Charged leptons don't mix. Why?

[fermion]

I have a question about the lepton mass mixing. We currently believe that the three generations of neutrinos mix with each other to produce three mass eigenstates, each of which is a linear (and orthonormal) combination of the weak interaction eigenstates of electron- muon- and tau- neutrinos. This mixing violates flavor symmetry. Such a mass matrix can be generated by Yukawa couplings of Higgs scalars to the leptons, which produce tree level mass terms for the leptons through their vacuum expectation values. The source of the non-diagonal mass matrix does not have to be based on Yukawa couplings, it may be something else altogether. The precise origin of the mass mixing probably does not matter for this question, which asks why the charged leptons do not mix in a similar way? They too couple to Higgs scalars (even though we have not found one yet), and they too should acquire a 3x3 flavor violating non-diagonal mass matrix, which produces a mixture of mass eigenstates. However, we know this never happens because the muon or the Tau (almost) never decay into an electron and a photon.

 

Why not? If we can have flavor violation in the neutral lepton sector, what symmetry prevents it in the charged sector? I have been thinking about the answer for a while, and finally I convinced myself that it cannot be based on an argument that you could redefine the fields such that only one side needs to mix. Such an argument can be successfully made for the quark mixing of Kobayashi-Maskawa type. But there, the mixing matrix connects two left handed flavor quark multiplets (u,c,t)_left and (d,s,b)_left. But the neutrino mass mixing (any fermion mass mixing) is between right and left handed neutrinos. What happens through that mixing does not involve the charged leptons at all. A mirror argument can be repeated for the charged leptons. Before I decided to ask this question, I also discounted non-symmetric solutions such as making right handed neutrinos to be of exotic types, such as Majorana fermions which may interact in unusual ways, while the right handed charged leptons remain as Dirac fermions. That is certainly possible, but utterly contrived.

 

Any elegant symmetry arguments?

[BenTheMan]

Well, on first thought, the tau is significantly more massive than the muon, which is significantly more massive than the electron. This means that they all decay (via weak interractions) to less massive leptons.

 

So my first guess is that the mass hierarchy (1:10^-3:10^-5) prevents the leptons from mixing. My first guesses are typically wrong, so I'll keep thinking

 

Does anyone agree? OR am I copletely wrong?

[fermion]

I don't think the mass hierarchy will forbid the mixing. It may suppress it, but it will not eliminate it. Given the experimental limits on the muon decay to an electron and a photon, this must be a very severe suppression, which needs to be no less than (me/mmu)**4, where me = electron mass, mmu= muon mass. Perhaps that's the answer: charged leptons mix, but very little.

 

I still wonder, however, if there is a symmetry argument which will prevent any amount of mixing.

[BenTheMan]

Actually I think it may, because the mixing goes like [math]e^{-\Delta m^2}[/math], right?

 

Symmetry argument, probably not, unless you look at putting in a discrete family symmetry or something. That's my guess, but I'll keep thinking about it.

[Severian]

It is just a definition. You can move the the PMNS matrix into the charged lepton sector if you really want to. After all, the neutrino interactions you are interested in in your neutrino detectors are always vertices which include the charged lepton. So it is just convention.

 

If you don't believe that, it is quite easy to think of a diagram for e to mu: e- emits a W- turning into a nu_e, nu_e oscillates to nu_mu, and nu_mu reabsorbs the W- to turn into mu-. In other words, your charged lepton 'oscillation' is fully described by the same quantities you use for neutrino oscillations (the PNMS matrix), so we just don't think of it as a separate phenomena (it isn't!) and don't worry about charged lepton oscillations.

[BenTheMan]

Ahh right. It's a question between mass eignestates and interaction eigenstates.

[fermion]

Severian,

 

The issue here is not charged lepton oscillations. If indeed charged leptons oscillate, the length of the oscillation scale would be incredibly short even for very relativistic charged leptons for us to observe that (This is due to the square-mass differences in the exponential.) The issue here is the muon decay into an electron and a photon, which we believe never happens. The experimental limits on the branching ratio for this process is below 10**(-8).

