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Can someone kindly help me to know how can i get electron momentum density for one orbital like home? what is the theory of electron momentum density? how can I derive electron momentum density from real electron density? how can i use Fourier transform?

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Coming from a particle physicist's perspective I find the terminology of this post difficult (different sub-fields/different vocabulary), but even leaving that aside the question is unclear. It seems you have the spacial distribution and know that the momentum distribution is connected to it by the Fourier transform. So what, exactly, is your question? –  dmckee Jul 23 '12 at 2:38
    
The momentum density of the electrons is zero in molecules in free space, because of time reversal symmetry. This means that the orbitals can be thought of as occupied states superposed with occupied time-reversed states. You can linearly combine complex orbitals which have momentum into real states, since every orbital comes with a complex conjugate. This makes the question ill posed. You can ask about electron momentum in a magnetic field, or for a given form of orbital, but if you use real orbitals, it will be zero by time-reversal invariance. –  Ron Maimon Jul 23 '12 at 7:13
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1 Answer 1

Electron momentum density $\rho(p)$ is Fourier transform of electron radial density $\rho(r)$.

$$\rho(p)=\sqrt(2/pi) (-i)^l \int j(pr) \rho(r) r^2 \mathrm{d}r $$

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Welcome, Julio. We have the MathJax rendering engine active on the site which means that you can write math at LaTeX. I've done my best to interpret your meaning here, but you should check my work. –  dmckee Aug 1 '12 at 19:06
    
This is assuming a spherically symmetric distribution of electrons. –  Ron Maimon Aug 2 '12 at 2:13
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