touqra said:
Why do only neutrino oscillates (showing massiveness of neutrino)? Why don't electron, muon and tau oscillates?
This is a wonderful question and shows that you are paying close attention to what you read, and not believing everything.
A neutron decays into three things, a proton, an electron, and an anti "neutrino". The usual description of this neutrino is an "electron anti neutrino" because it is emitted along with an electron. But this language is confusing to the student. The "electron neutrino" is a "flavor eigenstate" of the three neutrinos that are the "mass eigenstates".
A "mass eigenstate" means a particle state that has a definite mass. The electron, muon, and tau are mass eigenstates. The electron neutrino, the muon neutrino, and the tau neutrino are not mass eigenstates. The mass eigenstates of the neutrino are usually called \nu_1, \nu_2, \nu_3, and this is a better way of describing neutrinos.
A "flavor eigenstate" means a particle state that is an eigenstate of the weak force. The weak force mixes mass eigenstates of the three generations. As jtbell said, the way we talk about these is a little arbitrary. The way we treat the flavor eigenstates is to have the electron, muon and tau be flavor eigenstates, along with the mixtures of the three mass eigenstate neutrinos.
If we wanted to, we could modify the flavor eigenstate description so that \nu_1, \nu_2, \nu_3 were, call them "navor" eigenstates, and the electron, muon, and tau were not. We don't do it this way for two reasons.
The first is that historically, the neutrinos were not known to have mass. Particles that don't have mass don't need mass eigenstates, so there was no reason to define, for example, the electron neutrino as "the lightest mass eigenstate neutrino".
The second is that practically, we cannot create experiments where only one generation of neutrino is involved. Because of the small mass of the neutrino, and because the weak force mixes mass eigenstates, any experiment that has a 1st mass generation neutrino, that is, a \nu_1, will also have a contribution from a 2nd and 3rd generation neutrino.
To get a probability out of a quantum mechanics, one first writes down all the possible things that can happen. One then assigns a complex number to each of those things. Then you add up all the complex numbers. Then you take the resulting sum, a complex number, and compute its absolute value squared, a real number. This real number is your probability.
When you write down "all the possible things that can happen", you have to include all the things that are within the range of energies that your experiment can detect. This is what mjsd is talking about. For example, if you are detecting the decay of a neutron, you don't have to consider decays that are energetically impossible.
If you look up the masses of the neutron, proton, electron, and the three neutrinos, you will find that the three neutrinos are so close to zero that they don't even show up in the error bars of the neutron and proton masses. Thus any neutron decay experiment involves all three generations of neutrinos.
Now if you look up the masses of the electron, muon, and tau, you will find that a neutron cannot possibly decay into a proton, muon and anti neutrino. The reason is that the muon weighs too much.
If we were in a universe where the electron, muon, and tau masses were much smaller than the difference between the proton and neutron masses, then neutron decay would involve all three generations of charged leptons (i.e. electron, muon and tau), and it wouldn't be so natural to talk about "electron neutrinos".
In fact, just as the muon and tau are higher mass analogues of the electron (in the three generations 1, 2, 3), there exist higher mass analogues of the neutron. They are called "resonances", and sure enough, they can decay into a proton, muon, and anti neutrino. The resonances of the neutron that are particularly appropriate to this example are the P_{11} resonances, the N(939), N(1440), and the N(1710). The N(939) is just the neutron you're familiar with. The numbers in parentheses are the masses in MeV. The mass of the muon is 105MeV, so it can be created in the decay of a N(1440). The mass of the tau is 1776 MeV, so it is too heavy to be created in the decay of any of these resonances to a proton.
Here is where you can look up more incomprehensible information on these particle states:
http://pdg.lbl.gov/2007/listings/contents_listings.html
So when people talk about "neutrino oscillation" they are talking about oscillations between the flavor eigenstates. If you translate this into the mass eigenstate language, what is going on is interference between the emission of neutrinos of the three generations.
If you analyze decays of things like the N(1440) that have enough energy to create muons as well as electrons, you will find that these interfere just like the neutrinos do. If you were in the habit of labeling the flavor eigenstates so that the \nu_1, \nu_2, \nu_3 were flavor as well as mass eigenstates, and the electron, muon and tau were mass eigenstates but not flavor eigenstates, then in the case of the decay of the N(1440), you'd find a case of "electron oscillation", just like you're asking about.
And the oscillation would arise from exactly the same place that neutrino oscillation comes from. The mass eigenstates, the electron and muon, have different wavelengths. As a function of distance, they beat against each other just like any other wave interference does. And the result is that the probability of getting an electron, versus the probability of getting a muon, will depend on distance in the decay of the N(1440). This is the charged lepton analog to neutrino oscillation.