 

This branching ratio is in no danger if the charged leptons mix only through one loop corrections which include a W and and the three mixed neutrino states. This loop is suppressed by (mmu/mW)**2 in amplitude and (mmu/mW)**4 in decay rate. But the charged leptons can (in principle) mix at the tree level, not just through one-loop corrections through and neutrino emissions. The question I asked can be re-phrased then as "Why don't the charged leptons mix at the tree level? Is there a symmetry that prevents it?"

[Severian]

The issue is that the neutrino mass eigenstates and charged lepton mass eigenstates are not aligned. Look at the terms in the lagrangian which contribute to neutrino mixing. If I use the mass eigenbasis, then the neutrino mixing matrix is sitting in the charged current interaction term. It is not in the neutral current because it is unitary and will cancel.

 

Now what would happen if our charged lepton mass eigenbasis was not the same as the charged lepton interaction basis? Exactly the same thing - using the interaction basis, the mixing matrix would move into the charged current interaction - the same place that we moved the neutrino mixing matrix. We choose to call this matrix the neutrino mixing matrix, and define our states such that there is no charged lepton mixing.

 

To put it another way, if there were no charged leptons, there would be no neutrino oscillations, since we could just rotate it away. It is the presence of the charged leptons which prevent us from rotating the PNMS matrix away by redefining the fields, because the chaged lepton mass matrix and the neutrino mass matrix are not aligned.

[vincent]

It is just a definition.

 

I disagree. The underlying reason for the absence of mixing of the charged leptons is that neutrinos are (very nearly) massless. Consider the neutrinos in the charged weak current for leptons in terms of mass eigenstates. Since the different neutrinos are massless, they are degenerate so that any unitary transformed neutrino states can be taken as mass eigenstates. The unitary transformation can thus be taken as the identity matrix. Therefore significant lepton mixing via the charged weak current can never show up in any physical process.

[fermion]

Vincent: I do not understand your argument. (This is lkely to be my fault, but all the same, you lost me.)

 

Severian: You are making the same argument one makes with the Kobayashi-Maskawa mixing in the quark sector: it does not matter whether you rotate the (u,c,t) quarks or (d,s,b) quarks by the inverse matrix. The two sets are misaligned with each other, and it is relative. In my original posting I touched this point of field re-definition, but I will elaborate on it a bit better below.

 

I will ask the question again below, assuming your answer is the correct one:

To put it another way, if there were no charged leptons, there would be no neutrino oscillations, since we could just rotate it away. It is the presence of the charged leptons which prevent us from rotating the PNMS matrix away by redefining the fields, because the chaged lepton mass matrix and the neutrino mass matrix are not aligned.

 

Fine, then let's say that I choose to rotate the charged leptons instead of the neutrinos. The two should be equivalent right, just like the KM rotation? OK, say, we start from a pion, which decays into a charged lepton and a neutrino. Now the neutrino is a pure (redefined) lepton state but the charged lepton is a superposition of electron, muon, and tau. This superposition will immediately decay electromagnetically to the lowest mass eigenstate (with electron's mass, but itself a mixture). We don't have the same pion decay problem if we rotate the neutrinos. The heaviest neutrino may decay to a lighter one by some means, but it won't be observable (yet.)

 

Therefore either the two descriptions are not equivalent and you cannot simply choose to rotate whichever group you choose, or else you could still do that, but the electromagnetic amplitude for the mixed charged lepton state decaying into the lowest mass eigenstate is zero. Assuming the latter, then again I ask why? What makes that amplitude to be zero?

[Severian]

Think of it this way. There are two different ways of looking at things. So let's consider them one at a time.

 

 

1. Let's write the fields in the interaction term as the charged lepton mass eigenstates and let the neutrinos take all the PNMS matrix. This is the standard way of thinking, and since it is mass eigenstates which propagate, the charged leptons produced by the interaction will not oscillate but the neutrinos will.

 

Now, lets let a W decay via [math]W^- \to e^- \bar \nu_e[/math]. The e^- is a mass eigenstate so doesn't change but the neutrino oscillates, say to [math]\bar \nu_{\tau}[/math] before we detect it. So effectively we have seen a [math]W^-[/math] decay to [math]e^- \bar \nu_\tau[/math], and interpret that as a neutrino oscillation.

 

That was all standard.

