Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 58 (Quantum Information, Entanglement and Nonlocality)

Use base-two logarithms, so entanglement entropy is measured in bits. Write $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$ for the binary entropy, with $0{\log }_{2}0=0$. Taking a partial trace over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party reduced density matrices of the first pure state areEach division into two groups is one party versus the other two. The Schmidt coefficients are $\sqrt{p}$ and $\sqrt{1-p}$, so the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$. For the second pure state, set $p=1/2$: every one-party reduced density matrix is $I/2$, and all three entanglement entropies are one bit.

For every pair $XY$, the first pure state instead gives the reduced density matrixThis is a separable quantum state, explicitly a convex combination of two product states. Its relative entropy of entanglement is therefore zero: in $E_R(\rho )={\inf }_{\sigma \in \mathrm{Sep}}D(\rho \Vert \sigma )$, choose $\sigma =\rho$ to attain zero, while nonnegativity of quantum relative entropy excludes a negative value. The same argument with equal weights applies to the second pure state. ThusThe pairwise density matrices contain classical correlations, despite the entanglement of the global pure state; discarding a party loses the coherence between its two branches.

The entanglement entropy of $n$ independent source copies across any cut is $n{h}_2(p)$, whereas $f(n)$ copies of the target GHZ state have entanglement entropy $f(n)$. Average monotonicity of pure-state entanglement entropy under LOCC gives $P(n)f(n)\le n{h}_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n^{\prime }/n\to 1$. For asymptotically faithful outputs, a continuity bound for quantum conditional entropy with trivial conditioning changes these inequalities only by $o(n)$. Hence the reversible leading-order rate isThe zero pairwise relative entropies of entanglement are consistent with this conversion: no independent pairwise resource has to be supplied or removed.

There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the Schmidt rank of the exact state of $n^{\prime }$ source copies is $2^{n^{\prime }}$, whereas $f$ target GHZ states have Schmidt rank $2^f$. Every refined LOCC branch acts by a product of local linear operators and cannot increase this Schmidt rank. For $p\ne 1/2$, the displayed rate has $f<n^{\prime }$ for sufficiently large blocks when $n^{\prime }/n\to 1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The exact dilution obstruction from Schmidt rank is avoided in the usual asymptotic formulation by allowing outputs approaching the desired pure state. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known product state, the rate is zero, and local preparation supplies any desired number of source copies.

Here is a tripartite Schmidt projection entanglement concentration protocol. For $0<p<1$, set $q=1-p$. Regrouping the $n$ local qubits of each party giveswhere $|x|$ counts ones. Alice performs a projective measurement of that count, without learning the individual string. The orthogonal projector for outcome $k$ is ${\Pi }_k={\sum }_{|x|=k}|x⟩⟨x|$. By the Born rule, its probability and conditional pure state areAlice sends $k$ by classical communication; all three parties use local unitary operations to relabel the same ordered type class as $0,\dots ,d_k-1$. There are now $d_k$ equal common-label amplitudes. Measuring the individual qubits instead would have destroyed precisely this coherence.

To obtain an integer number $f$ of target GHZ states, choose $m=2^f\le d_k$. The following exact LOCC step handles type classes whose sizes are not powers of two. For every $m$-element subset $S\subseteq \{0,\dots ,d_k-1\}$, Alice uses the Kraus operatorEvery label occurs in $\left(\genfrac{}{}{0ex}{}{d_k-1}{m-1}\right)$ such subsets, proving ${\sum }_S{K}_S^{†}{K}_S=I$ on the occupied subspace. Thus this is a complete measurement in quantum mechanics. Each outcome has probability $m/[d_k(\genfrac{}{}{0ex}{}{d_k-1}{m-1})]=1/\left(\genfrac{}{}{0ex}{}{d_k}{m}\right)$, and leaves the normalized pure state $m^{-1/2}{\sum }_{r\in S}|r,r,r⟩$. After Alice communicates $S$, all parties relabel its elements in the same order. Binary expansion of the labels then factors the state into exactly $f$ independent copies of the target GHZ state. Local extensions of these operations handle unused basis vectors and do not affect the protocol.

For example, taking $f(k)=\lfloor {\log }_2{d}_k\rfloor$ gives an outcome-dependent yield without any failure branch. Since $K$ has a binomial distribution with parameters $(n,q)$, the weak law of large numbers gives $K/n\to q$ in probability. Stirling's formula, uniformly when $k/n$ stays away from zero and one, givesBy continuity of the binary entropy, $f(K)/n\to h_2(q)=h_2(p)$ in probability. The integer rounding and logarithmic correction have no first-order effect.

