In Mandl & Shaw's Quantum Field Theory (2nd edition p217), the radiative correction for the electron self-energy is:

$$ e_0^2 \Sigma(p) = \frac{\tilde{e_0}^2}{16\pi^2} (p\!\!/ -4m) \left(\frac{2}{\eta}-\gamma + \ln 4\pi\right) + e_0^2 \Sigma_c(p) $$ with $$ 16\pi^2 \Sigma_c(p) = (2m - p\!\!/) - 2\int_0^1 dz \left[p\!\!/(1-z) -2m\right] \ln \frac{m^2z -p^2 z(1-z)}{\mu^2} . $$

$\mu$ is here a mass term appearing during dimensional regularisation but is unphysical and should not be in the final result. In the book, it is said that there is no radiative correction for the external lines: $\Sigma_c( p\!\!/ = m) =0 $.

From this condition, I obtain that $\mu = e^{-2/3}m$, which means that $$ 16\pi^2 \Sigma_c(p) = (2m - p\!\!/) - \frac{4}{3}\int_0^1 dz \left[p\!\!/(1-z) -2m\right] \ln \left(z -z(1-z)\frac{p^2}{m^2}\right) . $$

Is this correct?

I am not sure that the condition $\Sigma_c( p\!\!/ = m) =0 $ is correct. As $\frac{\tilde{e_0}^2}{16\pi^2} (p\!\!/ -4m) \left(-\gamma + \ln 4\pi\right)$ is also finite, is it $\Sigma_c( p\!\!/ = m)+ \frac{\tilde{e_0}^2}{16\pi^2} (p\!\!/ -4m) \left(-\gamma + \ln 4\pi\right) =0$ instead?

  • 3
    $\begingroup$ No one scheme is more correct than another. $\endgroup$ Commented Aug 11, 2023 at 22:03

1 Answer 1


Your result is wrong. Perhaps the most important mistake is that the renormalization condition $\Sigma_c(m)=0$ is not a condition on $\mu$, but rather on the bare mass $m_0$ (and wave function renormalization). So you should not solve for $\mu$.

Your result also seems to be independent of any couplings, gauge parameters, etc. (the correct result is proportional to the fine structure constant $\alpha$, it should depend on the gauge fixing condition and on an infrared regulator, etc). So this is another point where your solution fails.

For the correct result, together with a detailed calculation, see ref 1.


  1. Itzykson C., Zuber J.-B. Quantum field theory, §7-1-2.
  • $\begingroup$ This result is obtained after setting the infrared regulator to 0. For the gauge fixing condition, do you mean the ultraviolet regulator? In the book, the Pauli-Villars regularization is used (ultraviolet cut-off). What is the problem with using the dimensional regularization? $\endgroup$
    – nomeruk
    Commented Aug 13, 2023 at 16:18
  • $\begingroup$ 1) you cannot set the IR regulator to zero, the self-energy has a log divergence as you take such limit. 2) For gauge fixing, I mean gauge fixing, not UV regulator. For example, gauge dependence is reflected in $\xi$ dependence in a particular choice of gauge, the so-called $R_\xi$ gauge (denoted by $\lambda$ in ref 1). 3) There is no problem in dim-reg. $\endgroup$ Commented Aug 13, 2023 at 23:37
  • $\begingroup$ Okay but after renormalization, your result should not depend on any gauge parameter, right? Does the renormalization that we can carry out later remove all gauge-dependent terms? $\endgroup$
    – nomeruk
    Commented Aug 14, 2023 at 14:43
  • $\begingroup$ The result should depend on the gauge parameter since $\Sigma$ is not gauge invariant (it is the two-point function of $\psi$, which transforms non-trivially under gauge transformations, see e.g. this PSE post). $\endgroup$ Commented Aug 14, 2023 at 19:28
  • $\begingroup$ Thank you, I think I understand it better now. So if I understood correctly, $\Sigma$ is not gauge invariant but it can decomposed into 2 terms: an infinite term that is gauge independent and will be included in the renormalized mass and charge and a finite term that is gauge independent and correspond to the actual radiative corrections, right? $\endgroup$
    – nomeruk
    Commented Aug 15, 2023 at 15:52

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