$\hat{S}_{13}$ and $\hat{S}_{24}$ are equal. The general scattering matrix relative to non-collinear angles is obtained from $S$ by rotating the spin quantization axis independently in the two conductors (note that this is possible because the two wires are decoupled). The “rotated” S-matrix is obtained [@brataas01] by the transformation $S_{\theta_1,\theta_2}={\cal U} S {\cal U}^{\dagger}$, where ${\cal U}$ is the rotation matrix given by: $${\cal U}=\left( \begin{array}{cccc} 
U_{\theta_1} & 0 &0 & 0\\
0&\mathbb{I}&0 & 0\\
0 & 0 & U_{\theta_2} & 0 \\
0&0&0& \mathbb{I}\end{array} \right)
\label{tra}$$ where $$U_{\theta}=\left( \begin{array}{cc} \cos{\theta\over 2}&\sin{\theta\over 2}\\
-\sin{\theta\over 2}&\cos{\theta\over 2}
\end{array} \right) .
\label{uteta}$$ The probability distributions are now given by the expressions in Eq. (\[stp\]) and Eq. (\[FCSt\]) where the state ${\left|\psi\right\rangle}$ is given by Eq. (\[st\]). In the case where both analyzers are present we set $\sigma=\sigma'=\uparrow$. The probability distribution when one of the analyzers is removed also possesses the structure of Eq.(\[FCSt\]) since, in this case, the correlators $\langle \hat{N}_{\text{O}}^{1\uparrow} \hat{N}_{\text{O}}^{1\downarrow} \rangle$ and $\langle \hat{N}_{\text{O}}^{1\uparrow} \hat{N}_{\text{O}}^{1\downarrow} 
\hat{N}_{\text{O}}^{2\uparrow} \rangle$ appearing in Eq.(\[chi\_s1\]) vanish. In particular when, for example, the upper analyzer in Fig. \[Ent\] is removed we need to replace $\hat{N}_{\text{O}}^{1\sigma}$ with $\hat{N}_{\text{O}}^{1\uparrow}+\hat{N}_{\text{O}}^{1\uparrow}$ and $\hat{N}_{\text{O}}^{2\sigma'}$ with $\hat{N}_{\text{O}}^{2\uparrow}$. For the other correlators one gets: 