University Physics V · The Hydrogen Atom · 11.6
Quantum Numbers & the Spin Tensor Factor
Students arrive able to recite four quantum numbers and unable to say what any of them is an eigenvalue of. This lesson pins each label to an operator, shows which one is a truncation index rather than an eigenvalue, and makes the doubling by spin visibly an import rather than a result.
Build the model
Connect the measurement to the mechanism.
A hydrogen eigenstate needs four labels because the space it lives in has three tensor slots and the Hamiltonian is blind to one of them. Write that space as L²((0,∞), dr) ⊗ L²(S²) ⊗ C². On the angular slot, L² and Lz hand over l and mₗ, and they would do so for any central potential.
On the radial slot, the demand that the Frobenius series terminate — otherwise u grows like e(+r/na₀) and leaves the Hilbert space — hands over an integer n = nᵣ + l + 1 with nᵣ ≥ 0 the node count, and that one equation is the entire content of the rule l ≤ n − 1. On the third slot nothing happens at all: the Schrödinger Hamiltonian is Horb ⊗ I₂, so it commutes with every component of S, Sz is free to join the commuting set, and each orbital level is doubled exactly — to 2n² states — without moving by a single meV. That is the honest shape of the label set: three labels earned by solving an equation, one imported by enlarging the space.
The price is that labels are only as good as the Hamiltonian certifying them. Switch on ξ(r)L⋅S and mₗ and mₛ cease to be eigenvalue labels, replaced by j and mⱼ, while the count 2n² survives untouched, because a change of basis rotates a degenerate subspace and never shrinks it.
- Simple definition
- A quantum number is the eigenvalue label a state carries for one member of a complete set of commuting observables, and for one electron in a central Coulomb field that set is (H, L², Lz, Sz), giving n, l, mₗ and mₛ.
- Example
- The ket |3 2 −1⟩ ⊗ |↓⟩ has H eigenvalue −13.606/3² = −1.51 eV, L² eigenvalue 2(2 + 1)ħ² = 6ħ², Lz eigenvalue −ħ and Sz eigenvalue −ħ/2; it is one of the 2 × 3² = 18 states sharing that energy.
One tensor slot per label group, so a basis vector is the product |n l⟩|l mₗ⟩|mₛ⟩ and the label list is just the slot list.
Dimensions multiply, never add: 2l + 1 angular states times 2 spin states is 2(2l + 1).
(H, L², Lz, Sz) commutes pairwise and is complete, so the four eigenvalues fix one ray in the space, not a subspace.
ħ = 1.0546 × 10⁻³⁴ J s; l and mₗ are dimensionless integers, mₛ = ±1/2. H supplies n through Hψ = Eₙψ.
The rule l ≤ n − 1 is not extra input — it is nᵣ ≥ 0 rewritten, since a series cannot stop after a negative number of terms.
nᵣ is the node count of u(r) in 0 < r < ∞, endpoints excluded; n and l integers, nᵣ integer.
The n² is rotational symmetry plus the accidental Coulomb degeneracy; the factor 2 is the C² slot and nothing else.
n = 1, 2, 3, 4 give 2, 8, 18, 32 states — the K, L, M and N shell capacities. A periodic-table block is one subshell wide, 2(2l + 1) = 2, 6, 10, 14.
Exact doubling at zero energy cost, which is why Eₙ carries no mₛ and why a doublet needs a term this H does not have.
S = (ħ/2)σ with σ the Pauli matrices; S² = (3/4)ħ²I₂ on every state, so s = 1/2 is fixed and never varies.
Goodness belongs to the pair (operator, Hamiltonian), never to the atom: mₗ and mₛ label eigenstates only while [H, Lz] and [H, Sz] vanish.
L⋅S = ½(J² − L² − S²), in J² s². Counting survives: 2(2l + 1) = (2l + 2) + 2l.
