Possibilty of flavor oscillation of an electron

  • #1

Summary:

Particles with mass close to 0 ( neutrinos ) oscillate at different flavors while propagating in space.Can charged leptons do the same thing?

Main Question or Discussion Point

Neutrinos oscillate at different flavors while propagating in space and this is due to their mass.Any particle being massless cannot oscillate between different flavors while leptons with the mass of the electron and above are very unlikely to change their flavor. Will we able to detect a change in flavor or the probability is so small well yeah we will never detect similar behavior.And what about virtual particles? When a particle is first created due to the Heisenberg principle we can find it to have different masses.As time passes the particle gets its 'natural' mass.Can a super light electron change its flavor and then what would happen?
 

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  • #2
PeroK
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The reason for neutrino oscillations is that, although they are produced in the weak interaction in a mass eigenstate (electron, muon or tau neutrino), those mass eigenstates are not eigenstates of the free Hamiltonian. From the perspective of the free Hamiltonian a mass eigenstate is a superposition of free Hamiltonian eigenstates - and the state oscillates harmonically. A neutrino measurement, however, collapses the state back into a mass eigenstate, where the usual quantum probabilities apply.

I believe in principle this could be the case for other particles. But, from experiment, the electron is an eigenstate of the free Hamiltonian and, therefore, is not going to oscillate between different electron, muon and tauon states.

PS I'm not sure whether there is a theoretical possibility of electron-muon oscillation, but I found this:

https://physics.stackexchange.com/q...ust-like-electron-neutrinos-into-muon-neutrin
 
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  • #3
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But, from experiment, the electron is an eigenstate of the free Hamiltonian
From experiment or from definition? We call the mass eigenstate "electron".
 
  • #4
PeroK
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From experiment or from definition? We call the mass eigenstate "electron".
The electron is defined as the mass eigenstate and is also an eigenstate of the free Hamiltonian (which in hindsight perhaps need not have been the case).

It looks so obvious that mass eigenstates are eigenstates of the free Hamiltonian that the neutrino oscillation was such a puzzle.
 
  • #5
vanhees71
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The reason for neutrino oscillations is that, although they are produced in the weak interaction in a mass eigenstate (electron, muon or tau neutrino), those mass eigenstates are not eigenstates of the free Hamiltonian. From the perspective of the free Hamiltonian a mass eigenstate is a superposition of free Hamiltonian eigenstates - and the state oscillates harmonically. A neutrino measurement, however, collapses the state back into a mass eigenstate, where the usual quantum probabilities apply.

I believe in principle this could be the case for other particles. But, from experiment, the electron is an eigenstate of the free Hamiltonian and, therefore, is not going to oscillate between different electron, muon and tauon states.

PS I'm not sure whether there is a theoretical possibility of electron-muon oscillation, but I found this:

https://physics.stackexchange.com/q...ust-like-electron-neutrinos-into-muon-neutrin
It's the other way around: In the weak interaction neutrino flavor eigenstates are produced (electron, muon, tau indicate flavor not mass eigenstates). The flavor eigenstates are superpositions of mass eigenstates and that's why the neutrinos we observe "oscillate" between different flavors, i.e., if you have e.g., an anti-electron neutrino from a decay of a neutron you can detect it after some distance as either of the three neutrino flavors.

A neutrino long-baseline experiment always consists of producing the neutrinos at one place A and then detect them at some distant place B. E.g., you can use neutrinos from decays of muons to produce them. Then you have a muon neutrino at place A (flavor eigenstate). At place B you can detect any other flavor of neutrino since on their way to B they "oscillate" because the different mass eigenstates in the superposition have different relative phases when propagating, and thus at place B you have a mixture of different flavor eigenstates, i.e., with some probability you can observe all flavors at place B.
 
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  • #6
Vanadium 50
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that's why the neutrinos we observe "oscillate" between different flavors
The more I think about it, the less I like the language of "oscillations". This is not oscillations. It's interference.

If we create a neutrino of a given flavor at A and detect one of a given (same or different) flavor at B, we don't know if in between it was a ν1, ν2 or ν3. It's the double-slit experiment (in this case, triple-slit) all over again.
 
