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Question on Time-independent perturbation theory: I am confused

  1. Sep 26, 2010 #1
    We all know from time-independent perturbation theory that if we have an atom in ground state [0>, and when a time-independent perturbation acts on it, the energy of the ground state gets shifted and the ground state wave function also gets modified. Using Time-independent Schroedinger eq.,

    [H0 + lambda . V] [0> = [E0] [0>, where V is the perturbation hamiltonian.

    Now we expand E0 as E0 = E0(0) + lambda . E0(1) + lambda^2. E0(2) +...
    and [0> as [0> = [0>(0) + lambda. [0>(1) + lambda^2. [0>(2) + ...

    Then we compare powers of lambda, left multiply with <0] to get the energy corrections and so on.

    My question is, is the average energy of the atom E0 = E0(0) + E0(1) + E0(2) + ....

    or is it [<0](0) + <0](1) + <0](2) + ....] [ H0 + lambda V] [[0>(0) + [0>(1) + [0>(2) + ....]

    Which one is it? They give different result.
  2. jcsd
  3. Sep 27, 2010 #2


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    They shouldn't be different to the order you calculated the wavefunction. However, it is a general theorem that the wavefunction of order n is sufficient to calculate the energy to order 2n+1 with the second formula you gave.
  4. Sep 27, 2010 #3
    Yeah I was hoping they should be equal. But in the second formula we get terms like <0(1)] H0 [0(1)>, which do not appear in the equation E = E0(0) + E0(1) + E0(2) +..., because

    E0(2) = <0] V [0(1)>
  5. Sep 27, 2010 #4


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    Be carefull, because your original wavefunction and the perturbed wavefunction do not have the same norm, i.e., although you may chose <0][sum_n O(n)>=1, <sum_n O(n)][sum_n O(n)> >1. That is you have to divide your last formula in #1 by <sum_n O(n)][sum_n O(n)>.
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