2p wave function radial distribution

The following graph shows the values of the wave function $\psi\_{210} $ (i.e. the $2\mathrm{p}\_z$ orbital) versus the radius (divided by $a\_{0}$).


![Text](https://www.researchgate.net/profile/Jose-Ramirez-19/publication/334400965/figure/fig1/AS:779781280894976@1562925753522/Wave-function-ps210-with-respect-to-r-a-of-different-HLA-with-atomic-number-Z-It-is-Graph.ppm)


I don't understand it, however, since the $2\mathrm{p}\_z$ orbital is not spherically symmetric:


1. do these represent the values along the $z$-axis (perpendicular to the angular nodal plane), or
2. is it the average of the values all around the sphere (some areas are zero-valued, some are non-zero-valued...)?


If (1) is the case, then why isn't it specified by the graph? They could have said $\theta = 0, \phi = \pi/2$.




What you're seeing is not a *wavefunction*, thus, the title of this graph is quite misleading. As you correctly pointed out, a wavefunction has an *angular dependency* as well; in fact, wavefunctions can generally be factorised as the following:


$$\psi\_{nlm}(r, \theta, \phi) = R\_{nl}(r) \cdot Y\_{lm}(\theta, \phi)$$


$R$ and $Y$ here are respectively called the radial and angular (components of the) wavefunction.


The graph that you've shown is actually only showing the $R$ component (technically speaking, it is showing $|R|^2 r^2$, which is what we call the **radial distribution function**).


Where does this expression $|R|^2r^2$ come from? Well, the electron density at any point is given by $|\psi|^2$ — we know this from quantum mechanics. The *total electron density*, let's call it $\rho\_\mathrm{tot}$, is obtained by integrating $|\psi|^2$ over all space:


$$\begin{align}
\rho\_\mathrm{tot} &= \int\_0^{2\pi} \int\_0^{\pi} \int\_0^\infty |\psi|^2 r^2 \sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi \\
&= \int\_0^{2\pi} \int\_0^{\pi} \int\_0^\infty |R(r)Y(\theta, \phi)|^2 r^2 \sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi \\
&= 1.
\end{align}$$


[the $r^2 \sin\theta$ factor arises naturally when performing [integration in spherical coordinates](https://en.wikipedia.org/wiki/Spherical_coordinate_system#Integration_and_differentiation_in_spherical_coordinates).] Of course, the total electron density in any single wavefunction is just 1, so that's not very interesting. What's more interesting is if we were to integrate out *only* the angular components, such that what's left only contains a radial dependence, and tells you how much electron density there is at a given $r$. Essentially, what happens is that all the angular bits above integrate out to 1 (by normalisation); and you're left with an integrand of $|R|^2 r^2$.


Because we are integrating over all angles, what you get at the end is pretty much an average (or more accurately a sum): so, to directly answer your question, it's more like the second option rather than the first.


For s-orbitals (i.e. $l = m = 0$), it turns out that the angular component is simply a constant: that is to say, there is no angular dependency, and hence we have $\psi \propto R$. In this case, it is OK to conflate the two. But for other orbitals this is not permissible.


I wrote more about this before, so for further information, please see: [What is the exact definition of the radial distribution function?](https://chemistry.stackexchange.com/questions/107097/what-is-the-exact-definition-of-the-radial-distribution-function)


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