Checkpoint: Violate a Bell Inequality

Time to put the module together. You will compute a CHSH value for a real entangled state and confirm it breaks the classical limit of $2$ — the experimental signature that local realism fails.

The state and the correlations

Use the Bell state

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

When both qubits are measured along directions in the $X$ – $Z$ plane — Alice at angle $\theta_a$ , Bob at $\theta_b$ — the joint outcomes ( $\pm 1$ ) have correlation

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

At equal angles $E = +1$ (perfectly correlated); this is the " $+$ " cousin of the singlet's anticorrelation.

The optimal settings

Choose the standard CHSH angles, all spaced by $45^\circ$ :

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

Then

\begin{aligned} E(a,b) &= \cos(-\tfrac{\pi}{4}) = +\tfrac{1}{\sqrt2}, & E(a,b') &= \cos(\tfrac{\pi}{4}) = +\tfrac{1}{\sqrt2}, \\ E(a',b) &= \cos(\tfrac{\pi}{4}) = +\tfrac{1}{\sqrt2}, & E(a',b') &= \cos(\tfrac{3\pi}{4}) = -\tfrac{1}{\sqrt2}. \end{aligned}

Assemble S

S = E(a,b) + E(a,b') + E(a',b) - E(a',b') = \tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2} - \big(-\tfrac{1}{\sqrt2}\big) = \frac{4}{\sqrt2} = 2\sqrt2.

So $|S| = 2\sqrt2 \approx 2.828 > 2$ . The classical bound is broken, and by exactly the amount Tsirelson allows.

Try it

This is a numerical exercise — your code should return a number. Define $E(\theta_a, \theta_b) = \cos(\theta_a - \theta_b)$ , evaluate the four correlations at the settings above, and return $|S|$ . The grader checks that you exceed $2$ and hit the quantum maximum $2\sqrt2$ .

Run your code to see the quantum state.