Entanglement Transformations

What may two distant parties do? Each may apply any channel or measurement to the systems in her own laboratory, including local ancillas. They may send each other classical messages, such as measurement results, and choose later operations based on them. This class is called local operations and classical communication, or LOCC.

What is excluded? Sending a quantum system from one laboratory to the other, and applying a joint gate such as a CNOT across the separation. With either of those, they could simply create a Bell pair, so the question would become trivial.

Local measurements can change the joint state differently in different branches. A conversion that works in one branch may fail in another. So any claim about LOCC must count the branch probabilities. We must also say whether success is required every time, may fail with a declared outcome, or only needs to hold approximately for many copies.

Which features of a pure joint state can no local unitary change? Suppose Alice applies \(U\). Her reduced state becomes \(U\rho_AU^\dagger\), which has the same eigenvalues, since \(U\) only rotates the eigenvectors. Bob's unitary does not affect \(\rho_A\) at all. So the eigenvalues of the reduced state are unchanged by local unitaries. For \((\sqrt3/2)\ket{00}+(1/2)\ket{11}\) they are \(3/4\) and \(1/4\).

These eigenvalues give a standard form for any pure joint state \(\ket\Psi\). Let \(\ket{u_j}\) be eigenvectors of \(\rho_A\) with eigenvalues \(\lambda_j\), and expand \(\ket\Psi=\sum_j\ket{u_j}\ket{w_j}\), where \(\ket{w_j}\) collects Bob's amplitudes attached to \(\ket{u_j}\). Are these Bob vectors related? Their inner products are entries of the reduced state: \(\braket{w_k}{w_j}=\bra{u_j}\rho_A\ket{u_k}=\lambda_j\) if \(j=k\), and 0 otherwise. So they are orthogonal, with squared lengths \(\lambda_j\). Normalizing them as \(\ket{v_j}=\ket{w_j}/\sqrt{\lambda_j}\) gives

\[\ket\Psi=\sum_j\sqrt{\lambda_j}\,\ket{u_j}\ket{v_j},\]

with orthonormal vectors on each side. This is the Schmidt decomposition. The numbers \(\sqrt{\lambda_j}\) are the Schmidt coefficients, and the \(\lambda_j\), which are probabilities, are the Schmidt probabilities. Local unitaries change the vectors but not the \(\lambda_j\). So it is natural to measure pure-state entanglement by \(S(\rho_A)=H(\lambda)\), the entropy of entanglement. A Bell pair has 1 ebit, and our example has \(h_2(1/4)\approx0.811\).

Which conversions can LOCC make with certainty? We quote Nielsen's theorem (1999). Sort the Schmidt probabilities of each state from largest to smallest. LOCC can turn \(\psi\) into \(\phi\) with certainty exactly when every partial sum of the largest probabilities of \(\psi\) is at most the corresponding sum for \(\phi\). This relation is called majorization and written \(\lambda(\psi)\prec\lambda(\phi)\). All the full sums equal one.

For two coefficients only the first sum matters: the largest probability may grow but not shrink. A Bell pair \((1/2,1/2)\) can become \((3/4,1/4)\), since \(1/2\leq3/4\). The reverse would need \(3/4\leq1/2\), which fails. So no LOCC protocol turns our pair into a Bell pair every time. A protocol that may fail, and says so, is a different task.

Alice wants a measurement outcome after which the amplitudes \(\sqrt p\) and \(\sqrt{1-p}\) are equal. Try a measurement operator \(K=\mathrm{diag}(k_0,k_1)\) on her qubit. It must shrink the larger amplitude: \(k_0\sqrt p=k_1\sqrt{1-p}\). Making success as likely as possible means taking \(k_1=1\), the largest value an effect allows, so

\[K=\mathrm{diag}\Bigl(\sqrt{\tfrac{1-p}p},\,1\Bigr),\qquad(K\otimes I)\bigl(\sqrt p\ket{00}+\sqrt{1-p}\ket{11}\bigr)=\sqrt{1-p}\,(\ket{00}+\ket{11}).\]

The squared length is \(2(1-p)\), so this outcome has probability \(2(1-p)\) and leaves a Bell pair. To complete the measurement we need \(L\) with \(K^\dagger K+L^\dagger L=I\). Take \(L=\mathrm{diag}\bigl(\sqrt{(2p-1)/p},0\bigr)\). It leaves \(\ket{00}\), a product state.