Four labels, four commuting operators
A quantum number is never just a counter; it is the eigenvalue a state carries for one operator in a commuting set. For one electron in a central field that set is (H, L², Lz, Sz). The first two brackets vanish for any V(r): [H, L²] = [H, Lz] = 0. The third is free, because S acts on a different tensor slot from everything orbital, so Sz commutes with H, with L² and with Lz identically. Complete means the four eigenvalues leave no residual degeneracy: fix Eₙ, l(l + 1)ħ², mₗħ and mₛħ and you have named one ray in the Hilbert space rather than a subspace. Notice what is missing from that list. S² belongs to the set too, but for a single electron S² = (3/4)ħ²I on every state, so s = 1/2 labels the representation the electron carries, not the state it is in; it never varies and therefore distinguishes nothing. Only mₛ does any work.
n is where the series has to stop
n is the odd one out: it is not read off an independent operator but forced by normalisability. Put u = rR, scale ρ = 2r/na₀, and strip the asymptotics with u = ρ(l+1)e(−ρ/2)v(ρ). The ratio of successive coefficients in the power series for v is (k + l + 1 − n)/[(k + 1)(k + 2l + 2)], which for large k behaves as 1/k — the coefficients of eρ. If the series never stops, v grows like eρ, u like e(+r/na₀), and the state leaves the Hilbert space. So the numerator must vanish at some integer k = nᵣ ≥ 0, giving n = nᵣ + l + 1. That one equation carries everything usually recited as separate rules: n starts at 1, l is at most n − 1, and the radial function of a given (n, l) has exactly nᵣ = n − l − 1 nodes. At n = 3 the subshells are 3s with two nodes, 3p with one and 3d with none — a nodeless radial function always sits at l = n − 1.
The angular labels belong to the sphere
l and mₗ come from L²(S²) and know nothing about the Coulomb potential. L² and Lz are built from angles alone, so their eigenbasis — the spherical harmonics, with ⟨n̂|l m⟩ = Yₗm(n̂) — is shared by the screened atom, the three-dimensional oscillator and the nuclear shell model alike. The ladder operators fix the range: L+ must annihilate the top state and L_− the bottom, which forces mₗ to run in integer steps from −l to +l, a multiplet of 2l + 1 states. Parity is (−1)l, and that is what later kills every dipole element with Δl = 0. The eigenvalues are also not a vector pointing somewhere: for l = 2, |L| = √6 ħ = 2.449ħ while the largest projection is only 2ħ, so cos θ = 2/√6 = 0.816 and θ = 35.3°. A non-zero angular momentum can never lie along its own quantisation axis, because Lₓ and Ly would then both be sharp and [Lₓ, Ly] = iħLz forbids it.
Spin is a tensor factor you add, not a solution you find
The Schrödinger equation is one scalar partial differential equation; separating it in a central field yields two separation constants, l and mₗ, on top of the energy eigenvalue — three labels in all, with no fourth hiding in the boundary conditions. Spin enters by enlarging the space: replace L²(ℝ³) with L²(ℝ³) ⊗ C² and postulate S = (ħ/2)σ, with σ the Pauli matrices. The old Hamiltonian becomes Horb ⊗ I₂. It commutes with all three components of S, its eigenvalues are exactly those of Horb, and every eigenspace is doubled with no energy shift whatever — hydrogen's n = 3 level goes from 9 states to 18 and stays at −1.51 eV. Two honest footnotes. The evidence is experimental: a silver beam splits in two, sodium's D line is a doublet, the Zeeman pattern is anomalous. The derivation is relativistic: the Dirac equation forces four components and delivers gₛ = 2. And the axis is a choice — mₛ is the Sz eigenvalue, but n̂⋅S serves equally well, with |n̂,+⟩ = (cos(θ/2), e(iφ)sin(θ/2)) in the z basis.
Count the level: n² orbital, 2n² in all
Fix n and sum the multiplets: 2l + 1 over l = 0 … n − 1 is 1 + 3 + 5 + … + (2n − 1), the sum of the first n odd numbers, which is n². Multiply — do not add — by the two states of the spin slot, and the level holds 2n². For n = 1, 2, 3, 4 that is 2, 8, 18, 32. The same number arrives from another direction in the next topic: rescaled on the negative-energy subspace, the Runge–Lenz vector closes so(4), and each level is one irreducible multiplet of dimension n². Read the count physically with care, because two other familiar sequences are not this one. The s, p, d and f blocks are 2, 6, 10 and 14 wide, since a block is one subshell's 2(2l + 1), not a whole shell's 2n². And the periodic table's rows run 2, 8, 8, 18, 18, 32, 32, because in a many-electron atom the l-degeneracy is already broken and 4s fills before 3d. So 2n² counts a hydrogenic level — it is the capacity of the K, L, M, N shells — and neither a block width nor a chemical period.
