A reduction of the separability problem to SPC states in the filter normal form

TL;DR

This paper demonstrates that solving the separability problem for SPC states in filter normal form would resolve the general problem, by constructing related states with the same Schmidt number and analyzing their properties.

Contribution

It shows how the separability problem reduces to SPC states in filter normal form and constructs states close to symmetric projections to analyze entanglement.

Findings

Constructed SPC states with the same Schmidt number as original states.

States near the symmetric projection have bounded Schmidt number.

Reducing the problem to SPC states could solve the general separability problem.

Abstract

It was recently suggested that a solution to the separability problem for states that remain positive under partial transpose composed with realignment (the so-called symmetric with positive coefficients states or simply SPC states) could shed light on entanglement in general. Here we show that such a solution would solve the problem completely. Given a state in $ \mathcal{M}_k\otimes\mathcal{M}_m$, we build a SPC state in $ \mathcal{M}_{k+m}\otimes\mathcal{M}_{k+m}$ with the same Schmidt number. It is known that this type of state can be put in the filter normal form retaining its type. A solution to the separability problem in $\mathcal{M}_k\otimes\mathcal{M}_m$ could be obtained by solving the same problem for SPC states in the filter normal form within $\mathcal{M}_{k+m}\otimes\mathcal{M}_{k+m}$. This SPC state can be built arbitrarily close to the projection on the symmetric subspace of $ \mathbb{C}^{k+m}\otimes\mathbb{C}^{k+m}$. All the information required to understand entanglement in $ \mathcal{M}_s\otimes\mathcal{M}_t$ $(s+t\leq k+m)$ lies inside an arbitrarily small ball around that projection. We also show that the Schmidt number of any state $\gamma\in\mathcal{M}_n\otimes\mathcal{M}_n$ which commutes with the flip operator and lies inside a small ball around that projection cannot exceed $\lfloor\frac{n}{2}\rfloor$.