# Motivation for the definition of k-distillability

Definition of k-distillability

For a bipartite state $$\rho$$, $$H=H_A\otimes H_B$$ and for an integer $$k\geq 1$$, $$\rho$$ is $$k$$-distillable if there exists a (non-normalized) state $$|\psi\rangle\in H^{\otimes k}$$ of Schimdt-rank at most $$2$$ such that,

$$\langle \psi|\sigma^{\otimes k}|\psi\rangle < 0, \sigma = \Bbb I\otimes T(\rho).$$

$$\rho$$ is distillable if it is $$k$$ for some integer $$k\geq 1.$$

Source

I get the mathematical condition but don't really understand the motivation for $$k$$-distillability in general, or more specifically the condition $$\langle \psi|\sigma^{\otimes k}|\psi\rangle < 0$$. Could someone explain where this comes from?

Remember that the partial transpose condition is generally good for detecting entanglement, i.e. a bipartite state $$\rho$$ is certainly entangled if the partial transpose is not non-negative. In other words, if there exists a state $$|\psi\rangle$$ such that $$\langle\psi|I\otimes\text{T}(\rho)|\psi\rangle<0,$$ then the state is certainly entangled.

If you want to be able to distil some entanglement from $$k$$ copies then, crudely, you'd like to look at $$k$$ copies of the partially transposed state, and if that has a negative eigenvalue, you would be able to extract some entanglement.

With that level of explanation, you'd ask why looking at more than one copy is any use -- the eigenvalues of many copies of $$\sigma$$ are easily related to the eigenvalues of a single copy. However, this is because of the extra condition that $$|\psi\rangle$$ must be Schmidt rank 1 or less. I presume that this is because you can give an explicit distillation protocol based on the properties of $$|\psi\rangle$$. Essentially, this is due to the fact that you're trying to project onto a Bell pair which, of course, is Schmidt rank 2.

For a better understanding that the very hand-wavy suggestions I've just given, you'd want to work through page 2 of https://arxiv.org/abs/quant-ph/9801069

• Isn't the overlap with an entangled two-qubit state an if and only if criterion for distillability? – Norbert Schuch Jul 21 '20 at 10:40
• @NorbertSchuch What do you mean by that? Even the maximally mixed state has overlap with the maximally entangled state. – DaftWullie Jul 21 '20 at 12:12
• IIRC a state is 1-distillable if (and only if, as I believe -- this was the question) it has a two-qubit subspace which is entangled -- which is what the above criterion says. I was indeed not very precise in what I wrote, but my point is that AFAIK said criterion is equivalent to distillability. – Norbert Schuch Jul 21 '20 at 13:13
• @NorbertSchuch Sure, but that maybe doesn't really touch the issue of motivation/insight, which I was trying to give. – DaftWullie Jul 21 '20 at 13:34
• I kind of think it does, since it explains why this is equivalent to the natural definition of k-distillability? – Norbert Schuch Jul 21 '20 at 13:39