Solve for Ground State Wave Function & Energy Eigenvalue: Hydrogen Atom TISE

[daveyman]

Problem:

In a practice test that I am studying, I am told to find the wave function \psi_{100} (r,\theta ,\phi) for the ground state of the hydrogen atom. This I can do.

Here is the second part of the question:
By direct substitution into the TISE, find the energy eigen value for this state.

Attempt at Solution:

Basically, all you have to do is plug the wave function into the TISE:

H\psi=E\psi

where

H=-\frac{\hbar^2}{2m}\nabla^2 + V_{hydrogen}

In the answer key, I noticed that only the r component of \nabla is considered. Why can the theta and phi terms be ignored?

Any help would be much appreciated. Thanks!

 

[Pacopag]

Look for the form of the SE in spherical coordinates (it should be in any textbook). It has a radial term and an angular term. When you see the equation, it is very clear why the angular part does not appear for the ground state.

 

[daveyman]

I'm sorry, I don't think I made my question very clear.

Yes, the TISE in spherical coordinates has both radial and angular terms, but it is not clear from the SE equation itself why the angular part is not a factor in the ground state.

I think you are referring to the fact that the first spherical harmonic does not have an angular term:

Y_0^0=(\frac{1}{4\pi})^{\frac{1}{2}}

Clearly if the angular terms of the wave function are zero (as they are in this case), you can simply ignore the angular terms in the TISE.

I guess my real question is regarding the first spherical harmonic, then. Why does it not depend on any angular terms (without going into complicated mathematics)?

My guess is that since this spherical harmonic is a "perfect" sphere, it doesn't matter where you are on that sphere.

Is this correct or naive?

 

[Pacopag]

Yes. You are right! So if you take the SE and do separation of variables, you already know the "constant" angular part of the solution. Go to the website
http://en.wikipedia.org/wiki/Particle_in_a_spherically_symmetric_potential
under the heading
2 Derivation of the radial equation
The third equation down gives the "radial" Schrödinger equation. In your case, l=m=0. It is a 1-D problem.

 

[daveyman]

Thanks!

 

[Pacopag]

Your answer to your question about the Y00 harmonic sounds reasonable. Unfortunately, I only have a mathematical answer (and it might be a load of bull&%#$). I can explain why there must be a constant spherical harmonic. We know that the spherical harmonics satisfy a completeness relation, thus they can be used as a basis in which to expand any function. But some functions are constant. So in order to represent a "constant" function in the spherical harmonic basis, then at least one of the spherical harmonics must be a constant.

 

[Pacopag]

That equation in the wiki site should be in your textbook, probably in the chapter on spherically symmetric potentials.