Improve knowledge base

7 files changed, 194 insertions, 136 deletions

diff --git a/source/know/concept/diffie-hellman-key-exchange/index.md b/source/know/concept/diffie-hellman-key-exchange/index.md
index 4735209..3525881 100644
--- a/source/know/concept/diffie-hellman-key-exchange/index.md
+++ b/source/know/concept/diffie-hellman-key-exchange/index.md
@@ -7,18 +7,19 @@ categories:
 layout: "concept"
 ---
 
-In cryptography, the **Diffie-Hellman key exchange** is a method
-for two parties to securely agree on an encryption key,
-when they can only communicate over an insecure channel.
+In cryptography, the **Diffie-Hellman key exchange** is a method for two parties,
+who can only communicate over an insecure channel,
+to securely agree on an encryption key.
 
 The fundamental assumption of the Diffie-Hellman scheme,
 upon which its security rests,
 is that the following function $$f(n)$$ is a **trapdoor function**,
 which means that calculating $$f$$ is easy,
-but its inverse $$f^{-1}$$ is extremely hard to find:
+but its inverse $$f^{-1}$$ is extremely hard:
 
 $$\begin{aligned}
-    f(n) = g^n \bmod p
+    f(n)
+    \equiv g^n \bmod p
 \end{aligned}$$
 
 Where $$n$$ is a natural number, and $$p$$ is a prime.
@@ -39,9 +40,11 @@ Alice and Bob each choose a secret number from $$\{0, ..., p \!-\! 2\}$$, respec
 and then privately calculate $$A$$ and $$B$$ as follows:
 
 $$\begin{aligned}
-    A = g^a \bmod p
-    \qquad \quad
-    B = g^b \bmod p
+    A
+    \equiv g^a \bmod p
+    \qquad \qquad
+    B
+    \equiv g^b \bmod p
 \end{aligned}$$
 
 Finally, they transmit these numbers $$A$$ and $$B$$
@@ -50,13 +53,16 @@ and then each side calculates $$k$$, which is the desired secret key:
 
 $$\begin{aligned}
     \boxed{
-        k = A^b \bmod p = B^a \bmod p = g^{ab} \bmod p
+        k
+        \equiv A^b \bmod p
+        = B^a \bmod p
+        = g^{ab} \bmod p
     }
 \end{aligned}$$
 
 The point is that $$k$$ includes both $$a$$ *and* $$b$$,
 but each side only needs to know *either* $$a$$ *or* $$b$$.
-And, due to the trapdoor assumption,
+Thanks to the trapdoor assumption,
 the eavesdropper knows $$A$$ and $$B$$,
 but cannot recover $$a$$ or $$b$$.
 
diff --git a/source/know/concept/fabry-perot-cavity/index.md b/source/know/concept/fabry-perot-cavity/index.md
index c88e62d..d5ea0ea 100644
--- a/source/know/concept/fabry-perot-cavity/index.md
+++ b/source/know/concept/fabry-perot-cavity/index.md
@@ -95,10 +95,10 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-We want non-trivial solutions, where we
-cannot simply satisfy the system by setting $$A_1$$, $$A_2$$, $$A_3$$ and
-$$A_4$$; this constraint will give us an equation for $$k_m$$. Therefore, we
-demand that the system matrix is singular, i.e. its determinant is zero:
+We do not want to simply satisfy this equation
+by 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:
 
 $$\begin{aligned}
     0 =
@@ -180,7 +180,7 @@ $$\begin{aligned}
     \end{bmatrix}
 \end{aligned}$$
 
-Again, we demand that the determinant is zero, in order to get non-trivial solutions:
+Again, we demand that the determinant is zero in order to get non-trivial solutions:
 
 $$\begin{aligned}
     0
@@ -225,8 +225,8 @@ $$\begin{aligned}
 
 Note that we have not demanded continuity of the electric field.
 This is because the mirrors are infinitely thin "magic" planes;
-had we instead used the full mirror structure,
-then we would have demanded continuity, as you maybe expected.
+had we instead included the full microscopic mirror structure,
+then we would have demanded continuity as before.
 