  • #7
vanhees71
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Sure, there's a huge unnecessary confusion in the (theoretical) literature. There's a long debate, how it works with energy and momentum conservation and all that. The solution is of course just using QFT.

Put in terms of Feynman diagrams the solution of these quibbles is very simple: You have to consider the neutrinos as a propgator line only, describing what you really observe, i.e. the reactions of the neutrinos in the detector material at the far-side detectors in long-baseline experiments and their creation in, e.g., an accelerator.

E.g. in the T2K experiment, by charged-pion decays mostly (branching ratio around 99.99%) like ##\pi^+ \rightarrow \mu^+ + \nu_{\mu}##. with the pions decaying in flight you have a nice neutrino beam which can then be detected at some distance at place B, where it reacts with the detector material (purified water at Kamikande in the T2K experiment, making reactions like ##\nu + X \rightarrow e^{-}+Y## and ##\nu+X \rightarrow \mu^{-} + Y##. In the first case you detect an electron neutrino in the second a muon neutrino. Kamiokande measures the fast elecetrons and muons analyzing their Cerenkov radiation. They can distinguish between electrons and muons and thus measure the amount neutrinos detected as electron or muon neutrinos.

Here's a paper where they describe an experiment measuring both "appearance" of electron neutrinos and "disappearance" of muon neutrinos:
https://arxiv.org/abs/1502.01550
 
  • #8
Vanadium 50
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I think that addresses a complication, "why doesn't the recoil kinematics select out a mass eigenstate" and the answer is that a decay pipe is a box and QM precludes this. My concern is that the "oscillation" view implies you create a νe that in flight turns into a νμ and back again - and at all times having a definite four-momentum and a definite (although time-varying) flavor. That's not a correct picture.
 
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  • #9
vanhees71
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The dilemma is that on the one hand a particle interpretation is only possible for asymptotic free mass eigenstates, but that neutrinos are not produced in such states but rather in a flavor eigenstate. Since these are not mass eigenstates it's not clear how to interpret them as external lines of a Feynman diagram.

The solution is to include the concrete production mechanism and the detection mechanism. Then you have no trouble but you get the probabilities for detecting the neutrino as either of the flavor eigenstates. There's no intraction that allows you to detect the mass eigenstates.

For a nice QFT treatment on neutrino oscillation experiments, see

https://arxiv.org/abs/1008.2077
 
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  • #10
Vanadium 50
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a particle interpretation is only possible for asymptotic free mass eigenstates, but that neutrinos are not produced in such states
Neither are the parents. One could say [itex]\pi \rightarrow \mu \nu[/itex] projects out a mass eigenstate if you measure the pion and muon four-momenta. But this isn't correct since the pion and muon are not free particles - i.e. because they are confined to a decay pipe they are not plane waves. They don't even have a continuum of allowed four-momenta.

This is just enough to ensure [itex]\pi \rightarrow \mu \nu[/itex] produces a single flavor eigenstate and not three mass eignestates.
 
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  • #11
vanhees71
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Yes, and the flavor eigenstate is a superposition of mass eigenstates!
 
  • #12
Vanadium 50
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The key word is "superposition". You get one decay to a flavor eigenstate and not three decays (with branching fractions adding up to unity) to three different mass eigenstates.
 
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  • #13
To solve the disagreement 3 things are true:

1)Neutrinos change their flavor while travelling in space

2)Neutrinos with specific flavor don't have a specific mass.

3)Neutrinos with specific mass don't have a specific flavor.
 
  • #14
Vanadium 50
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2 and 3 are true. However with 1, "change their flavor" implies they had a well-defined flavor to begin with.
 
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  • #15
2 and 3 are true. However with 1, "change their flavor" implies they had a well-defined flavor to begin with.
They do since neutrinos are produced in beta decay and the flavor must be conserved.
 
  • #16
Vanadium 50
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No, that does not put them in state of well-defined flavor. It puts them in a state with a non-zero projection of flavor at t = t0. The best you can do is have three probabilitues (which sum to unity) vs t.
 
  • #17
So you are saying that during a neutron decay to an electron proton and antineutrino we don't know the flavor of the antineutrino right after the decay.

Many textbooks would disagree with you.
 
