Pauli exclusion principle** is a theorem with
-profound consequences for how the world works.
+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:
+$$\ket{x_1}\ket{x_2} = \ket{x_1} \otimes \ket{x_2}$$,
+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:
 
 $$\begin{aligned}
-    \hat{P} \Ket{a}\Ket{b} = \Ket{b}\Ket{a}
+    \hat{P} \Ket{a}\Ket{b}
+    = \Ket{b}\Ket{a}
 \end{aligned}$$
 
-That is, it swaps the states of the particles. Obviously, swapping the
-states twice simply gives the original configuration again, so:
+That is, it swaps the states of the particles.
+Obviously, swapping the states twice simply gives the original configuration again, so:
 
 $$\begin{aligned}
-    \hat{P}^2 \Ket{a}\Ket{b} = \Ket{a}\Ket{b}
+    \hat{P}^2 \Ket{a}\Ket{b}
+    = \Ket{a}\Ket{b}
 \end{aligned}$$
 
-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 $$\lambda = -1$$:
+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 $$\lambda = -1$$:
 
 $$\begin{aligned}
-    \hat{P} \Ket{a}\Ket{b} = \lambda \Ket{a}\Ket{b}
+    \hat{P} \Ket{a}\Ket{b}
+    = \lambda \Ket{a}\Ket{b}
 \end{aligned}$$
 
-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:
+As it turns out, in nature, each type 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:
 
 $$\begin{aligned}
-    \hat{P}_f \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = - \Ket{a}\Ket{b}
-    \qquad
-    \hat{P}_b \Ket{a}\Ket{b} = \Ket{b}\Ket{a} = \Ket{a}\Ket{b}
+    \hat{P}_f \Ket{a}\Ket{b}
+    &= \Ket{b}\Ket{a}
+    = - \Ket{a}\Ket{b}
+    \\
+    \hat{P}_b \Ket{a}\Ket{b}
+    &= \Ket{b}\Ket{a}
+    = \Ket{a}\Ket{b}
 \end{aligned}$$
 
-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!
+Another fundamental fact 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,
+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 profound.
+Consider the following superposition state,
 where $$\alpha$$ and $$\beta$$ are unknown:
 
 $$\begin{aligned}
@@ -66,9 +78,10 @@ $$\begin{aligned}
     = \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:
+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:
 
 $$\begin{aligned}
     \hat{P} \Ket{\Psi(a, b)}
@@ -78,25 +91,28 @@ $$\begin{aligned}
     &= \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 $$\alpha = \beta$$:
+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 $$\alpha = \beta$$:
 
 $$\begin{aligned}
-    \Ket{\Psi(a, b)}_b = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
+    \Ket{\Psi(a, b)}_b
+    = C \big( \Ket{a}\Ket{b} + \Ket{b}\Ket{a} \big)
 \end{aligned}$$
 
-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 $$\alpha = -\beta$$:
+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 $$\alpha = -\beta$$:
 
 $$\begin{aligned}
-    \Ket{\Psi(a, b)}_f = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
+    \Ket{\Psi(a, b)}_f
+    = C \big( \Ket{a}\Ket{b} - \Ket{b}\Ket{a} \big)
 \end{aligned}$$
 
-This state is called **antisymmetric** under exchange: switching $$a$$ and $$b$$
-causes a sign change, as we would expect for fermions.
+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 $$C$$:
@@ -114,7 +130,8 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-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.
+And this is the Pauli exclusion principle:
+**fermions may never occupy the same quantum state**.
+One of the many notable consequences of this is
+that the shells of atoms only fit a limited number of electrons (which are fermions),
+since each must have a different quantum number.
diff --git a/source/know/concept/repetition-code/index.md b/source/know/concept/repetition-code/index.md
index fa039a3..ba83c1a 100644
--- a/source/know/concept/repetition-code/index.md
+++ b/source/know/concept/repetition-code/index.md
@@ -94,7 +94,7 @@ We could measure the state, but that would make it collapse,
 which is probably not what we want.
 
 The trick is to use operators called **stabilizers**,
-in this case for example $$ZZI = Z_1 \otimes Z_2 \otimes I_3$$,
+in this case $$ZZI = Z_1 \otimes Z_2 \otimes I_3$$,
 where $$I$$ is identity and $$Z$$ is the Pauli-$$Z$$ gate.
 The 3-qubit basis states are its eigenvectors:
 
@@ -127,7 +127,7 @@ $$\begin{alignedat}{2}
 We could measure $$ZZI$$ for $$\ket{\overline{\psi}}$$,
 and if the eigenvalue is $$-1$$,
 we know that a bit flip has occurred,
-whereas if the eigenvalue is $$+1$$,
+but if the eigenvalue is $$+1$$,
 there is *maybe* no error ($$\Ket{001}$$ and $$\Ket{110}$$ are false negatives).
 
