# Feynman diagram elementary vertex with 4 lines?

Are there processes that require vertices with 4 lines in a Feynman diagram? (And cannot be written as composition of 3-line vertices?)

If not, is it matter of models we use (where there are no fundamental couplings of higher order) or is it something more fundamental (e.g. all such coupling could be decomposed with a bosonic field, or something would be inconsistent)?

The number of lines in a given vertex is equal to the number of fields appearing in a term in the corresponding Lagrangian. In renormalizable Lagrangians, that means vertices have 3 or 4 lines. Since fermionic fields have dimension of $[E]^{3/2}$ while bosonic (scalar and vector) fields have dimension of $[E]$ (and the Lagrangian must have dimension of $[E]^4$), vertices with 4 lines must contain 4 bosonic fields.
• The Yang-Mills Lagrangian for non-Abelian gauge fields. In the case of the $\mathsf{SU}(3)_C$ strong symmetry, we have a vertex with 4 gluons. In the case of the $\mathsf{SU}(2)_L\times \mathsf{U}(1)_Y$ electroweak symmetry, things are a bit more messier, and we have a vertex with 4 $W$ bosons, a vertex with 2 $W$ and 2 $Z$ bosons, a vertex with 2 $W$ bosons and 2 photons, and a vertex with 2 $W$ bosons, one $Z$ boson and one photon.
• The kinetic term for the Higgs field, which gives us a vertex with 2 Higgs bosons and 2 $W$ bosons, and a vertex with 2 Higgs bosons and 2 $Z$ bosons.