University Physics V · Angular Momentum in Quantum Mechanics · 9.9

Tensor Products & Adding Angular Momenta

When an electron carries both orbital and spin angular momentum, neither mₗ nor mₛ survives the coupling — only their sum does. This is how you find the towers of total j hiding inside the product space, and how to make L⋅S diagonal without evaluating a single integral.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Give an electron an orbital state and a spin state and it lives on neither space but on their tensor product, of dimension (2l+1)(2s+1), where the natural operator is J = L ⊗ 1 + 1 ⊗ S. Operators on different factors commute, so the cross terms drop and J inherits the algebra untouched: [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ. The ladder argument therefore runs again, and the space must fall apart into complete su(2) towers.

Counting states fixes which ones: j runs from |l − s| to l + s in integer steps, and Σ(2j+1) reproduces (2l+1)(2s+1) exactly. The price of the new labels is the old ones. J² contains L₊S₋ + L₋S₊, which moves a quantum from one factor to the other, so [J², L_z] ≠ 0 and in the coupled basis mₗ and mₛ are gone — only mⱼ = mₗ + mₛ survives, while l and s survive because L² and S² commute with everything in sight.

What you buy is that L⋅S = (J² − L² − S²)/2 is now diagonal, with eigenvalue (ħ²/2)[j(j+1) − l(l+1) − s(s+1)], so the entire spin–orbit pattern — degeneracies, sign, interval ratios — falls out of algebra with no wavefunction anywhere. What you do not buy is the scale. That sits in a radial expectation value ζₙₗ, and the coupled basis is the right basis only while ζₙₗ beats every competing term in the Hamiltonian.

Simple definition: Adding two angular momenta means decomposing the tensor product of their state spaces into the total-J towers it contains: j takes every value from |j₁ − j₂| to j₁ + j₂ in integer steps, each tower carrying 2j+1 states.
Example: For a p electron, l = 1 and s = ½ give a six-dimensional space that splits into j = 3/2 with 4 states and j = 1/2 with 2 states: 4 + 2 = 3 × 2, so no state is created or lost.

Total angular momentum on the product spaceJ = J₁ ⊗ 1 + 1 ⊗ J₂, [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ

The sum obeys the same algebra as its parts, so every ladder result proved for one momentum applies to J unchanged.

J₁ and J₂ act on different factors, so they commute; every J carries units of ħ, that is J s.

Clebsch–Gordan series and the state countj = |j₁ − j₂|, …, j₁ + j₂ · Σⱼ (2j+1) = (2j₁+1)(2j₂+1)

Names every multiplet and audits it: if the dimensions do not add up, a tower has been dropped or double-counted.

j and m are dimensionless labels; the step in j is 1, and the total dimension is fixed before any physics enters.

Coupled state on the uncoupled basis|j m⟩ = Σ_(m₁+m₂=m) ⟨j₁m₁ j₂m₂|j m⟩ |j₁m₁⟩|j₂m₂⟩

The constraint m₁ + m₂ = m kills almost every term: for l ⊗ ½ at most two survive in any coupled state.

The coefficients are real and dimensionless in the Condon–Shortley convention, and the full matrix is unitary.

The dot product from the three squaresJ₁⋅J₂ = (J² − J₁² − J₂²)/2 → (ħ²/2)[j(j+1) − j₁(j₁+1) − j₂(j₂+1)]

Turns an operator product into three quantum numbers, so L⋅S is diagonalised without touching a radial wavefunction.

Eigenvalue in ħ², that is (J s)². Diagonal only in the coupled basis, where all three squares are simultaneously sharp.

Spin–orbit shift, and where its size comes fromΔE = (ζₙₗ/2)[j(j+1) − l(l+1) − s(s+1)], ζₙₗ = (ħ²/2mₑ²c²)⟨(1/r) dV/dr⟩

The bracket is free algebra; ζₙₗ is the radial integral that decides whether the doublet is µeV or eV wide.

ζₙₗ in joules or eV; for a hydrogen-like potential it scales as Z⁴ / [n³ l(l+½)(l+1)].

Landé interval ruleE(J) − E(J−1) = A⋅J, with A = ζₙₗ for a single electron outside closed shells

Interval ratios test an LS assignment straight off a spectrum: a ³P term, whose J runs 0, 1, 2, must give consecutive gaps in the ratio 1 : 2.

A is one fitted constant carrying the units of energy — the term constant, which equals the one-electron ζₙₗ only in the one-electron case; J is the larger of the two adjacent totals.

