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How is the violation of the usual CHSH inequality by a quantum state related to the entanglement of that quantum state?

Say we know that exist Hermitian and unitary operators $A_{0}$, $A_{1}$, $B_{0}$ and $B_{1}$ such that $$\mathrm{tr} ( \rho ( A_{0}\otimes B_{0} + A_{0} \otimes B_{1} + A_{1}\otimes B_{0} - A_{1} \otimes B_{1} )) = 2+ c > 2,$$ then we know that the state $\rho$ must be entangled. But what else do we know? If we know the form of the operators $A_{j}$ and $B_{j}$, then there is certainly more to be said (see e.g. http://prl.aps.org/abstract/PRL/v87/i23/e230402 ). However, what if I do not want to assume anything about the measurements performed?

Can the value of $c$ be used to give a rigourous lower bound on any of the familar entanglement measures, such as log-negativity or relative entropy of entanglement?

Clearly, one could argue in a slightly circular fashion and define an entanglement measure as the maximal possible CHSH violation over all possible measurements. But is there anything else one can say?

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Your questions is answered here: arXiv:0907.2170. BTW, device-independent is the key phrase to search for. –  Piotr Migdal Nov 18 '11 at 15:42
@PiotrMigdal Thanks for your comment. I had not thought of goggling for "device-independent" and was not aware of that paper. It seems to answer my question, though I'm still going through some of the details. –  Earl Nov 21 '11 at 10:37
@PiotrMigdal: Perhaps you should post that as an answer. –  Joe Fitzsimons Nov 21 '11 at 15:33

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up vote 5 down vote accepted

In a paper C.-E. Bardyn et al., PRA 80(6): 062327 (2009), arXiv:0907.2170, they discuss constrains on the state, given how much the CHSH equality is violated ($S=2+c$), but without putting any assumptions on the operator used.

In general people consider schemes, when operators (for a Bell-type measurement) are random or one or more parties cannot be trusted. One of the key phrases is device-independent and maybe also loophole-free (as even a slight misalignment of operators may change the results dramatically).

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