Neutrions are produced and detected as flavor eigenstates $\nu_{\alpha}$ with $\alpha=e, \mu, \tau$. These states have no fixed mass, but are the combinations of three mass eigenstates $\nu_{k}$ with $k=1, 2, 3$, with mass $m_1$, $m_2$ and $m_3$, respectively. My questions are:
a) do neutrinos travel from source to the detector as flavor eigenstates or mass eigenstates?
b) is it possible to know which mass eigenstate the neutrino is in?
(a) They start as a flavor eigenstate, which is a super position of mass eigenstates. The mass eigenstates have different time evolution, hence the state is, in general, a mixed state in either basis.
(b) No. As an analogy, consider polarized photons and Faraday rotation--it may start out + polarized, rotate to a mixture of + & - (with coefficients a & b) and then at your + detector you see it $a^2$ fraction of the time, and $b^2$ you don't. In either case, you can't say which state a particular photon was in.
(b') Can we detect a $\nu_e$ and know it's mass? Can it have the mass of a $\nu_{\tau}$? The $\nu_e$ doesn't have "a" mass, it has 3:
$|\nu_e\rangle=0.82|\nu_1\rangle+0.54|\nu_2\rangle-0.15|\nu_3\rangle$
while a tau-neutrino:
$|\nu_{\tau}\rangle=0.44|\nu_1\rangle-0.45|\nu_2\rangle-0.77|\nu_3\rangle$
So, "yes", if we measure it's mass, then it will have a mass that a tau neutrino mass measurement could yield.
In theory: it's not a sensible question to ask, since flavor eigenstates aren't mass eigenstates.
In practice: we do not know the masses of the mass eigenstates, and their differences are much less than an eV--so how are you going the measure that?
