# Perturbation Theory: Calculating for the correction on the ground state energy

1. Dec 1, 2012

### jhosamelly

1. The problem statement, all variables and given/known data

2. Relevant equations

$E_{1}=<ψ_{1}|V(r)|ψ_{1}>$

3. The attempt at a solution

That is equal to the integral ∫ψVψd^3r

So I'll just perform the integral, correct ? But r is not constant here right? So, I' ll keep it inside the integral? How should I continue? Please help. Thanks.

2. Dec 2, 2012

### TSny

Right, $r$ is a variable of integration.

Something to think about: In the integral should you use the potential energy $V(r)$ as stated in the problem or the perturbation $\delta V(r)$ of the potential energy (due to switching from the potential energy of a point nucleus to the potential energy of a finite-sized nucleus)?

3. Dec 2, 2012

### jhosamelly

I really don't get your point sorry. I'm guessing I should use the perturbation of the potential. But how can I get that?

4. Dec 2, 2012

### TSny

The hydrogen atom is usually solved treating the nucleus as concentrated in a point. The ground state wavefunction that you specified was derived under this assumption.

Now you want to treat the nucleus more realistically as having a finite size and calculate a correction to the ground state energy in going from a point nucleus to the finite nucleus. The Hamiltonian for a finite nucleus can be thought of as the Hamiltonian for the point nucleus plus a "perturbation". So, the perturbation is just the difference between the Hamiltonian for a finite nucleus and the Hamiltonian for a point nucleus. You should convince yourself that the perturbation is just the change $\delta V(r)$ in the potential energy function when going from the point nucleus to the finite nucleus.

Can you find a mathematical expression for $\delta V(r)$? The potential energy for a finite nucleus is given in the problem. So, you need to remember what the potential energy function is for a point nucleus.

Last edited: Dec 2, 2012
5. Dec 2, 2012

### andrien

you should just break up the integral at r=R.see if it works.