论文标题
完全不可证实的量子状态是可分离的
Fully undistillable quantum states are separable
论文作者
论文摘要
假设爱丽丝（Alice），鲍勃（Bob）和查理（Charlie）共享三方纯状态$ |ψ_{abc} \ rangle $。我们证明，如果爱丽丝无法使用$ |ψ_{abc} \ rangle $与鲍勃或查理的纠缠以及以下任何一种经典通信配置中的任何一种：$（a \ to b，a \ leftrightArrow c），（a \ leftrightArrow c） a \ leftrightArrow c）$，那么其他两个配置也是如此。此外，这正是在国家的降低$ AB $和$ AC $都可以分离的情况下，这是可以分开的，这进一步等同于PPT的减少。特别是，这意味着任何NPT两分状态都是国家本身或其补体都是2向蒸馏的。为了证明这些结果，我们首先在低等级两分状态的2向蒸馏纠缠上获得明确的下限。此外，我们表明，尽管并非所有低级状态都是1向蒸馏，但随机抽样的低级状态几乎肯定是1向蒸馏。
Assume that Alice, Bob, and Charlie share a tripartite pure state $|ψ_{ABC}\rangle$. We prove that if Alice cannot distill entanglement with either Bob or Charlie using $|ψ_{ABC}\rangle$ 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.
