For a density matrix $\hat \rho$, one can construct a "Bell operator" $\hat B$, such that the violation of the inequality $\langle \hat B \rangle \le 2$ is a clear indication of quantum correlations (entanglement) in the system.

Let's call the expectation of the Bell operator $(\langle \hat B \rangle)$ the "Bell value." The maximum Bell value possible is $2 \sqrt{2}$.

Suppose some state $\hat \rho$ has the Bell value, say $2.7$, and some other state $\hat \rho'$ has the Bell value $2.3$. What exactly can we conclude from the fact that the Bell value of one state is higher than the other? In particular, is there an advantage in using one over the other (potentially for some quantum communication task)?

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    $\begingroup$ Sure, if you're certifying randomness from the CHSH violation (as in a DIRNG protocol). Then more CHSH gives more randomness. $\endgroup$
    – Rammus
    Jan 12, 2023 at 13:50

1 Answer 1


Yes, the CHSH game/scenario is an example of a robust self-test.

What this means is that if you obtain a value that is close to the optimal Bell violation, then the state $\hat{\rho}$ and the measurement operators that were employed to obtain the correlations must be close to those that obtain the optimal value (up to the addition of local auxiliary systems that you cannot observe).

The advantage of being a robust self-test is that if you know the quantum state that achieves the optimal Bell violation is required to be a maximally-entangled state (like in CHSH), and you see a near-optimal violation, then you can certify with some degree of confidence (based on how far you are away from the optimal value) that the protocol did employed a maximally entangled state and the resulting error is simply attributable to noise.

Not all correlations, i.e. Bell scenarios, are robust self-tests (for states and measurements), and it's still an open question to classify which correlations have this remarkable property. For more on this I suggest looking at https://arxiv.org/abs/1904.10042 .


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