I'm studying one loop contribution for electron vertex function form Peskin and Schroeder's book " An introduction to quantum field theory " Section: 6.3. I have some troubles with Pauli- Villars regularization and getting the final results, so any help will be appreciated ..

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Starting from:

$I = \delta\Gamma^\mu(p',p)= 2 i e^2 \int \frac{d^4l}{(2\pi)^4} \int^1_0 dx dy dz \delta (x+y+z-1) \frac{2}{D^3} ~ \times \bar{u}(p') \Big[\gamma^\mu . \Big(-\frac{1}{2} l^2 +(1-x)(1-y)q^2+(1-4z+z^2)m^2\Big) + i \frac{\sigma^{\mu\nu}q_\nu}{2m} (2 m^2 z (1-z)) \Big]u(p)~~~~~~~~~(6.47)$

where $D= l^2-\Delta+i\epsilon, ~~~~~~~~\Delta=-xyq^2+(1-z)^2m^2$, while :

$ \Delta_\Lambda = -xyq^2+(1-z)^2m^2+ z \Lambda^2 $.

After momentum integration and Pauli- Villars regularization, this equals:

$I= \frac{\alpha}{2\pi} \int^1_0 dx dy dz \delta (x+y+z-1) ~ \times \bar{u}(p') \Big(\gamma^\mu . \Big[ \log \frac{z\Lambda^2}{\Delta} + \frac{1}{\Delta}\Big((1-x)(1-y)q^2+(1-4z+z^2)m^2\Big)\Big] + i \frac{\sigma^{\mu\nu}q_\nu}{2m\Delta} \Big[2 m^2 z (1-z)\Big] \Big)u(p)~~~~~~~~~(6.54)$

  • Here it's suppose it substitute by $\log\frac{\Delta_\Lambda}{\Delta} = \log\frac{-xy q^2+(1-z)^2m^2+z\Lambda^2}{\Delta}$ , but why instead it substitute only by $\log\frac{z\Lambda^2}{\Delta}$ ?

  • Then how can we reach for :

$ F_1(q^2) = 1 + \frac{\alpha}{2\pi} \int^1_0 dx dy dz \delta (x+y+z-1) ~ \times \Big[ \log \Big( \frac{m^2(1-z)^2}{m^2(1-z)^2-q^2xy}\Big) + \frac{m^2(1-4z+z^2)+q^2(1-x)(1-y)}{m^2(1-z)^2-q^2xy+\mu^2z} - \frac{m^2(1-4z+z^2)}{m^2(1-z)^2+\mu^2 z}\Big] ~~~~~~~ (6.56)$

In deed I'm little bit confused, how did we get $1$ term ?, where $\log z\Lambda^2$ had gone ? now $\log m^2(1-z)^2$ in the nominator, the part of $\Delta_\Lambda$ which didn't written in the previous equation , also in this $\log$ part of the nominator, where's the $q^2$ term ?

I tried to read the book explanation, but I can not understand too much so have any one made this exercise before ?

  • $\begingroup$ Could you please include all relevant equations into your post? Without looking up the section/equation numbers you mention, I'm not really able to follow what exactly the question is. $\endgroup$
    – ACuriousMind
    Commented Jan 26, 2017 at 0:28
  • $\begingroup$ @ ACuriousMind, of course, I added the whole question, i hope it's now more clear . $\endgroup$
    – S.S.
    Commented Jan 26, 2017 at 1:23

2 Answers 2


All they've done is taken the limit of large $\Lambda$. Then \begin{align} \Delta_\Lambda=-xyq^2+(1-z)^2m^2+z\Lambda^2\rightarrow z\Lambda^2 \end{align} and \begin{align} \log\left(\frac{\Delta_\Lambda}{\Delta}\right)\rightarrow \log\left(\frac{z\Lambda^2}{\Delta} \right)+\mathcal{O}(\Lambda^{-2}). \end{align} This is what they mean when they say just after equation (6.53) that "the convergent terms are modified by terms of $\mathcal{O}(\Lambda^{-2})$, which we ignore."

  • $\begingroup$ The $\mathcal{O}(\Lambda^{-2})$ mentioned, there are really two sources for that. I think in the book they are referring to the massive photon propagator taking on a $\Lambda^{-2}$ term. They really don't explain that the $\log(\frac{\Delta_{\Lambda}}{\Delta})$ term also leads to another $\Lambda^{-2}$ term. user213887 is almost right, but to see the $\Lambda^{-2}$ term, you first have to integrate out the log. $\endgroup$
    – Pecan Lim
    Commented Jan 7 at 8:59

I only recently stumbled over the same issue, so this answer might come a bit late:

The first term in 6.56 - the 1 - appears due to the fact, that here $F_1(q^2)$ includes all corrections (to all orders), indicated by the last term, which represents terms of second order and higher in the electric coupling constant by the Bachmann-Landau-Symbol O($\alpha^2$). Hence, the 1 represents the contribution of the first order of pertubation theory to the Form Factor, as shown on in section 6.2, page 184.

Now, in order to clear up the confusion around the devious log term, one only has to write out the executed subtraction $\delta F_1(q^2) - \delta F_1(0)$ as this will then give $$ \delta F_1(q^2) - \delta F_1(0) = log ( \frac{z \Lambda^2}{\Delta(q^2)}) - log ( \frac{z \Lambda^2}{\Delta(0)}) + rest= log (\frac{z \Lambda^2}{\Delta(q^2)}\cdot ( \frac{z \Lambda^2}{\Delta(0)})^{-1}) + rest = log ( \frac{\Delta(0)}{\Delta(q^2)}) + rest = log ( \frac{2m^2(1-z)^2}{m^2(1-z)^2-q^2xy} ) +rest \ \ \ \ . $$

  • 1
    $\begingroup$ But if anyone could clarify why we here only use $\Lambda$ in the log and not $\Lambda_{\mu}$, I would be very grateful. $\endgroup$ Commented Jul 25, 2018 at 14:53
  • 1
    $\begingroup$ I seem to remember that this is because the log term has no IR divergence, so you can take $\mu=0$. $\endgroup$ Commented Jan 5, 2020 at 13:13

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