University Physics V · The Hydrogen Atom · 11.8

Orbitals, Radial Densities & Probability Clouds

Everything measurable about where the electron is sits in |ψ|², and this is where you learn to read it: which density answers which question, how to count a state's nodes before plotting it, which radius the energy actually depends on, and which features of an orbital picture are physics rather than a choice of basis.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Solving the Coulomb eigenproblem hands you a vector; this topic is about reading it. The Born rule turns |n l m⟩ into a density that factorises, |ψ|² = |Rₙₗ(r)|²|Yₗ^m(θ, φ)|², and the object worth plotting is not |ψ|² but P(r) = r²|Rₙₗ|² = |u(r)|², the probability per unit radius, because the shell between r and r + dr carries a volume 4πr² dr that grows as r² — which is why the 1s peak sits at a₀ while |ψ|² is largest at the nucleus. The shape of P(r) is fixed before any integral is done: r^l suppresses the origin for l > 0, a Laguerre polynomial of degree n − l − 1 supplies exactly that many radial nodes, and e^{(−Zr/na₀)} sets the decay length na₀/Z and so the reach at about n²a₀/Z.

Moments then compress the curve into numbers that disagree on purpose: ⟨1/r⟩ = Z/n²a₀ carries no l-dependence and, through the virial pair ⟨V⟩ = 2E and ⟨T⟩ = −E, is what the energy is made of, while ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)] does depend on l and is what the size is made of. The cost is that half the familiar picture is convention: the radial profile belongs to (n, l) alone, but the lobed shapes belong to a basis — pₓ and p_y are a unitary rotation of m = ±1 inside a degenerate multiplet, no extra states and no extra information — and none of it is a trajectory, since the density is stationary and |ψ|² gives the statistics of a position measurement rather than a path followed between measurements.

Simple definition: The radial probability density of a hydrogenic eigenstate is P(r) = |u(r)|² = r²|Rₙₗ(r)|², the probability per unit radius of finding the electron in the shell between r and r + dr once every angle has been integrated over.
Example: For hydrogen's 1s state P(r) = (4/a₀³)r²e^(−2r/a₀), which peaks at exactly r = a₀ = 52.9 pm although |ψ|² is largest at the nucleus; 32.3% of the probability lies inside a₀, while ⟨r⟩ = 1.5a₀ = 79.4 pm.

Radial probability densityP(r) = r²|Rₙₗ(r)|² = |u(r)|², ∫₀^∞ P(r) dr = 1

Converts probability per unit volume into probability per unit radius — the only version whose peak answers the question "how far out is the electron?"

u = rR in m^(−1/2), P in m⁻¹; the r² is the Jacobian of the shell 4πr² dr, not a normalisation constant.

Node count from the labels aloneradial nodes = n − l − 1, angular = l, total = n − 1

Gives the shape of the curve before you plot it: 3s dips to zero twice, 3p once, 3d not at all — and it is the cheapest check on a numerical eigenvector.

Counted at 0 < r < ∞; r = 0 and r = ∞ never count. n − l − 1 is the degree of the Laguerre factor. The angular l is the REAL-orbital count (|m| planes + l − |m| cones); a complex Yₗ^m has only the l − |m| cones, since e^(imφ) never vanishes.

The two radii a hydrogenic state carries⟨1/r⟩ = Z/(n²a₀), ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)]

⟨1/r⟩ is what the energy is built from and 1/⟨1/r⟩ = n²a₀/Z; ⟨r⟩ is what the size is, and the two are never equal.

a₀ = 52.918 pm; ⟨1/r⟩ in m⁻¹, ⟨r⟩ in m. ⟨1/r⟩ has no l-dependence at all, ⟨r⟩ does.

Virial theorem for a 1/r potential2⟨T⟩ = −⟨V⟩ ⟹ ⟨T⟩ = −Eₙ, ⟨V⟩ = 2Eₙ

Hands you ⟨p²⟩ = 2mₑ⟨T⟩ with no integral, and hence vᵣₘₛ/c = Zα/n, the number that sizes every relativistic correction.

⟨T⟩ = +13.606 Z²/n² eV and ⟨V⟩ = −27.211 Z²/n² eV. Stationary states only; a wave packet obeys it only on time average.

Most probable radius of a circular statel = n − 1: P ∝ r⁽²ⁿ⁾ e^{(−2Zr/na₀)}, rₘₚ = n²a₀/Z

The one place Bohr's orbit radius survives verbatim — as the mode of a distribution rather than as a path.

