Time independent perturbation theory (QM)

[alkamid]

Homework Statement

I'm trying to derive the second-order correction of energy in time independent perturbation theory. My professor did it the Landau's way so I'd rather use his notation (without bra and kets). I already derived the first-order correction:
[tex]E_n^{(1)}=V_{nn}=\int \psi_n^{(0)\star} \hat{V} \psi_n^{(0)}[/tex]

Now, the idea is the same, only this time we expand:
[tex]E=E^{(0)}+E^{(1)}+E^{(2)}[/tex]
[tex]c_n= c_n^{(0)}+ c_n^{(1)}+ c_n^{(2)}[/tex]

What we obtain is (for n=k):
[tex](c_n^{(0)}+c_n^{(1)}+c_n^{(2)})(E_n^{(0)}+E_n^{(1)}+E_n^{(2)}-E_n^{(0)})=\sum_m(c_m^{(0)}+c_m^{(1)}+c_m^{(2)})V_{nm}[/tex]

The result should be:
[tex]E_n^{(2)}c_n^{(0)}=\sum_{m \neq n} V_{nm} c_m^{(1)}[/tex]

Homework Equations

The Attempt at a Solution

I don't quite understand what disappears and why in my equation. [tex]c_n^{(2)}[/tex] would create higher-order corrections so I'm not surprised it's gone, same with [tex]c_m^{(2)}[/tex] on RHS. What about the rest though? For the first-order we assumed [tex]c_n^{(0)}=1[/tex] and [tex]c_m^{(1)}=0[/tex] for [tex]m \neq n[/tex], although I'm not sure what those assumptions really mean. Moreover, I don't know if we do the same assumptions for the second-order correction! I wish Landau was more descriptive.

I'd appreciate any help!

 

[mark726]

Hello,

It seems like you are on the right track with your derivation. The reason why the terms with c_n^{(2)} and c_m^{(2)} disappear is because they represent higher-order corrections, as you mentioned. In time independent perturbation theory, we are only interested in the first and second-order corrections to the energy. Therefore, we can ignore these higher-order terms and focus on the first and second-order terms.

As for the assumptions for c_n^{(0)} and c_m^{(1)}, they represent the initial state and first-order perturbation, respectively. In other words, c_n^{(0)}=1 means that the initial state is the unperturbed state, and c_m^{(1)}=0 for m \neq n means that the first-order perturbation only affects the state n.

For the second-order correction, we make the same assumptions. c_n^{(0)}=1 and c_m^{(1)}=0 for m \neq n. This means that we are only considering the effect of the second-order perturbation on the unperturbed state n.

I hope this helps clarify things for you. Let me know if you have any other questions.