Observable Vs Projector in case of Neutrinos

For a spin $s$ system, we can define the observable $A=S_z$, such that $A(t) = S_z(t) = U^\dagger S_z U$, where $U$ is the unitary operator $e^{-iHt}$. We can always define a projector $\Pi_m = |s,m \rangle \langle s, m|$, such that $\Pi_m~(t) = U^\dagger(t) |s,m \rangle \langle s, m| U(t)$. So there is a clear distinction:

$S_z$ is the observable, and $\Pi_m = |s,m \rangle \langle s, m|$ is the projection operator.

Consider a similar situation in neutrino physics. We talk about the flavor projection operator $\Pi_{\alpha} = |\nu_\alpha \rangle \langle \nu_\alpha|$ which projects out a particular flavor state $| \nu_\alpha \rangle$. Also, $\Pi_\alpha(t) = U_f^\dagger(t) |\nu_\alpha \rangle \langle \nu_\alpha| U_f(t)$ where $U_f(t)$ is the flavor evolution operator.

My question: $\Pi_\alpha$ is similar to $\Pi_m$ for the spin, in the sense that both are projectors. What is the analogue of the observable $S_z$ in the case of neutrino?

Well, the spin-on-z operator "observed" is diagonal in the representation specified by a complete set of projectors, $$S_z=\sum_{m=-s}^{s} m ~\Pi_m =\sum_{m=-s}^{s} | s,m\rangle ~m~ \langle 2,m| ~,$$ and of course the evolution is a red herring: all operators evolve similarly by the respective hamiltonians. So, projectors specify the location of the observable's diagonal entries.
In neutrino physics, with flavor states predicated on the charged lepton flavor the neutrinos couple to, so that one observes suitably normalized numbers of charged leptons produced, $N_e, N_\mu, N_\tau$, folding all efficiencies, acceptances, etc..., per single incoming $\nu_e,\nu_\mu, \nu_\tau$, respectively.
So for $\Pi_{\alpha} = |\nu_\alpha \rangle \langle\nu_\alpha|$, the analogous observable is the relative adjusted distribution of charged lepton events in your detector, $$\sum_{\alpha=e,\mu,\tau} \Pi_{\alpha} N_\alpha ~,$$ or some less Platonic experimental object.
• Thank you @Cosmas Zachos. If I understand it right, then the observable you defined is $A = \sum\limits_{\alpha=e, \mu, \tau} \Pi_{\alpha} N_{\alpha}$, such that $A|\nu_e \rangle = N_e |\nu_e \rangle$ etc, gives the average number of neutrinos detected of type $\nu_e$. It looks fine. – Seeker Oct 3 '17 at 15:49
• Yes, actually the number of "primary" electron events , as a fraction of the incoming interacting $\nu_e$ s... In an idealized perfect detector, never missing neutrinos, the Ns would be all 1.... – Cosmas Zachos Oct 3 '17 at 16:15