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
(14!)3ϕ(x)ϕ(y)∫d4zϕϕϕϕ∫d4wϕϕϕϕ∫d4vϕϕϕϕ
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)⟩,
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: 4∗3∗4∗4∗2∗3∗2∗313!14!3=112
No comments:
Post a Comment