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The 1s orbital in polar coordinates is given by: $Ψ=2(1/a_0)^2*e^{-r/a_0}$ I am confused if $r^2$ here is $x^2+y^2$ or $x^2+y^2+z^2$(here z is nothing but $Ψ$ itself) .When my professor graphed this function,it looked like a sheet with a hump in the middle.If i substitute $r^2$=$x^2+y^2$ and plot the function, it matches the prfoessor's graph.Substituting the other option gives the wrong answer.But I learnt in the theory of spherical polar coordinates that $r^2$=$x^2+y^2+z^2$. Basically,I want to convert the polar system to a cartesian system.Where am I going wrong? enter image description here

Dániel G.
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coconutmercury
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  • This is basically the same question as the previous one. Your professor just chose to plot the xOy cross section of $\Psi$. In this plane $z=0$, but it still exists. In fact, it might just as easily be $xOz$ or $yOz$, since the function is spherically symmetric. – Ruslan Jan 27 '21 at 06:28
  • The professor didn't plot the cross section;He plotted the entire 3d function on a 3d graph plotter – coconutmercury Jan 27 '21 at 06:39
  • A plot of the entire function would be a 4D hypersurface. He's surely plotted a cross section, and either didn't declare it (thinking it's obvious), or you've missed it. In any case, (especially if you still don't believe me) just ask the professor. – Ruslan Jan 27 '21 at 06:43
  • i've added the picuture to the answer – coconutmercury Jan 27 '21 at 06:44
  • Yeah, this doesn't contradict what I said. In any cross section that goes through $(x,y,z)=(0,0,0)$ this function will look the same due to its symmetry. – Ruslan Jan 27 '21 at 06:46

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