University Physics V · The Hydrogen Atom · 11.7
Degeneracy, Runge–Lenz & the Hidden so(4)
Hydrogen's n²-fold degeneracy gets called accidental, and it is not. This lesson hands you the conserved vector operator behind it, the rescaling that turns it into a second su(2), and a checklist for deciding which degeneracies a real atom is still entitled to.
Build the model
Connect the measurement to the mechanism.
Two degeneracies sit inside hydrogen and they have different provenances. The 2l+1 states of fixed l share an energy because [H, L] = 0 for any V(r); that is so(3), and no atom escapes it. But hydrogen also puts l = 0 through n − 1 at one energy, and rotational symmetry says nothing about that.
The extra conservation law is the Runge–Lenz vector, classically A = p × L − μk r̂, which points along the major axis and stands still only when the force goes as exactly 1/r² — the same statement as a closed orbit. Symmetrise it into a Hermitian operator M and two identities do all the work: M⋅L = 0, and M² = (2H/μ)(L² + ħ²) + k². Restricted to a bound eigenspace, E < 0 lets you rescale M′ = √(−μ/2E) M so that [M′ᵢ, M′ⱼ] = iħ εᵢⱼₖ Lₖ, and L together with M′ closes so(4).
Split it as I = (L + M′)/2 and K = (L − M′)/2 and you have two commuting su(2)s whose Casimirs are forced equal by M′⋅L = 0; a single label j then delivers E = −μk²/2ħ²(2j+1)², dimension (2j+1)², and l running from 0 to 2j. So n = 2j+1 is a representation label, and the n² multiplet is one irreducible object rather than a coincidence. The cost is that this symmetry is fragile where so(3) is not: any departure from 1/r destroys it, and M, though conserved, does not commute with L², so it can never join n, l and mₗ in a label set.
- Simple definition
- The Runge–Lenz vector is a second conserved vector, over and above angular momentum, that exists only for a strictly 1/r potential; together with L it generates so(4), whose irreducible representations are hydrogen's n²-fold levels.
- Example
- For hydrogen n = 3 the algebra sets j = (n − 1)/2 = 1, so the level is one (2j+1)² = 9-dimensional multiplet holding l = 0, 1, 2 — the 1 + 3 + 5 = 9 orbital states that all sit at −1.512 eV.
A third conserved vector on top of E and L leaves the orbit no freedom to precess, which is why Kepler ellipses close.
k = Ze²/4πε₀ in J m, with e the elementary charge; A carries kg² m³ s⁻². Its magnitude is μk times the orbital eccentricity, and it points from the focus toward perihelion.
Orthogonality to L is what later forces the two su(2) Casimirs to be equal, collapsing two labels into the single j.
p × L is not Hermitian on its own, since (p × L)† = −L × p. M carries J m, and [H, M] = 0 exactly.
This single relation converts a representation label into an energy. No differential equation is solved anywhere in the derivation.
The classical A² = 2μEL² + μ²k² divided by μ², with L² shifted to L² + ħ² — the one purely quantum term.
Unrescaled, [Mᵢ, Mⱼ] = −(2iħ/μ) εᵢⱼₖ H Lₖ, which is no Lie algebra at all because an operator sits in the structure constant.
Legal only on a bound eigenspace, E < 0; M′ carries J s, the units of L. Above threshold the sign flips, closing so(3,1) instead.
Two independent spins make the level the (j, j) representation, of dimension (2j+1)², carrying l = 0 to 2j by Clebsch–Gordan.
so(4) ≅ su(2) ⊕ su(2). I² − K² = (L⋅M′ + M′⋅L)/2 = 0 forces one j, and su(2) allows j = 0, ½, 1, …
The integer a series truncation used to produce now arrives as a representation label, and l ≤ n − 1 follows from adding j to j.
n = 1, 2, 3, … because 2j+1 is. For Z = 1 with μ → mₑ that is −13.606 eV/n²; the true reduced mass trims it to −13.598 eV/n². Degeneracy n², or 2n² once spin is appended.
