Improve knowledge base

11 files changed, 143 insertions, 109 deletions

diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 9cb3bcd..5f4add0 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -65,7 +65,7 @@ But what about the collision term?
 Expressions for it exist, which are almost exact in many cases,
 but unfortunately also quite difficult to work with.
 In addition, $$f$$ is a 7-dimensional function,
-so the BTE is already hard to solve without collisions.
+so the BTE is already hard to solve without collisions!
 We only present the simplest case,
 known as the **Bhatnagar-Gross-Krook approximation**:
 if the equilibrium state $$f_0(\vb{r}, \vb{v})$$ is known,
@@ -314,7 +314,7 @@ For the sake of clarity, we write out the pressure term, including the outer div
 
 $$\begin{aligned}
     \nabla \cdot (\vb{V} \cdot \hat{P})
-    &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V}
+    &= (\nabla \cdot \hat{P}{}^\top) \cdot \vb{V}
     = \nabla \cdot
     \begin{bmatrix}
         P_{xx} & P_{xy} & P_{xz} \\
diff --git a/source/know/concept/boussinesq-wave-theory/index.md b/source/know/concept/boussinesq-wave-theory/index.md
index ad2fe4c..e5fd433 100644
--- a/source/know/concept/boussinesq-wave-theory/index.md
+++ b/source/know/concept/boussinesq-wave-theory/index.md
@@ -574,7 +574,7 @@ Because $$\phi_\xi$$ is real, we need the right-hand side to be positive,
 so $$w > 2 \phi$$; for $$\phi \to 0$$, this means that $$w > 0$$.
 This equation is similar to the one encountered when solving
 the [Korteweg-de Vries equation](/know/concept/korteweg-de-vries-equation/)
-and is integrated in the same way; refer there for details.
+and is integrated in the same way; look there for details.
 The result is:
 
 $$\begin{aligned}
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index d10d85d..3f9eafb 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -7,54 +7,59 @@ categories:
 layout: "concept"
 ---
 
-The **convolution theorem** states that a convolution in the direct domain
-is equal to a product in the frequency domain. This is especially useful
-for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an
-$$\mathcal{O}(n \log(n))$$ transform and product.
+The **convolution theorem** states that a convolution in the real domain
+is equal to a product in the frequency domain.
+This fact is especially useful for computation,
+as it allows replacing an $$\mathcal{O}(n^2)$$ convolution
+with an $$\mathcal{O}(n \log(n))$$ transform and product.
 
 
 
 ## Fourier transform
 
-The convolution theorem is usually expressed as follows, where
-$$\hat{\mathcal{F}}$$ is the [Fourier transform](/know/concept/fourier-transform/),
-and $$A$$ and $$B$$ are constants from its definition:
+The convolution theorem is usually expressed as follows,
+where $$\hat{\mathcal{F}}$$ is the [Fourier transform](/know/concept/fourier-transform/),
+and $$A$$ and $$B$$ are the constants from its definition:
 
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            A \cdot (f * g)(x) &= \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\} \\
-            B \cdot (\tilde{f} * \tilde{g})(k) &= \hat{\mathcal{F}}\{f(x) \: g(x)\}
+            A \cdot (f * g)(x)
+            &= \hat{\mathcal{F}}{}^{-1}\Big\{ \tilde{f}(k) \: \tilde{g}(k) \Big\}
+            \\
+            B \cdot (\tilde{f} * \tilde{g})(k)
+            &= \hat{\mathcal{F}}\Big\{ f(x) \: g(x) \Big\}
         \end{aligned}
     }
 \end{aligned}$$
 
 
 {% include proof/start.html id="proof-fourier" -%}
-We expand the right-hand side of the theorem and
-rearrange the integrals:
+We expand the right-hand side of the theorem and rearrange the integrals:
 
 $$\begin{aligned}
-    \hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
-    &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \Big) e^{-i s k x} \dd{k}
+    \hat{\mathcal{F}}{}^{-1}\Big\{ \tilde{f}(k) \: \tilde{g}(k) \Big\}
+    &= B \int_{-\infty}^\infty \tilde{f}(k) \bigg( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \bigg) e^{-i s k x} \dd{k}
     \\
-    &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \Big) \dd{x'}
+    &= A \int_{-\infty}^\infty g(x') \bigg( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \bigg) \dd{x'}
     \\
     &= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
-    = A \cdot (f * g)(x)
+    \\
+    &= A \cdot (f * g)(x)
 \end{aligned}$$
 
