Cite this problem as **Problem 21**.

**Problem**

Can one find bipartite density operators , neither of which violates any CHSH Bell inequality, with the property that does?

**Solution**

The problem was solved with an affirmative answer in [1]. There Navascués and Vértesi identify two-qubit states such that neither nor can violate any CHSH Bell inequality for any . However, if two parties were distributed the bipartite state , then they could violate the CHSH inequality by an amount of 2.023. The result relies on the following idea: if admit a 2-symmetric extension on different subsystems, then none of them (or any number of copies thereof) can violate a two-setting Bell inequality [2]. However, since their tensor product is in general no longer 2-symmetric extendable on any subsystem, it can potentially violate a given two-setting Bell inequality . The states and the measurement settings achieving a violation of can be found via standard see-saw numerical methods. In addition, given any such pair of states , one can construct a state which doesn’t violate any two-setting Bell inequality but such that violates [1].

**References**

[1] M. Navascués and T. Vértesi, Phys. Rev. Lett. 106, 060403 (2011).

[2] B. M. Terhal, A. C. Doherty and D. Schwab, Phys. Rev. Lett 90, 157903 (2003).