Any multipartite entangled state violating Mermin-Klyshko inequality can be distilled for almost all bipartite splits

Abstract

We study the explicit relation between violation of Bell inequalities and bipartite distillability of multi-qubit states. It has been shown that even though for $N\ge 8$ there exist $N$-qubit bound entangled states which violates a Bell inequality [Phys. Rev. Lett. {\bf 87}, 230402 (2001)], for all the states violating the inequality there exists at least one splitting of the parties into two groups such that pure-state entanglement can be distilled [Phys. Rev. Lett. {\bf 88}, 027901 (2002)]. We here prove that for all $N$-qubit states violating the inequality the number of distillable bipartite splits increases exponentially with $N$, and hence the probability that a randomly chosen bipartite split is distillable approaches one exponentially with $N$, as $N$ tends to infinity. We also show that there exists at least one $N$-qubit bound entangled state violating the inequality if and only if $N\ge 6$.

4 pages, 2 figures

References in corpus (2)

• Bound entanglement provides convertibility of pure entangled states
• Teleportation capability, distillability, and nonlocality on three-qubit states