Calculating Symmetry Factors for Graphs: Reasoning Behind Findings

[kau]

consider the following two graphs. I want to calculate the symmetry factors for them . I am using field contraction method for that. but I am getting 2 for the 2 vertex diagram and 1 for the zero vertex diagram where it should be 4 and two respectively. Can anyone do it for me. I am writing briefly the reason behind my findings. i am considering $$\phi^3$$ theory. So for vertex zero case i don't have any factors left. In case of two vertex case. say I am denoting two vertices as v and w and the sources as x and y. ok. now x can be combined with 6 possible ways and therefore y can be combined with 2 possible ways. now if i call that one v. then i am left with one field in v and 3 in w. they can combine 3 ways. so i got 6*3*2. now i have to divide it by 2*(3!*3!). and this would give me 1/2. so the symmetry factor is two. but that is a wrong answer. so where did i missed the point.??

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[kau]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

 

[kau]

i have found some logical explanation behind these things. actually what happen in case of drawing a feynman diagram is that if every term is properly normalized then all the factors get canceled appropriately. in that case we consider that permutations of propagators,vertices,externals lead to different diagram. but it turns out that some of these permutations are same. so we have over counted them,therefore we ave to divide the amplitude(we get following feynman rules) by that factor. like in the first diagram we can exchange the two sources and it is same as reversing the propagator,but we have considered both. so s=2. and in the second case you can exchange the derivatives at the vertex (connected to sources) and that is same as exchangin the propagators along with sources. so this gives 2. another two comes due to the identical figures due to exchange of two legs of the curved propagator. this gives another two. so we have total 4 as a symmetry factor.