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Say I have a bipartite state

$\rho = \sum_ip_i|\psi_{i}\rangle \langle \psi_{i}|_{AB}$

Where $\{|\psi_{i}\rangle_{AB}\}$ forms an orthonormal basis.

I now perform some local quantum operation on subsystem B, bringing my system to a new state:

$\rho' = \sum_iq_i|\phi_{i}\rangle \langle \phi_{i}|_{AB}$

Where, again, $\{|\phi_{i}\rangle_{AB}\}$ forms an orthonormal basis.

Of course, for any entanglement measure $E$ we must have $E(\rho') \leq E(\rho)$. But is it possible to have:

$\max\limits_i E(|\phi_{i}\rangle\langle \phi_{i}|_{AB}) > \max\limits_i E(|\psi_{i}\rangle\langle \psi_{i}|_{AB})$ ?

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2 Answers 2

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Yes, it's possible. This is a well known field of entanglement distillation, whereby one can probabilistically transform less entangled states into more entangled states (usually maximally entangled). However, the values of probabilities ensure that one cannot win. There have been calculations of thresholds of this probability (for instance http://arxiv.org/abs/quant-ph/0501105). I recommend reading Nielsen and Chuang, most of the basic quantum information is explained nicely there.

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In the generic case, it is impossible to increase the "maximum entanglement carried by one term". In special cases, this conclusion may change.

If by a "local quantum operation on subsystem B", you mean a unitary transformation $U_B$ that only acts on the B tensor factor of the Hilbert space, then $$ \rho' = U_B \rho U_B^{-1} $$ In fact, this simple transformation may be applied to each term in $\rho$ individually which simply means $$ q_i = p_i $$ for the most natural ordering of the eigenstates (the vectors in the terms are eigenstates of $\rho$ or $\rho'$, respectively) and $$ |\phi_i\rangle = U_B |\psi_i\rangle $$ The entanglement entropy of $|\phi_i\rangle\langle \phi_i|$ only gets extra factors of $U_B$ and $U_B^{-1}$ inside which cancel, by the cyclic property of the trace, so these individual values of $E$ are the same for each $i$. Because the set of numbers if we list the entries for all values of $i$ are the same, the maxima are equal, too.

Quite generally, there's a principle of "no creation by LOCC", see e.g. section 3.2 of this paper and the references therein. LOCC stands for "local operations and classical communication" and "no creation" refers to the fact that one can't create new entanglement.

The only loophole in the argument above may occur when $p_i=p_j$ at least for a pair of distinct indices $i,j$ (which is infinitely unlikely in a generic physics case – a measure zero subset of cases – but which is commonplace in quantum computation where we often talk about "exactly equal superpositions" of many states). In that case, the decomposition of $\rho$ into the terms isn't unique. If that is so, one may have $|\psi_i\rangle$ and $|\psi_j\rangle$ that carry different amounts of AB entanglement and because the ket-bra products carry the same coefficients, we may rotate them by unitary transformations and either concentrate or dilute the entanglement to a smaller or greater number of terms (values of $i$). But this is really about the choice of a basis or a decomposition; the operation done in B isn't needed. At most, it can make the new choice of the basis natural.

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Everything you write here is true, but you may have misunderstood the question (maybe OP can clarify). He's not asking if $E(\rho') < E(\rho)$, but whether a single component of $\rho'$ can have a greater entanglement than a single component of $\rho$. Imagine that $\rho$ is pure and you are doing probabilistic single-copy entanglement distillation. You can write the output as a mixed state, with one of the components having greater $E$, but the output state as a whole having less entanglement, so satisfying the LOCC requirements. –  SMeznaric Jun 28 '13 at 8:37
    
I did address these things. –  Luboš Motl Jun 28 '13 at 16:15

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