Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 65 (Quantum Information Theory)

Set $X=\rho -\sigma$. The trace distance is $D(\rho ,\sigma )=\frac{1}{2}\Vert X{\Vert }_1$, where $\Vert X{\Vert }_1=\mathrm{Tr}\sqrt{X^{†}X}$. Since $X$ is a Hermitian operator, its spectral decomposition is $X={\sum }_j{x}_j|j⟩⟨j|$. Define its positive part of a Hermitian operator and negative part of a Hermitian operator byThey are positive semidefinite operators, satisfy $X=Q-R$ and $QR=0$, and obey $|X|=Q+R$. ThereforeThe states have equal trace, so $\mathrm{Tr}X=0$ and $\mathrm{Tr}Q=\mathrm{Tr}R=D(\rho ,\sigma )$. This is the spectral-parts formula for trace distance.

For every orthogonal projector $P$, positivity givesThese inequalities do not require $P$ to commute with $Q$ or $R$: for example, $\mathrm{Tr}(PR)=\mathrm{Tr}(R^{1/2}P{R}^{1/2})\ge 0$, and apply the same observation to $I-P$ and $Q$.

Let $P_{+}$ project onto the positive spectral subspace of $X$. Then $P_{+}Q=Q$ and $P_{+}R=0$, so $\mathrm{Tr}(P_{+}X)=\mathrm{Tr}Q=D(\rho ,\sigma )$. Hence the variational characterization of trace distance over projectors isOne may include or exclude the zero eigenspace without changing the optimum.

The allowed operators are exactly the positive contractions $0\le T\le I$, or binary measurement effects. As in (ii),because $\mathrm{Tr}(TR)\ge 0$ and $\mathrm{Tr}[(I-T)Q]\ge 0$. The positive-spectral orthogonal projector $P_{+}$ is among these effects and attains the bound. ThusAllowing general binary effects instead of projectors does not improve the optimal trace distance test.

Use the unsquared convention for quantum fidelity:Choose a purification of a density operator $|\Psi {⟩}_{RA}$ of $\rho$, and put ${\tau }_{RA}=({\mathrm{id}}_R\otimes \Lambda )(|\Psi ⟩⟨\Psi |)$. By definition,The monotonicity of quantum fidelity under partial trace follows from Uhlmann's theorem: maximizing purifications of two joint states are also candidate purifications of their marginals, with the discarded subsystem included in the purifying system. The optimization for the marginals therefore cannot give a smaller overlap. Here those marginals are $\rho$ and $\Lambda (\rho )$. ConsequentlyThis entanglement fidelity bound by state fidelity distinguishes preserving a reference entanglement from merely reproducing the input marginal.

The $+1$ measurement effect of the Pauli operator ${\sigma }_x$ is $P_{+}=(I+{\sigma }_x)/2$. The Born rule and the Bloch vector representation giveThe other two Bloch vector components do not enter this projective measurement probability.

In the computational basis, writeThe Kraus representation of the amplitude damping channel givesReading off its Bloch vector gives the Bloch-vector map of amplitude damping:Here $0\le p\le 1$. The shift in $s_z$ shows that this quantum channel is not unital unless $p=0$; its fixed ground state lies at the north pole, rather than the center of the Bloch sphere.

Iterating the Bloch-vector map of amplitude damping givesThereforeThe repeated amplitude damping limit describes irreversible relaxation of the excited state $|1⟩$ into the ground state $|0⟩$, with energy transferred to the environment and no compensating thermal excitation. At $p=1$, the ground state is reached in one action. At the endpoint $p=0$, the quantum channel is the identity and the initial Bloch vector remains unchanged; the ground-state conclusion requires nonzero damping.

For Kraus operators $A_j$, the Kraus formula for entanglement fidelity isIndeed each Kraus contribution to the purified-state overlap is $|⟨\Psi |(I\otimes A_j)|\Psi ⟩{|}^2$, and the expectation in a purification is $\mathrm{Tr}(\rho A_j)$. In this case,The entanglement fidelity of amplitude damping is thusFor $p=0$, this is one. For a ground-state input it is one for every $p$, whereas for an excited-state input it is $1-p$. These checks agree with the physical relaxation mechanism.

Use base-two logarithms throughout, so information and entropy are in bits. The relative entropy of classical probability distributions isSet $0{\log }_2(0/q)=0$; if $p(x)>0$ and $q(x)=0$, define the value to be $+\infty$. In the finite case, $\ln u\le u-1$ givesHence $D(p\Vert q)\ge 0$, with equality exactly when $p=q$: equality in the logarithm inequality requires $q(x)=p(x)$ on the support, and equality in the last step leaves no mass outside it.

The quantum relative entropy is $S(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ when the support of a positive operator $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Write spectral decompositions $\rho ={\sum }_i{r}_i|i⟩⟨i|$ and $\sigma ={\sum }_j{t}_j|u_j⟩⟨u_j|$, and set $q_i=⟨i|\sigma |i⟩$.

The weights $w_{ij}=|⟨i|u_j⟩{|}^2$ sum to one. Concavity of the scalar logarithm gives the diagonal logarithm concavity boundFor $r_i>0$, support inclusion removes overlaps with zero eigenvalues of $\sigma$; alternatively take a positive regularization and pass to the limit. The two diagonal vectors $r$ and $q$ are probability distributions. ThereforeThis proves nonnegativity of quantum relative entropy without assuming the states commute. Equality in the strict logarithmic concavity condition makes each occupied $|i⟩$ an eigenvector of $\sigma$; classical equality gives $q_i=r_i$ and no weight outside the support, hence equality occurs exactly for $\rho =\sigma$.

