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Time-Independent Perturbation Theory

  1. Aug 30, 2006 #1

    eep

    User Avatar

    Hi,
    I'm working out the 2nd Edition of Quantum Mechanics by Bransden & Joachain and I'm a little puzzled by the sign of the last term in equation 8.30 on page 380, which reads...

    [tex]
    a_{nl}^{(2)} = \frac{1}{E_n^{(0)} - E_l^{(0)}}\sum_{k{\neq}n} \frac{H_{lk}^{'}H_{kn}^{'}}{E_n^{(0)} - E_l^{(0)}} - \frac{H_{nn}^{'}H_{ln}^{'}}{(E_n^{(0)} - E_l^{(0)})^2} - a_{nn}^{(1)}\frac{H_{ln}^{'}}{E_n^{(0)} - E_l^{(0)}}
    [/tex]

    They are deriving the formula for the coeffecients of the 2nd order correction of the wave function expanded into the basis of the unperturbed wavefunctions. That is,

    [tex]
    \psi_n^{(2)} = \sum_k a_{nk}\psi_k^{(0)}
    [/tex]

    We have derived that...

    [tex]
    a_{nl}^{(2)}(E_l^{(0)} - E_n^{(0)}) + \sum_kH_{lk}^{'}a_{nk}^{(1)} - E_n^{(1)}a_{nl}^{(1)} = 0
    [/tex]

    for l not equal to n.

    Ignoring the other parts of the equation, we move the sum over to the other side by subtracting it, and we know from before that...

    [tex]
    a_{nk}^{(1)} = \frac{H_{kl}^'}{E_n^{(0)} - E_k^{(0)}}
    [/tex]

    for k not equal to n. So for k equal to n we have the leftover term

    [tex]
    a_{nn}^{(1)}H_{ln}^'
    [/tex]

    which has a minus sign in front of it when moved to the other side, so we have

    [tex]
    a_{nl}^{(2)}(E_l^{(0)} - E_n^{(0)}) = -a_{nn}^{(1)}H_{ln}^{'} - \sum...
    [/tex]

    where I have ommited the rest of the sum and the other terms. Now, obviously, we can divide by [itex](E_l^{(0)} - E_n^{(0)})[/itex], pull out a negative sign, and switch the order of the subtraction. This makes the overall term positive, does it not? The rest of the summation is not negative so I don't understand why this term would be...
     
    Last edited: Aug 30, 2006
  2. jcsd
  3. Sep 1, 2006 #2

    eep

    User Avatar

    Well, some other people seem to think it's a plus as well, including the professor. So that's that solved.
     
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