Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 323 (Quantum Information Theory)

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

Use an orthonormal eigenbasis of the Hermitian operator $A$ and set $b_j=⟨{\alpha }_j|B|{\alpha }_j⟩$. Since $B$ is a positive contraction,For $0<r<d$, the missing weight among the first $r$ entries equals the weight in the remaining entries:The ordered eigenvalues satisfy ${\lambda }_j\ge {\lambda }_{r-1}$ for $j<r$ and ${\lambda }_j\le {\lambda }_{r-1}$ for $j\ge r$. ThereforeThis argument does not require the eigenvalues of $A$ to be positive. For $r=0$, positivity and zero trace give $B=0$; for $r=d$, the same argument applied to $I-B$ gives $B=I$. Thus in every caseThe orthogonal projection onto the leading $r$ eigenvectors attains equality. This is the Hermitian effect variational principle, extending the Ky Fan maximum principle to all positive contractions with the prescribed trace.

Choose Kraus representationsThe composite quantum channel has Kraus operators $T_{ba}=D_b{C}_a$. Each has matrix rank at most $k$, because it factors through the $k$-dimensional space $K$, andThe operation fidelity uses the unsquared quantum fidelity on a purification of a density operator, so its square is entanglement fidelity. The Kraus formula for entanglement fidelity givesIndeed each overlap $⟨{\Psi }_{\rho }|(I_R\otimes T_{ba})|{\Psi }_{\rho }⟩$ equals $\mathrm{Tr}({\rho }_Q{T}_{ba})$.

For any one of these Kraus operators $T$, let ${\Pi }_T$ be the orthogonal projection onto its image, of matrix rank $r_T\le k$. Since ${\Pi }_{T}T=T$, the Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives the rank bound for a weighted operator trace:The last step uses part (i) at $r_T$, followed by nonnegativity of the eigenvalues of the density operator ${\rho }_Q$. Summing over the Kraus operators and using their completeness relation proves the finite-dimensional quantum compression converse:No invertibility of ${\rho }_Q$, or restriction to an isometric decoder, was used.

The absolute value of an operator, trace norm, and operator norm are respectivelyThe positive square root of an operator makes $|L|$ well defined. Its eigenvalues $s_j\ge 0$ are the singular values of $L$, so $\Vert L{\Vert }_1={\sum }_j{s}_j$ and $\Vert L{\Vert }_{\mathrm{op}}={\max }_j{s}_j$.

Use the polar decomposition of a bounded operator $L=U|L|$, with the given unitary operator $U$. In an orthonormal eigenbasis $e_j$ for $|L|$,The operator norm assumption and unitarity imply $|⟨e_j|ZU|e_j⟩|\le \Vert Z{\Vert }_{\mathrm{op}}\le 1$. The triangle inequality therefore yields the required trace estimate:Choosing $Z=U^{†}$ attains equality, which also gives the finite-dimensional trace duality formula $\Vert L{\Vert }_1={\max }_{\Vert Z{\Vert }_{\mathrm{op}}\le 1}|\mathrm{Tr}(ZL)|$. A singular $L$ causes no difficulty: in a finite-dimensional square space the polar factor can be extended to a unitary operator on the complementary subspaces.

Let $\Gamma =T_B$ denote the partial transpose. The partial transpose of a maximally entangled projector iswhere $S$ is the swap operator, satisfying $S^{†}S=I$ and $\Vert S{\Vert }_{\mathrm{op}}=1$. If $\sigma$ has positive partial transpose, then ${\sigma }^{\Gamma }$ is a positive operator of trace one, hence $\Vert {\sigma }^{\Gamma }{\Vert }_1=1$.

For a pure state ${\varphi }^{+}=|\Phi ⟩⟨\Phi |$, the unsquared quantum fidelity obeys $F({\varphi }^{+},\sigma {)}^2=⟨\Phi |\sigma |\Phi ⟩$. Using the trace-adjoint identity for partial transpose and part (i),The overlap is nonnegative, so taking its square root proves the PPT maximally entangled overlap bound:The product state $|00⟩⟨00|$ attains the bound, so it is sharp.

In the computational basis ordered as $00,01,10,11$, the Hermitian matrix isIts trace is one, and its eigenvalues are $(1+\alpha )/2$, $(1-\alpha )/2$, $0$, and $0$. Thus the remaining density operator condition, positivity, holds exactly when both nonzero candidates are nonnegative:This is the admissible parameter interval for the dephased Bell-state mixture.

A density operator is a pure state exactly when its matrix rank is one. Within the admissible interval from part (a), one of $(1+\alpha )/2$ and $(1-\alpha )/2$ must vanish. Therefore the pure-state parameters and vectors arewhere the Bell states are $|{\Phi }^{\pm }⟩=(|00⟩\pm |11⟩)/\sqrt{2}$. Equivalently, the purity of a density operator is $\mathrm{Tr}{\rho }_{AB}^2=(1+{\alpha }^2)/2$, reaching one exactly at these two endpoints.

