5
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ related: physics.stackexchange.com/q/127301 $\endgroup$
    – zzz
    Jun 24, 2015 at 6:38
  • 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... $\endgroup$
    – romanovzky
    Jun 24, 2015 at 10:18
  • 1
    $\begingroup$ essentially a duplicate of: physics.stackexchange.com/q/189185 $\endgroup$
    – J-T
    Jun 24, 2015 at 10:27

1 Answer 1

1
$\begingroup$

As far as I know, I think you mean this diagram, enter image description here

where, enter image description here

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.