Calculating p^2/2m and -e^2/r for Hydrogen Atom in 3D

[silimay]

Homework Statement

I have a question on my quantum pset relating to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom (in 3D).

The Attempt at a Solution

I started out trying to calculate the averages with $$\psi$$ ... something like, for the ground state, $$\psi = \frac{e^{-2r/a_0}}{\sqrt{\pi * a_0^3}}$$.

But then I ran into problems (when I was trying to do the <-e^2/r>) when I came up with an integral involving a term $$\frac{e^{-2r/a_0}}{r}$$. As far as I could see, this integral sort of seems to explode at the origin / at infinity. I was wondering, should I use just the radial wavefunction part, since it has an extra r factor that would make this integral possible? I was just confused basically...

 

[nrqed]

Homework Statement

I have a question on my quantum pset relating to calculating <p^2/2m> and <-e^2/r> for the first two spherically symmetric states of the hydrogen atom (in 3D).

The Attempt at a Solution

I started out trying to calculate the averages with $$\psi$$ ... something like, for the ground state, $$\psi = \frac{e^{-2r/a_0}}{\sqrt{\pi * a_0^3}}$$.

But then I ran into problems (when I was trying to do the <-e^2/r>) when I came up with an integral involving a term $$\frac{e^{-2r/a_0}}{r}$$. As far as I could see, this integral sort of seems to explode at the origin / at infinity. I was wondering, should I use just the radial wavefunction part, since it has an extra r factor that would make this integral possible? I was just confused basically...

You are not giving the details of what you calculated so it's hard to answer but you should not have any problem because the volume element $$dV = r^2 \sin \theta ~dr d\theta d\phi$$ provides an extra factor of r^2. Did you take this into account?

 

[silimay]

mmm, I didn't :) Thanks so much :)

I know it was smth silly I wasn't paying attention to ^_^

It's not dependent on theta or phi, so I should just multiply it by 4 pi after doing the integral in r, right?

 

[nrqed]

mmm, I didn't :) Thanks so much :)

I know it was smth silly I wasn't paying attention to ^_^

It's not dependent on theta or phi, so I should just multiply it by 4 pi after doing the integral in r, right?

That's correct. The integral over phi and theta of the angular part of the volume element gives 4 pi (you should do it once explicitly to see how it works out!)

Glad I could help.