I want to set up an integral for the following diagram in $\phi^4$ theory

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my attempt is \begin{split} & 15(-i\lambda)^3\iint \frac{d^Dl}{(2\pi)^D}\frac{d^Dq}{(2\pi)^D}\frac{i}{l^2-m^2+i\epsilon}\frac{i}{q^2-m^2+i\epsilon}\frac{i}{(l+q)^2-m^2+i\epsilon} \end{split}

where there is zero external momentum.

Is the equation I wrote correct?

  • $\begingroup$ What is your question? $\endgroup$ – mike stone Jun 16 at 12:47
  • $\begingroup$ Is my approach to writing the integral representation of the loop correct? Are there other ways to write it? $\endgroup$ – sultana Jun 16 at 12:52
  • $\begingroup$ @sultana Your external legs do not carry momentum? $\endgroup$ – Prahar Jun 16 at 12:53
  • $\begingroup$ @PraharMitra yes, there is zero external momentum. $\endgroup$ – sultana Jun 16 at 12:55
  • 1
    $\begingroup$ If there is no external momenta, then all the loop propagators have momentum $\ell$ so the denominator should be $\frac{i^3 }{ [ \ell^2 - m^2 + i \epsilon ]^3 }$ $\endgroup$ – Prahar Jun 16 at 13:20

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