 These false negatives are fixed by including another stabilizer $$IZZ$$,
@@ -170,7 +170,7 @@ thanks to the eigenvalues:
 | $$I$$ | $$+1$$ | $$+1$$ |
 | $$X_1$$ | $$-1$$ | $$+1$$ |
 | $$X_2$$ | $$-1$$ | $$-1$$ |
-| $$X_1$$ | $$+1$$ | $$-1$$ |
+| $$X_3$$ | $$+1$$ | $$-1$$ |
 
 Where e.g. $$X_3$$ denotes that the 3rd qubit was flipped.
 The measurement outcomes on the last three rows are called **error syndromes**,
@@ -309,6 +309,9 @@ $$\begin{aligned}
     III \: XXX \: XXX
 \end{aligned}$$
 
+In this way, we are protected against all single-qubit errors,
+but at a significant physical cost.
+
 
 
 ## References
diff --git a/source/know/concept/second-quantization/index.md b/source/know/concept/second-quantization/index.md
index e446557..605ffd1 100644
--- a/source/know/concept/second-quantization/index.md
+++ b/source/know/concept/second-quantization/index.md
@@ -15,29 +15,26 @@ whether it is fermions or bosons that are being considered
 (see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)).
 
 Regardless of whether the system is fermionic or bosonic,
-the idea is to change basis to a set of certain many-particle wave functions,
-known as the **Fock states**, which are specific members of a **Fock space**,
-a special kind of [Hilbert space](/know/concept/hilbert-space/),
+the idea is to change basis to a set of many-particle wavefunctions
+known as the **Fock states**, which are specific members of a **Fock space**
+(a special kind of [Hilbert space](/know/concept/hilbert-space/))
 with a well-defined number of particles.
 
 For a set of $$N$$ single-particle energy eigenstates
-$$\psi_n(x)$$ and $$N$$ identical particles $$x_n$$, the Fock states are
-all the wave functions which contain $$n$$ particles, for $$n$$ going from $$0$$ to $$N$$.
-
-So for $$n = 0$$, there is one basis vector with $$0$$ particles,
-for $$n = 1$$, there are $$N$$ basis vectors with $$1$$ particle each,
-for $$n = 2$$, there are $$N (N \!-\! 1)$$ basis vectors with $$2$$ particles,
-etc.
+$$\psi_k(x)$$ and $$N$$ identical particles $$x_k$$,
+the Fock states are all the wavefunctions which contain $$n$$ particles,
+for $$n$$ going from $$0$$ to $$N$$.
 
 In this basis, we define the **particle creation operators**
 and **particle annihilation operators**,
 which respectively add/remove a particle to/from a given state.
-In other words, these operators relate the Fock basis vectors
+In other words, these operators relate the Fock basis states
 to one another, and are very useful.
 
-The point is to express the system's state in such a way that the
-fermionic/bosonic constraints are automatically satisfied, and the
-formulae look the same regardless of the number of particles.
+The idea is to express states in such a way
+that the fermionic/bosonic constraints are automatically satisfied,
+and that the formulas look the same regardless of the number of particles.
+
 
 
 ## Fermions
@@ -56,6 +53,8 @@ $$\begin{aligned}
             \\
             n &= 2:
             \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots
+            \\
+            &\:\:\vdots \qquad \qquad \qquad \vdots
         \end{aligned}
     }
 \end{aligned}$$
@@ -79,16 +78,17 @@ $$\begin{aligned}
 
 The creation operator $$\hat{c}_\alpha^\dagger$$ and annihilation
 operator $$\hat{c}_\alpha$$ are defined to live up to their name:
-they create or destroy a particle in the state $$\psi_\alpha$$:
+they create or destroy a particle in the state $$\psi_\alpha$$.
+Formally, this means:
 
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            \hat{c}_\alpha^\dagger \Ket{... (N_\alpha\!=\!0) ...}
-            &= J_\alpha \Ket{... (N_\alpha\!=\!1) ...}
+            \hat{c}_\alpha^\dagger \Ket{...0_\alpha...}
+            &= J_\alpha \Ket{...1_\alpha...}
             \\
-            \hat{c}_\alpha \Ket{... (N_\alpha\!=\!1) ...}
-            &= J_\alpha \Ket{... (N_\alpha\!=\!0) ...}
+            \hat{c}_\alpha \Ket{...1_\alpha...}
+            &= J_\alpha \Ket{...0_\alpha...}
         \end{aligned}
     }
 \end{aligned}$$
@@ -98,7 +98,8 @@ and is necessary here to enforce the fermionic antisymmetry,
 when creating or destroying a particle in the $$\alpha$$th state:
 
 $$\begin{aligned}
-    J_\alpha = (-1)^{\sum_{j < \alpha} N_j}
+    J_\alpha
+    = (-1)^{\sum_{j < \alpha} N_j}
 \end{aligned}$$
 
 So, for example, when creating a particle in state 4
@@ -110,7 +111,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The point of the Jordan-Wigner string
-is that the order matters when applying the creation and annihilation operators:
+is that the order matters when applying the creation and annihilation operators,
+so, for example:
 
 $$\begin{aligned}
     \hat{c}_1^\dagger \hat{c}_2 \Ket{0, 1}
@@ -124,14 +126,21 @@ $$\begin{aligned}
 
 In other words, $$\hat{c}_1^\dagger \hat{c}_2 = - \hat{c}_2 \hat{c}_1^\dagger$$,
 meaning that the anticommutator $$\{\hat{c}_2, \hat{c}_1^\dagger\} = 0$$.
-You can verify for youself that
+You can verify for yourself that
 the general anticommutators of these operators are given by:
 
 $$\begin{aligned}
     \boxed{
-        \{\hat{c}_\alpha, \hat{c}_\beta\} = \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\} = 0
-        \qquad \quad
-        \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\} = \delta_{\alpha\beta}
+        \begin{aligned}
+            \{\hat{c}_\alpha, \hat{c}_\beta\}
+            &= 0
+            \\
+            \{\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger\}
+            &= 0
+            \\
+            \{\hat{c}_\alpha, \hat{c}_\beta^\dagger\}
+            &= \delta_{\alpha\beta}
+        \end{aligned}
     }
 \end{aligned}$$
 
