Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 51 / 3

Put $s={\sin }^2\alpha$ and $t={\sin }^2\beta$. Then $0<s<t<1/2$, and the ordered squared Schmidt coefficients of the source and target are $(1-s,s)$ and $(1-t,t)$.

Nielsen's pure-state conversion theorem states that a deterministic exact conversion between bipartite pure states by LOCC exists precisely when the source's vector of squared Schmidt coefficients is majorized by the target's vector, with zero padding if needed. In the convention $x\prec y$, majorization requires ${\sum }_{j=1}^k{x}_j^{\downarrow }\le {\sum }_{j=1}^k{y}_j^{\downarrow }$ for every $k$, and equality of total sums. Here the first inequality would require $1-s\le 1-t$, but $s<t$ gives the reverse strict inequality. Thus exact conversion with certainty is impossible.

An upper bound follows directly from the Schmidt-tail entanglement monotonewhere ${\rho }_A^{\chi }$ is the reduced density matrix of a bipartite pure state. On these two-qubit states, this is the smaller squared Schmidt coefficient, so $E_2(\psi )=s$ and $E_2(\varphi )=t$.

Here is the average-monotonicity argument needed for all LOCC strategies. The largest eigenvalue is convex: for any unit vector $v$, $⟨v|{\sum }_r{p}_r{\rho }_r|v⟩\le {\sum }_r{p}_r{\lambda }_{\max }({\rho }_r)$, and maximizing the left side gives the claim. Hence $1-{\lambda }_{\max }$ is concave. If Alice makes a local measurement, Bob's conditional reductions average to his original reduction. Concavity therefore gives ${\sum }_r{p}_r{E}_2({\chi }_r)\le E_2(\chi )$. If Bob measures, use Alice's reductions instead. The two reductions of a pure state have the same nonzero eigenvalues by the Schmidt decomposition. Refining all measurement outcomes into individual Kraus operators keeps conditional states pure; iterating the inequality covers adaptive rounds of classical communication. The same bound persists for limits of such protocols.

If success produces the target with probability $p$, its contribution to the final average is $pt$. Failure branches have nonnegative $E_2$. ThusThis is a bound on arbitrary local protocols, not merely on a particular filtering measurement.

Alice makes a two-outcome local measurement in quantum mechanics with Kraus operatorsSince $0<\tan \alpha /\tan \beta <1$, these are valid and satisfy $K_{\mathrm{s}}^{†}{K}_{\mathrm{s}}+K_{\mathrm{f}}^{†}{K}_{\mathrm{f}}=I$. Alice sends Bob the success or failure outcome. The successful unnormalized branch isTherefore the normalized success state is exactly the target andBob needs no quantum correction in this known Schmidt basis. The failure branch is proportional to $|00⟩$, a product state, and has probability $1-s/t$. The successful branch saturates the previous bound: $p_{\mathrm{succ}}{E}_2(\varphi )=E_2(\psi )$ and failure contributes zero. This establishes optimal stochastic conversion of a two-qubit pure state.

Use the usual asymptotic convention: the output approaches the desired $m(n)$-copy pure target in trace distance, and any recorded failure probability tends to zero. Define the binary entropyThe entanglement entropies per source and target are $E_{\psi }=h_2(s)$ and $E_{\varphi }=h_2(t)$. The asymptotic pure-state entanglement conversion rate isBoth achievability and the upper bound matter here.

For entanglement concentration, expand $n$ sources in their correlated computational basis. Every pair of strings of Hamming weight $k$ has the same amplitude $(1-s{)}^{(n-k)/2}{s}^{k/2}$. Alice measures only this weight, without resolving the individual string. Bob's string has the same weight automatically. Conditional on $k$, the normalized state is maximally entangled on $D_k=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ correlated strings. Its outcome probability is $\left(\genfrac{}{}{0ex}{}{n}{k}\right)s^k(1-s{)}^{n-k}$, a binomial distribution. By the weak law of large numbers, $k/n\to s$ in probability; Stirling's formula then gives $n^{-1}{\log }_2{D}_k\to h_2(s)$. A uniform Schmidt vector on $D_k$ entries is majorized by a uniform vector on $2^{\lfloor {\log }_2{D}_k\rfloor }$ entries padded with zeros. Thus Nielsen's pure-state conversion theorem converts the conditional state into that many Bell pairs exactly. For any fixed $\delta >0$, one obtains at least $n(E_{\psi }-\delta )$ Bell pairs with probability tending to one.

