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During my QFT study I faced a problem of calculating amplitude of 3 to 3 scattering in massless $\phi^4$ theory in zero momenta limit at tree level. One of topologically distinct diagrams corresponding to this process gives a contribution of $\sim \frac{1}{(p_1+p_2+p_3)^2}$ (up to symmetry coefficients and coupling constant), where $p_1,p_2,p_3$ are momenta of incident particles. I have no idea how to take a limit $p_i \rightarrow 0$. Can anyone help?

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The process is indeed divergent in the IR (low energy) limit. There is no "trick" to take the limit as it is infinite. This is an idealization as in practice the amplitude would be cutoff by the mass of the particle and the diagram would go as $1/m^2$.

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  • $\begingroup$ In this theory we don't have any mass parameter because the theory is massless. The Lagrangian density is $\mathcal L = \frac{1}{2}(\partial_\mu \phi)^2 - \frac{\lambda}{4!}\phi^4$ $\endgroup$ – user43283 Mar 26 '14 at 15:02
  • $\begingroup$ Actually this problem arose from finding an effective low energy potential in theory with spontaneous symmetry breaking with Lagrangian density $\mathcal L = \vert \partial_\mu \xi \vert^2 - \frac{\lambda}{4} (\vert \xi \vert^2 -v^2)^2$. $\phi$ in the problem above is Goldstone field. $\endgroup$ – user43283 Mar 26 '14 at 15:10
  • $\begingroup$ Ah, I see. There may be a cancellation between all the diagrams that will give a finite final result (unless I'm mistaken there are many diagrams in this process). However, the amplitude you presented by itself is certainly divergent. $\endgroup$ – JeffDror Mar 26 '14 at 15:20
  • $\begingroup$ Unfortunately in my calculation there is no cancellation between all the diagrams but I don't exclude the possibility that I made miskate in calculation...Since I am novice in QFT I thought there might me some trick. Thank you for your help! $\endgroup$ – user43283 Mar 26 '14 at 15:27
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    $\begingroup$ I think that in this case, truncating to a tree diagram is not a good/consistent approximation, since higher loop results will also be divergent and hence non-negligible. In the simple-minded approach you need to "compute" and "resum" their effects. $\endgroup$ – Siva Mar 26 '14 at 18:35

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