The CHSH Game Strategy

We now construct the entangled strategy that beats the classical $3/4$ ceiling of the CHSH game, and compute exactly how often it wins. Everything follows from one shared Bell pair and four well-chosen measurement angles.

The shared state and the measurements

Alice and Bob share the Bell state

|\Phi^+\rangle = \frac{1}{\sqrt2}\big(|00\rangle + |11\rangle\big).

Each player measures spin in the $X$ – $Z$ plane of the Bloch sphere. A measurement at angle $\alpha$ projects onto $\cos\tfrac{\alpha}{2}|0\rangle \pm \sin\tfrac{\alpha}{2}|1\rangle$ , assigning outcome $\pm1$ . The key fact about $|\Phi^+\rangle$ is that the two outcomes are correlated through the angle difference only:

P(a=b) = \cos^2\!\frac{\alpha-\beta}{2}, \qquad E(\alpha,\beta) = \langle ab\rangle = \cos(\alpha-\beta).

The optimal angles

Choose

\alpha_0 = 0,\quad \alpha_1 = \frac{\pi}{2} \qquad\text{(Alice)},\qquad \beta_0 = \frac{\pi}{4},\quad \beta_1 = -\frac{\pi}{4} \qquad\text{(Bob)}.

Every relevant angle difference has cosine $\pm\tfrac{1}{\sqrt2}$ :

E(0,0)=\cos\tfrac{\pi}{4}=\tfrac{1}{\sqrt2},\;\; E(0,1)=\cos\tfrac{\pi}{4}=\tfrac{1}{\sqrt2},\;\; E(1,0)=\cos\tfrac{\pi}{4}=\tfrac{1}{\sqrt2},\;\; E(1,1)=\cos\tfrac{3\pi}{4}=-\tfrac{1}{\sqrt2}.

Hence $S = E(0,0)+E(0,1)+E(1,0)-E(1,1) = 4\cdot\tfrac{1}{\sqrt2} = 2\sqrt2$ .

Winning probability, term by term

The win rule needs equal outcomes when $x\wedge y=0$ (three inputs) and different outcomes when $x\wedge y=1$ (the single input $x=y=1$ ). With the angles above:

\begin{aligned} (x,y)=(0,0):&\quad P(a=b)=\cos^2\tfrac{\pi}{8},\\ (x,y)=(0,1):&\quad P(a=b)=\cos^2\tfrac{\pi}{8},\\ (x,y)=(1,0):&\quad P(a=b)=\cos^2\tfrac{\pi}{8},\\ (x,y)=(1,1):&\quad P(a\neq b)=\sin^2\tfrac{3\pi}{8}=\cos^2\tfrac{\pi}{8}. \end{aligned}

All four are equal, so the average is

p_{\text{win}}^{\,\text{quantum}} = \cos^2\frac{\pi}{8} = \frac{2+\sqrt2}{4} = \frac12 + \frac{\sqrt2}{4} \approx 0.8536.

This is consistent with $p_{\text{win}} = \tfrac12 + \tfrac{S}{8}$ at $S=2\sqrt2$ .

Try it

Average the quantum winning probability over the four equally likely inputs, using the angle differences above. You should obtain $\cos^2(\pi/8) = \tfrac12 + \tfrac{\sqrt2}{4}\approx 0.8536$ , comfortably above the classical $0.75$ .

Run your code to see the quantum state.