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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?

nomeruk
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1 Answers1

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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.

References

  1. Itzykson C., Zuber J.-B. Quantum field theory, ยง7-1-2.
AccidentalFourierTransform
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