 Solution

Write ${\mathbf{n}}_a=(\cos \alpha ,\sin \alpha )$ and ${\mathbf{n}}_b=(\cos \beta ,\sin \beta )$. Since ${\mathbf{b}}_1-{\mathbf{b}}_3=\sqrt{2}\hat{\mathbf{x}}$ and ${\mathbf{b}}_1+{\mathbf{b}}_3=\sqrt{2}\hat{\mathbf{y}}$, the CHSH inequality combination becomesThereforeParallel or antiparallel Eve axes attain the lower endpoint; orthogonal axes attain zero. This bound is specific to the two-sided projective attack and these planar settings. It is stronger than the generic local bound $|S|\le 2$ and should not be attributed to every possible eavesdropping strategy.

If Eve varies her axes independently of the users' still-secret random settings, the measured correlations average over her choices. The same linear combination obeys $S=-\sqrt{2}\mathbb{E}[{\cos }^2(\alpha -\beta )]$ and remains in the interval by convexity. Strategies correlated with disclosed measurement settings would violate that independence premise; the protocol must protect it.