the measurement directions, namely, $\vartheta_1 = \vartheta^{\prime}_{1} =\pi/2$, $\vartheta_2 = \vartheta^{\prime}_{2} =\pi/2$, $\varphi^{\prime}_{1} = \varphi_1 +\pi/2$, $\gamma\varphi^{\prime}_{2} = \gamma\varphi_2 +\pi/2$, and $\varphi_1 +\gamma\varphi_2 +z_1 z_3 \varphi_3 = 3\pi/4 +n\pi$ exhaust all the possibilities for a maximal violation of the CHSH inequality. The answer is certainly no. Indeed, there are infinitely many ways of achieving such a maximum. This follows directly from the fact that the Schmidt decomposition of a maximally entangled state is not unique. So, for example, consider the state (5a) with $c_1 = -c_2 = 1/\sqrt{2}$, $\vartheta_3 =\pi/2$, $\varphi_3 =0$, $z_1 =-z_2 =+1$, and $z_3 =+1$. The resulting singlet state can be equally expressed in an infinite number of alternative forms by replacing the quantization $z$ axis with any other unit vector ${\bf{\text n}}$ in three-dimensional space. Put it, for instance, as $$|\Psi^{+}_{12}\rangle = \frac{1}{\sqrt{2}} (|y+\rangle_1  |y-\rangle_2
- |y-\rangle_1 |y+\rangle_2 ),$$ where $|y+\rangle_1$ represents spin-up along the $y$ axis for particle 1. Hence, according to the previous paragraph, there must be measurement directions $\vec{e}_1$, $\vec{e}\,^{\prime}_{1}$, $\vec{e}_2$, and $\vec{e}\,^{\prime}_{2}$ in the $x$-$z$ plane (with $\vec{e}_1 \perp \vec{e}\,^{\prime}_{1}$ and $\vec{e}_2 \perp \vec{e}\,^{\prime}_{2}$) such that inequality (10) is maximally violated for the singlet state. As an example, take the choice $\vartheta_1 