in QED the corrections to electron propagator change the bare electron mass from $m_0$ to $m=m_0+δm=m_0+∑(\not{p}=m)$ (Peskin, formula 7.27). This is the consequence of the fact, that the quantum corrections change the propagator from $i/(\not{p}−m0)$ to $i/(\not{p}−m_0−∑(\not{p})))$. This part is written very well in Peskin's book, formula 7.23. Then (after fomula 7.30) Peskin goes on to explain that if the fermion was massless, then the quantum corrections would never give a mass to it, in other words, δm is 0 when $m_0$ is 0.

The question is the following. When $m_0$ is 0, the quantum corrections change the massless fermion propagator from $i/\not{p}$ to $i/(\not{p}−∑(\not{p}))$, and to me this 'corrected' propagator looks like massive. Seems like the quantum corrections gave mass to the massless fermion. To make the point clearer, the quantum corrections to photon propagator leave the denominator as $(1/q^2)$ and don't make it $1/(q^2-A)$, which is what happens to massless fermion propagator. My question is only from the point of view of diagrammatic approach (and not from the point of view of left and right spinors and the mass term connecting these two spinors).

Also, when we'll get to renormalization, seems like the fermionic field renormalization via $Z_2^{-1}=1-(d\Sigma/d\not{p})|_{\not{p}=m}$ will not be enough, we also would have to introduce mass renormalization because of that $\Sigma(\not{p})$ term in the 'corrected' propagator. Am I wrong?

  • $\begingroup$ You say that the propagator $1/(p - \Sigma(p))$ looks massive. What is the mass $m$ according to you? If in doubt, recall how you can extract the mass from a two-point function. $\endgroup$
    – Vibert
    Jan 22, 2014 at 18:39

1 Answer 1


We are looking to bare quantities $\psi_B, m_B$, such as the part of the Lagrangian which does not involve the electromagnetic field may be written :

$\mathcal L_B = i \bar \psi_B \not \partial\psi_B- m_B \bar \psi_B \psi_B \tag{1}$

Now, starting from physical quantities $\psi,m$, the inverse electron propagator may be written, in a dimensional regularization procedure (neglecting the finite terms) :
$\not{p}(1+\frac{\lambda ~\alpha}{\epsilon}) - m (1 + \frac{\mu~\alpha}{\epsilon}) \tag{2}$

where $\lambda, \mu$ are some constant, $\alpha = e^2$, and $\epsilon = d-4$.

So, we have a Lagrangian :

$\mathcal L_B = i \bar \psi \not \partial (1+\frac{\lambda ~\alpha}{\epsilon})\psi- m(1 + \frac{\mu~\alpha}{\epsilon}) \bar \psi \psi \tag{3}$

The $\frac{\lambda ~\alpha}{\epsilon}$ factor may "disappear" in the redefinition of the bare fermionic field : $\psi_B = \sqrt{Z_2}\psi= \sqrt{1- \frac{\lambda ~\alpha}{\epsilon}}~\psi = (1- \frac{\lambda ~\alpha}{2\epsilon} + O(\alpha^2))~\psi \tag{4}$ If $m=0$, this is enough.

Now, if $m \neq 0$, you need also a a redefinition of the bare mass : $m_B=Z_2^{-1}m(1+\frac{\mu~\alpha}{\epsilon}) = m (1+\frac{\lambda ~\alpha}{\epsilon})(1+\frac{\mu~\alpha}{\epsilon}) = m(1+\frac{(\lambda + \mu) ~\alpha}{\epsilon} + O(\alpha^2)) \\ = m +\delta m\tag{5}$.

  • $\begingroup$ Great answer and very interesting! In your derivation you made the assumption that the change in the mass term is proportional to the mass, $m$. Is this always going to be the case? $\endgroup$
    – JeffDror
    Feb 12, 2014 at 16:33
  • $\begingroup$ I am afraid I am not able to give you a solid answer. In this particular case (electron = massive QED fermion) you have to calculate the electron self-energy Feynman diagram correction (electron line with a virtual photon), this gives a correction with a $\not{p}$ term and a $m$ term. I have the feeling that it is the same logic for all massive fermions, but I am not able to give you a solid proof (maybe a new PSE question will be interesting). $\endgroup$
    – Trimok
    Feb 13, 2014 at 18:55
  • 1
    $\begingroup$ Yes I can see that to lowest order QED. I just posted a new question, thanks! $\endgroup$
    – JeffDror
    Feb 13, 2014 at 18:56

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