Quick Q: First order perturbation theory derivation

[RadonX]

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 |$$\psi^{1)_{n}$$> by$$\sum_{m}|\phi_{m}><\phi_{m}|$$ to solve for $$|\psi^{(1)}_{n}>$$ the 1st order correction to the wavefunction:

Homework Equations

NA

The Attempt at a Solution

Easy peasy... I start off with;

$$\sum_{m}|\phi_{m}><\phi_{m}|\psi^{(1)}_{n}> = |\psi^{(1)}_{n}>$$
$$\sum_{m}<\phi_{m}|\psi^{(1)}_{n}>|\phi_{m}> = |\psi^{(1)}_{n}>$$

Wait, then the solution suggests 'Multiply 1st order correction by $$<\phi_{m}|$$ :
So I do;

$$<\phi_{m}|H_{0}|\psi^{(1)}_{n}> + <\phi_{m}|W|\phi_{n}> = <\phi_{m}|E^{(0)}_{n}|\psi^{(1)}_{n}> + <\phi_{m}|E^{(1)}_{n}|\phi_{n}>$$

Which, if I've done that first step correctly, should be manipulated easily to their answer:

$$<\phi_{m}|\psi^{(1)}_{n}> = \frac{<\phi_{m}|W|\phi_{n}>}{E^{(0)}_{n} - E^{(0)}_{m}}$$

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

 

[diazona]

What does $|\phi_n\rangle$ represent? Are those the eigenfunctions of the unperturbed Hamiltonian?

 

[RadonX]

What does $|\phi_n\rangle$ represent? Are those the eigenfunctions of the unperturbed Hamiltonian?

Yep, I think so.
I think they could be expressed $$|\psi_n^{(0)}>$$
(That's if I'm understanding the perturbation theory correctly? That'd be the zeroth order approximation of the perturbed Hamiltonian, ie. UNperturbed yes?

 

[diazona]

OK, that makes sense. I wasn't sure if you were using that letter to mean something else.

Anyway, you got
$$\langle\phi_m|H_0|\psi^{(1)}_n\rangle + \langle\phi_m|W|\phi_n\rangle = \langle\phi_m|E^{(0)}_n|\psi^{(1)}_n\rangle + \langle\phi_m|E^{(1)}_n|\phi_n\rangle$$
by acting on the first-order terms in the perturbation expansion with $\langle\phi_m|$. Go through each of the four terms in that expression and simplify them by pulling constant factors out to the front. You should also be able to evaluate $\langle\phi_m|H_0$. Then rearrange to solve for $\langle\phi_m|\psi^{(1)}_n\rangle$, under the assumption $E_m \neq E_n$ (which implies $m\neq n$).

For the case $m = n$, youl need to use a different argument to show that $\langle\phi_n|\psi^{(1)}_n\rangle = 0$ (at least to first order).

 

[RadonX]

Thank you very much!
After your response I made a little headway on it late last night but in my tiredness began to worry myself over a problem that didn't exist!
For the first term, although I saw where operating $$H_0$$ on $$\langle\phi_m|$$ would give me my $$E_m^{(0)}$$ term, I mistakenly kept trying to write that if it operated on the $$|\psi_n^(1)\rangle$$ I would get an eigenvalue of $$E_n^{(0)}$$. Which would of course cancel with my right hand side term... causing all sorts of trouble.
Fortunately, with a bit of sleep and breakfast, I spotted that one!

So now I have;
$$E_m^{(0)}\langle\phi_m|\psi_n^{(1)}\rangle + \langle\phi_m|W|\phi_n\rangle = E_n^{(0)}\langle\phi_m|\psi_n^{(1)} + 0$$ <(as the last term is the inner product of orthogonal terms)
All of which is easily manipulated now to get my answer!

Thank you diazona!