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/larmor-precession/index.md | 32 +++++++++++++-------------
 1 file changed, 16 insertions(+), 16 deletions(-)

(limited to 'source/know/concept/larmor-precession')

diff --git a/source/know/concept/larmor-precession/index.md b/source/know/concept/larmor-precession/index.md
index 6b101e0..774af7b 100644
--- a/source/know/concept/larmor-precession/index.md
+++ b/source/know/concept/larmor-precession/index.md
@@ -10,18 +10,18 @@ layout: "concept"
 
 Consider a stationary spin-1/2 particle,
 placed in a [magnetic field](/know/concept/magnetic-field/)
-with magnitude $B$ pointing in the $z$-direction.
-In that case, its Hamiltonian $\hat{H}$ is given by:
+with magnitude $$B$$ pointing in the $$z$$-direction.
+In that case, its Hamiltonian $$\hat{H}$$ is given by:
 
 $$\begin{aligned}
     \hat{H} = - \gamma B \hat{S}_z = - \frac{\hbar}{2} \gamma B \hat{\sigma_z}
 \end{aligned}$$
 
-Where $\gamma = - q / m$ is the gyromagnetic ratio,
-and $\hat{\sigma}_z$ is the Pauli spin matrix for the $z$-direction.
-Since $\hat{H}$ is proportional to $\hat{\sigma}_z$,
-they share eigenstates $\Ket{\downarrow}$ and $\Ket{\uparrow}$.
-The respective eigenenergies $E_{\downarrow}$ and $E_{\uparrow}$ are as follows:
+Where $$\gamma = - q / m$$ is the gyromagnetic ratio,
+and $$\hat{\sigma}_z$$ is the Pauli spin matrix for the $$z$$-direction.
+Since $$\hat{H}$$ is proportional to $$\hat{\sigma}_z$$,
+they share eigenstates $$\Ket{\downarrow}$$ and $$\Ket{\uparrow}$$.
+The respective eigenenergies $$E_{\downarrow}$$ and $$E_{\uparrow}$$ are as follows:
 
 $$\begin{aligned}
     E_{\downarrow} = \frac{\hbar}{2} \gamma B
@@ -29,9 +29,9 @@ $$\begin{aligned}
     E_{\uparrow} = - \frac{\hbar}{2} \gamma B
 \end{aligned}$$
 
-Because $\hat{H}$ is time-independent,
-the general time-dependent solution $\Ket{\chi(t)}$ is of the following form,
-where $a$ and $b$ are constants,
+Because $$\hat{H}$$ is time-independent,
+the general time-dependent solution $$\Ket{\chi(t)}$$ is of the following form,
+where $$a$$ and $$b$$ are constants,
 and the exponentials are "twiddle factors":
 
 $$\begin{aligned}
@@ -40,9 +40,9 @@ $$\begin{aligned}
     \:+\: b \exp(- i E_{\uparrow} t / \hbar) \: \Ket{\uparrow}
 \end{aligned}$$
 
-For our purposes, we can safely assume that $a$ and $b$ are real,
-and then say that there exists an angle $\theta$
-satisfying $a = \sin(\theta / 2)$ and $b = \cos(\theta / 2)$, such that:
+For our purposes, we can safely assume that $$a$$ and $$b$$ are real,
+and then say that there exists an angle $$\theta$$
+satisfying $$a = \sin(\theta / 2)$$ and $$b = \cos(\theta / 2)$$, such that:
 
 $$\begin{aligned}
     \Ket{\chi(t)} = \sin(\theta / 2) \exp(- i E_{\downarrow} t / \hbar) \: \Ket{\downarrow}
@@ -50,7 +50,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Now, we find the expectation values of the spin operators
-$\expval{\hat{S}_x}$, $\expval{\hat{S}_y}$, and $\expval{\hat{S}_z}$.
+$$\expval{\hat{S}_x}$$, $$\expval{\hat{S}_y}$$, and $$\expval{\hat{S}_z}$$.
 The first is:
 
 $$\begin{aligned}
@@ -86,8 +86,8 @@ $$\begin{aligned}
     \matrixel{\chi}{\hat{S}_z}{\chi} = \frac{\hbar}{2} \cos(\theta)
 \end{aligned}$$
 
-The result is that the spin axis is off by $\theta$ from the $z$-direction,
-and is rotating (or **precessing**) around the $z$-axis at the **Larmor frequency** $\omega$:
+The result is that the spin axis is off by $$\theta$$ from the $$z$$-direction,
+and is rotating (or **precessing**) around the $$z$$-axis at the **Larmor frequency** $$\omega$$:
 
 $$\begin{aligned}
     \boxed{
-- 
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