(I attached the e-book link beneath)
First question is on P.220 the equation (7.27):
$$\delta m=m-m_0=\Sigma_2(p\!\!/=m)\approx\Sigma_2(p\!\!/=m_0).\tag{7.27}$$
Why taking $\Sigma_2(p\!\!/=m)\approx\Sigma_2(p\!\!/=m_0)$ in the last step? I do found some answer here: Order 1 Correction to the Electron Mass (Peskin 7.1) but I am still confused. Say, in the answer given by Prof.Legolasov, why $\Sigma(m) – \Sigma(m_0)=\mathcal{O}(e^2)$ ? I can't see how the $\Sigma(m) – \Sigma(m_0)$ will equal to something higher than order $e^2$ accroding to the answer.
Second question is on P.221 the first line of equation (7.31):
$$\delta Z_2=\frac{d\Sigma_2}{dp\!\!/}|_{p\!\!/=m}.\tag{7.31}$$
Where $\Sigma$ denotes 1PI diagram, and $\Sigma_2$ is the $\mathcal{O}(\alpha)$ (or $\mathcal{O}(e^2)$) 1PI diagram, then define: $\delta Z_2 \equiv Z_2-1$.
How can I derive (7.31) from: $$Z_2^{-1}=1-\frac{d\Sigma}{dp\!\!/}|_{p\!\!/=m}~ ? \tag{7.26}$$
Since $Z_2^{-1}=1-\frac{d\Sigma}{dp\!\!/}|_{p\!\!/=m}$ thus shouldn't there be $\delta Z_2=Z_2-1=\frac{\frac{d\Sigma_2}{dp\!\!/}|_{p\!\!/=m}}{1-\frac{d\Sigma_2}{dp\!\!/}|_{p\!\!/=m}}$? How to justify equation (7.31)? Am I miss some approximation?
The link of book: http://home.ustc.edu.cn/~gengb/200923/Peskin,%20An%20Introduction%20to%20Quantum%20Field%20Theory.pdf
Best Answer
I am sorry, you are not wasting my time, I was just busy and commented quickly.
The point is that you have some difference between $\Sigma$ and $\Sigma_2$. Indeed, while $\Sigma$ is the full 1PI propagator when multiplied by two bare propagators, $\Sigma_2$ represents only its first order in $\alpha$.
There is indeed a logarithmically divergent part in $\delta Z_2$, and this is what we seek. First, let us recall that: \begin{equation} (\not p-m_0)-\Sigma(p)\stackrel{p^2=m^2}{\leadsto}(\not p-m_0)-\Sigma(m)-\left.\frac{\partial}{\partial \not p}\Sigma(p)\right|_{p^2=m^2}(\not p-m)+\mathcal{O}((\not p-m)^2) \end{equation} Upon using the condition that the pole is centered in $m$ (the physical mass, not $m_0$), we find that $m_0+\Sigma(m)=m$. And this gives the desired result of $(7.26)$
But observe that this worked only because we use the form $\frac{i}{\not p-m_0-\Sigma(p)}$ of the propagator, which is the full propagator. To the order $\alpha$ we have the propagator: \begin{equation} \frac{i}{\not p-m_0-\Sigma_2(p)},\,\,\,\text{With}\,\,\Sigma_2(p) \propto \mathcal{O}(\alpha) \end{equation} This means we can use a series in the parameter $\alpha$ in the following expression: \begin{align} Z_2^{-1}=1-\left.\frac{\partial}{\partial \not p}\Sigma_2(p)\right|_{p^2=m^2} \Longrightarrow& Z_2 \simeq 1+\left.\frac{\partial}{\partial \not p}\Sigma_2(p)\right|_{p^2=m^2} +\mathcal{O}(\alpha) \\ \Longrightarrow& \delta Z_2 \simeq \left.\frac{\partial}{\partial \not p}\Sigma_2(p)\right|_{p^2=m^2} +\mathcal{O}(\alpha) \end{align} Note that we now have made a series in the variable $\alpha$. So yes, this term is divergent but we use a formal series in the variable $\alpha$, and not $\not p-m$.
I hope it is a bit more clear for you now that I've put the $\mathcal{O}$ symbols where they should be. Let me know if you something is unclear for you.