# Other oscillations

1. Nov 12, 2007

### Jim Kata

Is it possible that the other leptons like the electron, muon, and tauon could oscillate between one another, and maybe even the quarks oscillate between their generations? Sorry, I know this kind of a stoner question.

2. Nov 12, 2007

### blechman

quarks **do** mix with each other already in the SM through what is called the Cabibbo-Kobayashi-Maskawa (CKM) matrix. The charged leptons will not "mix" with each other, at least at tree-level (only the neutrinos do) - this is a mathematical fact from the way the fields are grouped together in the standard model. However, if the neutrinos mix, it *is* possible that leptons can decay to other leptons (for example, $\mu\rightarrow e\gamma$). It is also possible that you can generate new 4-fermion operators that violate individual lepton numbers; for example $(\mu^+ e^-)\rightarrow (\mu^- e^+)$. This is the lepton-version of what happens in $K\bar{K}$ mixing in the SM.

3. Nov 12, 2007

### CarlB

We say a "muon neutrino" is produced when a muon decays, and a "tau neutrino" is produced when a tau decays. An "electron neutrino" is produced when a (virtual) electron does a similar process, and the muon or tau could also be virtual.

But these types of neutrinos, i.e. the "muon neutrino", are not "particles" in the sense that they do not have exact masses. Technically, it is said that they are not "mass eigenfunctions".

The three neutrinos that are particles in the sense of having explicit masses are usually called the $$\nu_1, \nu_2, \nu_3$$ where the subscript gives the generation 1, 2, and 3. These have three masses. The physical problem is better described if it is put into the language of the neutrinos with mass.

A muon can decay into any of the three types of neutrinos, $$\nu_1, \nu_2, \nu_3$$. The neutrinos are so light that we cannot distinguish these three processes by energy considerations when we detect the reaction by absorbing the neutrino. Instead, we bundle all three processes into a single process and call it a "muon neutrino".

So in neutrino oscillation, what is going on physically is that the three types of neutrinos are interfering with each other. That is, you can take the physical problem of electron neutrinos, muon neutrinos, and tau neutrinos "oscillating into each other", and translate it into "$$\nu_1, \nu_2, \nu_3$$ interfering with each other".

Getting back to your original question, the electron, muon and tau cannot oscillate into each other because they are mass eigenstates. It is mass that determines the wavelength of a wave function (de Broglie's relation), and it is a difference in wavelength that causes oscillation. That is, to get oscillation, you have to have a quantum state that is composed of different mass eigenstates.

4. Nov 12, 2007

### malawi_glenn

CarlB: You are saying that it is a difference between mass- and flavor eigenstate?

blechman: You are saying that flavour- and colour eigenstates are mixed in the CKM- matrix, i.e. The flavour eigenstates (weak interaction) is a superposition of the coluor eigenstates and vice versa.

This is anyway what i have been taught in Elementary particle courses, and want to clarify this to the OP.

5. Nov 12, 2007

### CarlB

malawi_glenn;

Yes, it's better to talk about flavor eigenstates and mass eigenstates.

Flavor eigenstates oscillate but do not interfere, mass eigenstates interfere but do not oscillate. These are equivalent ways of describing the same problem. Superposition moves you from one to the other. The matrix that moves you from one to the other for the leptons is the "MNS" or "PMNS" matrix. My favorite theoretical guess for this matrix (which is measured through a long series of rather complicated experiments) is the "tribimaximal" matrix. It is well described on wikipedia:
http://en.wikipedia.org/wiki/Tribimaximal_mixing

The equivalent matrix for quarks is the CKM matrix. The CKM matrix is dominated by the Cabibbo angle, which Hans de Vrues has been describing lately in relation to the charged lepton masses (and I have promised to relate to the baryon resonance masses in 2008) here:

To get electrons, muons, and taus to interfere, you have to have an interaction where the nature of your measurement process is such that you can ignore the mass differences. A high energy virtual process will do that. With neutrinos, any experiment we can reasonably produce will have much higher energies than the neutrino masses so "oscillation" is a term that makes sense to use. However, I do think it is unnecessarily confusing to students and should not be taught that way. Instead, they should be taught that it is an interference between massive eigenstate propagators that can be described as an oscillation of flavor eigenstate propagators.