One space, and the only operator that makes sense on it

An electron with orbital state |l mₗ⟩ and spin state |s mₛ⟩ lives in neither space but in their tensor product, spanned by the (2l+1)(2s+1) product kets |l mₗ⟩ ⊗ |s mₛ⟩. For l = 1 and s = ½ that is six states, labelled (mₗ, mₛ) = (1,↑), (1,↓), (0,↑), (0,↓), (−1,↑), (−1,↓). The total angular momentum on that space is J = L ⊗ 1 + 1 ⊗ S, written L + S once the tensor factors are understood. Two facts earn it the definition. Operators on different factors commute, [Lᵢ, Sⱼ] = 0, so in [Jᵢ, Jⱼ] the cross terms vanish and the two algebras simply add: [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ. And J, not L, generates rotations of the whole electron, which is the symmetry a central Hamiltonian still has once spin–orbit coupling is switched on. Everything the ladder argument proved from the su(2) algebra therefore applies to J with no new work: its spectrum has to be ħ²j(j+1) and mⱼħ on towers of 2j+1 states.

Two complete sets, and the labels that do not survive

Six states need four commuting labels either way, and there are two natural choices. The uncoupled set (L², S², L_z, S_z) labels a state by (l, s, mₗ, mₛ). The coupled set (L², S², J², J_z) labels it by (l, s, j, mⱼ). L² and S² sit in both, so l and s are good in either basis — the common worry that coupling destroys l is misplaced. What fails is L_z. Expand J² = L² + S² + 2L_zS_z + L₊S₋ + L₋S₊: those last two terms lower one factor while raising the other, connecting (mₗ, mₛ) to (mₗ ± 1, mₛ ∓ 1), and a short commutator gives [J², L_z] = ħ(L₋S₊ − L₊S₋) ≠ 0. The sum is untouched, which is why J_z = L_z + S_z commutes with J² and why mⱼ = mₗ + mₛ is the one magnetic label common to both bases. The rule for practice: in the coupled basis quote j and mⱼ and never mₗ or mₛ. How widely a product ket spreads over j is set by how many towers reach its mⱼ, so the two-term answer below is special to l ⊗ ½ and must not be generalised — couple 1 ⊗ 1 instead and the single ket |m₁ = 1⟩|m₂ = −1⟩ already carries all three of L = 0, 1 and 2, with probabilities 1/3, 1/2 and 1/6.

Lower from the stretched state, then orthogonalise

The construction is mechanical. The state of largest mⱼ is unique, because only one product ket has mₗ + mₛ = l + s, so it must be the top of the tallest tower: |3/2, 3/2⟩ = |mₗ = 1⟩|↑⟩ for a p electron. Apply J₋ = L₋ + S₋ to both sides and use ħ√(j(j+1) − m(m−1)) on each ladder. On the left that factor is ħ√(15/4 − 3/4) = √3 ħ. On the right, L₋|1⟩ = √2 ħ|0⟩ and S₋|↑⟩ = ħ|↓⟩. Dividing out gives |3/2, 1/2⟩ = √(2/3)|0,↑⟩ + √(1/3)|1,↓⟩ — a superposition whose two Clebsch–Gordan coefficients are unequal because a ladder, not a symmetry, fixed them. Two product kets reach mⱼ = 1/2, so the coupled basis needs a second state there, and it is the normalised orthogonal combination |1/2, 1/2⟩ = −√(1/3)|0,↑⟩ + √(2/3)|1,↓⟩, the top of the shorter tower. Lower that in turn, and repeat at each mⱼ until the states are used up.

The same coefficients straight out of NumPy

Past two small momenta, stop doing it by hand. In the uncoupled basis of dimension (2j₁+1)(2j₂+1), build J₁z and J₂z as diagonal matrices of m₁ħ and m₂ħ and the ladder operators from ⟨m±1|J_±|m⟩ = ħ√(j(j+1) − m(m±1)), lift each to the product space with numpy.kron, and assemble J² = J₁² + J₂² + 2J₁z J₂z + J₁₊J₂₋ + J₁₋J₂₊. Then numpy.linalg.eigh returns eigenvalues ħ²j(j+1) with the right multiplicities — for l = 1 ⊗ s = ½, four values of 3.75ħ² and two of 0.75ħ² — and the eigenvector matrix is the Clebsch–Gordan transformation. One caution that costs people an afternoon: eigh fixes neither the sign of a column nor the mixing inside a degenerate eigenvalue, so the raw output is not the tabulated table. Because J² is block diagonal in mⱼ, diagonalise one mⱼ block at a time — here every block is 1×1 or 2×2 — and then only the phase convention is left to impose by hand.

