# Why can there be only 3-particle vertices in Feynman diagrams? [closed]

1. Why can there be only 3-particle vertices in Feynman diagrams? For example, when a particle and its anti-particle annihilate to form two photons, why is its Feynmann drawn the way it is in diagram 2 (3-particle vertices)? Why can't diagram 1 (4-particle vertices) be correct? Does diagram 1 violate some form of conservation laws?

2. Also, how could diagram 2's 3-particle vertices fully capture the reaction between a particle annihilating with its anti-particle to form 2 photons? Why is it that the particle can first emit the first photon, and then annihilate with its anti-particle to produce the second photon? It appears like the diagram 2 is a two-step process, of which the annihilation between a particle and its anti-particle only produces one photon (which cannot be right since momentum won't be conserved).

Actually there are 4 particle vertices e.g. 4-Higgs and 4-Gluon interactions. For particles including fermions however this is not possible. The reason for this is that each vertex i.e. each interaction corresponds to a term in the Standard Model lagrangian. Such an interaction Term is generally a Product of the interacting fields $\phi_i$ and a coupling constant g $$L_{int}=g\phi_1\phi_2...$$ The thing is now that only so-called renormalizable terms are allowed. That forbids all Terms in which the dimensions of the fields add up to more than 4. Since fermions fields have dimension 3/2 there cannot be 4 in such a Term. That means indeed Diagram 1 is forbidden because it is not renormalizable. However scalar fields like the Higgs or vector fields like gluons have dimension 1 and hence can form 4-particle interactions (as long as they are also invariant under the proper standard model symmetrys)
The reason why the second diagram does not violate momentum conservation is because the fermion in between the two vertices is a Virtual so-called off-shell particle which does not obey $$p^2=m^2$$