arXiv 1701.063212017-01quant-phcs.DS

量子纠缠、平方和与对数秩猜想

Quantum entanglement, sum of squares, and the log rank conjecture

Boaz Barak, Pravesh Kothari, David Steurer

备注：: 23 pages + 1 title-page + 1 table-of-contents

摘要：

对于每个$ε>0$，我们给出了$1$与$1-ε$ \emph{Best Separable State (BSS)}区分问题的$\exp(\tilde{O}(\sqrt{n}/ε^2))$ -时间算法，给定对应于量子测量的$n^2\times n^2$矩阵$\mathcal{M}$，在存在可分离（即，非纠缠）态$ρ$，$\mathcal{M}$以概率$1$接受，以及每个可分离态以概率至多$1-ε$接受的情况。等价地，我们的算法采用子空间$\mathcal{W} \subseteq \mathbb{F}^{n^2}$的描述（其中$\mathbb{F}$可以是实域或复域），并区分$\mathcal{W}$包含秩1矩阵的情况和每个秩1矩阵至少$ε$远离$\mathcal{W}$（在$\ell_2$距离中）的情况。据我们所知，这是第一次改进的暴力$\exp(n)$时间算法这个问题。我们的算法是基于\emph{sum-of-squares}层次和它的分析的启发Lovett的证明（STOC '14，JACM '16），每个秩- $n$布尔矩阵的通信复杂度是由$\tilde{O}(\sqrt{n})$的限制。

英文摘要：

For every $ε>0$, we give an $\exp(\tilde{O}(\sqrt{n}/ε^2))$-time algorithm for the $1$ vs $1-ε$ \emph{Best Separable State (BSS)} problem of distinguishing, given an $n^2\times n^2$ matrix $\mathcal{M}$ corresponding to a quantum measurement, between the case that there is a separable (i.e., non-entangled) state $ρ$ that $\mathcal{M}$ accepts with probability $1$, and the case that every separable state is accepted with probability at most $1-ε$. Equivalently, our algorithm takes the description of a subspace $\mathcal{W} \subseteq \mathbb{F}^{n^2}$ (where $\mathbb{F}$ can be either the real or complex field) and distinguishes between the case that $\mathcal{W}$ contains a rank one matrix, and the case that every rank one matrix is at least $ε$ far (in $\ell_2$ distance) from $\mathcal{W}$. To the best of our knowledge, this is the first improvement over the brute-force $\exp(n)$-time algorithm for this problem. Our algorithm is based on the \emph{sum-of-squares} hierarchy and its analysis is inspired by Lovett's proof (STOC '14, JACM '16) that the communication complexity of every rank-$n$ Boolean matrix is bounded by $\tilde{O}(\sqrt{n})$.

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摘要：

英文摘要：

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