# 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.

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