# How to compute amplitudes when Feynman diagrams have loops

I would appreciate a pedagogic walk through illustrating how to calculate the amplitude associated with a single Feynman diagram having a loop.

My example concerns the $\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}-\frac{\lambda}{4!}\phi^{4}$ theory and diagram:

Now, from my understanding of Feynman rules for this diagram, I would write

$$\begin{array}{cl} \left\langle out\left|S-1\right|in\right\rangle & =\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\left(-i\lambda\right)^{2}\left(2\pi\right)^{8}\frac{i}{k^{2}-m^{2}+i\varepsilon}\frac{i}{\left(p_{1}+p_{2}-k\right)^{2}-m^{2}+i\varepsilon}\times\\ & \qquad\qquad\times\delta\left(p_{1}+p_{2}-k-p_{1}-p_{2}+k\right)\delta\left(k+p_{1}+p_{2}-k-p'_{1}-p'_{2}\right) \end{array}$$

and extract the amplitude from

$$\left\langle out\left|S-1\right|in\right\rangle =i\mathcal{A}_{out,in}\left(2\pi\right)^{4}\delta^{4}\left(\sum_{i}p'_{i}-\sum_{i}p_{i}\right).$$

However, since I get two $\delta\left(0\right)$ inside the integral, I don't know how to proceed. Can you advise?

It’s easier than you think. If you had called the momenta of the propagators ${{k}_{1}}\And {{k}_{2}}$, you would need one of those delta functions (not both) to enforce momentum conservation, but since you have already enforced it by choosing ${{k}_{2}}={{p}_{1}}+{{p}_{2}}-{{k}_{1}}$, you are free to jettison the ${{(2\pi )}^{8}}\delta (...)\delta (...)$. You are left with a log-divergent integral, which is nasty enough, but you can get the coefficient of the infinite logarithm by inspection: ${{\pi }^{2}}{{(2\pi )}^{-4}}\log ({{\Lambda }^{2}}/s)$ where Mandelstam’s $s\equiv {{({{p}_{1}}+{{p}_{2}})}^{2}}$ . This would not be exact if $\Lambda$ were finite, but it’s pretty good if $s\gg {{m}^{2}}$.
• Sorry, this is just not very explanatory and if presented with a different diagram I'm afraid I wouldn't be able to compute it since I didn't understand the reason behind "you are free to jettison the...". Besides, when you say I can do that are you talking about computing $\left\langle out\left|S-1\right|in\right\rangle$ or $\mathcal{A}_{out,in}$? If you're talking about the former, it's not apparent to me right now now how to obtain the later from my last formula. Jul 2, 2018 at 18:42
• No. Your curly-A amplitude is just ${{\lambda }^{2}}\int{\tfrac{{{d}^{4}}k}{{{(2\pi )}^{4}}}}(propagator)(propagator)$. Jul 3, 2018 at 10:57