Good labels are good only against a Hamiltonian
Add ξ(r)L⋅S and the list changes. L⋅S contains L+S_− + L_−S+, which converts one unit of mₗ into one unit of mₛ, so [H, Lz] and [H, Sz] no longer vanish and mₗ and mₛ stop labelling eigenstates. What survives is J = L + S: the good set becomes (H, L², S², J², Jz) and the labels are n, l, j, mⱼ with j = l ± 1/2. Nothing is lost from the count, because 2(2l + 1) = (2l + 2) + 2l is exactly 2j + 1 summed over the two j values — six 2p states become four with j = 3/2 and two with j = 1/2. The scale is small: hydrogen's 2p interval is 10.97 GHz, or 4.54 × 10⁻⁵ eV, about 1.3 × 10⁻⁵ of |E₂| = 3.40 eV. Push a magnetic field past roughly 0.78 T, where μB B overtakes that interval, and the Paschen–Back regime hands mₗ and mₛ back. The label set is not a property of the atom; it is a property of the atom plus the terms you chose to keep.
Change one variable at a time
Make the relationship visible.
Drag n from 3 to 4 and a fourth bar appears: l = n − 1 = 3 is the largest the truncation n = nᵣ + l + 1 permits, and it is the nodeless one. Then drop the spin factor to 1 and every bar halves — that factor is the C² slot, imported, not produced by the radial equation.
ORBITAL STATES n²9 states
WITH SPIN sd × n²18 states
RINGED SUBSHELL6 states
RADIAL NODES n − l − 11 nodes
Live interpretationORBITAL STATES n²: 9 states. WITH SPIN sd × n²: 18 states. RINGED SUBSHELL: 6 states. RADIAL NODES n − l − 1: 1 nodes
Catch the common trap
Explain before calculating.
Neglecting fine structure, how many linearly independent states of hydrogen share the energy E₃ = −1.51 eV, and where does each factor in that count come from?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyWhich of these label sets (n, l, mₗ, mₛ) name a genuine hydrogen basis state, and which condition does each failure break? (a) (3, 2, −2, +1/2); (b) (2, 2, 0, −1/2); (c) (4, 3, 3, +1/2); (d) (3, 1, −2, +1/2); (e) (2, 0, 0, +1).
- These are not four conventions but four eigenvalue conditions. n ≥ 1 is an integer because the radial series must terminate; 0 ≤ l ≤ n − 1 restates nᵣ = n − l − 1 ≥ 0; mₗ runs in integer steps from −l to +l because L_± must annihilate the top and bottom of the multiplet; and mₛ = ±1/2 because Sz on C² has only those two eigenvalues.
- (a) (3, 2, −2, +1/2): l = 2 = n − 1 and |mₗ| = 2 = l. Allowed — a 3d state with nᵣ = 0, so its radial function has no nodes.
- (b) (2, 2, 0, −1/2): l = 2 exceeds n − 1 = 1, which would need nᵣ = n − l − 1 = −1. Forbidden: a power series cannot terminate after a negative number of terms.
- (c) (4, 3, 3, +1/2): l = 3 = n − 1 and mₗ = 3 = l. Allowed — the 4f state of maximum projection, again nodeless.
- (d) (3, 1, −2, +1/2): l = 1 is fine, but the l = 1 subspace is three-dimensional, with mₗ = −1, 0 or +1 only. Forbidden: |mₗ| ≤ l, or Lz would exceed |L| = √2 ħ.
- (e) (2, 0, 0, +1): mₛ = 1 is not an eigenvalue of Sz, whose spectrum on C² is ±ħ/2. Forbidden — s = 1/2 is fixed for an electron and only mₛ varies.
Answer(a) and (c) are allowed. (b) breaks l ≤ n − 1, (d) breaks |mₗ| ≤ l, and (e) breaks mₛ = ±1/2. Each failure is a violated eigenvalue condition, not a broken convention.
MediumFor the n = 4 level of hydrogen, neglecting fine structure: list the l multiplets and their sizes, give the total degeneracy, state the dimension of the E₄ eigenspace of H, and count how many of those states have mₗ = 0. Take E₁ = −13.606 eV.