A fixed yield with success probability tending to one is also explicit. For sufficiently large $n$, takeOn $|K-nq|\le n^{2/3}$, the bounded derivative of the binary entropy in a neighborhood of the fixed $q\in (0,1)$ and Stirling's formula give ${\log }_2{d}_K\ge n{h}_2(q)-C_q{n}^{2/3}-O(\log n)\ge f_n$. Chebyshev's inequality bounds the complementary probability by $q(1-q)n^{-1/3}$. Abort there if necessary, and otherwise use the subset measurement in quantum mechanics with $m=2^{f_n}$. ThusThis matches the first-order rate in part (i), with exact outputs on every successful forward branch. If $p=1/2$, simply retain the original $n$ target copies; if $p$ is an endpoint, output zero target copies.

The construction is a convex roof extension. First establish its convexity, without assuming that the minimizing quantum state ensemble exists. If $\tau ={\sum }_j{w}_j{\tau }_j$, choose a pure state ensemble for each ${\tau }_j$ whose average value of $\mu$ is within $\epsilon$ of $F({\tau }_j)$. Combining these ensembles, with their weights multiplied by $w_j$, gives a pure state ensemble for $\tau$. Taking arbitrarily small $\epsilon >0$ provesAlso, $F(\pi )=\mu (\pi )$ for a pure state $\pi$. Indeed, if $\pi ={\sum }_j{p}_j{\pi }_j$ and $v$ is orthogonal to its one-dimensional support, then $0={\sum }_j{p}_j⟨v|{\pi }_j|v⟩$. All terms are nonnegative, so every positive-weight ${\pi }_j$ has that same support and equals $\pi$.

Represent outcome $k$ of the operation by a linear unnormalized map ${\mathcal{I}}_k$, as for a quantum instrument. Normalization is performed only after applying this map: ${\mathcal{I}}_k(\rho )=P_k{\rho }_k$. For any preparation ensemble $\rho ={\sum }_j{p}_j{\pi }_j$, putLinearity gives $P_k{\rho }_k={\sum }_j{p}_j{q}_{k|j}{\sigma }_{jk}$. For $P_k>0$, Bayes' theorem identifies the conditional preparation weightsOutcomes with zero probability contribute nothing; normalized states on such branches need not be defined. The conditional states ${\sigma }_{jk}$ may be mixed, so convexity of the convex roof is essential here.

Applying that convexity separately for each recorded outcome, and then the assumed average monotonicity for each pure preparation, yieldsThe left side depends only on the initial density operator and the operation, not on the chosen preparation ensemble. Taking the infimum of the last expression over all its pure state ensembles provesEquivalently, choose an ensemble within $\epsilon$ of the infimum and then let $\epsilon \downarrow 0$. This is the monotonicity of a convex roof under a quantum instrument. It uses linearity of the unnormalized branches, not a false assumption that normalized conditional states depend linearly on the input. As usual, the averages are assumed well defined; finite-valued roofs and finitely many outcomes suffice. If an input roof is $-\infty$, taking ensembles with averages tending to $-\infty$ gives the same conclusion in the extended-real interpretation whenever the output average is defined.

The Schmidt decomposition theorem says that a normalized vector in the tensor product of two finite-dimensional Hilbert spaces can be writtenwhere both families are orthonormal. The numbers $\sqrt{{\lambda }_j}$ are its Schmidt coefficients. The Schmidt rank $r$ and coefficients, including multiplicities, are uniquely determined by the state; the local vectors may be changed within degenerate eigenspaces and by compensating phases. Zero coefficients may be appended to make $n_1$ terms, since $n_2\ge n_1$.

For a proof, form the reduced density matrix ${\rho }_1={\mathrm{Tr}}_2|\psi ⟩⟨\psi |$. It is a Hermitian matrix that is positive semidefinite, with trace one. The spectral theorem supplies an orthonormal basis $u_1,\dots ,u_{n_1}$ of eigenvectors with nonnegative eigenvalues ${\lambda }_j$. Define the unnormalized relative vectors in the second Hilbert space byCompleteness of the first orthonormal basis gives $|\psi ⟩={\sum }_j|u_j⟩|w_j⟩$. In any orthonormal basis $e_b$ of the second factor, the definition of the partial trace givesThus a zero eigenvalue gives $w_j=0$, and for each positive eigenvalue the vector $v_j=w_j/\sqrt{{\lambda }_j}$ has unit norm. Distinct such vectors are orthogonal by the same identity. Substituting them gives the required Schmidt decomposition, with at most $n_1$ nonzero terms.

