1
$\begingroup$

It is often claimed in QFT literature that infrared divergences of cross-sections of various processes can and should be compensated by taking into account processes of emission of soft photons in any order of perturbation theory.

However in the sources I have seen this is proven only in lowest orders for some concrete situations. An exception is Ch. 13 of Weinberg's book "The quantum theory of fields". However the discussion there is not detailed enough for me. Is there an alternative place to read about it? The case of QED would be sufficient for me.

$\endgroup$
4
  • 1
    $\begingroup$ Are you really saying that WEINBERG is not detailed enough for you?!?! I doubt you’ll find any textbook that’s more detailed. Also, I know for a fact that Weinberg does QED to all orders in perturbation theory - I can’t remember the chapter. You’ll have to explain what exactly you’re looking for. $\endgroup$
    – Prahar
    Mar 4, 2021 at 10:10
  • $\begingroup$ I checked and its chapter 13 as you mentioned. What part of the calculation do you think is not detailed enough? $\endgroup$
    – Prahar
    Mar 4, 2021 at 10:47
  • $\begingroup$ @PraharMitra: Ok, what I do not understand in Weinberg's book is formula (13.2.3). There the integration is over the region in the space-time such that $\lambda\leq |\mathbf{q}|\leq \Lambda$ while $q^0$ is arbitrary. At the same time the integrand takes the form like in (13.1.1), i.e. when $q\to 0$ (thus $q^0$ is also small). In other words (13.2.3) should look differently for large $q^0$, in my opinion. $\endgroup$
    – MKO
    Mar 4, 2021 at 10:47
  • $\begingroup$ answered below. $\endgroup$
    – Prahar
    Mar 4, 2021 at 11:00

1 Answer 1

2
$\begingroup$

In the comments, OP clarifies that the issue is with equation (13.2.3). I was confused about this part as well when I first learnt it. The proper way to perform the calculation is as follows.

  1. First, do the $q^0$ integral by contour integration. We assume that the numerator of the integrand (which involves the lower point amplitude, the numerator of propagators, vertex factors, etc.) does not contain any poles in $q^0$. Thus, all the poles arise from the zeros of the denominator so there are four poles in $q^0$.

  2. Note that some poles are $q^0 = {\cal O}(|\vec{q}|)$ and some poles are $q^0 = {\cal O}(1)$ in the small $|\vec{q}|$ limit.

  3. Show that the ${\cal O}(1)$ poles do not contribute to the infrared divergence. More precisely, the ${\cal O}(1)$ poles give an integral which is ${\cal O}(|\vec{q}|^{-1})$. The integral over $\vec{q}$ then gives a term that is ${\cal O}(\Lambda^2)$ which vanishes in the limit $\Lambda \to 0$.

  4. With the foresight that only poles which are ${\cal O}(|\vec{q}|)$ contribute to the infrared divergence, we can expand the entire integrand in small $q^\mu$ (not just small $\vec{q}$) and therefore arrive at eq. (13.2.3).

$\endgroup$
5
  • $\begingroup$ In 1. - I am not sure what integrand you mean: that of (13.2.3) or something else. Also what does it mean "the numerator of the integrand ... does not contain any poles of $q^0$? Thank you. $\endgroup$
    – MKO
    Mar 4, 2021 at 11:23
  • $\begingroup$ In fact I have a further question. The integrand in (13.2.3) should be multiplied by an expression corresponding to the rest of the diagram. This remaining expression also depends on $q$, in particular on $q^0$. In integrating over $q^0$ this dependence should be taken into account at least for large $q^0$. But in Ch. 13 this is completely ignorred. $\endgroup$
    – MKO
    Mar 4, 2021 at 11:51
  • $\begingroup$ Your second comment essentially answers your first. Even before arriving at eq. (13.2.3) you have an integral with a complicated integrand. You start with that integrand and perform the steps I mentioned in my answer and you will then arrive at eq. (13.2.3). $\endgroup$
    – Prahar
    Mar 4, 2021 at 12:22
  • $\begingroup$ Thank you. Could you please elaborate on part 3. More specifically on the claim that “the $\mathcal{O}(1)$ poles give an integral which is $\mathcal{O}(|\vec q|^{-1}$.” $\endgroup$
    – MKO
    Mar 5, 2021 at 12:11
  • $\begingroup$ I don't know what there is to elaborate honestly. You just have to perform the calculation to see it. It's a tedious calculation, but it has to be done. $\endgroup$
    – Prahar
    Mar 5, 2021 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.