 Then we do the same again,
 this time starting from a product in the $$x$$-domain:
 
 $$\begin{aligned}
-    \hat{\mathcal{F}}\{f(x) \: g(x)\}
-    &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \Big) e^{i s k x} \dd{x}
+    \hat{\mathcal{F}}\Big\{ f(x) \: g(x) \Big\}
+    &= A \int_{-\infty}^\infty f(x) \bigg( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \bigg) e^{i s k x} \dd{x}
     \\
-    &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \Big) \dd{k'}
+    &= B \int_{-\infty}^\infty \tilde{g}(k') \bigg( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \bigg) \dd{k'}
     \\
     &= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
-    = B \cdot (\tilde{f} * \tilde{g})(k)
+    \\
+    &= B \cdot (\tilde{f} * \tilde{g})(k)
 \end{aligned}$$
 {% include proof/end.html id="proof-fourier" %}
 
@@ -62,45 +67,51 @@ $$\begin{aligned}
 
 ## Laplace transform
 
-For functions $$f(t)$$ and $$g(t)$$ which are only defined for $$t \ge 0$$,
+For functions $$f(t)$$ and $$g(t)$$ that are only defined for $$t \ge 0$$,
 the convolution theorem can also be stated using
 the [Laplace transform](/know/concept/laplace-transform/):
 
 $$\begin{aligned}
-    \boxed{(f * g)(t) = \hat{\mathcal{L}}{}^{-1}\{\tilde{f}(s) \: \tilde{g}(s)\}}
+    \boxed{
+        (f * g)(t)
+        = \hat{\mathcal{L}}{}^{-1}\Big\{ \tilde{f}(s) \: \tilde{g}(s) \Big\}
+    }
 \end{aligned}$$
 
-Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is
-unpleasant, the theorem is often stated using the forward transform
-instead:
+Because the inverse Laplace transform $$\hat{\mathcal{L}}{}^{-1}$$ is usually difficult,
+the theorem is often stated using the forward transform instead:
 
 $$\begin{aligned}
-    \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)}
+    \boxed{
+        \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\}
+        = \tilde{f}(s) \: \tilde{g}(s)
+    }
 \end{aligned}$$
 
 
 {% include proof/start.html id="proof-laplace" -%}
 We expand the left-hand side.
 Note that the lower integration limit is 0 instead of $$-\infty$$,
-because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
+because we choose to set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
 
 $$\begin{aligned}
-    \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty g(t') \: f(t - t') \dd{t'} \Big) e^{-s t} \dd{t}
+    \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\}
+    &= \int_0^\infty \bigg( \int_0^\infty g(t') \: f(t - t') \dd{t'} \bigg) e^{-s t} \dd{t}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \Big) g(t') \dd{t'}
+    &= \int_0^\infty \bigg( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \bigg) \: g(t') \dd{t'}
 \end{aligned}$$
 
 Then we define a new integration variable $$\tau = t - t'$$, yielding:
 
 $$\begin{aligned}
-    \hat{\mathcal{L}}\{(f * g)(t)\}
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \Big) g(t') \dd{t'}
+    \hat{\mathcal{L}}\Big\{ (f * g)(t) \Big\}
+    &= \int_0^\infty \bigg( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \bigg) \: g(t') \dd{t'}
     \\
-    &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \Big) g(t') \: e^{-s t'} \dd{t'}
+    &= \int_0^\infty \bigg( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \bigg) \: g(t') \: e^{-s t'} \dd{t'}
     \\
     &= \int_0^\infty \tilde{f}(s) \: g(t') \: e^{-s t'} \dd{t'}
-    = \tilde{f}(s) \: \tilde{g}(s)
+    \\
+    &= \tilde{f}(s) \: \tilde{g}(s)
 \end{aligned}$$
 {% include proof/end.html id="proof-laplace" %}
 
diff --git a/source/know/concept/electric-dipole-approximation/index.md b/source/know/concept/electric-dipole-approximation/index.md
index 35cf00c..06f0f45 100644
--- a/source/know/concept/electric-dipole-approximation/index.md
+++ b/source/know/concept/electric-dipole-approximation/index.md
@@ -13,20 +13,22 @@ layout: "concept"
 
