Quantum Violation of CHSH

The classical CHSH bound is $|S| \le 2$ . Quantum mechanics can do better — and by a precise, finite amount. This violation is the experimental signature that local realism fails.

Quantum correlations of the singlet

Measure each particle of a spin singlet along a chosen axis, recording $\pm 1$ for spin up/down. Quantum mechanics predicts the correlation between Alice's axis (angle $\theta_a$ in some plane) and Bob's axis (angle $\theta_b$ ) is

E(\theta_a, \theta_b) = -\cos(\theta_a - \theta_b).

The minus sign reflects the singlet's anticorrelation: at equal angles, $E = -\cos 0 = -1$ , perfectly opposite outcomes.

The optimal settings

Choose Alice's two axes and Bob's two axes spaced by $45^\circ$ :

a = 0, \quad a' = \tfrac{\pi}{2}, \quad b = \tfrac{\pi}{4}, \quad b' = -\tfrac{\pi}{4}.

Every relevant angle difference is an odd multiple of $\pi/4$ , so each correlation is $\pm\cos(\pi/4) = \pm \tfrac{1}{\sqrt2}$ . Evaluate the four terms:

\begin{aligned} E(a,b) &= -\cos(0 - \tfrac{\pi}{4}) = -\tfrac{1}{\sqrt2}, \\ E(a,b') &= -\cos(0 + \tfrac{\pi}{4}) = -\tfrac{1}{\sqrt2}, \\ E(a',b) &= -\cos(\tfrac{\pi}{2} - \tfrac{\pi}{4}) = -\tfrac{1}{\sqrt2}, \\ E(a',b') &= -\cos(\tfrac{\pi}{2} + \tfrac{\pi}{4}) = +\tfrac{1}{\sqrt2}. \end{aligned}

Assembling S

S = E(a,b) + E(a,b') + E(a',b) - E(a',b') = -\tfrac{1}{\sqrt2} - \tfrac{1}{\sqrt2} - \tfrac{1}{\sqrt2} - \tfrac{1}{\sqrt2}.

The single minus sign in CHSH flips the last (positive) term so that all four contributions reinforce:

|S| = 4 \cdot \frac{1}{\sqrt2} = 2\sqrt2 \approx 2.828.

This exceeds the classical bound $2$ by a factor of $\sqrt 2$ . The number $2\sqrt2$ is famous: it is Tsirelson's bound, the largest CHSH value any quantum state can reach.

Try it

This is a numerical exercise — your code should return a number. Using the singlet correlation $E(x,y) = -\cos(x - y)$ and the four settings above, compute $|S|$ . You should obtain the quantum maximum $2\sqrt2 \approx 2.828$ .

Run your code to see the quantum state.