CHSH inequality is defined in the following way. Let $Q, R, S, T$ be two outcomes $\{\pm 1\}$ measurements. The measurements are chosen in a certain way, but that is not our concern right now. We know that a state $|\psi\rangle$ would not admit to any hidden variable representation if it violates the following inequality:

$$ \langle\psi| (\langle QS \rangle + \langle RS \rangle + \langle RT \rangle - \langle QT \rangle)|\psi\rangle \le 2, $$ where, $\langle . \rangle$ is defined as the expected value of those operators with respect to $|\psi \rangle$. My question is, why do we have a negative sign for $\langle QT \rangle$? Could we not have any other constant on the right and have all positive terms on the left? Why does it work?


Imagine you had a general formula $$ C=a_1QS+a_2RS+a_3RT+a_4QT. $$ Algebraically, we know that if $Q$, $S$, $R$ and $T$ are random variables with values $\pm 1$, then each term such as $QS\in\{\pm 1\}$. Hence, there is a trivial bound $$ C\leq |a_1|+|a_2|+|a_3|+|a_4| =C_\max. $$ This can never be beaten by any model, be it local hidden variable, quantum, post-quantum.... So, if there exists a local hidden variable model that achieves $C=C_\max$ for a particular set of $\{a_i\}$, then that set of $\{a_i\}$ is not very interesting to us because there's no possibility of getting a contradiction between the LHV prediction and the quantum case (for example).

In particular, if all the $a_i$ are positive, then the choices $Q=R=S=T=1$ saturate the bound. You can also check all cases where an even number of the $\{a_i\}$ are negative - you can always find a deterministic assignment to the random variables that saturates the $C_\max$ bound. So, for it to be interesting, we need an odd number of negative signs in the coefficients.

The particular choice of the numbers themselves is not so important, except that, ideally, you'd like as large a difference as possible between the value that an LHV can achieve and some other model can be predicted to achieve so that, experimentally, you've got the best chance of observing it. I won't go through the detail here but, given there's a large permutation symmetry, you might not be surprised to find that all the coefficients being equal fulfils this desire.

To give a vague illustration: let $a_1$, $a_2$, $a_3$ be positive, and $a_4$ be negative. $C_\max=a_1+a_2+a_3-a_4$. One choice of LHV is every variable being 1, so we can achieve the value $a_1+a_2+a_3+a_4$, making the gap between the two $2|a_4|$. So you'd think to make $|a_4|$ as large as possible. However, another choice is to set $T=-1$, in which case your LHV can achieve $a_1+a_2-a_3-a_4$, yielding a gap of $2|a_3|$. Which is bigger? Obviously whichever term is larger. The best balance is to set $|a_3|=|a_4|$. Repeat for other possibilities.


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