Fully undistillable quantum states are separable

Abstract

Assume that Alice, Bob, and Charlie share a tripartite pure state $|{\psi }_{ABC}⟩$. We prove that if Alice cannot distill entanglement with either Bob or Charlie using $|{\psi }_{ABC}⟩$ and local operations with any one of the following configurations for classical communication: $(A\to B,A\leftrightarrow C),(A\leftrightarrow B,A\to C),$ and $(A\leftrightarrow B,A\leftrightarrow C)$, then the same is also true for the other two configurations. Moreover, this happens precisely when the state is such that both its reductions on systems $AB$ and $AC$ are separable, which is further equivalent to the reductions being PPT. This, in particular, implies that any NPT bipartite state is such that either the state itself or its complement is 2-way distillable. To prove these results, we first obtain an explicit lower bound on the 2-way distillable entanglement of low rank bipartite states. Furthermore, we show that even though not all low rank states are 1-way distillable, a randomly sampled low rank state will almost surely be 1-way distillable.

v4: Closer to the published version

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Cited by in corpus (3)

• Estimate distillable entanglement and quantum capacity by squeezing useless entanglement
• Maximal extension on converse monogamy of entanglement for tripartite pure states
• Simultaneous superadditivity of the direct and complementary channel capacities