Last edited: Nov 12, 2007
6. Nov 12, 2007

### blechman

I hate to throw a wrench in the works, but there is a small flaw in your argument: why do the charged leptons have to be mass and flavor eigenstates and the neutrinos do not? Why couldn't it have been the other way around? The answer to that question is that it could be that way, and the only reason it isn't is because of the phase conventions of particle physics! That is, we choose the charged leptons to be both mass and flavor eigenstates, but then we no longer have the freedom to make the same alignment of bases in the neutrino sector. Exactly the same thing happens in the CKM, where we choose the Q=+2/3 quarks to be mass and flavor eigenstates, and then the Q=-1/3 quarks cannot be both. This is the subtle "mathematical fact" that I was alluding to in my previous post. It didn't have to be that way, but that's the convention set by our predecessors in the field, so that's what we stick with.

"color" eigenstate isn't the right word: "mass" eigenstate is better. Both sets of eigenstates are eigenstates of the color operator.

Continuing with CarlB's explanation, in the language of Feynman Diagrams:
MASS eigenstates PROPOGATE
FLAVOR eigenstates INTERACT

So when you write down a quark or neutrino in a Feynman diagram, the propogator is that of a mass eigenstate, but the vertex is then the interaction of a flavor eigenstate, which is why you need to include a CKM/MNS matrix element in the vertex, while the propogators still go like $i/(\slash\!\!\!p-m)$ where the m is unambiguous.

7. Nov 12, 2007

### CarlB

I don't see any flaw; but what you've said is otherwise true. I guess I could go on some more.

The mass eigenstates are not defined by phase conventions, they are defined by relativity, more or less. The flavor eigenstates are chosen by convention. The convention is that the heavier states (i.e. the electron, muon, and tau) are chosen to be both flavor and mass eigenstates. Once you've made that choice, you end up with a neutrino mixing angle matrix rather than a charged lepton mixing angle matrix. That matrix gives how to convert the lighter particles into the two bases.

The same convention is applied in the quarks. That is, the generally heavier up quarks, (charge +2/3), the u, c, t are kept as both mass and flavor eigenstates. The generally lighter down quarks (charge -1/3) d, c, and b are treated as mass eigenstates only. The CKM matrix showing how to convert between the flavor and mass eigenstates of the lighter quarks:
http://en.wikipedia.org/wiki/CKM_matrix

8. Nov 13, 2007

### smallphi

According to QM, a measurement of the energy must result in collapse of the system onto a Hamiltonian eigenstate. That's why measuring an atom results in transitions between energy eigenstates.

Now, how come weak interactions produce states that are not eigenstates of the Standard Model Hamiltonian?

Is that analogous to Rabi oscillations in quantum optics? How come in all other optical measurements, atoms make transition between energy eigenstates of the atomic hamiltonian (which doesn't include the radiation) only in Rabi oscillations atoms oscillate between states that are not eigenstates of the atomic hamiltonian? Hm?

Last edited: Nov 13, 2007
9. Nov 13, 2007

### CarlB

smallphi, excellent question. You are paying attention.

In QM, one must include all possible Feynman diagrams that can contribute to the interaction. For example, suppose you happened to have an experiment where a decaying neutron happened to be set up so that due to energy concerns, only the lowest mass neutrino could be emitted. Then, instead of having an anti-electron neutrino emitted in the decay, you could write the decay as that of an anti-$$\nu_1$$ and there could be no oscillation.

Another way of describing this situation is that in the normal case when a neutron decays, in modeling the process we do not put the neutrino onto its mass shell. If you wrote down the Feynman diagram for a neutron decay and assumed the neutrino was on its mass shell, you would get to choose a single neutrino generation and there would be no oscillation. But experimentally, we have no way of detecting neutrinos of a specific generation, so we instead have to sum over three generations, hence oscillation.

We can't force neutrinos onto their mass shell because they are so light compared to the other energies of the experiment. In other words, the reason that neutrinos do not obey the concept of collapsing onto an energy eigenstate is that our experiments are too coarse. If you managed to build an experiment that measured the neutron and proton, electron and anti-neutrino 4-momenta exactly, then you would find that energy and momentum was conserved exactly. You could then solve for the anti-neutrino mass and you could figure out which neutrino type was involved.

I guess I should point out that in neutrino oscillation, no mass eigenstate neutrinos are created or destroyed in the vacuum where the neutrinos are propagating. That is, the flux of neutrinos falls off with the usual 1/r^2 law if they are emitted spherically. What changes is the ratio of electron to muon to tau neutrinos. The ratios of $$\nu_1$$ to $$\nu_2$$ to $$\nu_3$$ do not change with distance. The massive neutrinos do not oscillate.

The difference between this and Rabi oscillations, I think, is that Rabi oscillations require a driving field oscillation. In that case, quantum states are not eigenstates of energy because the Hamiltonian is not constant. Another way of saying this is that energy is the result of the symmetry of a quantum system over time. When a driving field oscillates, time symmetry is reduced.

