University Physics IV · The Hydrogen Atom · 10.8

Degeneracy in ℓ and m, and What Lifts It

Hydrogen's n = 2 level holds eight states at one energy, and that is not arithmetic luck. One of its degeneracies belongs to every central potential; the other belongs to the inverse-square law alone. Tell them apart and you can say in advance what a given perturbation will split and what it cannot touch.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A hydrogen level labelled by n holds n² orbital states — 2n² once spin is counted — and the Schrödinger solution hands them all the same energy. Degeneracy on that scale is never coincidence; it is a conserved quantity you have not named. Two symmetries are at work, and they are not equally robust. Rotational invariance holds for every central potential: because H commutes with all three components of L, the 2ℓ+1 states of one ℓ must share an energy, and no reshaping of the radial physics can separate them.

The degeneracy across different ℓ at fixed n is a different animal. It survives only for V ∝ −1/r, where the Laplace-Runge-Lenz vector is conserved as well; L and that vector close on an SO(4) algebra whose irreducible representations have dimension exactly n². Calling that accidental is a confession, not an explanation. The cost of the model is that both symmetries are idealisations — a point nucleus, no relativity, no spin, no other electrons, no external field.

Every correction either deforms the potential away from 1/r or picks out a direction in space, and each therefore breaks one specific degeneracy by a predictable amount: screening by electronvolts, fine structure by tens of microelectronvolts, the Lamb shift by a few. Reading a level diagram is mostly reading which symmetry has just been broken.

Simple definition: Two states are degenerate when different quantum numbers give exactly the same energy; hydrogen's n² degeneracy exists because the Hamiltonian commutes with both the angular momentum L and the Runge-Lenz vector A.
Example: The n = 3 level has ℓ = 0, 1, 2 holding 1 + 3 + 5 = 9 orbital states, all at −13.6/9 = −1.51 eV. In sodium the same three sit 2.10 eV apart, because a screened core is no longer a 1/r potential.

State count of a levelg(n) = Σ(2ℓ+1) for ℓ = 0…n−1 = n² · 2n² with spin

n = 3 gives 9 orbital states; n = 4 gives 16, or 32 with spin — the length of the fourth period.

n = 1, 2, 3, …, and g is a pure count. Shell capacities run 2, 8, 18, 32.

Rotational symmetry protects m[H, L²] = [H, L_z] = 0 for any V(r)

The 2ℓ+1 values of m cannot split unless something in the problem picks out a direction in space.

L in J s. The commutators hold whenever the potential depends on r alone, screened or not.

Runge-Lenz vector: why ℓ is degenerateA = (p×L − L×p)/2μ − (e²/4πε₀) r̂, [H, A] = 0

Conserved only for a strict 1/r force. With L it generates SO(4), whose irreps have dimension n².

μ is the reduced mass in kg; A carries units of energy × length, J m.

Fine structure depends on n and j onlyE(n, j) = −(13.6 eV/n²)[1 + (α²/n²)(n/(j+½) − ¾)]

Splits 2p₃/₂ from 2p₁/₂ by 45.3 μeV, yet leaves 2s₁/₂ and 2p₁/₂ exactly together.

α = 1/137.036, so α² = 5.325×10⁻⁵; j = ℓ ± ½; energies in eV.

Quantum defect: screening restores ℓE(n,ℓ) = −13.6 eV / (n − δ_ℓ)²

One number per ℓ turns the hydrogen spectrum into an alkali one, and puts sodium's D line at 589 nm.

δ_ℓ is dimensionless and largest for penetrating low-ℓ orbits. Sodium: δₛ = 1.37, δₚ = 0.88, δ_d = 0.01.

Applied fields choose an axisΔE = g_J μ_B B mⱼ · ΔE = ±3 e a₀ E for n = 2

B breaks the m-degeneracy; the Stark shift is linear in E only because 2s and 2p₀ start degenerate.

μ_B = 57.9 μeV T⁻¹ and a₀ = 0.0529 nm, so 3ea₀E = 15.9 μeV at 10⁵ V m⁻¹.

Count the states before you explain them

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.

The m-degeneracy is rotational symmetry, and it is generic

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.

The ℓ-degeneracy belongs to the inverse-square force alone

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 third 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.

Screening is the largest splitting, and it is not relativity

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 = 1.37, δₚ = 0.88 and δ_d = 0.01, 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.

Fine structure, Lamb shift, hyperfine: in order of size

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.

Fields break the symmetry on purpose

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Magnetic field B along z0.60 T

Electric field E along z1.0 ×10⁵ V/m

Zero both sliders: four states, one line — that is the n² degeneracy. Raise B and only the m = ±1 pair moves; raise E and only the m = 0 pair moves, and it moves at all only because 2s and 2p₀ share an energy and have opposite parity.

ZEEMAN SPLIT34.7 μeV

STARK SPLIT15.9 μeV

FULL SPREAD69.5 μeV

DISTINCT LEVELS4

Live interpretationZEEMAN SPLIT: 34.7 μeV. STARK SPLIT: 15.9 μeV. FULL SPREAD: 69.5 μeV. DISTINCT LEVELS: 4

Catch the common trap

Explain before calculating.

