(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

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    $\begingroup$ For the first point, have you considered that maybe $\Sigma(m)=\Sigma(m_0)+\mathcal{O}(e^2)$? For the second point, maybe you can say that $1\gg \left.\frac{d\Sigma}{d \not p}\right|_{\not p = m}$? $\endgroup$ Commented Dec 12, 2022 at 16:10
  • $\begingroup$ @JeanbaptisteRoux Thank you so much for the comment! For the first point, could you please be more explicit about it? Via calculate I am sure about: $\Sigma(m) - \Sigma(m_0)=Constant\times \Sigma$ but I still have no idea why $Constant\times \Sigma$ can be neglected. Fot the second point. In the last paragraph of the P.221 P&S say: This expression (refer to 7.31)is also logarithmically ultraviolet divergent. Thus how can a logarithmically divergent number to be considered $\ll$ 1 ? $\endgroup$ Commented Dec 12, 2022 at 16:33
  • $\begingroup$ @JeanbaptisteRoux I'm so sorry for wasting your time, I am now get your first point. Let me take a moment to understand the second point. $\endgroup$ Commented Dec 13, 2022 at 5:27

2 Answers 2

  1. First question:

From P & S (7.19) we know that $$\Sigma_2(p) = \alpha \int_0^1 dx (2m_0-x\not p )log\left(\frac{\Lambda^2}{(1-x)m_0^2 +x\mu^2 -x(1-x)p^2}\right)$$

Therefore upon approximating $\Sigma_2(\not p =m)$ we can elaborate it a bit more:

$$\Sigma_2(\not p =m)\approx \Sigma_2(\not p =m_0) +\frac{d\Sigma_2(\not p )}{d\not p}|_{\not p =m_0} \delta m$$

We see here that the product of $\frac{d\Sigma_2(\not p )}{d\not p}|_{\not p =m_0}$ and $\delta m$ is $\alpha \times \alpha \equiv {\cal{O}}(e^4)$. Therefore assuming a finite cutoff $\Lambda$ the difference between $\Sigma_2(\not p =m)$ and $\Sigma_2(\not p =m_0)$ is already ${\cal{O}}(\alpha^2)$, whereas P & S only considers ${\cal{O}}(\alpha)$.

  1. Second question:

We have $$Z_2 = \frac{1}{ 1- \frac{d\Sigma_2(\not p )}{d\not p}|_{\not p =m} } \approx 1 + \frac{d\Sigma_2(\not p )}{d\not p}|_{\not p =m} + \ldots $$

where the approximate equal sign is a development in a geometrical series which again assumes a development up to order $\alpha$ and $ \frac{d\Sigma_2(\not p )}{d\not p}|_{\not p =m}\ll 1$ (and a finite cutoff) . Therefore to order $\alpha$:

$$\delta Z_2 = Z_2 -1 = \frac{d\Sigma_2(\not p )}{d\not p}|_{\not p =m}$$

A last word on dealing with cut-off dependent quantities in QED. To order $\alpha$ all renormalization constants only depend logarithmicly on the cutoff. This means that even for a large cut-off the logarithmic dependence makes the effect of that large number rather small.

Perturbation theory considers the appreciation of effects order by order of the coupling constant without considering the magnitude of the cut-off. This provides excellent agreement with the experiment. But it also shows that most QFTs, in particular QED, are only valid up to a certain high energy cut-off. Therefore we also speak of an effective theory. The theory which works on all energy scales is still not found. The amazing thing about renormalization is that it provides a tool to integrate out the high energy degrees of freedom without compromising the predictions for experiments at low energy.


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.

  • $\begingroup$ You are so good! When you say the phrase "formal series", it makes everything clear! $\endgroup$ Commented Dec 13, 2022 at 9:58

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