Last edited: Nov 13, 2007
10. Nov 13, 2007

### smallphi

So you are saying that not measuring the total energy exactly allows for production of linear combinations of energy eigenstates.

Sounds fine only inquiring minds would like to know then why it is not possible to produce oscillating electrons-muons-taons then by not measuring the energy exactly?

11. Nov 13, 2007

### CarlB

smallphi, electrons muons and taus are mass eigenstates so they don't oscillate. If a muon is created, you won't later find an electron (unless it decays), and you certainly don't find electrons turning into muons.

They do, howevrer, interfere, and you can make experiments where they will interfere. The one that comes to mind would be a decay of a tau into an electron or a muon, which is then absorbed in, say, a reverse beta decay.

For computing the amplitude for the whole process, you will have to sum over the electron and muon propagators, and then take absolute values and square. They will interfere, not oscillate.

12. Nov 13, 2007

### smallphi

If the experiment doesn't measure the energy precisely, according to what you said previously, a state that is a mixture of eigenstates (electron, muon, taon) will be produced. The eigenstates (electron,muon, taon) have different masses so they can interfere exacly like the neutrino energy eigenstates.

The only difference is that we always detect weak-interaction flavored neutrinos that are linear combinations of eigenstates.

For electrons-muons-taons, we always detect the eigenstates not combinations of them.

The question is why ? What in the detection aparatus chooses to detect eigenstates in one case and mixture of them in the other? Isn't the standard model hamiltonian always the same, then why the state collapses on eigenstate in one case and on mixture in the other?

Last edited: Nov 13, 2007
13. Nov 13, 2007

### CarlB

Again no. A state that is a mixture of mass eigenstates is always produced by a flavor changing interaction. What allows an oscillation is the fact that we (a) cannot distinguish between the neutrino mass eigenstates based on energy / mass measurement considerations, AND (b) we talk about neutrinos in their FLAVOR basis.

To get oscillation, you need BOTH these requirements. The first to force us to include all mass Feynman diagrams, the second so that we call what we observe "oscillation" instead of "interference". Oscillation requires that the quantum states be defined as flavor eigenstates instead of mass eigenstates. The electron, muon, and tau are mass eigenstates and so cannot oscillate. Instead, we call what they do "interference", and yes, they do interfere.

14. Nov 13, 2007

### Jim Kata

Ok, I think I got you Carl. Let me paraphrase what you said, and see if I'm correct. Now you can't make both neutrinos and the other leptons, electrons, muons, and taons, eigenvectors of the same mass operator. Maybe they don't commute? But, you can choose neutrinos to be eigenvectors of some flavour operator. Now if your talking about eigenvectors of the flavour operator you say oscillation, and if you are talking about eigenvectors of the mass operator you say interference. This seems like a bit of semantics. Do you use the eigenvectors of the mass matrix, electron, muon, and taon states to write the neutrino's mass states as superposition of them, but since it isn't a definite eigenvector state you say oscillate? Talking about something I even know less about, is it possible that that the electron, muon, taon do oscillate on some principal bundle, but when the symmetry is spontaneously broken to a smaller subgroup you can now a distinguish between the states that are eigenvectors of the mass matrix and the ones that are not?

15. Nov 13, 2007

### smallphi

Can this be DERIVED from the SM lagrangian or the lagragian corrected for neutrino oscillations without relying on any heuristic assumptions about 'the sensitivity of the experimental setup'?

16. Nov 14, 2007

### CarlB

Jim, let me define "oscillate" and "interfere", maybe that will help.

Oscillate means that a quantum object changes a quantum characteristic with time. Interfere means that two different quantum processes can contribute to the same measurement, and therefore when we calculate the probabilities the two processes interfere.

To get oscillation, you need to begin with just one quantum state.

To get interference, you need to begin with two quantum states.

There are three degrees of freedom in neutrino states (and three more for the anti neutrinos, but leave that off for now). We can describe those three degrees of freedom two different ways. This is not semantics, it is mathematics.

The natural and less confusing way to describe them is to use the mass eigenstates to separate the particles. Then the neutrinos are called "1", "2", and "3". If you do this, then a process that produces a "neutrino" in the sun will require modeling three neutrinos to model it. The overall interaction will be a sum of contributions from the 1, the 2, and the 3 neutrino. Because of the sum, these interactions can interfere. Because the three neutrinos have 3 different masses, they will have different wave lengths, and therefore there will be interference between the 3 neutrinos.

