Motivation for the definition of k-distillability

Question

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 $k\geq 1.$

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?

Answer 1

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