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I am trying to compute the cross-section for the diagram below with a divergent triangle loop: $\qquad\qquad\qquad\qquad\qquad$enter image description here

where $X^0$ and $X^-$ are some fermions with zero and negative charge respectively. I am interested in low energy limits, so you can consider W-propagator as $\frac{i\eta_{\mu\nu}}{M_w^2}$.

When computing the amplitude, ignoring the external wave functions, you end of with an integration of the form: $$ \int \frac {k_\mu \gamma^\mu +m_-} {k^2 -m_-^2 +i\epsilon} \frac {d^4 k} {(2\pi)^4} $$ where $m_-$ is mass of $X^-$.

Any ideas how to solve this integral in terms of kinematic parameters, masses etc?

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    $\begingroup$ Well, you notice yourself that the expression diverges. And I am not sure I understand how you get rid of the dependence of the W propagators on the loop momentum. The integrals are standard, look up Passarino-Veltman functions. $\endgroup$ – user178876 Oct 3 '18 at 3:23
  • $\begingroup$ @marmot, for simplicity, I just assumed that the loop momentum is less than W boson mass. Maybe, I oversimplified the integral as that is an internal momentum. $\endgroup$ – Ramtin Oct 3 '18 at 6:36
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    $\begingroup$ You cannot make this assumption since the loop momentum gets integrated all the way up to infinity, as your formula shows. $\endgroup$ – user178876 Oct 3 '18 at 14:08
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Agree with marnot that you can't get rid of W propagators in the loop.

That being said, the integral per se $$ \int \frac {k_\mu \gamma^\mu +m_-} {k^2 -m_-^2 +i\epsilon} \frac {d^4 k} {(2\pi)^4} $$ is a typical single fermion loop (e.g. a vacuum bubble diagram), where the first term related to $$ k_\mu \gamma^\mu $$ drops out since it's odd in $k_\mu$.

The second term related to $$ m_- $$ is quadratic divergent, which amounts to zero however in dimensional regularization (one of the peculiarities of DR).

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  • $\begingroup$ Why do you say the second term is zero? $\endgroup$ – CAF Oct 5 '18 at 20:42
  • $\begingroup$ It's one of the quirks of DR. $\endgroup$ – MadMax Oct 5 '18 at 20:45
  • $\begingroup$ It’s well understood and only vanishing for certain $d$ and if there’s no scale. Here there is a scale $m$_ present so you’re not dealing with a scaleless integral. $\endgroup$ – CAF Oct 5 '18 at 20:50

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