I have been calculating probability density functions for different states in the hydrogen atom, and want to plot them in 3D using a dot density plot like this one. However, I have not figured out how to convert my function into a list distribution that can be plotted using a density plot. Here are the three functions I've been working on.

{(E^(-2 Im[ArcTan[x, y]] - Re[Sqrt[x^2 + y^2 + z^2]]) Abs[x^2 + y^2])/(896 \[Pi]), 
(2 E^(2 Im[ArcTan[x, y]] - 2/3 Re[Sqrt[x^2 + y^2 + z^2]]) Abs[Sqrt[x^2 + y^2] z]^2)/(45927 \[Pi]),
(3 E^(-4 Im[ArcTan[x, y]] - 1/2 Re[Sqrt[x^2 + y^2 + z^2]]) Abs[(x^2 + y^2) (-12 + Sqrt[x^2 + y^2 + z^2])]^2)/(29360128 \[Pi])}

These are each for the states |211>, |32,-1>, |422>.

  • $\begingroup$ BTW, the three functions are normalized to their sum, so the first should be multiplied by 14, the second by 7/2, and the third by 14/9 to be normalized. $\endgroup$
    – Timevortex
    Jun 10, 2021 at 22:12
  • $\begingroup$ Related: mathematica.stackexchange.com/questions/223007/… $\endgroup$
    – MelaGo
    Jun 11, 2021 at 1:04
  • $\begingroup$ Is 14 probs[[1]] just equivalent to $\frac{\left(x^2+y^2\right) e^{-\sqrt{x^2+y^2+z^2}}}{64 \pi }$ ? In other words, is Im[ArcTan[x, y]] always equal to zero? $\endgroup$
    – JimB
    Jun 11, 2021 at 4:26


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