For entanglement dilution, consider the Schmidt strings of $m$ targets with probabilities $q_x$. The quantum typical subspace retains strings satisfyingThe retained probability $Q_m$ tends to one by the weak law applied to the independent variables $-{\log }_2{q}_{x_j}$. The number of retained strings is at most $2^{m(E_{\varphi }+\delta )}$, because their probabilities sum to at most one. Normalize the target after truncation to those strings. Its squared overlap with the full target is $Q_m\to 1$. A uniform Schmidt vector of dimension $2^L$, with $L=\lceil m(E_{\varphi }+\delta )\rceil$, is majorized by the truncated target's Schmidt vector padded with zeros: the sum of its largest $k$ entries is at least $k/2^L$. Hence $L$ Bell pairs prepare the truncated target exactly by LOCC. Its trace distance from the true target is $\sqrt{1-Q_m}\to 0$. Combining concentration and dilution achieves any rate $R<E_{\psi }/E_{\varphi }$. Choosing margins tending sufficiently slowly to zero attains the ratio as a limiting rate.

For the converse, average monotonicity of pure-state entanglement entropy follows from Concavity of Von Neumann entropy: a measurement by one party leaves the other party's reduction unchanged on average, so its average conditional entropy cannot increase. Refine local measurement records, including local discarded systems, to obtain pure output branches $\{p_r,|{\chi }_r⟩\}$ on the $m$ output qubits of each party. Iterating the concavity argument givesLet $|{\Phi }_m⟩=|\varphi {⟩}^{\otimes m}$ and let the squared-overlap error of the average output ${\rho }_{\mathrm{out}}$ be $\epsilon =1-⟨{\Phi }_m|{\rho }_{\mathrm{out}}|{\Phi }_m⟩\to 0$. Define $d_r=\sqrt{1-|⟨{\Phi }_m|{\chi }_r⟩{|}^2}$, the pure-state trace distance. Concavity of the square root gives ${\sum }_r{p}_r{d}_r\le \sqrt{\epsilon }$. Partial trace cannot increase trace distance, so the reduced states are also within $d_r$. The continuity bound for quantum conditional entropy, with trivial conditioning and local dimension $2^m$, impliesConsequentlywhich excludes a limiting rate exceeding $E_{\psi }/E_{\varphi }$.

The vanishing-error convention is essential. If exact deterministic conversion were required for every finite block, majorization would also require the largest squared Schmidt coefficient to satisfy $(1-s{)}^n\le (1-t{)}^m$, henceThis can be strictly smaller than the entropy ratio: at $s=0.1,t=0.2$, the two bounds are approximately $0.4722$ and $0.6496$. Thus the boxed rate answers the standard asymptotic task, rather than imposing unmentioned exact zero-error conditions on every finite block.

Applying the optimal single-copy filter independently gives a random number $M_n$ of exact successful targets with binomial distributionThe weak law of large numbers gives $M_n/n\to s/t$ in probability. In contrast, collective entanglement concentration and entanglement dilution yield asymptotically $n{h}_2(s)/h_2(t)$ targets with vanishing output error. The improvement is strict, not just nonnegative. Differentiating binary entropy givesSince $s<t$, it follows thatThis is collective advantage over independent two-qubit filtering. Independent filtering discards the entanglement in product-state failure branches; the collective reversible asymptotic procedure avoids that extensive loss. The separately filtered targets are exact conditional on success, whereas the collective statement uses the vanishing-error convention from the previous part. Here the counted outputs are target states $|\varphi ⟩$; that is also the notation in the original printed question.