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.