 
 
diff --git a/source/know/concept/ficks-laws/index.md b/source/know/concept/ficks-laws/index.md
index b205af9..8d5da7d 100644
--- a/source/know/concept/ficks-laws/index.md
+++ b/source/know/concept/ficks-laws/index.md
@@ -14,6 +14,7 @@ A diffusion process that obeys Fick's laws is called **Fickian**,
 as opposed to **non-Fickian** or **anomalous diffusion**.
 
 
+
 ## Fick's first law
 
 **Fick's first law** states that diffusing matter
@@ -21,13 +22,14 @@ moves from regions of high concentration to regions of lower concentration,
 at a rate proportional to the difference in concentration.
 
 Let $$\vec{J}$$ be the **diffusion flux** (with unit $$\mathrm{m}^{-2} \mathrm{s}^{-1}$$),
-whose magnitude and direction describe the "flow" of diffusing matter.
+whose magnitude and direction describes the "flow" of diffusing matter.
 Formally, Fick's first law predicts that the flux
 is proportional to the gradient of the concentration $$C$$ (with unit $$\mathrm{m}^{-3}$$):
 
 $$\begin{aligned}
     \boxed{
-        \vec{J} = - D \: \nabla C
+        \vec{J}
+        = - D \: \nabla C
     }
 \end{aligned}$$
 
@@ -43,6 +45,7 @@ but they say more about those systems
 than about diffusion in general.
 
 
+
 ## Fick's second law
 
 To derive **Fick's second law**, we demand that matter is conserved,
@@ -65,8 +68,7 @@ $$\begin{aligned}
     = - \int_V \nabla \cdot \vec{J} \dd{V}
 \end{aligned}$$
 
-For comparison, we differentiate the definition of $$M$$,
-and exploit that the integral ignores $$t$$:
+For comparison, we can also just differentiate the definition of $$M$$ directly:
 
 $$\begin{aligned}
     \dv{M}{t}
@@ -74,7 +76,8 @@ $$\begin{aligned}
     = \int_V \pdv{C}{t} \dd{V}
 \end{aligned}$$
 
-Both $$\idv{M}{t}$$ are equal, so stripping the integrals leads to this **continuity equation**:
+Above, we have two valid expressions for $$\idv{M}{t}$$,
+which must be equal, so stripping the integrals leads to this **continuity equation**:
 
 $$\begin{aligned}
     \pdv{C}{t}
@@ -101,6 +104,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## Fundamental solution
 
 Fick's second law has exact solutions for many situations,
@@ -110,11 +114,12 @@ where the initial concentration $$C(x, 0)$$ is
 a [Dirac delta function](/know/concept/dirac-delta-function/):
 
 $$\begin{aligned}
-    C(x, 0) = \delta(x - x_0)
+    C(x, 0)
+    = \delta(x - x_0)
 \end{aligned}$$
 
-According to Fick's second law,
-the concentration's time evolution of $$C$$ turns out to be:
+By solving Fick's second law with this initial condition,
+$$C$$'s time evolution turns out to be:
 
 $$\begin{aligned}
     H(x - x_0, t)
diff --git a/source/know/concept/gram-schmidt-method/index.md b/source/know/concept/gram-schmidt-method/index.md
index a62522e..1f5a0bd 100644
--- a/source/know/concept/gram-schmidt-method/index.md
+++ b/source/know/concept/gram-schmidt-method/index.md
@@ -58,8 +58,7 @@ into an orthonormal set $$\ket{n_1}, \ket{n_2}, ...$$ as follows
     to all previous $$\ket{n_m}$$ with $$m < k$$.
     This is known as the **modified Gram-Schmidt method**.
     
