I Do neutrinos ever stop oscillating?

1. Nov 28, 2016

Anypodetos

Neutrino oscillation seems to have as one of its prerequisites the fact that the flavour eigenstates differ from the mass eigenstates, so the time-dependent Schrödinger equation applies. Can a neutrino "drop" into the lowest mass eigenstate (which is also the lowest eigenstate of the Hamiltonian?) and stop oscillating?

I'm aware that this is a very unlikely process. The simplest Feynman diagram for the neutrino to get rid of the energy seems to involve three vertices, a W propagator and emission of a photon. Plus, there's a huge time dilation. But will it eventually happen? And could this be detected, in principle, by measuring flavour ratios from a faraway neutrino source?

2. Nov 28, 2016

ohwilleke

First of all (and anyone who disagrees feel free to correct me), I do not believe that neutrino oscillation is conceived as a virtual W boson Feynmann diagram process.

Second, I do think that in an extremely low energy system that mass-energy conservation could prevent a neutrino from oscillating from the lowest mass mass-eigenstate to higher mass mass-eigenstate. But, since the energy gap between two two closest mass-eigenstates is so small (ca. 7-8 meV/c^2), I think it might be a struggle to get a system cold enough for that to happen. (My struggle in being really confident about that answer is that kinetic energy is frame dependent and it should always be possible to choose a frame in which a neutrino is at rest and hence has no kinetic energy to convert to mass.)

3. Nov 28, 2016

ChrisVer

My thought: Well a "neutrino" can exist in any mass eigenstate (even the lowest mass one)... However that neutrino written in that basis is not the one that participate in the interactions (neutrinos interact weakly with their flavor states- for your Feynman diagram comment)...Since the flavor states are not equal to their mass eigenstates, it doesn't really matter.... Because if you call the lowest mass eigenstate $|1>$ then $|1> = a |\nu_e> + b|\nu_\mu> + c |\nu_\tau>$.
Then if the neutrino interacts it will have to fall into one of its flavor eigenstates (because it will interact with one lepton)... and you are back to $|\nu_a> = k_a |1> + m_a |2> + n_a |3>$....

4. Nov 28, 2016

Anypodetos

I didn't mean that. I was thinking of a process by which the neutrino could get rid of its energy when it dropped into a lower energy (eigen)state. Like when an electron precessing in a magnetic field emits a photon and drops into the lower energy eigenstate.

Also, I thought that oscillation happens between flavour eigenstates and energy eigenstates are never reached during oscillations. (Again, like a precessing electron stays away from the energy eigenstates until it emits a photon and stops precessing.)

5. Nov 28, 2016

Anypodetos

Yes, when the neutrino interacts, e.g. being measured, it will interact as a flavour eigenstate. But as the |νe> is (I think) somthing like 70-80% |1>, a bunch of |1> neutrinos will be measured as 70-80% electron neutrinos, but a bunch of oscillating neutrinos will be measured as 1/3 electron neutrinos. So the difference should be detectable, given enough neutrinos from the same source, and given that you measure my and tau neutrinos as well to get the statistics.

6. Nov 28, 2016

ChrisVer

Looking at the PMNS matrix the electron neutrino can be written roughly as 67%, 30%, 3% of 1,2,3 respectively...
I don't understand your idea after the "a bunch of |1>..." especially the bunch that is measured as 1/3 electron neutrinos...

7. Nov 28, 2016

Orodruin

Staff Emeritus
This is not correct. For oscillating neutrinos it depends on the baseline, energy, and initial flavour composition.

What you are describing is neutrino decay. There are several theories beyond the standard model where this might happen and some work is going into putting phenomenological bounds on neutrino decays.

It should also be noted that neutrinos will stop oscillating without decaying. Once the wave packets of the different mass eigenstates have separated enough, there will no longer be any interference terms and the transition probabilities will be constant. Effectively, this will occur when the kinematics of the experiment can be determined well enough to measure which eigenstate the neutrinos are in - thus removing the interference terms from the coherent sum - much like the double slit experiment is no longer showing you interference once you measure which slit the particle goes through.

8. Nov 29, 2016

Anypodetos

I was thinking about, say, neutrinos from the sun which are νe to begin with and have an energy spectrum broad enough so that 1/3 of each flavour reaches us on average. Please tell me if I'm wrong.

9. Nov 29, 2016

Orodruin

Staff Emeritus
No, this is not how solar neutrinos work. First of all, the flavour conversion of solar neutrinos is not due to oscillations in a strict sense, but to resonant flavour conversion based on the adiabatic propagation of neutrino matter eigenstates. If it was just regular oscillations, the average oscillation probability could not be lower than ca 0.5 due to the small mixing of the third mass eigenstate in the electron neutrino. In addition, the parameters for solar neutrinos are such that the wave packets do separate and so there is no oscillatory behaviour left.

