Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 (Quantum Information Theory)

A rate-$R$ block code consists of an encoder $E_n:J^n\to \{1,\dots ,M_n\}$ and decoder $D_n$, with $M_n\le 2^{nR}$; it is reliable when $\mathrm{Pr}[D_n(E_n(U^n))\ne U^n]\to 0$. Choose $\epsilon >0$ with $H(U)+\epsilon <R$. The typical set has probability tending to one and cardinality at most $2^{n(H(U)+\epsilon )}\le 2^{nR}$. Encode its words injectively and map every atypical word to a default codeword. The error probability is at most the atypical probability, so reliable compression exists.

A quantum rate-$R$ code uses a compression channel into a space of dimension at most $2^{nR}$ followed by a decompression channel; reliability means that its entanglement fidelity, and hence its average source-state fidelity, tends to one on ${\pi }^{\otimes n}$. Diagonalize $\pi ={\sum }_{u}p(u)|u⟩⟨u|$. The projector onto eigenvectors labelled by the classical typical set has dimension at most $2^{n(S(\pi )+\epsilon )}$ and expectation in ${\pi }^{\otimes n}$ tending to one. For $R>S(\pi )$ choose $\epsilon$ so this typical subspace fits in $2^{nR}$ dimensions, encode it isometrically, and send the orthogonal complement to a fixed state. The typical-subspace probability and the gentle-measurement estimate make the fidelity tend to one. This is Schumacher compression.

Every word in $C_n(r)$ has probability at least $2^{-nr}$, so normalization gives $|C_n(r)|\le 2^{nr}$. Choose $H(U)<r<R$ and then $\epsilon <r-H(U)$. Every $\epsilon$-typical word satisfies $p(u^n)\ge 2^{-n(H(U)+\epsilon )}\ge 2^{-nr}$, hence $T_{\epsilon }^{(n)}\subseteq C_n(r)$ and $\mathrm{Pr}[C_n(r)]\to 1$. Injectively encoding $C_n(r)$ and using a default codeword outside it gives rate at most $R$ and vanishing error.

The map $\Lambda$ is a completely positive map when $\Lambda \otimes {\mathrm{id}}_m$ maps positive operators to positive operators for every $m$. In a basis $\{|i⟩\}$ define the unnormalized Choi matrixIf $\Lambda$ is completely positive, $J(\Lambda )=(\mathrm{id}\otimes \Lambda )(|\Omega ⟩⟨\Omega |)\ge 0$, where $|\Omega ⟩={\sum }_i|ii⟩$. Conversely, decompose $J={\sum }_k|a_k⟩⟨a_k|$ and reshape each vector into an operator $A_k$. The Choi reconstruction formula gives $\Lambda (X)={\sum }_k{A}_{k}X{A}_k^{†}$, which is completely positive. Thus

For every ancillary system,when $X\ge 0$, proving complete positivity. Cyclicity of trace givesThis equals $\mathrm{Tr}\rho$ for every $\rho$ exactly whenthe trace-preserving condition in the Kraus representation.

With $P_i=|y_i⟩⟨y_i|$, the channel has Kraus form $Y(\rho )={\sum }_i{P}_i\rho P_i$ and ${\sum }_i{P}_i^{†}{P}_i=I$, so it is completely positive and trace preserving. It measures in the $y$ basis and prepares the observed basis state, hence is a measure-and-prepare channel. If $p_i=⟨y_i|\rho |y_i⟩$, then $Y(\rho )={\sum }_i{p}_i{P}_i$ andMoreover $\log Y(\rho )={\sum }_i(\log p_i)P_i$ on its support, so

Write $q_j=⟨z_j|\rho |z_j⟩$, $r_j=⟨z_j|Y(\rho )|z_j⟩$, and $c={\max }_{i,j}|⟨y_i|z_j⟩{|}^2$. Thenand thereforeThe data-processing inequality for quantum relative entropy applied to $Z$ and part c giveCombining them proves

Both states are block diagonal in $X$. On the support of each block,Substitution into quantum relative entropy givesTaking $-\mathrm{Tr}{\rho }_{XB}\log {\rho }_{XB}$ similarly yields the entropy of a classical-quantum state:

Data processing states $D(\Phi (\rho )\Vert \Phi (\sigma ))\le D(\rho \Vert \sigma )$ for every quantum channel $\Phi$. Joint convexity statesStrong subadditivity states $S(ABC)+S(B)\le S(AB)+S(BC)$.

For strong subadditivity, apply data processing under ${\mathrm{Tr}}_C$ to ${\rho }_{ABC}$ and $I_A/d_A\otimes {\rho }_{BC}$. Expanding both relative entropies cancels $\log d_A$ and gives precisely the stated inequality. For joint convexity, use flagged states $\hat{\rho }={\sum }_i{p}_i|i⟩⟨i|\otimes {\rho }_i$ and $\hat{\sigma }={\sum }_i{p}_i|i⟩⟨i|\otimes {\sigma }_i$. Part a gives $D(\hat{\rho }\Vert \hat{\sigma })={\sum }_i{p}_{i}D({\rho }_i\Vert {\sigma }_i)$; discarding the flag and applying data processing gives joint convexity of quantum relative entropy.