@@ -141,24 +150,29 @@ Note that these are *scalar* zeros:
 
 $$\begin{aligned}
     \boxed{
-        \hat{c}_\alpha^\dagger \Ket{... (N_\alpha\!=\!1) ...} = 0
-        \qquad \quad
-        \hat{c}_\alpha \Ket{... (N_\alpha\!=\!0) ...} = 0
+        \begin{aligned}
+            \hat{c}_\alpha^\dagger \Ket{...1_\alpha...}
+            &= 0
+            \\
+            \hat{c}_\alpha \Ket{...0_\alpha...}
+            &= 0
+        \end{aligned}
     }
 \end{aligned}$$
 
 Finally, as has already been suggested by the notation, they are each other's adjoint:
 
 $$\begin{aligned}
-    \matrixel{... (N_\alpha\!=\!1) ...}{\hat{c}_\alpha^\dagger}{... (N_\alpha\!=\!0) ...}
-    = \matrixel{...(N_\alpha\!=\!0) ...}{\hat{c}_\alpha}{... (N_\alpha\!=\!1) ...}
+    \matrixel{...1_\alpha...}{\hat{c}_\alpha^\dagger}{...0_\alpha...}
+    = \matrixel{...0_\alpha...}{\hat{c}_\alpha}{...1_\alpha...}^{*}
 \end{aligned}$$
 
 Let us now use these operators to define the **number operator** $$\hat{N}_\alpha$$ as follows:
 
 $$\begin{aligned}
     \boxed{
-        \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
+        \hat{N}_\alpha
+        = \hat{c}_\alpha^\dagger \hat{c}_\alpha
     }
 \end{aligned}$$
 
@@ -171,6 +185,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Bosons
 
 Bosons do not need to obey the Pauli exclusion principle, so multiple can occupy a single state.
@@ -188,8 +203,10 @@ $$\begin{aligned}
             n &= 2:
             \qquad \Ket{1, 1, 0, ...} \quad \Ket{1, 0, 1, ...} \quad \Ket{0, 1, 1, ...} \quad \cdots
             \\
-            &\qquad\:\:\:
+            &\qquad\:\,\,
             \qquad \Ket{2, 0, 0, ...} \quad \Ket{0, 2, 0, ...} \quad \Ket{0, 0, 2, ...} \quad \cdots
+            \\
+            &\:\:\vdots \qquad \qquad \qquad \vdots
         \end{aligned}
     }
 \end{aligned}$$
@@ -212,23 +229,31 @@ $$\begin{gathered}
         \end{aligned}
 }\end{gathered}$$
 
-Applying the annihilation operator $$\hat{c}_\alpha$$ when there are zero
-particles in $$\alpha$$ will quench the state:
+Applying the annihilation operator $$\hat{c}_\alpha$$
+when there are zero particles in $$\alpha$$ quenches the state:
 
 $$\begin{aligned}
     \boxed{
-        \hat{c}_\alpha \Ket{... (N_\alpha\!=\!0) ...} = 0
+        \hat{c}_\alpha \Ket{...0_\alpha...}
+        = 0
     }
 \end{aligned}$$
 
 There is no Jordan-Wigner string, and therefore no sign change when commuting.
-Consequently, these operators therefore satisfy the following:
+Consequently, these operators satisfy the following commutators:
 
 $$\begin{aligned}
     \boxed{
-        [\hat{c}_\alpha, \hat{c}_\beta] = [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] = 0
-        \qquad
-        [\hat{c}_\alpha, \hat{c}_\beta^\dagger] = \delta_{\alpha\beta}
+        \begin{aligned}
+            [\hat{c}_\alpha, \hat{c}_\beta]
+            &= 0
+            \\
+            [\hat{c}_\alpha^\dagger, \hat{c}_\beta^\dagger] 
+            &= 0
+            \\
+            [\hat{c}_\alpha, \hat{c}_\beta^\dagger]
+            &= \delta_{\alpha\beta}
+        \end{aligned}
     }
 \end{aligned}$$
 
@@ -237,90 +262,93 @@ ensure that $$\hat{N}_\alpha$$ keeps the same nice form:
 
 $$\begin{aligned}
     \boxed{
-        \hat{N}_\alpha = \hat{c}_\alpha^\dagger \hat{c}_\alpha
+        \hat{N}_\alpha
+        = \hat{c}_\alpha^\dagger \hat{c}_\alpha
     }
 \end{aligned}$$
 
 
+
 ## Operators
 
-Traditionally, an operator $$\hat{V}$$ simultaneously acting on $$N$$ indentical particles
-is the sum of the individual single-particle operators $$\hat{V}_1$$ acting on the $$n$$th particle:
+In the second quantization,
+changing between different bases of single-particle states
+is done in the usual way, where $$\alpha$$ and $$b$$ need not be in the same basis.
+Note that $$\Ket{0}$$ is the zero-particle Fock state,
+and $$\Ket{\alpha}$$ etc. are one-particle Fock states:
 
 $$\begin{aligned}
-    \hat{V}
-    = \sum_{n = 1}^N \hat{V}_1
+    \hat{c}_b^\dagger \Ket{0}
+    = \Ket{b}
+    = \sum_{\alpha} \Ket{\alpha} \inprod{\alpha}{b}
+    = \sum_{\alpha} \inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0}
 \end{aligned}$$
 
