Angular momentum operators and the rotation algebra

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.

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.

Write L̂² = L̂ⱼL̂ⱼ, summed over j, and commute it with L̂ᵢ using [ÂB̂, Ĉ] = Â[B̂,Ĉ] + [Â,Ĉ]B̂: [L̂², L̂ᵢ] = L̂ⱼ[L̂ⱼ, L̂ᵢ] + [L̂ⱼ, L̂ᵢ]L̂ⱼ = iħ εⱼᵢₖ (L̂ⱼL̂ₖ + L̂ₖL̂ⱼ). The bracket is symmetric under j ↔ k while εⱼᵢₖ is antisymmetric, so every term of the double sum meets its partner with the opposite sign and the whole thing vanishes — for all three values of i at once, with no component singled out. 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⟩.

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.

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.

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 the Heisenberg equation gives dL̂′/dt = −a × F̂, which for an electron reckoned about a point 1 Å off the proton is a nonzero operator: L̂′ is not conserved even though L̂ is. 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.