- l is bounded by the truncation condition nᵣ = n − l − 1 ≥ 0, so l = 0, 1, 2, 3 — the 4s, 4p, 4d and 4f subshells, with radial node counts 3, 2, 1 and 0.
- Each l supplies one 2l + 1 dimensional Lz multiplet: 1 + 3 + 5 + 7 = 16 = n². It is the sum of the first n odd numbers, which is exactly why the orbital count is a perfect square.
- The spin slot multiplies rather than adds: 16 × 2 = 32 = 2n². So the E₄ eigenspace of H is 32-dimensional, at E₄ = −13.606/4² = −0.8504 eV.
- mₗ = 0 occurs once in every multiplet, so four orbital states (4s, 4p₀, 4d₀, 4f₀), and eight once both mₛ values are counted.
- Cross-check by counting through mₗ instead of l: 4 states have mₗ = 0, 3 have each of mₗ = ±1, 2 have each of ±2 and 1 has each of ±3, giving 4 + 6 + 4 + 2 = 16 orbital states and 32 with spin.
Answerl = 0, 1, 2, 3 with 1, 3, 5 and 7 states; 16 = n² orbital and 32 = 2n² in all, so the E₄ = −0.8504 eV eigenspace is 32-dimensional. Eight of those states have mₗ = 0.
HardSwitch on HSO = ξ L⋅S inside the n = 2, l = 1 states of hydrogen. Show that the six states are relabelled rather than lost, find the L⋅S eigenvalue for each j, check that the centre of gravity does not move, and fix ξħ² from the measured 2p interval of 10.97 GHz. Then find the field at which a Zeeman term overtakes it. Take μB = 5.788 × 10⁻⁵ eV T⁻¹.
- L⋅S = ½(J² − L² − S²) commutes with J², Jz, L² and S², but not with Lz or Sz alone, because it contains L+S_− + L_−S+, which trades one unit of mₗ for one of mₛ. So mₗ and mₛ stop labelling eigenstates and j, mⱼ take over.
- Adding 1 ⊗ ½ gives j = 3/2 and j = 1/2, of dimensions 2j + 1 = 4 and 2. Then 4 + 2 = 6 = (2l + 1) × 2: the same six-dimensional eigenspace in a rotated basis. A change of basis creates and destroys nothing.
- Eigenvalues: L⋅S = (ħ²/2)[j(j + 1) − l(l + 1) − s(s + 1)]. For j = 3/2 that is (ħ²/2)(15/4 − 2 − 3/4) = +ħ²/2; for j = 1/2 it is (ħ²/2)(3/4 − 2 − 3/4) = −ħ².
- Centre of gravity: 4(+ħ²/2) + 2(−ħ²) = 0, so the perturbation is traceless on the multiplet and the weighted mean energy is unmoved. Degeneracy is redistributed, never destroyed.
- The gap is ξħ²(1/2 + 1) = (3/2)ξħ². For l = 1 the other relative-α² corrections are j-independent and the Darwin term acts only on l = 0, so the whole interval is spin-orbit. Measured: hν = (4.136 × 10⁻¹⁵ eV s)(1.097 × 10¹⁰ s⁻¹) = 4.54 × 10⁻⁵ eV, hence ξħ² = (2/3)(4.54 × 10⁻⁵ eV) = 3.02 × 10⁻⁵ eV. Sanity-check the size against the interval, not against ξħ²: 4.54 × 10⁻⁵ eV is 1.3 × 10⁻⁵ of |E₂| = 3.40 eV, the α² ≈ 5.3 × 10⁻⁵ order expected.
- A field restores the old labels once μB B exceeds the splitting: B ≈ 4.54 × 10⁻⁵ eV ÷ 5.788 × 10⁻⁵ eV T⁻¹ = 0.78 T. Below that mⱼ is the good label; well above it the Paschen–Back regime returns mₗ and mₛ.
Answerj = 3/2 with 4 states and j = 1/2 with 2 replace the six |mₗ mₛ⟩ kets; L⋅S = +ħ²/2 and −ħ², traceless. ξħ² = 3.0 × 10⁻⁵ eV from the 4.54 × 10⁻⁵ eV gap, which a Zeeman term overtakes near 0.78 T.