Degeneracy in ℓ and m, and what lifts it

Fix n. The radial truncation allows ℓ = 0, 1, …, n − 1, and each ℓ carries 2ℓ+1 values of m, so the level holds 1 + 3 + 5 + … + (2n − 1) orbital states. That is the odd-number series, and it sums to n²: nine for n = 3, sixteen for n = 4. Spin does not appear in the Schrödinger Hamiltonian at all, so every one of those states comes in two, giving 2n² — the shell capacities 2, 8, 18, 32 that set the lengths of the periods. Two remarks keep the count honest. It counts bound states only; the continuum above E = 0 is not quantised and is not in it. And it counts the states of an idealised Hamiltonian: one electron, a point charge, a strict 1/r potential, no relativity, no spin coupling, no field. Everything that follows is about which parts of that count survive when the idealisations are dropped one at a time.

Any potential depending on r alone leaves the Hamiltonian invariant under rotations, so H commutes with all three components of L. Two things follow. The energy cannot depend on m, because the ladder operators L₊ and L₋ change m by one without touching H, carrying any eigenstate onto a degenerate partner; a level of orbital angular momentum ℓ therefore holds 2ℓ+1 states at one energy. And no change to the radial problem can undo it. Screen the nucleus with fifty electrons, give it a finite radius, make the potential exponential — while V is a function of r, the 2ℓ+1 degeneracy stands. Switch on spin-orbit coupling and L alone stops being conserved, but J = L + S is conserved and generates the same rotations, so the multiplicity simply becomes 2j+1. Only something carrying a direction — an applied electric or magnetic field, a neighbouring atom, a crystal lattice — lifts an m-degeneracy.

None of that explains why 2s and 2p share an energy: they have different ℓ, and rotations do not connect them. The classical clue is Bertrand's theorem — only the 1/r and the harmonic potentials give closed bounded orbits — and what closes a Kepler ellipse is a second conserved vector, the Laplace-Runge-Lenz vector A, which points along the major axis and fixes the orbit's orientation. Quantum mechanically A survives as a Hermitian operator commuting with H. Suitably scaled, L and A satisfy the commutation relations of SO(4) rather than SO(3), and for bound states that algebra splits into two independent spin-like pieces with j₁ = j₂ = (n − 1)/2. The irreducible representation has dimension (2j₁ + 1)(2j₂ + 1) = n²: the whole level, every ℓ included, in one multiplet. The degeneracy was never accidental — it is a symmetry acting in a space larger than ordinary rotations. Move the exponent of r off −1 and A stops being conserved, and the ℓ levels fan apart at once.

Give the atom more electrons and the valence electron no longer sees a bare 1/r field. Close in it feels the full nuclear charge Z; well outside the core it feels roughly one unit of charge. Low-ℓ orbitals penetrate the core, because the centrifugal term ℓ(ℓ+1)ħ²/2μr² keeps high-ℓ states out of it, so an s electron spends more time where the effective charge is large and is bound more tightly. Spectroscopy absorbs the whole effect into one number per ℓ, the quantum defect: E = −13.6 eV/(n − δ_ℓ)². Sodium has δₛ = 1.373, δₚ = 0.883 and δ_d = 0.010, putting 3s at −5.14 eV and 3p at −3.03 eV. Hydrogen holds those two at the same energy; sodium separates them by 2.10 eV, and the transition between them is the 589 nm yellow of a street lamp. The m-degeneracy is untouched, because a closed core is spherical on average. The same ℓ-dependence is why 4s fills before 3d.

In hydrogen itself three corrections of order α⁴mc² arrive together: the relativistic p⁴ correction to the kinetic energy, the spin-orbit term, and the Darwin term that acts only on ℓ = 0. Their sum is remarkable — it depends on n and j and not on ℓ. So fine structure sorts n = 2 into j = ½ and j = 3/2, separating 2p₃/₂ from 2p₁/₂ by 4.53×10⁻⁵ eV, or 10.9 GHz, while leaving 2s₁/₂ and 2p₁/₂ exactly degenerate. Removing that last remnant of the SO(4) degeneracy took quantum electrodynamics: the Lamb shift lifts 2s₁/₂ above 2p₁/₂ by 1057 MHz, or 4.37 μeV, a tenth of the fine structure. Comparable in size is hyperfine structure, the coupling to the proton's moment, worth 1420 MHz or 5.87 μeV in the ground state — the 21 cm line radio astronomy runs on. Each mechanism resolves a degeneracy the one above it left standing, and the ladder 10.2 eV, then 45 μeV, then a few μeV is the ranking to carry in your head.

Put the atom in a magnetic field and you have chosen an axis. In the weak-field regime a level of given j fans into 2j+1 sublevels spaced by g_J μ_B B, and μ_B B is 57.9 μeV at 1 T — comparable with n = 2's 45 μeV of fine structure, which is why hydrogen's Paschen-Back crossover sits near one tesla. An electric field is the more revealing case, because it reads the ℓ-degeneracy directly. First-order perturbation theory gives no shift for a non-degenerate state of definite parity, since ⟨z⟩ = 0 there. But 2s and 2p₀ are degenerate and of opposite parity, so degenerate perturbation theory diagonalises a 2×2 block and returns shifts of ±3ea₀E: 15.9 μeV at 10⁵ V m⁻¹, linear in the field, with the m = ±1 states unmoved. Sodium shows no linear Stark effect at all, because screening has already separated 3s from 3p, so its shift is quadratic. The linear Stark effect in hydrogen is the accidental degeneracy made visible on a bench.