One-particle irreducibles in P&S

[auditor]

[SOLVED] One-particle irreducibles in P&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
<br /> -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}}<br />
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)
<br /> -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}<br />

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,
<br /> \int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{\ell^{\mu}}{D^{3}}=0<br />
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!

 

[kdv]

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
<br /> -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}}<br />
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)
<br /> -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}<br />

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,
<br /> \int\frac{d^{4}\ell}{\left(2\pi\right)^{4}}\frac{\ell^{\mu}}{D^{3}}=0<br />
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!

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 l_\mu \rightarrow - l_\mu so it must vanish for the same reason that the integral <br /> <br /> \int dx~ \frac{x}{1+ x^3} from minus infinity to plus infinity vanishes.

 

[auditor]

Thanks kdv! It's all clear now.