University Physics V · Angular Momentum in Quantum Mechanics · 9.2

Angular Momentum Operators & the Rotation Algebra

Once r × p becomes an operator, its three components stop being three separate numbers. They close on each other under commutation, and everything that follows — the Casimir, the labels, the rotation unitary, the one condition on V that makes any of it conserved — is read off that closure.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The model is a promotion. Take the classical Noether charge of rotational invariance, L = r × p, replace r and p by the operators of the canonical pair, and ask what [x̂ᵢ, p̂ⱼ] = iħδᵢⱼ forces. Two structural facts come back, and neither needed a wavefunction. The components close on themselves, [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ, so the three are one object — the Lie algebra so(3), identical in form to su(2) — and commuting any two manufactures nothing new.

And the quadratic combination L̂² commutes with every component, so the algebra has one Casimir and rank one: you may diagonalise L̂² together with exactly one component, never two. The pair of labels (l, m) is fixed there, before any equation is solved. Exponentiating the Hermitian generator gives the unitary Û(θ, n̂) = exp(−iθ n̂⋅L̂/ħ), under which r̂ transforms as R(θ)r̂; the failure of two such unitaries about different axes to commute is not a nuisance but the commutator itself, appearing at second order in the angles. The cost comes in two clauses.

L̂ is defined about a chosen origin — displace it by a and L̂ becomes L̂ − a × p̂ — so angular momentum is never a property of the particle alone. And [Ĥ, L̂] = 0 is not free: it holds when V depends on r = |r| alone. Break the sphere down to an axis and only the generator of that axis survives.

Simple definition: The angular momentum operators are the three Hermitian generators L̂ᵢ = εᵢⱼₖ x̂ⱼ p̂ₖ, defined about a chosen origin, whose commutators close on themselves as [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ, making exp(−iθ n̂⋅L̂/ħ) the unitary that rotates a state by θ about n̂.
Example: Take ψ = x + iy = r sinθ e^(iφ). Then L̂_z ψ = −iħ ∂ψ/∂φ = +ħψ and L̂²ψ = 2ħ²ψ, both sharp; but L̂ₓ ψ is a different function, with ⟨L̂ₓ⟩ = 0 and ΔL̂ₓ = ħ/√2 = 7.5 × 10⁻³⁵ J s.

Hermitian promotion of r × pL̂ = r̂ × p̂, L̂ᵢ = εᵢⱼₖ x̂ⱼ p̂ₖ

L̂ is Hermitian exactly as written — unlike r̂⋅p̂, which must be symmetrised to (r̂⋅p̂ + p̂⋅r̂)/2.

Units J s. In every term the position index differs from the momentum index, so [x̂ⱼ, p̂ₖ] = 0.

The rotation algebra[L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ

Closure: commuting two components hands back a third, never a new operator. That is what makes L̂ one object.

ε₁₂₃ = +1 and cyclic; both sides carry (J s)². The same relations define so(3) and su(2).

L̂² is the CasimirL̂² = L̂ₓ² + L̂_y² + L̂_z², [L̂², L̂ᵢ] = 0

Buys a shared eigenbasis of L̂² and one component — the labels l and m, and rank one means nothing more.

Proved from [Â², B̂] = Â[Â, B̂] + [Â, B̂]Â; the two anticommutator terms cancel in pairs.

Vector and scalar operators under L̂[L̂ᵢ, V̂ⱼ] = iħ εᵢⱼₖ V̂ₖ, [L̂ᵢ, Ŝ] = 0

Classify each piece of Ĥ this way and [Ĥ, L̂] falls out in one line, with no wavefunction written down.

V̂ is any vector operator (r̂, p̂, L̂ itself); Ŝ any rotational scalar such as r̂⋅r̂, p̂⋅p̂ or r̂⋅p̂.

The rotation unitaryÛ(θ, n̂) = exp(−i θ n̂⋅L̂ / ħ), Û† r̂ Û = R(θ) r̂

L̂ generates rotations, so the algebra is measurable: two Û about different axes fail to commute at order θ².

n̂ a unit vector, θ in radians; unitary because n̂⋅L̂ is Hermitian and θ is real.

Origin dependence and conservationL̂′ = L̂ − a × p̂[Ĥ, L̂] = 0 ⟺ V = V(r)

Names the price: L̂ is defined about a point, and its conservation is a claim about the potential, not a law.

a is the displacement of the new origin; r = |r| means the whole sphere, not merely one symmetry axis.

