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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

    Hepth

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    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.)
     
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