Two degeneracies, two different warrants
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, Ly] = [H, Lz] = 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 δd = 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.
What the classical vector conserves, and when it stops
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.
Symmetrising, and the two identities that carry the argument
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.
Rescale, then split into two independent spins
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.
Reading the spectrum straight off the representation
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.
What breaks it, in what order, and what that costs
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.
Change one variable at a time
Make the relationship visible.
Leave ε at 0 and the four arrows lie exactly on top of one another: A does not move, the orbit closes, and every l at one n is degenerate. Nudge ε to 0.02 and a single orbit hides the drift — drag N to 6 and the fan opens to 22°.
FORCE EXPONENT 2 + ε2.00
APSIDAL ANGLE Φ180.0 deg
A TURN PER ORBIT Δφ0.00 deg
TOTAL TURN AFTER N ORBITS0.0 deg
Live interpretationFORCE EXPONENT 2 + ε: 2.00. APSIDAL ANGLE Φ: 180.0 deg. A TURN PER ORBIT Δφ: 0.00 deg. TOTAL TURN AFTER N ORBITS: 0.0 deg
Catch the common trap
Explain before calculating.
The symmetrised Runge–Lenz operator M commutes with the hydrogen Hamiltonian, so it is a constant of the motion. Why can its eigenvalues nevertheless not be added to n, l and mₗ as a fourth label for hydrogen states?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFor the hydrogen level n = 5, find the su(2) ⊕ su(2) label j, the dimension of the representation and the l values it contains. Check that dimension against the direct sum over l, give the level energy for Z = 1 with an infinitely heavy nucleus, and say how many distinct energies survive once the potential is screened.
- The algebra gives n = 2j + 1, so j = (n − 1)/2 = 2. Both su(2) factors carry the same j, forced by M′⋅L = 0.
- The representation is (j, j), of dimension (2j + 1)² = 5² = 25 = n². Every one of those states shares an energy while M is conserved.
- Its orbital content is the Clebsch–Gordan series for 2 ⊗ 2: l = 0, 1, 2, 3, 4, that is l = 0 up to 2j = n − 1.
- Direct count: Σ(2l + 1) over l = 0…4 is 1 + 3 + 5 + 7 + 9 = 25, matching. Appending the spin factor doubles it to 2n² = 50.
- Energy: E₅ = −13.6057 eV / 5² = −13.6057/25 = −0.5442 eV, the same for all 25 orbital states.
- Screen the potential and M dies while L survives: the 25 states split into five energies, one per l, of degeneracies 1, 3, 5, 7 and 9, and no (2l+1) block splits.
Answerj = 2, a 25-dimensional (2, 2) multiplet holding l = 0, 1, 2, 3, 4, all at −0.5442 eV, or 50 states with spin. Screening leaves five energies of degeneracy 1, 3, 5, 7, 9 — so(3) survives and so(4) does not.
MediumStarting from I² = K² = ħ²j(j+1), M′⋅L = 0 and M′² = −(L² + ħ²) − μk²/2E, derive Eₙ = −μk²/2ħ²n² with n = 2j + 1. Then evaluate the n = 2 level of He⁺, taking μ/mₑ = 0.999863 for a ⁴He nucleus, and compare it with the n = 1 level of hydrogen, where μ/mₑ = 0.999456.
- I² − K² = (L⋅M′ + M′⋅L)/2 = 0, so the two Casimirs are equal: I² = K² = (L² + M′²)/4, one label j for both factors.
- Substitute the rescaled Casimir identity: L² + M′² = L² − (L² + ħ²) − μk²/2E = −ħ² − μk²/2E, with no L² left.
- Hence 4ħ²j(j+1) = −ħ² − μk²/2E, that is ħ²[4j(j+1) + 1] = ħ²(2j+1)² = −μk²/2E.
