# Basic question about Feynman Diagrams

1. Feb 9, 2012

### 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.

2. Feb 9, 2012

### thedemon13666

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

3. Feb 9, 2012

### 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 WμνWμν where Wμν = ∂μWν - ∂νWμ - gWμ x Wν, which leads among other things to a 4-particle WWWW vertex. Also there is a WWhh vertex from its interaction with the Higgs.

4. Feb 9, 2012

### 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.

5. Feb 9, 2012

### bapowell

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.

6. Feb 9, 2012

Staff Emeritus
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.

7. Feb 9, 2012

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 ($F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}$)).

8. Feb 9, 2012

### McLaren Rulez

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