Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 28 insertions, 28 deletions

diff --git a/source/know/concept/matsubara-sum/index.md b/source/know/concept/matsubara-sum/index.md
index 45381ba..8b903d4 100644
--- a/source/know/concept/matsubara-sum/index.md
+++ b/source/know/concept/matsubara-sum/index.md
@@ -18,18 +18,18 @@ $$\begin{aligned}
     = \frac{1}{\hbar \beta} \sum_{i \omega_n} g(i \omega_n) \: e^{i \omega_n \tau}
 \end{aligned}$$
 
-Where $i \omega_n$ are the Matsubara frequencies
-for bosons ($B$) or fermions ($F$),
-and $g(z)$ is a function on the complex plane
+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/)
 except for a known set of simple poles,
-and $\tau$ is a real parameter
+and $$\tau$$ is a real parameter
 (e.g. the [imaginary time](/know/concept/imaginary-time/))
-satisfying $-\hbar \beta < \tau < \hbar \beta$.
+satisfying $$-\hbar \beta < \tau < \hbar \beta$$.
 
 Now, consider the following integral
-over a (for now) unspecified counter-clockwise contour $C$,
-with a (for now) unspecified weighting function $h(z)$:
+over a (for now) unspecified counter-clockwise contour $$C$$,
+with a (for now) unspecified weighting function $$h(z)$$:
 
 $$\begin{aligned}
     \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
@@ -37,14 +37,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where we have applied the residue theorem
-to get a sum over all simple poles $z_p$
-of either $g$ or $h$ (but not both) enclosed by $C$.
+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 $i \omega_n$.
+if we choose $$h$$ such that it has poles at $$i \omega_n$$.
 
-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/):
+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/):
 
 $$\begin{aligned}
     h(z)
@@ -59,11 +59,11 @@ $$\begin{aligned}
     = \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
-Matsubara frequencies $i \omega_n$ of bosons and fermions,
+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
+Matsubara frequencies $$i \omega_n$$ of bosons and fermions,
 and the residues are:
 
 $$\begin{aligned}
@@ -84,10 +84,10 @@ $$\begin{aligned}
     = - \frac{1}{\hbar \beta}
 \end{aligned}$$
 
-In the definition of $h$, the sign flip for $\tau \le 0$
+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:
+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:
 
 $$\begin{aligned}
     \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \dd{z}
@@ -98,8 +98,8 @@ $$\begin{aligned}
     \pm \frac{1}{\hbar \beta} \sum_{i \omega_n} 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 $S_{F,B}$,
+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:
 
 $$\begin{aligned}
@@ -108,10 +108,10 @@ $$\begin{aligned}
     \pm \oint_C \frac{g(z) h(z)}{2 \pi i} e^{z \tau} \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
+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:
@@ -126,7 +126,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We thus arrive at the following results
-for bosonic and fermionic Matsubara sums $S_{B,F}$:
+for bosonic and fermionic Matsubara sums $$S_{B,F}$$:
 
 $$\begin{aligned}
     \boxed{