QFT Peskin Errata: Pauli Vilars Regularization & Ward Takahashi Identity

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The forum discussion focuses on the Pauli Villars regularization technique and its consistency with the Ward-Takahashi identity as presented in Chapter 7 of Peskin & Schroeder. A user encountered an issue where their calculations yielded a result of α/4π instead of zero, prompting a request for confirmation of their approach. Another participant clarified that the error stemmed from not splitting the logarithm correctly during integration, which is crucial for achieving the expected result of zero.

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simic4
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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
 
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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.

\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]
=\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]

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

(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}

Now split the log up

\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]
=\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]

Plugging that back in gives

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

As

\int^1_0dz(1-z)=-\int^1_0dz(1-2z)\log z=\frac{1}{2}

I presumably did the same as you first time, as I got \alpha/4\pi, 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:
Hey thanks a million.

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

sim.
 
Qutie alright
 

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