Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 (Quantum Information Theory)

A quantum channel $T:\mathcal{B}(\mathcal{H})\to \mathcal{B}(\mathcal{K})$ is a linear, completely positive, trace-preserving map. Complete positivity means $T\otimes {\mathrm{id}}_n$ is positive for every ancillary dimension $n$.

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The Kraus representation isThe second identity is precisely trace preservation. Different Kraus families related by an isometry represent the same channel.

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Since $|\varphi ⟩⟨\varphi |\ge 0$ and a channel is completely positive,Writingand tracing over the output system givesThuswhere the factor $1/d$ follows from the normalized maximally entangled state.

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For $T(\rho )={\sum }_k{p}_k{U}_k\rho U_k^{†}$,Each $(U_k\otimes I)|\varphi ⟩$ is a maximally entangled state, so the Choi matrix is a convex combination of maximally entangled pure states.

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Every random unitary channel satisfiesIt is therefore a unital quantum channel.

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The normalized Choi matrix of the trace-to-identity map is $I\otimes I/d$, while that of transposition is the flip operator $F/d$. Hence the Werner–Holevo channel hasthe normalized projector onto the antisymmetric subspace.

If the channel were random unitary, part (i) would express $C_T$ as a mixture of maximally entangled vectors. Every vector in such a mixture must lie in the support of $C_T$, hence in the antisymmetric subspace. Under vectorization, an antisymmetric vector corresponds to a skew-symmetric matrix $A$, while maximal entanglement requires $A{A}^{†}$ to be proportional to $I$. In odd dimension, $\det A=\det (-A^T)=-\det A$, so $\det A=0$; such an $A$ cannot be proportional to a unitary. Thus for odd $d\ge 3$, and in particular $d=3$, this unital channel is not random unitary, disproving the converse. The odd-dimensional qualification matters because antisymmetric maximally entangled vectors can exist in even dimension.

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A bipartite density operator is a separable quantum state whenfor probabilities $p_j$ and local states. If no such convex decomposition exists, it is an entangled state.

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The positive partial transpose criterion says separability implies ${\rho }_{AB}^{T_B}\ge 0$. A nonpositive partial transpose therefore proves entanglement. Positive partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, but not in general higher dimensions.

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After permuting the computational basis, $\rho$ is the direct sum ofThe stated diagonal and trace conditions already give Hermiticity and trace one. Each $2\times 2$ block is positive semidefinite exactly when its determinant is nonnegative. Thus $\rho$ is a density matrix precisely when

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Partial transposition interchanges the positions occupied by $x$ and $y$. Positivity of ${\rho }^{T_B}$ therefore requiresBecause the system is $2\otimes 2$, the positive partial transpose criterion is necessary and sufficient. Combining these inequalities with validity of the original state gives

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For a bipartite input,Each second factor is positive and has trace equal to the probability of measurement outcome $k$. Dividing nonzero factors by their traces therefore writes the output as a convex combination of product states. Hence every measure-and-prepare channel is an entanglement-breaking channel.

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If $T$ is entanglement breaking, applying it to one half of $|\varphi ⟩⟨\varphi |$ immediately shows that its Choi matrix is separable.

Conversely supposeis separable. The Choi reconstruction formula for the normalized convention isThe trace-preserving condition ${\mathrm{Tr}}_1{C}_T=I/d$ implies ${\sum }_k{M}_k=I$, so $\{M_k\}$ is a POVM. Part (i) now proves that $T$ is entanglement breaking. Thus

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In quantum binary hypothesis testing, hypothesis zero supplies ${\rho }_0$ with prior $p$ and hypothesis one supplies ${\rho }_1$ with prior $1-p$. A two-outcome POVM $\{Q,I-Q\}$ decides zero on outcome $Q$. The conditional errors areSymmetric testing minimizes the prior-weighted average error $p\alpha +(1-p)\beta$, equivalently maximizing the average success probability.

