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 $a_{0}$).


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 $\theta = 0, \phi = \pi/2$.


1 Answer 1


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.] 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?

  • $\begingroup$ I don't understand how we can have the choice to normalise the angular wave function? Is it a choice we have when solving the Shrödinger equation ? $\endgroup$
    – niobium
    Apr 5, 2022 at 14:21
  • $\begingroup$ Yes, it's conventional to separately normalise both the radial and the angular components (which then leads to the entire wavefunction being normalised). $\endgroup$ Apr 5, 2022 at 14:26

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