For an ensemble $\{p_x,{\rho }_x\}$ and measurement outcome $Y$, the Holevo bound iswhere $\bar{\rho }={\sum }_x{p}_x{\rho }_x$. Form the classical-quantum state ${\omega }_{XB}={\sum }_x{p}_x|x⟩⟨x|\otimes {\rho }_x$. Part a givesThe measurement is a quantum channel from $B$ to a classical register $Y$. Data processing for relative entropy, applied to the mutual-information representation, gives $I(X:Y)\le I(X:B)$ and proves the bound.

The quantity ${\chi }^{\ast }(\Lambda )$ is the one-use Holevo information of the channel: by the Holevo--Schumacher--Westmoreland theorem, its regularization gives the classical capacity, and without entangled inputs across uses it is the asymptotically achievable classical rate using collective output measurements.

If an ensemble contains a mixed ${\rho }_x={\sum }_j{q}_{j|x}|{\psi }_{xj}⟩⟨{\psi }_{xj}|$, refine its label to $(x,j)$ with probability $p_x{q}_{j|x}$. The average channel output is unchanged, while Concavity of Von Neumann entropy givesThus refinement cannot decrease the Holevo quantity, so the maximum may be restricted to pure input states.

For Hermitian $X$, $\Vert X{\Vert }_1={\max }_{-I\le H\le I}\mathrm{Tr}(HX)$. The adjoint of a trace-preserving completely positive map is unital and positive, so $-I\le H\le I$ implies $-I\le {\Lambda }^{\ast }(H)\le I$. ThereforeHence trace distance is contractive under quantum channels.

Let $X=\rho -\sigma =X_{+}-X_{-}$ be its positive-negative spectral decomposition and let $P$ project onto the support of $X_{+}$. Since $\mathrm{Tr}X=0$, $\mathrm{Tr}{X}_{+}=\mathrm{Tr}{X}_{-}=\Vert X{\Vert }_1/2$. Thus

Choose a basis $|a⟩$ of $H$ and define Kraus operators from $H$ to ${\mathbb{C}}^m$ byThenandThe displayed map is therefore a completely positive trace-preserving measure-and-prepare channel.

For any POVM, part c followed by trace-distance contraction givesFor the binary POVM $\{P,I-P\}$ from part b, equality holds. Consequently

For outcome probabilities $p_k=\mathrm{Tr}({\Lambda }_k\rho )$ and $q_k=\mathrm{Tr}({\Lambda }_k\sigma )$, the hint givesThe left side is at most $\Vert \rho -\sigma {\Vert }_1/2$ by part d. Choose a POVM attaining the stated minimum characterization of quantum fidelity; then

The Schmidt decomposition is $|\psi ⟩={\sum }_j{s}_j|a_j⟩|b_j⟩$. Its reduced states are ${\rho }_A={\sum }_j{s}_j^2|a_j⟩⟨a_j|$ and ${\rho }_B={\sum }_j{s}_j^2|b_j⟩⟨b_j|$. They have the same nonzero eigenvalues, so

For the pure state ${\sigma }_{RQE}$ induced by a purification and Stinespring dilation, the coherent information is

Choose any pure input $\rho =|\varphi ⟩⟨\varphi |$. Its reference system may be one-dimensional, and the Stinespring output on $QE$ is pure. Its two reduced states have equal entropy by part a, hence

Because ${\sigma }_{RQE}$ is pure, $S(RQ)=S(E)$ and $S(RE)=S(Q)$. ThereforeThus

For an anti-degradable quantum channel, a channel from $E$ simulates $Q$. The data-processing inequality for quantum relative entropy applied to quantum mutual information gives $I(R:Q)\le I(R:E)$. Part d then implies $I_c(\Lambda ,\rho )\le 0$ for every input. Part c supplies a pure input attaining zero, so

Assume an anti-degradable channel transmits perfectly with encoder $\mathcal{E}$ and decoder $\mathcal{D}$. Apply its Stinespring isometry to $\mathcal{E}(\rho )$, producing receiver system $Q$ and environment $E$. The receiver obtains $\rho$ by $\mathcal{D}$; anti-degradability lets the environment simulate the receiver output using $\mathcal{N}$, and then $\mathcal{D}\circ \mathcal{N}$ also produces $\rho$.

Apply these two local decoding channels simultaneously to $Q$ and $E$. Each marginal of the resulting bipartite state is the original pure state $\rho$. A bipartite state with a pure marginal is a product, so the joint output is $\rho \otimes \rho$. We have therefore constructed one quantum channel mapping every pure $\rho$ to $\rho \otimes \rho$, contradicting the no-cloning theorem for two pure states. Hence no anti-degradable channel can transmit arbitrary quantum information perfectly in one use.