In the standard model, the baryon number $B$ and the lepton number$L$ are conserved perturbatively. However, the B+L quantum number is anomalous and is violated via non-perturbative effects while keeping $B-L$ anomaly-free and conserved. Therefore, it is the $B-L$ quantum number which is conserved.

But it is written in the paragraph above Eq. (5) of this reference that it is the combination $\frac{1}{3}B-L_\alpha$ ($L_\alpha$ is the lepton number for lepton flavor $\alpha$) which remains conserved in the Standard model interactions. Where does this factor of $1/3$ come from?


It's the same thing, with a different convention for the assignment of baryon numbers $B$. If you say that leptons have $L= 1$ and protons have $B=1$ then quarks have $B=1/3$ and $B-L$ is conserved. If instead you say that quarks have $B=1$ then $\frac{1}{3}B-L$ is conserved instead.

  • $\begingroup$ If quarks have $B=1$, wouldn't it be $3B-L$? $\endgroup$ – Kosm May 23 '17 at 3:07
  • $\begingroup$ The thing that is conserved is the number of baryons minus the number of leptons, but $B$ doesn't necessarily mean the number of baryons. If quarks have $B=1/3$ then $B$ is the same as the number of baryons. If quarks have $B=1$ then baryons have $B=3$ so the number of baryons is $B/3$. $\endgroup$ – Luke Pritchett May 23 '17 at 12:03

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