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

Given this Feynman diagram

one loop propagator

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?

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

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.

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