 Coplanar qubit realization of the chained Bell inequality

On $|{\Phi }^{+}⟩$, measure ${\sigma }_z\cos \theta +{\sigma }_x\sin \theta$, assigning bit $a$ to eigenvalue $(-1{)}^a$. Choose Alice's angles $2(i-1)\delta$ and Bob's angles $(2i-1)\delta$, with $\delta =\pi /(2N)$. The Bell state correlation is $\cos (\theta -\varphi )$. Each adjacent disagreement term and the last agreement term therefore has probability ${\sin }^2(\pi /(4N))$. For $N\ge 2$, their total violates the chained modular Bell inequality bound one, and it tends to zero as $N\to \infty$. At $N=1$ the value is exactly one.

// Target: probability-and-statistics.bigb