In Peskin and Schroeder's QFT book, on chapter 6, the book tell us how to deal with the Infrared Divergence (IR).

Here is the book's logic:

  1. For the process of Soft Bremsstrahlung, $$ \begin{aligned} \frac{d \sigma}{d \Omega}\left(p \rightarrow p^{\prime}+\gamma\left(k<E_{\ell}\right)\right)&=\left(\frac{d \sigma}{d \Omega}\right)_{0}\frac{\alpha}{2 \pi} \mathcal{I}\left(\mathbf{v}, \mathbf{v}^{\prime}\right)\log \left(\frac{E_{\ell}^{2}}{\mu^{2}}\right) \\ &=\left(\frac{d \sigma}{d \Omega}\right)_{0}\frac{\alpha}{\pi} f_{\mathrm{IR}}\left(q^{2}\right)\log \left(\frac{E_{\ell}^{2}}{\mu^{2}}\right) \end{aligned} $$ where $E_{\ell}$ is the detector limiting energy, $\mu$ is the virtual photon mass that we introduced to avoid divergence.

  2. For the process $\mathbf{p} \rightarrow \mathbf{p}^{\prime}$: $$ \begin{aligned} \frac{d \sigma}{d \Omega}\left(p \rightarrow p^{\prime}\right)&=\left(\frac{d \sigma}{d \Omega}\right)_{0}F_{1}\left(q^{2}\right)^2\\ &=\left(\frac{d \sigma}{d \Omega}\right)_{0}\left[1-\frac{\alpha}{\pi} f_{\mathrm{IR}}\left(q^{2}\right) \log \left(\frac{-q^{2} \text { or } m^{2}}{\mu^{2}}\right)\right]. \end{aligned} $$

However, in our physical process, what we measure is: $$ \frac{d \sigma}{d \Omega}\left(p \rightarrow p^{\prime}\right)+\frac{d \sigma}{d \Omega}\left(p \rightarrow p^{\prime}+\gamma\left(k<E_{\ell}\right)\right) \equiv\left(\frac{d \sigma}{d \Omega}\right)_{\text {measured }}. $$ So above two process's IR can cancel with each other to some extent.

Here comes to my question: why in the correction process of $\mathbf{p} \rightarrow \mathbf{p}^{\prime}$, $\textbf{we only use the formal factor $F_1(q^2)$, not using $F_2(q^2)$}$?

Because when we do the vertex loop calculation, we can get both correction to $F_1(q^2)$ or $F_2(q^2)$, and $F_2(q^2)$ is related with anomalous magnetic moment.

Here is some of my thought:

  1. Since their don't have divergence in $F_2(q^2)$ when we perform vertex loop correction, so we don't consider $F_2(q^2)$;

  2. Since $\Gamma^{\mu}\left(p^{\prime}, p\right)=\gamma^{\mu} F_{1}\left(q^{2}\right)+\frac{i \sigma^{\mu \nu} q_{\nu}}{2 m} F_{2}\left(q^{2}\right)$, the coefficients before $F_1(q^2)$ is $\gamma^{\mu}$, which may related with the $\mathbf{p} \rightarrow \mathbf{p}^{\prime}$ process, $\sigma^{\mu \nu} q_{\nu}$ not contribute?


1 Answer 1


You are right in pointing out the book's logic [Steps 1 & 2 in your post]. You are also right in deducing that the form factor $F_2(q^2)$ does not contribute to the infrared divergences of the loop corrected $\textbf{p}\rightarrow\textbf{p}'$ graphs and hence, the authors of P&S discard it for the study of the infrared divergent part of the loop corrections.

Remember that what we wish to study here is the infrared behavior of loop corrections in amplitudes that correspond to a fermion, scatterred off by some external potential. The tree level graph for that is simple. Adding virtual particles to the aforementioned graph can occur in two ways:

  1. Adding a virtual fermion: such graphs do not produce infrared divergences
  2. Adding a virtual photon: if the photon is SOFT (i.e. low energetic), such graphs correspond to infrared divergences, and hence, those graphs are the graphs we wish to study.

In the latter case (of the additional virtual photon being soft), the effect of adding the additional soft virtual photon in any tree level graph, corresponds to multiplying the amplitude corresponding to that tree level graph by a factor. This factor is precisely the for factor $F_1(q^2)$. $F_1(q^2)$ is the leading term in an expansion with respect to the soft momentum, carried by the additional virtual photon. The effect of adding virtual photons that are not soft, or the effects coming from adding the remaining terms of the soft expansion (that do not yield any infrared divergences), are accounted for in corrections to the tree-level amplitude that do not factor out, like $F_2(q^2)$. I suggest reading the paper of Weinberg, called "Infrared Photons and Gravitons", or Chap.13 from vol. I of its book on QFT.

Apart from the vertex corrections, there also exist the fermion self-energy graphs that may occur by adding a virtual photon onto an external fermion. Those also contain infrared divergences, but they are not considered here, as their effect is completely cancelled by renormalization (see Weinberg's vol.I in QFT, chapter 13.2 for more and see also a question I had asked about infrared divergences and self-energy graphs: Renormalization and virtual soft divergences).

Now that we have studied all the possible ways in which an infrared virtual divergence can occur, we must also study the emission of a single photon from a tree level $\textbf{p}\rightarrow\textbf{p}'$ graph. The amplitudes corresponding to that emission are also infrared divergent and we observe that the cross section corresponding to that amplitude exactly cancels the infrared divergences associated with the addition of a virtual soft photon into the tree level $\textbf{p}\rightarrow\textbf{p}'$ graph. The remaining (finite) part of the sum of the two cases is what we observe in experiments due to the finite resolution of the detectors (the detectors can not observe infinitely small frequencies/energies of the emitted photons).

So, to sum up, you are right in pointing out that $F_2(q^2)$ does not contain any infrared divergences, and is, therefore, beyond the scope of the analysis of sub-chapter 6.4 in P&S. I have just provided a more spherical, in my view, picture of what is going on here. I hope it helps. If anything is unclear, please comment.


Your Answer

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

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