Every separable quantum state has positive partial transpose. Here the partial transpose iswith eigenvalues $1/2,1/2,\alpha /2,-\alpha /2$. Any nonzero $\alpha$ therefore gives a negative eigenvalue, excluding a separable quantum state. At $\alpha =0$, the explicit decompositionis a mixture of product states. ConsequentlyOnly necessity of the positive partial transpose criterion is needed here; sufficiency at zero follows directly from the decomposition.

For any normalized pure state on finite-dimensional Hilbert spaces ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$, the Schmidt decomposition givesThe $u_j$ and $v_j$ form orthonormal sets in their respective spaces. The positive numbers $\sqrt{p_j}$ are the Schmidt coefficients, and $r$ is the Schmidt rank. Taking a partial trace gives the reduced density matricesThus the nonzero eigenvalues of the two reduced density matrices agree.

One construction is a singular value decomposition of the coefficient matrix $M$ in $|\psi ⟩={\sum }_{a,b}{M}_{ab}|a⟩|b⟩$. If $M=U\Sigma V^{†}$, the columns of $U$ give $u_j$, the complex conjugates of the columns of $V$ give $v_j$, and the singular values give $\sqrt{p_j}$. This also explains why the Schmidt coefficients are real and nonnegative.

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

Write $|\zeta ⟩={\sum }_{i,j}{M}_{ij}|i{⟩}_A|j{⟩}_B$. Normalization says $\Vert M{\Vert }_{\mathrm{HS}}=1$, and the Schmidt rank equals the matrix rank of $M$. Its nonzero singular values are therefore $s_1,\dots ,s_r$ with ${\sum }_j{s}_j^2=1$.

The overlap with the maximally entangled state is $⟨{\varphi }^{+}|\zeta ⟩=\mathrm{Tr}(M)/\sqrt{d}$. The trace norm estimate of Q3(i), with test linear operator $I$, followed by the Cauchy-Schwarz inequality, givesFor two pure states, the unsquared quantum fidelity is the absolute value of their overlap. Hence the maximally entangled overlap bound from Schmidt rank isThe bound is attained by $|\zeta ⟩=r^{-1/2}{\sum }_{j=0}^{r-1}|jj⟩$. Thus uniform Schmidt coefficients on $r$ aligned basis pairs give the greatest possible overlap for that Schmidt rank.

Use logarithms to base two, so Von Neumann entropy and quantum relative entropy are measured in bits. For density operators $\rho ,\sigma$,The quantum relative entropy formula applies when the support of a positive operator $\rho$ is contained in that of $\sigma$; otherwise $D(\rho \Vert \sigma )=+\infty$. Use $0{\log }_{2}0=0$ in the Von Neumann entropy formula.

The binary density operator $\omega [p]$ is diagonal with eigenvalues $1-p,p$. Its Von Neumann entropy is the binary entropy:The two diagonal density operators commute, so their quantum relative entropy isA term with zero numerator has value zero; a positive numerator and zero denominator give $+\infty$. In particular, $D(\omega [0]\Vert \omega [0])=D(\omega [1]\Vert \omega [1])=0$, whereas $q=0,p>0$ or $q=1,p<1$ gives $+\infty$.

Choose $P_X$ uniform on ${\mathcal{A}}_M$ and zero outside it, so the marginal density operator of $Q$ is ${\bar{\rho }}_Q=k^{-1}{\sum }_m\rho (m{)}_Q$. Define the binary test for quantum decoding success byEach $E(m)$ is a positive contraction. The computational basis blocks of $F(0)$ are these effects on ${\mathcal{A}}_M$ and zero elsewhere. Thus $0\le F(0)\le I$, making $F$ a POVM on the whole space, including labels outside ${\mathcal{A}}_M$.

On the correlated classical-quantum state, its acceptance probability isOn the product of the marginal density operators,since ${\sum }_{m}E(m)=I_Q$. The two requested probabilities are therefore $1-ϵ$ and $1/k$. In the notation for binary output density operators, the measurement channel sends the joint input to $\omega [ϵ]$ and the product input to $\omega [1-1/k]$; their parameters describe outcome one, not outcome zero.

For the distribution and POVM from part 2, apply the data-processing inequality for quantum relative entropy to the measurement channel. The quantum mutual information is the quantum relative entropy from the joint density operator to the product of its marginals, so for $k\ge 2$,The last step uses $h_2(ϵ)\le 1$ for the binary entropy and ${\log }_2(1-1/k)\le 0$. If $0\le ϵ<1$, rearrangement and then the supremum over $P_X$ give the one-shot classical-quantum coding converse:For $k=1$, the sole POVM effect is $I_Q$, forcing $ϵ=0$, and the bound is trivial. For $ϵ=1$, division by $1-ϵ$ is undefined: the undivided inequality remains valid and gives no size restriction. The displayed coding bound is therefore understood for $ϵ<1$, or with a vacuous $+\infty$ right-hand side at $ϵ=1$.

An alternative uses the original decoder directly. Let $M$ be the uniform message and $\hat{M}$ its measured estimate. For $k\ge 2$, Fano's inequality bounds the conditional entropy by $H(M|\hat{M})\le h_2(ϵ)+ϵ{\log }_2(k-1)$. The Holevo bound givesThus the same coding bound follows by bounding the classical mutual information obtainable from the POVM.