Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 58 / 1

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.