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The optical theorem in Quantum Mechanics states: $$Im(f(\theta=0))= \frac{k}{4 \pi} \sigma_{out}.$$ Where $f$ is the scattering amplitude, in this case for an angle $\theta = 0$, i.e. in forward direction. $\sigma_{out}$ is the total scattering cross section.

I know that there have already been several posts about this topic but none of them explicitly explains how we can physically motivate that the imaginary part of the scattering amplitude corresponds to the scattered part of the wave. I can understand the derivation but it usually directly starts with "Let's look at the the imaginary part of ..." and then some steps lead to the theorem stated above. So why do we take the imaginary part of $f$?

Quasar
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    I think it is clear in this page https://en.wikipedia.org/wiki/Optical_theorem – Lê Dũng Apr 14 '17 at 08:01
  • https://scipp.ucsc.edu/~haber/ph216/opticalth.pdf notes on the Optical Theorem by Howard Haber. See also https://youtu.be/MtK9rIbdlis?feature=shared (MIT OpenCourseWare: Optical theorem). Closely related: https://physics.stackexchange.com/q/55063/226902 https://physics.stackexchange.com/q/200368/226902 – Quillo Feb 12 '24 at 15:50

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