Why is atomic orbital one electron wave function? Why distance from centre is proportional to angular wave function?

Consider the following statements:



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> An atomic orbital is one electron wave function $\psi(r,\theta,\varphi)$ obtained from the solution to the Schrödinger equation.
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There two electrons in an atomic orbital, so shouldn't atomic orbital be two electron wave function? Or since we get three eigenvalues from Schrödinger's equation, not four, so atomic orbital is one electron wave function?



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> In angular wave function plots, the distance from the centre is proportional to numerical values of $\Theta\Phi$ in that direction and is not the distance from centre of the nucleus.
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Distance is from which *centre* if not from centre of nucleus? Also why distance from centre is proportional to $\Theta\Phi$?


Lastly, I guess, it is impossible to plot the variation of $\psi$ with $r,\theta,\varphi$, because it would require 4 axis (or four-dimension graph). Am I right?




OK, let's address these one by one.


1. Yes, it should be a two-electron wave function, i.e., the one that depends on **6** spatial coordinates. Just imagine the amount of trouble required to simply write it down, let alone produce a meaningful plot or do anything else with it. So we go with two one-electron functions instead (see [Hartree-Fock method](https://en.wikipedia.org/wiki/Hartree%E2%80%93Fock_method)). That's an approximation, of course, but it would suffice for now.
2. Yes, in angular wave function plots, it is the numerical value of $\Theta\Phi$ that is actually plotted by means of the distance from the origin. There is no physical reasons behind this. It is not a physics question at all. It is a question about drawings made by humans on pieces of paper. We may plot whatever we like against whatever else we like. If you find the type of plot which gives you better intuition of what the [spherical harmonics](https://en.wikipedia.org/wiki/Spherical_harmonics) look like, you are welcome to use those. (Hint: people have been using these for quite a while now.)  

The *"not the distance"* part of the quote is cryptic to me. I can only guess it is a botched attempt to point out that the plot is **not related at all** to the radial part of the wave function, and hence to the distance *from the electron* to the origin.
3. Yes, that would require a four-dimensional graph, which we can't do on paper. Wait, we can't even do a **three**-dimensional graph on paper. That's why we resort to cross-sections and other weird plots in the first place.


So it goes.


 2


> 
> There two electrons in an atomic orbital, so shouldn't atomic orbital be two electron wave function?
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No, it should not. The thing is that the actual one-electron wave function is the so-called *spin orbital* which is a function of not only spatial coordinates of a single electron $(r,\theta,\phi)$ but also of its spin coordinate ($m\_s$). Spin orbitals are constructed from spatial orbitals by using the so-called "spin up" and "spin down" functions defined as follows:
\begin{equation}
 \alpha(m\_{s}) =
 \begin{cases}
 1, & m\_{s} = +1/2 \\
 0, & m\_{s} = -1/2
 \end{cases} \, ,
 \quad \quad \quad
 \beta(m\_{s}) =
 \begin{cases}
 0, & m\_{s} = +1/2 \\
 1, & m\_{s} = -1/2
 \end{cases} \, ,
\end{equation}
and in general chemistry it is assumed that spin orbitals always come in pairs - from each spatial orbital $\chi$ two spin orbitals are constructed:
\begin{equation} \label{eq:rhf\_spin\_orbitals}
 \begin{aligned}
 \psi\_{\alpha}(r,\theta,\phi,m\_s) &= \chi(r,\theta,\phi) \alpha(m\_s) \, , \\ 
 \psi\_{\beta}(r,\theta,\phi,m\_s) &= \chi(r,\theta,\phi) \beta(m\_s) \, . 
 \end{aligned}
\end{equation}
Thus, both electrons that occupy the same spatial orbital (say, atomic one) are described by the wave functions (spin orbitals) that share exact same spatial part and this spatial part (spatial orbital) still is a one-electron function in a sense that it depends on spatial coordinates of a single electron only. Note that this apparent paradox arises because calling spatial orbitals wave functions is strictly speaking incorrect - the spin orbitals are the actual wave functions, not spatial ones.



> 
> Distance is from which centre if not from centre of nucleus? Also why
>  distance from centre is proportional to ΘΦ?
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This is tough one *without the context*. Can share more text with us or reference a book you are apparently reading?



> 
> Lastly, I guess, it is impossible to plot the variation of 𝜓 with
>  𝑟,𝜃,𝜑, because it would require 4 axis (or four-dimension graph).
>  Am I right?
>  This is obviously true. Being limited with plane sheets of paper we can only draw two-dimensional graphs.
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