# Can the entangelement of basis vectors increase under local operations?

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:

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

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.
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.
• 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