-This can be rewritten using the second quantization operators as follows:
+With this, we define the **field operators**,
+which create or destroy a particle at a position $$\vb{r}$$:
 
 $$\begin{aligned}
     \boxed{
-        \hat{V}
-        = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}_1}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta
+        \hat{\Psi}^\dagger(\vb{r})
+        = \sum_{\alpha} \inprod{\alpha}{\vb{r}} \hat{c}_\alpha^\dagger
+        \qquad \qquad
+        \hat{\Psi}(\vb{r})
+        = \sum_{\alpha} \inprod{\vb{r}}{\alpha} \hat{c}_\alpha
     }
 \end{aligned}$$
 
-Where the matrix element $$\matrixel{\alpha}{\hat{V}_1}{\beta}$$ is to be
-evaluated in the normal way:
-
-$$\begin{aligned}
-    \matrixel{\alpha}{\hat{V}_1}{\beta}
-    = \int \psi_\alpha^*(\vec{r}) \: \hat{V}_1(\vec{r}) \: \psi_\beta(\vec{r}) \dd{\vec{r}}
-\end{aligned}$$
-
-Similarly, given some two-particle operator $$\hat{V}$$ in first-quantized form:
+By the same basis-changing principle,
+any single-particle (non-interacting) operator $$\hat{V}$$ can be translated
+to its second-quantized $$N$$-particle version as follows:
 
 $$\begin{aligned}
     \hat{V}
-    = \sum_{n \neq m} v(\vec{r}_n, \vec{r}_m)
+    &= \sum_{\alpha, \beta} \ket{\alpha} \matrixel{\alpha}{\hat{V}}{\beta} \bra{\beta}
+    = \sum_{\alpha, \beta} \ket{\hat{c}_\alpha^\dagger 0} \matrixel{\alpha}{\hat{V}}{\beta} \bra{\hat{c}_\beta^\dagger 0}
 \end{aligned}$$
 
-We can rewrite this in second-quantized form as follows.
-Note the ordering of the subscripts:
+We take out the creation operators,
+which allows us to generalize to multi-particle states:
 
 $$\begin{aligned}
     \boxed{
         \hat{V}
-        = \sum_{\alpha, \beta, \gamma, \delta}
-        v_{\alpha \beta \gamma \delta} \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma
+        = \sum_{\alpha, \beta} \matrixel{\alpha}{\hat{V}}{\beta} \hat{c}_\alpha^\dagger \hat{c}_\beta
     }
 \end{aligned}$$
 
-Where the constant $$v_{\alpha \beta \gamma \delta}$$ is defined from the
-single-particle wave functions:
+Where the matrix element $$\matrixel{\alpha}{\hat{V}}{\beta}$$
+is to be evaluated in the normal way:
 
 $$\begin{aligned}
-    v_{\alpha \beta \gamma \delta}
-    = \iint \psi_\alpha^*(\vec{r}_1) \: \psi_\beta^*(\vec{r}_2)
-    \: v(\vec{r}_1, \vec{r}_2) \: \psi_\gamma(\vec{r}_1)
-    \: \psi_\delta(\vec{r}_2) \dd{\vec{r}_1} \dd{\vec{r}_2}
+    \matrixel{\alpha}{\hat{V}}{\beta}
+    = \int \psi_\alpha^*(\vb{r}) \: \hat{V}(\vb{r}) \: \psi_\beta(\vb{r}) \dd{\vb{r}}
 \end{aligned}$$
 
-Finally, in the second quantization, changing basis is done in the usual way:
+In the same way, a two-particle interaction operator $$\hat{W}$$
+can be rewritten in the form below.
+Note the ordering of the operators' subscripts:
 
 $$\begin{aligned}
-    \hat{c}_b^\dagger \Ket{0}
-    = \Ket{b}
-    = \sum_{\alpha} \Ket{\alpha} \Inprod{\alpha}{b}
-    = \sum_{\alpha} \Inprod{\alpha}{b} \hat{c}_\alpha^\dagger \Ket{0}
+    \boxed{
+        \hat{W}
+        = \sum_{\alpha, \beta, \gamma, \delta}
+        W_{\alpha \beta \gamma \delta} \: \hat{c}_\alpha^\dagger \hat{c}_\beta^\dagger \hat{c}_\delta \hat{c}_\gamma
+    }
 \end{aligned}$$
 
-Where $$\alpha$$ and $$b$$ need not be in the same basis.
-With this, we can define the **field operators**,
-which create or destroy a particle at a given position $$\vec{r}$$:
+Where the constant $$W_{\alpha \beta \gamma \delta}$$
+is defined from the single-particle wavefunctions like so:
 
 $$\begin{aligned}
-    \boxed{
-        \hat{\Psi}^\dagger(\vec{r})
-        = \sum_{\alpha} \Inprod{\alpha}{\vec{r}} \hat{c}_\alpha^\dagger
-        \qquad \quad
-        \hat{\Psi}(\vec{r})
-        = \sum_{\alpha} \Inprod{\vec{r}}{\alpha} \hat{c}_\alpha
-    }
+    W_{\alpha \beta \gamma \delta}
+    \equiv \iint \psi_\alpha^*(\vb{r}_1) \: \psi_\beta^*(\vb{r}_2)
+    \: W(\vb{r}_1, \vb{r}_2) \: \psi_\gamma(\vb{r}_1)
+    \: \psi_\delta(\vb{r}_2) \dd{\vb{r}_1} \dd{\vb{r}_2}
 \end{aligned}$$
 
