University Physics IV · Angular Momentum and Spin · 9.1

Angular-Momentum Operators & Commutators

Before any spectrum, any orbital, any Zeeman pattern, you have to know which angular-momentum questions a single state can answer at once. Three lines of commutator algebra settle that, and almost everything later in the unit is bookkeeping laid on top of them.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Angular momentum enters quantum mechanics with no new postulate: take the classical L = r × p and promote every factor to an operator. What comes out is not classical. Because a coordinate fails to commute with its own momentum, the three components of L fail to commute with one another — [Lₓ, L_y] = iℏL_z, and cyclically — so no state carries sharp values of two components at once.

Meanwhile L² = Lₓ² + L_y² + L_z² commutes with all three, so a magnitude and one chosen projection can be sharp together. That single algebraic fact is the architecture of everything that follows: it is why an angular state carries exactly two labels rather than three, and why the spectrum can be extracted by ladder operators without ever solving a differential equation. The cost is the classical picture.

There is no angular-momentum vector pointing in a definite direction that we merely fail to resolve; non-commuting components share no joint probability distribution, so the direction of L is a question the theory declines to answer. Robertson's inequality prices the refusal, ΔLₓ ΔL_y ≥ ½ℏ|⟨L_z⟩|, and the only full escape is l = 0, where L² has eigenvalue zero and all three components are certainly zero at once.

Simple definition: The angular-momentum commutation relations say that the three components of L = r × p do not commute with one another, [Lₓ, L_y] = iℏL_z and cyclically, while L² commutes with each of them.
Example: In the state |l = 1, m = 1⟩ the floor is ½ℏ|⟨L_z⟩| = ½ℏ², and the actual spread product is ΔLₓ ΔL_y = ½[1(2) − 1]ℏ² = ½ℏ². The two are equal, so neither Lₓ nor L_y has a value even though L² and L_z both do.

Angular momentum as an operatorL = r × p, Lₓ = y p_z − z p_y (cycle x → y → z → x)

Everything below follows from this plus [x, pₓ] = iℏ; no separate postulate about angular momentum is needed.

L in J s; r in m; p = −iℏ∇ in kg m s⁻¹, with ℏ = 1.055 × 10⁻³⁴ J s.

The cyclic commutator[Lₓ, L_y] = iℏL_z, [L_y, L_z] = iℏLₓ, [L_z, Lₓ] = iℏL_y

Non-zero, so two components are never sharp at once — unless the right-hand side has zero expectation in that state.

Compactly [Lᵢ, Lⱼ] = iℏ εᵢⱼₖ Lₖ, summed over k. Each side has the units of L², J² s².

L² commutes with every componentL² = Lₓ² + L_y² + L_z², [L², Lₓ] = [L², L_y] = [L², L_z] = 0

This is why an angular state carries a magnitude label and one projection label — two numbers, never three.

Units J² s². The proof needs only the identity [A, BC] = [A, B]C + B[A, C].

Spherical-polar formsL_z = −iℏ ∂/∂φ, L² = −ℏ²[(1/sinθ) ∂/∂θ (sinθ ∂/∂θ) + (1/sin²θ) ∂²/∂φ²]

L² is the angular part of ∇², so one angular solution serves every central potential and fixes no energy.

θ polar and φ azimuthal, both in radians; r has dropped out of both operators entirely.

Robertson bound applied to two componentsΔA ΔB ≥ ½|⟨[A, B]⟩| → ΔLₓ ΔL_y ≥ ½ℏ|⟨L_z⟩|

Converts the commutator into a measurable floor — and shows that floor collapsing to zero whenever ⟨L_z⟩ = 0.

ΔA = √(⟨A²⟩ − ⟨A⟩²), in the unit of A. The bound is a property of the state, not just of the operators.

Spreads inside a state of definite l and m⟨Lₓ⟩ = ⟨L_y⟩ = 0, ΔLₓ ΔL_y = ½[l(l+1) − m²] ℏ²

Saturates the floor at m = ±l, and still leaves ½l(l+1)ℏ² of spread at m = 0, where the floor is zero.

