Defining total energy in weak interaction involving a neutrino This question was inspired by the brief introduction to neutrino oscillations given in a Particle Physics course I took recently and has been bothering me for a while now. I hope to find some clarification here.
To the best of our knowledge mass (energy) eigenstates for neutrinos are distinct from flavour eigenstates, in the sense that the states in one representation are always a superposition of at least two states in the alternative representation. This means that when we talk about, say, an electron neutrino, the particle's energy is not well defined. 
Now, neutrinos only interact via weak force. I assume that in order for the associated vertex to make sense (e.g. in a Feynman Diagram representation), the incoming neutrino must exist in a well defined flavour eigenstate. This however would imply that its energy is NOT well defined. 
If the above is correct, does it mean we can no longer define a total energy for the interaction?
A similar question can be asked for quarks but the fact that neutrinos only interact weakly makes it clearer in my opinion.
 A: This might be for K McDonald, 2016 to answer properly, by balancing exuberant jitterbugging angels on the head of a pin. I'll just let you remind yourself of the 11 orders of magnitude of irrelevance of your question, which I have to assume your instructor on neutrino oscillations emphasized to you before he went on to recondite and subtler phenomena. 
A wave packet of comoving neutrino mass eigenstates hits your detector at E ~ 5MeV; let's take only two species, with masses m and m', and energies E and E', mixing maximally, and take a common momentum p for simplicity. Your un-normalized minimal wavepacket then is just
$$
e^{ipx -iEt }  + e^{ipx -iE't } = e^{ip(x-t)} (e^{-it\frac{m^2}{2p}}+e^{-it\frac{m'^2}{2p}}),
$$ 
where we have expanded $E\sim p+m^2/2p$ for relativistic neutrinos.
So, yes, there is a "slop" of $\Delta E \sim (m^2-m'^2)/2E ~$ in the 
wavepacket, of order $10^{-4} \cdot 10^{-7}$ eV, for typical $\Delta m^2\sim 10^{-4}$eV$^2$. The packet's energy spread is  thus $10^{-11}$eV.
This is what you wish to probe by multiplying it by Ls of hundreds and hundreds of kilometers, and monitoring changes in the cosine envelope of the wavepacket; these huge distances are there for a reason; so, ipso facto, it is nothing you could resolve in centimeters or meters, nay, 10s of meters in your detector, if you imagined you'd meaningfully capture it, somehow. 
You recall that pre-fab terms such as "electron neutrino" are merely terms of convenience: you may write your fundamental vertices without ever naming, or knowing about, an electron neutrino. Sure, go ahead, "properly" compute three separate vertices in 3 separate Feynman diagrams with 3 different neutrino mass eigenstates and energies, and fold them into the PMNS matrix coefficients, and sum them, integrate them over suitable ranges, etc... but why? Part of what you presumably learned in your course is to choose your battles wisely and just do the simplest possible calculation, but not simpler. That's why most/all people opt for the above convenience. Why would you fuss about $10^{-11}$eVs?
Nevertheless, Kirk's paper mentioned gives you a magnificent bibliography of papers that dared fuss... 
