# Symmetry factor for 1PI Feynman diagrams in $\phi^4$ theory

I am trying to understand the various factors that the Feynman amplitude will carry corresponding to the Feynman diagrams of Fig. 1 of this reference. I understand that the $n^{th}$ diagram containing $n$ propagators will contribute a factor $i^n(p^2-m^2+i\epsilon)^{-n}$. $n$ vectices will contribute a factor of $(-i\lambda)^n$.

1. They explained that $n$ external lines contribute $(\phi_c)^{2n}$. But in a scalar field theory, the external lines do not contribute anything in a Feynman diagram. Am I missing something?

2. The symmetry factor $S$ associated with such a diagram, as given in Cheng and Li's book on Gauge theory of elementary particle physics (section 6.4, page 193) is $$S=\frac{(2n)!}{2^n(2n)}.$$ If possible can someone explain this symmetry factor? In particular, why is it that the factor $(2n)!$ lives in the numerator and the factors $2^n$ and $2n$ in the denominator?

• "There is a global symmetry factor of $\frac{1}{2n}$, where $\frac{1}{n}$ comes from the symmetry of the diagram under the discrete group of rotations $\mathbf{Z}_n$ and $\frac{1}{2}$ from the symmetry of the diagram under reflection". Is it this statement you're having a problem with? – Demosthene Mar 19 '17 at 17:47
• @Demosthene I would like to understand why is it that the factor $(2n)!$ goes in the numerator and $2^n(2n)$ in the denominator. – SRS Mar 19 '17 at 19:25