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...

<br /> a_{nl}^{(2)} = \frac{1}{E_n^{(0)} - E_l^{(0)}}\sum_{k{\neq}n} \frac{H_{lk}^{&#039;}H_{kn}^{&#039;}}{E_n^{(0)} - E_l^{(0)}} - \frac{H_{nn}^{&#039;}H_{ln}^{&#039;}}{(E_n^{(0)} - E_l^{(0)})^2} - a_{nn}^{(1)}\frac{H_{ln}^{&#039;}}{E_n^{(0)} - E_l^{(0)}}<br />

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,

<br /> \psi_n^{(2)} = \sum_k a_{nk}\psi_k^{(0)}<br />

We have derived that...

<br /> a_{nl}^{(2)}(E_l^{(0)} - E_n^{(0)}) + \sum_kH_{lk}^{&#039;}a_{nk}^{(1)} - E_n^{(1)}a_{nl}^{(1)} = 0<br />

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...

<br /> a_{nk}^{(1)} = \frac{H_{kl}^&#039;}{E_n^{(0)} - E_k^{(0)}}<br />

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

<br /> a_{nn}^{(1)}H_{ln}^&#039;<br />

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

<br /> a_{nl}^{(2)}(E_l^{(0)} - E_n^{(0)}) = -a_{nn}^{(1)}H_{ln}^{&#039;} - \sum...<br />

where I have ommited the rest of the sum and the other terms. Now, obviously, we can divide by (E_l^{(0)} - E_n^{(0)}), 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.