Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

We can seperate the wave function of an hydrogen atom in a radial and an angle part: $$ \phi_{n,l,m} (\mathbf{r}) = R_{n,l,m}(r) Y_{l,m}(\vartheta,\varphi) \, , $$ where $Y_{l,m}$ are the spherical harmonics.
My question is: How does this look like in momentum space? Is the general form preserved? Do we get as well a radial and an angle dependent part?

share|improve this question
    
related: physics.stackexchange.com/questions/137796/… ; see Lombardi, Phys Rev A 22 (1980) 797, forum.sci.ccny.cuny.edu/Members/lombardi/publications/… –  Ben Crowell Nov 7 at 17:38

1 Answer 1

up vote 3 down vote accepted

To get it in the momentum representation, one has to do the Fourier transform of this function. This reference can be useful:

http://forum.sci.ccny.cuny.edu/Members/lombardi/publications/MOMREP-H-atom.pdf/view

At the end, separation of variables after transformation to the momentum space is not trivial, and the mixing of quantum number is presented.

share|improve this answer
    
I am not convinced. The Fourier transform contains $\exp (- \mathrm{i} \mathbf{k} \cdot \mathbf{r})$ which mixes the integration of the angles and the radius. –  DaP May 2 '13 at 8:45
    
I think the momentum operator should not be necessary in Cartesian coordinates. I added a reference relating to your question. –  freude May 2 '13 at 8:53
    
Ok, so the 'forminvariance' is a consequence of the Hamiltonian in the Schrödinger equation. I guess it is not easy (and though not a good idea) to show this using a Fourier transformation. –  DaP May 2 '13 at 9:07
    
Yes, I guess you are right. Now I see from that paper, that separation of variables is a bit tricky since the mixing of quantum numbers is presented. –  freude May 2 '13 at 9:10

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.