Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Coupling order in Fenman diagrams

  1. Feb 14, 2012 #1
    Sorry if this appears a basic question,
    but could you pls advise me is it possible to have a two identical vertexes but with different strength coupling?

    I have some toy Lagrangian and when I calculate Feynman rules I get for one vertex following expression:

    (M1+M2)*(combination of fields )+M1(combination of fields).

    Where M1 and M2 are coupling strength, so if I want to calculate some process in some order I should consider diagrams of the same order with respect to the coupling strenght. And in my case I am calculating diagrams with respect to M2 strength coupling.... so does it mean that for my vertex I should extract only M2 coupling part from that general vertex which wrote above: (M1+M2)*(combination of fields ) - > (M2)*(combination of fields ) ?
  2. jcsd
  3. Feb 15, 2012 #2
    Just forgotten to mention (combination of fields) exactly the same . . . so identical field combinations in two vertex but different coupling strength . . .
  4. Feb 15, 2012 #3
    If I understand correctly you have two identical terms in the lagrangian, just different coefficients( coupling constants). Just add the terms into one, and you have one vertex with a coupling constant that is the sum of the two.
  5. Feb 17, 2012 #4


    User Avatar
    Gold Member

    Think about it as a taylor expansion. What is small?

    If you're expanding in the fact that M2 is small, like you would in QED , then remember the leading order vertex is just M1, and the subleading the combination.

    But remember! Just because a VERTEX is subleading does not mean the DIAGRAM is subleading. (If you're expanding in coupling strength I believe its the same, but if you end up expanding in mass or momenta you can end up with powers canceling due to different vertices as well as propagators.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook