# Mass eigenstates of the leptons

1. Nov 22, 2013

### BruceW

Hi everyone,

my question seems pretty simple, but I couldn't find any answers. OK, so the neutrino flavour eigenstates are different to the neutrino mass eigenstates. And this is why neutrino oscillation is possible. But for the charged leptons (the electron, muon and tauon) the mass eigenstates are the same as the flavour eigenstates. I understand that this is a good thing, because experimentally we don't observe oscillations between the electron muon and tauon. But I was hoping for some theoretical reason for why the mass eigenstates must be the same as the flavour eigenstates for the charged leptons. Otherwise, it just seems like a weird coincidence that the mixing matrix for the charged leptons is diagonal, while the neutrino mixing matrix is not diagonal.

I was thinking maybe it has something to do with the fact that the electron, muon and tauon all have charge (and maybe this somehow means their mixing matrix must be diagonal). But I can't think of why this would be true. Or maybe the mixing matrix for the charged leptons is not diagonal, but because the masses are so great, the effect of oscillations of the charged leptons is negligible anyway. (This is compared to the neutrinos, which have much smaller masses, and we can observe their oscillations between different flavours).

Anyway, thanks for reading, hopefully someone out there knows the answer :)

2. Nov 22, 2013

### Bill_K

What you're asking is basically, can a muon decay into an electron. And the answer is yes, of course, it always does that. So the question is really can it decay into an electron without at the same time emitting those neutrinos.

And the answer to that is, it can't in the Standard Model but it could in principle, and so people have looked for it, but they haven't found it. According to Wikipedia, the current limit on neutrinoless muon decay is around 10-13.

3. Nov 22, 2013

### BruceW

no, it's not decay I'm thinking about. I mean oscillation, just like neutrino oscillation. For example, if we send a beam of only electrons all with the same momentum (for simplicity), then the initial state will just be the electron eigenstate, which can be written as a superposition of the mass eigenstates:
$$|e\rangle = a |1\rangle +b |2\rangle +c |3\rangle$$
where 'e' is the electron eigenstate, 1,2,3 represent the 3 mass eigenstates and a,b,c are complex coefficients. So then, at some time and position further down the beam, the probability to find an electron would be given by the mod square of the following expression:
$$a \exp(-i xp_1) + b \exp(-i xp_2) + c \exp(-i xp_3)$$
where $x$ is the position 4-vector and $p_n$ is the 4-momentum associated with the n'th mass eigenstate. So, from this, the probability to find an electron further down the beam is not equal to 1. There is some probability of finding muons and taons instead. Also, if you keep moving down the beam, and wait for the right time, then you find that again there is probability 1 that you detect electron, and there is zero probability to detect muon or taon.

Of course, no-one ever talks about electron oscillation. Only neutrino oscillation. Which is equivalent to saying that (for example) a=1, b=0, c=0, so that the electron is in fact a mass eigenstate. But I was wondering why the electron is a mass eigenstate but the neutrinos are not... and I couldn't find any definite answer.

edit: of course, I'm assuming the beam is moving through free space, which is why the mass eigenstates evolve in time and space with the nice and simple phase factor $\exp(-ixp)$. Also, I forgot to say, this means taking the inner product of $xp$, so I should really say $x_\mu p^\mu$ if I was writing it all out properly.

Last edited: Nov 22, 2013
4. Nov 22, 2013

### Bill_K

They would both involve the same off-diagonal matrix element connecting muon to electron.

What makes it "oscillation" in the case of neutrinos is the incredibly small mass differences involved, and so when neutrino A decays into neutrino B, it's remains coherent and can turn back into neutrino A. This would not be the case with muon and electron.

5. Nov 22, 2013

### Meir Achuz

Neutrino (and K_1 and K_2) oscillations are seen because the mass differences are small.
In principle, the e and the mu could oscillate to a high order in the weak interaction, but this is not seen because their mass difference is large. The oscillation angle goes like the e=mu matrix element over the mass difference.

6. Nov 22, 2013

### BruceW

oh I see. that's pretty cool. so the electron could exhibit oscillations, but it would require something like a really large beam, and maybe used in space so there was enough room for the electron to oscillate and without the beam becoming totally decoherent.

