I am using the book QFT by Srednicki, Chapter 9. From the rules that he mentions, one can count the number of terms that correspond to the same Feynman diagram $(t_i)$. But for the case $V=2, P=3$, the diagrams are

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For both of the diagrams I am getting the same number: $(3!)^V$ for the vertices and $V!$ for the permutations of the vertices and $P!$ for the permutations of all the propagators. This is $t_1=t_2=(3!)^2\times2!\times3!=432$. So, $t_1+t_2=864$. But $t_1+t_2$ should be equal to $\frac{(2P)!}{(2P-3V)!}=720$ in my understanding. What am I doing wrong here?


1 Answer 1

  1. In OP's example we have $V=2$ cubic vertices $Y$ with in total 6 derivatives $\frac{\delta}{\delta J}$ attached, and $P=3$ propagators with in total 6 sources $J$ attached, cf. eq. (9.11) in Ref. 1. The derivatives and sources can be contracted in $6!=720=288+432$ ways, i.e. the number of permutations.

  2. One may show that $6\cdot 4\cdot 2\cdot 6=288$ contractions$^1$ lead to the sunset diagram $\theta$ and $6\cdot 4\cdot3\cdot6=432$ contractions lead to the dumbbell diagram $O\!\!-\!\!O$.

  3. If we divide with the normalization $$(3!)^V\cdot V! \cdot(2!)^P\cdot P!~=~(3!)^2\cdot 2! \cdot(2!)^3\cdot 3!~=~3456,$$ we get the reciprocal symmetry factors $\frac{1}{12}$ and $\frac{1}{8}$ for the 2 diagrams, respectively.


  1. M. Srednicki, QFT, 2007; equation (9.11) and figure 9.1. A prepublication draft PDF file is available here.


$^1$ Sketched proof for the sunset diagram $\theta$: Consider the 3 derivatives $\frac{\delta}{\delta J}$ on the 1st cubic vertex $Y$. The 1st derivative can be contracted in 6 ways. The 2nd derivative can only be contracted in 4 ways, since the 5th possibility would create a self-loop. Similarly, the 3rd derivative can only be contracted in 2 ways. Next consider the 3 derivatives $\frac{\delta}{\delta J}$ on the 2nd cubic vertex $Y$. Here all remaining contractions $3!=6$ work. In total there are $6\cdot 4\cdot 2\cdot 6=288$ contractions. $\Box$


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