Exact only for l = n − 1, where R has no radial node. For smaller l the outermost peak sits well BEYOND n²a₀/Z: 5.24a₀ for 2s against 4a₀, and 13.07a₀ for 3s against 9a₀.

Real orbitals as a rotation of the m basis|pₓ⟩ = (|1,−1⟩ − |1,1⟩)/√2, |p_y⟩ = i(|1,−1⟩ + |1,1⟩)/√2

Same subspace, same energy, same P(r): a different picture of the same three states, never three more of them.

Unitary within the fixed n = 2, l = 1 triple. Both have ⟨L_z⟩ = 0 with ΔL_z = ħ, and both share the 2p radial density.

Three densities, and only one of them peaks at a₀

Three densities live inside one state and answer three different questions. |ψ(r)|² = |Rₙₗ|²|Yₗ^m|² is probability per unit volume — what a point detector of volume dV registers. Integrating over angles gives P(r) = r²|Rₙₗ(r)|² = |u(r)|², probability per unit radius, which is what a spherical shell of thickness dr collects. The r² is a Jacobian, not a fudge: the shell at 2a₀ has four times the volume of the shell at a₀. For 1s, |ψ|² = e^(−2r/a₀)/πa₀³ falls monotonically from 2.15 × 10³⁰ m⁻³ at the nucleus, while P(r) = (4/a₀³)r²e^(−2r/a₀) starts at zero, peaks at exactly a₀, and encloses 1 − 5e⁻² = 32.3% of the probability inside a₀ and 1 − 13e⁻⁴ = 76.2% inside 2a₀. Both densities are physical, and the value at the origin is not an artefact: |ψ(0)|² is precisely what the Fermi contact term of hyperfine structure integrates against, and it is non-zero only for l = 0.

Count the nodes before you plot anything

The qualitative curve follows from the labels. Write Rₙₗ(r) = r^l · L(r) · e^{(−Zr/na₀)}, with L an associated Laguerre polynomial of degree n − l − 1. The exponential never vanishes and r^l vanishes only at the origin, so every zero at finite r > 0 comes from L: exactly n − l − 1 radial nodes. Concretely, with ρ = r/a₀ and Z = 1: R₃₀ ∝ (27 − 18ρ + 2ρ²)e^(−ρ/3) has roots at ρ = (9 ± 3√3)/2 = 1.90 and 7.10; R₃₁ ∝ ρ(6 − ρ)e^(−ρ/3) has one, at ρ = 6; R₃₂ ∝ ρ²e^(−ρ/3) has none. The angular count needs care about which basis you are in. The polar factor Pₗ^|m|(cos θ) has l − |m| zeros in 0 < θ < π, which are nodal cones; the azimuthal factor e^(imφ) of a complex Yₗ^m has no zeros at all, so |2,1,±1⟩ carries no nodal surface whatever. The familiar |m| planes through the z-axis belong to the REAL combinations, where cos(mφ) and sin(mφ) genuinely vanish; there the angular count is |m| planes plus l − |m| cones = l, and the textbook total n − 1 holds across a shell. The radial count, which is basis-independent, is also the cheapest numerical diagnostic available. Diagonalise the radial Hamiltonian on a grid with u(0) = 0 and count sign changes in each eigenvector: the k-th eigenvector at fixed l, counting from k = 0, must show exactly k of them, and if it does not, the grid is too coarse or the box too small.

One state, several radii, and they disagree on purpose

A density has no single size. ⟨1/r⟩ = Z/n²a₀ follows in one line from Hellmann–Feynman: ∂H/∂Z = −e²/4πε₀r, so differentiating E = −Z² × 13.606 eV/n² with respect to Z returns ⟨1/r⟩ directly, and no l appears anywhere. ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)] does depend on l: for n = 3 it is 13.5a₀ = 714 pm, 12.5a₀ = 661 pm and 10.5a₀ = 556 pm for l = 0, 1, 2 — a 22% spread across states of identical energy, because the centrifugal barrier ħ²l(l+1)/2μr² keeps high-l clouds out of the small-r region while low-l states put weight there and pay for it further out. Meanwhile 1/⟨1/r⟩ = n²a₀/Z = 9a₀ for all three, equal to ⟨r⟩ for none of them; Jensen's inequality guarantees ⟨1/r⟩ ≥ 1/⟨r⟩ for any spread density. Higher inverse moments are more l-sensitive still: ⟨1/r²⟩ = Z²/[a₀²n³(l + ½)] and ⟨1/r³⟩ = Z³/[a₀³n³l(l + ½)(l + 1)], the latter defined only for l ≥ 1 because the true integral ∫|R|²r⁻¹dr diverges logarithmically when R(0) ≠ 0. Be careful what those moments buy you. ⟨1/r³⟩ is what the spin–orbit term is built from, and that term is l-dependent piece by piece, but in hydrogen it combines with the relativistic kinetic correction so that the total order-α⁴ fine structure depends only on n and j: 2s₁/₂ and 2p₁/₂ stay degenerate at that order, and only the Lamb shift separates them. What actually lifts the l-degeneracy in a many-electron atom is screening, not spin–orbit.