Promoting r × p needs no ordering choice

Write L̂ᵢ = εᵢⱼₖ x̂ⱼ p̂ₖ, so L̂ₓ = ŷp̂_z − ẑp̂_y, L̂_y = ẑp̂ₓ − x̂p̂_z and L̂_z = x̂p̂_y − ŷp̂ₓ. In every one of those six terms the position index differs from the momentum index, and [x̂ⱼ, p̂ₖ] = iħδⱼₖ vanishes when j ≠ k. So ŷ and p̂_z commute, (ŷp̂_z)† = p̂_zŷ = ŷp̂_z, and the operator is Hermitian exactly as the classical expression is written. That is not automatic. The classical scalar r⋅p becomes x̂p̂ₓ + ŷp̂_y + ẑp̂_z, whose adjoint is p̂ₓₓ̂ + p̂_yŷ + p̂_zẑ = r̂⋅p̂ − 3iħ; the naive product is not Hermitian and has to be replaced by (r̂⋅p̂ + p̂⋅r̂)/2. Quantisation by the rule {A, B} → [Â, B̂]/iħ is a postulate carrying ordering ambiguities, and angular momentum is one of the places where the ambiguity happens not to bite. Everything downstream leans on that one operator being self-adjoint, because only then does it have real eigenvalues and generate a unitary.

One commutator, computed from [x̂ᵢ, p̂ⱼ] = iħδᵢⱼ

Expand [L̂ₓ, L̂_y] = [ŷp̂_z − ẑp̂_y, ẑp̂ₓ − x̂p̂_z]. Four cross terms; two die at once, because [ŷp̂_z, x̂p̂_z] and [ẑp̂_y, ẑp̂ₓ] each involve only mutually commuting factors. The survivors need the ẑ–p̂_z pair: [ŷp̂_z, ẑp̂ₓ] = ŷ[p̂_z, ẑ]p̂ₓ = −iħ ŷp̂ₓ, and [ẑp̂_y, x̂p̂_z] = x̂p̂_y[ẑ, p̂_z] = +iħ x̂p̂_y. Assembling with the signs the bracket demands gives −iħŷp̂ₓ + iħx̂p̂_y = iħ(x̂p̂_y − ŷp̂ₓ) = iħL̂_z. Cycling x → y → z → x supplies the other two, so [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ. Read what that says rather than filing it: commuting two components returns a third, so the set closes and is the Lie algebra so(3). Because the same relations define su(2), the algebra by itself cannot tell an orbital l from a half-integer spin — that distinction has to come from somewhere else.

L̂² commutes with every component

L̂² = L̂ₓ² + L̂_y² + L̂_z², and [L̂_z², L̂_z] = 0, so testing [L̂², L̂_z] leaves two terms. Use [Â², B̂] = Â[Â, B̂] + [Â, B̂]Â. The algebra gives [L̂ₓ, L̂_z] = iħ ε₁₃₂ L̂_y = −iħL̂_y and [L̂_y, L̂_z] = iħ ε₂₃₁ L̂ₓ = +iħL̂ₓ, so [L̂ₓ², L̂_z] = −iħ(L̂ₓL̂_y + L̂_yL̂ₓ) while [L̂_y², L̂_z] = +iħ(L̂_yL̂ₓ + L̂ₓL̂_y). The two anticommutators are the same operator carrying opposite signs, and they cancel exactly. Nothing in that argument distinguished z, so L̂² commutes with all three components. L̂² is therefore the Casimir of the algebra, and so(3) has rank one: exactly one component may be diagonalised alongside it. That is the whole origin of the two-label scheme |l, m⟩ — fixed before any differential equation is written, and the reason there is no such thing as a state |l, mₓ, m_y, m_z⟩.

Exponentiating the generator: the rotation unitary

Rotate a spinless state by δθ about z: ψ′(r) = ψ(R⁻¹r), which for small δθ is ψ − δθ ∂ψ/∂φ = (1̂ − (i/ħ)δθ L̂_z)ψ, using L̂_z = −iħ∂/∂φ. Compose N of these with δθ = θ/N and let N → ∞: Û(θ, ẑ) = exp(−iθL̂_z/ħ), and for a general axis Û(θ, n̂) = exp(−iθ n̂⋅L̂/ħ). It is unitary because n̂⋅L̂ is Hermitian and θ real, so norms and probabilities survive the rotation. In Heisenberg form Û†r̂Û = R(θ)r̂: the operator components turn the way the classical vector does. The payoff is that the algebra becomes measurable. Rotations about different axes do not commute, and the leading failure is second order: Û_y(−α)Ûₓ(−α)Û_y(α)Ûₓ(α) = exp(+iα²L̂_z/ħ) + O(α³). With α = 0.100 rad the loop misses closing by a rotation of 0.0100 rad, or 0.573°, about z. That residue is [L̂ₓ, L̂_y] = iħL̂_z, read off a protractor.