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Let $\Delta =p{\rho }_0-(1-p){\rho }_1$. The success probability of $Q$ isWrite the spectral decomposition $\Delta ={\Delta }_{+}-{\Delta }_{-}$. For every effect $0\le Q\le I$,with equality when $Q$ projects onto the positive spectral subspace, with arbitrary action on the kernel. Since $\mathrm{Tr}\Delta =2p-1$ and $\Vert \Delta {\Vert }_1=\mathrm{Tr}{\Delta }_{+}+\mathrm{Tr}{\Delta }_{-}$, the Holevo–Helstrom theorem follows:

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Put $c=\cos (\theta /2)$, $s=\sin (\theta /2)$, and $\zeta =e^{2\pi i/3}$. Averaging the three states cancels the off-diagonal phases:For $c,s>0$, the pretty good measurement isbecause ${\bar{\rho }}^{-1/2}|{\psi }_k⟩=|0⟩+{\zeta }^k|1⟩$. The matrices are positive and ${\sum }_k{M}_k=I$. At the endpoint values of $\theta$, the same formula is understood on the support of $\bar{\rho }$ and may be completed arbitrarily on its kernel.

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The Holevo optimality conditions say a POVM is optimal whenis Hermitian and $\Gamma -p_k{\rho }_k\ge 0$ for every $k$. Hereandwhose eigenvalues are $0$ and $2cs/3$. The pretty good measurement is therefore optimal. Its success probability is

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For a density operator $\rho$ on a finite-dimensional Hilbert space, the Von Neumann entropy iswith $0\log 0=0$. Its concavity says that, for $0\le p\le 1$,

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Put $\rho =p{\rho }_1+(1-p){\rho }_2$. The operator inequality $\rho \ge p{\rho }_1$ and the operator monotonicity of logarithm give, on the support of ${\rho }_1$,Consequently,The corresponding inequality from $\rho \ge (1-p){\rho }_2$ yieldsAdding these inequalities and using $S(\rho )=-\mathrm{Tr}(\rho \log \rho )$ proves the entropy bound for a binary mixture:Singular states follow by adding a positive multiple of the identity and taking a limit; the endpoint cases use $0\log 0=0$.

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If $\epsilon =0$, then ${\rho }_{AB}={\sigma }_{AB}$ and the desired continuity bound is immediate, so assume $\epsilon >0$. Apply the positive-negative decomposition of a Hermitian operator toBecause $\mathrm{Tr}X=0$ and $\Vert X{\Vert }_1=2\epsilon$, one hasThus ${\Delta }_{AB}=X_{+}/\epsilon$ is positive with trace one, hence is a density operator. DefineIt is a convex combination of states. The equation $\epsilon {\Delta }_{AB}^{\prime }=(1+\epsilon ){\omega }_{AB}-{\rho }_{AB}$ giveswhich is likewise positive and has trace one.

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The two forms of ${\omega }_{AB}$ found in part (i) giveTherefore, in the Löwner order,

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Set $q=(1+\epsilon {)}^{-1}$, so $1-q=\epsilon /(1+\epsilon )$ and $\omega =q\rho +(1-q){\Delta }^{\prime }$. For every state ${\xi }_B$, the entropy bound for a binary mixture givesTaking the minimum over ${\xi }_B$ and using the variational characterization of quantum conditional entropy on each term givesSince binary entropy satisfies $H(q)=H(1-q)$, this is

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The other convex decomposition is $\omega =q\sigma +(1-q)\Delta$. Applying the concavity of quantum conditional entropy, which follows from the Strong subadditivity of Von Neumann entropy, yields

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Combining parts (iii) and (iv), then multiplying by $1+\epsilon$, givesThe dimension bound for quantum conditional entropy isIndeed, Subadditivity of Von Neumann entropy gives $H(A|B{)}_{\tau }\le S({\tau }_A)\le \log d_A$, while the Araki–Lieb inequality gives $H(A|B{)}_{\tau }\ge -S({\tau }_A)\ge -\log d_A$. HenceInterchanging $\rho$ and $\sigma$ proves the continuity bound for quantum conditional entropy:

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