Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Off-Diagonal Hamiltonian elements

  1. Dec 15, 2009 #1

    I just have a quick question about Quantum Mechanics. It's probably a bit basic but I'm trying to get my head around the off-diagonal Hamiltonian elements of a perturbation. We can assume the unperturbed Hamiltonian to be degenerate.

    If I have a Hamiltonian
    where the perturbation contains off-diagonal elements what physical meaning does that have?

    edit: When i say the off-diagonal elements I mean the off-diagonal elements of:
    <nr|H'|ns> where r,s represent the degeneracy. Sorry!

    I know not having the off-diagonal matrix elements in the perturbation remove the degeneracy (at least to first order in perturbation theory).

    If they remain what is the physical meaning? I think it has to do with the various states interating in some way but I'm not sure. Any help on this would be great!
    Last edited: Dec 15, 2009
  2. jcsd
  3. Dec 15, 2009 #2
    Actually I've just been thinking about it and I think it just has to do with interaction energies between states. Is this correct?
  4. Dec 15, 2009 #3

    You can think of it as the "mixing" between your pure energy eigenstates. If there's coupling between degenerate states, as you show in your Hamiltonian, this means that there will be a coupling between these degenerate states.

    A good physical example is spin-orbit coupling where the degenerate spin levels are coupled via Rashba Hamiltonian for instance...

    Other off-diagonal elements may correspond to scattering elements in the Hamiltonian, but of course the mixing need not be between different levels, it could be between degenerate levels (most commonly spin) also, just as in your example.
  5. Dec 15, 2009 #4
    Thanks for the reply I think It's all slowly falling into place.

    I've not heard of the Rashba Hamiltonian but I'll try find it in a text book somewhere.

    Thanks again for the help!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook