Angular-momentum operators and commutators

Angular momentum is not handed a new postulate. Take L = r × p, write out the Cartesian components, and replace every factor by its operator: Lₓ = y p_z − z p_y, L_y = z pₓ − x p_z, L_z = x p_y − y pₓ. The only extra input is the canonical relation [x, pₓ] = iℏ, with every other position–momentum pair commuting. Now expand [Lₓ, L_y] = [y p_z − z p_y, z pₓ − x p_z] into four commutators. Two die immediately, because they pair coordinates with momenta belonging to different axes. Of the survivors, [y p_z, z pₓ] = y [p_z, z] pₓ = −iℏ y pₓ, and [z p_y, x p_z] = x p_y [z, p_z] = iℏ x p_y. Adding them gives [Lₓ, L_y] = iℏ(x p_y − y pₓ) = iℏ L_z. Nothing was assumed about a potential, a state, or an apparatus — this is pure operator algebra, and cycling x → y → z → x supplies the other two relations.

Build L² = Lₓ² + L_y² + L_z² and test it against L_z with [A, BC] = [A, B]C + B[A, C]. The L_z² term is trivially zero. For the first term, [Lₓ², L_z] = Lₓ[Lₓ, L_z] + [Lₓ, L_z]Lₓ = −iℏ(LₓL_y + L_yLₓ); for the second, [L_y², L_z] = +iℏ(L_yLₓ + LₓL_y). The two combinations are identical and the signs are opposite, so the sum vanishes exactly: [L², L_z] = 0, and by symmetry the same holds for Lₓ and L_y. The consequence is structural rather than computational. A state may be a simultaneous eigenstate of L² and one component — conventionally L_z, though which axis you call z is your choice — but never of two components. That is why every angular state in this course carries two labels, l and m, and not three: the algebra supplies exactly one commuting pair, and a third label would need an operator commuting with both, which the cyclic relation forbids.

Convert the Cartesian derivatives to spherical polars and the pieces simplify dramatically. L_z becomes −iℏ ∂/∂φ, a generator of rotation about the z axis and nothing more, while L² becomes −ℏ²[(1/sinθ) ∂/∂θ (sinθ ∂/∂θ) + (1/sin²θ) ∂²/∂φ²]. Set that beside the Laplacian, ∇² = (1/r²) ∂/∂r (r² ∂/∂r) − L²/(ℏ²r²): the angular part of ∇² is −L²/(ℏ²r²), so the kinetic-energy operator already contains angular momentum inside it. Two things follow that are easy to miss. The radial coordinate has dropped out of both operators completely, so the angular eigenvalue problem is the same whatever V(r) is — solve it once and reuse it for hydrogen, for a three-dimensional oscillator, for a nucleus. And because the angular equation never sees V(r), it fixes no energy: an angular quantum number by itself tells you a shape and a degeneracy, never a level.

Robertson's general result says that for any two observables, ΔA ΔB ≥ ½|⟨[A, B]⟩|. Feed in the commutator and you get ΔLₓ ΔL_y ≥ ½ℏ|⟨L_z⟩|, with the cyclic partners alongside. Read the right-hand side carefully: it is an expectation value, so the bound depends on the state and not only on the operators. Inside |l, m⟩ the ladder operators shift m, which forces ⟨Lₓ⟩ = ⟨L_y⟩ = 0, and symmetry about z forces ⟨Lₓ²⟩ = ⟨L_y²⟩. Since ⟨Lₓ²⟩ + ⟨L_y²⟩ = ⟨L²⟩ − ⟨L_z²⟩ = [l(l+1) − m²]ℏ², the actual product is ½[l(l+1) − m²]ℏ² against a floor of ½|m|ℏ². For |1, 1⟩ both come to ½ℏ² — the stretched state saturates the bound exactly. For |3, 1⟩ the product is ½(12 − 1)ℏ² = 5.5ℏ² against a floor of 0.5ℏ², eleven times the minimum. The relation gives a floor, never a prediction.

It is tempting to justify the bound by saying that measuring Lₓ knocks L_y about. That story is not what the algebra says, and it makes wrong predictions. ΔLₓ and ΔL_y are computed from a single state vector before any apparatus appears: prepare a million copies of |2, 2⟩, measure Lₓ on every one of them and never once touch L_y, and the Lₓ results still scatter with ΔLₓ = ℏ, because ⟨Lₓ²⟩ = ½(6 − 4)ℏ² = ℏ². The spread belongs to the state in the same way that a wave packet has a width whether or not anyone looks at it. The deeper point is that quantum mechanics assigns non-commuting observables no joint probability distribution at all: within any l ≥ 1 multiplet Lₓ and L_y share no eigenstate, so no state makes 'Lₓ = a and L_y = b' a definite fact, and the formalism attaches no probability to that pair. So the question 'which way is L really pointing?' is not defeated by experimental clumsiness — it has no answer inside the theory.

There is exactly one orbital state in which all three components are simultaneously certain: l = 0. There ⟨L²⟩ = 0, and since L² is a sum of squares of Hermitian operators, each ⟨Lᵢ²⟩ must vanish on its own, so Lₓ, L_y and L_z are all certainly zero. A spherically symmetric state has no orientation to be uncertain about. Do not confuse that with m = 0 at l ≠ 0, where the floor ½ℏ|⟨L_z⟩| also vanishes but nothing else does. In |2, 0⟩ the floor is zero while ⟨Lₓ²⟩ = ⟨L_y²⟩ = ½(6)ℏ² = 3ℏ², so ΔLₓ = ΔL_y = √3 ℏ ≈ 1.73ℏ and their product is 3ℏ². A vanishing bound is permission for sharpness, never a guarantee of it — the inequality runs one way only, and just computing the variances is what settles what the state actually does.