In the Standard Model, the baryon number is not exactly conserved due to anomaly but the decay rate is extraordinarily small at ordinary temperatures. Does this make free protons unstable in the Standard model itself (with no new physics and no new particles)? If the free proton is really unstable, what would be its dominant mode of $B$-violating decay?
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1$\begingroup$ See physics.stackexchange.com/questions/288892/… $\endgroup$– FrodCubeCommented Jun 10, 2022 at 16:47
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$\begingroup$ The answer by Thomas, in the linked post above, points out that "Electroweak instantons violate baryon number (and lepton number) by three units (all three generations participate in the 't Hooft vertex). As a result, the proton is absolutely stable in the standard model." I could not follow the consequence of proton stability in the SM (last sentence) from the reason given (first sentence). $\endgroup$– SRSCommented Jun 10, 2022 at 17:16
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1$\begingroup$ The B violating transitions can only change B by 3. The proton state is B = 1, while the decay product state has B = 0 (since there are no lighter baryons that the proton could decay into). You can't go from 1 to 0 by adding or subtracting 3. $\endgroup$– FrodCubeCommented Jun 10, 2022 at 18:09
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$\begingroup$ Here is a nice review of the 6-quark effective operator flipping chirality. You add the leptons in the vertex, and tie the particles up with suitable allowed EW interactions, to lead to a proton and suitable leptons. The onus is on you, but the rate is unforgiving. $\endgroup$– Cosmas ZachosCommented Jun 10, 2022 at 19:58
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