
#1
Feb1412, 11:54 PM

P: 133

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
Feb1512, 12:07 AM

P: 133

Just forgotten to mention (combination of fields) exactly the same . . . so identical field combinations in two vertex but different coupling strength . . .




#3
Feb1512, 06:55 AM

P: 82

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.




#4
Feb1712, 12:10 AM

PF Gold
P: 445

Coupling order in Fenman diagrams
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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