 Suppose that an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
 is travelling through an atom, and affecting the electrons.
-The general Hamiltonian of an electron in such a wave is given by:
+The general Hamiltonian of an electron in an electromagnetic field is:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \varphi
+    &= \frac{(\vu{P} - q \vb{A})^2}{2 m} + q \Phi
     \\
-    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \varphi
+    &= \frac{\vu{P}{}^2}{2 m} - \frac{q}{2 m} (\vb{A} \cdot \vu{P} + \vu{P} \cdot \vb{A}) + \frac{q^2 \vb{A}^2}{2m} + q \Phi
 \end{aligned}$$
 
-With charge $$q = - e$$,
-canonical momentum operator $$\vu{P} = - i \hbar \nabla$$,
-and magnetic vector potential $$\vb{A}(\vb{x}, t)$$.
-We reduce this by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$,
-so that $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$:
+Where $$q < 0$$ is the electron's charge,
+$$\vu{P} = - i \hbar \nabla$$ is the canonical momentum operator,
+$$\vb{A}$$ is the magnetic vector potential,
+and $$\Phi$$ is the electric scalar potential.
+We start by fixing the Coulomb gauge $$\nabla \cdot \vb{A} = 0$$
+such that $$\vu{P}$$ and $$\vb{A}$$ commute;
+let $$\psi$$ be an arbitrary test function:
 
 $$\begin{aligned}
     \comm{\vb{A}}{\vu{P}} \psi
@@ -35,20 +37,29 @@ $$\begin{aligned}
     &= -i \hbar \vb{A} \cdot (\nabla \psi) + i \hbar \nabla \cdot (\vb{A} \psi)
     \\
     &= i \hbar (\nabla \cdot \vb{A}) \psi
-    = 0
+    \\
+    &= 0
+\end{aligned}$$
+
+Meaning $$\vb{A} \cdot \vu{P} = \vu{P} \cdot \vb{A}$$.
+Furthermore, we assume that $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible,
+so the Hamiltonian is reduced to:
+
+$$\begin{aligned}
+    \hat{H}
+    &\approx \frac{\vu{P}{}^2}{2 m} - \frac{q}{m} \vu{P} \cdot \vb{A} + q \Phi
 \end{aligned}$$
 
-Where $$\psi$$ is an arbitrary test function.
-Assuming $$\vb{A}$$ is so small that $$\vb{A}{}^2$$ is negligible, we split $$\hat{H}$$ as follows,
-where $$\hat{H}_1$$ can be regarded as a perturbation to $$\hat{H}_0$$:
+We now split $$\hat{H}$$ like so,
+where $$\hat{H}_1$$ can be regarded as a perturbation to the "base" $$\hat{H}_0$$:
 
 $$\begin{aligned}
     \hat{H}
     = \hat{H}_0 + \hat{H}_1
-    \qquad \quad
+    \qquad\qquad
     \hat{H}_0
-    \equiv \frac{\vu{P}{}^2}{2 m} + q \varphi
-    \qquad \quad
+    \equiv \frac{\vu{P}{}^2}{2 m} + q \Phi
+    \qquad\qquad
     \hat{H}_1
     \equiv - \frac{q}{m} \vu{P} \cdot \vb{A}
 \end{aligned}$$
@@ -56,14 +67,16 @@ $$\begin{aligned}
 In an electromagnetic wave, $$\vb{A}$$ is oscillating sinusoidally in time and space:
 
 $$\begin{aligned}
-    \vb{A}(\vb{x}, t) = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t)
+    \vb{A}(\vb{x}, t)
+    = \vb{A}_0 \sin(\vb{k} \cdot \vb{x} - \omega t)
 \end{aligned}$$
 
