How to calculate symmetry factor of Feynman graphs?

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SUMMARY

The discussion focuses on calculating symmetry factors for Feynman graphs in the context of interacting scalar fields, as outlined in Srednicki's book. The standard procedure involves recognizing that interchanging two legs of vertices contributes a factor of 2 for each matching swap. The participants emphasize the importance of understanding Wick's theorem and field contractions to derive these factors naturally. Specific examples from attached diagrams illustrate how to identify symmetry factors, with one diagram noted to have a symmetry factor of 2 due to equivalent interaction terms.

PREREQUISITES
  • Understanding of Feynman diagrams and their components
  • Familiarity with Wick's theorem and field contractions
  • Knowledge of scalar field theory
  • Basic grasp of quantum field theory terminology
NEXT STEPS
  • Study Wick's theorem in detail to understand field contractions
  • Learn about symmetry factors in quantum field theory
  • Explore examples of Feynman diagrams with varying symmetry factors
  • Review scalar field theory concepts and their applications in Feynman graphs
USEFUL FOR

Students and researchers in theoretical physics, particularly those studying quantum field theory and Feynman diagrams, will benefit from this discussion.

kau
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I am reading Srednicki's book and I am stuck at this point of calculating symmetry factors for Feynman's Graph in the context of dealing with interacting scalar field. First of all my question is what is the standard procedure to calculate it. The way Srednicki has talked about it is that if interchanging any two things match with any other swapping then we should take into account factor 2 for each of that case. First of all I didn't understand what does it mean to interchange two legs of vertices. which part is called leg of a vertex? alright then consider the four diagrams I have attached below. And explain me the factors in each case using the kind of arguments like swapping this and that and so on. then answer these ques.. In the first diagram can I swap all three sources?? in the third diagram can I swap the lower two parts of the big bubble which is holding the lower loop?? Please help me. thanks.
e 3.jpg

e 4.jpg
e2.jpg
e 3 s.jpg
 
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For example the:
https://www.physicsforums.com/attachments/e-4-jpg.73744/
has symmetry factor =2
One way to determine symmetry factors is by using the Wick's theorem and field contractions. The rules then come out naturally after you get used in using that correctly...
You can understan it's 2 because 2 diagrams give exactly the same interaction term...
if you name the vertices A and B in that picture, whether you bring the two left external legs to A and the right to B, or bring the left to B and the right to A you get the same term...

I don't know the way of path integrals... but you can search for "vertices derivatives and symmetry factors" and find some results that can help you personally.
 
Last edited:
kau said:
I am reading Srednicki's book and I am stuck at this point of calculating symmetry factors for Feynman's Graph in the context of dealing with interacting scalar field. First of all my question is what is the standard procedure to calculate it. The way Srednicki has talked about it is that if interchanging any two things match with any other swapping then we should take into account factor 2 for each of that case. First of all I didn't understand what does it mean to interchange two legs of vertices. which part is called leg of a vertex? alright then consider the four diagrams I have attached below. And explain me the factors in each case using the kind of arguments like swapping this and that and so on. then answer these ques.. In the first diagram can I swap all three sources?? in the third diagram can I swap the lower two parts of the big bubble which is holding the lower loop?? Please help me. thanks.
This is extreamly tedious business to explain in here. See the pdf below. If this does not help you, then you should ask your instructor to explain it to you. One really needs 2-3 blackboards for that :)
 

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