L⋅S diagonalised without a single integral

The payoff of the coupled basis is the identity L⋅S = (J² − L² − S²)/2, whose eigenvalue is (ħ²/2)[j(j+1) − l(l+1) − s(s+1)]. For s = ½ it collapses to ⟨L⋅S⟩ = +ħ²l/2 when j = l + ½ and −ħ²(l+1)/2 when j = l − ½, so the two eigenvalues differ by ħ²(l + ½) — and the two energy levels by ζₙₗ(l + ½), once a radial constant is supplied to convert them. The upper level, though it holds more states, moves the smaller distance. The asymmetry is not an accident: weight each shift by its degeneracy and they cancel exactly, (2l+2)(l/2) + (2l)(−(l+1)/2) = 0, so the centre of gravity of the multiplet stays where the unsplit level was. Use that as an arithmetic check on every fine-structure calculation. Two spin-halves behave the same way: ⟨S₁⋅S₂⟩ = +ħ²/4 for the triplet and −3ħ²/4 for the singlet, a gap of ħ², which is the skeleton of both the exchange splitting in helium and hydrogen's 21 cm hyperfine line.

What the algebra will not tell you

Everything above is a unitary change of basis and costs nothing. It also predicts nothing about size. The spin–orbit term is H = (ζₙₗ/ħ²) L⋅S with ζₙₗ = (ħ²/2mₑ²c²)⟨(1/r) dV/dr⟩, a radial expectation value you have to compute or measure. Its value separates cases that look identical on paper: hydrogen's 2p doublet splits by 4.5 × 10⁻⁵ eV, about 11 GHz, while sodium's 3p doublet — the D lines at 589.0 and 589.6 nm — splits by 2.13 meV, or 17.2 cm⁻¹, some fifty times wider, because ζ climbs steeply with the charge the electron sees. The scale also decides whether the coupled basis is the right one at all. Switch on a magnetic field and two terms compete; the coupled basis wins while ζₙₗ ≫ μ_B B, and the uncoupled basis returns in the Paschen–Back limit. With ζ₃p = 1.42 meV sodium needs roughly 25 T to cross over, but hydrogen's 2p, with ζ ≈ 3.0 × 10⁻⁵ eV, crosses near 0.5 T — a field you can reach on a bench.

Change one variable at a time

Make the relationship visible.

Interactive model

Orbital quantum number l2

Spin–orbit constant ζ0.6 meV

Push l from 1 to 4 and watch the two shifts approach each other in size while the gap grows as ζ(l + ½): the upper level always holds more states and always moves less, which is exactly what pins the dashed centre of gravity in place. ζ only rescales the picture.

SHIFT j = l + ½0.60 meV

SHIFT j = l − ½-0.90 meV

SPLITTING1.50 meV

WEIGHTED SUM OF SHIFTS0.00 meV

Live interpretationSHIFT j = l + ½: 0.60 meV. SHIFT j = l − ½: −0.90 meV. SPLITTING: 1.50 meV. WEIGHTED SUM OF SHIFTS: 0.00 meV

Catch the common trap

Explain before calculating.

Misconception check

Picturing coupling as two arrows being added invites two errors at once. The first is arithmetic on lengths: |L| = ħ√(l(l+1)) and |S| = ħ√(s(s+1)), so for l = 1 and s = ½ the arrows have lengths 1.414ħ and 0.866ħ and students expect a maximum of 2.28ħ, when the stretched state actually has |J| = ħ√15/2 = 1.94ħ. The vectors never line up, because their transverse components cannot both be sharp. The second error is worse, because it survives into calculations: assuming that inside |j, mⱼ⟩ the electron still has definite mₗ and mₛ that add to mⱼ. It does not. J² contains L₊S₋ + L₋S₊, which shuffles a quantum between the factors, so [J², L_z] ≠ 0. Only the sum is a good quantum number. Note what does survive: l and s, because L² and S² commute with J² and J_z alike.

A p electron (l = 1, s = ½) is prepared in the coupled state |j = 3/2, mⱼ = 1/2⟩, and L_z is then measured. What does the measurement give?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo p electrons in different shells have orbital angular momenta l₁ = 1 and l₂ = 1. List the allowed total L, then check the state count two independent ways.

The range rule gives L from |l₁ − l₂| = 0 up to l₁ + l₂ = 2 in integer steps, so L = 0, 1, 2.
Coupled count: each tower holds 2L+1 states, so the total is 1 + 3 + 5 = 9.
Uncoupled count: the product space has (2l₁+1)(2l₂+1) = 3 × 3 = 9 product kets. The two agree, so no tower has been dropped.
Cross-check a single M. In the uncoupled basis M = 1 comes from (m₁, m₂) = (1, 0) and (0, 1), giving 2 states; in the coupled basis the L = 1 and L = 2 towers each supply one M = 1 state, again 2.
Note how much richer this is than l ⊗ ½: three towers reach M = 0, so a product ket there is a three-term superposition, not a two-term one.

