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Homework Help: Is Schiff's Quantum Mechanics wrong? Degenerate stationary perturbation theory.

  1. Nov 6, 2014 #1
    1. The problem statement, all variables and given/known data

    However incorrect the text seems to me, I suspect there's something I'm missing, since it's a renowned text: Schiff - Quantum Mechanics 3rd edition 1968.

    The topic is degenerate stationary perturbation theory. In this example there's only two eigenfunctions associated with the same eigenvalue of the unperturbed Hamiltonian. The perturbed energy states are found by solving the secular equation we know well from the general theory.

    Schiff says this at page 249:


    The problem lies in the minus sign within the discriminant:

    <m|H'|m> - <l|H'|l>

    According to me, it should be plus.

    2. Relevant equations

    ( <m|H'|m> - W_1 ) a_m + <m|H'|l> a_l = 0
    <l|H'|m> a_m + ( <l|H'|l> - W_1 ) a_l = 0

    The discriminant seems to have (<m|H'|m> + <l|H'|l>)^2 as the first term instead of (<m|H'|m> - <l|H'|l>)^2 .

    Thus, for the degeneracy to be NOT removed at the first order, the matrix elements must be <l|H'|m> = - <m|H'|l> , and not the same as the text states. Unless I'm wrong, which I suspect.

    3. The attempt at a solution

    Clearly the term <l|H'|m><m|H'|l> within the discriminant is ignored as it is second-order.

    However I put it, the Hamiltonian matrix elements and the perturbed energy value to be solved appear with the same sign respectively, in the secular equation. Moreover, as stated in the text, the diagonal values of the perturbed Hamiltonian matrix are real, and they are the values we are looking at.

    Didn't use latex because the equations are very basic. Thanks!
  2. jcsd
  3. Nov 6, 2014 #2


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    Staff: Mentor

    There is no error. Redo the algebra carefully, and you will see that no term is dropped and that Schiff uses
    \left(H_{mm}' + H_{ll}' \right)^2 - 4 H_{mm}' H_{ll}' = \left( H_{mm}' - H_{ll}' \right)^2
  4. Nov 6, 2014 #3
    Ah, damn, I had a prejudice on the dropping of that term. Thanks.
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