Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 (Quantum Information Theory)

All entropies and information rates below use base-two logarithms. The quantum fidelity uses the unsquared convention.

A bipartite density operator ${\rho }_{AB}$ has positive partial transpose when ${\rho }_{AB}^{T_B}\ge 0$ in a product basis. Positivity is independent of the choice of local bases, although the matrix representing the partial transpose changes.

If the state is a separable quantum state, write ${\rho }_{AB}={\sum }_j{p}_j{\rho }_j^A\otimes {\rho }_j^B$. Taking its partial transpose givesIndeed, the matrix transpose preserves the nonnegative eigenvalues of each local density operator; a tensor product of positive semidefinite operators is positive, and so is their nonnegative sum. This proves the necessary direction of the positive partial transpose criterion.

Use two precise finite-dimensional results. The positive-map separability criterion says that a state on ${\mathbb{C}}^2\otimes {\mathbb{C}}^3$ is separable exactly when $({\mathrm{id}}_2\otimes \Phi )(\sigma )\ge 0$ for every positive linear map $\Phi :M_3(\mathbb{C})\to M_2(\mathbb{C})$. The Størmer-Woronowicz decomposability theorem says that every such map is a decomposable positive map, sowith ${\Phi }_1,{\Phi }_2$ completely positive maps and $T$ the matrix transpose on $M_3$.

If $\sigma$ has positive partial transpose, both $\sigma$ and ${\sigma }^{T_B}$ are positive. HenceThe positive-map separability criterion now proves that $\sigma$ is a separable quantum state. The dimension-specific decomposability theorem is essential: the conclusion does not extend to arbitrary bipartite dimensions.

The partial transpose of the spin singlet state projector is the Hermitian operatorFor every separable quantum state $\tau$, the positive partial transpose criterion and the trace-adjoint identity for partial transpose giveFor the specified Bell state, however,Thus $W$ is an entanglement witness detecting $|{\Phi }^{+}⟩$. In fact, it detects any state whose overlap with that Bell state exceeds $1/2$.

Use the normalized Choi state $C_{\Lambda }=(\Lambda \otimes \mathrm{id})(|{\Phi }_d⟩⟨{\Phi }_d|)$, with $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|ii⟩$. If $\Lambda$ is an entanglement-breaking channel, its action on this particular bipartite input makes $C_{\Lambda }$ a separable quantum state.

Conversely, suppose $C_{\Lambda }={\sum }_a{p}_a{\sigma }_a\otimes {\tau }_a$, with local density operators ${\sigma }_a,{\tau }_a$. The Choi reconstruction formula givesBecause $\Lambda$ is trace preserving, ${\mathrm{Tr}}_A{C}_{\Lambda }=I/d$, so ${\sum }_a{E}_a=I$. Thus the $E_a$ form a POVM and the channel is a measure-and-prepare channel.

For any bipartite input ${\rho }_{AR}$, define the positive, possibly unnormalized reference operatorsIts output is ${\sum }_a{\sigma }_a\otimes R_a$. Since ${\sum }_a\mathrm{Tr}{R}_a=1$, normalizing each nonzero $R_a$ expresses this as a convex combination of product states. Hence every output is separable, proving the separable Choi-state criterion for entanglement breaking.

Apply the measure-and-prepare channel locally to an arbitrary bipartite density operator ${\rho }_{AR}$. The resulting state iswhereTo see that this is the correct output, the partial trace is cyclic for operators acting only on the traced subsystem, so the displayed expression equals ${\mathrm{Tr}}_A[(E_a\otimes I){\rho }_{AR}]$. Put $q_a=\mathrm{Tr}{R}_a$. The POVM completeness relation implies ${\sum }_a{q}_a=1$, and terms with $q_a=0$ vanish. Thereforeis a separable quantum state. Thus the channel is an entanglement-breaking channel, for every reference-system dimension.

Expanding the normalized maximally entangled state in the definition of the Choi state givesHere the bar denotes componentwise complex conjugation in the basis defining the Choi state. Every term is a positive tensor-product operator, and the whole operator has trace one because $\tilde{\Lambda }$ is assumed to be a quantum channel. It is therefore a separable quantum state, proving entanglement breaking by the separable Choi-state criterion for entanglement breaking.

If the vectors are unit vectors, the displayed $p_j$ are already the product-state weights. If they are not normalized, absorb their squared norms into the weights and normalize the nonzero vectors. Trace preservation requiresthe probability-distribution condition alone would not guarantee this. Equivalently, the Kraus operators are $K_j=\sqrt{d{p}_j}|a_j⟩⟨b_j|$, exhibiting the rank-one Kraus representation of an entanglement-breaking channel.

Use base-two logarithms. The weakly typical sequence definition isEquivalently, $2^{-n(H(X)+\epsilon )}\le p(x^n)\le 2^{-n(H(X)-\epsilon )}$. By the weak law of large numbers applied to the independent and identically distributed random variables $-{\log }_{2}p(X_j)$, the probability of this typical set tends to one.

This captures the usual probability scale of a long sample, but it need not make every symbol frequency representative. For a fair binary source, every word has probability $2^{-n}$ and $H(X)=1$, so even the all-zero word is weakly typical. The strongly typical sequence definition additionally controls empirical symbol frequencies and excludes this word for small tolerance. Thus weak typicality agrees with the high-probability-set intuition, while allowing individually unrepresentative members.

