From 9d9693af6fb94ef4404a3c2399cb38842e5ca822 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 12 May 2023 21:19:19 +0200
Subject: Improve knowledge base

---
 source/know/concept/bloch-sphere/index.md | 58 ++++++++++++++++++++-----------
 1 file changed, 38 insertions(+), 20 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 99ac45d..b348d22 100644
--- a/source/know/concept/bloch-sphere/index.md
+++ b/source/know/concept/bloch-sphere/index.md
@@ -21,23 +21,23 @@ and their extremes are the eigenstates of the Pauli matrices:
 
 $$\begin{aligned}
     \hat{\sigma}_z
-    \to \{\Ket{0}, \Ket{1}\}
+    \to \{\ket{0}, \ket{1}\}
     \qquad
     \hat{\sigma}_x
-    \to \{\Ket{+}, \Ket{-}\}
+    \to \{\ket{+}, \ket{-}\}
     \qquad
     \hat{\sigma}_y
-    \to \{\Ket{+i}, \Ket{-i}\}
+    \to \{\ket{+i}, \ket{-i}\}
 \end{aligned}$$
 
-Where the latter two states are expressed as follows in the conventional $$z$$-basis:
+Where the latter two pairs are expressed as follows in the conventional $$z$$-basis:
 
 $$\begin{aligned}
-    \Ket{\pm}
-    = \frac{\Ket{0} \pm \Ket{1}}{\sqrt{2}}
-    \qquad \quad
-    \Ket{\pm i}
-    = \frac{\Ket{0} \pm i \Ket{1}}{\sqrt{2}}
+    \ket{\pm}
+    = \frac{\ket{0} \pm \ket{1}}{\sqrt{2}}
+    \qquad \qquad
+    \ket{\pm i}
+    = \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}}
 \end{aligned}$$
 
 More generally, every point on the surface of the sphere
@@ -45,11 +45,12 @@ describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
 respectively the elevation and azimuth:
 
 $$\begin{aligned}
-    \Ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \Ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \Ket{1}
+    \ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \ket{1}
 \end{aligned}$$
 
-We can generalize this further by describing points using the **Bloch vector** $$\vec{r}$$,
-with radius $$r \le 1$$:
+Another way to describe states is the **Bloch vector** $$\vec{r}$$,
+which is simply the $$(x,y,z)$$-coordinates of a point on the sphere.
+Let the radius $$r \le 1$$:
 
 $$\begin{aligned}
     \boxed{
@@ -60,8 +61,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 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.
+but rather a description of one.
 The main point of the Bloch vector is that it allows us
 to describe the qubit using a [density operator](/know/concept/density-operator/):
 
@@ -73,7 +73,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 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
+Now, we know that a density matrix represents a pure ensemble
 if and only if it is idempotent, i.e. $$\hat{\rho}^2 = \hat{\rho}$$:
 
 $$\begin{aligned}
@@ -82,8 +82,8 @@ $$\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 $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
+we get $$\hat{\rho}$$ again, so the state is pure:
 
 $$\begin{aligned}
     (\vec{r} \cdot \vec{\sigma})^2
@@ -109,11 +109,28 @@ else if $$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 $$\vec{r}$$:
 
 $$\begin{aligned}
-    \Expval{\hat{\sigma}_z}
+    \boxed{
+        \begin{aligned}
+            \expval{\hat{\sigma}_{x}}
+            &= r_{x}
+            \\
+            \expval{\hat{\sigma}_{y}}
+            &= r_{y}
+            \\
+            \expval{\hat{\sigma}_{z}}
+            &= r_{z}
+        \end{aligned}
+    }
+\end{aligned}$$
+
+This is a consequence of the above form of the density operator $$\hat{\rho}$$.
+For example for $$\hat{\sigma}_z$$:
+
+$$\begin{aligned}
+    \expval{\hat{\sigma}_z}
     &= \Tr(\hat{\rho} \hat{\sigma}_z)
     = \frac{1}{2} \Tr\!\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
     = \frac{1}{2} \Tr\!\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
@@ -124,6 +141,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 
+
 ## References
 1.  N. Brunner,
     *Quantum information theory: lecture notes*,
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