 

2. Now let's imaging writing the fields in the interaction as being the neutrino mass eigenstate and but the PNMS matrix into the definition of the charged leptons. This is not the way it is usually thought of: now the neutrinos don't oscillate but the charged leptons do.

 

This time, let the W decay via [math]W^- \to \tau^- \bar \nu_\tau[/math]. The [math]\bar \nu_\tau[/math] is a mass eigenstate so doesn't change but the charged lepton oscillates, say to [math]e^-[/math] before we detect it. So effectively we have seen a [math]W^-[/math] decay to [math]e^- \bar \nu_\tau[/math], exactly as in case 1!

 

How do we interpret this? We could interpret it as a changed lepton oscillation, but that is not the usual way of looking at things. We normally interpret it as a neutrino oscillation. In other words, the language that we have conventionally use to write the lagrangian in makes us interpret this mixing only as neutrino oscillations and never charged lepton oscillations.

[fermion]

Severian,

You chose not to answer the question I posed, but you answered another one: what happens when W decays leptonically. That's fine, but W decay is not a very severe test of the equivalency of rotating the leptons. Pion decay, and the subsequent decay of the muon resulting from the pion decay is a much better test. I gave this example, but you ignored it. Perhaps you will like it better if I describe it both ways, which is a good idea:

 

1) The standard description: we rotate the neutrinos. In this scheme, a pion decays into a muon and a neutrino. The muon is a mass eigenstate, and it is pure. The neutrino is in the muon neutrino state, which is a mixture of three mass eigenstates. Later this neutrino may ocscillate, but we don't care, because we are not looking at the neutrino. We are looking at what happens to the charged lepton, the pure muon in this case. The muon cannot decay into an electron and a photon, and it sits there forever, until it can decay weakly eons later. In this wiew the rate of muon decay to an electron and a photon is zero.

 

2) Complimentary(?) view: we rotate the charged leptons. In this view, a pion decays into a muon neutrino and a charged lepton. The muon neutrino is a mass eigenstate, it does not mix. But the charged lepton is a mixture of three mass eigenstates, predominantly the second heaviest one, but it also contains some amplitude of the lightest and the heaviest charged leptons as well. Similarly, the lightest charged lepton which is a mass eigenstate (predominantly electron) has some (small but non-negligible) muon component. Therefore the matrix element that connects this lightest charged lepton and the initial muon is non zero. Of course, you have to emit a photon to conserve the four momentum and to put the lightest final lepton on its mass shell. In other words, the muon will decay into a photon and into the lightest charged mass eigenstate lepton. You may wonder why I started from a pion here, when the real question is the electromagnetic muon decay. The pion is relevant here only to the extent that it produces (by its decay) a charged lepton state that is a pure flavor state, but that is a mixture of mass eigenstates. If instead you started from the second heaviest charged lepton mass eigenstate, then it won't decay by photon emission. That's where the pion comes in: in this second view, the pion decay does not produce the second heaviest charged lepton mass eigenstate, it produces the pure muon eigenstate that is a mixture of the mass eigenstates. To get a non-zero matrix element, it is essential that both the initial and final lepton not be in the same type of state (such as both mass eigenstates or such as both flavor eigenstates.) Pion decay provides this beautifully!

 

It appears that the two views predict two differents outcomes for what happens to the charged lepton produced in the pion decay. How could that be? Either the two views must not be equivalent, or else, the computation of the amplitude for muon decay into an electron and a photon is faulty or incomplete (such as missing a term that cancels it identically.) Having failed to find fault on the computation of this amplitude, I concluded that the two views must not be equivalent. (Please see my original post for that statement.) But two views can be equivalent if you re-calculate the amplitude correctly (presuming that I did not) and show that it is indeed identically zero. Can you do that? If you choose to answer, please answer this before making other statements. I'll be very happy to see that I computed it incorrectly and the amplitude indeed vanishes.

[BenTheMan]

Now I have fully convinced myself that I was wrong.

 

If it was truly the case that the mixing was supressed by the large mass differences, then the quarks wouldn't mix. But they do. So I was wrong.

[Severian]

Sorry, I missed fermion's reply to this.

 

Severian,

You chose not to answer the question I posed, but you answered another one: what happens when W decays leptonically.

