English (United Kingdom)

https://curated-unify.zendy.io/wp-json/zendy-region/v1/featured_content/oa?rat=en

https://curated-unify.zendy.io/wp-json/zendy-region/v1/highlighted_journal/

Global: Zendy Plus annual

Presents the access of premium content as premium feature

Premium Content

Presents the keyphrase highlighting as premium feature

Keyphrase Highlighting

Presents the summarisation as premium feature

Summarisation

Insights

Presents the pdf analysis as premium feature

PDF Analysis

Presents the zaia usage as premium feature

ZAIA

Global: Zendy Plus monthly

Global: Zendy Tools annual

Global: Zendy Tools monthly

Global: Zendy Free Plan

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, givenan $n^2\times n^2$ 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 everyseparable 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 orcomplex field) and distinguishes between the case that $\mathcal{W}$ contains arank 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 thebrute-force $\exp(n)$-time algorithm for this problem. Our algorithm is basedon the \emph{sum-of-squares} hierarchy and its analysis is inspired by Lovett'sproof (STOC '14, JACM '16) that the communication complexity of every rank-$n$Boolean matrix is bounded by $\tilde{O}(\sqrt{n})$.

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