Phi^4 theory two-loop contributions

[gobbles]

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 expression for it is
$\int\frac{d^4k}{k^2-m^2}\frac{d^4q}{q^2-m^2}$
with a high degree of divergence ($\Lambda^4$). Am I correct?

 

[The_Duck]

Depending on the renormalization scheme, this diagram might be completely canceled by a counterterm diagram. Recall that there is a counterterm introduced to cancel the one-loop quadratic divergence in the propagator. When computing the two-loop propagator you have to include a diagram like yours but with the upper loop replaced by this counterterm. This might completely cancel your diagram.

 

[vanhees71]

Yes, but it's a nice very simple example for renormalization. Of course, it's just the tadpole diagram squared. So you can calculate the tadpole diagram (say in dimensional regularization). Then you'll see that there are divergences going like $A/\epsilon^2$, where $A$ is some constant. This you can subtract. It's the overall divergence. But then there are terms like $B/\epsilon \ln(m^2/\mu^2)$, where $m$ is the mass of the particle and $\mu$ the renormalization scale (entering the dim.-reg. regularization scheme to keep the coupling constant dimensionless at all space-time dimensions). These you get rid of by using the one-loop counter terms for the two sub-divergences, where one counterterm is the tadpole loop on top of the "8" and the other the coupling-constant counter term from the lower loop of the "8". After this you should get rid of all negative powers of eps, and thus you are at a finite result. For the on-shell scheme, where you consider $m$ as the renormalized mass of the particle, you get 0 for the "double tadpole" as for the "single tadpole" diagram.