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

TL;DR

This paper demonstrates that multipartite entangled states violating Mermin-Klyshko inequalities are almost always distillable across bipartite splits as the number of qubits increases, revealing a strong link between Bell inequality violation and distillability.

Contribution

It proves that all N-qubit states violating the inequality have exponentially increasing distillable bipartite splits, and characterizes the existence of bound entangled states violating the inequality.

Findings

Number of distillable bipartite splits increases exponentially with N

Probability of a random bipartite split being distillable approaches one as N grows

Existence of bound entangled states violating the inequality for N ≥ 6

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 $N\ge 6$.