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Basic question about Feynman Diagrams

  1. Feb 9, 2012 #1
    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. jcsd
  3. Feb 9, 2012 #2
    You can indeed. For instance gluons can make a 4 vertex.
     
  4. Feb 9, 2012 #3

    Bill_K

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    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.
     
  5. Feb 9, 2012 #4
    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.
     
  6. Feb 9, 2012 #5

    bapowell

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    In QED, the interaction term from the Lagrangian is [itex]e\bar{\psi}\gamma^\mu A_\mu \psi[/itex]. Since three fields take part in the interaction, these are 3-vertices.
     
  7. Feb 9, 2012 #6

    Vanadium 50

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    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.
     
  8. Feb 9, 2012 #7
    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 [itex]\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi[/itex] or [itex]F^{\mu\nu}F_{\mu\nu}[/itex]), or
    • a mass term, if not (eg [itex]m\bar{\psi}_L\psi_R[/itex]).
    ([itex]F^{\mu\nu}F_{\mu\nu}[/itex] does have differential operators but they are "hidden" in its definition ([itex]F^{\mu\nu} = \partial^{\mu}A^{\nu} - \partial^{\nu}A^{\mu}[/itex])).
     
  9. Feb 9, 2012 #8
    Thank you! I think I have a clearer picture in my head now.
     
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