So I guess I mean, in principle the electron muon and tauon do oscillate in the same way as the neutrinos, but because of the (Edit: I said small here, but I meant large) mass difference, it would be very difficult to make an experiment to test this.

And you say the mixing matrix for charged leptons is the same as the mixing matrix for neutrinos? Is there a reason for this? (sorry for the extra question I've snuck in here).

7. Nov 22, 2013

### Bill_K

No, I meant to say that muon-electron decay and oscillation would both involve the same element.

8. Nov 22, 2013

### BruceW

ah, I think I get what you're saying. The neutrinoless (and photonless) decay of a muon to an electron is essentially the same thing as oscillation. Is that what you mean? Also, I just thought of something that doesn't make sense. the mass of an electron is measured to a fairly high accuracy, right? good enough accuracy that they know it is not the same as the mass of a muon or a tauon. So, it doesn't make sense to me that the electron could be a linear superposition of all 3 mass eigenstates... Unless it is almost completely in one mass eigenstate, so only once in a blue moon would you measure an electron to have one of the other masses. That seems odd though.

9. Nov 22, 2013

Staff Emeritus
This is being made more complicated than it has to be.

We label the three mass eigenstates e, mu and tau. That is, we define flavor to map onto the mass eigenstates.

10. Nov 22, 2013

### BruceW

that does make sense to me. So, instead we could have used the three neutrino mass eigenstates as definition of the three flavour eigenstates. And then the e, mu and tau would no longer be mass eigenstates. But in any case, what we commonly see is the charged lepton mass eigenstates (because they have large mass it is difficult to get them to superpose). Therefore, the thing orbiting the nucleus would still be a mass eigenstate with mass 0.51MeV/c2 but it would be a superposition of flavour eigenstates.

And similarly, normally we see interactions where one of the charged lepton mass eigenstates is emitted from the atom (like in beta decay). So if we define the neutrino mass eigenstates to be the flavour eigenstates, then the neutrino that is emitted in beta decay would be a superposition of flavour eigenstates (as well as the charged lepton that is emitted). Therefore the emitted neutrino would still do neutrino oscillations, same as it would under the old definitions.

So yes, I think I am pretty well convinced that the charged lepton eigenstates are equal to the mass eigenstates simply because we define them to be that way. And the only reason we need to think about flavour is because of weak isospin which couples each of the charged leptons to one of the neutrinos. And it is not surprising that this coupling does not couple mass eigenstate to mass eigenstate.

11. Sep 4, 2014

### arivero

Is this usual terminology, by the way?

It seems clear to speak of the mass eigenstates, on one side, and of the weak eigenstates, on the other, the later ones being the chiral eigenstastes. I thought that the flavour eigenstates were a rotation of weak eigenstates in generation space, not changing the chirality, are they?

In this point of view, for a dirac particle, mass eigenstates are a equal-weight combination of |R> and |L> weak eigenstates.

12. Sep 7, 2014

### BruceW

uh... I really don't know much about this area, so I don't know for sure. But I just found the book "fundamentals of neutrino physics and astrophysics" by Giunti and Kim, and they say on page 187:

"An interesting question is why the mixing is always applied, as we have done, to the neutrinos, whereas the charged leptons are treated as particles with definite mass. The reason is that the only characteristic that distinguishes the three charged leptons is their mass and the flavor of a charged lepton is identified by measuring its mass. The mass determines its kinematical properties and its decay modes, which can be measured directly through long-range electromagnetic interactions. Hence, charged leptons with a definite flavor are, by definition, particles with definite mass. On the other hand, neutrinos can be detected only indirectly by identifying the charged particles produced in weak interactions and the flavor of a neutrino created or destroyed in a a charged-current weak interaction process is, by definition, the flavor of the associated charged lepton. Therefore, flavor neutrinos are not required to have a definite mass and the mixing implies that they are superpositions of neutrinos with definite masses."

So I think this is at least the practical idea of why we have the mixing matrix defined in that way. Anyway, sorry for going off on a tangent there. Specifically about what you were saying, I am not 100% sure what you mean. The chirality of a free massive Dirac particle will oscillate with time, if I understand correctly. So are you saying that the electron does undergo flavour oscillations? Also, are you sure that the weak eigenstates are the same thing as the chirality eigenstates? I don't know about rotations through generation space, unfortunately.