# Hydrogen wave function in momentum space

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?

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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 '14 at 17:38

## 1 Answer

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.

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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