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

For every constant ε>0, we give an exp(Õ(∞n))-time algorithm for the 1 vs 1 - ε Best Separable State (BSS) problem of distinguishing, given an n2 x n2 matrix ℳ corresponding to a quantum measurement, between the case that there is a separable (i.e., non-entangled) state ρ that 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 𝒲 ⊆ 𝔽n2 (where 𝔽 can be either the real or complex field) and distinguishes between the case that contains a rank one matrix, and the case that every rank one matrix is at least ε far (in 𝓁2 distance) from 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 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 Õ(√n).

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