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---
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:
{% include image.html file="sketch-full.png" width="67%"
alt="Bloch sphere" %}
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 pairs are expressed as follows in the conventional $$z$$-basis:
$$\begin{aligned}
\ket{\pm}
= \frac{\ket{0} \pm \ket{1}}{\sqrt{2}}
\qquad \qquad
\ket{\pm i}
= \frac{\ket{0} \pm i \ket{1}}{\sqrt{2}}
\end{aligned}$$
More generally, every point on the surface of the sphere
describes a pure qubit state in terms of the angles $$\theta$$ and $$\varphi$$,
respectively the elevation and azimuth:
$$\begin{aligned}
\ket{\Psi} = \cos\!\Big(\frac{\theta}{2}\Big) \ket{0} + \exp(i \varphi) \sin\!\Big(\frac{\theta}{2}\Big) \ket{1}
\end{aligned}$$
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 $$r \le 1$$:
$$\begin{aligned}
\boxed{
\vec{r}
= \begin{bmatrix} r_x \\ r_y \\ r_z \end{bmatrix}
= \begin{bmatrix} r \sin\theta \cos\varphi \\ r \sin\theta \sin\varphi \\ r \cos\theta \end{bmatrix}
}
\end{aligned}$$
Note that $$\vec{r}$$ is not actually a qubit state,
but rather 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/):
$$\begin{aligned}
\boxed{
\hat{\rho}
= \frac{1}{2} \Big( \hat{I} + \vec{r} \cdot \vec{\sigma} \Big)
}
\end{aligned}$$
Where $$\vec{\sigma} = (\hat{\sigma}_x, \hat{\sigma}_y, \hat{\sigma}_z)$$ is the Pauli "vector".
Now, we know that a density matrix 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 $$(\vec{r} \cdot \vec{\sigma})^2 = \hat{I}$$,
we get $$\hat{\rho}$$ again, so 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 $$\vec{r}$$:
$$\begin{aligned}
\boxed{
\begin{aligned}
\expval{\hat{\sigma}_{x}}
&= r_{x}
\\
\expval{\hat{\sigma}_{y}}
&= r_{y}
\\
\expval{\hat{\sigma}_{z}}
&= r_{z}
\end{aligned}
}
\end{aligned}$$
This is a consequence of the above form of the density operator $$\hat{\rho}$$.
For example for $$\hat{\sigma}_z$$:
$$\begin{aligned}
\expval{\hat{\sigma}_z}
&= \Tr(\hat{\rho} \hat{\sigma}_z)
= \frac{1}{2} \Tr\!\big(\hat{\sigma}_z + (\vec{r} \cdot \vec{\sigma}) \hat{\sigma}_z \big)
= \frac{1}{2} \Tr\!\big( (r_x \hat{\sigma}_x + r_y \hat{\sigma}_y + r_z \hat{\sigma}_z) \hat{\sigma}_z \big)
\\
&= \frac{1}{2} \Tr\!\big( r_x \hat{\sigma}_x \hat{\sigma}_z + r_y \hat{\sigma}_y \hat{\sigma}_z + r_z \hat{\sigma}_z^2 \big)
= \frac{1}{2} \Tr\!\big( r_z \hat{I} \big)
= r_z
\end{aligned}$$
## References
1. N. Brunner,
*Quantum information theory: lecture notes*,
2019, unpublished.
2. J.B. Brask,
*Quantum information: lecture notes*,
2021, unpublished.
|
