The spatial distribution of probability density

The last section ended with a provocative hint: “the place where $\lvert\psi\rvert^2$ peaks is the radius where the electron is most likely to be found” — that statement is wrong. This section explains exactly where it goes wrong. This is not hair-splitting: the orbital pictures in chemistry books and the “electron clouds” of popular science are among the most misread images in all of quantum mechanics, and reading them right takes nothing more than telling two questions apart.

A concrete contradiction

Put the ground state on the table. Two sections ago we found

$$\lvert\psi_{100}\rvert^2=\frac{1}{\pi a^3}\,\ee^{-2r/a}\tag{6.5.1}$$

This is a monotonically decreasing function — the probability density is largest at $r=0$, dead centre on the nucleus.

Yet a numerical exercise in that same section found the probability of the electron being inside the sphere of radius $a_0$ to be only 32%, and every textbook says “the electron is most likely to be found one Bohr radius from the nucleus”. Density peaks on the nucleus, most probable radius at $a_0$ — a contradiction?

No. These are two different questions:

• Question A: “Near which point is the probability density of finding the electron, per small volume, the largest?” — answered by $\lvert\psi\rvert^2$: at the centre.
• Question B: “What is the electron’s most likely distance $r$ from the nucleus?” — this asks which spherical shell holds the most total probability, and a different function answers it.

The radial distribution function: where the r² factor comes frombasic~5 minexpand

Step 1: write the probability that the nuclear distance lies between $r$ and $r+\dd r$. That means integrating the probability density over the whole shell:

$$P(r)\,\dd r=\int_\text{shell}\lvert\psi_{nlm}\rvert^2\,\dd^3r =\lvert R_{nl}(r)\rvert^2\,r^2\dd r\int\lvert Y_l^m\rvert^2\dd\Omega\tag{6.5.2}$$

The angular integral equals 1 (spherical harmonics are normalised), so

$$P(r)=r^2\,\lvert R_{nl}(r)\rvert^2\tag{6.5.3}$$

$P(r)$ is called the radial distribution function (in plain words: the probability distribution of the electron’s distance from the nucleus).

Step 2: see where the $r^2$ comes from. A shell of radius $r$ and thickness $\dd r$ has volume $4\pi r^2\dd r$ — bigger radius, bigger shell. Distance distribution = density × shell volume, a competition between density and geometric weight:

• small $r$: the density is high, but the shell is pitifully small (volume $\to0$ as $r\to0$), so $P(r)\to0$;
• large $r$: the shell is big, but the density dies exponentially, so $P(r)\to0$;
• somewhere in between, a peak.

Step 3: find the ground state’s peak. $P(r)\propto r^2\ee^{-2r/a}$; set the derivative to zero:

$$\frac{\dd}{\dd r}\left(r^2\ee^{-2r/a}\right)=\left(2r-\frac{2r^2}{a}\right)\ee^{-2r/a}=0 \;\Longrightarrow\; r_\text{mp}=a=a_0\tag{6.5.4}$$

The most probable radius is exactly the Bohr radius. The afterimage of Bohr’s circular orbit at $r=a_0$ in the true theory is the peak of this distribution curve.

Step 4: compute the mean while we’re at it. Peak is not mean: the distribution has a tail toward large $r$, dragging the average rightward:

$$\langle r\rangle_{1s}=\int_0^\infty r\,P(r)\,\dd r=\frac{4}{a^3}\int_0^\infty r^3\ee^{-2r/a}\dd r=\frac{4}{a^3}\cdot\frac{3!\,a^4}{2^4}=\frac{3a}{2}\tag{6.5.5}$$

Peak at $a$, mean $1.5a$, median about $1.34a$ — three “typical radii”, three different numbers. Before quoting one, say which you mean.

Nodes: an orbital’s fingerprint

Sketching the first few $P(r)$ in words (all $r$ in units of $a$; peak positions from setting derivatives to zero — check them yourself):

Orbital	Radial nodes $n-l-1$	Shape of $P(r)$	Main peak	$\langle r\rangle$
1s	0	single peak	$1.0\,a$	$1.5\,a$
2s	1	small peak — zero (at $r=2a$) — main peak	$5.24\,a$	$6\,a$
2p	0	single peak	$4\,a$	$5\,a$
3s	2	peak — zero — peak — zero — main peak	$13.1\,a$	$13.5\,a$
3d	0	single peak	$9\,a$	$10.5\,a$

The patterns leap out:

• Radial node count = $n-l-1$ (the series termination order, a product of section 6.3): there $R_{nl}=0$, the probability of that exact nuclear distance is zero, and the distribution is sliced into concentric shells;
• Angular node count = $l$: $Y_l^m$ has $l$ nodal surfaces (planes or cones) — the “waist” of a p orbital and the “cross” of a d orbital are angular nodal surfaces;
• the two add up to exactly $n-1$ — reconnecting with the old one-dimensional rule that “the $k$-th excited bound state has $k$ nodes”.

