- #1
maverick280857
- 1,789
- 5
Quantum Mechanical Scattering (3D) -- What about forward scattering?
Hi
In nonrelativistic QM, the standard way to solve the 3D scattering problem is to consider an incident plane wave [itex]\psi(z) = Ae^{ikz}[/itex] which encounters a scattering potential, producing an outgoing spherical wave having the asymptotic form
[tex]\psi(r, \theta, \phi) = A\left\{e^{ikz} + f(\theta,\phi)\frac{e^{ikr}}{r}\right\}[/tex]
This of course corresponds to
[tex]\frac{d\sigma(\theta,\phi)}{d\Omega} = |f(\theta, \phi)|^2[/tex]
My question is: in the "forward" direction ([itex]\theta = 0[/itex]) we have both the scattered wave as well as some of the incident wave that hasn't been scattered. But the differential cross section above only corresponds to the scattered wave. How does one mathematically account for the ticks a detector placed in the forward direction will register due to the un-scattered incident wave?
Thanks.
Hi
In nonrelativistic QM, the standard way to solve the 3D scattering problem is to consider an incident plane wave [itex]\psi(z) = Ae^{ikz}[/itex] which encounters a scattering potential, producing an outgoing spherical wave having the asymptotic form
[tex]\psi(r, \theta, \phi) = A\left\{e^{ikz} + f(\theta,\phi)\frac{e^{ikr}}{r}\right\}[/tex]
This of course corresponds to
[tex]\frac{d\sigma(\theta,\phi)}{d\Omega} = |f(\theta, \phi)|^2[/tex]
My question is: in the "forward" direction ([itex]\theta = 0[/itex]) we have both the scattered wave as well as some of the incident wave that hasn't been scattered. But the differential cross section above only corresponds to the scattered wave. How does one mathematically account for the ticks a detector placed in the forward direction will register due to the un-scattered incident wave?
Thanks.