Scalars, vectors, and the one condition on V

Sort the pieces of Ĥ by how they transform. A rotational scalar Ŝ — anything built from r̂⋅r̂, p̂⋅p̂ or r̂⋅p̂ — satisfies [L̂ᵢ, Ŝ] = 0. A vector operator V̂ satisfies [L̂ᵢ, V̂ⱼ] = iħ εᵢⱼₖ V̂ₖ, and r̂, p̂ and L̂ itself are all vectors. Now take Ĥ = p̂²/2m + V(r̂) with r̂ = √(r̂⋅r̂). Both terms are scalars, so [Ĥ, L̂ᵢ] = 0 for all three components at once, and not one wavefunction has been written. Break the symmetry and the same classification says exactly what survives. Add a uniform electric field along z, so Ĥ → Ĥ + eEẑ. Now ẑ is one component of a vector: [L̂_z, ẑ] = 0 still holds, but [L̂ₓ, ẑ] = −iħŷ and [L̂², ẑ] ≠ 0. Full rotational symmetry has collapsed to rotations about z alone, so mₗ stays a good quantum number and l does not — which is precisely why the linear Stark effect mixes hydrogen's 2s and 2p₀ states.

What the commuting set buys, and what the origin costs

For a spinless particle in a central potential, (Ĥ, L̂², L̂_z) is a complete set of commuting observables: three mutually commuting Hermitian operators whose joint eigenvalues label a state uniquely as |n, l, m⟩. Rotational invariance alone then forces a degeneracy. L̂_± = L̂ₓ ± iL̂_y commute with Ĥ but shift m by one, so the 2l+1 states of a given l must share an energy. Hydrogen's extra degeneracy in l is not part of this — it needs the Runge–Lenz vector and the 1/r potential specifically. Two costs deserve stating plainly. First, L̂ is defined about a chosen origin: shift it by a and L̂′ = L̂ − a × p̂, so d⟨L̂′⟩/dt = −a × ⟨F̂⟩, which for an electron reckoned about a point 1 Å off the proton is not zero. Angular momentum is never a property of the particle alone. Second, [Ĥ, L̂] = 0 is a statement about V, not a law: strictly central, or the vector conservation fails.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotation about x, angle a30 °

Rotation about y, angle b30 °

Set either angle to zero and the loop closes exactly: the residue is the product ab, not a sum. Push both to 60 deg and the gap falls well short of the −ab prediction, because that is only the leading term. The length readout never moves - no rotation can touch L-squared.

EXACT GAP ANGLE-14.50 °

PREDICTION −ab-15.71 °

OUT-OF-PLANE z0.058

VECTOR LENGTH1.000

Live interpretationEXACT GAP ANGLE: −14.50 °. PREDICTION −ab: −15.71 °. OUT-OF-PLANE z: 0.058. VECTOR LENGTH: 1.000

Catch the common trap

Explain before calculating.

An electron sits in a central potential V(r). A uniform electric field along ẑ is switched on, adding eEẑ to the Hamiltonian. Which of L̂², L̂ₓ, L̂_y and L̂_z still commute with the new Ĥ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUsing L̂_z = x̂p̂_y − ŷp̂ₓ and [x̂ᵢ, p̂ⱼ] = iħδᵢⱼ, evaluate [L̂_z, x̂], [L̂_z, ŷ] and [L̂_z, ẑ]. Check the answers against the vector-operator rule [L̂ᵢ, V̂ⱼ] = iħ εᵢⱼₖ V̂ₖ, and say what they mean geometrically.

[L̂_z, x̂] = [x̂p̂_y, x̂] − [ŷp̂ₓ, x̂]. The first bracket vanishes, since x̂ and p̂_y both commute with x̂. The second is ŷ[p̂ₓ, x̂] = ŷ(−iħ) = −iħŷ, and it enters with a minus sign, so [L̂_z, x̂] = +iħŷ.
[L̂_z, ŷ] = [x̂p̂_y, ŷ] − [ŷp̂ₓ, ŷ] = x̂[p̂_y, ŷ] − 0 = x̂(−iħ) = −iħx̂.
[L̂_z, ẑ] = 0: every factor in L̂_z is built from x̂, ŷ, p̂ₓ and p̂_y, and all four commute with ẑ.
Rule check with i = 3: [L̂₃, x̂₁] = iħ ε₃₁₂ x̂₂ = iħŷ, [L̂₃, x̂₂] = iħ ε₃₂₁ x̂₁ = −iħx̂, and [L̂₃, x̂₃] = 0 because ε₃₃k = 0. All three agree.
Geometry: to first order, Û†(θ,ẑ) x̂ Û = x̂ + (iθ/ħ)[L̂_z, x̂] = x̂ − θŷ and Û†ŷÛ = ŷ + θx̂, while ẑ is untouched. That is exactly R(θ) about z acting on the coordinate operators.

