Time independent perturbation theory

1. Jan 17, 2016

Plaetean

In my course notes for atomic physics, looking at time independent perturbation for the non-degenerate case, we have the following:

http://i.imgur.com/ao4ughk.png

However I am confused about the equation 5.1.6. We know that < phi n | phi m > = 0 for n =/= m, so shouldn't this mean that < phi n | V | phi m > = 0 as well? Would someone be able to explain why this is not the case? I feel like I must be missing some important aspect of Dirac notation here.

Thanks

2. Jan 17, 2016

blue_leaf77

$|\phi_m\rangle$ is the eigenstate of the unperturbed Hamiltonian, that is, the original Hamiltonian before the presence of the perturbation $V$. This means $|\phi_m\rangle$ doesn't necessarily happen to be the eigenstate of $V$ as well.

3. Jan 17, 2016

Plaetean

Could you elaborate a bit?

4. Jan 17, 2016

blue_leaf77

First, I would like to ask why do you think that $\langle \phi_n|V| \phi_m \rangle = 0$ for $m\neq n$?

5. Jan 17, 2016

Plaetean

Because different states are orthogonal, so $\langle \phi_n | \phi_m \rangle = 0$ for $m\neq n$

6. Jan 17, 2016

blue_leaf77

That's the hole where you fall, $V$ is an operator, it's not just number. Operating $V$ on $|\phi_m\rangle$ will generally result in another state, i.e. $V|\phi_m\rangle = |\psi\rangle$, where $|\psi\rangle$ may not very often be connected with $|\phi_m\rangle$ by a mere constant prefactor.

7. Jan 17, 2016

Plaetean

Excellent, thank you very much, that makes sense now. So the operator can change the basis vectors themselves?

8. Jan 17, 2016

blue_leaf77

Yes, because $V$ can be any operator.