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Time independent perturbation theory

  1. Jan 17, 2016 #1
    In my course notes for atomic physics, looking at time independent perturbation for the non-degenerate case, we have the following:

    http://i.imgur.com/ao4ughk.png

    However I am confused about the equation 5.1.6. We know that < phi n | phi m > = 0 for n =/= m, so shouldn't this mean that < phi n | V | phi m > = 0 as well? Would someone be able to explain why this is not the case? I feel like I must be missing some important aspect of Dirac notation here.

    Thanks
     
  2. jcsd
  3. Jan 17, 2016 #2

    blue_leaf77

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    ##|\phi_m\rangle## is the eigenstate of the unperturbed Hamiltonian, that is, the original Hamiltonian before the presence of the perturbation ##V##. This means ##|\phi_m\rangle## doesn't necessarily happen to be the eigenstate of ##V## as well.
     
  4. Jan 17, 2016 #3

    Could you elaborate a bit?
     
  5. Jan 17, 2016 #4

    blue_leaf77

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    First, I would like to ask why do you think that ##\langle \phi_n|V| \phi_m \rangle = 0## for ##m\neq n##?
     
  6. Jan 17, 2016 #5
    Because different states are orthogonal, so ##\langle \phi_n | \phi_m \rangle = 0## for ## m\neq n##
     
  7. Jan 17, 2016 #6

    blue_leaf77

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    That's the hole where you fall, ##V## is an operator, it's not just number. Operating ##V## on ##|\phi_m\rangle## will generally result in another state, i.e. ##V|\phi_m\rangle = |\psi\rangle##, where ##|\psi\rangle## may not very often be connected with ##|\phi_m\rangle## by a mere constant prefactor.
     
  8. Jan 17, 2016 #7
    Excellent, thank you very much, that makes sense now. So the operator can change the basis vectors themselves?
     
  9. Jan 17, 2016 #8

    blue_leaf77

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    Yes, because ##V## can be any operator.
     
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