Answer[L̂_z, x̂] = iħŷ, [L̂_z, ŷ] = −iħx̂, [L̂_z, ẑ] = 0, matching [L̂ᵢ, V̂ⱼ] = iħ εᵢⱼₖ V̂ₖ. L̂_z turns x̂ and ŷ into each other and leaves its own axis alone: it is the generator of rotations about z.

MediumShow that [L̂², L̂_z] = 0 using the algebra alone, with no wavefunction and no differential operator. Then state exactly how many independent labels the algebra supplies for an angular-momentum state, and why not one more.

L̂² = L̂ₓ² + L̂_y² + L̂_z², and [L̂_z², L̂_z] = 0, so only two terms can contribute.
Use [Â², B̂] = Â[Â, B̂] + [Â, B̂]Â. From [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ: [L̂ₓ, L̂_z] = iħ ε₁₃₂ L̂_y = −iħL̂_y, and [L̂_y, L̂_z] = iħ ε₂₃₁ L̂ₓ = +iħL̂ₓ.
So [L̂ₓ², L̂_z] = L̂ₓ(−iħL̂_y) + (−iħL̂_y)L̂ₓ = −iħ(L̂ₓL̂_y + L̂_yL̂ₓ), and [L̂_y², L̂_z] = L̂_y(iħL̂ₓ) + (iħL̂ₓ)L̂_y = +iħ(L̂ₓL̂_y + L̂_yL̂ₓ).
The two anticommutators are the same operator with opposite signs, so they cancel exactly and [L̂², L̂_z] = 0. Note that the argument never assumed the components commute — only the algebra was used.
Nothing singled out z, so the same three lines give [L̂², L̂ₓ] = [L̂², L̂_y] = 0. But [L̂ₓ, L̂_z] = −iħL̂_y ≠ 0, so at most one component may join L̂² in a commuting set.
Hence exactly two labels: l from the Casimir L̂², and m from whichever single component you diagonalise. A third would need a fourth commuting operator, and a rank-one algebra has none to offer.

Answer[L̂², L̂_z] = −iħ(L̂ₓ, L̂_y) + iħ(L̂ₓ, L̂_y) = 0, and likewise for L̂ₓ and L̂_y. Since no two components commute, the largest commuting set inside the algebra is (L̂², one component): two labels, l and m, and no more.

HardAct on a state with four rotations in order — Ûₓ(α), then Û_y(α), then Ûₓ(−α), then Û_y(−α), each Ûₙ̂(θ) = exp(−iθ n̂⋅L̂/ħ), with α = 0.100 rad. Show the sequence is not the identity, identify the leading residue and its size, and say which states can register it.

Set Â = −iαL̂ₓ/ħ and B̂ = −iαL̂_y/ħ. With the rightmost operator acting first, the composite is Ĉ = e^(−B̂) e^(−Â) e^(B̂) e^(Â).
Baker–Campbell–Hausdorff gives e^(X̂) e^(Ŷ) e^(−X̂) e^(−Ŷ) = exp([X̂, Ŷ] + O(cubic)). Matching X̂ = −B̂ and Ŷ = −Â reproduces Ĉ term for term, so Ĉ = exp([−B̂, −Â]) = exp([B̂, Â]).
[Â, B̂] = (−iα/ħ)²[L̂ₓ, L̂_y] = (−α²/ħ²)(iħL̂_z) = −iα²L̂_z/ħ, so [B̂, Â] = +iα²L̂_z/ħ.
Hence Ĉ = exp(+iα²L̂_z/ħ) = Û_ẑ(−α²): a rotation about z through α² = (0.100)² = 0.0100 rad, or 0.573°. The loop closes only at α = 0, and the residue is second order because the commutator is bilinear in the two angles.
Registering it: on |0,0⟩ we have L̂_z|0,0⟩ = 0, so Ĉ = 1̂ exactly and a rotationally invariant state cannot see the failure. On |1,+1⟩ it is the global phase e^(+0.0100i), unobservable alone; but on a superposition of m = +1 and m = −1, whose angular pattern picks out a direction in the xy-plane, the two components pick up opposite phases and the pattern turns bodily through 0.0100 rad about z.

AnswerĈ = exp(+iα²L̂_z/ħ) = Û_ẑ(−α²), a rotation about z through 0.0100 rad (0.573°) — not the identity. The failure to close is [L̂ₓ, L̂_y] = iħL̂_z measured directly, and it vanishes only on l = 0 states.