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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:

enter image description here

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?

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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 $s\gg {{m}^{2}}$.

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  • $\begingroup$ 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. $\endgroup$ – johani Jul 2 '18 at 18:42
  • $\begingroup$ 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 $\endgroup$ – user198207 Jul 2 '18 at 21:48
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    $\begingroup$ @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. $\endgroup$ – Bert Barrois Jul 2 '18 at 23:40
  • $\begingroup$ Do I need to divide this diagram's amplitude by some symmetry factor? $\endgroup$ – johani Jul 3 '18 at 0:06
  • $\begingroup$ No. Your curly-A amplitude is just ${{\lambda }^{2}}\int{\tfrac{{{d}^{4}}k}{{{(2\pi )}^{4}}}}(propagator)(propagator)$. $\endgroup$ – Bert Barrois Jul 3 '18 at 10:57

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