For the neutrinos with energies above a few MeV, the solar neutrino flavour conversion is mainly attributed to the MSW effect because neutrinos are produced above the MSW resonance. This means that they arrive at the Earth as essentially pure second mass eigenstate. The mixing between electron neutrinos and the second mass eigenstate is therefore what is going to determine your measured fraction (the strongest bounds on the mixing angle $\theta_{12}$ are based on this).

At lower energies (sub MeV), neutrinos are produced below the MSW resonance and you essentially have averaged out oscillations - implying that (due to the small mixing with the third eigenstate) this probability has to be larger than 0.5. This is indeed also what is observed. But as I said, for the high energy neutrinos where the ratio is ca 1/3, the flavour conversion is due to the MSW effect and not actual oscillations.

That solar neutrino flavour conversion is due to adiabatic transitions rather than oscillations is discussed by Alexei Smirnov (the S of MSW) here and I tend to agree with him on this issue. Still, some people are calling it oscillations in the literature but it is a misnomer.

10. Nov 29, 2016

nikkkom

Please, anyone who knows this stuff well, if you see a mistake, correct me.

Oscillations are the result of the fact that neutrinos are very light, and thus all three mass states travel almost with th same velocity, near light speed. Therefore, their wavefunctions overlap and produce interference pattern.

If you are far enough away from the source, the velocity difference becomes important. IOW: the neutrino is created in a flavor eigenstate (say, e-neutrino), but if you know its energy, and you observe it in your detector and measure the time it took to cover the distance, you will see that it traveled as one of the mass eigenstates. Not a mix of them.

Quarks in fact work the same way. What we call u,c,t quarks are not analogous to e,mu,tau-neutrinos. They are analogous to *mass* eigenstates of neutrinos, not flavors. The "correct" flavor eigenstates of up-type quarks are (u' c' t') = Mckm (u c t).

So up-quarks, too, "oscillate". But since their masses are very different, for quarks it would be quite hard to detect it experimentally. The "large distance" for them to become observationally separate is microscopically small.

Moreover, it's possible to reformulate everything so that up-type leptons and quarks have coinciding mass and flavor eigenstates, and down-type lepton and quarks would "oscillate". This would be a more complicated picture mathematically (neutrinos wouldn't oscillate. Electrons _would_), but the predictions of it will be the same.

IIUC heavier neutrino can decay into a lighter one (virtual W emitting a photon), but the probability is very low.

11. Nov 30, 2016

Orodruin

Staff Emeritus
No, it is the result of neutrino masses being very degenerate.

This would make no sense. Sure, you can define that new basis and call it "flavour", but how are you going to measure the flavour of a neutrino?

You would still find that neutrino processes at large distance produce different mass eigenstate leptons.

The entire reason to work in the quark mass eigenstate bases (even though weak interactions become non-diagonal) is that the mass eigenstates are the physical states. This is true also for neutrinos, but for neutrinos you often have coherence between the different mass eigenstates.

12. Nov 30, 2016

vanhees71

Indeed, a particle interpretation in the sense of asymptotic free states is only possible in the mass eigenstates, but in fact you can only observe neutrinos by reactions with your detector material, and the weak interaction, responsible for these interactions, couples to the flavor eigenstates. Taken into account both the production of neutrinos (e.g., by pion decay) and the detection of the neutrinos, you can describe everything by the usual QFT formalism (S matrix), and the neutrino always occurs as internal line of the corresponding Feynman diagram. So you never measure neutrinos other than via its propator within a Feynman diagram, describing the entire process of producing and detecing the neutrino. So there is no need to speculate about how to define neutrinos as asymptic states, and there's no problem with energy-momentum conservation, etc. The great confusion about neutrino oscillations is that many textbook writers make their explanations "simpler than possible" (against Einstein's clear advice to make things as simple as possible but not simpler). The problem is that in the usual S-matrix formalism of scattering processes you don't take into account the finite distance between production and detection, but that's crucial to describe the long-baseline experiments to measure neutrino oscillations. So the usual S-matrix formalism has to be adapted to this situation, and one cannot work with plane-wave asymptotic states as in usual scattering theory but with wave packets reflecting the local nature of the production and detection processes at the (far distant) places. A good description can be found in

https://arxiv.org/abs/1008.2077

Note also Ref. [1] by the same authors:

https://arxiv.org/abs/0905.1903