- #1
RadonX
- 8
- 0
Homework Statement
Going over and over the perturbation theory in various textbooks, I feel that I've NEARLY cracked it. However, in following a particular derivation I fail to understand a particular step. Could anyone enlighten me on the following?
Multiply |[tex]\psi^{1)_{n}[/tex]> by[tex]\sum_{m}|\phi_{m}><\phi_{m}|[/tex] to solve for [tex]|\psi^{(1)}_{n}>[/tex] the 1st order correction to the wavefunction:
Homework Equations
NA
The Attempt at a Solution
Easy peasy... I start off with;
[tex]\sum_{m}|\phi_{m}><\phi_{m}|\psi^{(1)}_{n}> = |\psi^{(1)}_{n}>[/tex]
[tex]\sum_{m}<\phi_{m}|\psi^{(1)}_{n}>|\phi_{m}> = |\psi^{(1)}_{n}>[/tex]
Wait, then the solution suggests 'Multiply 1st order correction by [tex]<\phi_{m}|[/tex] :
So I do;
[tex]<\phi_{m}|H_{0}|\psi^{(1)}_{n}> + <\phi_{m}|W|\phi_{n}> = <\phi_{m}|E^{(0)}_{n}|\psi^{(1)}_{n}> + <\phi_{m}|E^{(1)}_{n}|\phi_{n}>[/tex]
Which, if I've done that first step correctly, should be manipulated easily to their answer:
[tex]<\phi_{m}|\psi^{(1)}_{n}> = \frac{<\phi_{m}|W|\phi_{n}>}{E^{(0)}_{n} - E^{(0)}_{m}}[/tex]
But how? There's something about when m = n, terms cancel out. But I feel this is the important bit I'm not grasping.
I should perhaps also point out that obviously H_0 is the unperturbed hamiltonian while W is the perturbation hamiltonian but I'm sure that if you're in a position to answer this you'd already have figured that one out! :D