 Mathematically, it is more convenient to represent this with a complex exponential,
 whose real part should be taken at the end of the calculation:
 
 $$\begin{aligned}
-    \vb{A}(\vb{x}, t) = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
+    \vb{A}(\vb{x}, t)
+    = - i \vb{A}_0 \exp(i \vb{k} \cdot \vb{x} - i \omega t)
 \end{aligned}$$
 
 The corresponding perturbative [electric field](/know/concept/electric-field/) $$\vb{E}$$ is then given by:
@@ -75,10 +88,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $$\vb{E}_0 = \omega \vb{A}_0$$.
-Let us restrict ourselves to visible light,
-whose wavelength $$2 \pi / |\vb{k}| \sim 10^{-6} \:\mathrm{m}$$.
-Meanwhile, an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$,
-so $$\vb{k} \cdot \vb{x}$$ is negligible:
+Light in and around the visible spectrum
+has a wavelength $$2 \pi / |\vb{k}| \sim 10^{-7} \:\mathrm{m}$$,
+while an atomic orbital is several Bohr radii $$\sim 10^{-10} \:\mathrm{m}$$,
+so $$\vb{k} \cdot \vb{x}$$ is very small. Therefore:
 
 $$\begin{aligned}
     \boxed{
@@ -96,14 +109,13 @@ and the electron quantum-mechanically.
 
 Next, we want to rewrite $$\hat{H}_1$$
 to use the electric field $$\vb{E}$$ instead of the potential $$\vb{A}$$.
-To do so, we use that $$\vu{P} = m \: \idv{\vu{x}}{t}$$
+To do so, we use that momentum $$\vu{P} \equiv m \: \idv{\vu{x}}{t}$$
 and evaluate this in the [interaction picture](/know/concept/interaction-picture/):
 
 $$\begin{aligned}
     \vu{P}
-    = m \idv{\vu{x}}{t}
+    &= m \dv{\vu{x}}{t}
     = m \frac{i}{\hbar} \comm{\hat{H}_0}{\vu{x}}
-    = m \frac{i}{\hbar} (\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0)
 \end{aligned}$$
 
 Taking the off-diagonal inner product with
@@ -111,34 +123,37 @@ the two-level system's states $$\Ket{1}$$ and $$\Ket{2}$$ gives:
 
 $$\begin{aligned}
     \matrixel{2}{\vu{P}}{1}
-    = m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
-    = m i \omega_0 \matrixel{2}{\vu{x}}{1}
+    &= m \frac{i}{\hbar} \matrixel{2}{\hat{H}_0 \vu{x} - \vu{x} \hat{H}_0}{1}
+    \\
+    &= m i \omega_0 \matrixel{2}{\vu{x}}{1}
 \end{aligned}$$
 
-Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$,
-where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap,
+Where $$\omega_0 \equiv (E_2 \!-\! E_1) / \hbar$$ is the resonance of the energy gap,
 close to which we assume that $$\vb{A}$$ and $$\vb{E}$$ are oscillating, i.e. $$\omega \approx \omega_0$$.
-We thus get:
+Therefore, $$\vu{P} / m = i \omega_0 \vu{x}$$, so we get:
 
 $$\begin{aligned}
     \hat{H}_1(t)
     &= - \frac{q}{m} \vu{P} \cdot \vb{A}
-    = - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t)
     \\
-    &\approx - q \vu{x} \cdot \vb{E}_0 \exp(- i \omega t)
-    = - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
+    &= - (- i i) q \omega_0 \vu{x} \cdot \vb{A}_0 \exp(- i \omega t)
+    \\
+    &\approx - \vu{d} \cdot \vb{E}_0 \exp(- i \omega t)
 \end{aligned}$$
 
-Where $$\vu{d} \equiv q \vu{x} = - e \vu{x}$$ is
+Where $$\vu{d} \equiv q \vu{x}$$ is
 the **transition dipole moment operator** of the electron,
-hence the name **electric dipole approximation**.
+hence the name *electric dipole approximation*.
 Finally, we take the real part, yielding:
 
 $$\begin{aligned}
     \boxed{
-        \hat{H}_1(t)
-        = - \vu{d} \cdot \vb{E}(t)
-        = - q \vu{x} \cdot \vb{E}_0 \cos(\omega t)
+        \begin{aligned}
+            \hat{H}_1(t)
+            &= - \vu{d} \cdot \vb{E}(t)
+            \\
+            &= - q \vu{x} \cdot \vb{E}_0 \cos(\omega t)
+        \end{aligned}
     }
 \end{aligned}$$
 
diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md
index 9951883..758e12f 100644
--- a/source/know/concept/ghz-paradox/index.md
+++ b/source/know/concept/ghz-paradox/index.md
@@ -11,13 +11,13 @@ layout: "concept"
 
