I have a question on page 655 of Peskin and Schroeder.
The second equation of (19.23) is discussed here.
But the first equation of (19.23) is still a mystery.
$$ \underset{\epsilon \to 0}{\text{symm lim}}=\left\{\frac{\epsilon^{\mu}}{\epsilon^2}\right\} =0 $$ How can we understand this?


1 Answer 1


Look at (19.27). $$ \bar\psi(x+\varepsilon/2)\,\Gamma\,\psi(x-\varepsilon/2) = \frac{-i}{2\pi} \mathrm{tr} \left[ \frac{\gamma^{\alpha}\epsilon_{\alpha}}{\epsilon^2} \Gamma \right]\tag{19.27} $$ where the two fermion fields are contracted.
And note the first sentence of the paragraph just below (19.27) :

Because the contraction of fermion fields is singular as $\epsilon \to 0$, the terms of order $\epsilon$ in the last line of (19.25) can give a finite contribution.

i.e. When one put $\Gamma =I$ in (19.27), one should get a divergent quantity.
So the first expression in (19.23) is misprinted. it should be replaced by $$ \underset{\epsilon \to 0}{\text{symm lim}}=\Bigl\{\frac{\epsilon^{\mu}}{\epsilon^2}\Bigr\} \to \infty$$

  • $\begingroup$ I think this is the correct expression. Any comments will be welcomed. $\endgroup$
    – GotchaP
    Nov 22, 2016 at 9:16
  • $\begingroup$ By the way, the reason for posting this question is that I cannot add a comment to this post. For I don't have 50 reputation to comment. $\endgroup$
    – GotchaP
    Nov 22, 2016 at 9:20
  • $\begingroup$ There seem to be many misprints in this section 19.1 which is not still listed in the errata. See here and perhaps more. $\endgroup$
    – GotchaP
    Nov 22, 2016 at 15:13
  • 1
    $\begingroup$ If you put $\Gamma=I$ in 19.27, you get zero because the trace of $\gamma^\mu$ is zero. If you put $\Gamma= \gamma^\nu$ you get the symmetric limit of $\epsilon^\nu/\epsilon^2$ which is zero because averaging $ \epsilon^\nu$ over all directions (this is what the "symmetric limit" means) of the $\epsilon$ vector gives zero also. $\endgroup$
    – mike stone
    May 20, 2019 at 17:54

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