Wherever I see calculations of two-loop contributions to the $\phi^4$ propagator (such as Peskin, page 328, on the bottom), only the sunset diagram (aka the Saturn diagram) is considered, but not, say, the two-loop diagram involving a loop on top of a loop (looks like this: _8_). Does it not contribute? As far as I can tell, it does and the loop integral for it is $$\int\frac{d^4k}{(2\pi)^4}\frac{d^4q}{(2\pi)^4}\frac{1}{(k^2−m^2)^2}\frac{1}{q^2−m^2}$$ with a high degree of divergence ($Λ^2$). Am I correct or am I missing something?
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$\begingroup$ related: physics.stackexchange.com/q/127301 $\endgroup$– zzzJun 24, 2015 at 6:38
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4$\begingroup$ They do contribute, but they are easy to compute. Due to energy-momentum conservation there is no momentum flowing between them or between the external legs, so each integral will be independent (as seen above by your expression) and reduces to a simple gamma-function result. I find it odd that P&S book would not mention that they are easy to compute and hence not discussed in the text... $\endgroup$– romanovzkyJun 24, 2015 at 10:18
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1$\begingroup$ essentially a duplicate of: physics.stackexchange.com/q/189185 $\endgroup$– J-TJun 24, 2015 at 10:27
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As far as I know, I think you mean this diagram,
And the divergent part of $I_1$ is $I_1^{\text{div}}=-\frac{m^2}{8\pi^2\epsilon}$, where $\epsilon$ is from $d=4-\epsilon$. At least we can see that $I_1^{\text{div}}\frac{\partial I_1^{\text{div}}}{\partial m^2}$ will have a divergent term like $1/\epsilon^2$. Thus, we can see this $\_8\_$ diagram will contribute to the two-loop for the correction of mass.