# Bell violation by tensoring

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# 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 by Navascués and Vértesi in [1]. They constructed an explicit example according to the following neat idea. If $\rho_1, \rho_2$ admit a 2-symmetric extension on different subsystems, then none of them can violate any CHSH inequality. But since their tensor product is in general no longer 2-symmetric extendable on any subsystem it can potentially violate one. They even showed that there exist states $\rho$ which doesn't violate any CHSH Bell inequality but $\rho \otimes \rho$ does.

Remark: As a further enhancement of the statement, Liang proposed to consider the following stronger problems:

• Q1a: Find a density operator $\rho$ which does not violate any Bell inequality, but where $\rho^{\otimes N}$ does violate some Bell inequality for some large enough N.
• Q1b: Find two density operators $\rho_1$ and $\rho_2$, both of which do not violate any Bell inequality, but $\rho_1\otimes\rho_2$ does violate some Bell inequality.

Note that Q1a would imply Q1b. The difference to the posed problem on this page is that one demands the starting states not to violate any Bell inequality for any number of measurements and settings and not just for the CHSH setting with two binary measurements per side.

# Literature

1. M. Navascués and T. Vértesi, »CHSH Activation« arXiv:1010.5191 (2010)