A deterministic quantum operation here is a CPTP map. By Stinespring dilation, there is an isometry $V:B\to B^{\prime }E$ with $\Lambda (\tau )={\mathrm{Tr}}_E(V\tau V^{†})$. Let ${\omega }_{A{B}^{\prime }E}=(I\otimes V){\rho }_{AB}(I\otimes V^{†})$. An isometry preserves all nonzero eigenvalues, soThe quantum mutual information is $I(A:B)=S(A)+S(B)-S(AB)$. The difference after discarding the environment iswhere the last inequality is Strong subadditivity of Von Neumann entropy, in the form $S(A{B}^{\prime })+S(B^{\prime }E)\ge S(B^{\prime })+S(A{B}^{\prime }E)$. Thus data processing for quantum mutual information givesSubsystem $A$ is unchanged; its prime simply labels the output. Keeping the environment would preserve the mutual information, while tracing it out cannot increase it.

Put $u=\epsilon /2$. The state is well defined for $0\le \epsilon \le 2$, with eigenvalues $1-u,u$. The first state is pure and has zero Von Neumann entropy, so the exact difference isThis is the entropy of a binary diagonal qubit mixture. Its trace distance from the pure state is $u$, not $\epsilon$. An explicit elementary upper bound follows from $-(1-u)\ln (1-u)\le u$:At zero the limit is zero. For small $\epsilon$, the sharp exact answer behaves as $(\epsilon /2){\log }_2(2/\epsilon )+\epsilon /(2\ln 2)+O({\epsilon }^2)$, displaying the logarithmic continuity of entropy near a pure state.

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.

The Bell basis consists ofLet $Z={\sigma }_z$, $X={\sigma }_x$. The Pauli operators $Z\otimes Z$ and $X\otimes X$ commute: each of the two local anticommutations contributes a minus sign, so they cancel. Their eigenvalue table isTo make the bit values literally zero or one rather than signed eigenvalues, the commuting Bell parity and phase observables areThey distinguish all four states jointly. The parity bit records whether the computational bits agree; the phase bit records the relative sign.

Apply a Hadamard gate to qubit $A$, followed by a controlled-NOT gate with $A$ controlling $B$. The resulting Bell-basis conversion circuit is $U=\mathrm{CNOT}(H\otimes I)$ and givesThus $00,01,10,11$ map respectively to ${\Phi }^{+},{\Psi }^{+},{\Phi }^{-},{\Psi }^{-}$. The first input bit becomes the phase bit and the second becomes the parity bit.

Both gates are self-inverse, but inversion of their product reverses the order:So the same two gates convert the Bell basis back to the computational basis when applied in reverse order: first the controlled-NOT, then the Hadamard. Using the forward order twice does not generally work, since these gates do not commute. For example, $U|{\Phi }^{+}⟩=(|00⟩+|01⟩-|10⟩+|11⟩)/2$, which is not a computational-basis state.

Write a bipartite pure state in fixed product bases as $|\psi ⟩={\sum }_{ij}{M}_{ij}|i⟩|j⟩$. The Schmidt decomposition theorem, equivalently the singular value decomposition of $M$, says that its Schmidt rank equals $\mathrm{rank}M$. A branch of local operations is a product operator $A\otimes B$, so its coefficient matrix isThe elementary rank inequality $\mathrm{rank}(AM{B}^{\mathsf{T}})\le \mathrm{rank}M$ proves Schmidt-rank contraction under product operators. Normalizing a nonzero branch does not change matrix rank.

For an adaptive LOCC protocol, a complete classical transcript selects one local Kraus operator at each stage. Multiplying the local operators along that transcript still gives one product $A_t\otimes B_t$. ConsequentlyClassical communication changes which product operator is chosen, not this rank bound. If the overall output is a pure state, every nonzero branch must be proportional to that same vector, so its Schmidt rank also cannot increase. If the outcome is discarded and the output is mixed, pure-state Schmidt rank is not defined; the induced decomposition instead shows that its Schmidt number is at most the original rank. This includes probabilistic filtering as well as deterministic pure-state conversion.

The final printed direction is reversed under the majorization convention stated in the same part. A completely depolarizing unital quantum channel sends a pure qubit state to $I/2$, giving spectra $r=(1,0)$ and $s=(1/2,1/2)$. The proposed $r\prec s$ fails already at the first partial sum, since $1>1/2$. The valid conclusion isHere is its general proof. Choose eigenbases $\rho ={\sum }_i{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and define the eigenvalue mixing matrix of a unital quantum channelIts entries are nonnegative by positivity. Trace preservation gives ${\sum }_j{A}_{ji}=1$ for every column. Unitality gives ${\sum }_i{A}_{ji}=⟨v_j|\Lambda (I)|v_j⟩=1$ for every row. Thus $A$ is a doubly stochastic matrix, and linearity givesThe supplied doubly stochastic characterization of majorization then yields $s\prec r$, completing the intended corrected result. This proves spectral majorization under a unital quantum channel. A unital quantum channel can make a spectrum more uniform; it cannot generally make the output spectrum majorize the input. The counterexample resolves the false printed assertion rather than treating the reversed inequality as a convention change.