University Physics V · Angular Momentum in Quantum Mechanics · 9.1

Angular Momentum as a Rotation Generator

Before any operator appears, angular momentum is already a generator: the charge rotational symmetry conserves, and the function whose Poisson bracket rotates everything else. This topic gets the εᵢⱼₖ algebra out of classical mechanics and the SO(3) matrices, so that when ħ arrives you can see exactly what was postulated.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Angular momentum enters classical mechanics twice, and the two entrances have to be recognised as one. First as a quantity: L = r × p, the Noether charge of a Lagrangian left unchanged by rigid rotations, conserved exactly when the potential depends on |r| alone. Second as an operation: L is the function whose Poisson bracket generates rotations, δf = δθ (f, n̂⋅L), so that rotating a system and taking a bracket with n̂⋅L are the same act.

The second reading is the one that survives quantisation, and it forces an algebra. Compose two small rotations in both orders and they disagree at second order, by a rotation about the third axis; in bracket form that is {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ, and the identical structure constants sit in the 3×3 antisymmetric generators of SO(3), where [Tᵢ, Tⱼ] = εᵢⱼₖ Tₖ with no mechanics in sight. The algebra is therefore a property of rotations in three dimensions, not of any particular particle.

What it costs is honesty about the last step. Dirac's rule {A, B} → [Â, B̂]/iħ turns it into [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ, but the ħ is inserted by hand: the rule is a postulate, Groenewold and van Hove proved no consistent version covers all polynomials, and L_z escapes the ordering ambiguity only because x and p_y happen to commute. Classical mechanics hands over the shape of the algebra, never the spectrum, and it never produces a half-integer.

Simple definition: Classical angular momentum L = r × p is the conserved charge of rotational symmetry and, equivalently, the generator of rotations in phase space: bracketing any function with n̂⋅L rotates it about n̂.
Example: For a 0.20 kg puck at r = (0.30, −0.40, 0) m with p = (0, 1.0, 0.40) kg m s⁻¹, L = r × p = (−0.16, −0.12, 0.30) kg m² s⁻¹, of magnitude 0.36 kg m² s⁻¹ — some 3.4 × 10³³ ħ, which is why nothing quantum shows.

Angular momentum as a Noether chargeL = r × p, Lᵢ = εᵢⱼₖ xⱼ pₖ

Rigid rotation leaves the Lagrangian invariant ⇒ n̂⋅L is conserved for every fixed axis n̂.

x in m, p in kg m s⁻¹, so L carries kg m² s⁻¹ = J s, the unit of ħ itself

The Poisson bracket{A, B} = Σᵢ (∂A/∂xᵢ ∂B/∂pᵢ − ∂A/∂pᵢ ∂B/∂xᵢ)

Bilinear, antisymmetric, and a derivation in each slot — enough to get {Lᵢ, Lⱼ} without differentiating anything.

sum over the three Cartesian pairs; {xᵢ, pⱼ} = δᵢⱼ is the only non-zero canonical bracket

The rotation algebra, classically{Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ, (L², Lᵢ) = 0

Non-zero brackets mean the Lᵢ fail to commute as flows, not as measurements — no ħ has appeared yet.

εᵢⱼₖ = +1 cyclic, −1 anticyclic, 0 if any index repeats; both sides carry J s

L generates rotations in phase spaceδf = δθ (f, n̂⋅L) · {Lᵢ, xⱼ} = εᵢⱼₖ xₖ

Bracketing with n̂⋅L is what rotating about n̂ does, so dL/dt = {L, H} makes conservation a symmetry statement.

δθ in radians; order matters, since the bracket is antisymmetric: δx = δθ {x, L_z} = −δθ y

SO(3) generators carry the same constantsR(n̂, θ) = exp(θ n̂⋅T), (Tᵢ)ⱼₖ = −εᵢⱼₖ, [Tᵢ, Tⱼ] = εᵢⱼₖ Tₖ

The structure constants are geometry, not mechanics: anything that rotates must reproduce εᵢⱼₖ.

Tᵢ are real antisymmetric 3×3 matrices; θ in radians, and R is orthogonal with det R = +1

Dirac's quantisation rule — a postulate{A, B} → [Â, B̂] / iħ ⇒ [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ

It buys the quantum algebra in one line — and Groenewold–van Hove says no such map works for every polynomial.

ħ = 1.055 × 10⁻³⁴ J s supplies the dimensions, so Ĵ carries J s and j, m stay pure numbers

Rotational symmetry, and the charge it conserves

Rotate the whole system rigidly about a fixed axis n̂ through the origin: each position shifts by δrₐ = δθ (n̂ × rₐ), each velocity by δṙₐ = δθ (n̂ × ṙₐ). The kinetic energy Σ ½mₐ|ṙₐ|² is untouched, because a rotation preserves lengths, so the Lagrangian is invariant precisely when the potential is — true for V(|r|) about a force centre, and for V(|rₐ − r_b|) between particles. Noether's theorem then hands over the conserved charge Q = Σₐ pₐ⋅(∂rₐ/∂θ) = Σₐ pₐ⋅(n̂ × rₐ) = n̂⋅Σₐ (rₐ × pₐ) = n̂⋅L, where the middle step is the scalar triple product a⋅(b × c) = c⋅(a × b). The axis n̂ was arbitrary, so all three components of L = Σₐ rₐ × pₐ are separately conserved. Two riders. The origin is a choice: shift it by a and L → L − a × P, so a system can conserve angular momentum about one point and not another. And a uniform gravitational field leaves only rotations about the vertical invariant, so only L_z survives — which is why a spinning top precesses instead of holding L fixed.