When neutrinos were first discovered, it was believed that there was only one. When it was discovered that there was more than one type, it was assumed that they were all massless. If neutrinos really were massless, it would make no sense to call them "1", "2" and "3" according to their mass eigenstates, so instead they called them "electron", "muon", and "tau" neutrinos according to the charged lepton that decayed to produce them.

This led to the less natural and more confusing, but historical way to describe the three degrees of freedom. This is called "flavor eigenstates", but what is really meant is flavor relative to the charged leptons, that is, relative to the electron, muon, and tau. With that method, the decay of a muon in the sun proceeds not through three neutrinos with three different masses, but instead with a single neutrino, the "muon anti neutrino" whose mass is not sharply defined.

The problem with the less natural method is that it has to give the same results as the more natural, mass method. But the three massive neutrinos interfere with each other. To match the results, you have to invent something called "neutrino oscillation".

It might help to look at what happens when the "muon anti neutrino" gets absorbed by a lepton on earth.

The earth has a lot of charged leptons, almost all of which are electrons. Accordingly, the easiest lepton to see absorbing that muon anti neutrino would be an electron. And if an electron absorbed a muon anti neutrino, we would expect it to turn into a muon.

Just exactly this activity is observed on the earth. However, the rate at which it occurs is about 1/3 that expected from an analysis of the number of muons expected to decay in the sun. (Forgive me if I screw the details on this up, I study elementary particles, not astrophysics.) The deficiency was discoved quite some time ago at the Homestake mine in South Dakota. The modern explanation is "neutrino oscillation".

So it's not so much semantics as mathematics. The early names for the neutrinos stuck and you now have to deal with it. If you instead called them the 1, 2, and 3 neutrino, your life would be easier and all you would have to deal with is interference.

The best write-up I've seen on neutrino oscillations on this subject (at the beginning grad student level) is that of Smirnov, of the MSW effect, but I can't find it on the web. I'm the "C A Brannen" mentioned in the following review of neutrino masses and quark / lepton complementarity:
http://arxiv.org/abs/hep-ph/0603118

17. Nov 14, 2007

### smallphi

Oscillation or interferece, what is weird is that electrons are always produced as energy eigenstates and neutrinos as mixture of such states. My question was if that follows from the SM lagrangian corrected to mix neutrinos.

More speicifically, does the SM lagrangian (corrected or not) imply that the matrix element for the beta decay

n-> p + e(-) + nu(e, bar)

is zero if the neutrino is in an exact energy eigenstate and not a mixture?

18. Nov 14, 2007

### CarlB

The weirdness is due only to the historical way that the existence of neutrinos were exposed to physicists. If we'd known all along that there was three of them and that they had mass, they would be treated identically and both would be treated as eigenstates of mass.

If the SM is not corrected for neutrino masses, then the neutrino mass eigenstates are degenerate (all the same mass = 0), and there is no reason to treat the mass eigenstates differently from the flavor eigenstates. So in that case the above matrix element is non zero.

If the SM is corrected, then the above matrix element is treated identically. The correction applies when the neutrino is detected, and consists of either (1) applying neutrino oscillation, or (2) breaking the nu(e,bar) into three states, nu(1,bar), nu(2,bar), and nu(3,bar), which are three mass eigenstates with three different masses. For any reasonable experiment detecting neutrino decay, these will interfere. The calculation will give the same result as neutrino oscillation to the flavor states.

19. Nov 15, 2007

### smallphi

So the corrected SM lagrangian doesn't impose in any way that neutrino should be detected as mixture and electrons as eigenstates.

It would be interesting to do an experiment where a mixture of electron-muon-taon states is produced and study that state. It won't be an 'oscillation' because our devices don't read off mixed states like in the case of neutrino but it will prove that electron-muon-taons are not special compared to neutrinos.

20. Nov 15, 2007

### CarlB

yeah, the Lagrangian itself can be arranged to treat electrons and neutrinos very equally. That we model an "electron" as a mass eigenstate, but model a "electron neutrino" as a superposition of three mass eigenstates (which we call a flavor eigenstate) is only an accident of how we learned about neutrinos. It makes sense only because neutrinos are so so light compared to the charged leptons.

By the way, "taon" is not a very common spelling for the heaviest charged lepton. The standard spelling is tau. The second most common spelling is tauon, but in the industry, everybody I know calls it the "tau", which rhymes with "cow". I realize that this is contrary to how the mu becomes a muon, but that's the way it is, another historical oddity of some sort.

Here's the google hit counts on arXiv:

taon (16):