In Standard Model, there are vertices with 3 legs (example: $W l\nu$) and with 4 legs (example: $WWWW$).

Is there a fundamental reason why in the Standard Model, there is no Feynman rule where a vertex has more than 4 legs?

Is this possible in Beyond-the-Standard-Model theories?


1 Answer 1


Fundamental? Power counting. The SM is renormalizable, that is, without dimensionful couplings. The lagrangian must have dimension 4, and any vertex with 5 legs or more would dictate a coupling of dimension -1 or less. So the gauge couplings g,g', the Yukawa y, etc, are all dimensionless.

(You do have Feynman diagrams of higher dimension, like a 4-fermion diagram with exchange of a vector boson: four fermions are of dimension 6. So in the effective theory for that, it is summarized by 4-Fermi coupling with a dimensionful (-2) effective $G_F$.)

Anything is possible in BSM theories, but, normally, one looks for renormalizable theories as well—otherwise one simply tacks on unrenormalizable effective interactions to the SM, which afford little "explanation" of the energy behavior of the amplitudes.


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