Bell violation by tensoring – Open Quantum Problems

Cite this problem as Problem 21.

Problem

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?

Solution

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].

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).