I rederived the formula for vertex correction in QED following Srednicki and got two different factors with his result (62.49) which started with the integral (62.40). Page 383. \begin{aligned} V^\mu_\text{1-loop}(p',p )= \frac{e^3}{8\pi^2} \bigg[ \bigg( \frac{\color{red}{1}}{\varepsilon} - \color{red}{1} - \frac{1}{2} \int d F_3 \, \log (D/\mu^2) \bigg) \, \gamma^\mu + \frac{1}{4} \int d F_3 \, \frac{\tilde{N}^\mu}{D} \bigg] \end{aligned} where $\int d F_3 = 2 \int_0^1 dx_1 dx_2 dx_3 \delta(x_1 + x_2 + x_3 -1)$

The following is my results, from the the first order vertex correction, we has an integral (62.40), using the Feynman parameters to connect the three fractions into single one, and using the integral formula in d-dimensional Euclidean space \begin{aligned} \int \frac{\mathrm{d}^d \bar q}{(2\pi)^2}\, \frac{(\bar q^2)^a}{(\bar q^2 + D)^b} = \frac{\Gamma(b-a -\frac{d}{2}) \Gamma(a +\frac{d}{2})}{(4 \pi)^{d/2} \, \Gamma(b) \Gamma(d/2)} \, D^{-(b-a -\frac{d}{2})} \end{aligned} Where in this case $a = 1, d =4$ and $b=3$.

Next, I performed the dimensional regularization

  1. $d \to 4 - \varepsilon$;

  2. $e \to e \tilde{\mu}^{\varepsilon/3}$

The firs part in that integral reads \begin{align} \tilde{\mu}^\varepsilon \frac{\Gamma(\frac{\varepsilon}{2}) \Gamma(3-\frac{\varepsilon}{2})}{(4 \pi)^{2 -\frac{\varepsilon}{2}} \, \Gamma(3) \Gamma(2 - \frac{\varepsilon}{2})} \left(\frac{1}{D}\right)^{\frac{\varepsilon}{2}} &= \frac{1}{2\cdot 16 \pi^2} \left(2 - \frac{\varepsilon}{2}\right) \, \Gamma\left(\frac{\varepsilon}{2}\right) \, \bigg[ \frac{4 \pi \tilde{\mu}^2}{D} \bigg]^{\frac{\varepsilon}{2}} \\ \color{green}{\Gamma\left(\frac{\varepsilon}{2}\right) = \frac{2}{\varepsilon} - \gamma + O(\frac{\varepsilon}{2})} &\Downarrow \color{green}{A^{\varepsilon/2} = 1 + \frac{\varepsilon}{2} \log A + O(\varepsilon) } \\ &= \frac{2 - \frac{\varepsilon}{2}}{32 \pi^2} \bigg[ \frac{2}{\varepsilon} -\gamma + O(\frac{\varepsilon}{2}) \bigg]\bigg[1 + \frac{\varepsilon}{2} \log \bigg(\frac{4 \pi \tilde{\mu}^2}{D} \bigg) + O(\varepsilon) \bigg] \\ &\Downarrow \text{New Definition: } \mu^2 := e^{-\gamma} \tilde{\mu}^2 4\pi \\ &= \frac{1}{32 \pi^2} \bigg[ \frac{4}{\varepsilon} - \color{red}{1} + 2 \log\bigg(\frac{\mu^2}{D} \bigg) + O(\varepsilon) \bigg] \\ &= \frac{1}{8 \pi^2} \bigg[ \frac{1}{\varepsilon} - \color{red}{\frac{1}{4}} - \frac{1}{2} \log\bigg(\frac{D}{\mu^2} \bigg) + O(\varepsilon) \bigg] \end{align} Put the above result into the integral (62.41), and the Feynman integral over the first two terms only contribute a factor $2$, and finally \begin{align} V^\mu_\text{1-loop} (p',p) = \frac{e^3}{8\pi^2} \bigg[\frac{\color{red}{2}}{\varepsilon} -\color{red}{\frac{1}{2}} - \frac{1}{2} \int d F_3 \log\bigg(\frac{D}{\mu^2} \bigg) \cdots \bigg] \end{align}

where the red parts highlight the difference between my calculation and Srednicki's textbook. Please comment the difference and correct my calculation if I got some mistakes.

  • 3
    $\begingroup$ Nice use of LaTeX :) $\endgroup$ – JamalS Nov 13 '16 at 10:05
  • $\begingroup$ I don't think that this is what's going on here, but I want to point it out anyway: some people use $d=4-\epsilon$ and some others use $d=4-2\epsilon$. This may be the origin of $1/\epsilon$ vs. $2/\epsilon$ (but I doubt so). On the other hand, Srednicki might have implicitly absorbed some finite part into the counterterms (note that a finite part $1/2$ is to all practical purposes equivalent to any other numerical constant, such as $1$ or anything else). Yet again, I don't think that this is what's going on here (the issue must lie somewhere else...) $\endgroup$ – AccidentalFourierTransform Nov 13 '16 at 12:21

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