[morten925]

Good evening,

I am looking into the proof of the RS perturbation theory and I do not feel comfortable with the fact that multiplying

(E_i⁰ - H₀|ψ_i⁽¹⁾>=(V-E_i¹)|i> =(V - <i|V|i>)|i>,

where V = perturbation, i = |Ψ_i⁰>, E_i⁰ is the zeroth-order energy term and H₀ is the zeroth-order hamiltonian which eigenfunctions and eigenvalues are considered known, by

<n|,

where <n| comes from the expansion of the first order wavefunction into a linear combination of n atomic orbitals and coefficients,
 
gives

(E_i⁰-E_n⁰)<n|Ψ_i¹> = <n|V|i>


Does the integral <i|V|i> equal zero? Does the E_n⁰ come from the zeroth order hamiltonian? As obvious, I am not comfortable with RSPT yet, that is why I ask on this forum.

[morten925]

It is because V is hermitian - is it not?

[mjc123]

Does the integral <i|V|i> equal zero?

No, it equals the first-order energy correction E_i¹.

Does the En0 come from the zeroth order hamiltonian?

Yes

I think you've missed out a term on the RHS of your equation

(Ei0-En0)<n|Ψi1> = <n|V|i>

If you multiply (V-E_i¹)|i>  by <n| you get
<n|V|i> - E_i¹<n|i>  = <n|V|i> - E_i¹δ_ni
So when n = i the LHS = 0 and <i|V|i> - E_i¹ = 0

[morten925]

So when n = i the LHS = 0 and <i|V|i> - E_i¹ = 0

Thanks mjc

Isn't when n≠i the Kronecker delta gives 0?

Best Regards Morten VS

[mjc123]

correct