# Why is the CHSH inequality defined with a minus sign?

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?

• – glS May 13 at 14:54

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.
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.