Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 3

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

Assume a finite-dimensional memoryless quantum channel, no shared entanglement, and a code of $M_n$ messages with $n^{-1}{\log }_2{M}_n\to R$. Use a uniform message $M$ to relate average and maximum error. Each permitted input is a product state ${\rho }_m^{(n)}={⨂}_{i=1}^n{\rho }_{m,i}$, giving output ${\sigma }_m^{(n)}={⨂}_i\Lambda ({\rho }_{m,i})$. The factors may depend on both the message and the channel use.

Let ${\chi }^{\ast }(\Lambda )$ denote the one-use Holevo capacity in the question. The Subadditivity of Von Neumann entropy for the average output, and entropy additivity for each product output, implyThis is the key step of the product-input classical-capacity converse; it does not assume that the average output itself is a product.

Bob may measure all outputs jointly. The Holevo bound still gives $I(M:\hat{M})\le {\chi }_n$. Let $P_e$ be his average message-error probability. Fano's inequality givesSince the maximum error $P_{\max }$ is at least $P_e$,when $R>{\chi }^{\ast }(\Lambda )$. Thus the maximum error cannot vanish asymptotically.

Put $p=⟨\psi |M_1^2|\psi ⟩\ge 1-\epsilon$. Since $M_1$ is a positive contraction, the successful post-measurement state is the pure state represented by $|{\psi }^{\prime }⟩=M_1|\psi ⟩/\sqrt{p}$. Its squared quantum fidelity with the input isThe finite-dimensional spectral theorem and $0\le M_1\le I$ give $M_1^2\le M_1$, so $⟨M_1⟩\ge p$. Therefore the pure-state gentle measurement bound isThe positive-operator assumption matters: an arbitrary unitary measurement operator can have success probability one while rotating the state to an orthogonal vector.