Time independent perturbation theory

[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

 

[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.

 

[Plaetean]

$|\phi_m\rangle$ 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.

Could you elaborate a bit?

 

[blue_leaf77]

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

 

[Plaetean]

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

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

 

[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.

 

[Plaetean]

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.

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

 

[blue_leaf77]

So the operator can change the basis vectors themselves?

Yes, because $V$ can be any operator.