I'm working with two theories.

Theory A: $H_{int} =\int d^3x \frac{Mg}{2}\phi\varphi^2$

Theory B: $\phi^4$-interaction: $H_{int} = \int d^3 x \frac{\lambda}{4!}\phi(x)^4$

Where $M$ is the mass associated to the field $\phi$.

I have to calculate the $n$-points Green-function at tree-level for these theories, $G^A(p_1,…,p_n)$ , $G^B(p_1,…,p_n)$, considering the limit $M\rightarrow + \infty$ *for the scattering of scalar particles of $\varphi$ field*

*In every following diagrams, $v$ is the total number of vertices and the external propagators (indicated by the arrows) are $n$.*

**Feynman diagram for theory A**

Let's work with the theory A, trying to calculate the symmetry factor of this diagram.

If we have $n$ external propagators, then $n = v+2$. The number of propagators of field $\phi$ is $\#_{\Delta_\phi} = v/2$, while the number of $\Delta_\varphi$ is $\#_{\Delta_\varphi} = v/2-1$ and so the total number of propagator is $\#_\Delta = v-1 = n-3$.

I calculate the symmetry factor as

$$

\frac{1}{S!} = \frac{1}{2^v}2^2 3^{\#_{\Delta_\varphi}}2^{\#_{\Delta_\varphi}} = \frac{1}{2^v}2^2 6^{\frac{v}{2}-1} = \left( \frac{3}{2}\right)^{\frac{v}{2}-1}

$$

because the initial and final vertices have 2 possibles equals configurations (point $p_1$ can be contracted in two different way with the initial vertex, same thing for $p_n$) and then they contribute with a $2^2$ factor. Then, for each internal $\varphi$-propagator we have $3\cdot 2$ ways to contract external momenta and vertex legs to obtain such diagram.

*Is it true or am I missing something?*

**Feynman diagram for theory B**

In this case $n = 2(v+1)$, $\#_{\Delta_\varphi} = v-1 = (n-4)/2$.

I would say $1/S! = 1$ because for each vertex there are $4!$ possible ways to contract legs and external momenta and so,

$$

\frac{1}{S!} = \frac{1}{(4!)^v}(4!)^v = 1

$$

But I'm not sure this is correct…

*Which is the symmetry factor of such diagram? How to derive it?*

## Best Answer

The symmetry factor should be 1 in both theories. First a few remarks. If you are computing amputated diagrams the external propagators are removed from the evaluation of the amplitude, however it is unnatural (to me) to remove them from the count when considering symmetry factors. In Theory A I would say that $\#\Delta_{\varphi}=\frac{3v}{2}+1$ and in Theory B that $\#\Delta_{\varphi}=3v+1$. BTW in Theory A $n$ is $v+2$ and not $v-2$.

Wick contraction calculation in Theory A:The symmetry factor is $\frac{1}{v!}\frac{1}{2^v}C$ where $C$ is the number of Wick contractions which produce the desired shape. You have to choose which internal vertex attaches to $p_1$, $p_2$ (gives a factor $v$) then which vertex attaches to the previous one etc. So you get a factor of $v!$. Then you contract the $\phi$'s for which there are no choices. Finally, you contract the $\varphi$'s. The extreme vertices give $2^2$ as you rightly noted, but so do the vertices internal to the chain. This is because for each of the two $\varphi$ legs you need to choose who connects to the exernal leg and who connects to another chain vertex. So altogether you have $C=v!\times 2^v$.

Wick contraction calculation in Theory B:Same reasoning.

You can learn more about symmetry factors in my answer Problem understanding the symmetry factor in a feynman diagram or for the systematic theory using Joyal's theory of combinatorial species in my article "Feynman diagrams in algebraic combinatorics" but that's not an easy read.