Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 33 / 5 / a

Use base-two logarithms, so the Von Neumann entropy is $S(\rho )=-\mathrm{Tr}(\rho {\log }_2\rho )$, with $0\log 0=0$. The quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$; this is the quantity denoted by $S({\rho }_A:{\rho }_B)$ in the question, using the given joint state.

Strong subadditivity of Von Neumann entropy states that for every tripartite density operator,Equivalently the conditional quantum mutual information $I(A:C\mid B)$ is nonnegative. We apply this property to the original state or to a dilation of the relevant quantum channel.

Expand the difference between the two quantum mutual informations:This is precisely the nonnegative expression in Strong subadditivity of Von Neumann entropy. ThusThe reduced state on $AB$ is obtained by the partial trace over $C$, so this proves that discarding a system cannot increase its quantum mutual information with the retained reference.

Here a quantum operation is interpreted as a deterministic quantum channel, so it is completely positive and matrix trace preserving. Write its Kraus representation with ${\sum }_k{A}_k^{†}{A}_k=I$ and define the isometry in its Stinespring dilationThen $V^{†}V=I$ and $\Phi (\rho )={\mathrm{Tr}}_E(V\rho V^{†})$. Acting on $B$ gives ${\sigma }_{A{B}^{\prime }E}=(I_A\otimes V){\rho }_{AB}(I_A\otimes V^{†})$. An isometry preserves the nonzero eigenvalues of a density operator. Hence $S(A{B}^{\prime }E)=S(AB)$, $S(B^{\prime }E)=S(B)$, and $S(A)$ is unchanged. It follows that $I(A:B^{\prime }E{)}_{\sigma }=I(A:B{)}_{\rho }$.

Now discard $E$ and apply part 1. This proves data processing for quantum mutual information:The matrix trace-preserving convention matters. Postselection can increase conditional quantum mutual information: take a uniform classical bit $A$, copy it to $B$ with probability $\epsilon$, and otherwise put $B$ in an erasure state independent of $A$. Initially $I(A:B)=\epsilon$ bits. Projecting onto the nonerased sector and conditioning on success gives a perfectly correlated bit pair with quantum mutual information one. That success branch is matrix trace decreasing, and its renormalized action is not a deterministic channel. Retaining the full outcome flag and averaging all branches restores the ordinary channel inequality.