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/ehrenfests-theorem/index.md | 28 ++++++++++++-------------
 1 file changed, 14 insertions(+), 14 deletions(-)

(limited to 'source/know/concept/ehrenfests-theorem')

diff --git a/source/know/concept/ehrenfests-theorem/index.md b/source/know/concept/ehrenfests-theorem/index.md
index 4d96989..fba0192 100644
--- a/source/know/concept/ehrenfests-theorem/index.md
+++ b/source/know/concept/ehrenfests-theorem/index.md
@@ -9,10 +9,10 @@ layout: "concept"
 ---
 
 In quantum mechanics, **Ehrenfest's theorem** gives a general expression for the
-time evolution of an observable's expectation value $\expval{\hat{L}}$.
+time evolution of an observable's expectation value $$\expval{\hat{L}}$$.
 
 The time-dependent Schrödinger equation is as follows,
-where prime denotes differentiation with respect to time $t$:
+where prime denotes differentiation with respect to time $$t$$:
 
 $$\begin{aligned}
     \Ket{\psi'} = \frac{1}{i \hbar} \hat{H} \Ket{\psi}
@@ -20,8 +20,8 @@ $$\begin{aligned}
     \Bra{\psi'} = - \frac{1}{i \hbar} \Bra{\psi} \hat{H}
 \end{aligned}$$
 
-Given an observable operator $\hat{L}$ and a state $\Ket{\psi}$,
-the time-derivative of the expectation value $\expval{\hat{L}}$ is as follows
+Given an observable operator $$\hat{L}$$ and a state $$\Ket{\psi}$$,
+the time-derivative of the expectation value $$\expval{\hat{L}}$$ is as follows
 (due to the product rule of differentiation):
 
 $$\begin{aligned}
@@ -48,12 +48,12 @@ the last term often vanishes.
 
 As a interesting side note, in the [Heisenberg picture](/know/concept/heisenberg-picture/),
 this relation proves itself,
-when one simply wraps all terms in $\Bra{\psi}$ and $\Ket{\psi}$.
+when one simply wraps all terms in $$\Bra{\psi}$$ and $$\Ket{\psi}$$.
 
-Two observables of particular interest are the position $\hat{X}$ and momentum $\hat{P}$.
-Applying the above theorem to $\hat{X}$ yields the following,
-which we reduce using the fact that $\hat{X}$ commutes
-with the potential $V(\hat{X})$,
+Two observables of particular interest are the position $$\hat{X}$$ and momentum $$\hat{P}$$.
+Applying the above theorem to $$\hat{X}$$ yields the following,
+which we reduce using the fact that $$\hat{X}$$ commutes
+with the potential $$V(\hat{X})$$,
 because one is a function of the other:
 
 $$\begin{aligned}
@@ -76,7 +76,7 @@ $$\begin{gathered}
     }
 \end{gathered}$$
 
-Next, applying the general formula to the expected momentum $\expval{\hat{P}}$
+Next, applying the general formula to the expected momentum $$\expval{\hat{P}}$$
 gives us:
 
 $$\begin{aligned}
@@ -86,8 +86,8 @@ $$\begin{aligned}
     = \frac{1}{i \hbar} \Expval{[\hat{P}, V(\hat{X})]}
 \end{aligned}$$
 
-To find the commutator, we go to the $\hat{X}$-basis and use a test
-function $f(x)$:
+To find the commutator, we go to the $$\hat{X}$$-basis and use a test
+function $$f(x)$$:
 
 $$\begin{aligned}
     \Comm{- i \hbar \dv{}{x}}{V(x)} \: f(x)
@@ -113,8 +113,8 @@ $$\begin{gathered}
 \end{gathered}$$
 
 There is an important consequence of Ehrenfest's original theorems
-for the symbolic derivatives of the Hamiltonian $\hat{H}$
-with respect to $\hat{X}$ and $\hat{P}$:
+for the symbolic derivatives of the Hamiltonian $$\hat{H}$$
+with respect to $$\hat{X}$$ and $$\hat{P}$$:
 
 $$\begin{gathered}
     \boxed{
-- 
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