0

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

$$ \Delta B = 3ยท\Delta n_{CS}, \quad n_{CS} \in \mathbb{Z}\ ({\rm Chern-Simons\ index\ for\ vacuum}) \tag1$$ 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?

Vicky
  • 1,675

1 Answers1

2

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.

rob
  • 96,301