Energy derivation step with first-order perturbation correction

[rwooduk]

In class we went through the derivation of the energy of a perturbed system, I've dug my old notes out and found a bra ket derivation of the same thing, there's just one step that doesn't look right and was wondering if someone could tell me if its a misprint or actually correct (and why).

$$E_{n}^{1}\left \langle \Psi _{k}^{0} |\Psi _{n}^{0} \right \rangle = E_{n}^{0}\delta _{kn}$$

im assuming its a misprint, but then its presented in the final solution.i.e. you need everything in terms of $$E_{n}^{0}$$ for it to work.

 

[Orodruin]

In class we went through the derivation of the energy of a perturbed system, I've dug my old notes out and found a bra ket derivation of the same thing, there's just one step that doesn't look right and was wondering if someone could tell me if its a misprint or actually correct (and why).

$$E_{n}^{1}\left \langle \Psi _{k}^{0} |\Psi _{n}^{0} \right \rangle = E_{n}^{0}\delta _{kn}$$

im assuming its a misprint, but then its presented in the final solution.i.e. you need everything in terms of $$E_{n}^{0}$$ for it to work.

Indeed, $\left \langle \Psi _{k}^{0} |\Psi _{n}^{0} \right \rangle = \delta _{kn}$ so what you say should hold. Without seeing the full derivation and knowing exactly what is being argued, it is difficult to tell though.

 

[rwooduk]

Thanks for the reply! I managed to borrow my freinds scanner and have attached the derivation. I've put a red arrow (2nd file) which indicates the problem / misprint. As you can see you need it as printed to get alpha k n.

edit yes its a misprint, the term goes to zero in the following step, must be going blind! thanks for the help anyway!