Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 5

First prove monotonicity for a partial trace. Let $r=\dim {\mathcal{H}}_B$ and define the Heisenberg-Weyl twirling channelThe supplied identity on individual operators extends to bipartite operators by expanding them in a tensor-product basis. The joint convexity of quantum relative entropy and its unitary invariance implyThe additivity of quantum relative entropy makes the left side $D({\rho }_A\Vert {\sigma }_A)$, proving the partial-trace case.

For a general deterministic quantum operation, meaning a quantum channel, use a Stinespring dilation $\Lambda (X)={\mathrm{Tr}}_E(VX{V}^{†})$ with $V^{†}V=I$. An isometric embedding preserves quantum relative entropy, because restricting the output to the common image of $V$ preserves the eigenvalues and trace formula. Applying the partial-trace result givesThe other properties used are unitary invariance and invariance under adjoining an identical ancillary state; both follow directly from the relative-entropy trace formula. This is the Lindblad-Uhlmann monotonicity theorem. The proof applies to deterministic channels; normalized postselection is not such a linear channel.

Write a qubit density operator in Bloch vector form, $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. The quantum depolarizing channel maps $\mathbf{r}$ to $p\mathbf{r}$, so its output eigenvalues are $(1\pm p|\mathbf{r}|)/2$. The output Von Neumann entropy is minimized on pure inputs, where it equals $h_2((1+p)/2)$.

For any input ensemble, its output Holevo quantity is at most the maximum qubit entropy minus this minimum output entropy:Equality is attained by two equiprobable orthogonal pure inputs: their average output is $I/2$, while each output has the minimum entropy. ThusThis uses the question's retention parameter $p$; in the usual convex-mixture range, $0\le p\le 1$. The expression also holds throughout the full completely positive qubit range $-1/3\le p\le 1$, since $h_2((1+p)/2)=h_2((1+|p|)/2)$.

The supplied additivity of Holevo capacity gives ${\chi }^{\ast }({\Lambda }_{\mathrm{dep}}^{\otimes n})=n{\chi }^{\ast }({\Lambda }_{\mathrm{dep}})$. The Holevo-Schumacher-Westmoreland theorem expresses the unassisted classical capacity of a quantum channel as the regularized Holevo capacity, soTherefore entangled inputs across channel uses cannot increase this classical capacity.

Let $\overline{\omega }={\sum }_j{p}_j{\omega }_j$. When the relative entropies are finite, expanding their trace definitions givesThus Donald's identity isFor every $p_j>0$, the support of a positive operator ${\omega }_j$ lies in that of $\overline{\omega }$, so the first sum on the right is finite in finite dimension. If the support of $\overline{\omega }$ is not contained in the support of $\rho$, at least one positive-weight ${\omega }_j$ also violates the support condition. Both sides of the identity are then $+\infty$. This extends the conclusion without subtracting infinite quantities.

By Donald's identity, the objective equalsThe nonnegativity of quantum relative entropy makes the last term at least zero, with equality exactly when $\rho =\overline{\omega }$. Hence the unique minimizing density operator is the ensemble average andFor the classical-quantum state ${\sigma }_{XB}$, its marginal on $X$ has entropy $H(p)$, its marginal on $B$ is $\overline{\omega }$, and its joint entropy is $H(p)+{\sum }_j{p}_{j}S({\omega }_j)$. Substituting in the definition of quantum mutual information givesThis is the relative-entropy barycenter of a quantum ensemble: the average state minimizes mean distinguishability from the ensemble. Its minimal value is exactly the Holevo quantity.