The Exercise

Three quantum systems $A$, $B$ and $C$, or four with a system $D$, share a joint state $\rho$. For a group $X$ of them, $S(X)$ is the von Neumann entropy of the reduced state of $X$, and the entropy of $X$ conditioned on a second group $Y$ is $$S(X|Y) = S(X,Y) - S(Y),$$(1) with $S(X,Y)$ the entropy of the two groups together.

Theorem 11.16 says that when one argument of a conditional entropy is a pair of systems, splitting the pair cannot lower it: the conditional entropy with the pair is at most the sum of the two in which one member of the pair takes its place. The pair may stand in either argument, as $AB$ given $C$ or $A$ given $BC$. With four systems both arguments split at once, and $AB$ given $CD$ is compared with $A$ given $C$ and $B$ given $D$.

Systems, States and Entropy

A quantum system $X$ is described by its state space, a finite dimensional complex Hilbert space.^✦ The composite system $X \otimes Y$ of two quantum systems $X$ and $Y$ is the quantum system whose state space is the tensor product of their state spaces.^✦ A state $\rho$ of $X$, pure or mixed, is a positive operator of unit trace on the state space of $X$^✦, and its von Neumann entropy, in nats, is $$S(\rho) = -\operatorname{tr}(\rho \log \rho).$$(2)^✦

For a state $\rho$ of $X \otimes Y$, the reduced state $\rho^X = \operatorname{tr}_Y \rho$ of $X$ is the partial trace of $\rho$ over $Y$, taken in an orthonormal basis of the state space of $Y$.^✦ The reduced state $\rho^Y = \operatorname{tr}_X \rho$ of $Y$ is likewise the partial trace over $X$, taken in an orthonormal basis of the state space of $X$.^✦

Regrouping the Systems

An identification $e$ of a quantum system $X$ with a quantum system $Y$ is a linear isometric bijection from the state space of $X$ onto that of $Y$^✦; along it, a state $\rho$ of $X$ is relabelled as the state $e \rho e^{-1}$ of $Y$.^✦ The regrouping $\alpha_{X,Y,Z}$ identifies $X \otimes (Y \otimes Z)$ with $(X \otimes Y) \otimes Z$ by $|u\rangle \otimes (|v\rangle \otimes |w\rangle) \mapsto (|u\rangle \otimes |v\rangle) \otimes |w\rangle$, and the swap $\mathrm{swap}_{X,Y}$ identifies $X \otimes Y$ with $Y \otimes X$ by $|u\rangle \otimes |v\rangle \mapsto |v\rangle \otimes |u\rangle$.^✦ The identity $1_X$ identifies $X$ with itself by the identity map of its state space.^✦ Two identifications, $e$ of $X$ with $X'$ and $f$ of $Y$ with $Y'$, give their product $e \otimes f$, the identification of $X \otimes Y$ with $X' \otimes Y'$ by the tensor product of the two maps.^✦

The reduced state of a group of systems is taken by relabelling along these identifications until the group is one factor of the composite, then tracing out the rest. For a state $\rho$ of $A \otimes (C \otimes B)$, the reduced states $\rho^{CB} = \operatorname{tr}_A \rho$ and $\rho^{C} = \operatorname{tr}_B \rho^{CB}$ need no relabelling, while $$\rho^{AC} = \operatorname{tr}_B\bigl(\alpha_{A,C,B}\,\rho\,\alpha_{A,C,B}^{-1}\bigr).$$(3) For a state of $A \otimes (B \otimes C)$, the pair $AC$ is first brought together by $1_A \otimes \mathrm{swap}_{B,C}$ and then regrouped by $\alpha_{A,C,B}$.

The Claim

For a state $\rho$ of the composite system of all the systems, $S(\rho)$ is its von Neumann entropy, and a difference $S(\rho^{XY}) - S(\rho^{Y})$ of reduced states is the conditional entropy $S(X|Y)$ (1).

What Is Checked

The three inequalities are stated over states of composite quantum systems, their reduced states and von Neumann entropies, and states relabelled along identifications of systems. The library proves of these constructions that the partial trace of a density operator is a density operator, and that a state relabelled along an identification is again a state, since conjugation by an isometry preserves positivity and trace.