Undergrad Quantum Scattering Differential Probability

[CDL]

I am reading Griffiths' Introduction to Quantum Mechanics, specifically the chapter on scattering. He is discussing the scenario where an incoming beam of particles scatter off an azimuthally symmetric target.

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.

 

[azntoon]

A:It's not true in general that the probability of passing through an area $d\sigma$ is equal to the probability of scattering into a solid angle $d\Omega$. That would only be true if the cross section for the scattering process was independent of the direction of the scattered particle, i.e. $\frac{d\sigma}{d\Omega}$ was independent of $\theta$. This is obviously not true in general, since even for a simple elastic scattering process, the magnitude of the scattering angle depends on the relative momentum of the incoming and outgoing particles.However, Griffiths' statement is true in the case of azimuthally symmetric targets. In this case, the scattering cross section does not depend on the direction of the scattered particle, and thus the probability of passing through an area $d\sigma$ is equal to the probability of scattering into a solid angle $d\Omega$. This can be seen by noting that the probability of scattering into a solid angle $d\Omega$ is proportional to the product $\frac{d\sigma}{d\Omega} \times d\Omega$, where $\frac{d\sigma}{d\Omega}$ is independent of $\theta$. Since the area element $d\sigma$ is related to the solid angle element $d\Omega$ by $d\sigma = r^2 d\Omega$, the two probabilities must be equal.