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What exactly is a lepton number of a particle? With the charge (eg proton is just 1, not the exact charge), I can understand because it's a physical property, put a particle with charge + next to another particle with charge + and they will repel. What is the lepton number in similar terms? Or is it just a convention that worked in the observable particle interactions (that it is conserved, like charge)

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See also baryon number, hypercharge, isospin, etc... – dmckee Nov 5 '11 at 23:27
@dmckee, sorry if this is too basic and the teacher said we are starting with lepton number because it is the simplest. I didn't have time to ask the teacher before this weekend,and sites like Wikipedia just give a chicken and egg type cycle – Jonathan. Nov 5 '11 at 23:30
They are all quantum numbers. They each label the eigenstates of some operator (sometimes a abstract operator, and sometime approximate eigenstates). Some of them appear in other operations. Some of them are conserved in certain classes of interactions. These facts were discovered experimentally (though often classified phenomenologically) and you'll have to find a way of remembering all this stuff. Hang on and you'll be given some structures to hand it all on. – dmckee Nov 5 '11 at 23:35
People first join different particles in "families" (leptons, for example), and then they need a feature to distinguish them within this family. So electron and muon have different lepton numbers. It prevents us from making a superposition of electron states and muon states because they are never observed in such superpositions. – Vladimir Kalitvianski Nov 7 '11 at 10:44

Electric charge is a "special" kind of physical property because it corresponds to a very simple physical effect. But that's not true of most physical properties. The lepton number doesn't have any force associated with it, the way electric charge does, because it's not a coupling constant.

Lepton number is just a mathematical expression of what it means to be a lepton. The quantum fields which correspond to the particles we call leptons (electron, muon, tau, and their corresponding neutrinos) each have a lepton number of 1, and the fields corresponding to their antiparticles each have a lepton number of -1. Lepton number is considered to be a useful property because it is conserved in all observed reactions.

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The lepton number is just a conserved quantity in all observed processes, with the almost certain exception of neutrino oscillations. The electron number is just the number of electrons plus neutrinos, minus the number of antielectrons and antineutrinos. The muon number and tau number are defined similarly. The total lepton number is the sum of the electron, muon, and tau number.

The reason for the conservation of lepton number is only because the low-energy interactions are renormalizable---- which implies that only two fermions at most can take part in an interaction, along with one boson (a scalar or a vector). Electromagnetic processes either produce or remove electron/positron pairs (which is net zero lepton number change), while weak interactions will produce electrons together with electron-antineutrinos (again, zero lepton number change). That's it for the renormalizable lepton interactions in the standard model.

The observation of neutrino masses, however, means that the standard model is incomplete. By far the most likely explanation is that the masses are small non-renormalizable corrections to the standard model, due to neutrinos scattering off two Higgs bosons at once, a non-renormalizable two-fermion-two-scalar interaction. This leads to a mass which is suppressed by the scale at which the two Higgs boson absorptions are not simultaneous, and if this scale is about $10^{16}$ GeV, the mass of the neutrinos turns out to be $.01 eV$, which is what is observed. The observed tiny neutrino mass matches with this prediction so well, that it is all but certain that this explanation is correct.

But this mechanism violates lepton number, because a neutrino field has only one chirality, so that the massless neutrino has one helicity, while an antineutrino has an opposite helicity. You can chase a massive neutrino faster and faster, until you see it stopped in your frame, then going the other way faster and faster. The momentum flips, but the spin stays the same, so the helicity is opposite. So the neutrino and the antineutrino are not separate objects, they are two helicities which make up one massive particle, so lepton number cannot be conserved.

The nuclear physicists are searching for what is called "neutrinoless double-beta decay", a process where two neutrons decay in a nucleus nearly simultaneously. The first decays to a proton and an electron and an electron anti-neutrino (conserving lepton number), then the anti-neutrino flips to a neutrino (by neutrino mass), and the neutrino is absorbed by the second neutron, causing it to decay into a proton and an electron. This process would be a direct experimental confirmation of the above theory of the origin of neutrino masses.

In theory, it is impossible for Lepton number to be exactly conserved, because unlike electric charge, it does not come with an electric field associated to it. The only quantities which are conserved in black hole formation and evaporation have to be visible outside the black hole, and only quantities like the electric and magnetic field, gauge fields, poke out of a black hole to affect its decay.

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Chirality is Lorentz invariant and does not flip if you move faster than a neutrino – Tim Apr 15 '15 at 14:07
@Tim: Chirality is only Lorentz invariant for a massless neutrino, it's the spin in the direction of motion. Your comment is incorrect. There is no chirality for a stationary neutrino, there is only a spin. Depending on which way you boost to infinite momentum, you get opposite chirality. – Ron Maimon Apr 15 '15 at 14:51
that is wrong, sorry. You're confusing helicity and chirality. Chirality is lorentz invariant, helicity is not. And of course there is chirality for any neutrino, because this is what tells us under what representation the corresponding spinor transforms – Tim Apr 15 '15 at 15:05
@Tim: Oops, I meant helicity, sorry, will fix. – Ron Maimon Apr 15 '15 at 15:12
@Tim: To explain more properly, the two-component Lagrangian mass term for a chiral field $\psi_i$ is $\psi_i \psi_j \epsilon^{ij} + \bar{\psi_\dot{i}}\bar{\psi_\dot{j}} \epsilon^{\dot{i}\dot{j}}$. The field $\psi$ has one chirality with one helicity when massless, the field $\bar\psi$ transforms as the conjugate of opposite chirality and has opposite helicity, and creates an antineutrino, and the mass term violates lepton number. By chasing a nearly massless neutrino, you reverse direction of motion, and you convert it to an antineutrino, violating Lepton number by boosting. – Ron Maimon Apr 15 '15 at 15:18

Lepton number, that in some formulations is the sum of three different numbers, one for each generation, is an intriguing beast because, differently to electric charge, is not associated to a gauge field; it is just there, preserved because there is not way in the Standard Model to violate it.

In some Beyond Standard Model theories, the lepton number contributes indirectly one half to the electric charge. In this way it is easier to understand the isospin charge: positron is one half from lepton number, one half from isospin, and neutrino is the difference instead of the sum. In the Standard Model, this contribution comes from the hypercharge, which is a truly gauge field.

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by indirectly, I mean that usually this comes from a field B-L, baryon minus lepton numbers. Of course, this also means that baryon number contributes another half inside baryons, or one sixth in each quark. – arivero Nov 6 '11 at 2:03

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