# 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?

## 1 Answer

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
• My guess is that the jettison remark refers to the external lines, which enforce the conservation of momentum and energy, but you are calculating the amplitude within them....sorry that I lack thebackground to be more specific...I have calculated simple scattering but not yet covered loops
– user198207
Jul 2, 2018 at 21:48
• @johani -- When I say jettison the factors, I mean omit them. And I don't mean than you may omit them, but that you must. The arguments of both of delta functions are exactly zero, so they are giving you nonsense. Jul 2, 2018 at 23:40
• Do I need to divide this diagram's amplitude by some symmetry factor? Jul 3, 2018 at 0:06
• 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