Basic question about Feynman Diagrams

[McLaren Rulez]

Can a vertex of a Feynman diagram have more than three particles going in or out from it? Assuming all other conservation laws are obeyed, of course. I haven't seen this being explicitly stated but all the Feynman vertices I have seen have three arms attached. Thank you.

 

[thedemon13666]

You can indeed. For instance gluons can make a 4 vertex.

 

[Bill_K]

An interaction term in the Lagrangian which is trilinear leads to a 3-particle vertex, while one which is quadrilinear leads to a 4-particle vertex. These things occur for example in electroweak theory. The kinetic energy term for the W boson is 1/4 W_μνW^μν where W_μν = ∂_μW_ν - ∂_νW_μ - gW_μ x W_ν, which leads among other things to a 4-particle WWWW vertex. Also there is a WWhh vertex from its interaction with the Higgs.

 

[McLaren Rulez]

Thank you for the quick replies. I do not know about the interaction terms in the Lagrangian. I am just learning to draw Feynman Diagrams for basic QED and Weak Force processes. So, in these cases, are my vertices 3 vertices or 4 vertices or are both okay? Thank you.

 

[bapowell]

Thank you for the quick replies. I do not know about the interaction terms in the Lagrangian. I am just learning to draw Feynman Diagrams for basic QED and Weak Force processes. So, in these cases, are my vertices 3 vertices or 4 vertices or are both okay? Thank you.

In QED, the interaction term from the Lagrangian is e\bar{\psi}\gamma^\mu A_\mu \psi. Since three fields take part in the interaction, these are 3-vertices.

 

[Vanadium 50]

I do not know about the interaction terms in the Lagrangian. I am just learning to draw Feynman Diagrams

But that's what Feynman diagrams are.

They are not little cartoons describing collisions between billiard-ball like particles. They are a calculational shorthand for terms in a Lagrangian.

 

[AdrianTheRock]

The simpler electroweak processes are all three-vertices:

• two fermions interacting with a W boson;
• two fermions interacting with a Z boson; or
• two fermions or a W boson interacting with a photon.

In general, the number of fields multiplied together in each Lagrangian term imply interactions with that number of particpants. If there are only two - in which case they will always be a field and conjugate of either the same field or its chiral partner, the term represents

• a kinetic term, if it has one or more partial differential operators (eg \bar{\psi}\gamma^{\mu}\partial_{\mu}\psi or F^{\mu\nu}F_{\mu\nu}), or
• a mass term, if not (eg m\bar{\psi}_L\psi_R).

(F^{\mu\nu}F_{\mu\nu} does have differential operators but they are "hidden" in its definition (F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu})).

 

[McLaren Rulez]

Thank you! I think I have a clearer picture in my head now.