Time-Independent Perturbation Theory

[eep]

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...

 

[eep]

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