Is Lepton Flavour Universality an accidental symmetry of the Standatd Model? If it is, why? How does it emerge from the Standard Model?
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2 Answers
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It is not accidental, at least in the sense that this term is used.
The accidental symmetry is the symmetry that is forced on you by the renormalizability - i.e. for a given field set you can't write the interaction that would violate such symmetry. On the other hand nonrenormalizable interactions or renormalizable interactions with extra fields (these two options are related) may violate symmetry. The example is the baryonic number conservation which is easily violated in many extensions of the standard model.
On the other hand the lepton universality is the property of the non-abelian gauge invariance. In qft the non-abelian gauge interaction is determined solely by the representation of the field and the coupling constant, common for all the fields. I.e. you can't have $SU(2)$ doublet interacting with $W$ boson 1% stronger than another doublet.
Of course this constraint is weaker for $U(1)$ interaction, such as hypercharge field. But if you change somewhat $Y$ for some lepton, this may ruin the gauge invariance of the mass terms and the triangle anomaly cancellation.
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$\begingroup$ Thanks @OON .One thing I don't understand is why lepton flavour universality is assumed in the SM. (probably I need more theoretical knowledge). Why it is assumed if there is evidence that is not "conserved"? Why the leptons coupling to W,Z must be the same? If evidence turns into discovery, why people talk about possible new particles instead of changing the couplings? $\endgroup$ May 7, 2022 at 17:48
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1$\begingroup$ @AbelGutiérrez choosing lepton couplings to Z W be the same for the three flavors is compatible with the data and thus no change is needed to the standard model, except if the data will show differences between expected calculations from the SM and the data, which is what is being checked at the LHC. $\endgroup$– anna vMay 7, 2022 at 18:21
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$\begingroup$ I understand that there are some observables that preserve lepton flavour universality, but others seem to do not. Is that right? That is my confusion. $\endgroup$ May 9, 2022 at 7:26
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In the Standard Model of particle physics the three charged leptons are identical copies of each other, apart from mass differences, and the electroweak coupling of the gauge bosons to leptons is independent of the lepton flavour. This prediction is called lepton flavour universality (LFU) and is well tested. In tree level decays, any violation of LFU would be a clear sign of physics beyond the Standard Model
The standard model developed slowly over the last century by continually fitting the data, using axiomatic assumptions and testing them against data. One of these axiomatic assumptions was LFU .
LFU is not an accidental symmetry, it is the result of the assumed couplings within the standard model.
From the introduction in the link
In the Standard Model of particle physics (SM), the electroweak gauge bosons Z and W ± have identical couplings to all three lepton flavours. This means that branching fractions of decays involving different lepton families do not depend on lepton flavour but differ only by phase space and helicity-suppressed contributions
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2$\begingroup$ Dear, Anna v. I will repeat my criticism for some other answers of yours. You again answer the theoretical question with "the standard model was developed to fit the data, if it would not fit the data you would change the model". The aim of the theoretical question is rather to understand how some particular properties of the model are related to each other. And what exactly should be changed if this property of the model does not agree with experiment. $\endgroup$– OONMay 7, 2022 at 12:46
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$\begingroup$ @OON In this particular answer there is the theoretical part in the second quote, "it is the identical couplings to all three lepton flavors for the verti ces of Z and W exchange" that imposes the universality lfu to the SM. $\endgroup$– anna vMay 7, 2022 at 14:35
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$\begingroup$ @OON you say "The aim of the theoretical question is rather to understand how some particular properties of the model are related to each other" and I accept it, but the aim of main stream physics is to model mathematically nature with validated models, which is something theoretically inclined people tend to ignore, prefering to play with the mathematics. if LFU was not evident in the data , a different mathematically model would be necessary. That is what makes experiments look for violations of LFU, to find new physics. $\endgroup$– anna vMay 7, 2022 at 15:43
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$\begingroup$ I prefer @OON answer but agree with anna v last comment. I look at Physics as a model to describe nature based on mathematics... but maybe today mathematical speculations become useful in future (or in a parallel universe, idk). Anyway, anna v maybe you can contribute to my comments at OON answer. $\endgroup$ May 7, 2022 at 17:59