# What do entanglement cost and distillable entanglement have to do with measuring entanglement?

So far what I have learned is that von-Neumann entropy is a tool to measure or quantify information and therefore entanglement for a given pure state system. However, similar concepts emerge from the task of measuring entanglement in general, which would be entanglement cost and distillable entanglement. My question is what are the similarities and divergences of these concepts with von-Neumann entropy?

The reference from which I spotted this question is from this article.

• it would be nice if you could add references to where you heard of these concepts, to provide more context – glS Mar 7 '19 at 13:06

• "Entanglement cost: Let us assume that Alice and Bob share many EPR pairs, say $|\phi^+_{AB} \rangle$, and that they wish to prepare a large number n of copies of a given bipartite "pure" state $|\psi_{AB} \rangle$, using only local operations and classical communication. If we call kmin the minimum number of EPR pairs necessary to accomplish this task, we define the entanglement cost as the limiting ratio kmin/n, for n → ∞. ... – EnthusiastiC Mar 8 '19 at 23:58
• Distillable entanglement: Let us consider the reverse process; that is, Alice and Bob share a large number n of copies of a "pure" state $|\psi_{AB} \rangle$ and they wish to concentrate entanglement, again using only local operations supplemented by classical communication. If k'max denotes the maximum number of EPR pairs that can be obtained in this manner, we define the distillable entanglement as the ratio kmax k'max /n in the limit n → ∞." The text is describing pure states only and this is confusing from what you said. – EnthusiastiC Mar 8 '19 at 23:58
• However, they coincide $\lim _ { n \rightarrow \infty } \frac { k _ { \min } } { n } = \lim _ { n \rightarrow \infty } \frac { k _ { \max } ^ { \prime } } { n } = S \left( \rho _ { A } \right) = S \left( \rho _ { B } \right)$ in the asymptotic limit, that is, from what I understood, when $n$ becomes sufficiently large and of course von-Neumann entropy is applicable for pure states since it works for single-qubit state. – EnthusiastiC Mar 9 '19 at 0:07