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Within SM, it is know that baryon number is not preserved and changes as

$$ \Delta B = 3·\Delta n_{CS}, \quad n_{CS} \in \mathbb{Z}\ ({\rm Chern-Simons\ index\ for\ vacuum}) \tag1$$ Then, its minimum value is $\Delta B = -3 = B(f) - B(i)$ for a reaction $i \rightarrow f$. Looking for $B$ violation in nuclei, you have to take $i$ as a set of neutrons and protons. I have tried to use 3 nucleons, so $B(i) = 3$ and therefore $B(f) = 0$ but I couldn't find any way to obtain mesons from the valence quarks of the nucleons.

Do you know any examples of nuclei decays with $B$ violation?

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    $\begingroup$ related: physics.stackexchange.com/q/70034 . According to Michael Brown's answer, baryon number nonconservation occurs in the standard model, but only at temperatures $\gtrsim$ 100 GeV. This is above my pay grade in QFT, but I believe this explains why we don't actually see the kind of processes you're talking about. $\endgroup$
    – user4552
    May 2, 2019 at 15:14
  • $\begingroup$ @BenCrowell I think that I understand that we have some $T$ limit between the situation with $\Delta n_{CS} = 0$ and $\Delta n_{CS} \neq 0$, but if the proper SM, whithout nothing else, has that anomaly, why their own Feynman rules aren't allowing me to see that kind of process? Is that due to Feynman rules comes out of Lagrangian that, as the classical one, has baryon number preserved? How can we deal with these 2 ideas that are so opposed? $\endgroup$
    – Vicky
    May 3, 2019 at 3:18
  • $\begingroup$ See physics.stackexchange.com/a/288913/8554 $\endgroup$
    – Thomas
    May 4, 2019 at 2:15

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No.

Baryon number violation is one of the Sakharov conditions for transitioning from a universe with only radiation to a universe with more baryonic matter than baryonic antimatter, along with CP violation and thermal disequilibrium. However, there are so far no observed processes that change the baryon number of a closed system.

If you're thinking of baryon-number violation in sphaleron processes, that's only a high-temperature process, above hundreds of GeV. A nucleus becomes unstable against nucleon emission at temperatures of tens of MeV, and even the nucleon-QGP phase transition is far, far below the temperature at which sphalerons are predicted to become observable. If it's hot enough that sphalerons matter, there aren't any nuclei around.

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  • $\begingroup$ Then what you are telling me is that despite baryon number is violated in SM, I can't find examples of it? I don't understand that. If we have a formula as Eq. (1) on my post that asserts to exist B violation within SM, I should be able to get it via SM interactions, no? I think I'm not understanding something $\endgroup$
    – Vicky
    May 2, 2019 at 7:53
  • $\begingroup$ @Vicky I've edited the answer to address your comment. $\endgroup$
    – rob ♦
    May 2, 2019 at 15:05
  • $\begingroup$ That formula came from triangle Feynman diagrams, that are the usual things used to find anomalies at quantum level. It can be shown that within SM baryon and lepton number conservation is a symmetry broken at quantum level giving that variation. Look at the answer giving by Michael Brown in physics.stackexchange.com/questions/70034/… (recommedation given by Ben Crowell in a comment above) $\endgroup$
    – Vicky
    May 3, 2019 at 3:06
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    $\begingroup$ Oh, sphalerons. Those are sufficiently far away from the experimentally tested parts of the SM that it's easy to forget the process is a Standard Model prediction. Not observed, though. Re-edited. $\endgroup$
    – rob ♦
    May 3, 2019 at 12:00
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    $\begingroup$ @Vicky It’s a non-perturbative effect, it cannot be described by the Feynman rules of the perturbation theory. As ‘t Hooft explains in his paper we can describe the effect by the Feynman rules of an effective Lagrangian ( now usually called the ‘t Hooft vertex). It contains all the left handed fermions of the SM $\endgroup$
    – Thomas
    May 4, 2019 at 13:35

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