The Exercise

Two quantum systems $A$ and $B$ are prepared together in one of finitely many joint states $\rho_i$, the $i$th with probability $p_i$. An observer who does not know which preparation was made assigns the pair the mixture of these states. Corollary 11.13 compares the conditional entropy of $A$ given $B$ in that mixture with its values in the states $\rho_i$, and finds it a concave function of the joint state.^✦

Systems and States

A quantum system $Q$ is a physical system with a finite dimensional complex Hilbert space as its state space.^✦ Two systems $A$ and $B$ together form the composite system $A \otimes B$, the quantum system whose state space is the tensor product of the state spaces of $A$ and $B$.^✦ A state $\rho$ of $Q$ is a positive operator of unit trace on the state space of $Q$.^✦

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

The reduced state $\rho^B$ of a state $\rho$ of $A \otimes B$ is the state of $B$ obtained by tracing out $A$: $$\rho^B = \operatorname{tr}_A \rho .$$(2)^✦

Entropies

The von Neumann entropy $S(\rho)$ of a state $\rho$ of $Q$ is the Shannon entropy of the eigenvalues of $\rho$, measured in nats: $$S(\rho) = -\operatorname{tr}(\rho \log \rho) .$$(3)^✦

The conditional entropy $S(A|B)_\rho$ of a state $\rho$ of $A \otimes B$ is the entropy of $\rho$ less the entropy of its reduced state on $B$: $$S(A|B)_\rho = S(\rho) - S(\rho^B) .$$(4)^✦

The Claim

What Is Checked

The claim is stated over states of the composite system $A \otimes B$, their mixture (1) and the conditional entropy (4), which takes the von Neumann entropy (3) of a joint state and of its reduced state (2) on $B$.