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A question about the Hamiltonian of the fission of Uranium

  1. Dec 4, 2014 #1
    so im 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 particale , but the detector can only detect particales 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?
  2. jcsd
  3. Dec 9, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. Dec 9, 2014 #3


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    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)).
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