 
+
 ## References
 1.  L.E. Ballentine,
     *Quantum mechanics: a modern development*, 2nd edition,
diff --git a/source/know/concept/shors-algorithm/index.md b/source/know/concept/shors-algorithm/index.md
index 678d5d2..bab88a3 100644
--- a/source/know/concept/shors-algorithm/index.md
+++ b/source/know/concept/shors-algorithm/index.md
@@ -29,10 +29,6 @@ This is a so-called *hidden subgroup problem* for a *finite Abelian group*.
 With minimal modifications,
 Shor's algorithm can solve practically every such problem.
 
-
-
-## Integer factorization
-
 Originally, Shor's algorithm was designed to factorize an integer $$N$$.
 For reasons explained later,
 this means our goal is to find the period $$s$$ of
@@ -40,7 +36,7 @@ the modular exponentiation function $$f$$:
 
 $$\begin{aligned}
     f(x)
-    = a^x \bmod N
+    \equiv a^x \bmod N
 \end{aligned}$$
 
 For a given $$a$$ and $$N$$.
@@ -72,7 +68,8 @@ $$\begin{aligned}
     = \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{0}^{\otimes q}
 \end{aligned}$$
 
-Where $$Q = 2^q$$, and $$\Ket{x}$$ is the computational basis state $$\Ket{x_1} \cdots \Ket{x_q}$$.
+Where $$Q \equiv 2^q$$,
+and $$\Ket{x}$$ is the computational basis state $$\Ket{x_1} \cdots \Ket{x_q}$$.
 Moving on to $$U_f$$:
 
 $$\begin{aligned}
@@ -81,32 +78,41 @@ $$\begin{aligned}
     \frac{1}{\sqrt{Q}} \sum_{x = 0}^{Q - 1} \Ket{x} \Ket{f(x)}
 \end{aligned}$$
 
-Then we measure $$f(x)$$, causing it collapse as follows
-for an unknown arbitrary value of $$x_0$$:
+Then we measure $$f(x)$$, causing it collapse
+for an unknown arbitrary value of $$x_0$$.
+Let $$L$$ be number of periods that fit in the available qubits,
+then we know that:
 
 $$\begin{aligned}
-    f(x_0) = f(x_0 + s) = f(x_0 + 2s) = \cdots = f(x_0 + (L-1) s)
+    f(x_0)
+    = f(x_0 + s)
+    = \cdots
+    = f(x_0 + \ell s)
+    = \cdots
+    = f(x_0 + (L-1) s)
 \end{aligned}$$
 
 Due to [entanglement](/know/concept/quantum-entanglement/),
-the unmeasured (top $$q$$) qubits change state into a superposition:
+the unmeasured (top $$q$$) qubits change state, so we now have:
 
 $$\begin{aligned}
-    \frac{1}{\sqrt{L}} \sum_{\ell = 0}^{L - 1} \Ket{x_0 + \ell s}
+    \frac{1}{\sqrt{L}} \sum_{\ell = 0}^{L - 1} \Ket{x_0 + \ell s} \Ket{f(x_0)}
 \end{aligned}$$
 
 Clearly, there is a periodic structure here,
 but we cannot measure it directly,
-because we do not know the value of $$x_0$$,
-which, to make matters worse, changes every time we run the algorithm.
-This is where the QFT comes in, which outputs the following state:
+since we do not know the value of $$x_0$$,
+which, to make matters worse, changes every time we run the algorithm!
+But now we apply the QFT, which outputs the state below,
+where $$\omega_Q$$ is a $$Q$$th root of unity.
+From now on, we no longer write the measured qubits $$\Ket{f(x_0)}$$, leaving:
 
 $$\begin{aligned}
     \frac{1}{\sqrt{QL}} \sum_{k = 0}^{Q - 1} \bigg( \sum_{\ell = 0}^{L - 1} \omega_Q^{(x_0 + \ell s) k} \bigg) \Ket{k}
 \end{aligned}$$
 
-Where $$\omega_Q$$ is a $$Q$$th root of unity.
-Measuring this state yields a $$\Ket{k}$$, with a probability $$P(k)$$:
+Measuring this state causes a collapse into some $$\Ket{k}$$
+with a probability $$P(k)$$:
 
 $$\begin{aligned}
     P(k)
@@ -116,9 +122,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The last step holds because $$|\omega_Q| = 1$$.
-Surprisingly, this implies that we did not need
-to perform the measurement of $$f(x)$$ earlier!
-This makes sense: the period $$s$$ does not depend on $$x_0$$,
+Surprisingly, $$x_0$$ has disappeared,
+implying that we did not need to perform the measurement of $$f(x)$$ earlier!
+This makes some sense: the period $$s$$ does not depend on $$x_0$$,
 so why would we need an implicit $$x_0$$ to determine $$s$$?
 