-3.  Normalize the resulting ortho*gonal* vector $$\ket{g_j}$$ to make it
-    ortho*normal*:
+3.  Normalize the resulting ortho*gonal* vector $$\ket{g_j}$$ to make it ortho*normal*:
     
     $$\begin{aligned}
         \ket{n_j}
diff --git a/source/know/concept/legendre-transform/index.md b/source/know/concept/legendre-transform/index.md
index 51c003e..c4fdeb4 100644
--- a/source/know/concept/legendre-transform/index.md
+++ b/source/know/concept/legendre-transform/index.md
@@ -9,76 +9,90 @@ layout: "concept"
 ---
 
 The **Legendre transform** of a function $$f(x)$$ is a new function $$L(f')$$,
-which depends only on the derivative $$f'(x)$$ of $$f(x)$$, and from which
-the original function $$f(x)$$ can be reconstructed. The point is,
-analogously to other transforms (e.g. [Fourier](/know/concept/fourier-transform/)),
-that $$L(f')$$ contains the same information as $$f(x)$$, just in a different form.
+which depends only on the derivative $$f'(x)$$ of $$f(x)$$,
+and from which the original function $$f(x)$$ can be reconstructed.
+The point is 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 has
-a slope $$f'(x_0)$$ and intersects the $$y$$-axis at $$-C$$:
+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 $$-C$$:
 
 $$\begin{aligned}
-    y(x) = f'(x_0) (x - x_0) + f(x_0) = f'(x_0) x - C
+    y(x)
+    = f'(x_0) (x - x_0) + f(x_0)
+    = f'(x_0) \: x - C
 \end{aligned}$$
 
-The Legendre transform $$L(f')$$ is defined such that $$L(f'(x_0)) = C$$
-(or sometimes $$-C$$) for all $$x_0 \in [a, b]$$,
-where $$C$$ corresponds to the tangent line at $$x = x_0$$. This yields:
+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 $$L(f'(x_0)) = C$$
+(some authors use $$-C$$ instead).
+Renaming $$x_0$$ to $$x$$:
 
 $$\begin{aligned}
-    L(f'(x)) = f'(x) \: x - f(x)
+    L(f'(x))
+    = f'(x) \: x - f(x)
 \end{aligned}$$
 
-We want this function to depend only on the derivative $$f'$$, but
-currently $$x$$ still appears here as a variable. We fix that problem in
-the easiest possible way: by assuming that $$f'(x)$$ is invertible for all
-$$x \in [a, b]$$. If $$x(f')$$ is the inverse of $$f'(x)$$, then $$L(f')$$ is
-given by:
+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:
+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:
 
 $$\begin{aligned}
     \boxed{
-        L(f') = f' \: x(f') - f(x(f'))
+        L(f')
+        = f' \: x(f') - f(x(f'))
     }
 \end{aligned}$$
 
 The only requirement for the existence of the Legendre transform is thus
-the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$, which can
-only be true if $$f(x)$$ is either convex or concave, i.e. its derivative
-$$f'(x)$$ is monotonic.
+the invertibility of $$f'(x)$$ in the target interval $$[a,b]$$,
+which can only be true if $$f(x)$$ is either convex or concave,
+meaning its derivative $$f'(x)$$ is monotonic.
 
-Crucially, the derivative of $$L(f')$$ with respect to $$f'$$ is simply
-$$x(f')$$. In other words, the roles of $$f'$$ and $$x$$ are switched by the
-transformation: the coordinate becomes the derivative and vice versa.
-This is demonstrated here:
+The derivative of $$L(f')$$ with respect to $$f'$$ is simply $$x(f')$$.
+In other words, the roles of $$f'$$ and $$x$$ are switched by the transformation:
+the coordinate becomes the derivative and vice versa:
 
 $$\begin{aligned}
     \boxed{
-        \dv{L}{f'} = \dv{x}{f'} \: f' + x(f') - \dv{f}{x} \dv{x}{f'} = x(f')
+        \dv{L}{f'}
+        = f' \dv{x}{f'} + x(f') - f' \dv{x}{f'}
+        = x(f')
     }
 \end{aligned}$$
 
-Furthermore, Legendre transformation is an *involution*, meaning it is
-its own inverse. Let $$g(L')$$ be the Legendre transform of $$L(f')$$:
+Furthermore, Legendre transformation is an *involution*,
+meaning it is its own inverse.
+To show this, let $$g(L')$$ be the Legendre transform of $$L(f')$$:
 
 $$\begin{aligned}
-    g(L') = L' \: f'(L') - L(f'(L'))
-    = x(f') \: f' - f' \: x(f') + f(x(f')) = f(x)
+    g(L')
+    = L' \: f'(L') - L(f'(L'))
+    = x(f') \: f' - f' \: x(f') + f(x(f'))
+    = f(x)
 \end{aligned}$$
 
