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In the standard model Lagrangian, the electric charges of the particles are the coefficients of the interaction terms (e.g. $(-2/3e)u'Au$ for the up quark shows it's charge is $(2/3)e $)

How can we see what the baryon numbers and lepton numbers for each particle are just by looking at the Lagrangian terms?

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  • $\begingroup$ In theory one gets it from the resulting EOMs, but with enough experience staring at the Lagrangian makes the outcome of doing that obvious. $\endgroup$
    – J.G.
    Commented Feb 29 at 19:21
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    $\begingroup$ The baryon and lepton numbers of the Lagrangian of fermion bilinears are zero. They are not gauged, so there is nothing altering baryon or lepton number in the SM. You have to move on to GUTs and anomalies to incorporate these low energy experimental facts in the lagrangian (B-L conservation). $\endgroup$ Commented Feb 29 at 23:14

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As with any conservation law, you can see get it from Noether's theorem by noticing a symmetry in the Lagrangian. By looking at how the symmetry acts on the fields, you can deduce the charge.

For the baryon number, it's the $U(1)$ symmetry obtained by multiplying the quark fields by the same phase (and the conjugate fields by the conjugate phase). The corresponding current is the sum of each individual quark current. Since you never observe an individual quark, it's natural to divide it by $3$ so that you can interpret it as a baryon number. By the action of the group, all quarks have baryon number $1/3$ (what matters is that it is equal), their antiparticles $-1/3$ (what matters is the relative sign) and the rest $0$.

Similarly, for the lepton number, it's the $U(1)$ symmetry obtained by multiplying all the lepton fields by the same phase (and the conjugate fields by the conjugate phase). The corresponding current is the sum of each individual lepton current. By the action of the group, all leptons have lepton number $1$ (what matters is that it is equal), their antiparticles $-1$ (what matters is the relative sign) and the rest $0$.

Hope this helps.

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