 

Actually I did answer your question, and everything you should need to satisfy you is in my post. Semi-leptonic pion decay is the decay of a W. How else are you going to make the leptons? The charged pion, for example, is [math]u \bar d[/math] which decays into a [math]W^+[/math] and subsequantly to mainly [math]\mu^+ \nu_\mu[/math] (since the initial state is spin-0, we need a heavish object in the final stat to give a spin flip, so the electron final state is suppressed, and the tau is too heavy).

 

So everything I posted earlier was relevant to charged pion decays.

 

But I am happy to go through it in your scenarios.

 

1) The standard description: we rotate the neutrinos. In this scheme, a pion decays into a muon and a neutrino. The muon is a mass eigenstate, and it is pure. The neutrino is in the muon neutrino state, which is a mixture of three mass eigenstates. Later this neutrino may ocscillate, but we don't care, because we are not looking at the neutrino. We are looking at what happens to the charged lepton, the pure muon in this case. The muon cannot decay into an electron and a photon, and it sits there forever, until it can decay weakly eons later. In this wiew the rate of muon decay to an electron and a photon is zero.

That is fair enough, except that I say that we do care about the neutrino, since it is the mismatch between charge lepton and neutrino flavours which is important in any leptonic oscillation.

 

2) Complimentary(?) view: we rotate the charged leptons. In this view, a pion decays into a muon neutrino and a charged lepton. The muon neutrino is a mass eigenstate, it does not mix. But the charged lepton is a mixture of three mass eigenstates, predominantly the second heaviest one, but it also contains some amplitude of the lightest and the heaviest charged leptons as well. Similarly, the lightest charged lepton which is a mass eigenstate (predominantly electron) has some (small but non-negligible) muon component. Therefore the matrix element that connects this lightest charged lepton and the initial muon is non zero.

Up to here I am with you.

 

Of course, you have to emit a photon to conserve the four momentum and to put the lightest final lepton on its mass shell. In other words, the muon will decay into a photon and into the lightest charged mass eigenstate lepton.

I disagree here slightly. The photon is not required. I can conserve energy and momentum perfectly well by keeping intermediate particles off-shell. The rate for [math]\pi^+ \to l_1^+ \nu_\mu[/math] (where [math]l_1[/math] is the lightest charged lepton mass eigenstate) is non-zero even without an extra photon (though it is very suppressed because of the small mass of [math]l_1[/math]). But this isn't very relevant to the discussion.

 

You may wonder why I started from a pion here, when the real question is the electromagnetic muon decay. [snip] That's where the pion comes in: in this second view, the pion decay does not produce the second heaviest charged lepton mass eigenstate, it produces the pure muon eigenstate that is a mixture of the mass eigenstates.

This was also true for W decays. In fact, it is true for pions, only because of the W in between...

 

It appears that the two views predict two differents outcomes for what happens to the charged lepton produced in the pion decay. How could that be?

The problem is that you are not comparing like with like. Both of the above are completely fine, but they are different effects that each can happen.

 

Let us assume you measure the final state particle flavours, rather than their masses (we could do their masses later if you like).

 

For example 1, there are 3 possible final states (since the neutrino oscillates): [math]\mu^+ \nu_\mu[/math] or [math]\mu^+ \nu_e[/math] or [math]\mu^+ \nu_\tau[/math], but for any individual event you only see one of these.

 

For example 2, there are also 3 possible final states (since the charged lepton oscillates): [math]\mu^+ \nu_\mu[/math] or [math]e^+ \nu_\mu[/math] or [math]\tau^+ \nu_\mu[/math], but again for any individual event you only see one of these.

 

So the final states in your examples only match for [math]\mu^+ \nu_\mu[/math] so that is the only one you can compare, and since the oscillation in each involves the same entry in the PNMS matrix they will have identical cross-sections and you have your equivalence.

 

In other words, when thinking in philosophy 2, you see e.g. [math]e^+ \nu_\mu[/math], how are you going to persuade your philosophy 1 minded friend that this wasn't just the production of [math]e^+ \nu_e[/math] followed by the oscillation [math]\nu_e \to \nu_\mu[/math]? (Notice both processes have the same rate.)