Concavity of the von Neumann entropy

Let $\rho$ be a density operator on a finite-dimensional Hilbert space. Its von Neumann entropy is $S(\rho) = -\mathrm{Tr}(\rho \log \rho)$.

Statement

For any density operators $\rho_1,\dots,\rho_n$ and any probabilities $p_1,\dots,p_n$ with $\sum_i p_i = 1$, define the mixture $\bar\rho = \sum_{i=1}^n p_i \rho_i$. Then

Key hypotheses

• Each $\rho_i$ is positive semidefinite and has unit trace: $\rho_i \succeq 0$ and $\mathrm{Tr}(\rho_i)=1$.
• The coefficients form a probability vector: $p_i \ge 0$ and $\sum_i p_i = 1$.

Key conclusion

• The map $\rho \mapsto S(\rho)$ is concave on the convex set of density operators. In particular, classical randomization (“forgetting which state was prepared”) cannot decrease entropy.

Proof idea / significance

A standard route uses monotonicity of quantum relative entropy .

Introduce a classical register $X$ with orthonormal basis $\{|i\rangle\}$ and form the block-diagonal (classical–quantum) state $\omega_{XQ} = \sum_i p_i\, |i\rangle\langle i| \otimes \rho_i$, whose $Q$-marginal is $\omega_Q = \bar\rho$. One then checks the identity

where $D(\cdot\|\cdot)$ denotes quantum relative entropy. The right-hand side is nonnegative, yielding the concavity inequality.

Thermodynamically, this expresses the idea that coarse-graining or mixing increases uncertainty: entropy is increased by ignoring classical information about the preparation.