$\vartheta^{\prime}_{1} = \vartheta_1$ and $\vartheta^{\prime}_{2} = \vartheta_2$, a maximal violation of the CHSH inequality occurs provided that (i) $|c_1|=|c_2|=1/\sqrt{2}$, and (ii) $\vartheta_1 =\vartheta_2 =\vartheta_3 =\pi/2$. Clearly, condition (ii) entails that all measurement directions $\vec{e}_1$, $\vec{e}\,^{\prime}_{1}$, $\vec{e}_2$, $\vec{e}\,^{\prime}_{2}$, and $\vec{e}_3$ must lie in the $x$-$y$ plane.

A few remarks should be added here. In the first place, it is to be noted that the requirements $|c_1|=|c_2|=1/\sqrt{2}$ and $\vartheta_3 =\pi/2$ together imply that either state (5a) or (5b) is maximally entangled. This is consistent with the fact according to which a maximally entangled state not only gives the maximum violation of the CHSH inequality but also gives the [*largest*]{} violation attainable for any pairs of four spin observables $\sigma(\vec{e}_1)$, $\sigma(\vec{e}\,^{\prime}_{1})$, $\sigma(\vec{e}_2)$, and $\sigma(\vec{e}\,^{\prime}_{2})$ (provided $\vec{e}_1 \!\nparallel \! \vec{e}\,^{\prime}_{1}$, and $\vec{e}_2 \!\nparallel \! \vec{e}\,^{\prime}_{2}$) [@Kar]. Therefore, both condition (i) and the requirement $\vartheta_3 = \pi/2$ turn out to be absolutely necessary in order to achieve the largest violation, no matter what the orientation of $\vec{e}_1$, $\vec{e}\,^{\prime}_{1}$, $\vec{e}_2$, and $\vec{e}\,^{\prime}_{2}$ may be. However, for the special case leading to Eq. (14) we have that the two measurement directions $\vec{e}_i$ and $\vec{e}\,^{\prime}_{i}$ for particle $i$ ($i=1,2$) giving the maximal violation are perpendicular between themselves. That 