If we have a one-loop diagram in $\phi ^ 3$ scalar field theory with $n$ external lines, then what is its symmetry factor?
I have drawn the diagram I am looking for, but instead of $6$ external lines, I want the diagram to have $n$ external lines. Please ignore the arrows in my diagram and assume that the external points are held fixed.
Cheng and Li's appendix gives the generic symmetry factor $S^{-1}$ with $$S=g\prod_{n\geq 2}2^{\beta}(n!)^{\alpha_n},$$ where $\alpha_n$ are the number of pairs of vertices connected by $n$ identical self-conjugate lines, $\beta$ is the number of lines connecting a vertex with itself, and $g$ is the number of permutations of vertices that leave the diagram unchanged with fixed external lines.
For your diagram, as long as the number of vertices $N>2$, all of the $\alpha_n=0$ (I suppose $\alpha_1=N$, but this doesn't affect the symmetry factor). You also have no tadpoles, so $\beta=0$. Finally, $g=1$ since you can't permute the vertices without changing the connectivity of the external lines. So the symmetry factor of the diagram is just one.
That is not to say that there aren't many ($(N-1)!$ in fact) other diagrams with the same kinematic structure that might need to be included in a final calculation of scattering amplitudes, just with permuted vertices.
