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 $n^2times n^2$ 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 Lovetts 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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