Feynman rules in momentum space?

[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 p_k each one gets a factor of [itex]\frac{i}{p_k^2 - m^2 +iε}[/itex].

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

[tex]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} [/tex]
[tex] = -\frac{\lambda}{4!}\frac{1}{(p_1^2 - m^2 +iε)^2}∫\frac{d^4k}{k^2 - m^2 +iε}[/tex] 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.

 

[Nate810]

Yes, you are on the right lines. The combinatorial factor 12 comes from the fact that there are 6 ways to connect the two external lines to each other and two ways to connect them to the internal line. This means that there are 12 possible diagrams for this particular configuration. To understand this better, it might be useful to draw out all of the possible diagrams and calculate the corresponding expressions. That way, you can see how the combinatorial factor works in practice. As for resources, I would recommend the book "Quantum Field Theory and the Standard Model" by Matthew D. Schwartz. It has a lot of example problems with solutions which should help you get a better grasp of these concepts.