1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
|
---
title: "Bloch sphere"
sort_title: "Bloch sphere"
date: 2021-03-09
categories:
- Quantum mechanics
- Quantum information
- Two-level system
layout: "concept"
---
In quantum mechanics, particularly quantum information,
the **Bloch sphere** is an invaluable tool to visualize qubits.
All pure qubit states are represented by a point on the sphere's surface:
<a href="bloch.jpg">
<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,
and their extremes are the eigenstates of the Pauli matrices:
$$\begin{aligned}
\hat{\sigma}_z
\to \{\Ket{0}, \Ket{1}\}
\qquad
\hat{\sigma}_x
\to \{\Ket{+}, \Ket{-}\}
\qquad
\hat{\sigma}_y
\to \{\Ket{+i}, \Ket{-i}\}
\end{aligned}$$
Where the latter two states are expressed as follows in the conventional $$z$$-basis:
$$\begin{aligned}
\Ket{\pm}
= \frac{\Ket{0} \pm \Ket{1}}{\sqrt{2}}
\qquad \quad
\Ket{\pm i}
= \frac{\Ket{0} \pm i \Ket{1}}{\sqrt{2}}
\end{aligned}$$
More generally, every point on the surface of the sphere
describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
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 $$r \le 1$$:
$$\begin{aligned}
\boxed{
\vec{r}
= \begin{bmatrix} r_x \\ r_y \\ r_z \end{bmatrix}
= \begin{bmatrix} r \sin\theta \cos\varphi \\ r \sin\theta \sin\varphi \\ r \cos\theta \end{bmatrix}
}
\end{aligned}$$
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.
The main point of the Bloch vector is that it allows us
to describe the qubit using a [density operator](/know/concept/density-operator/):
$$\begin{aligned}
\boxed{
\hat{\rho}
= \frac{1}{2} \Big( \hat{I} + \vec{r} \cdot \vec{\sigma} \Big)
}
\end{aligned}$$
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. $$\hat{\rho}^2 = \hat{\rho}$$:
$$\begin{aligned}
\hat{\rho}^2
&= \frac{1}{4} \Big( \hat{I}^2 + 2 \hat{I} (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
= \frac{1}{4} \Big( \hat{I} + 2 (\vec{r} \cdot \vec{\sigma}) + (\vec{r} \cdot \vec{\sigma})^2 \Big)
\end{aligned}$$
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:
$$\begin{aligned}
(\vec{r} \cdot \vec{\sigma})^2
&= (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z)^2
\\
&= r_x^2 \hat{\sigma}_x^2 + r_x r_y \hat{\sigma}_x \hat{\sigma}_y + r_x r_z \hat{\sigma}_x \hat{\sigma}_z
+ r_x r_y \hat{\sigma}_y \hat{\sigma}_x + r_y^2 \hat{\sigma}_y^2
\\
&\quad + r_y r_z \hat{\sigma}_y \hat{\sigma}_z + r_x r_z \hat{\sigma}_z \hat{\sigma}_x
+ r_y r_z \hat{\sigma}_z \hat{\sigma}_y + r_z^2 \hat{\sigma}_z^2
\\
&= r_x^2 \hat{I} + r_y^2 \hat{I} + r_z^2 \hat{I}
+ r_x r_y \{ \hat{\sigma}_x, \hat{\sigma}_y \}
+ r_y r_z \{ \hat{\sigma}_y, \hat{\sigma}_z \}
+ r_x r_z \{ \hat{\sigma}_x, \hat{\sigma}_z \}
\\
&= (r_x^2 + r_y^2 + r_z^2) \hat{I}
= r^2 \hat{I}
\end{aligned}$$
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 $$\hat{\sigma}_z$$:
$$\begin{aligned}
\Expval{\hat{\sigma}_z}
&= \Tr(\hat{\rho} \hat{\sigma}_z)
= \frac{1}{2} \Tr\!\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
= \frac{1}{2} \Tr\!\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
\\
&= \frac{1}{2} \Tr\!\big( r_x \hat{\sigma}_x \hat{\sigma}_z + r_y \hat{\sigma}_y \hat{\sigma}_z + r_z \hat{\sigma}_z^2 \big)
= \frac{1}{2} \Tr\!\big( r_z \hat{I} \big)
= r_z
\end{aligned}$$
## References
1. N. Brunner,
*Quantum information theory: lecture notes*,
2019, unpublished.
2. J.B. Brask,
*Quantum information: lecture notes*,
2021, unpublished.
|
