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Hydrogen Orbitals: Spherical Harmonics and the Shapes of s, p, d States

Why the hydrogen atom's Schrödinger equation splits cleanly into a radial piece and an angular piece, and how that split hands us the familiar s, p and d orbital shapes on every chemistry textbook cover.

mysimulator teamUpdated July 2026≈ 10 min read▶ Open the simulation

A spherically symmetric potential separates cleanly

The hydrogen atom is one electron bound to a proton by the Coulomb potential V(r) = −e²/(4πε₀r), which depends only on distance r, never on direction. That spherical symmetry means the Hamiltonian commutes with the total angular momentum operators L² and L_z, so all three share a common set of eigenstates. Looking for solutions of the product form ψ(r,θ,φ) = R(r)·Y(θ,φ) then splits the full Schrödinger equation into one purely angular equation and one purely radial equation, tied together only by a separation constant l(l+1).

The angular part: spherical harmonics

The angular equation is solved by the spherical harmonics Ylm(θ,φ) — the same family of functions that describes multipole fields in electromagnetism and the vibration modes of a sphere. Each one factors into a θ-dependent associated Legendre polynomial and a φ-dependent phase eimφ. The quantum number l = 0, 1, 2, 3… is labelled s, p, d, f by 1890s spectroscopic convention (sharp, principal, diffuse, fundamental), and for each l, m ranges over 2l+1 integer values from −l to +l — the origin of "one s orbital, three p orbitals, five d orbitals."

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The radial part and the −13.6 eV/n² energy ladder

Solving the radial equation for the Coulomb potential requires substituting R(r) = e−r/na₀·(r/a₀)l·L(r), where a₀ is the Bohr radius. Demanding that the wavefunction stay finite as r → ∞ forces L(r) to be an associated Laguerre polynomial — and that same finiteness condition fixes the energy:

Eₙ = −(m e⁴) / (2(4πε₀)²ℏ²n²) = −13.6 eV / n²

Remarkably, this depends only on the principal quantum number n, never on l or m — an "accidental" degeneracy unique to the pure 1/r potential. It vanishes for any other central potential, and it is lifted in real multi-electron atoms by electron-electron screening, which is exactly why heavier atoms need the full periodic-table structure rather than hydrogen's simple ladder.

Quantum numbers n, l, m and orbital shapes

The three quantum numbers fully label an orbital: n (1, 2, 3…) sets the energy shell, l (0 to n−1) sets the shape, and m (−l to +l) sets the orientation. s orbitals (l=0) are spherically symmetric with non-zero density right at the nucleus; p orbitals (l=1) are two lobes of opposite sign split by a nodal plane; d orbitals (l=2) form four-lobed cloverleaf shapes with two angular nodal surfaces. Every orbital has n−1 total nodes split between radial shells and angular planes — the same standing-wave counting that governs the harmonics of a vibrating string, just in three dimensions and bound by the Coulomb attraction instead of a fixed length.

Frequently asked questions

Why does the hydrogen wavefunction split into a radial part and an angular part?

The Coulomb potential depends only on distance r, not direction, so the Hamiltonian commutes with the total angular momentum operators L² and L_z. That shared symmetry lets you look for solutions of the product form ψ = R(r)·Y(θ,φ), which splits the Schrödinger equation into one purely radial equation and one purely angular equation, joined only by the separation constant l(l+1).

Why does hydrogen's energy depend only on n, not on l or m?

Solving the radial equation with the pure 1/r Coulomb potential forces the radial function to be an associated Laguerre polynomial, and the finiteness condition on that polynomial fixes the energy as Eₙ = −13.6 eV/n², with no l or m dependence. This is a special accidental degeneracy of the 1/r potential alone; it disappears for any other central potential and is lifted in real multi-electron atoms by electron screening.

What do the quantum numbers n, l and m actually control?

The principal number n (1, 2, 3, …) sets the energy shell. The azimuthal number l (0 to n−1) sets the orbital's shape and is labelled s, p, d, f by historical spectroscopic convention. The magnetic number m (−l to +l) sets the orbital's orientation in space. For each n there are n² distinct orbitals, which doubles to 2n² once electron spin is included.

Try it live

Everything above runs in your browser — open Hydrogen Orbital, dial in n, l and m, and watch the probability cloud reshape between s, p and d as the sliders change.

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