I've been reading a book "Why Chemical Reactions happen" which devotes a chapter to the concept that each of hydrogen's orbitals can be expressed with a wavefunction, which is a function of $(x,y,z)$, or of $(r, \theta, \phi)$ when spherical polar co-ordinates are used - the book sticks to the latter.

The book has a section about the radial distribution function, which by my understanding, tells us the probability that the electron is found in a thin shell of radius r. I have a feeling the book skips over some important details to simplify the maths, but it defines the radial distribution function as:

$$\text{RDF} = |\psi|^24\pi r^2$$

This made sense for the wavefunctions that only have a radial part (since they are spherically symmetrical), but I couldn't understand why, for wavefunctions that have an angular part, we discard this angular part when calculating the radial distribution function. Surely this radial distribution function needs to take into account the fact that the wavefunction doesn't have a constant value for all points on a thin spherical shell of radius $r$?

I apologise if my writing is a little disorganised, I am still very new to quantum mechanics.


In your formula for the RDF $$|\psi|^2\cdot 4\pi r^2$$

The $\psi$ is assumed to only depend on $r$. In other words it should be $$|\psi_r|^2\cdot 4\pi r^2$$

Where the entire wavefunction is assumed to be a product of functions: $$\psi(r,\phi,\theta)=\psi_r(r)\psi_{\theta,\psi}(\theta,\phi)$$

So the RDF is just focused on the radial part. The entire wavefunction can depend on the angles, but if we are just interested in the radial part then we can define the RDF as above.

To go deeper into this, the probability to find the electron in some spherical shell means that we do not care what angle we find it at. Therefore, the probability of finding the electron at any angles $\phi$ (the azimuthal angle) and $\theta$ (the polar angle from the z-axis) within some shell between $r$ and $r+dr$ is given by integrating the probability distribution function over the volume. (the $\frac{1}{4\pi}$ factor will be explained):

$$\begin{align}P(r,r+dr) & =\frac{1}{4\pi}\int_r^{r+dr} \rho ^2\space d\rho \int_0^\pi \sin\theta\space d\theta\int_0^{2\pi}d\phi\space |\psi(\rho,\phi,\theta)|^2 \\ & = \frac{1}{4\pi}\int_r^{r+dr} \rho ^2\space d\rho \int_0^\pi \sin\theta\space d\theta\int_0^{2\pi}d\phi\space |\psi_r(\rho)|^2,|Y_l^m(\theta,\phi)|^2 \\ & = \frac{1}{4\pi}\left (\int_r^{r+dr}|\psi_r(\rho)|^2\rho^2\space d\rho\right )\cdot \left( \int_0^\pi\int_0^{2\pi}|Y_l^m(\theta,\phi)|^2\sin\theta\space d\theta\space d\phi \right) \end{align}$$

Now, the first integral of this is the radial part of the distribution function: $$r^2|\psi_r(r)|^2dr$$ (Based on the comments it seems like you might be unfamiliar with the math. The integral "goes away" because this integral is basically just the area of a rectangle of height $r^2|\psi_r(r)|^2$ and a width $dr$ if you look at the integral as a Riemann sum).

The integrand of the second integral comes from the solution to the Schrodinger equation mentioned above. (i.e. $\psi_{\theta,\phi}(\theta,\phi)=Y_l^m(\theta,\phi)$). These functions are called "Spherical Harmonics". There are actually many different normalization conventions that are used for this second integral. It looks like your book uses the convention $$\int_0^\pi\int_0^{2\pi}|Y_l^m(\theta,\phi)|^2\sin\theta\space d\theta\space d\phi=4\pi$$ Typically in physics the spherical harmonics are normalized to $1$, but everything is fine as long as all probabilities are normalized to $1$. This is why I had that factor of $\frac{1}{4\pi}$ at the beginning of the integrals, to make sure all probabilities sum to $1$.

Therefore, our probability becomes: $$P(r,r+dr)=\frac{1}{4\pi}\cdot|\psi_r|^24\pi r^2 \space dr$$

Or if you want to write out just the distribution function to integrate over $$RDF=|\psi_r|^24\pi r^2$$ $$P(r,r')=\frac{1}{4\pi}\int_r^{r'}|\psi_r(\rho)|^24\pi \rho^2\space d\rho$$ Where $P(r,r')$ Is the probability of finding the electron between $r$ and $r'$.

So we get to what I mentioned above and in the comments. We lose information about the angles because we are just interested in the radial part. In terms of probabilities we are basically saying "What is the probability of finding the electron within this fixed shell and these angles OR the fixed shell and these angles OR...." Which is why we integrate (add) over all angles, which then removes that information. We haven't "thrown out" the angles. You are still able to go back to the original integral and not integrate over all angles if needed. But if you just want to look at the overall radial behavior, you can jump to the RDF.

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  • $\begingroup$ So why can we actually discard the angular part of the wavefunction? Surely if we are trying to find the probability of an electron being found in a thin shell of radius r, we need to take into account the fact that the wavefunction isn't constant across the entire shell - this is what I was trying to convey in the last paragraph of my question. $\endgroup$ – Kevin Gu Sep 4 '18 at 23:55
  • $\begingroup$ @KevinJu if you are just looking at a shell of radius $r$ then you will have to average over the angles, and you will lose information about how the probability distribution varies with the angles. This is true of any multivariable PDF where you want to know about just one variable independent of the others. $\endgroup$ – BioPhysicist Sep 5 '18 at 2:23
  • $\begingroup$ @KevinJu On second thought, my answer has some flaws in it. Let me get back to you when I wake up in the morning. $\endgroup$ – BioPhysicist Sep 5 '18 at 3:29
  • $\begingroup$ I was thinking what maybe could be done is to take an example of a wave function that has both a radial and angular part, integrate the square of this wave function over the volume of a thin shell of radius R (by setting the limits of the r integral as R and R+dr), and then equate the result of the integral to the probability that the electron is found in this thin shell of radius R. We could then try and rearrange the result of this integral to get the square of the radial part of the wave function multiplied by 4 $\pi R^2$ multiplied by the thicknesss of the shell, dr. This would verify the $\endgroup$ – Kevin Gu Sep 5 '18 at 7:19
  • $\begingroup$ definition of the RDF, however I’m not quite sure how to do the maths yet (preuniversity at the moment). $\endgroup$ – Kevin Gu Sep 5 '18 at 7:19

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