# Is entanglement nonincreasing on average by local operations for all possible ensemble decompositions?

We know for a pure state conversion $$|\psi \rangle \rightarrow_\textrm{LOCC} |\phi \rangle$$ via local operation and classical communication (LOCC), an entanglement monotone should not increase, that is, $$E(|\psi \rangle) \geq E(|\phi \rangle)$$.

Say for a pure to mixed state conversion $$|\psi \rangle \rightarrow_\textrm{LOCC} \rho$$ given by the unique ensemble $$l = \{ q_l, |\xi_l \rangle \} \in \mathcal{D}$$ fixed by pointer states (which are omitted), where $$\mathcal{D}$$ is the set of all decompositions for $$\rho$$.

Naturally, the convex roof extension $$E_\textrm{min}(\rho) = \min_\mathcal{D} \Sigma_i q_i |\xi_i \rangle \langle \xi_i|$$ should not increase, nor should that of the obtained ensemble $$l$$ on average, i.e., $$E(|\psi \rangle) \geq E_l(\rho) \geq E_\textrm{min}(\rho)$$.

So, is it for certain there is not some decomposition $$j = \{ q_j, |\xi_j \rangle \}$$ that violates the nonincreasing criterion for $$E$$ on average, i.e., $$E_j(|\rho \rangle) \geq E(\psi \rangle)$$?

## 1 Answer

You don't even define $$E_j(\rho)$$. This makes your question hard to answer.

But if this is the average entanglement over that decomposition: Of course this can violate it. For instance, you can map any state to a maximally mixed state by LOCC, and the maximally mixed state can be written as a convex combination of the four Bell states, which are all maximally entangled.

• and that's why such a conversion cannot occur, so obvious, thank you – 2ub Sep 12 '20 at 19:38
• What cannot occur? – Norbert Schuch Sep 12 '20 at 19:50
• $|\psi \rangle \rightarrow_\textrm{LOCC} \mathbb{I}/d$ where $\mathbb{I}/d$ is given by maximally entangled states ensemble $\{ p_i, |\phi \rangle_i \}$ marked by pointer states. – 2ub Sep 13 '20 at 3:38
• @updraft If there are pointer states this is no longer the maximally mixed state. But if you mean that the measurement outcomes of the LOCC cannot correspond to maximally mixed states: yes. – Norbert Schuch Sep 13 '20 at 10:01
• @updraft My point is that these "pointers" only make sense if you interpret them as the different outcomes of a measurement in the LOCC. Is that what yo mean? In that sense, indeed an LOCC protocol yields an ensemble over (possibly mixed) states, rather than just a density matrix. – Norbert Schuch Sep 13 '20 at 15:35