 So, what does the above probability $$P(k)$$ work out to?
@@ -134,12 +140,13 @@ $$\begin{alignedat}{2}
 
 Where the latter case was evaluated as a geometric series.
 The condition $$\omega_Q^{sk}\!=\!1$$ is equivalent to asking
-if $$sk$$ is a multiple of $$Q$$, i.e. if $$sk = cQ$$, for an integer $$c$$.
+if $$sk$$ is a multiple of $$Q$$.
+i.e. if $$sk = cQ$$, for an integer $$c$$.
 
 Recall that $$L$$ is the number of times that $$s$$ fits in $$Q$$,
 so $$L\!=\!\lfloor Q / s \rfloor$$.
-Assuming $$Q/s$$ is an integer, then $$L\!=\!Q/s$$ and $$Q\!=\!s L$$,
-which tells us that
+For now, let us assume that $$Q/s$$ is an integer,
+then $$L\!=\!Q/s$$ and $$Q\!=\!s L$$, which tells us that
 $$\omega_Q^{sk}\!=\!\omega_{s L}^{s k}\!=\!\omega_L^k$$.
 This implies that if $$k$$ is a multiple of $$L$$ (i.e. $$k\!=\!c L$$),
 then $$\omega_L^k\!=\!1$$, so $$P(k) = L / Q$$,
@@ -245,9 +252,11 @@ $$\begin{aligned}
     = a^x \bmod N
 \end{aligned}$$
 
-$$N$$ is the number to factorize, and $$a$$ is a random integer *coprime* to $$N$$,
+$$N$$ is the number to factorize, and $$a$$ is an integer that we can choose.
+For this to work, we should pick an $$a$$ that is *coprime* to $$N$$,
 meaning $$\gcd(a, N) = 1$$.
-The fact that $$s$$ is the period of $$f$$ for a certain $$a$$-value, implies that:
+Then the fact that $$s$$ is the period of $$f$$
+for a certain $$a$$-value implies that:
 
 $$\begin{aligned}
     a^x
@@ -257,7 +266,7 @@ $$\begin{aligned}
     = a^s \bmod N
 \end{aligned}$$
 
-Suppose that $$s$$ is even. In that case,
+For now, suppose that $$s$$ is even. In that case,
 we can rewrite the above equation as follows:
 
 $$\begin{aligned}
@@ -273,7 +282,8 @@ $$\begin{aligned}
     = 0 \bmod N
 \end{aligned}$$
 
-Because $$s$$ is even by assumption, the two factors on the left are integers,
+Because $$s$$ is even by assumption (for now),
+the two factors on the left are integers,
 and as just mentioned, their product is a multiple of $$N$$.
 Then we only need to calculate:
 
diff --git a/source/know/concept/sokhotski-plemelj-theorem/index.md b/source/know/concept/sokhotski-plemelj-theorem/index.md
index 445b029..e139954 100644
--- a/source/know/concept/sokhotski-plemelj-theorem/index.md
+++ b/source/know/concept/sokhotski-plemelj-theorem/index.md
@@ -10,7 +10,7 @@ layout: "concept"
 ---
 
 The goal is to evaluate integrals of the following form,
-where $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$:
+where $$f(x)$$ is real and continuous in the integration interval $$[a, b]$$:
 
 $$\begin{aligned}
     \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
@@ -56,7 +56,7 @@ $$\begin{aligned}
     &= \lim_{m \to +\infty} \frac{\pi}{\pi} \int_a^b \frac{m}{1 + m^2 x^2} f(x) \dd{x}
 \end{aligned}$$
 
-The expression $$m / \pi (1 + m^2 x^2)$$ is a so-called *nascent delta function*,
+The expression $$m / (\pi (1 + m^2 x^2))$$ is a so-called *nascent delta function*,
 meaning that in the limit $$m \to +\infty$$ it converges to
 the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$:
 
diff --git a/source/know/concept/thermodynamic-potential/index.md b/source/know/concept/thermodynamic-potential/index.md
index b15c011..60eee78 100644
--- a/source/know/concept/thermodynamic-potential/index.md
+++ b/source/know/concept/thermodynamic-potential/index.md
@@ -12,17 +12,17 @@ layout: "concept"
 whose minima or maxima represent equilibrium states of a system.
 Such functions are either energies (hence *potential*) or entropies.
 
-Which potential (of many) decides the equilibrium states for a given system?
-That depends which variables are assumed to already be in automatic equilibrium.
-Such variables are known as the **natural variables** of that potential.
-For example, if a system can freely exchange heat with its environment,
-and is consequently assumed to be at the same temperature $$T = T_{\mathrm{env}}$$,
+Of the many options, which potential decides the equilibrium state for a given system?
+It depends on which variables are assumed to be in automatic equilibrium.
+Such variables are called the **natural variables** of that potential.
+For example, if a system can exchange heat with its environment,
+and is consequently at the same temperature $$T = T_{\mathrm{env}}$$,
 then $$T$$ must be a natural variable.
 
 The link from natural variables to potentials
 is established by [thermodynamic ensembles](/know/category/thermodynamic-ensembles/).
 
