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QFT Peskin Errata

  1. Jul 24, 2006 #1
    Hi,

    This is regarding showing, in ch.7, around p.220, that the Pauli Vilars regularization technique is consistent with the ward takahashi identity.

    I cannot get the following to work:

    I add eq. 7.31 to eq. 7.32 and do not get zero. I get alpha over 4 pi.
    (I am left with integral ( 1 - z) * alpha over 2 pi )


    we are supposed to show it is zero. i ve checked it and some of the preceding results a few times but cannot get it.

    what am i missing. can anyone confirm the problem?

    Id really appreciate it! its making me a little nuts.

    thanks!

    simic
     
  2. jcsd
  3. Jul 24, 2006 #2
    I shall give it a go for you, it's pretty straightforward, you've probably just made some small cock up somewhere, I do it all the time.

    [tex]\delta Z_2+\delta F_1(0)=\frac{\alpha}{2\pi}\int^1_0dz\left[-z\log\frac{z\Lambda^2}{(1-z)^2m^2+z\mu^2}+2(2-z)\frac{z(1-z)m^2}{(1-z)^2m^2+z\mu^2}+(1-z)\log\frac{z\Lambda^2}{(1-z)^2m^2+z\mu^2}+(1-z)\frac{(1-4z+z^2)m^2}{(1-z)^2m^2+z\mu^2}\right][/tex]
    [tex]=\frac{\alpha}{2\pi}\int^1_0dz\left[(1-2z)\log\frac{z\Lambda^2}{(1-z)^2m^2+z\mu^2}+\frac{(1-z^2)(1-z)m^2}{(1-z)^2m^2+z\mu^2}\right][/tex]

    Because, as I'm sure you've already worked out,

    [tex](1-z)\frac{(1-4z+z^2)m^2}{(1-z)^2m^2+z\mu^2}+2(2-z)\frac{z(1-z)m^2}{(1-z)^2m^2+z\mu^2}=\frac{(1-z^2)(1-z)m^2}{(1-z)^2m^2+z\mu^2}[/tex]

    Now split the log up

    [tex]\int^1_0dz(1-2z)\log\frac{z\Lambda^2}{(1-z)^2m^2+z\mu^2}=\int^1_0dz\left[(1-2z)\log\frac{\Lambda^2}{(1-z)^2m^2+z\mu^2}+(1-2z)\log z\right][/tex]
    [tex]=\int^1_0dz\left[(1-z)-\frac{(1-z^2)(1-z)m^2}{(1-z)^2m^2+z\mu^2}+(1-2z)\log z\right][/tex]

    Plugging that back in gives

    [tex]\delta Z_2+\delta F_1(0)=\frac{\alpha}{2\pi}\int^1_0dz\left[(1-z)+(1-2z)\log z\right]=0[/tex]

    As

    [tex]\int^1_0dz(1-z)=-\int^1_0dz(1-2z)\log z=\frac{1}{2}[/tex]

    I presumably did the same as you first time, as I got [itex]\alpha/4\pi[/itex], I forgot the extra logarithm you're left over with at the end, or you just didn't notice that P&S had split it up in the first place (if you don't split it up, i.e. leave the z in the numerator of the log, the integration by parts they performed for you diverges).
     
    Last edited: Jul 24, 2006
  4. Jul 24, 2006 #3
    Hey thanks a million.

    I forgot to split up the log, completely missed it :).

    sim.
     
  5. Jul 24, 2006 #4
    Qutie alright
     
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