The Exercise

A system is prepared in the state $\rho_i$ with probability $p_i$ , for finitely many $i$ , so that its state is the mixture of the $\rho_i$ with weights $p_i$ . The von Neumann entropy is concave: the entropy of the mixture is never less than the average of the entropies of the $\rho_i$ , taken with the same weights. The exercise is to show that the two are equal exactly when all the states $\rho_i$ with $p_i > 0$ are one and the same state.^✦ States of zero weight do not enter the mixture and are left free.

A Quantum System

The library states the equivalence for the states of one quantum system. A quantum system $Q$ is a physical system whose state space $\mathcal{H}_Q$ is a finite dimensional complex Hilbert space.^✦

A state $\rho$ of $Q$ is a positive operator of unit trace on the state space of $Q$ .^✦ Pure and mixed states are both states in this sense.

Mixture and Entropy

The mixture of a finite family of states $\rho_i$ of $Q$ , indexed by $i$ in a finite set $I$ , with weights $p_i \geq 0$ summing to one, is the state of $Q$ whose density operator is the convex combination $\sum_{i \in I} p_i \, \rho_i .$ (1)^✦

The von Neumann entropy of a state $\rho$ of $Q$ is the real number $S(\rho) = -\operatorname{tr}(\rho \log \rho),$ (2)^✦ the Shannon entropy of the eigenvalues of $\rho$ , with the natural logarithm, so that it is measured in nats.

The Claim

Theorem 1. Let $\rho_i$ , for $i$ in a finite set $I$ , be states of a quantum system $Q$ , and let $p_i$ be nonnegative weights with $\sum_i p_i = 1$ . The von Neumann entropy of their mixture equals the average of their entropies, $S\Bigl(\sum_{i \in I} p_i \, \rho_i\Bigr) = \sum_{i \in I} p_i \, S(\rho_i),$ (3)^✦ if and only if $\rho_i = \rho_j$ for all $i$ and $j$ with $p_i > 0$ and $p_j > 0$ .

What Is Checked

The claim is stated over the mixture (1) of states of a quantum system and the von Neumann entropy (2) of a state. About these constructions the library states that the mixture is a state of $Q$ .