AnswerL = 0, 1 and 2, with 1 + 3 + 5 = 9 states = 3 × 3.

MediumA d electron (l = 2, s = ½) shows a spin–orbit doublet whose measured splitting is 2.50 meV. Find the two j values and their degeneracies, the L⋅S eigenvalues in units of ħ², the spin–orbit constant ζ₃d, and the shift of each level from the unsplit position.

j runs from |l − s| = 3/2 to l + s = 5/2, so j = 5/2 and j = 3/2, holding 2j+1 = 6 and 4 states; 6 + 4 = 10 = 5 × 2, the full product dimension.
⟨L⋅S⟩ = (ħ²/2)[j(j+1) − l(l+1) − s(s+1)] with l(l+1) = 6 and s(s+1) = 0.75. For j = 5/2: (ħ²/2)[8.75 − 6 − 0.75] = +1.00 ħ². For j = 3/2: (ħ²/2)[3.75 − 6 − 0.75] = −1.50 ħ².
Those eigenvalues differ by 2.50 ħ², which is ħ²(l + ½) as it must be, so the energy gap is ΔE = ζ₃d(l + ½) = 2.5 ζ₃d and ζ₃d = 2.50 meV ÷ 2.5 = 1.00 meV.
Shifts: E(5/2) = ζ₃d × (+1.00) = +1.00 meV and E(3/2) = ζ₃d × (−1.50) = −1.50 meV.
Centre-of-gravity check: 6 × (+1.00 meV) + 4 × (−1.50 meV) = 6.00 − 6.00 = 0, so the weighted mean has not moved.

Answerj = 5/2 (6 states) and 3/2 (4 states); ⟨L⋅S⟩ = +1.00ħ² and −1.50ħ²; ζ₃d = 1.00 meV; shifts +1.00 meV and −1.50 meV.

HardFor a p electron (l = 1, s = ½), build |j = 3/2, mⱼ = 1/2⟩ from the stretched state by lowering. Then find ⟨L_z⟩, ⟨S_z⟩ and the probability of measuring S_z = −ħ/2 in that state, and evaluate ⟨L⋅S⟩ twice: once from the quantum numbers, once by acting with L⋅S term by term.

The stretched state is unique: |3/2, 3/2⟩ = |mₗ = 1⟩|↑⟩. Applying J₋ on the left gives ħ√(15/4 − 3/4)|3/2, 1/2⟩ = √3 ħ|3/2, 1/2⟩.
On the right, L₋|1⟩ = ħ√(2 − 0)|0⟩ = √2 ħ|0⟩ and S₋|↑⟩ = ħ|↓⟩, so the right side is √2 ħ|0,↑⟩ + ħ|1,↓⟩. Divide by √3: |3/2, 1/2⟩ = √(2/3)|0,↑⟩ + √(1/3)|1,↓⟩.
⟨L_z⟩ = (2/3)(0) + (1/3)(ħ) = ħ/3, and ⟨S_z⟩ = (2/3)(+ħ/2) + (1/3)(−ħ/2) = ħ/3 − ħ/6 = ħ/6. Their sum is ħ/2 = mⱼħ, as it must be. P(S_z = −ħ/2) = |√(1/3)|² = 1/3.
Route one, from the quantum numbers: ⟨L⋅S⟩ = (ħ²/2)[15/4 − 2 − 3/4] = (ħ²/2)(1) = ħ²/2, which is the ħ²l/2 expected for j = l + ½.
Route two, term by term with L⋅S = L_zS_z + ½(L₊S₋ + L₋S₊). The diagonal piece gives (2/3)(0)(ħ/2) + (1/3)(ħ)(−ħ/2) = −ħ²/6.
The flip-flop piece connects the two terms: ½L₊S₋|0,↑⟩ = (√2/2)ħ²|1,↓⟩ and ½L₋S₊|1,↓⟩ = (√2/2)ħ²|0,↑⟩, so its expectation is 2 × √(2/3) × √(1/3) × (√2/2) ħ² = (2/3)ħ². Total: −ħ²/6 + 2ħ²/3 = ħ²/2, agreeing with route one.

Answer|3/2, 1/2⟩ = √(2/3)|0,↑⟩ + √(1/3)|1,↓⟩; ⟨L_z⟩ = ħ/3, ⟨S_z⟩ = ħ/6, P(S_z = −ħ/2) = 1/3; ⟨L⋅S⟩ = ħ²/2 by both routes.