Holds in |l, m⟩ only, where symmetry forces ⟨Lₓ²⟩ = ⟨L_y²⟩ = ½[l(l+1) − m²]ℏ².

Promote r × p and turn the crank

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.

L² commutes with all three, and that fixes the labels

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.

The same operators in spherical polars

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 divided by 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.

What the uncertainty relation is a statement about

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.

Incompatibility is not disturbance

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 non-commuting observables have no joint probability distribution at all: no table of pairs (Lₓ, L_y) with probabilities reproduces the two separate distributions. So the question 'which way is L really pointing?' is not defeated by experimental clumsiness — it has no answer inside the theory.

Where the obstruction really relaxes

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Angular momentum number l3

Projection m (clamped to |m| ≤ l)1

Hold l at 3 and push m out to 3: the product bar falls until it lands exactly on the dashed floor, because the stretched state |l, l⟩ saturates the bound. Then set l = 0 and both bars vanish — the one case where Lₓ, L_y and L_z are all certain together.

ΔLₓ ΔL_y5.50 ℏ²

FLOOR ½ℏ|⟨L_z⟩|0.50 ℏ²

GAP ABOVE FLOOR5.00 ℏ²

ΔLₓ = ΔL_y2.35 ℏ

Live interpretationΔLₓ ΔL_y: 5.50 ℏ². FLOOR ½ℏ|⟨L_z⟩|: 0.50 ℏ². GAP ABOVE FLOOR: 5.00 ℏ². ΔLₓ = ΔL_y: 2.35 ℏ

Catch the common trap

Explain before calculating.

Misconception check

The commutator is routinely explained as disturbance: measure Lₓ and you jog L_y, so you can never pin both down. That reading fails on its own terms. ΔLₓ and ΔL_y are computed from a single state vector by arithmetic alone, with no apparatus anywhere in the calculation, and they are already there in a state nobody has touched. It also predicts the wrong shape of answer: disturbance would give a fixed constant, whereas the real bound is ½ℏ|⟨L_z⟩|, a state-dependent number that drops to zero in every m = 0 state. And it quietly restores the classical picture — a vector pointing somewhere definite that has merely been knocked out of view. There is no such vector. Non-commuting components share no joint probability distribution, so 'which way is L pointing' is a question the formalism declines, not a fact the apparatus hides.

A hydrogen electron occupies a state with l = 2 and m = 0, so ⟨L_z⟩ = 0 and the right-hand side of ΔLₓ ΔL_y ≥ ½ℏ|⟨L_z⟩| vanishes. Which statement about the other two components is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyStarting from L_z = x p_y − y pₓ and Lₓ = y p_z − z p_y, and using only [x, pₓ] = [y, p_y] = [z, p_z] = iℏ with every other position–momentum pair commuting, evaluate [L_z, Lₓ].

Expand the bracket into four terms: [L_z, Lₓ] = [x p_y, y p_z] − [x p_y, z p_y] − [y pₓ, y p_z] + [y pₓ, z p_y].
Kill the middle two. In [x p_y, z p_y] the operators are x, p_y, z, p_y, and no coordinate ever meets its own momentum, so it vanishes; [y pₓ, y p_z] vanishes for the same reason.
First survivor: only p_y and y fail to commute, so [x p_y, y p_z] = x [p_y, y] p_z = x(−iℏ)p_z = −iℏ x p_z.
Second survivor: only y and p_y fail to commute, so [y pₓ, z p_y] = z [y, p_y] pₓ = iℏ z pₓ.
Add them: [L_z, Lₓ] = −iℏ x p_z + iℏ z pₓ = iℏ(z pₓ − x p_z) = iℏ L_y, since L_y = z pₓ − x p_z.

Answer[L_z, Lₓ] = iℏ L_y — the cyclic partner of [Lₓ, L_y] = iℏ L_z, obtained from the canonical relations alone.

MediumAn electron is in the state |l = 3, m = 3⟩. (a) Find ΔLₓ and ΔL_y. (b) Evaluate the floor ½ℏ|⟨L_z⟩|. (c) Say whether the bound is met with slack or exactly. (d) Quote ΔLₓ in SI units, taking ℏ = 1.055 × 10⁻³⁴ J s.

