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I have read that because of the conservation of the leptonic number, a neutron should decay into $p + e^- + \overline{\nu}_e$. I don't understand this argument because I have also learnt that the leptonic number may not be conserved.

Since the flavour eigenstates of a neutrino are not the same as the propagating eigenstates, it seems to me that we should also consider muonic or tauic neutrinos for this decay (or rather the mass eigenstates), because when one computes the cross section, one assumes that the neutrino is observed a long time after the interaction had taken place, so that flavour oscillating neutrino may occur.

What do you think ?

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You do understand that short-range neutrino measurements show the expected flavor mix, right?

That is if we set up a detector a few meters from a nuclear core we observe neutrino interactions involving electrons. If we set one up just downstream of a neutrino beam-line we see mostly interactions involving muons (with the expected admixture of those involving electrons and anti-muons).

These fact are re-tested sporadically in each new generation of machines because precision neutrino work requires a near detector.

Consequently the flavor states are defined as the states that participates in weak interactions with a flavored lepton. The neutrino-state in a weak interaction involving an electron is what we call "an electron (anti-)neutrino".

So the short answer is, that no, there is no chance that the thing we define to be an electron neutrino is the (different) thing that we define to be a moun neutrino.

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  • $\begingroup$ Thanks for your answer! How do you detect a neutrino, and specifically a electronic neutrino (which reaction is involved for such a detection) ? $\endgroup$ – user1162101 Nov 9 '14 at 18:59
  • $\begingroup$ The usual diagnostic reaction is charged-current quasi-elastic scattering: $\nu_l + n \rightarrow l^- + p$ where $l$ represents a charged lepton ($e$, $\mu$, or $\tau$). $\endgroup$ – dmckee Nov 9 '14 at 19:57

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