# One-particle irreducibles in P&amp;S

1. Feb 24, 2008

### auditor

[SOLVED] One-particle irreducibles in P&amp;S

I'm going through the derivation of the $$-i\Sigma_2 (p)$$ correction in Peskin & Shcroeder. On the top of page 218, eq. 7.17 they say
$$-i\Sigma_{2}\left(p\right) = -e^{2}\int_{0}^{1}dx\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{-2x pslash+4m_{0}}{\left[\ell^{2}-\Delta+i\epsilon\right]^{2}}$$
where the denominator is O.K. Have that $$\ell \equiv k-xp$$.

Before the Feynman parameter was introduced, the correction had the appearance (p. 217, eq. 7.16)
$$-i\Sigma_{2}\left(p\right)=\left(-ie\right)^{2}\int\frac{d^{4}k}{\left(2\pi\right)^{4}}\gamma^{\mu}\frac{i\left(kslash+m_{0}\right)}{k^{2}-m_{0}^{2}+i\epsilon}\gamma_{\mu}\frac{-i}{\left(p-k\right)^{2}-\mu^{2}+i\epsilon}$$

To make a long story short: I get the numerator to be $$-2\gamma^{\mu}\ell_{\mu}-2x pslash+4m_{0}$$ as I suspect the authors did to. But they're dropping the linear terms in $$\ell$$, and I think the reason is to be found on p. 191, eq. 6.45.

"... This task is simplified by noting that since $$D$$ depends only on the magnitude of $$\ell$$,
$$\int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{\ell^{\mu}}{D^{3}}=0$$
The (...) identity follows from symmetry."
(They have defined $$D \equiv \ell^2 - \Delta +i\epsilon$$, so in my case the numerator will be $$D^2$$)

I don't really see how this follows from symmetry. Any suggestions?

Btw: sorry about the kslash/pslash notation.

Thanks!

2. Feb 24, 2008

### kdv

You are integrating over all possible values of all the components of l. So if you have a term linear in $$l_\mu [/itex], all the positive contributions will cancel out all the negative contributions. In other words, the integrand is odd under [tex] l_\mu \rightarrow - l_\mu$$ so it must vanish for the same reason that the integral

$$\int dx~ \frac{x}{1+ x^3}$$ from minus infinity to plus infinity vanishes.

3. Feb 26, 2008

### auditor

Thanks kdv! It's all clear now.