From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/dyson-equation/index.md | 62 ++++++++++++++---------------
 1 file changed, 31 insertions(+), 31 deletions(-)

(limited to 'source/know/concept/dyson-equation')

diff --git a/source/know/concept/dyson-equation/index.md b/source/know/concept/dyson-equation/index.md
index 962b64a..ae9eb35 100644
--- a/source/know/concept/dyson-equation/index.md
+++ b/source/know/concept/dyson-equation/index.md
@@ -9,7 +9,7 @@ layout: "concept"
 ---
 
 Consider the time-dependent Schrödinger equation,
-describing a wavefunction $\Psi_0(\vb{r}, t)$:
+describing a wavefunction $$\Psi_0(\vb{r}, t)$$:
 
 $$\begin{aligned}
     i \hbar \pdv{}{t}\Psi_0(\vb{r}, t)
@@ -18,50 +18,50 @@ $$\begin{aligned}
 
 By definition, this equation's
 [fundamental solution](/know/concept/fundamental-solution/)
-$G_0(\vb{r}, t; \vb{r}', t')$ satisfies the following:
+$$G_0(\vb{r}, t; \vb{r}', t')$$ satisfies the following:
 
 $$\begin{aligned}
     \Big( i \hbar \pdv{}{t}- \hat{H}_0(\vb{r}) \Big) G_0(\vb{r}, t; \vb{r}', t')
     = \delta(\vb{r} - \vb{r}') \: \delta(t - t')
 \end{aligned}$$
 
-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 $$\hat{G}{}_0^{-1} G_0 = \delta(\vb{r} \!-\! \vb{r}') \: \delta(t \!-\! t')$$:
 
 $$\begin{aligned}
     \hat{G}{}_0^{-1}(\vb{r}, t)
     &\equiv i \hbar \pdv{}{t}- \hat{H}_0(\vb{r})
 \end{aligned}$$
 
-Note that $\hat{G}{}_0^{-1}$ is an operator, while $G_0$ is a function.
+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 $t'$:
+the operator $$\hat{G}_0(\vb{r}, t)$$
+as a multiplication by $$G_0$$
+and integration over $$\vb{r}'$$ and $$t'$$:
 
 $$\begin{aligned}
     \hat{G}_0(\vb{r}, t) \: f
     \equiv \iint_{-\infty}^\infty G_0(\vb{r}, t; \vb{r}', t') \: f(\vb{r}', t') \: \dd{\vb{r}}' \dd{t'}
 \end{aligned}$$
 
-For an arbitrary function $f(\vb{r}, t)$,
-so that $\hat{G}{}_0^{-1} \hat{G}_0 = \hat{G}_0 \hat{G}{}_0^{-1} = 1$.
+For an arbitrary function $$f(\vb{r}, t)$$,
+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 $\hat{G}{}_0^{-1}$:
+using $$\hat{G}{}_0^{-1}$$:
 
 $$\begin{aligned}
     \hat{G}{}_0^{-1}(\vb{r}, t) \: \Psi_0(\vb{r}, t)
     = 0
 \end{aligned}$$
 
-Let us assume that $\hat{H}_0$ is simple,
-such that $G_0$ and $\hat{G}{}_0^{-1}$ can be found without issues
+Let us assume that $$\hat{H}_0$$ is simple,
+such that $$G_0$$ and $$\hat{G}{}_0^{-1}$$ can be found without issues
 by solving the defining equation above.
 
 Suppose we now add a more complicated and
-possibly time-dependent term $\hat{H}_1(\vb{r}, t)$,
+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:
+$$G(\vb{r}, \vb{r}', t, t')$$ satisfies:
 
 $$\begin{aligned}
     \delta(\vb{r} - \vb{r}') \: \delta(t - t')
@@ -71,8 +71,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This equation is typically too complicated to solve,
-so we would like an easier way to calculate this new $G$.
-The perturbed wavefunction $\Psi(\vb{r}, t)$
+so we would like an easier way to calculate this new $$G$$.
+The perturbed wavefunction $$\Psi(\vb{r}, t)$$
 satisfies the Schrödinger equation:
 
 $$\begin{aligned}
@@ -80,9 +80,9 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-We know that $\hat{G}{}_0^{-1} \Psi_0 = 0$,
+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:
+and then we apply $$\hat{G}_0$$ in front:
 
 $$\begin{aligned}
     \hat{G}_0^{-1} \Psi - \hat{H}_1 \Psi
@@ -110,7 +110,7 @@ $$\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 $$\hat{G}$$ and write the so-called **Dyson equation**:
 
 $$\begin{aligned}
     \boxed{
@@ -119,9 +119,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Such an iterative scheme is excellent for approximating $\hat{G}(\vb{r}, t)$.
+Such an iterative scheme is excellent for approximating $$\hat{G}(\vb{r}, t)$$.
 Once a satisfactory accuracy is obtained,
-the perturbed wavefunction $\Psi$ can be calculated from:
+the perturbed wavefunction $$\Psi$$ can be calculated from:
 
 $$\begin{aligned}
     \boxed{
@@ -131,10 +131,10 @@ $$\begin{aligned}
 \end{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:
+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:
 
 $$\begin{aligned}
     \iint \big( \hat{G}{}_0^{-1} \!-\! \hat{H}_1 \big) G(\vb{r}', t') \: f(\vb{r}', t') \dd{\vb{r}'} \dd{t'}
@@ -142,9 +142,9 @@ $$\begin{aligned}
     = f
 \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 $t'$):
+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 $$t'$$):
 
 $$\begin{aligned}
     \hat{G}_0 f
@@ -154,8 +154,8 @@ $$\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, as expected, the operator $$\hat{G}$$
+is defined as multiplication by the function $$G$$ followed by integration:
 
 $$\begin{aligned}
     \hat{G}(\vb{r}, t) \: f(\vb{r}, t)
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