-Moreover, the inverse of a (forward) transform always exists, because
-the Legendre transform of a convex function is itself convex. Convexity
-of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$, which yields
-the following proof:
+Moreover, a Legendre transform is always invertible,
+because the transform of a convex function is itself convex.
+Convexity of $$f(x)$$ means that $$f''(x) > 0$$ for all $$x \in [a, b]$$,
+so a proof is:
 
 $$\begin{aligned}
     L''(f')
+    = \dv{}{f'} \Big( \dv{L}{f'} \Big)
     = \dv{x(f')}{f'}
     = \dv{x}{f'(x)}
     = \frac{1}{f''(x)}
     > 0
 \end{aligned}$$
 
+And an analogous proof exists for concave functions where $$f''(x) < 0$$.
+
 Legendre transformation is important in physics,
 since it connects [Lagrangian](/know/concept/lagrangian-mechanics/)
 and [Hamiltonian](/know/concept/hamiltonian-mechanics/) mechanics to each other.
diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md
index aef8379..0e04455 100644
--- a/source/know/concept/matsubara-sum/index.md
+++ b/source/know/concept/matsubara-sum/index.md
@@ -14,60 +14,88 @@ which notably appears as the inverse
 [Matsubara Green's function](/know/concept/matsubara-greens-function/):
 
 $$\begin{aligned}
-    S_{B,F}
-    = \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
+    \boxed{
+        S_{B,F}
+        \equiv \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau}
+    }
 \end{aligned}$$
 
-Where $$i \omega_n$$ are the Matsubara frequencies
-for bosons ($$B$$) or fermions ($$F$$),
-and $$g(z)$$ is a function on the complex plane
-that is [holomorphic](/know/concept/holomorphic-function/)
+$$g(z)$$ is a *meromorphic* function on the complex frequency plane,
+i.e. it is [holomorphic](/know/concept/holomorphic-function/)
 except for a known set of simple poles,
-and $$\tau$$ is a real parameter
-(e.g. the [imaginary time](/know/concept/imaginary-time/))
-satisfying $$-\hbar \beta < \tau < \hbar \beta$$.
+and $$\tau \in [-\hbar \beta, \hbar \beta]$$ is a real parameter.
+The Matsubara frequencies $$i \omega_n$$ are defined as follows
+for bosons (subscript $$B$$) or fermions (subscript $$F$$):
+
+$$\begin{aligned}
+    \omega_n \equiv
+    \begin{cases}
+        \displaystyle\frac{2 n \pi}{\hbar \beta}
+        & \mathrm{bosons}
+        \\
+        \displaystyle\frac{(2 n + 1) \pi}{\hbar \beta}
+        & \mathrm{fermions}
+    \end{cases}
+\end{aligned}$$
 
-Now, consider the following integral
-over a (for now) unspecified counter-clockwise contour $$C$$,
-with a (for now) unspecified weighting function $$h(z)$$:
+How do we evaluate Matsubara sums?
+Given a counter-clockwise closed contour $$C$$,
+recall that the [residue theorem](/know/concept/residue-theorem/)
+turns an integral over $$C$$ into a sum of the residues
+of all the integrand's simple poles $$p_g$$ that are enclosed by $$C$$:
 
 $$\begin{aligned}
-    \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
-    = \sum_{z_p} e^{z_p \tau} \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) h(z) \big)
+    \oint_C \frac{g(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
+    = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: e^{z \tau} \Big\}
+    = \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: e^{p_g \tau}
 \end{aligned}$$
 
-Where we have applied the [residue theorem](/know/concept/residue-theorem/)
-to get a sum over all simple poles $$z_p$$
-of either $$g$$ or $$h$$ (but not both) enclosed by $$C$$.
-Clearly, we could make this look like a Matsubara sum,
-if we choose $$h$$ such that it has poles at $$i \omega_n$$.
+Now, the trick is to manipulate this relation
+until a Matsubara sum appears on the right.
 
