# Symmetric limit in Peskin and Schroeder

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

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 $$\underset{\epsilon \to 0}{\text{symm lim}}=\Bigl\{\frac{\epsilon^{\mu}}{\epsilon^2}\Bigr\} \to \infty$$

• I think this is the correct expression. Any comments will be welcomed. – GotchaP Nov 22 '16 at 9:16
• 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. – GotchaP Nov 22 '16 at 9:20
• There seem to be many misprints in this section 19.1 which is not still listed in the errata. See here and perhaps more. – GotchaP Nov 22 '16 at 15:13
• 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. – mike stone May 20 '19 at 17:54