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I don't understand why the propagation of neutrino mass eigenstates are given by planewave solutions as expressed in this Wikipedia article.

In addition to not being used to thinking in the Schrodinger picture when it comes to quantum field theories, I think I'm misunderstanding what 'mass eigenstates' and 'flavor eigenstates' mean. By 'mass eigenstates' I presume that they mean that these are eigenstates of the free neutrino theory. Then the flavor eigenstates are eigenstates of the electro-weak theory, I think without the interaction with the charged leptons.

But, given this understanding, I'm not sure why the mass eigenstates propagate as planewaves instead of as something much more complicated because they are not eigenstates of the interacting Hamiltonian.

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  • $\begingroup$ I think it's to do with the speed varying slightly. Check out photon effective mass, and have a look at the breathers here. Focus on the red line. There's a concertina-like lengthening and shortening going on. $\endgroup$
    – John Duffield
    Commented Mar 14, 2016 at 23:40
  • $\begingroup$ A proper description of neutrino propagation (and hence neutrino oscillation) requires to use the wave packet formalism (a superposition of plane waves) to avoid inconsistencies like the impossibly to conserve both energy and momentum for an entangled state. It is properly described in Giunti's book (chapter 8). $\endgroup$
    – Paganini
    Commented Mar 16, 2016 at 8:58

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You are clear on the meaning of mass eigenstate and flavor eigenstate. That's a good place to start.

Now, recall that every state can be written in terms of any basis at any time.

So, treat the processes of emission, propagation and interaction (detection in an experimental setting) thus:

  • A weak interaction produces a well defined flavor state with a particular momentum thanks to the creation operator.

  • That same state (with it's existing momentum) is also described in the mass basis (but as an admixture, rather than a eigenstate), and as the particle is now in free motion it is this basis we use to describe the time evolution of the state. The different states have different frequencies, which gives rise to a oscillation in the mass-basis content of the state.

  • The neutrino may interact with any matter it encounters along the way, but must do so in a flavor state and the cross-section for doing so depends on the amplitude for the appropriate flavor state(s), which in turn depends on the current admixture of mass states. In any case the neutrino is observed to have the original momentum.

This should give rise to some questions, the foremost of which is 'How can the momentum be the same for all the different mass states?', which is not a trivial thing. I'm now officially out of my depth, but I think that it helps that the neutrinos we experiment on are all ultra-relativistic.

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  • $\begingroup$ The part that I don't understand is the one where you assume free motion from the point where the neutrino originates (wherever the beam starts) to where you observe it in your detector. That seems crazy to me: it's QFT. Everything is always interacting. Further, I'm still not really clear on why this is the appropriate phase factor for free motion. The state is a spinor, aren't its free solutions different from the Klein-Gordon ones? $\endgroup$
    – Kevin Driscoll
    Commented Mar 15, 2016 at 1:55
  • $\begingroup$ There are a couple of things here. Sure, it's a quantum field theory, but the only interaction is the weak force which has negligible range at MeV and GeV energies, so you don't worry much about it, but there is the "matter effect" which accounts for that---it really is a small correction in most experimental contexts so treating it as a free particle is the right place to start. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 15, 2016 at 2:18
  • $\begingroup$ Secondly, the state may be Dirac or it may be Majorana but in either case it is ultra-relativistic (in experimental contexts anyway) so the spin doesn't really get involved. Also, I should warn you that I am a greasy-handed, solder-fume addled experimenter, not a theorist, so I tend to take a rough and ready approach to these things until someone tells me I have to pay attentions. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 15, 2016 at 2:18
  • $\begingroup$ Alright, thanks for your help. As a theorist, it's not that I think what you're saying in dubious. Quite the opposite, it has a very reasonable physical basis. But, I generally only feel as though I've understood these things when I can show somewhat rigorously that this approximate framework works out. And honestly, I'm surprised I'd even have to do that here: it was presented as an almost trivial thing. I thought I was having a brain fart and missing some little 3 line derivation. $\endgroup$
    – Kevin Driscoll
    Commented Mar 15, 2016 at 3:56
  • $\begingroup$ No hassle. I figured that out eventually. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 15, 2016 at 14:58

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