I have been very confused by the radial distribution function which is often used in chemistry to predict the probability of finding an electron at a distance from the nucleus.

From Atkins' Physical Chemistry:

Now consider the total probability of finding the electron anywhere between the two walls of a spherical shell of thickness $\mathrm dr$ at radius $r$. The... volume of the shell... is now $4\pi r^2 \mathrm dr$. ... The probability that the electron will be found within the inner and outer surfaces of the shell is the probability density at the radius $r$ multiplied by volume... i.e $|\psi|^2\times 4 \pi r^2 \mathrm dr$. $$P(r) = 4\pi r^2|\psi(r)|^2$$ The function $P(r)$ is called the radial distribution function (in this case, for an s orbital).

Immediately after this, I find:

It is also possible to devise a more general expression which applies to orbitals that are not spherically symmetrical. ... $$P(r)=r^2R(r)^2$$

From Housecroft and Sharpe's Inorganic Chemistry:

A useful way of depicting the probability density is to plot the radial distribution function ... $$\text{Radial distribution function} = 4\pi r^2R(r)^2$$

Again, from this page, the probability density turns out to be (for the 1s orbital) $$P(r)=4\pi r^2|R(r)|^2$$ However, the integral that the answerer calculated $\bigg(\int \limits _{\theta =0}^{\pi }\int \limits _{\varphi =0}^{2\pi }|Y(\theta ,\varphi)|^2\sin \theta \,\mathrm {d} \theta \,\mathrm {d} \varphi \bigg)$ to get $4\pi$ (for 1s orbital) is calculated to be 1 (for all orbitals) in Atkins', as "the spherical harmonics are normalized" (I don't know what that signifies).

My confusion is compounded by the fact that there is another function called the probability density which is $|\psi|^2$. This function is maximum at the nucleus and falls exponentially (for an 1s orbital). The radial distribution function, however, is maximum at a certain distance from the nucleus.

So, my questions are:

  • What is the physical significance of probability distribution and that of radial distribution function?

There is a similar question here but the answer does not satisfy me. If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?

  • What is the correct expression for calculating radial distribution function?

Which begs the question, what is the definition of radial distribution function?

  • 1
    $\begingroup$ "If there is no physical difference between these two functions, then why do we use the second to find out where the electron is most likely present? Why not the first function?": The former is $p(\mathbf{r})$ and the latter is $p(r) = p(|\mathbf{r}|)$; usually the latter is more relevant. $\endgroup$ Dec 26, 2018 at 22:23

1 Answer 1


The atomic orbitals (wavefunctions) $\psi(r,\theta,\phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(\theta,\phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and angular wavefunctions, and the angular wavefunctions are also called spherical harmonics.

As a concrete example, the $2\mathrm p_z$ orbital ($\{n,l,m\} = \{2,1,0\}$) has the radial component $R_{2,1}$ and the angular component $Y_{1,0}$:

$$R_{2,1} = \sqrt{\frac{1}{24}}\left(\frac{Z}{a}\right)^{3/2}\rho\mathrm e^{-\rho/2} \qquad Y_{1,0} = \sqrt{\frac{3}{4\pi}}\cos\theta \tag{1}$$

where $a$ is a collection of constants ($a \equiv 4\pi\varepsilon_0\hbar^2/\mu e^2$), and $\rho$ is related to $r$ by another collection of constants ($\rho \equiv 2Zr/na$). The product of these two expressions is the wavefunction corresponding to the $2\mathrm p_z$ orbital.

The atomic orbitals themselves are normalised in the sense that

$$\int_{\text{all space}} |\psi(r,\theta,\phi)|^2 \mathrm dV = \int_0^{2\pi} \!\!\!\!\int_0^\pi\!\!\!\int_0^\infty |R(r)|^2|Y(\theta,\phi)|^2 r^2\sin\theta\,\mathrm dr\,\mathrm d\theta\,\mathrm d\phi = 1 \tag{2}$$

Conventionally, both the radial and angular wavefunctions are separately normalised. This means that

$$\int_0^\infty |R(r)|^2 r^2\mathrm dr = 1; \qquad \int_0^{2\pi}\!\!\!\!\int_0^\pi |Y(\theta,\phi)|^2 \sin\theta\,\mathrm d\theta\,\mathrm d\phi = 1 \tag{3}$$

(by multiplying these two together, you recover the original normalisation of the entire atomic orbital.)

Normally, since $\psi = \psi(r,\theta,\phi)$, the probability density $|\psi|^2$ depends on all three coordinates. The probability density (technically, $|\psi|^2\,\mathrm dV$) gives you the probability of finding an electron in an infinitesimal cube frustum-like shape,† located at the point $(r,\theta,\phi)$. By summing these $|\psi|^2\,\mathrm dV$ over every single point in space, i.e. by integrating over all three variables, we get the total probability of finding an electron in all space, which is $1$.

