before we compute the density matrix of the last spin by tracing out the rest $$\label{rho_n_ud_t}
 \rho_N^{\alpha}(t)=Tr_{\widehat{N}} |\Psi^\alpha (t)\ra \la \Psi^\alpha (t)| \ \ \ \text{for $\alpha=\uparrow,\downarrow$}.$$ To quantify the quality of the state transfer we compute the fidelity as $$\label{fid_r_ud_t}
 F_r^\alpha(t)= \la \psi^{(+1)}|  \rho_N^{\alpha}(t) |\psi^{(+1)}\ra.$$ Unlike the fidelity $F_u^\pm (t)$ for unrestricted measurement basis the $F_r^\alpha(t)$ depends on input parameters $\theta$ and $\phi$. To have an input independent quantity one may compute the average fidelity for all possible pure input states on the surface of the Bloch sphere $$\label{Fav_ud_t}
 F_{av}(t)= \frac{1}{4\pi} \int{F_r^\alpha(t) \sin(\theta) d\theta d\phi}.$$ Using a little bit of maths one can show that

$$\begin{aligned}
 \label{Fav_t}
 F_{av}(t)&=& \frac{1}{6} \left\{ \la\downarrow\Downarrow|e^{+iHt}  |\uparrow_N\ra \la\uparrow_N|  e^{-iHt} |\downarrow\Downarrow\ra
           +                  \la\uparrow\Downarrow|e^{+iHt} |\downarrow_N\ra \la\downarrow_N|  e^{-iHt} |\uparrow\Downarrow\ra  \right\} \cr
          &+& \frac{1}{3} \left\{  \la\uparrow\Downarrow|e^{+iHt} |\uparrow_N\ra \la\uparrow_N| e^{-iHt} |\uparrow\Downarrow\ra
          +                 \la\downarrow\Downarrow|e^{+iHt} |\downarrow_N\ra \la\downarrow_N| e^{-iHt} |\downarrow\Downarrow\ra \right\}  \cr
          &+& \frac{1}{3} abs\left\{ \la\downarrow\Downarrow|e^{+iHt} |\downarrow_N\ra \la\uparrow_N| e^{-iHt} |\uparrow\Downarrow \ra \right\}.\end{aligned}$$

where in the above formula it is assumed that the outcome of the measurement is spin up and to have the formula for the outcome spin down one has to only replace $\Downarrow$ with $\Uparrow$ in Eq. (\[Fav\_t\]). In fact, due to the symmetries of the system $F_{av}(t)$ is identical 