The virial theorem turns ⟨1/r⟩ into kinetic energy

In a stationary state d⟨r⋅p⟩/dt = 0, and evaluating the commutator gives 2⟨T⟩ = ⟨r⋅∇V⟩. The Coulomb potential is homogeneous of degree −1, so r⋅∇V = −V and 2⟨T⟩ = −⟨V⟩. With E = ⟨T⟩ + ⟨V⟩ this fixes both pieces from the eigenvalue alone: ⟨T⟩ = −Eₙ = +13.606 Z²/n² eV and ⟨V⟩ = 2Eₙ = −27.211 Z²/n² eV, with no integral performed. The cross-check is ⟨V⟩ = −(Ze²/4πε₀)⟨1/r⟩ = −27.211 Z²/n² eV, the same number. Two payoffs follow. First ⟨p²⟩ = 2mₑ⟨T⟩ = ħ²Z²/n²a₀², so √⟨p²⟩ = ħZ/na₀, and for 1s the radial spread is Δr = √(⟨r²⟩ − ⟨r⟩²) = (√3/2)a₀, giving a product 0.87ħ — the atom's size is visibly an uncertainty-principle compromise rather than an orbit. (Note this is an estimate built from √⟨p²⟩ and a coordinate spread, not the canonical Δr Δpᵣ relation: radial momentum has no self-adjoint extension on the half-line and is not an observable.) Second vᵣₘₛ/c = √(2⟨T⟩/mₑ)/c = Zα/n, which is 7.30 × 10⁻³ for hydrogen's ground state: that is why the non-relativistic treatment works, and why corrections enter at relative order (Zα/n)² = 5.33 × 10⁻⁵ — a scale of α²|E₁| = 7.2 × 10⁻⁴ eV, of which the actual 1s fine-structure shift is about a quarter, −1.81 × 10⁻⁴ eV. For a 1s electron in gold, Zα = 0.58, and the same estimate says this model has run out.

Real orbitals are a rotation inside a degenerate multiplet

The textbook pₓ, p_y and d_xy pictures are not extra states. Inside the degenerate 2p triple, |pₓ⟩ = (|1,−1⟩ − |1,1⟩)/√2 and |p_y⟩ = i(|1,−1⟩ + |1,1⟩)/√2 with |p_z⟩ = |1,0⟩ — a unitary change of basis in a three-dimensional subspace, so three states before and three after. Each remains an eigenstate of H and of L² with unchanged eigenvalues, and each has the identical radial density P(r), because P depends on n and l only. What changes is L_z: pₓ is not an L_z eigenstate, and measuring L_z on it returns +ħ or −ħ with equal probability, giving ⟨L_z⟩ = 0 and ΔL_z = ħ. The physical consequence is a current. For |n l m⟩ the probability current is azimuthal, j_φ = (ħmₗ/μr sinθ)|ψ|², carrying an orbital moment μ_z = −mₗμ_B; the real orbitals, being real up to a global phase, carry no current and no moment. Which basis is right is decided by whatever breaks the degeneracy — a field along z makes the m states the eigenstates of the perturbation, a bond or a crystal field along x makes the lobes.

What the cloud is, and what a grid calculation draws

Read the picture carefully. |ψ|² is the density for a position measurement on an ensemble of identically prepared atoms: the stippled cloud is ten thousand atoms measured once each, not one electron smeared into a fog. Yet −e|ψ|² is also exactly the mean charge density that sources the Hartree potential in a self-consistent field and that X-ray scattering measures as a form factor, so the charge-cloud reading is legitimate for mean quantities and wrong for fluctuations. And nothing in the picture moves: ∂|ψ|²/∂t = 0 for any eigenstate, which is why an atom in an orbital does not radiate, while a superposition of two levels has a density oscillating at ω = ΔE/ħ, and that is what makes a dipole radiate. Numerically the whole topic is one script: build the radial Hamiltonian on a grid with u(0) = 0, take eigenvectors from scipy.linalg.eigh_tridiagonal, normalise by trapezoid, and plot u². Then check three diagnostics — node count n − l − 1, ⟨1/r⟩ = Z/n²a₀, and ⟨T⟩ = −E. A box that is too small truncates the tail, which raises E and shrinks ⟨r⟩, and it bites the high-n states first.

