Checkpoint: Win the CHSH Game

This checkpoint pulls the module's core thread together: build the CHSH Bell operator explicitly, evaluate it on a Bell state to get the quantum value $S$, and convert that into the probability of winning the CHSH game. You will reproduce the optimal quantum strategy purely by linear algebra.

The plan

The CHSH value is the expectation of the Bell operator in the shared state:

Take the optimal observables in the $X$–$Z$ plane,

and the maximally entangled state $|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle+|11\rangle)$. Because everything is real and lies in the $X$–$Z$ plane, you can work with real $2\times2$ and $4\times4$ matrices throughout — no complex arithmetic needed.

Building the operator

Form the two single-party combinations $B_0+B_1 = \sqrt2\,Z$ and $B_0-B_1 = \sqrt2\,X$, then take Kronecker products with Alice's observables and add:

On $|\Phi^+\rangle$, both $Z\otimes Z$ and $X\otimes X$ act as $+1$ (it is a $+1$ eigenstate of each), so $\langle\mathcal B\rangle = \sqrt2\,(1+1) = 2\sqrt2$. The exercise computes this by explicit matrix–vector arithmetic rather than by spotting the eigenvalues, so the answer is forced by the linear algebra.

From $S$ to the win probability

Using $p_{\text{win}} = \tfrac12 + \tfrac{S}{8}$ from the game lesson,

Try it

Construct the $4\times4$ Bell operator with Kronecker products, evaluate $S=\langle\Phi^+|\mathcal B|\Phi^+\rangle$ by direct summation, and return $p_{\text{win}} = \tfrac12 + S/8$. A correct implementation yields $\cos^2(\pi/8)\approx 0.8536$ — beating the classical ceiling of $0.75$ and winning the CHSH game.

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