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

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.