Change one variable at a time

Make the relationship visible.

Interactive model

Principal quantum number n3

Orbital number l (capped at n−1)0

Hold n at 3 and step l from 0 to 2: the two radial nodes vanish, ⟨r⟩ pulls in from 13.5a₀ to 10.5a₀, and at l = 2 the single peak lands exactly on the 1/⟨1/r⟩ = 9a₀ marker — while the energy readout never moves.

RADIAL NODES n−l−12

MEAN RADIUS ⟨r⟩13.50 a₀

1/⟨1/r⟩ = n²a₀9 a₀

ENERGY E = −⟨T⟩-1.512 eV

Live interpretationRADIAL NODES n−l−1: 2. MEAN RADIUS ⟨r⟩: 13.50 a₀. 1/⟨1/r⟩ = n²a₀: 9 a₀. ENERGY E = −⟨T⟩: −1.512 eV

Catch the common trap

Explain before calculating.

Misconception check

The lobes are the trap. Students treat pₓ, p_y and p_z as "the three 2p states" and then reason about them as though each had a definite L_z, or as though the m = ±1 donuts were a rival set competing with them. Neither holds. The 2p subspace is three-dimensional in both bases, and the real set is a unitary rotation of the complex one, so no state is created or lost; pₓ is a superposition of m = +1 and m = −1, and measuring L_z on it returns +ħ or −ħ, each half the time, never 0. Nor is the dumbbell the atom's true shape: an isolated atom has no preferred direction, so no member of a degenerate multiplet is singled out. The lobes become the physically distinguished states only when something breaks the symmetry along x or y — a bond, a crystal field — and if the perturbation is a magnetic field along z, the m states are the ones that diagonalise it.

Hydrogen's ground state is ψ = (πa₀³)^(−1/2) e^(−r/a₀). Three claims can be made about where the electron is: where |ψ|² is largest, where the radial probability density P(r) = r²|R|² peaks, and what ⟨r⟩ equals. Which set of three is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHydrogen's 2p state has R₂₁(r) ∝ r e^(−r/2a₀). Count its nodes, find the most probable radius from P(r), and compare that with ⟨r⟩ = (a₀/2)[3n² − l(l+1)]. Take a₀ = 52.918 pm.

Radial nodes: n − l − 1 = 2 − 1 − 1 = 0, so R₂₁ never crosses zero for r > 0. In the real basis the angular factor supplies l = 1 nodal surface — for p_z (m = 0) the xy-plane — giving n − 1 = 1 node in total. For the complex |2,1,±1⟩ there is no nodal surface at all, since e^(±iφ) never vanishes.
Radial density: P(r) = r²|R₂₁|² ∝ r² · r²e^(−r/a₀) = r⁴e^(−r/a₀). Note the exponent doubles, since |e^(−r/2a₀)|² = e^(−r/a₀).
Differentiate: dP/dr ∝ (4r³ − r⁴/a₀)e^(−r/a₀) = 0 gives 4 = r/a₀, so rₘₚ = 4a₀ = 4 × 52.918 = 211.7 pm. This is the circular case l = n − 1, where rₘₚ = n²a₀ exactly.
Mean radius: ⟨r⟩ = (a₀/2)[3(4) − 1(2)] = (a₀/2)(10) = 5a₀ = 264.6 pm, a quarter beyond the peak because the tail is one-sided.
Third radius: ⟨1/r⟩ = Z/(n²a₀) = 1/(4a₀), so 1/⟨1/r⟩ = 4a₀ = 211.7 pm — equal to rₘₚ here only because l = n − 1.

AnswerNo radial nodes and (in the real basis) one angular node; rₘₚ = 4a₀ = 211.7 pm, ⟨r⟩ = 5a₀ = 264.6 pm, and 1/⟨1/r⟩ = 4a₀ = 211.7 pm.

MediumTake hydrogen's n = 3 shell, with ρ = r/a₀ and R₃₀ ∝ (27 − 18ρ + 2ρ²)e^(−ρ/3), R₃₁ ∝ ρ(6 − ρ)e^(−ρ/3), R₃₂ ∝ ρ²e^(−ρ/3). Locate every radial node, then compare ⟨r⟩ and the energy across the three states.

