# Why is baryon number conservation an accidental symmetry

I have to write a report surrounding the subject of baryogenesis and I wanted to start this report off with explaining how the first Sakharov condition: Baryon number violation is possible within the Standard Model. I, however, can't seem to find a good explanation as to why in the classical sense the baryon number is conserved and why this is a so-called accidental symmetry. Most papers I read about baryogenesis state that due to a general Lagrangian being invariant under the gauge group: SU(3) x SU(2) x U(1) that it automatically has an global symmetry around U(1) with which lepton and baryon numbers can be identified. But how?

I think your misunderstanding is precisely that you think that $$U(1)$$ gauge symmetry in SM can be associated to any quantum number such as baryon or lepton number. No, it can not. The $$U(1)$$ coming from $$SU(3)_C\otimes SU(2)_L\otimes U(1)_Y$$ is related to a quantum number called hypercharge, $$Y$$

We say that baryon and lepton numbers are symmetries in a 'classical' sense since they are preserved at Lagrangian level. This is, the Lagrangian is invariante under the change

$$\phi \rightarrow e^{-i\epsilon N}\phi$$

Where $$\phi$$ is any field, $$N$$ can be either baryon ($$B$$) or lepton ($$L$$) numbers and $$\epsilon$$ the group parameter.

Nevertheless, as in QFT books is proven via triangle diagrams you can see that at quantum level (not Lagrangian level) baryon and lepton numbers are not preserved, i.e.,

$$\partial_\mu j^\mu_{N} = n_{CS} \Rightarrow \Delta N \neq 0 \tag1$$

with $$j^\mu_{N}$$ the Noether current related to $$N = B, L$$ and $$n_{CS}$$ the Chern-Simmons index. But since this index is the same for $$B$$ and $$L$$, from Eq. (1) you can deduce that

$$\partial_\mu (j^\mu_{B} - j^\mu_{L}) = 0 \Rightarrow \Delta(B - L) = 0$$

The accidental (which means not pre-impossed in Lagrangian) quantum symmetry is not for $$B$$ or $$L$$ but for $$B - L$$.