Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 65 / 4

For a finite-dimensional memoryless quantum channel $\Lambda$, the Holevo-Schumacher-Westmoreland theorem identifies its product-state classical capacity with the optimized output Holevo quantity:With product codewords and a collective measurement on the outputs, every rate below this value is achievable with vanishing error; no larger rate can be reliable under that product-input restriction. For a fixed classical-quantum output alphabet the corresponding optimized entropy difference gives its classical coding capacity. If entangled inputs across channel uses are also allowed, the general unassisted capacity is the regularized value $C(\Lambda )={\lim }_{n\to \infty }{n}^{-1}{\chi }^{\ast }({\Lambda }^{\otimes n})$. The requested calculation below concerns the single-use optimized product-input expression, so no unproved additivity assumption is needed.

Conjugation by a Pauli operator preserves its matching Bloch vector component and reverses the other two. Summing all three conjugations therefore sends $\mathbf{s}$ to $-\mathbf{s}$. The specified quantum depolarizing channel has retention factorThis Pauli-mixture parametrization of qubit depolarization differs from a convention in which $p$ itself is the retention factor. For $0\le p\le 1$, $-1/3\le \eta \le 1$. A pure input gives output eigenvalues $(1\pm |\eta |)/2$, and hence entropy $h_2((1+\eta )/2)$ by symmetry of binary entropy. Mixed inputs have a smaller output radius and at least this much entropy. Every average output has entropy at most one. ThusTwo equiprobable orthogonal pure inputs have antipodal output Bloch vectors, maximally mixed average output, and this same minimal individual output entropy. They attain the bound. Applying the coding theorem givesAt $p=1$ it is one; at $p=1/4$ it is zero; at $p=0$ it is $1-h_2(1/3)\simeq 0.0817$. Negative retention reverses the labels of the two signals but does not eliminate their distinguishability.

Encode the ensemble label in an orthogonal classical register:The classical-quantum state has $S(XB)=H(p)+{\sum }_x{p}_{x}S({\rho }_x)$, $S(X)=H(p)$ and $S(B)=S(\bar{\rho })$, where $\bar{\rho }={\sum }_x{p}_x{\rho }_x$. HenceApplying ${\mathrm{id}}_X\otimes \Lambda$ changes the conditional states to the output ensemble. The same identity gives $I(X:B^{\prime })=\chi ({\mathcal{E}}^{\prime })$. Data processing for quantum mutual information, proved in Question 3, therefore yields the Holevo quantity under a quantum channel:This concerns the information available in the whole ensemble, not the change in the entropy of each individual state.

Diagonalize the source average $\pi ={\sum }_j{\lambda }_j|j⟩⟨j|$, using its positive eigenvalues. For fixed $\delta >0$, let ${\Pi }_{n,\delta }$ project onto product eigenvectors whose eigenvalues obeyThe typical subspace theorem states that, for every $ϵ>0$ and all sufficiently large $n$,The probability statement is the weak law of large numbers applied to $-{\log }_2{\lambda }_j$; the dimension bounds follow by summing the typical eigenvalue bounds. This explains why the quantum typical subspace retains almost all probability using about $nS(\pi )$ qubits.

For $R>S(\pi )$ choose $0<\delta <R-S(\pi )$. Measure $\{{\Pi }_{n,\delta },I-{\Pi }_{n,\delta }\}$. On success, encode the projected state isometrically into a space of dimension $\dim {\mathcal{T}}_{n,\delta }$; on failure, output a separate fixed flag. Decode the successful sector by the inverse isometry, and map the flag to a fixed state $|{\phi }_0⟩$. The compressed dimension is at most $2^{n(S(\pi )+\delta )}+1$, hence fits within rate $R$ for all sufficiently large $n$. Both maps are trace-preserving quantum channels, not merely successful postselected operations.

Writing $\Pi ={\Pi }_{n,\delta }$, the composite channel isFor a pure source signal $|{\Psi }_k⟩$, set $a_k=⟨{\Psi }_k|\Pi |{\Psi }_k⟩$. Its squared quantum fidelity after the composite channel is at least $a_k^2$. Since the source is memoryless, its average state on $n$ uses is ${\pi }^{\otimes n}$. Convexity of the square givesThe Kraus formula for entanglement fidelity gives the same lower bound for $F_e({\pi }^{\otimes n},{\mathcal{N}}_n)$, because one Kraus operator is $\Pi$ and the other terms are nonnegative. Taking $ϵ\to 0$ proves reliable compression at every $R>S(\pi )$, in both average pure-signal fidelity and the stronger entanglement-fidelity sense. This is typical-subspace compression with a failure flag.