 The **Greenberger-Horne-Zeilinger** or **GHZ paradox**
 is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/)
-that does not use inequalities,
-but the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$ instead,
+that does not use inequalities, but instead
+the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$:
 
 $$\begin{aligned}
     \boxed{
         \ket{\mathrm{GHZ}}
-        = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big)
+        \equiv \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big)
     }
 \end{aligned}$$
 
@@ -49,8 +49,8 @@ $$\begin{aligned}
 
 In other words, the GHZ state is a simultaneous eigenstate of these composite operators,
 with eigenvalues $$+1$$ and $$-1$$, respectively.
-Let us introduce two more operators in the same way,
-so that we have a set of four observables,
+Let us do the same for two more operators,
+so that we have a set of four observables
 for which $$\ket{\mathrm{GHZ}}$$ gives these eigenvalues:
 
 $$\begin{aligned}
diff --git a/source/know/concept/hydrogen-atom/index.md b/source/know/concept/hydrogen-atom/index.md
index a8443dd..9692d10 100644
--- a/source/know/concept/hydrogen-atom/index.md
+++ b/source/know/concept/hydrogen-atom/index.md
@@ -525,7 +525,7 @@ $$\begin{aligned}
     \boxed{
         \frac{1}{\lambda_0}
         = \frac{\omega}{2 \pi c}
-        = \mathcal{R}\Big( \frac{1}{n_i^2} - \frac{1}{n_f^2} \Big)
+        = R_\mathrm{H} \bigg( \frac{1}{n_i^2} - \frac{1}{n_f^2} \bigg)
     }
 \end{aligned}$$
 
@@ -554,7 +554,7 @@ and is sometimes used as a unit in calculations:
 
 $$\begin{aligned}
     \mathrm{Ry}
-    = 2 \pi \hbar c R_\mathrm{H}
+    \equiv 2 \pi \hbar c R_\mathrm{H}
     = |E_1|
     = \frac{\mu}{2 \hbar^2} \bigg( \frac{q^2}{4 \pi \varepsilon_0} \bigg)^2
     \approx 13.61 \:\mathrm{eV}
diff --git a/source/know/concept/laser-rate-equations/index.md b/source/know/concept/laser-rate-equations/index.md
index 1f42f73..c81f02b 100644
--- a/source/know/concept/laser-rate-equations/index.md
+++ b/source/know/concept/laser-rate-equations/index.md
@@ -97,14 +97,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Typically, $$\gamma_\perp$$ is much larger than the rate of any other decay process,
-in which case $$\ipdv{}{\vb{P}0^{+}\!}{t}$$ is negligible compared to $$\gamma_\perp \vb{P}_0^{+}$$.
+in which case $$\ipdv{\vb{P}_0^{+}\!}{t}$$ is negligible compared to $$\gamma_\perp \vb{P}_0^{+}$$.
 Effectively, this means that the polarization $$\vb{P}_0^{+}$$
 near-instantly follows the electric field $$\vb{E}^{+}\!$$.
-Setting $$\ipdv{}{\vb{P}0^{+}\!}{t} \approx 0$$, the second MBE becomes:
+Setting $$\ipdv{\vb{P}_0^{+}\!}{t} \approx 0$$, the second MBE becomes:
 
 $$\begin{aligned}
     \vb{P}^{+}
-    = -\frac{i |g|^2}{\hbar (\gamma_\perp + i (\omega_0 \!-\! \omega))} \vb{E}^{+} D
+    = -\frac{i |g|^2}{\hbar (\gamma_\perp + i (\omega_0 - \omega))} \vb{E}^{+} D
     = \frac{|g|^2 \gamma(\omega)}{\hbar \gamma_\perp} \vb{E}^{+} D
 \end{aligned}$$
 
