Why Does the Helium Wavefunction Require 9 Coordinates Instead of 6?

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SUMMARY

The wavefunction of Helium requires 9 coordinates due to the inclusion of both electrons' positions and the necessity to account for the nucleus's position. Each electron contributes three spatial coordinates (X, Y, Z), resulting in 6 coordinates, but the additional 3 coordinates arise from the need to describe the relative motion of the two electrons and the nucleus. This is distinct from the hydrogen atom, where the center of mass can simplify the problem to just one electron's motion relative to a stationary nucleus. The approximation of a stationary nucleus in Helium introduces a minor error, calculated as ##\frac{m_e}{m_\alpha} \approx 0.01\%##.

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sidnake
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I'm doing about wavefunctions for my course, I'm a bit confused as to why the wavefunction of Helium has 9 coordinates and time and not 6 coordinates and time. As far as I was aware the wave function was used to describe the movement of electrons around the nucleus of an atom, and it was assumed the nucleus was stationary due to the high speed of the electrons. If the electrons can move in three dimensions, X,Y and Z then there should be three coordinates for one electron. Helium has two electrons so there should be 6 coordinates. Why is it that the wavefunction of Helium actually has 9 coordinates? Do you infact need to include coordinates of the nucleus? If anyone knows where I'm going wrong, please point it out to me.
 
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As far as I was aware the wave function was used to describe the movement of electrons around the nucleus of an atom, and it was assumed the nucleus was stationary due to the high speed of the electrons.
For hydrogen, you don't need that assumption: You can separate the problem into a motion of the center of mass (and ignore this) and a relative motion between nucleus and electron - similar to the Kepler problem in classical physics. With 2 electrons, this is no longer possible. I would expect a good approximation with the assumption that the nucleus does not move (with an error of ##\frac{m_e}{m_\alpha} \approx 0.01\%##), however.
 

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