# Calculating the symmetry factor of the sunset Feynman diagram

Edit: This is in $$\phi^4$$ theory.

Given this Feynman diagram

And using this formula to calculate the symmetry factor

$$S = v\prod_{k}(k!)^{\pi _{k}}$$

I calculate: $$v = 1$$, as you can only change the vertices labels once and retain the diagram's topology.

$$\pi _{0} = 0$$ as there are 0 pairs of vertices connected by 0 identical propagators.

$$\pi _{1} = 0$$ similarly

$$\pi _{2} = 1$$, as the pair are connected by 2 identical propagators.

$$\pi _{3}$$ = $$\pi _{4} = 0$$ as above.

This gives an overall symmetry factor of 2.

However, in the solution, $$\pi _{2} = 0$$, and $$\pi _{3} = 1$$, giving an overall symmetry factor of 6.

From my understanding of $$\pi _{k}$$, I don't see how this can be, as there aren't 3 identical propagators connecting this pair. What makes them identical?

• Could You please write down the anwer for symmetric factor and clarify does theory consist factor $1/4!$ in interaction term? – Artem Alexandrov Jan 11 at 13:58

You can interchange the 3 lines connecting the 2 vertices in $$3!=6$$ different ways. This argument is the same as counting how many different ways 3 people can sit in 3 seats. The complete rules for calculating the symmetry factors of Feynman diagrams can be found in Chapter 4 of Peskin & Schroeder.