this orthogonality condition is not incidental can be seen by computing the square of the Bell operator (11). This is given by [@Kar] $$B^{2}_{\text{CHSH}} = 4(I+\sin\theta_1 \sin\theta_2 \, \sigma_{\perp 1} \otimes
\sigma_{\perp 2}),$$ where $\theta_i$ is the angle included between the vectors $\vec{e}_i$ and $\vec{e}\,^{\prime}_{i}$, and $\sigma_{\perp i}$ denotes the spin operator for particle $i$ along the direction perpendicular to the plane containing $\vec{e}_i$ and $\vec{e}\,^{\prime}_{i}$. From Eq. (15) it follows that the largest eigenvalue of $B_{\text{CHSH}}$ is (in terms of the absolute value) $\lambda_l = 2(1+|\sin\theta_1 \sin\theta_2|)^{1/2}$, so that a maximum violation is obtained when both $\theta_1$ and $\theta_2$ are $\pi/2$ (mod $\pi$). The eigenvector associated with $\lambda_l$ will consist of a superposition of the states $|\sigma_{\perp 1}\rangle|\sigma_{\perp 2}\rangle$ and $|-\sigma_{\perp 1}\rangle|-\sigma_{\perp 2}\rangle$ with an appropriate relative phase between them, where $|\sigma_{\perp i}\rangle$ ($|-\sigma_{\perp i}\rangle$) denotes the eigenvector of $\sigma_{\perp i}$ with eigenvalue $\sigma_{\perp i}=\pm 1$ ($-\sigma_{\perp i}=\mp 1$), and where $\text{sgn}\sigma_{\perp 1} =\text{sgn}\sigma_{\perp 2}$ ($\text{sgn}\sigma_{\perp 1} \neq \text{sgn}\sigma_{\perp 2}$) for $\text{sgn}(\sin\theta_1) = \text{sgn}(\sin\theta_2)$ \[$\text{sgn}(\sin\theta_1) \neq \text{sgn}(\sin\theta_2)$\]. As was mentioned, such a superposition state has to be completely entangled in order to get the largest possible violation [@Kar].

A question that naturally arises is whether the above conditions on 