# Is it advantageous for a state to have higher Bell CHSH violation?

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)?

• Sure, if you're certifying randomness from the CHSH violation (as in a DIRNG protocol). Then more CHSH gives more randomness. Jan 12 at 13:50

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).