Degeneracy, Runge–Lenz and the hidden so(4)

Count the hydrogen level n = 4: l = 0, 1, 2, 3 gives 1 + 3 + 5 + 7 = 16 = n² orbital states, and 32 once spin is appended. Two separate facts produced that number, and they fail under different provocations. Each (2l+1) block is degenerate because [H, Lₓ] = [H, L_y] = [H, L_z] = 0 for any V(r) whatsoever, so L_± walks you through the multiplet without touching E — pure so(3). That warrant is geometric, and it holds for every rotationally invariant Hamiltonian: hydrogen, helium, sodium, any ion with a central core. It is not unbreakable, only well defended — it fails the moment something picks out an axis, which is what a magnetic or electric field does when it splits m, and what a molecule does permanently, since an electronic Hamiltonian built on two or more fixed nuclei keeps only an axial subgroup. What a central field does not protect is the equality of the four l blocks. Put the same electron in sodium's screened field and that equality is gone: with quantum defects δₛ = 1.373, δₚ = 0.883 and δ_d = 0.010 in E = −13.606/(n − δₗ)² eV, the n = 3 terms sit at −5.14, −3.04 and −1.52 eV. That is a 2.10 eV gap between 3s and 3p — the sodium D lines near 589 nm, worth 2.105 eV — while each (2l+1) block inside it stays degenerate to the accuracy of the central-field model, with only the far smaller spin–orbit term splitting 3p into the doublet those two D lines actually are. The m-degeneracy and the l-degeneracy are therefore not one phenomenon, and only one of them is protected by rotations.

The two-body problem with V = −k/r conserves more than it has any right to. Beyond E and the three components of L there is A = p × L − μk r̂. Differentiate it: with ṗ = −k r̂/r² the two terms cancel identically — but only because the force went as exactly 1/r². A lies in the orbital plane, since A⋅L = 0, points from the focus toward perihelion, and has magnitude μke with e the orbital eccentricity, so it fixes both the orientation and the shape of the ellipse. Take away that exact power and A starts to turn. For a nearly circular orbit under F ∝ r^−(2+ε) the apsidal angle is Φ = π/√(1 − ε), so the perihelion advances by 2π(1/√(1 − ε) − 1) per revolution, and ε = 0 is the only value returning zero. General relativity adds an effective −GML²/c²r³ term, and Mercury's perihelion creeps forward 43 arcseconds a century. Bertrand's theorem names the harmonic force as the only other law that closes orbits, and it too hides a symmetry — su(3), which is where the isotropic oscillator's own extra degeneracy comes from.

Promoting A to an operator needs one repair. p × L is not Hermitian, because (p × L)† = −L × p, so the symmetric combination M = (1/2μ)(p × L − L × p) − k r̂ is the operator you want; the two orderings differ at order ħ, which is where the quantum corrections below originate. Three facts then follow by direct computation and they are all you need. First [H, M] = 0, so M is conserved. Second M⋅L = L⋅M = 0, the operator form of the statement that A lies in the orbital plane. Third, the Casimir identity M² = (2H/μ)(L² + ħ²) + k², which is the classical A² = 2μEL² + μ²k² divided by μ² with L² shifted to L² + ħ². That lone ħ² is not decoration. Keep it and the spectrum goes as 1/(2j+1)² = 1/n²; drop it and the same algebra returns 1/4j(j+1) = 1/(n² − 1), which sends the ground state to −∞. Notice what has not been used: no Laguerre polynomial, no series truncation, no asymptotic analysis at large r.

The bracket [Mᵢ, Mⱼ] = −(2iħ/μ) εᵢⱼₖ H Lₖ is not a Lie algebra: its structure constant is an operator. On a single bound eigenspace, though, H acts as the number E, and for E < 0 the factor −μ/2E is positive, so M′ = √(−μ/2E) M is well defined and Hermitian, with [M′ᵢ, M′ⱼ] = iħ εᵢⱼₖ Lₖ. Now L and M′ close: six generators obeying the commutation relations of so(4), the rotation algebra of four-dimensional Euclidean space — with no suggestion that the electron moves in four dimensions. The standard move is I = (L + M′)/2 and K = (L − M′)/2, which gives [Iᵢ, Iⱼ] = iħ εᵢⱼₖ Iₖ, the same for K, and [Iᵢ, Kⱼ] = 0: two independent angular momenta. Their Casimirs differ by I² − K² = (L⋅M′ + M′⋅L)/2, which vanishes because M′ is orthogonal to L, so one number j labels both, and su(2) permits j = 0, ½, 1, 3/2, … . The half-integers are bookkeeping for the hidden algebra and have nothing to do with spin: L = I + K adds two equal j's and is therefore always integer, as an orbital angular momentum must be.

Everything from here is algebra. Set I² = K² = ħ²j(j+1) and use the Casimir identity in rescaled form, M′² = −(L² + ħ²) − μk²/2E, so that L² + M′² = −ħ² − μk²/2E with no L² left standing. Since I² = K² = (L² + M′²)/4, this reads 4ħ²j(j+1) = −ħ² − μk²/2E, that is ħ²(2j+1)² = −μk²/2E. Write n = 2j+1 and Eₙ = −μk²/2ħ²n², which for Z = 1 and μ → mₑ is −13.606 eV/n². The representation is the (j, j) of su(2) ⊕ su(2), of dimension (2j+1)² = n², and its orbital content follows from adding j to j: l = 0, 1, …, 2j = n − 1, with Σ(2l+1) = n², matching the direct count exactly. Three results the series solution produces separately have arrived together — the integer n, the level formula, and the range l ≤ n − 1. And the labels change meaning: for n = 3, j = 1, and 3s, 3p and 3d stop being three unrelated solutions of a differential equation. They are the l = 0, 1 and 2 pieces of a single irreducible object, which is why nothing inside the Coulomb problem can tell them apart.

Rank the breakings by size, because each removes a different piece of the algebra. Screening is by far the largest: any V ≠ −k/r destroys M outright, so the l-degeneracy goes while every (2l+1) block survives, and sodium's 2.10 eV 3s–3p gap stands about 4.6 × 10⁴ times above hydrogen's n = 2 fine structure. That fine structure is the next term down, entering at relative order α² = 5.3251 × 10⁻⁵: the p⁴, spin–orbit and Darwin corrections depend on j rather than l, so they split n = 2 into j = ½ and j = 3/2 while leaving 2s½ and 2p½ exactly on top of one another — Dirac's Coulomb solution preserves that last shard of the old symmetry, and only QED removes it, the Lamb shift lifting 2s½ above 2p½ by roughly a further factor of ten below the fine structure. An external field keeps only the generators it respects, and hydrogen's linear Stark effect exists at all because the surviving l-degeneracy puts opposite-parity states at one energy, so ⟨2s|z|2p⟩ acts already in first order; the Hard example below puts all three scales in eV and finds the field at which the Stark term overtakes the fine structure. Above threshold nothing is broken, only changed: the same rescaling with E > 0 gives [M″ᵢ, M″ⱼ] = −iħ εᵢⱼₖ Lₖ, the Lorentz algebra so(3,1), whose unitary representations are infinite-dimensional — the continuum.