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I'm slightly befuddled by is what it means when I'm asked to

Draw the Feynman diagram in momentum space for the two point function of $\frac{\lambda}{3!}\phi^3$ theory for order $O(\lambda^2).$

I can draw Feynman diagrams, and I thought two-point function meant

$$\langle0\|\phi(x)\phi(y)\|0\rangle$$

and what I know about $ O(\lambda^2)$ is that it will have more diagrams than $ O(\lambda).$

Other than that, I'm a bit lost. I mean, I'm not even sure if this is a really simple calculation or quite a long one.

Apologies to myself if anything I've written above is embarrassing.

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  • $\begingroup$ Nothing to be embarrassed about. It means draw all Feynman diagrams with two external legs and two interaction vertices as each vertex contributes a power of $\lambda$. $\endgroup$ – suresh May 18 '14 at 0:42
  • $\begingroup$ @suresh Ok I know why we need two external legs, but how many of these diagrams will there be? And does $\phi^3$ theory just mean each vertex will have 3 legs attached? $\endgroup$ – Phibert May 19 '14 at 12:38
  • $\begingroup$ @Phibert Do you know the answer now? $\endgroup$ – Nanashi No Gombe Apr 25 '17 at 17:32
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Order $O(\lambda^2)$ means that your diagram includes two such $\lambda\phi^3/3!$ vertices. Since overall you would have 6 legs of which 2 are the external (you are calculating a two point function with just two external legs) you have to contract four of them. This gives you a loop diagram (Well, there is more than one loop diagram but only one type is 1PI)

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  • $\begingroup$ Following Srednicki? $\endgroup$ – rainman May 17 '14 at 18:46
  • $\begingroup$ @TwoBs Where do I start then? $\endgroup$ – Phibert May 18 '14 at 9:08

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