However, often we are not interested in the angular dependency, and so what we can do is to sum up all these bits which have one value of $r$, but different values of $\theta$ and $\phi$. That would give us the probability of finding the electron in a tiny spherical shell of radius $r$. Mathematically, we need to integrate over $\theta$ and $\phi$. By virtue of the angular wavefunction being normalised, the integral over $\theta$ and $\phi$ evaluates to $1$, and hence we have

$$|\psi|^2\,\mathrm dV = (|R|^2r^2\,\mathrm dr)\cdot (|Y|^2\sin\theta\,\mathrm d\theta\,\mathrm d\phi) \overset{\text{integrate over }\theta,\phi}{\longrightarrow} |R|^2r^2\,\mathrm dr \tag{4}$$

So, this quantity $|R|^2r^2\,\mathrm dr$ gives us the probability of finding the electron in a spherical shell of radius $r$. The corresponding probability density function is the function which controls how this probability varies with $r$, and we give it a special name, the radial distribution function (RDF); hence Atkins defines the RDF to be equal to $|R|^2r^2$.

This expression is valid regardless of the form of the angular wavefunction, as no matter what it is, when integrating over $\theta$ and $\phi$, it comes out to be $1$. The widely-cited alternative definition $4\pi r^2 |\psi|^2$ is only really relevant for spherically symmetric orbitals (in other words, s-orbitals). In this case, we have $\{l, m\} = \{0,0\}$ and

$$Y_{0,0} = \sqrt{\frac{1}{4\pi}} \tag{5}$$

(you may check that this is normalised according to equation $(3)$.) Hence

$$|\psi|^2 = |R|^2|Y^2| = \frac{1}{4\pi}|R^2| \tag{6}$$

and it therefore follows that $|R|^2r^2 = 4\pi r^2|\psi|^2$, so both definitions are equivalent in the case of spherically symmetric orbitals (s-orbitals). Notice that in the other linked question it is an s-orbital under consideration, and so the second definition $4\pi r^2|\psi|^2$ is used (or $4\pi r^2 \rho$, where the probability density $\rho$ is simply $|\psi|^2$). However, for anything that is not spherically symmetric, these are not the same and the more general expression $|R|^2r^2$ should be used.

Based on the above arguments this is my take on the matter:

  • $|R^2|r^2$ is the best definition of the RDF;
  • $4\pi r^2|\psi|^2$ is equivalent to that for spherically symmetric orbitals and agrees with a-cyclohexane-molecule's answer here, but does not work for other orbitals (p, d, ...);
  • I still don't fully understand why Housecroft has an additional factor of $4\pi$ – something which I've noticed many years ago. There is probably some motivation behind it, but I don't know what it is. In Feodoran's answer $Y_{0,0}$ is taken to be equal to $1$, which leads to an extra factor of $4\pi$. However, I think it is more conventional to have $Y_{0,0}$ normalised as $\sqrt{1/(4\pi)}$.

Thanks to Ruslan for pointing this out! If you are curious as to what this looks like, there is a good diagram on LibreTexts, which I have re-uploaded to SE here. Note, however, that the convention in this diagram is $(\rho,\phi,\theta)$ instead of $(r,\theta,\phi)$.

  • 2
    $\begingroup$ Since you're working in spherical coordinates, $|\psi|^2\,\mathrm dV$ represents probability not in a cube: instead it's in a frustum-like shape, which is especially different from a cube near the origin ($r\approx0$). $\endgroup$
    – Ruslan
    Dec 27, 2018 at 6:08
  • $\begingroup$ @Ruslan Thank you for the correction! I never knew there was a term for it. $\endgroup$
    – orthocresol
    Dec 27, 2018 at 13:45
  • $\begingroup$ Thanks! But I still don't understand why we use Radial Distribution function as a measure of probability of finding electron if $\mathrm{\psi^2 dV}$ gives the same thing? Also, it seems that the volume of the frustum will go to 0 near the centre of the sphere. In that case, why does the plot of $\mathrm{\psi^2 dV}$ show a maximum at 0? $\endgroup$
    – S R Maiti
    Dec 28, 2018 at 16:59
  • 1
    $\begingroup$ @ShoubhikRajMaiti it is not really the same thing, as the RDF depends only on $r$, whereas $|\psi|^2$ depends on $(r,\theta,\phi)$. They do both describe how electron density is distributed, but from a different perspective. As for why you would use one over the other, if you look in an inorganic textbook, there is usually some discussion of concepts such as shielding and penetration. Although I find these to be somewhat unconvincing, they rely on using and interpreting the RDFs, as trying to directly interpret $|\psi|^2$ would be incredibly complicated due to the angular dependency. $\endgroup$
    – orthocresol
    Dec 28, 2018 at 17:05
  • $\begingroup$ @ShoubhikRajMaiti these are not plots of the probability $|\psi|^2\mathrm dV$, they are plots of the probability density $|\psi|^2$, which (only for some orbitals, mind you) is a maximum at the nucleus. Now, whatever number $|\psi|^2$ may take on at the nucleus, if you multiply that by $\mathrm dV$ you get zero, so the probability ($= |\psi|^2\mathrm dV$) of finding it there is zero. $\endgroup$
    – orthocresol
    Dec 28, 2018 at 17:08

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