Why does uranium fission with alpha particle detector give V as momentum perturbation?

[raeed]

so I am taking a quantum mechanics course, we started taking about dispersion.
so he the lecturer gave us an example about the fission of uranium by alpha ray... he said that we should place a detector in order to detect the alpha particlee , but the detector can only detect particlees with specific energies and and also specific angles. so he said the Hamiltonian will be H=H0+V, and we will look at V as a perturbation, and thus V=∫([p[2]2m) |p)(p| [d][3]p
he said that its true because the normal hamiltonian H0 contains both continuous states and discreet states.
so my question is how did he get to V and why the H0 contiains continuous and discreet states?

 

[dextercioby]

The H₀ is the (probably simplified) Hamiltonian of the nucleus. Usually one needs a nuclear model: http://en.wikipedia.org/wiki/Atomic_nucleus#Nuclear_models to write it down. This model: http://en.wikipedia.org/wiki/Nuclear_shell_model resembles the atomic model (for the electron shells and subshells). Atomic hamiltonians are notorious to have also a continuous part of the spectrum which would correspond to the scattering states of the free electrons. His V looks like a free particle Hamiltonian (probably for the alpha particle seen as a different subsystem (the other subsystem would be the nucleus/disintegrated nucleus)).