# Nuclei violating B number

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?

• 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. – Ben Crowell May 2 at 15:14
• @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? – Vicky May 3 at 3:18
• – Thomas May 4 at 2:15