# Homework Help: Feynman rules in momentum space?

1. Dec 3, 2011

### jeebs

I'm trying to do the following problem:

ie. I'm trying to use the Feynman rules for momentum space to write down the mathematical expression that the diagram is supposed to represent. However, I don't feel very confident in what I've managed so far.

Now as I understand it, these diagrams represent expressions that are equal to n-point Green's functions, where n is the number of external lines. So, looking at that diagram, I would think that the 2 straight lines coming off that single vertex are the external lines, so n=2.

So, according to the Feynman rules, I assign them each a momentum pk each one gets a factor of $\frac{i}{p_k^2 - m^2 +iε}$.

Also, another of the rules states that each internal line (which I take it means that single loop here) has a momentum kj and contributes a factor of $∫\frac{d^4k_j}{2\pi^4}\frac{i}{k_j^2 - m^2 +iε}$, and the vertex (the dot, right?) contributes a factor $\frac{-iλ}{4!}2\pi\delta(\sum momenta)$. This gives:

$$G_2(p_1,p_2) = \frac{i}{p_1^2 - m^2 +iε}\frac{i}{p_2^2 - m^2 +iε}∫\frac{d^4k}{2\pi^4}\frac{i}{k^2 - m^2 +iε}\frac{-iλ}{4!}2\pi\delta(p1+p2) = -\frac{\lambda}{4!}\frac{\delta(p_1 + p_2)}{(p_1^2 - m^2 +iε)(p_2^2 - m^2 +iε)}∫\frac{d^4k}{k^2 - m^2 +iε}$$
$$= -\frac{\lambda}{4!}\frac{1}{(p_1^2 - m^2 +iε)^2}∫\frac{d^4k}{k^2 - m^2 +iε}$$ since for the delta function to be equal to 1, we need p2 = -p1 (which ensures momentum is conserved at each vertex).

Now this looks OK to me so far, but apparently I am supposed to also multiply this by a "combinatorial factor", which I have heard is supposedly 12, but I do not know how to get this, or what its significance is.
Am I on the right lines here or what?

Also, if anyone knows any good resources on the net (or elsewhere) that could help me understand this stuff better, I would appreciate it - the notes I'm learning from aren't the greatest. Ideally I could do with seeing some example problems with solutions, if you happen to know where any might be found.

Thanks.