# Differential cross-section derivation from S-matrix

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?

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