@@ -137,7 +137,7 @@ $$\begin{aligned}
     &= i (\omega - \Omega) \vb{E}_0^{+} + i \frac{|g|^2 \omega \gamma(\omega)}{2 \hbar \varepsilon_0 \gamma_\perp n^2} \vb{E}_0^{+} D
 \end{aligned}$$
 
-Next, we insert our ansatz for $$\vb{E}^{+}\!$$ and $$\vb{P}^{+}\!$$
+Next, we insert our ansatz for $$\vb{E}^{+}$$ and $$\vb{P}^{+}$$
 into the third MBE, and rewrite $$\vb{P}_0^{+}$$ as above.
 Using our identity for $$\gamma(\omega)$$,
 and the fact that $$\vb{E}_0^{+} \cdot \vb{E}_0^{-} = |\vb{E}|^2$$, we find:
@@ -293,11 +293,13 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             \pdv{N_p}{t}
-            &= - (\gamma_\mathrm{out} + \gamma_\mathrm{abs} + \gamma_\mathrm{loss}) N_p + \gamma_\mathrm{spon} N_e + G_\mathrm{stim} N_p N_e
+            &= - (\gamma_\mathrm{out} + \gamma_\mathrm{abs} + \gamma_\mathrm{loss}) N_p
+            + \gamma_\mathrm{spon} N_e + G_\mathrm{stim} N_p N_e
             \\
             \pdv{N_e}{t}
             &= R_\mathrm{pump} + \gamma_\mathrm{abs} N_p
-            - (\gamma_\mathrm{spon} + \gamma_\mathrm{n.r.} + \gamma_\mathrm{leak}) N_e - G_\mathrm{stim} N_p N_e
+            - (\gamma_\mathrm{spon} + \gamma_\mathrm{n.r.} + \gamma_\mathrm{leak}) N_e
+            - G_\mathrm{stim} N_p N_e
         \end{aligned}
     }
 \end{aligned}$$
diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
index c4fdeb4..d09613f 100644
--- a/source/know/concept/legendre-transform/index.md
+++ b/source/know/concept/legendre-transform/index.md
@@ -10,48 +10,49 @@ layout: "concept"
 
 The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$,
 which depends only on the derivative $$f'(x)$$ of $$f(x)$$,
-and from which the original function $$f(x)$$ can be reconstructed.
+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/).
 
 Let us choose an arbitrary point $$x_0 \in [a, b]$$ in the domain of $$f(x)$$.
 Consider a line $$y(x)$$ tangent to $$f(x)$$ at $$x = x_0$$,
-which must have slope $$f'(x_0)$$, and intersects the $$y$$-axis at $$-C$$:
+which has slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$y = -C$$:
 
 $$\begin{aligned}
     y(x)
-    = f'(x_0) (x - x_0) + f(x_0)
-    = f'(x_0) \: x - C
+    &= f'(x_0) (x - x_0) + f(x_0)
+    \\
+    &= f'(x_0) \: x - C
 \end{aligned}$$
 
 Where $$C \equiv f'(x_0) \: x_0 - f(x_0)$$.
-We now define the Legendre transform $$L(f')$$ such that
-for all $$x_0 \in [a, b]$$ we have $$L(f'(x_0)) = C$$
+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 $$x$$:
 
 $$\begin{aligned}
     L(f'(x))
-    = f'(x) \: x - f(x)
+    &= f'(x) \: x - f(x)
 \end{aligned}$$
 
 We want this function to depend only on the derivative $$f'$$,
 but currently $$x$$ still appears here as a variable.
-We fix this problem in the easiest possible way:
+We solve this problem in the easiest possible way:
 by assuming that $$f'(x)$$ is invertible for all $$x \in [a, b]$$.
 If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is given by:
 
 $$\begin{aligned}
     \boxed{
         L(f')
-        = f' \: x(f') - f(x(f'))
+        = f' \: x(f') - f\big(x(f')\big)
     }
 \end{aligned}$$
 
 The only requirement for the existence of the Legendre transform is thus
 the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$,
-which can only be true if $$f(x)$$ is either convex or concave,
+which is only satisfied if $$f(x)$$ is either convex or concave,
 meaning its derivative $$f'(x)$$ is monotonic.
 