Put in \(p=3/4\). Then \(K=\mathrm{diag}(1/\sqrt3,1)\), and the success branch \((\ket{00}+\ket{11})/2\) has squared length \(1/2\). The failure branch is \(\ket{00}/\sqrt2\), also with squared length \(1/2\). Half the trials give one ebit and half give none, so the expected yield is \(1/2\) ebit per pair. The original pair had \(0.811\) ebits of entanglement entropy. Reporting only the successful branch would overstate the gain.

The filter wastes entanglement. Can many copies, processed together, do better? Take \(n\) copies of \(\sqrt p\ket{00}+\sqrt{1-p}\ket{11}\). Expanding the product, every term pairs a string \(x\) on Alice's side with the same string on Bob's:

\[\sum_xp^{(n-k)/2}(1-p)^{k/2}\ket x\ket x,\qquad k=\text{number of 1s in }x.\]

All strings with the same \(k\) have the same coefficient. So Alice measures \(k\) alone, a projective measurement that does not reveal which string. The result is an equal superposition of \(\binom nk\) terms \(\ket x\ket x\). Its Schmidt probabilities are all equal, so it carries \(\log_2\binom nk\) ebits, which further LOCC steps can turn into nearly that many Bell pairs.

How large is \(\log_2\binom nk\)? The likely values of \(k\) are near \(n(1-p)\), the typical strings, and then \(\log_2\binom nk/n\) approaches \(h_2(p)\). For \(p=3/4\) and \(k=n/4\), this ratio is about 0.55 at \(n=10\), 0.78 at \(n=100\) and 0.81 at \(n=1000\), approaching \(h_2(1/4)\approx0.811\). Processing copies together beats the one-copy filter's 0.5.

This is entanglement concentration (Bennett, Bernstein, Popescu and Schumacher, 1996). The reverse task, dilution, prepares many copies of the weak pair from Bell pairs. We quote that it needs Bell pairs at the same rate, \(h_2(p)\) per copy, with classical communication and error tending to zero. So for pure states the entropy of entanglement is an exchange rate that works both ways, in the limit of many copies. That asymptotic reversibility does not contradict the single-copy restriction. It is not an exact conversion of a finite block, and it never produces a fractional Bell pair.

Consider \(\rho_{\rm c}=\tfrac12(\ket{00}\bra{00}+\ket{11}\bra{11})\). Its Z results agree perfectly. Can Alice and Bob prepare it with a coin and a phone call? Yes. Alice tosses a fair coin, prepares her qubit in the coin's value, and tells Bob the value, and he prepares the same. No Bell pair is used.

The same recipe works for any average of product states. If a classical random choice \(k\), made with probability \(p_k\), selects local preparations \(\rho_k^A\) and \(\sigma_k^B\), the result is

\[\rho_{AB}=\sum_kp_k\,\rho_k^A\otimes\sigma_k^B.\]

A weighted average like this, with probabilities as weights, is called a convex mixture. A state of this form is separable, and a mixed state that is not separable is entangled. For pure states, separable just means product, so this agrees with the earlier definition. The Bell pair is not separable, because its X correlations, which the coin recipe cannot produce, come from its coherence.

For entangled mixed states we ask two rate questions about many independent copies. The distillable entanglement is the best rate of Bell pairs that LOCC can extract, with error tending to zero. The entanglement cost is the smallest rate of Bell pairs from which LOCC can prepare the copies. For pure states both equal the entropy of entanglement.

Dephase a Bell pair, discarding its phase record. The result is \(\rho_{\rm c}\): the Z correlations survive and the entanglement is gone. Can LOCC bring it back?

No. LOCC cannot create entanglement from a separable state. Local operations turn each product term into another product term, and classical messages only correlate the random choice \(k\). So a separable state stays separable. The resource was lost when the record went to an environment that Alice and Bob cannot reach.

For mixed states the two rates need not agree. We quote a result of the Horodeckis (1998): there are bound entangled states, which are entangled, so preparing them costs Bell pairs, yet no LOCC distillation extracts any Bell pairs from them. Preparing and extracting entanglement can be irreversible. These are theorem-level existence statements; they do not follow from the two-coefficient filter.

So "more entangled" is not one complete ordering of states. Every conversion claim must say whether it is deterministic, may fail with a declared outcome, is approximate, or holds only for many copies, and whether classical messages go one way or both.

Entanglement was destroyed because an environment kept a record. Noise does this all the time. Can quantum information be stored so that the environment learns nothing about it?

Sources and further study: Nielsen, Conditions for entanglement transformations; Bennett et al., Concentrating partial entanglement; Horodecki et al., Mixed-state entanglement and irreversibility.