Quantum Computing Asked on August 20, 2021

Definition of k-distillabilityFor a bipartite state $rho$, $H=H_Aotimes H_B$ and for an integer $kgeq 1$, $rho$ is $k$-distillable if there exists a (non-normalized) state $|psiranglein H^{otimes k}$ of Schimdt-rank at most $2$ such that,

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

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

^{Source}

I get the mathematical condition but don’t really understand the motivation for $k$-distillability in general, or more specifically the condition $langle psi|sigma^{otimes k}|psirangle < 0$. 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 $|psirangle$ such that $$ langlepsi|Iotimestext{T}(rho)|psirangle<0, $$ 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 $|psirangle$ 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 $|psirangle$. 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

Correct answer by DaftWullie on August 20, 2021

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