fermion

I don't think there is any mathematical argument for unitarity (i.e. for conservation of probability.) We insist on unitarity based on experimental observation. Among other things, for example, the fundamental particles do not disappear into nothing occasionally, or appear from nowhere. All conservation laws (both continuous and discrete) have some connection to unitarity, and in the absence of unitarity they all get modified in strange ways. While we don't know for sure we live in a unitary universe, no observation has been reported to the contrary. So, I think you can build non-unitary theories:-) , but you will have a very difficult time to keep the predicted rates of the unwanted (more precisely yet unobserved) processes under control:-( . You may be able to tune some parameters to suppress some of them, but probably not all.

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.

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: 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, 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?"

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.

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?

Thank you for three useful replies . I think I agree that the quantization is the key issue here as all three replies focused on it. However, there may be some subtle weakness in the quantization argument. Before I get to that, I would suggest BenTheMan not to worry about the Yukawa sector: there are no fermions in this model. (With Fermions, the scaling is still possible, but it involves the simultaneous rescaling of fermions as well.) We can also avoid the complications of the vacuum expectation value of fi (both classical and renormalized) by choosing the sign of m**2 appropriately so that the vacuum expectation value is zero. Back to the quantization argument: Typically the free field fi (with lambda=0) can be expanded in terms of (momentum dependent) creation and annihilation operators, and these are assumed to satisfy the standard canonical quantization rules. (See the book by Peskin and Schroeder for example.) All is well at this point. The next step is to turn lambda on. Now, at the Lagrangian level, the field fi can no longer be expanded in terms of the creation and annihilation operators (with or without attempting to quantize the creation and annihilation operators.) At this level, therefore, it is impossible to assert that the quantization sets a well defined scale for lambda, because we don't even know what we are quantizing. The statement could still be true, but it is a leap of faith, not a rigorous argument. But there is a way out. We turn lambda on, but we keep it small (as small as possible, assuming that a small values of lambda are stable and do not grow too big by renormalization. This is another leap of faith, but it sounds plausible.) With this small lambda, we intend to do perturbative renormalization. To this end, we assume that the expansion of fi in terms of creation and annihilation operators is still valid, and we quantize it as we did for the free field case. Based on this free field quantization condition, we get the usual quadratically divergent renormalization counter-terms for the mass and lambda, which are handled within a well defined and self-consistent renormalization prescription. This of course prevents rescaling of lambda as pointed out by two of the replies. The weakness in this argument is in the second sentence of this paragraph: "... but keep it small...". Obviously the assumption that lambda remains small is contrary to the statement that lambda can be scaled away. If scaling was possible there would be no difference between lambda=0.001 and lambda=100. Now, only if we could make the argument in the above paragraph without relying on the fact that the (renormalized) values for lambda remain small if the bare coupling is small, then I think I will be convinced.

Today there is a major enrollment problem in CS, and in fact in most branches of engineering in the US. This statement applies to men and women almost equally. It stems from the fact that the engineering is considered very hard education and very hard work (after graduation) with poor pay. The average engineer (any kind) does not earn as much as an average real estate agent, let alone comparisons to doctors and lawyers. Young people are well aware of this fact: males and females both. If the average engineer's salary was $200K, the college kids will be flocking to engineering programs in most universities with good reputation. Again this applies to males and females both! I have teenage sons (who are very good students), but I cannot convince them to study engineering in the future. I am not sure if I can recommend it to them in good conscience given the reality of the financial world out there. How is it possible that we have a shortage of engineers, and it is likely to get worse in the near future, and yet the engineering wages are not rising at a rapid pace? What happened to supply and demand? The US corporations solve the supply problem by importing engineers from abroad. These engineers do not make significantly less money than those who are US citizens, but their sheer numbers depress the wages for everybody. In CS and CIS, easily more than half of all practicing engineers are not US citizens (at least around where I work.)

Lagrangian density for a self interacting real scalar field in four space-time dimensions is: L = (1/2)* [((del-mu)fi) * ((del-mu)fi) - (m**2) * (fi**2)] + (1/4) * lambda * (fi**4), where fi=real scalar field, m=mass, lambda=quartic self coupling constant, and (del-mu)fi=derivative of the field fi with respect to the space-time index mu, with mu = 0,1,2,3. If lambda=0, we have a free scalar field. With lambda > 0, we can scale it away with a simple transform: psi = fi * sqrt(lambda). If we substitute psi for fi in L we get: L = (1/lambda) * { (1/2)*[((del-mu)psi) * ((del-mu)psi) - (m**2) * (psi**2)] + (1/4)*(psi**4) } Expressed for the new scaled field psi, (1/lambda) is an overall numerical constant factor multiplying the Lagrangian density, which has no effect on the equations of motion. In other words lambda is scaled away. Does that mean that all non-zero values of lambda are equivalent? Is this interpretation correct? Is this a well known result???