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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$\begingroup$ related: physics.stackexchange.com/questions/137796/… ; see Lombardi, Phys Rev A 22 (1980) 797, forum.sci.ccny.cuny.edu/Members/lombardi/publications/… $\endgroup$– user4552Commented Nov 7, 2014 at 17:38
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1 Answer
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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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$\begingroup$ 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. $\endgroup$– DaPCommented May 2, 2013 at 8:45
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$\begingroup$ I think the momentum operator should not be necessary in Cartesian coordinates. I added a reference relating to your question. $\endgroup$– freudeCommented May 2, 2013 at 8:53
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$\begingroup$ 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. $\endgroup$– DaPCommented May 2, 2013 at 9:07
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$\begingroup$ 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. $\endgroup$– freudeCommented May 2, 2013 at 9:10