-Once enough natural variables have been found,
+Once the natural variables have been determined,
 the appropriate potential can be selected from the list below.
 All non-natural variables can then be calculated
 by taking partial derivatives of the potential
@@ -48,8 +48,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It is a function of the entropy $$S$$, volume $$V$$, and particle count $$N$$:
-these are its natural variables.
-An infinitesimal change $$\dd{U}$$ is as follows:
+these are its natural variables,
+so an infinitesimal change $$\dd{U}$$ is as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -59,7 +59,7 @@ $$\begin{aligned}
 
 The non-natural variables are
 temperature $$T$$, pressure $$P$$, and chemical potential $$\mu$$.
-They can be recovered by differentiating $$U$$
+These can be recovered by differentiating $$U$$
 with respect to the natural variables $$S$$, $$V$$, and $$N$$:
 
 $$\begin{aligned}
@@ -92,8 +92,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It is a function of the entropy $$S$$, pressure $$P$$, and particle count $$N$$:
-these are its natural variables.
-An infinitesimal change $$\dd{H}$$ is as follows:
+these are its natural variables,
+so an infinitesimal change $$\dd{H}$$ is as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -103,7 +103,7 @@ $$\begin{aligned}
 
 The non-natural variables are
 temperature $$T$$, volume $$V$$, and chemical potential $$\mu$$.
-They can be recovered by differentiating $$H$$
+These can be recovered by differentiating $$H$$
 with respect to the natural variables $$S$$, $$P$$, and $$N$$:
 
 $$\begin{aligned}
@@ -132,8 +132,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It depends on the temperature $$T$$, volume $$V$$, and particle count $$N$$:
-these are natural variables.
-An infinitesimal change $$\dd{H}$$ is as follows:
+these are its natural variables,
+so an infinitesimal change $$\dd{H}$$ is as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -142,8 +142,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The non-natural variables are
-entropy $$S$$, pressure $$P$$, and chemical potential $$\mu$$.
-They can be recovered by differentiating $$F$$
+the entropy $$S$$, pressure $$P$$, and chemical potential $$\mu$$.
+These can be recovered by differentiating $$F$$
 with respect to the natural variables $$T$$, $$V$$, and $$N$$:
 
 $$\begin{aligned}
@@ -171,8 +171,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It depends on the temperature $$T$$, pressure $$P$$, and particle count $$N$$:
-they are natural variables.
-An infinitesimal change $$\dd{G}$$ is as follows:
+they are its natural variables,
+so an infinitesimal change $$\dd{G}$$ is as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -181,7 +181,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The non-natural variables are
-entropy $$S$$, volume $$V$$, and chemical potential $$\mu$$.
+the entropy $$S$$, volume $$V$$, and chemical potential $$\mu$$.
 These can be recovered by differentiating $$G$$
 with respect to the natural variables $$T$$, $$P$$, and $$N$$:
 
@@ -210,8 +210,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It depends on temperature $$T$$, volume $$V$$, and chemical potential $$\mu$$:
-these are natural variables.
-An infinitesimal change $$\dd{\Omega}$$ is as follows:
+these are its natural variables,
+so an infinitesimal change $$\dd{\Omega}$$ is as follows:
 
 $$\begin{aligned}
     \boxed{
@@ -239,7 +239,8 @@ $$\begin{aligned}
 ## Entropy
 
 The **entropy** $$S$$, in units of energy over temperature,
-is an odd duck, but nevertheless used as a thermodynamic potential.
+is an odd duck, but nevertheless used as a thermodynamic potential,
+to be maximized instead of minimized.
 It is given by:
 
 $$\begin{aligned}
@@ -249,8 +250,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 It depends on the internal energy $$U$$, volume $$V$$, and particle count $$N$$:
-they are natural variables.
-An infinitesimal change $$\dd{S}$$ is as follows:
+they are its natural variables,
+so an infinitesimal change $$\dd{S}$$ is as follows:
 
 $$\begin{aligned}
     \boxed{
diff --git a/source/know/concept/wkb-approximation/index.md b/source/know/concept/wkb-approximation/index.md
index ef57a3b..fb04414 100644
--- a/source/know/concept/wkb-approximation/index.md
+++ b/source/know/concept/wkb-approximation/index.md
@@ -8,24 +8,25 @@ categories:
 layout: "concept"
 ---
 
-In quantum mechanics, the **Wentzel-Kramers-Brillouin** or simply the **WKB
-approximation** is a technique to approximate the wave function $$\psi(x)$$ of
-the one-dimensional time-independent Schrödinger equation. It is an example
-of a **semiclassical approximation**, because it tries to find a
-balance between classical and quantum physics.
+In quantum mechanics, the **Wentzel-Kramers-Brillouin**
+or simply the **WKB approximation**
+is a technique to approximate the wavefunction $$\psi(x)$$
+of the 1D 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:
 
 $$\begin{aligned}
-    E = \frac{1}{2} m \dot{x}^2 + V(x)
-    \quad \implies \quad
+    E = \frac{1}{2} m (x')^2 + V(x)
+    \qquad \implies \qquad
     m^2 (x')^2 = 2 m (E - V(x))
 \end{aligned}$$
 
 The left-hand side of the rearranged version is simply the momentum squared,
-so we define the magnitude of the momentum $$p(x)$$ accordingly:
+so we know that the magnitude of the momentum $$p(x)$$ is:
 
 $$\begin{aligned}
     p(x) = \sqrt{2 m (E - V(x))}
@@ -38,8 +39,9 @@ We rewrite the Schrödinger equation:
 
 $$\begin{aligned}
     0
-    = \dvn{2}{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
-    = \dvn{2}{\psi}{x} + \frac{p^2}{\hbar^2} \psi
+    &= \dvn{2}{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi
+    \\
+    &= \dvn{2}{\psi}{x} + \frac{p^2}{\hbar^2} \psi
 \end{aligned}$$
 
