1 files changed, 82 insertions, 55 deletions
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}$$