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تسجيل مستخدم جديدAny multipartite entangled state violating Mermin-Klyshko inequality can be distilled for almost all bipartite splits
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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 $Nge 8$ 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 $Nge 6$.
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