 If $$V(x)$$ were constant, and by extension $$p(x)$$ too, then the solution
@@ -50,20 +52,20 @@ $$\begin{aligned}
     = \psi(0) \exp(\pm i p x / \hbar)
 \end{aligned}$$
 
-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
+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:
+$$2 \pi / k(x)$$, 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 $$x$$.
+$$\chi(\xi)$$ 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:
 
 $$\begin{aligned}
@@ -71,111 +73,118 @@ $$\begin{aligned}
     = \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.
+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:
 
 $$\begin{aligned}
-    \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x)
-    \quad \implies \quad
-    \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)}
+    \psi'(x)
+    = \pm \frac{i}{\hbar} \chi(x) \psi(x)
 \end{aligned}$$
 
-Next, we insert this ansatz for $$\psi(x)$$ into the Schrödinger equation
-to get:
+We insert this ansatz for $$\psi(x)$$ into the Schrödinger equation to get:
 
 $$\begin{aligned}
     0
     &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi
-    = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi
-    = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
+    \\
+    &= \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi
+    \\
+    &= \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi
 \end{aligned}$$
 
-Dividing out $$\psi$$ and rearranging gives us the following, which is
-still exact:
+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:
+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:
 
 $$\begin{aligned}
-    \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ...
+    \chi(x)
+    = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + \cdots
+\end{aligned}$$
+
+The heart of the WKB approximation is its assumption that quantum effects
+are sufficiently weak that we only need to consider
+the first two terms of this expansion,
+i.e. $$\hbar^2$$ is so small that it is negligible.
+Therefore, our approximated wavefunction $$\psi(x)$$ now looks like this:
+
+$$\begin{aligned}
+    \psi(x)
+    &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
 \end{aligned}$$
 
-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:
+Inserting the expansion's first two terms into our equation for $$\chi(x)$$ gives:
 
 $$\begin{aligned}
     \pm \frac{\hbar}{i} \chi_0'
     &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1
 \end{aligned}$$
 
-Where we have discarded all terms containing $$\hbar^2$$. At order
-$$\hbar^0$$, we then get the expected classical result for $$\chi_0(x)$$:
+Where we have discarded all terms containing $$\hbar^2$$.
+At order $$\hbar^0$$, we then get the expected classical result for $$\chi_0(x)$$:
 
 $$\begin{aligned}
     0 = p^2 - \chi_0^2
-    \quad \implies \quad
-    \chi_0(x) = p(x)
+    \qquad \implies \qquad
+    \chi_0(x)
+    = p(x)
 \end{aligned}$$
 
-While at order $$\hbar$$, we get the following quantum-mechanical
-correction:
+While at order $$\hbar$$, we get the following quantum-mechanical correction:
 
 $$\begin{aligned}
     \pm \frac{\hbar}{i} \chi_0'
     = - 2 \frac{\hbar}{i} \chi_0 \chi_1
-    \quad \implies \quad
-    \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
+    \qquad \implies \qquad
+    \chi_1(x)
+    = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)}
 \end{aligned}$$
 
-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:
+We can use this to simplify the latter exponential in $$\psi(x)$$
+using integration by substitution:
 
 $$\begin{aligned}
     \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big)
     &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big)
-    = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big)
+    \\
+    &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big)
     \\
     &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big)
-    = \frac{1}{\sqrt{\chi_0(x)}}
-    = \frac{1}{\sqrt{p(x)}}
+    \\
+    &= \frac{1}{\sqrt{\chi_0(x)}}
 \end{aligned}$$
 
-In the WKB approximation for $$E > V$$, the solution $$\psi(x)$$ is thus
-given by:
+In the WKB approximation for $$E > V$$,
+the solution $$\psi(x)$$ is therefore given by:
 
 $$\begin{aligned}
     \boxed{
-        \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
+        \psi(x)
+        \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big)
     }
 \end{aligned}$$
 
-What if $$E < V$$? In classical mechanics, this is just not allowed; a ball
-cannot simply go through a potential bump without the necessary energy.
+What if $$E < V$$? In classical mechanics, this is not allowed:
+a ball cannot simply go through or over a potential bump without the necessary energy.
 On the other hand, in quantum physics, particles can **tunnel** through barriers.
 
 Luckily, the only thing we need to change for the WKB approximation
 is to let the momentum take imaginary values:
 
 $$\begin{aligned}
-    p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)}
+    p(x)
+    = \sqrt{2 m (E - V(x))}
+    = i \sqrt{2 m (V(x) - E)}
 \end{aligned}$$
 
 And then take the absolute value in the appropriate place in front of $$\psi(x)$$:
@@ -186,12 +195,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-In the classical region ($$E > V$$), the wave function oscillates, and
-in the quantum-physical region ($$E < V$$) it is exponential.
+In the classical region ($$E > V$$), the wavefunction oscillates,
+and in the quantum-physical region ($$E < V$$) it is exponential.
 Note that for $$E \approx V$$ the approximation breaks down,
 because of the appearance of $$p(x)$$ in the denominator.
 
 
+
 ## References
 1.  D.J. Griffiths, D.F. Schroeter,
     *Introduction to quantum mechanics*, 3rd edition,