shall address the question of whether this also produces a suppression of ${\cal S}_{CH}$ in case of violation.

Let us now analyze the possibility of violation of the CH inequality for different values of $Q_1$ and $Q_2$. First consider the situation where the entangler emits a single entangled pair of electrons in which case $P^{\theta_1,\theta_2}(1,1)={\left\langle\psi\right|}
\hat{N}_{\text{O}}^{1\uparrow} \hat{N}_{\text{O}}^{2\uparrow}{\left|\psi\right\rangle}$, $P^{-,\theta_2}(1,1)={\left\langle\psi\right|}(
\hat{N}_{\text{O}}^{1\uparrow}+\hat{N}_{\text{O}}^{1\downarrow})
\hat{N}_{\text{O}}^{2\uparrow}{\left|\psi\right\rangle}$ and $P^{\theta_1,-}(1,1)={\left\langle\psi\right|}
\hat{N}_{\text{O}}^{1\uparrow} (\hat{N}_{\text{O}}^{2\uparrow}+
\hat{N}_{\text{O}}^{2\uparrow}){\left|\psi\right\rangle}$. We find that the CH inequality is maximally violated for the following choice of angles: $\theta_2-\theta_1=\theta'_2-\theta'_1=3\pi /4$. More precisely we obtain: $${\cal S}_{CH}= T^2 \frac{\sqrt{2}-1}{2}$$ which is equal to the result obtain for an entangled pair of photons [@mandel95], where $T$ plays the role of the quantum efficiency of the photon detectors. In the more general case of $Q_1=Q_2=M$, for $M\gg 1$, we have $$\begin{aligned}
P^{\theta_1,\theta_2}(M,M)&=&\frac{T^{2M}}{2^M} 
\left [\sin^2 \left ( \frac{\theta_1 \pm \theta_2}{2} \right ) \right ]^M \; \nonumber \\
P^{\theta_1,-}(M,M)&=& P^{-,\theta_2}(M,M)=\frac{T^{2 M}}{2^M}  \;\label{prob}\end{aligned}$$ so that the no-enhancement assumption is always satisfied and the quantity ${\cal S}_{CH}$ can be easily evaluated: $${\cal S}_{CH}=\frac{T^{2M}}{2^M} \left[ \sin^{2M}\frac{\theta_1\pm \theta_2}{2}-
\sin^{2M}\frac{\theta_1\pm \theta'_2}{2} +
\sin^{2M}\frac{\theta'_1\pm \theta_2}{2} +
\sin^{2M}\frac{\theta'_1\pm \theta'_2}{2} -2
\right] .$$ The rotational invariance makes $P^{\theta_1,-}$ and $P^{-,\theta_2}$ independent of angles, and $P^{\theta_1,\theta_2}$ dependent on the angles through $\frac{\theta_1\pm 