Probability density “AT” the nucleus?

Just had a conceptual question from radial wave functions.

I wanted to know if probability density AT the nucleus is maximum, or zero.

From the attached images, it is clear that the function is at zero for p and d orbitals, but TENDS to infinity as approaching r = 0 for the s-orbitals. Is the value of the function 0 at r = 0?

Intutively, as the distance from the nucleus is the least, there should be maximum probability density.

Additionally, can there be some other interpretation for the phrase "at the nucleus"? It is quite confusing.

Thanks a lot.

EDIT: Apologies, don't know what was going through my head while posting - I initially posted something about PROBABILITY of finding an electron at the nucleus (I know the answer to that, oops).

I have changed it to what I meant to ask. • Probability density of s orbitals at the nucleus is not 0, nor is it infinite. – Ivan Neretin Aug 5 at 11:21
• @IvanNeretin Wouldn't you say that $\int_0^0 \phi^2(\vec{r}) \mathrm{d}\vec{r}=0$? – Martin - マーチン Aug 5 at 17:06
• @Martin-マーチン Nucleus isn't really a point. – Mithoron Aug 5 at 19:12
• @Martin That's true. But then again, that would be equally true at any point. – Ivan Neretin Aug 6 at 7:53
• @Martin-マーチン The integral of the square modulus of the wavefunction over any range is a probability. The square modulus itself is a probability density. They are not the same thing - they have different units for starters. – Ian Bush Aug 6 at 12:37

Hydrogenic Wavefunctions:

A hydrogenic wavefunction, $$\psi_{nlm}(\boldsymbol{r})$$, can be written as the product $$\psi_{nlm}(r,\theta,\phi) = R_{nl}(r) Y_{lm}(\theta,\phi)$$ where $$R_{nl}$$ is a radial wavefunction and $$Y_{lm}$$ is a spherical harmonic.

• $$R_{nl}$$ is a radial wavefunction
• $$Y_{lm}$$ is a spherical harmonic
• $$n$$ is the principal quantum number, $$n = 1,2,3\dots$$
• $$l$$ is the angular quantum number, $$l = 0=s, 1=p, 2=d,\dots,n-1$$
• $$m$$ is the magnetic quantum number $$m = -l, -l+1,\dots,l-1,l$$

Probability/Electron Density

The probability density, $$\rho$$, is equal to the square modulus of the wavefunction: \begin{align} \rho_{nlm}(\boldsymbol{r}) &= |\psi_{nlm}(\boldsymbol{r})|^2 \\ \rho_{nlm}(r,\theta,\phi)&= |R_{nl}(r)|^2 \cdot |Y_{lm}(\theta,\phi)|^2 \end{align}

The spherical harmonics $$Y_{lm}$$ have angular nodes through the nuclei for all values of $$l$$ and $$m$$ except $$l =m =0$$, so we can say that for all $$l>0$$ orbitals, i.e. all orbitals except s-orbitals, the nuclei will have $$0$$ probability density.

S-orbitals

Consider s-orbitals, in atomic units and substituting in the solution of the radial wavefunctions in generalized Laguerre polynomials $$L^\alpha_k$$: \begin{align} \rho_{n00}(\boldsymbol{r})&= |R_{n0}(r)|^2 \cdot |Y_{00}(\theta,\phi)|^2\\ &= |R_{n0}(r)|^2 \frac{1}{4\pi} \\ \rho_{n00}(r)&= \frac{Z^3}{\pi n^5} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\ &= A(n) \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \end{align}

For all values of $$n$$, $$L^1_{n-1}\left(0\right) = n$$, so we know that the nucleus never has zero probability density for s-orbitals, and has probability density $$\rho_{n00}(r=0) = \frac{Z^3}{\pi n^4}$$

Stationary Points

But we can differentiate with respect to $$r$$ to find stationary points in the radial probability density.

\begin{align} \frac{d}{dr}\rho_{n00}(r)&= A(n) \frac{d}{dr} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\ &= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) + \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\ \text{if}\;n=1: \frac{d}{dr}\rho_{000}(r)&= -\frac{Z^4}{\pi n^6} \exp\left(-Zr/n\right) \\ &> 0 \;\forall\;r \end{align} Hence the probability density is constantly decreasing away from the nucleus. So as you previously assumed the maximum for probability density for the 1s orbital is at the nucleus.

$$n>1$$:

\begin{align} \text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) + \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\ &= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) - L^2_{n-2}\left(2Zr/n\right) \right] \end{align}

Setting the expression above equal to zero and solving for $$r$$ will give the positions for the radial nodes and maxima inbetween for a given value of $$n$$. I am not aware of a general method/expression in n.

As you asked about nucleus, we can also consider the point $$r=0$$:

\begin{align} \text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= - \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} L^1_{n-1}\left(0\right) + L^2_{n-2}\left(0\right) \right] \\ &= - \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} n +\frac{n^2-n}{2}\right] \\ &= - \frac{Z^4}{\pi n^5} - \frac{Z^3 \left(n-1\right)}{2\pi n^4} < 0 \end{align}

Conclusion/TL;DR

Hence the nuclei is always a local maximum in the probability density for an s-orbital. You would expect the exponential decay in the radial wavefunction to dominate over the Laguerre polynomial term, so would expect local maxima further out to be smaller that the maxima at the nuclei and hence that the point of highest probability density is at the nucleus.

This makes physical sense, as classically the negatively charged electron wants to be as close as possible to the positively charged nucleus to minimise its potential energy, but the quantum mechanical nature of the electron prevents it from tumbling directly into the nucleus.

• All the math detail is nice, but the simple answer is that the functions simply don't consider that the nucleus has a finite volume. You'd need a more complicated model to account for that fact. – MaxW Aug 6 at 7:59
• In an otherwise excellent answer there is a small problem with "The spherical harmonics Ylm have angular nodes through the nuclei for all values of l and m except l=m=0, so we can say that for all l>0 orbitals, i.e. all orbitals except s-orbitals, the nuclei will have 0 probability density." The spherical harmonics have no radial dependence so we can't say anything from them about the behaviour at any value of r and r=0 in particular. The point is rather that Rnl is zero for r=0 for all l>0 (i.e. not an s type orbital) from the Laguerre polynomials – Ian Bush Aug 7 at 9:47