# 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?
Thanks.

• – Qmechanic Aug 16 '17 at 11:25

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