2
$\begingroup$

In Section 7.1 in Peskin and 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, $$[p\!\!/-m_0-\Sigma(p\!\!/)]|_{p\!\!/=m}=0$$ ($m_0$ 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, $$\delta m=m-m_0=\Sigma_2(p\!\!/=m)\approx\Sigma_2(p\!\!/=m_0)$$ 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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.