Purity and Tr(ρ²)

How "mixed" is a state? A single number answers this: the purity.

Definition

The purity of a density operator $\rho$ is

It ranges between two bounds set by the dimension $d$ of the Hilbert space:

The two extremes have clean meanings:

• $\operatorname{Tr}(\rho^2) = 1$ if and only if $\rho$ is pure ($\rho = |\psi\rangle\langle\psi|$).
• $\operatorname{Tr}(\rho^2) = 1/d$ if and only if $\rho$ is maximally mixed, $\rho = I/d$.

For a single qubit ($d = 2$), purity runs from $1/2$ (maximally mixed) up to $1$ (pure).

Why purity detects mixedness

Use the spectral decomposition $\rho = \sum_k \lambda_k |k\rangle\langle k|$. Then $\rho^2 = \sum_k \lambda_k^2 |k\rangle\langle k|$, so

The eigenvalues form a probability distribution ($\lambda_k \ge 0$, $\sum_k \lambda_k = 1$). A pure state has one eigenvalue equal to $1$ and the rest $0$, giving $\sum_k \lambda_k^2 = 1$. Spreading the distribution out lowers $\sum_k \lambda_k^2$ — it is minimized when all $d$ eigenvalues equal $1/d$, giving $d \cdot (1/d)^2 = 1/d$. So purity is large when probability is concentrated (little classical uncertainty) and small when it is spread (much classical uncertainty).

A direct test for purity

The relation $\operatorname{Tr}(\rho^2) = 1 \iff \rho^2 = \rho$ gives a quick check: a density operator is pure exactly when it is idempotent, $\rho^2 = \rho$ (it is then a projector onto a single ket). Any deviation of $\operatorname{Tr}(\rho^2)$ below $1$ signals genuine mixedness.

Try it

Compute the purity $\operatorname{Tr}(\rho^2)$ of the mixed state $\rho = \tfrac12|0\rangle\langle0| + \tfrac12|+\rangle\langle+| = \begin{pmatrix}3/4 & 1/4\\ 1/4 & 1/4\end{pmatrix}$. You should get a value strictly between $1/2$ and $1$, confirming the state is mixed but not maximally so.

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