# Actions and Cross Sections

1. Jun 14, 2015

### Cluelessluke

Can someone point towards how to derive that the cross section is proportional to the imaginary part of the Action? Also, I thought the Action was a real number?

Thanks!

2. Jun 14, 2015

### fzero

You are probably referring to the Optical Theorem. In that case, $S$ is not the action but the scattering matrix (S-matrix), which is basically $S = e^{iHt}$. An explanation of the scattering matrix and Optical Theorem can be found in http://www.itp.phys.ethz.ch/research/qftstrings/archive/12HSQFT1/Chapter10.pdf [Broken].

Last edited by a moderator: May 7, 2017
3. Jun 14, 2015

### Cluelessluke

Thanks for the reply! To be more specific, I'm referring to equation (14) in http://arxiv.org/pdf/1206.5311v2.pdf.

They have an e^{-2Im(S)} contribution in their cross section (where I believe S in the action not the S-matrix) and I'm having a hard time seeing where it comes from.

Last edited: Jun 14, 2015
4. Jun 14, 2015

### fzero

Their equation (3) expresses the cross-section in terms of the S-matrix and they credit reference [7] with a calculation in the path-integral formalism that introduces the action. It is natural in the path-integral formulation that the action would appear,. Afterwards, they suggest that the expression is dominated by a saddle-point in a certain limit that takes $g\rightarrow 0$. This saddle-point approximation is closely related to the WKB approximation that should be familiar from ordinary QM. What is happening is that, in this limit, the classical paths (critical points of the action) dominate the path integral, so the path integral expression can be approximated by their result $\exp W$. As to why the action can be complex, I would suggest looking at their references for the details that they're clearly leaving out. There is some discussion of working in the Euclidean formalism, but I can't follow them well enough to give a concrete explanation.

You should try to understand the details of their arguments (perhaps some of their references might give further details), but you should know that the fact that they can express the cross section in terms of the imaginary part of the action is not a general rule. The Optical Theorem is general, but the expression from this paper relies on this physical problem having the correct properties to allow the saddle point approximation to work. There are many examples of physics problems where the saddle point approximation is useful, so it's worth learning why it works here. However the statement you present in your OP is most definitely not true in general.

5. Jun 14, 2015

### Cluelessluke

Great! Thanks so much for your help!