The Exercise

Three quantum systems $A$, $B$ and $C$ share a joint state. Tracing out the rest gives a state of each system alone and of each pair, and each of these states, like the joint one, has a von Neumann entropy: $S(A,B,C)$ for the whole, $S(A,B)$ for the pair $AB$, $S(B)$ for $B$ alone, and so on. The theorem compares the sum $S(A,B) + S(B,C)$ for the two pairs that overlap in $B$ with the sum $S(A,B,C) + S(B)$ for the whole and the middle system, and with the sum $S(A) + S(C)$ for the two outer systems. The first comparison is the strong subadditivity of the entropy, the second its weak monotonicity.

The library states both comparisons for a joint state $\rho$ of the three systems, taken as one composite system with $B$ and $C$ grouped first and $A$ joined to the pair, written $A \otimes (B \otimes C)$.

Systems, States and Entropy

A quantum system $Q$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ The composite system $Q \otimes R$ of two quantum systems $Q$ and $R$ is the system whose state space is the tensor product of their state spaces.^✦ A state $\rho$ of $Q$ is a positive operator of unit trace on the state space of $Q$^✦, and its von Neumann entropy is $$S(\rho) = -\operatorname{tr}(\rho \log \rho),$$(1)^✦ the Shannon entropy of the eigenvalues of $\rho$, measured in nats.

The reduced state $\rho^Q$ of a state $\rho$ of $Q \otimes R$ is the state $\operatorname{tr}_R\,\rho$ of $Q$, the partial trace of $\rho$ over $R$^✦, and its reduced state $\rho^R$ is the state $\operatorname{tr}_Q\,\rho$ of $R$, the partial trace over $Q$.^✦

An identification $e$ of a system $Q$ with a system $R$ is a linear isometric bijection from the state space of $Q$ onto that of $R$.^✦ Along it a state $\rho$ of $Q$ is relabelled as the state $e \rho e^{-1}$ of $R$.^✦ The regrouping $\alpha$ of three systems $A$, $B$ and $C$ is the identification of $A \otimes (B \otimes C)$ with $(A \otimes B) \otimes C$ that moves the brackets, $$\alpha\bigl(|u\rangle \otimes (|v\rangle \otimes |w\rangle)\bigr) = (|u\rangle \otimes |v\rangle) \otimes |w\rangle,$$(2)^✦ for vectors $|u\rangle$, $|v\rangle$ and $|w\rangle$ of the state spaces of $A$, $B$ and $C$.

The Claim

For a state $\rho$ of $A \otimes (B \otimes C)$, the reduced states on the single systems and on the pairs are $$\rho^{A} = \operatorname{tr}_{BC}\,\rho, \quad \rho^{BC} = \operatorname{tr}_{A}\,\rho, \quad \rho^{B} = \operatorname{tr}_{C}\,\rho^{BC}, \quad \rho^{C} = \operatorname{tr}_{B}\,\rho^{BC}, \quad \rho^{AB} = \operatorname{tr}_{C}\bigl(\alpha \rho \alpha^{-1}\bigr),$$(3) where the pair $AB$, which is not a factor of $A \otimes (B \otimes C)$, gets its state from $\rho$ relabelled along the regrouping. The entropies of the exercise are $S(A,B,C) = S(\rho)$ and $S(X) = S(\rho^X)$ for each system or pair $X$.

What Is Checked

The claim is stated over the von Neumann entropy (1) of a state of the composite system $A \otimes (B \otimes C)$ and of the states of its parts (3), which are partial traces over one factor of a composite and the relabelling of $\rho$ along the regrouping (2). About these constructions the library proves that the partial trace of a density operator is again a density operator, so that every reduced state is a state, and that conjugation by an isometric bijection preserves positivity and trace, so that the relabelled state $\alpha \rho \alpha^{-1}$ is a state of $(A \otimes B) \otimes C$.