One more look at the angular part. $\lvert Y_l^m\rvert^2$ does not depend on $\phi$ (the phase $\ee^{\ii m\phi}$ disappears on taking the modulus), so the electron cloud of any strict $(n,l,m)$ eigenstate is rotationally symmetric about the $z$ axis: $\lvert Y_1^0\rvert^2\propto\cos^2\theta$ is a dumbbell along $z$, and $\lvert Y_1^{\pm1}\rvert^2\propto\sin^2\theta$ is a doughnut around the equator.

The three commonest misreadings

One day you will spin all these shapes around with your own hands — the 3D orbital visualiser for this chapter is on the roadmap. For now, the node-counting drill plus the fact-check list above are enough to read any textbook figure correctly.

Calibrating your sense of scale

Close with real numbers. The ground-state hydrogen “90% probability radius” is about $2.7a_0\approx0.14$ nm — this is where the chemical size of atoms (about 0.1 nm) comes from — while the nuclear radius is about $10^{-6}$ nm: one part in $10^{15}$ of the atom’s volume is nucleus; all the rest is the probability cloud’s thin frontier. Blow the nucleus up to a marble at the centre of a stadium, and the “electron” at the most probable radius sits in the stands — but remember, it still has a 32% chance of being anywhere between the marble and the stands, including inside the marble itself. That last point is not rhetoric: the electron’s small but non-zero probability of being inside the nucleus is the precondition for electron capture (a nucleus absorbing an inner electron), and the physical root of the next section’s Darwin term.

∎Key formulas

• Radial distribution function

  $$P(r)=r^2\lvert R_{nl}(r)\rvert^2,\qquad\int_0^\infty P(r)\,\dd r=1$$

  r² is the shell-volume weight; answers "what is the most likely nuclear distance"

• Three ground-state radii

  $$r_{\text{mp}}=a_0,\quad\langle r\rangle=1.5\,a_0,\quad P(r<a_0)=0.32$$

  Peak = Bohr radius; yet the density maximum is at r = 0

• Mean radius

  $$\langle r\rangle_{nl}=\frac{a}{2}\left[3n^2-l(l+1)\right]$$

  Size ∝ n²; within one n, larger l sits slightly closer in on average

• Node counting

  $$(n-l-1)_{\text{radial}}+l_{\text{angular}}=n-1$$

  Reading any orbital picture starts with counting nodes

?Self-check4 questions

1. 1.

   In ground-state hydrogen, where is the maximum of the probability density |ψ|², and where is the most probable nuclear distance?

   • Density peaks at r = 0 (on the nucleus); most probable nuclear distance at r = a₀
   • Both at a₀
   • Both at r = 0
   • Density peaks at a₀; most probable nuclear distance at r = 0

   Submit
2. 2.

   How many radial nodes and how many angular nodal surfaces does the 3p orbital have?

   • 1 radial node (a sphere), 1 angular nodal surface (a plane)
   • 2 radial nodes, 1 angular nodal surface
   • 1 radial node, 0 angular nodal surfaces
   • 0 radial nodes, 2 angular nodal surfaces

   Submit
3. 3.

   Which statements about the chemistry books' p_x and p_y orbitals are correct? (Select all that apply.)

   Select all that apply

   • They are superpositions of the m = +1 and m = −1 eigenstates
   • They describe the same degenerate subspace as the m eigenstates, just in a different basis
   • They are more "real" orbitals than Y₁^{±1}
   • They are real functions, more convenient for discussing bond directions

   Submit
4. 4.

   Find the most probable nuclear distance for the 2p orbital, in nm. (P(r) ∝ r⁴e^{−r/a}, a₀ = 0.0529 nm)

   nm1% relative tolerance

   Submit

What comes next

The Coulomb-model hydrogen atom has now been squeezed dry: levels, degeneracies, lines, shapes — all in hand. But turn the spectrometer’s resolution up one more notch and the story turns a page: back in 1887 Michelson found that $\mathrm H\alpha$ is not one line at all, but two, about 0.016 nm apart. Coulomb potential plus Schrödinger equation cannot produce that splitting. The three missing pieces — relativity, spin–orbit coupling, and a correction with no classical counterpart — are collectively called fine structure, and they enter next.