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One-particle irreducibles in P&S

  1. Feb 24, 2008 #1
    [SOLVED] One-particle irreducibles in P&S

    I'm going through the derivation of the [tex]-i\Sigma_2 (p)[/tex] 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 [tex] \ell \equiv k-xp [/tex].

    Before the Feynman parameter was introduced, the correction had the appearance (p. 217, eq. 7.16)

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

    "... This task is simplified by noting that since [tex]D[/tex] depends only on the magnitude of [tex]\ell[/tex],
    The (...) identity follows from symmetry."
    (They have defined [tex]D \equiv \ell^2 - \Delta +i\epsilon[/tex], so in my case the numerator will be [tex]D^2[/tex])

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

    Btw: sorry about the kslash/pslash notation.

  2. jcsd
  3. Feb 24, 2008 #2


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    You are integrating over all possible values of all the components of l. So if you have a term linear in [tex] 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 [/tex] so it must vanish for the same reason that the integral

    [tex] \int dx~ \frac{x}{1+ x^3} [/tex] from minus infinity to plus infinity vanishes.
  4. Feb 26, 2008 #3
    Thanks kdv! It's all clear now.
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