3s: solve 2ρ² − 18ρ + 27 = 0, giving ρ = [18 ± √(324 − 216)]/4 = (9 ± 3√3)/2 = 1.902 and 7.098, i.e. 100.6 pm and 375.6 pm. Two radial nodes, matching n − l − 1 = 3 − 0 − 1 = 2.
3p: ρ(6 − ρ) vanishes at ρ = 0 and ρ = 6, but the root at the origin is the r^l prefactor, not a node — every l ≥ 1 state vanishes there. So one radial node, at 6a₀ = 317.5 pm, matching n − l − 1 = 3 − 1 − 1 = 1.
3d: ρ²e^(−ρ/3) has no zero for ρ > 0, so no radial nodes, matching 3 − 2 − 1 = 0. In the real basis its l = 2 angular factor carries two nodal surfaces, so all three states still reach the textbook total n − 1 = 2.
Sizes: ⟨r⟩ = (a₀/2)[27 − l(l+1)] gives 13.5a₀ = 714 pm for 3s, 12.5a₀ = 661 pm for 3p and 10.5a₀ = 556 pm for 3d — the 3d cloud is 22% tighter than the 3s.
Energies: ⟨1/r⟩ = Z/(n²a₀) = 1/(9a₀) for all three regardless of l, so ⟨V⟩ = −27.211/9 = −3.023 eV and E = ½⟨V⟩ = −1.512 eV for each. Three visibly different clouds, one energy — the accidental l-degeneracy of the pure 1/r potential.

Answer3s: radial nodes at 1.90a₀ and 7.10a₀ (100.6 pm, 375.6 pm); 3p: one at 6a₀ = 317.5 pm; 3d: none. ⟨r⟩ = 13.5a₀, 12.5a₀, 10.5a₀, yet E = −1.512 eV for all three.

HardHe⁺ (Z = 2) is prepared in the state |n = 2, l = 1, m = +1⟩. Find E, ⟨T⟩, ⟨V⟩, ⟨r⟩, rₘₚ and the rms speed. The state is then re-expressed in the real basis as |2pₓ⟩ = (|2,1,−1⟩ − |2,1,1⟩)/√2: say which of those numbers change, and what does.

Energy: E = −13.606 Z²/n² eV = −13.606 × 4/4 = −13.606 eV, so He⁺ in n = 2 is degenerate with hydrogen's ground state — Z²/n² is all the eigenvalue knows.
Virial: for a 1/r potential 2⟨T⟩ = −⟨V⟩, so ⟨T⟩ = −E = +13.606 eV and ⟨V⟩ = 2E = −27.211 eV. Cross-check with the moment: ⟨1/r⟩ = Z/(n²a₀) = 1/(2a₀), and ⟨V⟩ = −(Ze²/4πε₀)⟨1/r⟩ = −27.211 × Z²/n² = −27.211 eV ✓.
Radii: ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)] = (a₀/4)(12 − 2) = 2.5a₀ = 132.3 pm, and since l = n − 1 the peak is exact: rₘₚ = n²a₀/Z = 2a₀ = 105.8 pm. Both are half the hydrogen 2p values — the cloud contracts as 1/Z while the energy grows as Z².
Speed: ⟨T⟩ = ⟨p²⟩/2mₑ = 13.606 eV = 2.180 × 10⁻¹⁸ J, so vᵣₘₛ = √(2 × 2.180 × 10⁻¹⁸/9.109 × 10⁻³¹) = 2.19 × 10⁶ m s⁻¹, i.e. v/c = 7.30 × 10⁻³ = Zα/n. Relativistic corrections therefore enter at relative order 5.3 × 10⁻⁵ of E.
Basis change: |2pₓ⟩ is a superposition inside the same degenerate l = 1 triple, so H, L² and every function of r are untouched — E, ⟨T⟩, ⟨V⟩, ⟨r⟩ and rₘₚ are all unchanged, because P(r) depends on n and l alone.
What does change is the angular factor and L_z: ⟨L_z⟩ falls from +ħ to 0, with the outcome now spread 50:50 over ±ħ so that ΔL_z = ħ, and the azimuthal current j_φ ∝ mₗ|ψ|² vanishes, so the orbital magnetic moment μ_z = −mₗμ_B goes from −μ_B to zero.

AnswerE = −13.606 eV, ⟨T⟩ = +13.606 eV, ⟨V⟩ = −27.211 eV, ⟨r⟩ = 2.5a₀ = 132.3 pm, rₘₚ = 2a₀ = 105.8 pm and vᵣₘₛ = 2.19 × 10⁶ m s⁻¹. Rotating to 2pₓ changes none of them; it only sends ⟨L_z⟩ from +ħ to 0 with ΔL_z = ħ, and kills the orbital current together with its −μ_B moment.

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