-Therefore, we choose the weighting function $$h(z)$$ as follows,
-where $$n_B(z)$$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/),
-and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/):
+Let us introduce a (for now) unspecified weight function $$h(z)$$,
+which crucially does not share any simple poles with $$g(z)$$,
+so $$\{p_g\} \cap \{p_h\} = \emptyset$$.
+This constraint allows us to split the sum:
+
+$$\begin{aligned}
+    \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
+    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \: h(z) \: e^{z \tau} \Big\}
+    + \sum_{p_h} \underset{z \to p_h}{\mathrm{Res}}\Big\{ g(z) \: h(z) \: e^{z \tau} \Big\}
+    \\
+    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: h(p_g) \: e^{p_g \tau}
+    + \sum_{p_h} g(p_h) \: \underset{z \to p_h}{\mathrm{Res}}\Big\{ h(z) \Big\} \: e^{p_h \tau}
+\end{aligned}$$
+
+Here, we could make the rightmost term look like a Matsubara sum
+if we choose $$h$$ such that it has poles at $$i \omega_n$$.
+We make the following choice,
+where $$n_B(z)$$ is the [Bose-Einstein distribution](/know/concept/bose-einstein-distribution/) for bosons,
+and $$n_F(z)$$ is the [Fermi-Dirac distribution](/know/concept/fermi-dirac-distribution/) for fermions:
 
 $$\begin{aligned}
     h(z)
-    =
+    \equiv
     \begin{cases}
         n_{B,F}(z) & \mathrm{if}\; \tau \ge 0
         \\
         -n_{B,F}(-z) & \mathrm{if}\; \tau \le 0
     \end{cases}
-    \qquad \qquad
-    n_{B,F}(z)
-    = \frac{1}{e^{\hbar \beta z} \mp 1}
 \end{aligned}$$
 
-The distinction between the signs of $$\tau$$ is needed
-to ensure that the integrand $$h(z) e^{z \tau}$$ decays for $$|z| \to \infty$$,
-both for $$\Real(z) > 0$$ and $$\Real(z) < 0$$.
-This choice of $$h$$ indeed has poles at the respective
+The distinction between the signs of $$\tau$$ is necessary
+to ensure that $$h(z) \: e^{z \tau} \to 0$$ for all $$z$$ when $$|z| \to \infty$$
+(take a moment to convince yourself of this).
+The sign flip for $$\tau \le 0$$ is also needed,
+as negating the argument negates the residues
+$$\mathrm{Res}\{ n_{B,F}(-i \omega_n) \} = -\mathrm{Res}\{ n_{B,F}(i \omega_n) \}$$.
+
+Indeed, this choice of $$h$$ has poles at the respective
 Matsubara frequencies $$i \omega_n$$ of bosons and fermions,
-and the residues are:
+and the residues are given by:
 
 $$\begin{aligned}
-    \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_B(z) \big)
+    \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_B(z) \Big\}
     &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} - 1} \bigg)
     = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} - 1} \bigg)
     \\
@@ -75,7 +103,7 @@ $$\begin{aligned}
     = \lim_{\eta \to 0}\!\bigg( \frac{\eta}{1 + \hbar \beta \eta - 1} \bigg)
     = \frac{1}{\hbar \beta}
     \\
-    \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_F(z) \big)
+    \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_F(z) \Big\}
     &= \lim_{z \to i \omega_n}\!\bigg( \frac{z - i \omega_n}{e^{\hbar \beta z} + 1} \bigg)
     = \lim_{\eta \to 0}\!\bigg( \frac{i \omega_n + \eta - i \omega_n}{e^{i \hbar \beta \omega_n} e^{\hbar \beta \eta} + 1} \bigg)
     \\
@@ -84,37 +112,36 @@ $$\begin{aligned}
     = - \frac{1}{\hbar \beta}
 \end{aligned}$$
 
