QFT Peskin Errata: Pauli Vilars Regularization & Ward Takahashi Identity

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Discussion Overview

This discussion revolves around the Pauli Vilars regularization technique and its consistency with the Ward-Takahashi identity as presented in chapter 7 of a textbook. Participants are examining specific equations and integrals to verify their results and address discrepancies in calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant expresses difficulty in showing that the sum of two equations results in zero, instead obtaining a non-zero value involving alpha.
  • Another participant provides a detailed breakdown of the equations and integrals involved, suggesting that the issue may stem from a missed step in splitting a logarithmic term.
  • The second participant outlines the integration steps and the necessary adjustments to arrive at zero, indicating a common mistake that can occur in these calculations.
  • A later reply acknowledges the oversight regarding the logarithmic term and expresses gratitude for the clarification.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial calculations, but there is agreement on the importance of correctly handling the logarithmic term in the integration process. The discussion reflects a shared understanding of the mathematical steps required, though initial interpretations varied.

Contextual Notes

Participants reference specific equations and integrals from the textbook, indicating that their discussion is contingent on these definitions and steps. The resolution of the problem relies on careful manipulation of logarithmic terms and integration techniques.

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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