Purity and the Bloch ball

Intuition

How mixed is a state? Its purity, $\operatorname{tr}\hat{\rho}^{2}$ , says: 1 for a pure state, less for a mixture, down to $1/d$ for the completely mixed state of a $d$ -level system, the identity divided by $d$ . For a spin $\tfrac12$ every density matrix is $\hat{\rho}=\tfrac12(\hat{1}+\mathbf{r}\cdot\vec{\sigma})$ with a Bloch vector $\mathbf{r}$ of length at most 1. Pure states lie on the surface of the Bloch sphere, mixtures fill the ball inside it, and the completely mixed state sits at the centre. The purity is $\tfrac12(1+|\mathbf{r}|^{2})$ : distance from the centre measures how pure the state is.

A beam of light can be fully polarised, unpolarised, or partly polarised, with a degree of polarisation between 0 and 1. For a spin that degree is the length of its Bloch vector.

A cut through the Bloch ball. Pure states lie on the sphere, $|\mathbf{r}|=1$ ; a mixed state has its Bloch vector $\mathbf{r}$ inside, here of length $0.54$ and purity $\tfrac12(1+0.29)\approx0.65$ ; every state on the dashed circle is equally mixed, and the centre is the completely mixed state.

The purity of $\hat{\rho}$ , and the form of every density matrix of a spin $\tfrac12$ :

\begin{aligned}\operatorname{tr}\hat{\rho}^{2}&\le1,\qquad \operatorname{tr}\hat{\rho}^{2}=1\iff\hat{\rho}\ \text{pure}\\ \hat{\rho}&=\tfrac12\bigl(\hat{1}+\mathbf{r}\cdot\vec{\sigma}\bigr),\qquad |\mathbf{r}|\le1\end{aligned}

Properties

$\operatorname{tr}\hat{\rho}^{2}=\sum_{i}\lambda_{i}^{2}$ over the eigenvalues; its least value, , belongs to .

The purity of a spin \tfrac12

\hat{\rho}=\tfrac12\bigl(\hat{1}+\mathbf{r}\cdot\vec{\sigma}\bigr)\ \Longrightarrow\ \operatorname{tr}\hat{\rho}^{2}=\tfrac12\bigl(1+|\mathbf{r}|^{2}\bigr)

Square the density matrix. The square of $\mathbf{r}\cdot\vec{\sigma}$ is $|\mathbf{r}|^{2}$ times the identity, by the multiplication rule of the Pauli matrices, and the Pauli matrices have zero trace. What is left is one half of one plus the squared length.

Proof steps

\hat{\rho}=\tfrac12\bigl(\hat{1}+\mathbf{r}\cdot\vec{\sigma}\bigr)

Every Hermitian $2\times2$ matrix of trace one, written with the Pauli matrices.

\hat{\rho}^{2}=\tfrac14\bigl(\hat{1}+2\,\mathbf{r}\cdot\vec{\sigma}+(\mathbf{r}\cdot\vec{\sigma})^{2}\bigr)

Multiply out.

(\mathbf{r}\cdot\vec{\sigma})^{2}=|\mathbf{r}|^{2}\,\hat{1}

From $\sigma_{i}\sigma_{j}=\delta_{ij}\hat{1}+i\epsilon_{ijk}\sigma_{k}$ : the antisymmetric part cancels in the symmetric sum .

\operatorname{tr}\hat{\rho}^{2}=\tfrac14\bigl(2+2|\mathbf{r}|^{2}\bigr)=\tfrac12\bigl(1+|\mathbf{r}|^{2}\bigr)

$\operatorname{tr}\hat{1}=2$ and each $\sigma_{i}$ has trace zero.

\operatorname{tr}\hat{\rho}^{2}\le1\iff|\mathbf{r}|\le1

With equality, a pure state, exactly on the sphere.

Applications

Practice

Purity

The purity is the trace of the density matrix squared, which is the sum of the squares of its eigenvalues.

\operatorname{tr}\hat{\rho}^{2}=\sum_{i}\lambda_{i}^{2}

Try it

What is the purity of $\hat{\rho}=\operatorname{diag}(0.75,0.25)$ ?