Sunday, 16 October 2016

homework and exercises - Is this symmetry factor in Peskin wrong?


I am trying to compute the symmetry factor of a Feynman diagram in ϕ4 but i do not get the result Peskin Claims. This is the diagram I am considering


enter image description here


(14!)3ϕ(x)ϕ(y)d4zϕϕϕϕd4wϕϕϕϕd4vϕϕϕϕ


my attempt is the following: there are 4 ways to join ϕ(x) with ϕ(z). There are then 3 ways to connect ϕ(y) with ϕ(z). Then, there are 8 ways to connect ϕ(z) with ϕ(w) and 4 ways to contract the remainning ϕ(w) with ϕ(v). Finally the there are 6 ways to contract the ϕ(w) and the three ϕ(v) in pairs


(14!)3˙4˙3˙8˙4˙6=16


but the result claimed in Peskin (page 93) is 1/12. What am I doing wrong?



Answer





What am I doing wrong?



The expansion of ex is: ex=1+x+x2/2+x3/3!+


From expanding the expression: ϕxϕyexp(λ4!dzϕ4z),

the third order term is: ϕxϕy13!(λ4!)3dzdwdvϕzϕzϕzϕzϕwϕwϕwϕwϕvϕvϕvϕv


There are four (4) ways to connect x to z and then three (3) ways to connect y to z. There are four ways (4) to connect one of the remaining zs to a w and four ways to connect the other remaining z to a v (4), this can be done for either of the two remaining zs (2), i.e., the "third" z can connect to w or the "fourth" z can connect to w. There are six (3!) ways to connect up the remaining ws and vs. And finally, there is nothing special about "z", I can treat "w" the same way as "z" or "v" the same way as "z", so that gives another factor of three (3).


So the overall symmetry factor is: 4344232313!14!3=112


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