- So E = −μk²/2ħ²(2j+1)². With j = 0, ½, 1, … the quantity 2j+1 runs over the positive integers; call it n and Eₙ = −μk²/2ħ²n².
- Scaling: k = Ze²/4πε₀ and Eₙ ∝ μk², so Eₙ = −13.6057 eV × Z²(μ/mₑ)/n². For He⁺ at n = 2: E = −13.6057 × 4 × 0.999863 / 4 = −13.6038 eV.
- Hydrogen at n = 1: E = −13.6057 × 0.999456 = −13.5983 eV. The two differ by 5.5 meV, entirely from the reduced masses — Z²/n² alone would make them identical.
AnswerEₙ = −μk²/2ħ²n² with n = 2j + 1 an integer. He⁺ n = 2 lies at −13.6038 eV and H n = 1 at −13.5983 eV, 5.5 meV apart, the whole gap coming from the reduced mass.
HardFollow hydrogen's n = 2 level as the so(4) symmetry is dismantled. Treat the four orbital states as degenerate at −3.4014 eV, then (a) find the fine-structure gap from E_{n, j} = −(13.6057/n²)[1 + (α²/n²)(n/(j+½) − ¾)] with α² = 5.3251 × 10⁻⁵, in eV and in GHz; (b) compare it with the 1057.8 MHz Lamb shift; (c) find the field at which the linear Stark shift 3ea₀ℰ matches the fine structure, and say why sodium has no linear Stark effect at all.
- Unbroken: n = 2 holds (2j+1)² = 4 orbital states, 2s and 2p, at E₂ = −13.6057/4 = −3.4014 eV. So(4) alone would keep them exactly level.
- Fine structure, j = ½: the bracket is (2/1 − 0.75) = 1.25, so the shift is 3.401425 × (5.3251 × 10⁻⁵/4) × 1.25 = 3.401425 × 1.331275 × 10⁻⁵ × 1.25 = 5.660 × 10⁻⁵ eV downward.
- Fine structure, j = 3/2: the bracket is (2/2 − 0.75) = 0.25, a shift of 3.401425 × 1.331275 × 10⁻⁵ × 0.25 = 1.132 × 10⁻⁵ eV. The gap is 5.660 × 10⁻⁵ − 1.132 × 10⁻⁵ = 4.528 × 10⁻⁵ eV; dividing by h = 4.1357 × 10⁻¹⁵ eV s gives 10.95 GHz.
- That correction depends on j alone, so 2s½ and 2p½ stay degenerate — Dirac leaves one piece of the old degeneracy standing. QED removes it: the Lamb shift is 1057.8 MHz = 1.0578 × 10⁹ × 4.1357 × 10⁻¹⁵ = 4.375 × 10⁻⁶ eV, which is 1057.8/10949 = 9.7% of the fine-structure gap.
- Stark: the linear n = 2 shift is ±3ea₀ℰ, so it matches the fine structure when 3a₀ℰ = 4.528 × 10⁻⁵ V, that is ℰ = 4.528 × 10⁻⁵ / (3 × 5.2918 × 10⁻¹¹) = 2.85 × 10⁵ V m⁻¹. Well below that field the fine structure dominates and the shift is quadratic; well above it, linear.
- The linear term exists only because 2s and 2p, of opposite parity, share an energy, so ⟨2s|z|2p⟩ acts in first order — a degeneracy handed over by so(4). In sodium 3s and 3p are 2.10 eV apart, so degenerate perturbation theory has nothing to mix and the Stark effect is quadratic.
AnswerFine structure splits n = 2 by 4.528 × 10⁻⁵ eV, or 10.95 GHz; the Lamb shift removes the last 2s½–2p½ degeneracy at 1057.8 MHz = 4.375 × 10⁻⁶ eV, 9.7% of that; and 3ea₀ℰ matches the fine structure at ℰ = 2.85 × 10⁵ V m⁻¹. Sodium has no linear effect because screening has already split 3s from 3p by 2.10 eV.