Taking the other partial trace now givesTherefore the two reduced density matrices have identical nonzero eigenvalues, namely the squared Schmidt coefficients. This also proves uniqueness of the coefficient multiset and shows that the pure state is a product state exactly when $r=1$.

Let $Q$ be Alice's unknown qubit, with pure state $|\varphi {⟩}_Q=\alpha |0⟩+\beta |1⟩$. Alice and Bob initially share the Bell state $|{\Phi }^{+}{⟩}_{AB}=(|00⟩+|11⟩)/\sqrt{2}$. Alice holds $Q,A$ and Bob holds $B$. Define the Bell basisExpanding in this orthonormal basis gives the Bell-basis teleportation identityFor example, the last branch on $B$ is $\alpha |1⟩-\beta |0⟩$, which is $XZ|\varphi ⟩$; the ordering of the Pauli operators therefore matters.

Alice performs a projective measurement in the Bell basis on her two local qubits. Each outcome has probability $1/4$, independent of the unknown amplitudes. She encodes the four possible outcomes as two classical bits and sends them to Bob. For outcomes ${\Phi }^{+},{\Phi }^{-},{\Psi }^{+},{\Psi }^{-}$, respectively, Bob applies the local unitary operationsThese are inverses of the corresponding branch Pauli operators, so Bob obtains exactly the unknown input qubit in every branch. Using $XZ$ instead of $ZX$ as the last correction also succeeds up to an irrelevant overall phase. Alice need not know $\alpha$ or $\beta$. Every quantum operation is local to one of the parties, and the only transmitted information is the two-bit classical outcome. The shared Bell pair is consumed.

Before the message arrives, Bob's density operator isso no unknown-state information is available without the classical communication. Alice's original qubit has been included in the destructive Bell-basis measurement; the protocol creates no extra copy. Since the corrected transformation is the identity on every input vector and is linear, it also preserves entanglement with an external reference and transfers mixed inputs. Thus this is deterministic quantum teleportation, using precisely one shared maximally entangled Bell pair, two classical bits, and local operations.

The EPR criterion of reality proposes that a quantity has an element of physical reality if its value can be predicted with certainty without disturbing the system. To use remote predictions this way, one additionally assumes locality: the choice and execution of a measurement in quantum mechanics on separated particles cannot disturb the local physical properties of the remaining particle. The contradiction below is with this combination of the criterion and locality, rather than with a definition of reality alone.

Use the computational basis $|0⟩=|\uparrow ⟩$, $|1⟩=|\downarrow ⟩$ and the conventional Pauli matricesFor each site $j$, let $O_j$ be the tensor product of $X$ at that site and $Y$ at all other sites. For $N=4M+3$, the number of $Y$ factors is $N-1=4M+2$, soThe minus sign in the given GHZ state then impliesHence every one-$X$, remaining-$Y$ product has measurement outcome $+1$ with certainty, while the all-$X$ product has outcome $-1$ with certainty. These are experimentally separate local Pauli measurements on identically prepared copies, not joint measurements of $X$ and $Y$ on the same particle.

For a chosen site, the other $N-1$ sites can determine its $X$ outcome with certainty by measuring $Y$ and using the corresponding $O_j$ relation. They can determine its $Y$ outcome with certainty by selecting an $O_{\ell }$ with $\ell \ne j$, measuring $X$ at $\ell$ and $Y$ at all other remote sites. Since $N\ge 3$, both choices are available. Locality and the EPR criterion of reality therefore assign each site definite values $x_j,y_j\in \{+1,-1\}$, independent of which remote choice is made, even though quantum theory does not jointly measure the local $X$ and $Y$.

The certain one-$X$ correlations require those assigned values to obeyMultiplying these scalar equations givessince $N-1$ is even and $y_k^2=1$. Thus the all-$X$ product would necessarily be $+1$. But its quantum value is $-1$, givingThe equations cannot hold even for one assignment of values, so allowing a probability distribution over assignments cannot remove the contradiction. This is the GHZ contradiction for N congruent to 3 modulo 4 for every $M\ge 0$. Multiplying classical assigned values is crucial: one must not commute the incompatible local Pauli operators as though they were ordinary numbers. Their noncommutativity is exactly what permits the quantum correlations to evade the classical product argument.