Every member of the typical set has probability at least $2^{-n(H(X)+\epsilon )}$. Summing these probabilities givessoThe upper bound itself holds at every block length. For sufficiently large $n$, the high-probability property also gives the companion lower boundhence $|T_{\epsilon }^{(n)}|\ge (1-\delta )2^{n(H(X)-\epsilon )}$. Together these are the typical-set cardinality bounds underlying Shannon source coding theorem.

Memorylessness means that the ensemble's average density operator is ${\rho }^{(n)}={\pi }^{\otimes n}$. Its spectral decomposition isThe product eigenvalues giveZero eigenvalues contribute zero to the Von Neumann entropy and are excluded from logarithmic typicality tests.

The quantum typical subspace is the span of eigenvectors whose eigenvalues satisfyIts orthogonal projection $P_{\epsilon }^{(n)}$ selects precisely the classical typical set for the eigenvalue distribution $q$. The typical-set cardinality bounds and the weak law of large numbers therefore givefor any fixed $\delta >0$ and all sufficiently large $n$.

A compression-decompression scheme consists of quantum channels ${\mathcal{E}}_n:\mathcal{D}({\mathcal{H}}^{\otimes n})\to \mathcal{D}({\mathcal{K}}_n)$ and ${\mathcal{D}}_n:\mathcal{D}({\mathcal{K}}_n)\to \mathcal{D}({\mathcal{H}}^{\otimes n})$, where ${\mathrm{limsup}}_n{n}^{-1}{\log }_2\dim {\mathcal{K}}_n\le R$. For the emitted pure-state ensemble, reliability means that the mean squared quantum fidelity tends to one:The average is taken over the actual source distribution, rather than the worst possible input vector.

A standard stronger formulation of reliable quantum source compression requireswhere entanglement fidelity tests preservation of a purification and its reference-system correlations. It implies the mean-fidelity condition for every pure-state ensemble of ${\pi }^{\otimes n}$. Schumacher compression achieves this at any rate $R>S(\pi )$: choose $0<\epsilon <R-S(\pi )$, encode the quantum typical subspace, and map atypical outcomes to a fallback state. The composed channel has a Kraus term $P_{\epsilon }^{(n)}$, so its entanglement fidelity is at least $[\mathrm{Tr}({\pi }^{\otimes n}{P}_{\epsilon }^{(n)}){]}^2\to 1$.

Let $Z$ indicate whether the outcome differs from $x_1$. Then $\mathbb{P}(Z=1)=q$. The chain rule for information entropy givessince $X=x_1$ is determined when $Z=0$. The remaining conditional distribution has at most $m-1$ outcomes, so the maximum entropy on a finite alphabet is ${\log }_2(m-1)$. Therefore the Shannon source coding theorem limit obeysHere $h_2$ is the binary entropy function; equality holds when the rare outcomes are equiprobable. A convenient explicit small-$q$ bound isbecause $-(1-q){\log }_2(1-q)\le q{\log }_{2}e$. For a fixed alphabet, this upper bound tends to zero as $q\to 0$.

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.

Let $\Delta =\rho -\sigma ={\Delta }_{+}-{\Delta }_{-}$ be its positive-negative decomposition. Since $\mathrm{Tr}\Delta =0$, the positive and negative parts have equal trace, namely $D(\rho ,\sigma )$. For every positive contraction $M$,Here both products have nonnegative trace, and $M\le I$ bounds the first term. Equality is attained by the orthogonal projection $P_{+}$ onto the positive spectral subspace of $\Delta$, since $P_{+}{\Delta }_{+}={\Delta }_{+}$ and $P_{+}{\Delta }_{-}=0$. Thus the variational characterization of trace distance is

Choose an orthonormal basis $\{|j⟩{\}}_{j=1}^d$ of the input space and define operators from the input to the $k$-dimensional outcome register byThese give a Kraus representation:The POVM completeness relation givesTherefore $\Phi$ is a completely positive map and is trace preserving: it is a quantum channel representing a deterministic measurement channel.

Let $P_{+}$ be the maximizing projection in the variational characterization of trace distance. Use the binary POVM $\{P_{+},I-P_{+}\}$ and its measurement channel. The difference between its two output probability vectors iswhere $D=D(\rho ,\sigma )$ and the second equality uses $\mathrm{Tr}(\rho -\sigma )=0$. Since trace distance between diagonal density operators is half the L1 norm of their probability-vector difference,Thus a binary quantum measurement can preserve the distinguishability of this particular pair exactly. This is the trace-distance-preserving binary measurement, which may depend on the two states.

Choose the trace-distance-preserving binary measurement from the previous part. Let $p,q$ be its output probability distributions, so $\Vert p-q{\Vert }_1=2D(\rho ,\sigma )$. The data-processing inequality for quantum relative entropy givesFor diagonal density operators, the quantum expression is the classical Kullback-Leibler divergence. Applying the supplied Pinsker's inequality givesConsequently the quantum Pinsker inequality isThe $\ln 2$ factor converts natural-logarithm relative entropy to bits. If the relative entropy is infinite because its support condition fails, the inequality holds automatically.

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.