# Order 1 Correction to the Electron Mass (Peskin & Schroeder Section 7.1)

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?

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.