The CHSH Inequality

Clauser, Horne, Shimony, and Holt (CHSH, 1969) recast Bell's theorem into a form that real experiments can use. It needs only two settings per party and tolerates imperfect correlations — exactly what detectors deliver.

The CHSH quantity

Alice chooses between two measurements with outcomes $A, A' \in \{+1, -1\}$; Bob chooses between two with outcomes $B, B' \in \{+1, -1\}$. Define the four correlation functions $E(A,B) = \langle AB \rangle$, and assemble the CHSH combination

Note the single minus sign on the last term — it is essential.

The classical (local-realistic) bound

In any local hidden-variable model, each run carries definite values $A, A', B, B' \in \{\pm 1\}$, fixed before measurement. For a single run the quantity

Here is the key observation: $B$ and $B'$ are each $\pm 1$, so exactly one of $B + B'$ and $B - B'$ is $0$ and the other is $\pm 2$. Whichever survives is multiplied by $A$ or $A'$, each $\pm 1$. Therefore

Averaging over the hidden-variable distribution $\rho(\lambda)$ can only shrink an average of numbers bounded by $2$, so the CHSH inequality is

Any local-realistic theory must obey it, no matter how complicated $\rho(\lambda)$, $A(\cdot,\lambda)$, and $B(\cdot,\lambda)$ are.

Why two settings, not three

Bell's original inequality assumed perfect anticorrelation, which no lab achieves. CHSH drops that requirement entirely: it only assumes $\pm 1$ outcomes and local realism. That robustness is why every modern Bell test — from Aspect's 1982 photon experiments to the 2015 loophole-free tests — reports a CHSH value and compares it against $2$.

Try it

This is a numerical exercise — your code should return a number. Brute-force all $2^4 = 16$ sign assignments of $(A, A', B, B')$, compute $|S|$ for each, and return the maximum. You should recover the classical CHSH bound of $2$.

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