 The derivative of $$L(f')$$ with respect to $$f'$$ is simply $$x(f')$$.
@@ -72,9 +73,11 @@ To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$:
 
 $$\begin{aligned}
     g(L')
-    = L' \: f'(L') - L(f'(L'))
-    = x(f') \: f' - f' \: x(f') + f(x(f'))
-    = f(x)
+    &= L' \: f'(L') - L(f'(L'))
+    \\
+    &= x(f') \: f' - f' \: x(f') + f(x(f'))
+    \\
+    &= f(x)
 \end{aligned}$$
 
 Moreover, a Legendre transform is always invertible,
@@ -84,7 +87,7 @@ so a proof is:
 
 $$\begin{aligned}
     L''(f')
-    = \dv{}{f'} \Big( \dv{L}{f'} \Big)
+    &= \dv{}{f'} \Big( \dv{L}{f'} \Big)
     = \dv{x(f')}{f'}
     = \dv{x}{f'(x)}
     = \frac{1}{f''(x)}
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index a7ce0bf..b2db490 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -17,9 +17,11 @@ where $$A$$, $$B$$, and $$s$$ are constants from the FT's definition:
 $$\begin{aligned}
     \boxed{
         \begin{aligned}
-            \inprod{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+            \inprod{f(x)}{g(x)}
+            &= \frac{2 \pi B^2}{|s|} \inprod{\tilde{f}(k)}{\tilde{g}(k)}
             \\
-            \inprod{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \Inprod{f(x)}{g(x)}
+            \inprod{\tilde{f}(k)}{\tilde{g}(k)}
+            &= \frac{2 \pi A^2}{|s|} \inprod{f(x)}{g(x)}
         \end{aligned}
     }
 \end{aligned}$$
@@ -71,8 +73,8 @@ $$\begin{aligned}
 
 
 For this reason, physicists like to define the Fourier transform
-with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely
-conserves the functions' normalization.
+with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$,
+because then it nicely preserves the functions' normalization.
 
 
 
diff --git a/source/know/concept/salt-equation/index.md b/source/know/concept/salt-equation/index.md
index 77f4755..d7f8ef3 100644
--- a/source/know/concept/salt-equation/index.md
+++ b/source/know/concept/salt-equation/index.md
@@ -268,9 +268,10 @@ it emits photons without any significant light amplification taking place.
 Upon gradually increasing the pump $$D_0$$ in the active region,
 all $$\Imag(k_n)$$ become less negative,
 until one hits the real axis $$\Imag(k_n) = 0$$,
-at which point that mode starts lasing,
-i.e. the Light gets Amplified by [Stimulated Emission](/know/concept/einstein-coefficients/) (LASE).
-After that, $$D_0$$ can be increased even further until some other $$k_n$$ become real.
+at which point that mode starts *lasing*:
+its Light gets Amplified by [Stimulated Emission](/know/concept/einstein-coefficients/) of Radiation (LASER).
+After that, $$D_0$$ can be increased even further until some other $$k_n$$ become real,
+so there are multiple active modes competing for charge carriers.
 
 Below threshold (i.e. before any mode is lasing), the problem is linear in $$\Psi_n$$,
 but above threshold it is nonlinear via $$h(\vb{x})$$.
diff --git a/source/know/concept/sturm-liouville-theory/index.md b/source/know/concept/sturm-liouville-theory/index.md
index d7984b5..117d893 100644
--- a/source/know/concept/sturm-liouville-theory/index.md
+++ b/source/know/concept/sturm-liouville-theory/index.md
@@ -151,9 +151,9 @@ $$\begin{aligned}
     \boxed{
         \begin{aligned}
             \hat{L}_\mathrm{SL} \{u(x)\}
-            &= \hat{L}_\mathrm{SL}^\dagger \{u(x)\}
-            \\
             &= \Big( p(x) \: u'(x) \Big)' + q(x) \: u(x)
+            \\
+            &= \hat{L}_\mathrm{SL}^\dagger \{u(x)\}
         \end{aligned}
     }
 \end{aligned}$$