-In the definition of $$h$$, the sign flip for $$\tau \le 0$$
-is introduced because negating the argument also negates the residues,
-i.e. $$\mathrm{Res}\big( n_F(-z) \big) = -\mathrm{Res}\big( n_F(z) \big)$$.
-With this $$h$$, our contour integral can be rewritten as follows:
+With this, our contour integral can now be rewritten as follows:
 
 $$\begin{aligned}
-    \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
-    &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
-    + \sum_{i \omega_n} e^{i \omega_n \tau} g(i \omega_n) \: \underset{z \to i \omega_n}{\mathrm{Res}}\!\big( n_{B,F}(z) \big)
+    \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
+    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
+    + \sum_{i \omega_n} g(i \omega_n) \underset{z \to i \omega_n}{\mathrm{Res}}\!\Big\{ n_{B,F}(z) \Big\} \: e^{i \omega_n \tau}
     \\
-    &= \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
-    \pm \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
+    &= \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
+    \pm \frac{1}{\hbar \beta} \sum_{n = -\infty}^\infty g(i \omega_n) \: e^{i \omega_n \tau}
 \end{aligned}$$
 
-Where $$+$$ is for bosons, and $$-$$ for fermions.
-Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$,
-for which we isolate, yielding:
+Where the top sign ($$+$$) is for bosons,
+and the bottom sign ($$-$$) is for fermions.
+Here, we recognize the last term as the Matsubara sum $$S_{F,B}$$.
+Isolating for that yields:
 
 $$\begin{aligned}
     S_{B,F}
-    = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{z \to z_p}{\mathrm{Res}}\big( g(z) \big)
-    \pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
+    = \mp \sum_{p_g} \underset{z \to p_g}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
+    \pm \oint_C \frac{g(z) \: h(z) \: e^{z \tau}}{i 2 \pi} \dd{z}
 \end{aligned}$$
 
-Now we must choose $$C$$. Assuming $$g(z)$$ does not interfere,
-we know that $$h(z) e^{z \tau}$$ decays to zero
-for $$|z| \to \infty$$, so a useful choice would be a circle of radius $$R$$.
-If we then let $$R \to \infty$$, the contour encloses
-the whole complex plane, including all of the integrand's poles.
-However, thanks to the integrand's decay,
-the resulting contour integral must vanish:
+Now we must choose $$C$$.
+Earlier, we took care that $$h(z) \: e^{z \tau} \to 0$$ for $$|z| \to \infty$$,
+so a good choice would be a circle of radius $$R$$.
+If $$R \to \infty$$, then $$C$$ encloses the whole complex plane,
+including all of the integrand's poles.
+However, because the integrand decays for $$|z| \to \infty$$,
+we conclude that the contour integral must vanish
+(also for other choices of $$C$$):
 
 $$\begin{aligned}
     C
@@ -131,7 +158,7 @@ for bosonic and fermionic Matsubara sums $$S_{B,F}$$:
 $$\begin{aligned}
     \boxed{
         S_{B,F}
-        = \mp \sum_{z_p} e^{z_p \tau} n_{B,F}(z_p) \: \underset{ {z \to z_p}}{\mathrm{Res}}\big(g(z)\big)
+        = \mp \sum_{p_g} \underset{ {z \to p_g}}{\mathrm{Res}}\Big\{ g(z) \Big\} \: n_{B,F}(p_g) \: e^{p_g \tau}
     }
 \end{aligned}$$
 
diff --git a/source/know/concept/sokhotski-plemelj-theorem/index.md b/source/know/concept/sokhotski-plemelj-theorem/index.md
index 66e89bc..445b029 100644
--- a/source/know/concept/sokhotski-plemelj-theorem/index.md
+++ b/source/know/concept/sokhotski-plemelj-theorem/index.md
@@ -9,8 +9,8 @@ categories:
 layout: "concept"
 ---
 
-The goal is to evaluate integrals of the following form, where $$a < 0 < b$$,
-and $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$:
+The goal is to evaluate integrals of the following form,
+where $$f(x)$$ is assumed to be continuous in the integration interval $$[a, b]$$:
 
 $$\begin{aligned}
     \lim_{\eta \to 0^+} \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
@@ -22,12 +22,14 @@ into its real and imaginary parts (limit hidden):
 $$\begin{aligned}
     \int_a^b \frac{f(x)}{x + i \eta} \dd{x}
     &= \int_a^b \frac{x - i \eta}{x^2 + \eta^2} f(x) \dd{x}
-    = \int_a^b \bigg( \frac{x}{x^2 + \eta^2} - i \frac{\eta}{x^2 + \eta^2} \bigg) f(x) \dd{x}
+    \\
+    &= \int_a^b \frac{x}{x^2 + \eta^2} f(x) \dd{x} - i \int_a^b \frac{\eta}{x^2 + \eta^2} f(x) \dd{x}
 \end{aligned}$$
 