In any |l, m⟩ the raising and lowering operators shift m, so the diagonal elements of Lₓ and L_y vanish: ⟨Lₓ⟩ = ⟨L_y⟩ = 0, and therefore ΔLₓ² = ⟨Lₓ²⟩.
Use the sum rule: ⟨Lₓ²⟩ + ⟨L_y²⟩ = ⟨L²⟩ − ⟨L_z²⟩ = [l(l+1) − m²]ℏ² = (12 − 9)ℏ² = 3ℏ².
Symmetry about the z axis makes the two equal, so ⟨Lₓ²⟩ = ⟨L_y²⟩ = 1.5ℏ², giving ΔLₓ = ΔL_y = √1.5 ℏ = 1.225ℏ and a product of 1.5ℏ².
The floor is ½ℏ|⟨L_z⟩| = ½ℏ(3ℏ) = 1.5ℏ², identical to the product, so this stretched state saturates the inequality.
In SI units: ΔLₓ = 1.225 × 1.055 × 10⁻³⁴ J s = 1.29 × 10⁻³⁴ J s.

AnswerΔLₓ = ΔL_y = √1.5 ℏ = 1.29 × 10⁻³⁴ J s; the product 1.5ℏ² equals the floor 1.5ℏ² exactly, so |3, 3⟩ is a minimum-uncertainty state for Lₓ and L_y.

HardTake |ψ⟩ = (|1, 1⟩ + |1, −1⟩)/√2. Find ⟨L_z⟩ and ΔL_z, then ⟨Lₓ⟩ and ΔLₓ, then ⟨L_y⟩ and ΔL_y, and test ΔLₓ ΔL_z ≥ ½ℏ|⟨L_y⟩|. Use L₊|1, m⟩ = ℏ√(2 − m(m+1))|1, m+1⟩ and L₋|1, m⟩ = ℏ√(2 − m(m−1))|1, m−1⟩. Then say what the vanishing bound does and does not permit.

L_z is diagonal: ⟨L_z⟩ = ½(+ℏ) + ½(−ℏ) = 0, and ⟨L_z²⟩ = ½ℏ² + ½ℏ² = ℏ², so ΔL_z = ℏ.
For l = 1 every non-zero ladder step is ℏ√2. With Lₓ = (L₊ + L₋)/2, both Lₓ|1, 1⟩ and Lₓ|1, −1⟩ equal (ℏ/√2)|1, 0⟩.
So Lₓ|ψ⟩ = (1/√2)(2 × ℏ/√2)|1, 0⟩ = ℏ|1, 0⟩. Since |ψ⟩ has no |1, 0⟩ component, ⟨Lₓ⟩ = 0, while ⟨Lₓ²⟩ is the norm-squared of ℏ|1, 0⟩, namely ℏ². Hence ΔLₓ = ℏ.
With L_y = (L₊ − L₋)/2i the two contributions carry opposite signs and cancel: L_y|ψ⟩ = 0. So |ψ⟩ is an eigenstate of L_y with eigenvalue zero, giving ⟨L_y⟩ = 0 and ΔL_y = 0.
Test the inequality: ΔLₓ ΔL_z = ℏ × ℏ = ℏ² against a floor of ½ℏ|⟨L_y⟩| = 0, satisfied with ℏ² to spare. Consistency check: ⟨Lₓ²⟩ + ⟨L_y²⟩ + ⟨L_z²⟩ = ℏ² + 0 + ℏ² = 2ℏ² = l(l+1)ℏ², as required.
So a vanishing floor permits a sharp component but never forces one. Here it is L_y that is sharp, and the price is that Lₓ and L_z are each spread by ℏ. Only l = 0 makes all three certain at once.

Answer⟨L_z⟩ = 0 with ΔL_z = ℏ; ⟨Lₓ⟩ = 0 with ΔLₓ = ℏ; ⟨L_y⟩ = 0 with ΔL_y = 0, so |ψ⟩ is the L_y = 0 eigenstate. ΔLₓ ΔL_z = ℏ² against a floor of 0 — satisfied with room to spare.