In ordinary beta decay, an electron and an anti-neutrino, together with a proton, are emitted. The proton has zero lepton number, the electron +1 lepton number, and, it is said, the neutrino-type particle has -1 lepton number, so is an anti-neutrino. Is there any independent experimental reason for deciding that it is an anti-neutrino rather than a "regular" neutrino, or is it termed an anti-neutrino just to make the law of lepton number conservation be true? If the neutrinos from ordinary beta-decay could be made, or were observed to, collide with other neutrinos that were independently known, by some different type of criterion, to be "ordinary" neutrinos, annihilating each other and producing two photons, this would be reason to call the neutrinos from beta-decay "anti-" neutrinos, but as far as I know, this isn't the case. I have read about considerations related to the spin direction of the neutrino relative to its direction of motion, but these seem to be dependent on whether the neutrino is in fact massless, and so travels at the speed of light, which I gather is still uncertain.
2 Answers
There are infinitely many quantities that you could define. Some of them are conserved (or approximately conserved), and hence useful, so we give them names.
For example, we could have defined the "volume" of a cube of water to be the square of its height times the cosine of its length divided by the logarithm of its width. But then it wouldn't stay the same if we poured the water from one container to another. It's much more useful to consider height times length times width, which is why we call that "volume" instead.
Similarly, the number of electrons (counting positrons as negative) plus the number of anti-electron neutrinos isn't remotely conserved, so it's not useful. But the number of electrons plus the number of electron neutrinos is approximately conserved, so we decide to call that "electron number". And summing that over families gives what we call the "lepton number". (This is conserved within the SM to extreme, though not perfect accuracy, with tiny violations due to electroweak sphalerons and (if they exist) Majorana neutrino masses.)
Now you might worry: doesn't this mean the conservation of lepton number is "fake"? Absolutely not, for two reasons. First, within the Standard Model, it's a useful tool to help us know what to expect in experiments. And second, more exotic physics, such the effects of new particles, could change the lepton number, so we can search for such physics by looking for violations of lepton number.
In fact, in a general theory, it's not even guaranteed that you can define any number like lepton number that is conserved. For example, it could have been the case that beta decay can produce an electron and an electron-anti-neutrino half the time, and an electron and an electron-neutrino the other half. Then there would be no reasonable way to define anything like lepton number that's even remotely conserved. It's a nontrivial statement about the Standard Model that you can.
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$\begingroup$ I agree with this answer, but it seems to somewhat contradict your statement "If you lose electron number conservation, this distinction doesn't work anymore" in your comment below this other answer of yours. $\endgroup$– tparkerCommented May 16, 2020 at 1:58
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2$\begingroup$ Also, I respectfully think that this answer is kind of confusing, because you first imply that lepton number conservation is only approximate, but then you say that in "physics beyond the Standard Model, it could be that this quantity is no longer conserved," which implies that it is (exactly) conserved in the Standard model. $\endgroup$– tparkerCommented May 16, 2020 at 2:03
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1$\begingroup$ Isn't the correct statement that whether or not total lepton number (summed across generations) is exactly conserved depends on whether or not you include Majorana mass terms in your extension of the "traditional" massless Standard Model (see my comment to the other answer)? $\endgroup$– tparkerCommented May 16, 2020 at 2:04
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$\begingroup$ @tparker Damn it, you're keeping me too honest! I was trying to keep a little detail under the rug. I'll edit to make the answer really true. (And as long as we're on the subject, total lepton number would also be violated by electroweak sphalerons, which are part of the vanilla SM.) $\endgroup$– knzhouCommented May 16, 2020 at 2:21
Actually, there is no fundamental reason for lepton number to be conserved. However, the SM Lagrangian respects the total lepton number symmetry. At the moment, there are several experiments, which search for the lepton number violating processes, namely, neutrinoless $2 \beta$ decay.
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$\begingroup$ Surely this means that a non-conservation of lepton number would be a major departure from the SM? $\endgroup$ Commented May 15, 2020 at 20:50
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$\begingroup$ This answer is oversimplified. If you don't include any sterile neutrino fields as part of your postulated matter fields (as the original Standard Model does not), then there absolutely is a fundamental reason for lepton number to be conserved - it's required by electroweak gauge symmetry. But if you add in sterile neutrino fields, which is the most natural way to get massive neutrinos, then you are correct that there's no natural reason to conserve lepton number. $\endgroup$– tparkerCommented May 16, 2020 at 1:38
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$\begingroup$ @AndrewSteane It depends on exactly how you define "the Standard Model". Strictly speaking, "the" Standard Model formulated by Weinberg, Salaam, Glashow has completely massless neutrinos and conserves lepton number even within each generation - but this Standard Model was experimentally falsified back in 2001 with the discovery of neutrino oscillations. I believe that these days, many people loosely use the term "Standard Model" to mean the original SM plus sterile neutrino fields, since that's the simplest extension of the original SM that yields massive neutrinos. Whether or not these ... $\endgroup$– tparkerCommented May 16, 2020 at 1:45
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$\begingroup$ ... sterile neutrino fields are coupled to the active neutrinos via Majorana mass terms remains an open question - there's no reason to omit these terms on any gauge symmetry grounds, but we haven't yet observed the neutrinoless double $\beta$-decay that they would induce. So the loose modern sense of "Standard Model" (i.e. the minimal model that incorporates all experimentally confirmed non-gravitational physics) isn't actually fully well-defined - you can choose whether or not to include Majorana mass terms. $\endgroup$– tparkerCommented May 16, 2020 at 1:46
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1$\begingroup$ @tparker thanks---I think your comments amount to an answer, and a much more helpful one than the one we were commenting on! $\endgroup$ Commented May 16, 2020 at 10:09