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

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.

Let $A$ be a register with orthonormal basis $\{|x⟩\}$ for Alice's classical symbol, $Q$ the transmitted system, and $B$ a register initially in a fixed blank state. The initial classical-quantum state isA POVM fixes probabilities but not a unique conditional state of $Q$. Choose the Lüders rule instrument $M_y=\sqrt{E_y}$, which satisfies ${\sum }_y{M}_y^{†}{M}_y=I$. Record outcome $y$ in orthogonal states of $B^{\prime }$ and retain the corresponding quantum output $Q^{\prime }$. The final nonselective state isThe positive operators in this sum are unnormalized: their traces already include the outcome probabilities. If $\mathrm{Tr}(E_y{\rho }_x)>0$, the normalized conditional state of $Q^{\prime }$ is $M_y{\rho }_x{M}_y^{†}/\mathrm{Tr}(E_y{\rho }_x)$.

More generally, an instrument may use operators $M_{y\mu }$ with ${\sum }_{\mu }{M}_{y\mu }^{†}{M}_{y\mu }=E_y$, replacing $M_y{\rho }_x{M}_y^{†}$ by ${\sum }_{\mu }{M}_{y\mu }{\rho }_x{M}_{y\mu }^{†}$. All subsequent classical information bounds are unchanged. This explicitly accounts for the fact that a POVM does not determine the post-measurement state of a retained quantum system.

The initially blank register $B$ is pure and independent, so adding it changes neither relevant Von Neumann entropy difference: $I(A:QB)=I(A:Q)$. Bob's measurement and recording procedure, with all outcomes retained, is a matrix trace-preserving quantum channel from $QB$ to $Q^{\prime }{B}^{\prime }$. Part (a) therefore gives $I(A^{\prime }:Q^{\prime }{B}^{\prime })\le I(A:QB)$. Discarding $Q^{\prime }$ gives $I(A^{\prime }:B^{\prime })\le I(A^{\prime }:Q^{\prime }{B}^{\prime })$. Combining the two inequalities provesAlice's register is not changed by Bob's local operation; the primes label its place in the final joint state rather than a change in its classical symbol.

Trace out $Q^{\prime }$ in the final state. With $q(x,y)=p(x)\mathrm{Tr}(E_y{\rho }_x)$, the remaining state is diagonal in the product classical basis:The Von Neumann entropy of a classical diagonal density operator equals the Shannon entropy of its diagonal probabilities, by its spectrum. Therefore $I(A^{\prime }:B^{\prime })=H(X)+H(Y)-H(X,Y)=I(X:Y)$, the mutual information between source symbol and measurement outcome.

For the input, put $\bar{\rho }={\sum }_{x}p(x){\rho }_x$. The marginal on $Q$ is $\bar{\rho }$, the marginal on $A$ has Von Neumann entropy $H(X)$, and the entropy of a classical-quantum state isTo justify the last formula, diagonalize each ${\rho }_x$ with eigenvalues ${\lambda }_{xj}$. The joint block-diagonal state has eigenvalues $p(x){\lambda }_{xj}$; expanding $-{\sum }_{x,j}p(x){\lambda }_{xj}{\log }_2[p(x){\lambda }_{xj}]$ yields exactly the displayed Von Neumann entropy. Thus $I(A:Q)=S(\bar{\rho })-{\sum }_{x}p(x)S({\rho }_x)$, the Holevo quantity of the ensemble. Part (ii) becomes the Holevo boundIt holds for every POVM, and hence also for the accessible information obtained by maximizing the mutual information over all measurements. The only extra Von Neumann entropy identities used were the classical-diagonal and classical–quantum block-spectrum formulas, both justified above; the information inequality itself followed from the requested strong-subadditivity argument.