Stronger Bell Inequalities for Werner states?

Cite this problem as Problem 19.

Problem

Find Bell Inequalities which are stronger than the CHSH inequalities in the sense that they are violated by a wider range of Werner states.

Background

Werner states are $d\times d$ bipartite states of the form:

$\rho_{AB}=p\frac{2P_{asym}}{d(d-1)}+(1-p)\frac{\mathbb{I}}{d^2}$ ,

where $P_{asym}$ , is the projector onto the anti-symmetric space in $\mathbb{C}^d\otimes \mathbb{C}^d$ . In the qubit case ( $d=2$ ), the definition reduces to

$\rho_{AB}=p|\psi^{-}\rangle\langle \psi^{-}|+(1-p)\frac{\mathbb{I}}{4}$ ,

with $|\psi^{-}\rangle=\frac{1}{\sqrt{2}}(|0,1\rangle-|1,0\rangle)$ .

The above state violates the CHSH inequality as long as $p>\frac{1}{\sqrt{2}}$ . As noted in [1], when we restrict to Bell inequalities for dichotomic measurements, then the two-qubit Werner state violates a Bell inequality iff $p>\frac{1}{K_G(3)}$ , where $K_G(3)$ is Grothendieck’s constant of order $3$ [2]. Unfortunately, the exact value of $K_G(3)$ is unknown, and, at the time this problem was posed, the best lower bound on $K_G(3)$ was $K_G(3)\geq \sqrt{2}$ . This, once again, gives the critical visibility $p>\frac{1}{\sqrt{2}}$ .

Attempts at using other facets for the local polytope with a higher number of settings failed to improve this value. In 2003, Daniel Collins and Nicolas Gisin [3] found a Bell Inequality and states which violate the new but not the CHSH inequalities. Alas, the range of Werner states violating the new inequality is smaller than that for the CHSH setting.

Solution

The problem was solved by Tamás Vértesi in [4], where he presents a family of Bell inequalities of the form:

$I_{n,m}=\sum_{x=1}^{n}\sum_{y=1}^{m}\langle A_xB_y\rangle +\sum_{1\leq y\leq y'\leq m}\langle A_{yy'}B_{y}\rangle-\langle A_{yy'}B_{y'}\rangle+$ $\sum_{1\leq x\leq x'\leq n}\langle B_{xx'}A_{x}\rangle-\langle B_{xx'}A_{x'}\rangle$ .

Here $A_x, A_{yy'}$ ( $B_y, B_{xx'}$ ) denote Alice’s (Bob’s) dichotomic observables. On one hand, the maximum classical value of $I_{n,n}$ can be shown equal to $n^2$ . On the other hand, using see-saw optimization methods, it is possible to find numerical lower bounds for the maximum quantum value achievable for $I_{n,n}$ in a singlet state. Doing this for $n=100$ and dividing the result by $100^2$ , we obtain the improved lower bound $K_G(3)\geq 1.417241>\sqrt{2}$ . Note that, for $n=100$ , the Bell inequality $I_{n,n}$ has 465 settings.