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I am trying to derive the usual expression for the differential scattering cross section:

$\frac{d\sigma}{d\Omega} = \frac{q_f}{q_i}|f(\vec q_f,\vec q_i)|^2.$

I am familiar with the derivation which follows from writing the in and out wave functions as

$\psi_\mathrm{in} = e^{i\vec q_i\cdot \vec r} \\ \psi_\mathrm{out} = f(\vec q_f,\vec q_i)\frac{e^{iq_fr}}{r}$

and then equating in incident and outgoing probability currents and etc...

But I am curious if there is a way to derive the DCS using arguments relating to transition probabilities without making reference to the form of the wave functions.

For example, the transition probability in terms of the $S$ matrix is

$P(\vec q_f\leftarrow \vec q_i) = |S(\vec q_f,\vec q_i)|^2$

and I choose the normalisation where the scattering amplitude is related to the $S$ matrix by

$S(\vec q_f,\vec q_i) = \delta(\vec q_f-\vec q_i)+\frac{1}{2\pi i}f(\vec q_f,\vec q_i),$

so that when I make the usual assumption of not taking measurements in the $\theta=0$ direction we have

$P(\vec q_f\leftarrow\vec q_i) \propto |f(\vec q_f,\vec q_i)|^2,$

which is pretty close to the DCS already, so I am sure there is some argument that can be made about the transition probability $P(\vec q_f\leftarrow\vec q_i)$ which will result in the correct DCS formula. I'm just not sure how to make the connection. Does anyone have any insight?

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  • $\begingroup$ In my opinion, in and out states are exactly states and when You writing down expressions like $\exp(iqr)$, You formally writes $\langle\psi|x\rangle$ $\endgroup$ – Artem Alexandrov Jan 7 at 14:56
  • $\begingroup$ Yes of course, I have edited the post to say wave functions instead of states. $\endgroup$ – quixedjetr Jan 7 at 15:02

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