# 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

The entanglement entropy (what you call "von Neumann entropy") is a good measure for entanglement of pure states in the asymptotic setting, i.e. when one is dealing with many copies. However, it is not a good measure for mixed states.

Distillable entanglement and entanglement cost are entanglement measures which apply to both pure and mixed states. Moreover, for pure states, all three measures are equal, which is desirable due to the special status of the entanglement entropy. Finally, entanglement cost and distillable entanglement are the largest and smallest entanglement measure which coincides with entanglement entropy for pure states, and which is operationally meaningful in an asymptotic scenario.

• What I had read from "Giuliano Benenti, Giulio Casati, Giuliano Strini - Principles of quantum computation and information. Volume 2" is ... Mar 8 '19 at 23:57
• "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 → ∞. ... 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. 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. Mar 9 '19 at 0:07
• and sorry about my last question above "this is confusing from what you said" because it becomes clear for me after writing the last comment and it's too late to edit the comment. Mar 9 '19 at 0:09