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I would like to better understand how neutrino oscillations are consistent with conservation of momentum because I'm encountering some conceptual difficulties when thinking about it. I do have a background in standard QM but only rudimentary knowledge of particle physics.

If the velocity expectation value of a neutrino in transit is constant, then it would appear to me that conservation of momentum could be violated when the flavor eigenstate at the location of the neutrino source is different from that at the location of the interaction, since they are associated with different masses.

For this reason I would think that the velocity expectation value changes in transit (for instance, in such a way to keep the momentum expectation value constant as the neutrino oscillates), but then it seems to me that the neutrino is in effect "accelerating" without a "force" acting on it (of course, since the momentum expectation value is presumed constant, there may not be a real problem here, but it still seems strange).

Any comments?

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Related: physics.stackexchange.com/q/2949/2451 –  Qmechanic Feb 23 '12 at 9:09

1 Answer 1

If by "they are associated with different masses" you mean that the flavor eigenstates have different masses then you are working from a misconception. Those states are not eigenstates of the free Hamiltonian so they don't have a mass as such. (They do have a mean expectation if you could weight a bunch of them, but it does not apply to any given neutrino.)


Update 30 April 2012 I had a talk with Fermilab theorist Boris Kayser today after he gave a colloquium and he squared me away on a few things.

  1. This question is one that has been considered many times by many people in many ways.

  2. Not only is what I had written originally not rigorous, but attempts to make it rigorous run into real trouble and get a result at odds with the conventional formalism and inconsistent with experiment.

  3. There is a way to make a rigorous analysis (whole thing at arXiv:1110.3047), and it ends up agreeing with the usual formulation at first order in $\Delta m^2_{i,j}$. It requires that you consider an experiment in the rest frame of the particle that decays to produce the neutrino (and a charged lepton). You define that decay as occurring at space time point $(0,0)$ and compute the amplitude for a neutrino in mass state $i$ to be detected at space time point $(x_\nu, t_\nu)$ in coincidence with the charged lepton being detected at space time point $(x_l, t_l)$ (both also written in the rest frame of the decay particle). Then you notice that the propagators for the two leptons are kinematically entangled. Sum the amplitudes coherently (because the mass state of the neutrino is unobserved). Using the fact that the neutrinos are ultra-relativistic approximate to first order in $\delta m^2_{i,j}$ and drop all terms that don't affect the phase-differences (because neutrino mixing only depends on the phase differences). Somewhere in there was a boost back to the lab frame and a cute calculation of how L-over-E is invariant under the boost: $\frac{L^0}{E^0} = \frac{L}{E}$. The result should be the one we usually give, only now we've dealt with this cute little puzzle.

    Here's an image from an earlier talk he gave on the same subject which shows the whole process on which the calculation is performed: enter image description here

So, long story short: Good question, the usual formalism doesn't seem to have a good answer, but a rigorous calculation can be made and at to leading order agrees with the usual formalism.

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I deleted an answer which was on the lines of the K0long K0short explanation, when I realized the "equal masses", also stated in the question above. I feel this entanglement business is a hand waving. The statement that at the production vertex and the interaction vertex, the neutrino is not in a mass eigenstate,only on its path, means that if I had an accurate experiment of the pi to mu nu decay, and plotted the missing mass I would get three masses? The mind boggles. –  anna v May 1 '12 at 8:57
    
@dmckee You ask for an experiment taking place on the rest frame of the decaying particle. Isn't that necessarily a massless $W^\pm$ boson, so that the rest frame moves at the speed of light? Or am I doing something silly? –  Emilio Pisanty May 1 '12 at 20:04
    
@episanty: The decaying pion or muon. The intermeidate-$W$ is necessarily very far off-shell. I think that Boris must have published a proper paper on this, and ought to dig through his inspire entries some more to find it as I haven't done the description proper justice. –  dmckee May 1 '12 at 20:12
    
@dmckee: ah, yes. Or, if I read you right, it can even be the associated lepton itself (i.e. $\mu^-\rightarrow \overline{\nu}_\mu e^-\overline{\nu}_e$ and track the $\overline{\nu}_\mu$)? –  Emilio Pisanty May 1 '12 at 20:17

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