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On page 183 of Peskin and Schroeder, we have the following scattering cross section

$$\tag{6.23}d\sigma(p\rightarrow p'+\gamma)=d\sigma(p\rightarrow p')\cdot \int\,\frac{d^3k}{(2\pi)^3}\frac{1}{2k}\sum_{\lambda=1,2}e^2\bigg|\frac{p'\cdot\epsilon^{(\lambda)}}{p'\cdot k}-\frac{p\cdot\epsilon^{(\lambda)}}{p\cdot k}\bigg|$$

My question is that $\frac{1}{2k}$ makes no sense because $k$ is a 4 vector. Should it be $\frac{1}{2k^0}$ instead?

Then Peskin and Schroeder claims that the differential probability of radiating a photon with momentum $k$, given that the electron scatters from $p$ to $p'$ is

$$\tag{6.24}d(\text{prob})=\frac{d^3k}{(2\pi)^3}\sum_{\lambda}\frac{e^2}{2k}\bigg|\boldsymbol{\epsilon_\lambda}\cdot\Big(\frac{\boldsymbol{p'}}{p'\cdot k}-\frac{\boldsymbol{p}}{p\cdot k}\Big)\bigg|^2$$

How does 6.23 give 6.24? In 6.23 we have four vector dot products $p'\cdot \epsilon^{(\lambda)}$ whereas in 6.24 we have three vector dot products.

Peskin and Schroeder then goes on to say that when we integrate over the photon momentum, we have

$$\tag{6.25} \text{Total probability}\approx\frac{\alpha}{\pi}\int_0^{|\boldsymbol{q}|}dk\frac{1}{k}I(\boldsymbol{v},\boldsymbol{v'})$$ where $I(\boldsymbol{v},\boldsymbol{v'})$ does not depend on $k$. But how does 6.25 follow from 6.24? In 6.24 the denominators have $p'\cdot k$.

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I will try to give some answers and I hope they are satisfactory.

With the label $k$ in the factor $\frac{1}{2k}$, I think that the authors refer to the magnitude of the three-momentum $\textbf{k}$. I think the reason there is no $k^0$ there is simply because there is $k^0$ has a specific value and it is given by $k^0=k$, where $k$ is the magnitude of $\textbf{k}$. The authors, in my opinion, try to save using the label $k^0$ for cases in which there exists an integral over the latter and hence it has not yet been identified by a specific value (such as $|\textbf{k}|$). There is also Eq. (6.25), which can be seen as evident for interpreting $k$ as the magnitude of the three-momentum $\textbf{k}$, since there the integration limits start from zero to some cutoff that is also a three-momentum magnitude...

As far as Eq. (6.24) is concerned, the latter is actually equal to $$\frac{d^3k}{(2\pi)^3}\frac{1}{2k}\sum_{\lambda=1,2}e^2\bigg|\frac{p'\cdot\epsilon^{(\lambda)}}{p'\cdot k}-\frac{p\cdot\epsilon^{(\lambda)}}{p\cdot k}\bigg|$$ with the polarization vectors corresponding to the emission of (real) photons being chosen such that their timelike components vanish (see for example the first unlabelled equation in p. 160). The authors, however, choose to represent the differential probability in the form of Eq. (6.24) instead of the form I have written here, such that there exists some similarity with the formulas describing the classical case.

I hope this helps...

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    $\begingroup$ The factor $\frac{1}{2k}$ come from phase space integration, but since we are now summing photon polarization states shouldn't it give another $\frac{1}{2}$ factor? $\endgroup$ Nov 23, 2022 at 3:50
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    $\begingroup$ Hi and thanks for the comment. I think we are summing with respect to the photon polarization, not averaging over them. So, why would we include another factor of 2 in the denominator? $\endgroup$
    – schris38
    Nov 23, 2022 at 4:58
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    $\begingroup$ Thanks a lot! It's very helpful, and by the way, I just discussed this question with my classmates, they give me this PSE question link: physics.stackexchange.com/questions/137188/… this answer perfectly solve my question. $\endgroup$ Nov 23, 2022 at 6:32
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I think when he write (6.23) he is referring / using idea of (6.19), where he insert $k$ inside (6.18), he said $k$ is energy, while the integration variable $k$ is 3vector.

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