condition, however, is not a sufficient one, as is clear from the preceding example (indeed, for this example, in addition to this condition it is necessary that $\vartheta_2 =\vartheta_1 +\pi/4$). Note also that the directions giving the largest violation of the CHSH inequality for the singlet state must satisfy in all cases the constraint $\text{sgn}(\sin\theta_1) \neq \text{sgn}(\sin\theta_2)$. This is because the singlet state is an eigenvector of the operator $\sigma_{n1}\otimes\sigma_{n2}$ (where $\sigma_{ni}$ denotes the spin operator along the ${\bf{\text n}}$ axis for particle $i$) with eigenvalue $-1$, and so, from Eq. (15), the factors $\sin\theta_1$ and $\sin\theta_2$ must be opposite in sign in order to get the largest violation.

Consider now the state (5a) with $c_1 = c_2 = 1/\sqrt{2}$, $\vartheta_3 =\pi/2$, $\varphi_3 =0$, $z_1 =-z_2 =+1$, and $z_3 =+1$. Unlike the singlet state, the resulting triplet state is not rotationally invariant. However, it can also be written in an infinity of equivalent biorthogonal forms. Put it, for instance, as $$|\Psi^{+}_{12}\rangle = \frac{1}{\sqrt{2}} (|x+\rangle_1  |x+\rangle_2
- |x-\rangle_1 |x-\rangle_2 ),$$ where $|x+\rangle_1$ represents spin-up along the $x$ axis for particle 1. So, there will be measurement directions $\vec{e}_1$, $\vec{e}\,^{\prime}_{1}$, $\vec{e}_2$, and $\vec{e}\,^{\prime}_{2}$ (with $\vec{e}_1 \perp \vec{e}\,^{\prime}_{1}$ and $\vec{e}_2 \perp \vec{e}\,^{\prime}_{2}$) in 