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/bloch-sphere/index.md | 30 +++++++++++++++---------------
 1 file changed, 15 insertions(+), 15 deletions(-)

(limited to 'source/know/concept/bloch-sphere/index.md')

diff --git a/source/know/concept/bloch-sphere/index.md b/source/know/concept/bloch-sphere/index.md
index 5d747f7..2cb7742 100644
--- a/source/know/concept/bloch-sphere/index.md
+++ b/source/know/concept/bloch-sphere/index.md
@@ -17,7 +17,7 @@ All pure qubit states are represented by a point on the sphere's surface:
 <img src="bloch-small.jpg" style="width:60%">
 </a>
 
-The $x$, $y$ and $z$-axes represent the components of a spin-1/2-alike system,
+The $$x$$, $$y$$ and $$z$$-axes represent the components of a spin-1/2-alike system,
 and their extremes are the eigenstates of the Pauli matrices:
 
 $$\begin{aligned}
@@ -31,7 +31,7 @@ $$\begin{aligned}
     \to \{\Ket{+i}, \Ket{-i}\}
 \end{aligned}$$
 
-Where the latter two states are expressed as follows in the conventional $z$-basis:
+Where the latter two states are expressed as follows in the conventional $$z$$-basis:
 
 $$\begin{aligned}
     \Ket{\pm}
@@ -42,15 +42,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 More generally, every point on the surface of the sphere
-describes a pure qubit state in terms of the angles $\theta$ and $\varphi$,
+describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
 respectively the elevation and azimuth:
 
 $$\begin{aligned}
     \Ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \Ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \Ket{1}
 \end{aligned}$$
 
-We can generalize this further by describing points using the **Bloch vector** $\vec{r}$,
-with radius $r \le 1$:
+We can generalize this further by describing points using the **Bloch vector** $$\vec{r}$$,
+with radius $$r \le 1$$:
 
 $$\begin{aligned}
     \boxed{
@@ -60,7 +60,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Note that $\vec{r}$ is not actually a qubit state,
+Note that $$\vec{r}$$ is not actually a qubit state,
 but rather an implicit description of one,
 meaning that it does not need to be normalized.
 The main point of the Bloch vector is that it allows us
@@ -73,9 +73,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$ is the Pauli "vector".
-Now, we know that $\hat{\rho}$ represents a pure ensemble
-if and only if it is idempotent, i.e. $\hat{\rho}^2 = \hat{\rho}$:
+Where $$\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$$ is the Pauli "vector".
+Now, we know that $$\hat{\rho}$$ represents a pure ensemble
+if and only if it is idempotent, i.e. $$\hat{\rho}^2 = \hat{\rho}$$:
 
 $$\begin{aligned}
     \hat{\rho}^2
@@ -83,8 +83,8 @@ $$\begin{aligned}
     = \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
 \end{aligned}$$
 
-You can easily convince yourself that if $(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$,
-then we get $\hat{\rho}$ again, and the state is pure:
+You can easily convince yourself that if $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
+then we get $$\hat{\rho}$$ again, and the state is pure:
 
 $$\begin{aligned}
     (\vec{r} \cdot \vec{\sigma})^2
@@ -105,13 +105,13 @@ $$\begin{aligned}
     = r^2 \hat{I}
 \end{aligned}$$
 
-Therefore, if the radius $r = 1$, the ensemble is pure,
-else if $r < 1$ it is mixed.
+Therefore, if the radius $$r = 1$$, the ensemble is pure,
+else if $$r < 1$$ it is mixed.
 
 Another useful property of the Bloch vector
 is that the expectation value of the Pauli matrices
-are given by the corresponding component of $\vec{r}$,
-for example for $\hat{\sigma}_z$:
+are given by the corresponding component of $$\vec{r}$$,
+for example for $$\hat{\sigma}_z$$:
 
 $$\begin{aligned}
     \Expval{\hat{\sigma}_z}
-- 
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