Bell violation by tensoring

Cite this problem as Problem 21.


Can one find bipartite density operators \rho_1, \rho_2, neither of which violates any CHSH Bell inequality, with the property that  \rho_1\otimes\rho_2 does?


The problem was solved with an affirmative answer in [1]. There Navascués and Vértesi identify two-qubit states \rho_1,\rho_2 such that neither \rho^{\otimes N}_1 nor \rho^{\otimes N}_2 can violate any CHSH Bell inequality for any N. However, if two parties were distributed the bipartite state \rho_1\otimes\rho_2, then they could violate the CHSH inequality by an amount of 2.023. The result relies on the following idea: if \rho_1, \rho_2 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 B. The states \rho_1,\rho_2 and the measurement settings achieving a violation of B can be found via standard see-saw numerical methods. In addition, given any such pair of states \rho_1,\rho_2, one can construct a state \rho which doesn’t violate any two-setting Bell inequality but such that \rho^{\otimes 2} violates B [1].


[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).