Misconception check

The standard error is to read "accidental degeneracy" as coincidence, and then to assume that any perturbation splits it. Both halves are wrong. Fine structure does sort n = 2 into j = ½ and j = 3/2, and students conclude that 2s and 2p are now apart — but the fine-structure energy depends on n and j alone, so 2s₁/₂ and 2p₁/₂ stay exactly degenerate through order α⁴mc². Separating them took the Lamb shift, 1057 MHz, and a quantum field theory. The mirror error runs the other way: the 2.10 eV gap between sodium's 3s and 3p gets blamed on spin-orbit coupling, when spin-orbit splits the 3p term itself by only 2.1 meV, a thousand times less. Screening does that work, and it is not relativistic at all. Name the symmetry a mechanism breaks before you estimate its size.

Sodium's valence electron moves in a screened, non-Coulomb central field. Compared with hydrogen's n = 3 level, what happens to the nine orbital states 3s, 3p and 3d?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyHow many distinct orbital states (n, ℓ, m) does hydrogen's n = 4 level hold, how many including spin, and how many values of m does the largest-ℓ subshell carry?

For a fixed n the radial truncation allows ℓ = 0, 1, …, n − 1, so here ℓ = 0, 1, 2, 3 — the s, p, d and f subshells.
Each ℓ carries m = −ℓ up to +ℓ, that is 2ℓ+1 values: 1, 3, 5 and 7.
Add them: 1 + 3 + 5 + 7 = 16 = 4². Consecutive odd numbers always sum to a square, which is why the level count is exactly n².
Spin doubles every orbital state without entering the Hamiltonian, giving 2n² = 32. The largest ℓ is 3, with 7 values of m running from −3 to +3.

Answer16 orbital states, 32 including spin; the ℓ = 3 (f) subshell holds 7 of the 16.

MediumUse E(n, j) = −(13.6 eV/n²)[1 + (α²/n²)(n/(j+½) − ¾)] with α² = 5.325×10⁻⁵ to find the 2p₃/₂–2p₁/₂ splitting in eV and in GHz. Does the same expression separate 2s₁/₂ from 2p₁/₂?

Gross term for n = 2: −13.6/4 = −3.400 eV. The correction prefactor is α²/n² = 5.325×10⁻⁵/4 = 1.331×10⁻⁵.
For j = ½: n/(j+½) − ¾ = 2/1 − 0.75 = 1.25, so the shift is −3.400 × 1.331×10⁻⁵ × 1.25 = −5.658×10⁻⁵ eV.
For j = 3/2: 2/2 − 0.75 = 0.25, so the shift is −3.400 × 1.331×10⁻⁵ × 0.25 = −1.132×10⁻⁵ eV.
The splitting is the difference: 5.658×10⁻⁵ − 1.132×10⁻⁵ = 4.53×10⁻⁵ eV = 45.3 μeV. As a frequency, ΔE/h = 4.53×10⁻⁵ / 4.136×10⁻¹⁵ = 1.09×10¹⁰ Hz.
The expression contains n and j and no ℓ. Both 2s₁/₂ and 2p₁/₂ have n = 2 and j = ½, so they take identical shifts and stay degenerate: fine structure breaks the ℓ-degeneracy only halfway.

Answer45.3 μeV, or 10.9 GHz. 2s₁/₂ and 2p₁/₂ remain degenerate at this order; the 1057 MHz Lamb shift, a QED effect outside the formula, is what finally separates them.

HardSodium's valence terms follow E = −13.6 eV/(n − δ_ℓ)² with δₛ = 1.373 and δₚ = 0.883. Find the 3s and 3p energies and the 3s→3p wavelength, then compare that splitting with hydrogen's n = 3 level and with the 2.1 meV spin-orbit splitting of the 3p term.

3s: n − δₛ = 3 − 1.373 = 1.627, and 1.627² = 2.647, so E(3s) = −13.6/2.647 = −5.138 eV.
3p: n − δₚ = 3 − 0.883 = 2.117, and 2.117² = 4.482, so E(3p) = −13.6/4.482 = −3.035 eV.
The gap is 5.138 − 3.035 = 2.103 eV, and λ = 1239.84 eV nm ÷ 2.103 eV = 590 nm — the yellow sodium D line.
In hydrogen both states sit at −13.6/9 = −1.511 eV with no gap at all. The whole 2.10 eV is the failure of the 1/r potential: the 3s orbital penetrates the neon-like core and sees an effective charge well above one, while 3p barely does.
Rank the two mechanisms: screening 2.10 eV, spin-orbit 2.1 meV. Screening is a thousand times larger, so the D line's position measures screening and only its 0.6 nm doublet splitting measures a relativistic effect.

AnswerE(3s) = −5.14 eV, E(3p) = −3.03 eV, ΔE = 2.10 eV, λ = 590 nm. Hydrogen's gap is exactly zero, and spin-orbit supplies only 0.1% of sodium's.