- #1

CDL

- 20

- 1

At large separation ##r## from the scattering centre, the wavefunction for incoming particles is $$\psi(r, \theta) \approx A \left[e^{ikz} + f(\theta) \frac{e^{ikr}}{r}\right] = \psi_\text{incident} + \psi_\text{scattered}$$ He says,

"The probability that the incident particle, traveling at speed v, passes through the infinitesimal area ##d \sigma##, in time ##dt## is equal to the probability that the particle scatters into the corresponding solid angle ##d \Omega##."

He equates ##dP =|\psi_{\text{incident}}|^2 dV = |A|^2 (v dt) d\sigma## to ##dP = |\psi_{\text{scattered}}|^2 dV = \frac{|A|^2 |f|^2}{r^2} (vdt) r^2 d \Omega##

This is used to derive the relation ##\frac{d \sigma}{ d \Omega} = | f (\theta) | ^2##.

Is this the quantum analog of the classical scenario, where when a particle passes through an area ##d \sigma## it is guaranteed to pass through the solid angle ##d \Omega##? Why should this be true? I can't wrap my head around why this claim would be made. What happens, say if we know that a particle passed through an area ##d \sigma##. Would that affect the probability of it passing through the corresponding solid angle ##d \Omega##? To me it seems like the claim is plausible, but a more in depth and hopefully intuitive explanation would be appreciated.