In Section 7.1 in Peskin & Schroeder, (pp. 270), the first order correction to the electron mass is calculated. They define (eq. 7.24) the physical mass $m$ of the electron as the solution to, $$[p\!\!/-m_0-\Sigma(p\!\!/)]|_{p\!\!/=m}=0\tag{7.24}$$ ($m_0$ being the bare mass, $\Sigma$ being the sum of all 1PI diagrams) and calculate (eq. 7.27) the $\mathcal{O}(\alpha)$ shift in mass as, $$\delta m=m-m_0=\Sigma_2(p\!\!/=m)\approx\Sigma_2(p\!\!/=m_0)\tag{7.27}$$ and show it to have UV divergence. Here $\Sigma_2(p)$ is the $\mathcal{O}(\alpha)$ 1PI diagram with external momentum $p$.

My question is, how do I understand the justification for the last step, i.e., taking $$\Sigma_2(p\!\!/=m)\approx\Sigma_2(p\!\!/=m_0),$$ since the difference between $m$ and $m_0$ is divergent?


1 Answer 1


It is justified because we are working to the second order in the QED coupling constant $e$.

$\Sigma(p\!\!/)$ is already of order $e^2$, thus the difference between $m$ and $m_0$ (which is also of order $e^2$) makes the induced difference $\Sigma(m) - \Sigma(m_0)$ of a higher order in $e$ than 2.

All these calculations only make sense provided a regularization parameter $\Lambda$, ofcourse, so all quantities are finite. We only take the $\Lambda \rightarrow \infty$ limit afterwards.


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