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This question is related to this question Proof of geometric series two-point function.

Suppose we have a graph $A$ with a symmetry factor $s_1$. According to Srednicki (chapter 9, eq. (9.13)) for a graphic $A^n$ we have an addition symmetry factor $n!$ so that the total symmetry factor of the graph $A^n$ is

$$s_n=n!s_1^n.$$

Now to proof that $$G_c^{(2)}(x_1,x_2)=[p^2-m^2+\Pi(p)]^{-1}$$ whe have to show that

$$G_c^{(2)}(x_1,x_2)=G_0 + G_0\Pi G_0+G_0\Pi G_0\Pi G_0+G_0\Pi G_0\Pi G_0\Pi G_0+...$$

$\Pi$ is the some of all irreducible graph so $\Pi$ is given by

$$\Pi= A+B+C+...$$

with $A,B,C,...$ irreducible graphs.

Now if we focus only on the graph $A$ we would have (including the symmetry factors)

$$G_c^{(2)}(x_1,x_2)=G_0 + G_0(\frac{AG_0}{s_1})+\frac{1}{2!}G_0(\frac{AG_0}{s_1})^2+\frac{1}{3!}G_0(\frac{AG_0}{s_1})^3+...$$ which is a not a geometric series.

What am I doing wrong here?

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    $\begingroup$ There is no additional $n!$. I think Srednicki was talking of $A^n$ representing $n$ diagrams not connected together. Here the diagrams are connected as a chain by free propagators in between. Graph automorphism counted by the symmetry factor must leave the external legs fixed. So by induction that fixes the neighbor of say the first leg, and then the neighbor of that neighbor, etc. So you can't permute the $A$'s. $\endgroup$ – Abdelmalek Abdesselam Nov 16 '18 at 15:12
  • $\begingroup$ @Abdelmalek Abdesselam: That seems like an answer. $\endgroup$ – Qmechanic Nov 17 '18 at 14:20

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