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When calculating the self-energy correction of a massless quark up to one loop, I get $$i\Sigma(p)=i\frac{\alpha_s}{4\pi}C_F/\!\!\!{p}\left[\frac{1}{\varepsilon_{\text{UV}}}-\gamma+\ln(4\pi)+1+\ln(\frac{-\mu^2}{p^2})\right]+i/\!\!\!{p}\delta_2.$$ I assume it is easier to work with the on-shell (OS) scheme, since the diagrams with self-energy insertions on the external lines vanish that way. But when I try to set $\Sigma(p^2=0)$ to determine the OS counter-terms, the logarithm on the right obviously diverges. The same thing happens with the $\frac{\text{d}\Sigma}{\text{d}p\!\!/}|_{p^2=0}$. Is OS scheme even possible in the massless case? Or are other schemes better suited?

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Well, if the quark is massless then the self-energy of the form

one-loop quark self-energy

is scaleless (and thus vanishing in dimensional regularization) since the gluon (or the photon) is massless, the quark is massless and the on-shell condition is given by $p^2 = 0$. Thus the field renormalization constant and the mass renormalization constant vanish.

The problem you are facing is that the expansion in $\epsilon_\text{UV}$ (for $\epsilon_\text{UV} \approx 0$) does not commute with the limit $p^2 \to 0$. Actually this is a quite common problem when computing loop corrections. Whenever you go into a critical kinematical limit you have to start again from the full integral (Or you compute the integral with full $\epsilon_\text{UV}$ dependence take the kinematic limit and then expand in $\epsilon_\text{UV}$). And actually when you go back to your computation of the loop integral for $p^2 \neq 0$ you will find that at some point you actually needed this assumption in order to arrive at your expression.

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  • $\begingroup$ I understand, but how is this even possible analytically? In the expression I gave above, we have a clear UV divergence, and a logarithm term which diverges for $p^2\to0$. However, I have used the Feynman gauge. Is it possible that what you mean is only possible in Landau gauge? $\endgroup$
    – Ozzy
    Feb 19 at 14:09
  • $\begingroup$ See the edit. Furthermore this has nothing to do with the used gauge! $\endgroup$ Feb 19 at 14:17
  • $\begingroup$ I see, that was really helpful. Thank you very much! $\endgroup$
    – Ozzy
    Feb 19 at 18:48

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