-To evaluate the real part,
-we notice that for $$\eta \to 0^+$$ the integrand diverges for $$x \to 0$$,
-and thus split the integral as follows:
+In the real part, notice that the integrand diverges
+for $$x \to 0$$ when $$\eta \to 0^+$$;
+more specifically, there is a singularity at zero.
+We therefore split the integral as follows:
 
 $$\begin{aligned}
     \lim_{\eta \to 0^+} \int_a^b \frac{x f(x)}{x^2 + \eta^2} \dd{x}
@@ -56,7 +58,7 @@ $$\begin{aligned}
 
 The expression $$m / \pi (1 + m^2 x^2)$$ is a so-called *nascent delta function*,
 meaning that in the limit $$m \to +\infty$$ it converges to
-the [Dirac delta function](/know/concept/dirac-delta-function/):
+the [Dirac delta function](/know/concept/dirac-delta-function/) $$\delta(x)$$:
 
 $$\begin{aligned}
     \lim_{\eta \to 0^+} \int_a^b \frac{\eta \: f(x)}{x^2 + \eta^2} \dd{x}
@@ -66,7 +68,8 @@ $$\begin{aligned}
 
 By combining the real and imaginary parts,
 we thus arrive at the (real version of the)
-so-called **Sokhotski-Plemelj theorem** of complex analysis:
+**Sokhotski-Plemelj theorem** of complex analysis,
+which is important in quantum mechanics:
 
 $$\begin{aligned}
     \boxed{
@@ -82,29 +85,33 @@ This awkwardly leaves $$\mathcal{P}$$ behind:
 
 $$\begin{aligned}
     \frac{1}{x + i \eta}
-    = \mathcal{P} \Big( \frac{1}{x} \Big) - i \pi \delta(x)
+    = \mathcal{P} \frac{1}{x} - i \pi \delta(x)
 \end{aligned}$$
 
-The full, complex version of the Sokhotski-Plemelj theorem
-evaluates integrals of the following form
-over a contour $$C$$ in the complex plane:
+That was the real version of the theorem,
+which is a special case of a general result in complex analysis.
+Consider the following function:
 
 $$\begin{aligned}
-    \phi(z) = \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta}
+    \phi(z) = \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta}
 \end{aligned}$$
 
 Where $$f(z)$$ must be [holomorphic](/know/concept/holomorphic-function/).
-The Sokhotski-Plemelj theorem then states:
+For all $$z$$ not on $$C$$, this $$\phi(z)$$ exists,
+but not for $$z \in C$$, since the integral diverges then.
+However, in the limit when approaching $$C$$, we can still obtain a value for $$\phi$$,
+with a caveat: the value depends on the direction we approach $$C$$ from!
+The full Sokhotski-Plemelj theorem then states, for all $$z$$ on the closed contour $$C$$:
 
 $$\begin{aligned}
     \boxed{
-        \lim_{w \to z} \phi(w)
-        = \frac{1}{2 \pi i} \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \frac{f(z)}{2}
+        \lim_{y \to z} \phi(y)
+        = \mathcal{P} \oint_C \frac{f(\zeta)}{\zeta - z} \dd{\zeta} \pm \: i \pi f(z)
     }
 \end{aligned}$$
 
-Where the sign is positive if $$z$$ is inside $$C$$, and negative if it is outside.
-The real version follows by letting $$C$$ follow the whole real axis,
-making $$C$$ an infinitely large semicircle,
-so that the integrand vanishes away from the real axis,
+Where $$\pm$$ is $$+$$ if $$C$$ is approached from the inside, and $$-$$ if from outside.
+The above real version follows by making $$C$$ an infinitely large semicircle
+with its flat side on the real line:
+the integrand vanishes away from the real axis,
 because $$1 / (\zeta \!-\! z) \to 0$$ for $$|\zeta| \to \infty$$.