  • #18
Vanadium 50
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So you are saying that during a neutron decay to an electron proton and antineutrino we don't know the flavor of the antineutrino right after the decay.
"right after" is a complicated business. The state is evolving because the three amplitudes are interfering. See message 6.
 
  • #19
vanhees71
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As I said, all these quibbles are solved using the neutrinos only as inner propagator lines when calculating S-matrix elements. According to the usual lepton couplings neutrinos are produced in flavor eigenstates, determined by how they are produced. E.g. if you use ##\pi^0 \rightarrow \mu \nu## decays the ##\nu## is ##\nu_{\mu}## when produced at a given place and time (in the usual QFT sense of course!). See @Vanadium 50 's statement in #10.

Now the handwaving argument brought forward in many textbooks and also in papers is that this flavor eigenstate is no mass eigenstate and thus no energy eigenstate. That's why as a function of time (in the Schrödinger picture) the state of the "propagating free neutrino" is a superposition of different flavor eigenstates. The precise "initial flavor state" in terms of superpositions of mass eigenstates is determined by the neutrino mixing matrix (PMNS matrix named after Pontecorvo, Maki, Nakagawa, and Sakata). So depending on at "which time" you detect a neutrino you get with the corresponding probability again one of the possible neutrino flavor states (because you can also only detect flavor eigenstates depening on the reaction used for detection, e.g., if the detection is something like ##\nu + X \rightarrow \mu +X## the neutrino was detected as a muon neutrino, and that happens with a certain probability.

Now this naive picture gives rise to a long debate about the physics of neutrino oscillations, particularly on the question, what about energy-momentum conservation. For me the solution is of course the application of relativistic QFT describing the entire experiment from the "creation" of the "neutrino" in the initial and the "detection/destruction" of the neutrino in the final state, measured by the detector. Of course for the asymptotic free in and out states total (sic!) energy and momentum are conserved as always in QFT.

The upshot is: to describe a typical long-baseline neutrino oscillation experiment you have to invoke the wave-packet formalism for the asymptotic free states in the in and out channel to set clear spacetime information to establish the "geometry" of the production and detection place to establish the (probabilistic!) prediction about what's expected to be measured via the corresponding S-matrix treatment. A nice review is given in

https://arxiv.org/abs/1008.2077
 
  • #20
hilbert2
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How would it even be possible for an electron to convert to a muon or tau with both energy and momentum being conserved if there's no other particle involved in the process? If you consider it in a coordinate system where the initial electron is at rest, it looks quite impossible.
 
  • #21
How would it even be possible for an electron to convert to a muon or tau with both energy and momentum being conserved if there's no other particle involved in the process? If you consider it in a coordinate system where the initial electron is at rest, it looks quite impossible.
They could travel as mass eigenstates without a specific flavor.
 
  • #22
Vanadium 50
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How would it even be possible for an electron to convert to a muon or tau with both energy and momentum being conserved if there's no other particle involved in the process?
It can't.
 
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  • #23
It can't.
It could if it travelled in the form of a mass eigenstate but without a specific flavor.
 
  • #24
Vanadium 50
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The upshot is: to describe a typical long-baseline neutrino oscillation experiment you have to invoke the wave-packet formalism for the asymptotic free states in the in and out channel to set clear spacetime information to establish the "geometry" of the production and detection place to establish the (probabilistic!) prediction about what's expected to be measured via the corresponding S-matrix treatment
I agree that you can do it this way, and it is perfectly correct. I'm not sure I agree you "have to", and I am not even sure this is the most enlightening even if it is the most rigorous. I think the ordinary QM interference picture I posted captures most of the physics without being misleading (otherwise I wouldn't have posted it).

I think the oscillation picture is just bad. It suggests questions like "what would I see if traveling next to a muon neutrino" and "what is the flavor of this neutrino when I am not measuring it". Both are ill-defined.
 
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  • #25
Vanadium 50
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It could if it travelled in the form of a mass eigenstate but without a specific flavor.
No, it can't. For changed leptons mass eigensttaes are flavor eigenstates. "Mass eigenstate but without a specific flavor" describes something that doesn't exist, by definition. Like a four-sides triangle.
 
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