The Exercise

Three quantum systems $A$, $B$ and $C$ share a joint state. How strongly $A$ is correlated with $B$ is measured by the quantum mutual information $S(A:B)$ of the reduced state of the pair $AB$, and its correlation with $C$ by $S(A:C)$. The exercise bounds the two together by the entropy of $A$ alone: their sum never exceeds $2S(A)$.^✦ It then asks for a state of two systems in which a single mutual information is larger than the entropy, $S(A:B) > S(A)$, something classical correlations cannot do, since the mutual information of two random variables is at most the entropy of either.

Systems, States and Entropy

A quantum system $A$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ The composite system $A \otimes B$ of quantum systems $A$ and $B$ is the quantum system whose state space is the tensor product of their state spaces.^✦ A state $\rho$ of a quantum system $A$ is a positive operator of unit trace on the state space of $A$.^✦

A state $\rho$ of a composite system $A \otimes B$ has two reduced states: the state $\rho_A = \operatorname{Tr}_B \rho$ of $A$, its partial trace over $B$^✦, and the state $\rho_B = \operatorname{Tr}_A \rho$ of $B$, its partial trace over $A$.^✦

The von Neumann entropy of a state $\rho$ is the Shannon entropy of the eigenvalues of $\rho$, measured in nats: $$S(\rho) = -\operatorname{tr}(\rho \log \rho).$$(1)^✦ The quantum mutual information of a state $\rho$ of a composite system $A \otimes B$ is $$S(A:B) = S(\rho_A) + S(\rho_B) - S(\rho).$$(2)^✦

Regrouping Three Systems

An identification $e$ of a quantum system $A$ with a quantum system $A'$ is a linear isometry of the state space of $A$ onto that of $A'$, under which the two count as the same system up to reorganisation.^✦ The relabelling of a state $\rho$ of $A$ along $e$ is the state $e \rho e^{-1}$ of $A'$.^✦

The regrouping $\alpha_{A,B,C}$ of quantum systems $A$, $B$ and $C$ is the identification of $A \otimes (B \otimes C)$ with $(A \otimes B) \otimes C$^✦, and the swap $\mathrm{swap}_{B,C}$ of $B$ and $C$ is the identification of $B \otimes C$ with $C \otimes B$.^✦ On product vectors they act as $$\alpha_{A,B,C}\colon u \otimes (v \otimes w) \mapsto (u \otimes v) \otimes w, \qquad \mathrm{swap}_{B,C}\colon v \otimes w \mapsto w \otimes v.$$(3) The identity $\mathrm{id}_A$ of $A$ is its identification with itself by the identity map of its state space.^✦ For identifications $e$ of $A$ with $A'$ and $f$ of $B$ with $B'$, the product $e \otimes f$ is the identification of $A \otimes B$ with $A' \otimes B'$ by the tensor product of the two isometries.^✦

For a state $\rho$ of $A \otimes (B \otimes C)$, the reduced states of the pairs $AB$ and $AC$ are taken after the factors are regrouped, $$\rho_{AB} = \operatorname{Tr}_C\bigl(\alpha_{A,B,C}\,\rho\,\alpha_{A,B,C}^{-1}\bigr), \qquad \rho_{AC} = \operatorname{Tr}_B\bigl(\beta\,\rho\,\beta^{-1}\bigr), \quad \beta = \alpha_{A,C,B}\,(\mathrm{id}_A \otimes \mathrm{swap}_{B,C}),$$(4) and $\rho_A = \operatorname{Tr}_{BC}\,\rho$ is the reduced state of $\rho$ on $A$.

The Claim

What Is Checked

Both claims are stated over states of composite quantum systems, their reduced states, and the entropy (1) and mutual information (2) built from them; the bound is stated in addition over the regrouping, the swap, the identity and their product in (4), with the relabelling of a state along each.

The library proves that the partial trace of a state over either factor is again a state, and that a state relabelled along an identification is a state, since conjugation by an isometry preserves positivity and trace.