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

Abstract

For every $\epsilon>0$, we give an $\exp(\tilde{O}(\sqrt{n}/\epsilon^2))$-time algorithm for the $1$ vs $1-\epsilon$ \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 $\rho$ that $\mathcal{M}$ accepts with probability $1$, and the case that every separable state is accepted with probability at most $1-\epsilon$. 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 $\epsilon$ 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 $\tilde{O}(\sqrt{n})$.Comment: 23 pages + 1 title-page + 1 table-of-content