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.