---
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.