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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?

Qmechanic
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GotchaP
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1 Answers1

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

GotchaP
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