From the bracket to the algebra, in four terms

The bracket {A, B} = Σᵢ (∂A/∂xᵢ ∂B/∂pᵢ − ∂A/∂pᵢ ∂B/∂xᵢ) is bilinear, antisymmetric, and obeys Leibniz, {AB, C} = A{B, C} + {A, C}B. So once you know {xᵢ, pⱼ} = δᵢⱼ and {xᵢ, xⱼ} = {pᵢ, pⱼ} = 0, you never differentiate again. Take Lₓ = y p_z − z p_y and L_y = z pₓ − x p_z and expand into four brackets. Two vanish: {y p_z, x p_z} and {z p_y, z pₓ} each need only brackets of a coordinate with a momentum of a different axis, all zero. The survivors keep one δ each: {y p_z, z pₓ} = y{p_z, z}pₓ = −y pₓ, and {z p_y, x p_z} = x{z, p_z}p_y = +x p_y. Adding, {Lₓ, L_y} = x p_y − y pₓ = L_z, and cyclic relabelling gives {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ. One more bracket matters: (L², L_z) = 2Lₓ{Lₓ, L_z} + 2L_y{L_y, L_z} = −2LₓL_y + 2L_yLₓ = 0. That vanishing is the classical ancestor of the compatibility of L² with any one component.

A bracket is a flow, and this one is a rotation

A Poisson bracket is not only a product; it is a flow. For any phase-space function G, δf = ε{f, G} is the change of f under the canonical transformation G generates, and time evolution ḟ = {f, H} is that same statement with G = H and ε = dt. Put G = L_z and test it on the coordinates. {L_z, x} = y and {L_z, y} = −x, so by antisymmetry δx = ε{x, L_z} = −εy and δy = ε{y, L_z} = +εx — exactly the first-order form of x′ = x cos ε − y sin ε, y′ = x sin ε + y cos ε. The same bracket rotates p the same way, and leaves the rotational invariants r⋅r, p⋅p and r⋅p untouched, as it must. Now compose. Rotating about x and then about y, versus about y and then about x, gives results differing at order ε₁ε₂, and the residual is a rotation about z whose generator is precisely {Lₓ, L_y} = L_z. Read that way the algebra is not a fact about r × p at all: it is the non-abelian character of the rotation group written in phase-space language.

The same constants, read straight off SO(3)

Drop mechanics entirely and look at the matrices. A rotation about z by θ has rows (cos θ, −sin θ, 0), (sin θ, cos θ, 0), (0, 0, 1). Differentiate at θ = 0 and the generator T_z is all zeros except (T_z)₁₂ = −1 and (T_z)₂₁ = +1; in general (Tᵢ)ⱼₖ = −εᵢⱼₖ, real and antisymmetric, with R(n̂, θ) = exp(θ n̂⋅T) for θ in radians. Multiply out TₓT_y and T_yTₓ — each has a single non-zero entry, in slot 21 and slot 12 respectively — and the difference is [Tₓ, T_y] = T_z, so [Tᵢ, Tⱼ] = εᵢⱼₖ Tₖ. The structure constants match {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ exactly, with no particle, no mass and no Hamiltonian anywhere in the derivation. You can see it without matrices too: turn 5° about x, then 5° about y, then undo each in the same order, and what is left over is a turn of roughly (5°)(5°) in radians, 0.0076 rad or 0.44°, about z. The commutator of two small rotations is a third rotation whose angle is the product, and εᵢⱼₖ is the table of which axis it is about.

Where the classical story stops: the quantisation rule

Dirac's rule is one line: replace each classical variable by a Hermitian operator and each bracket by {A, B} → [Â, B̂]/iħ. Apply it to {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ and out comes [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ, the relation the rest of this unit is built on. Three things about that line deserve saying out loud. It is a postulate, justified by the classical limit and by results, not proved: Groenewold and van Hove showed that no map from classical polynomials to operators can respect every bracket while keeping x̂ and p̂ canonical, so the rule is exact only on the low-order functions we actually use. Ordering is genuinely ambiguous in general — classical xp could become x̂p̂, p̂x̂ or the symmetric ½(x̂p̂ + p̂x̂), and only the last is Hermitian — but L̂_z = x̂p̂_y − ŷp̂ₓ is spared, because x and p_y belong to different axes and their operators commute. And the iħ is inserted by hand: it carries the dimensions, since [x̂, p̂] must come out in J s, and it is what converts a statement about flows into a statement about measurements. Nothing classical fixes its value.

What the algebra decides, and what it leaves open

Two boundaries are worth marking before the ladder argument arrives. First, a non-zero bracket is not an uncertainty relation. A classical state is a single point of phase space, and at that point Lₓ, L_y and L_z all have exact values at once; {Lₓ, L_y} = L_z says only that the flow generated by Lₓ moves L_y, which is a claim about what rotations do, not about what can be known. Joint values are obstructed only after the bracket becomes an operator commutator carrying iħ, because Robertson's inequality σ_Aσ_B ≥ ½|⟨[Â, B̂]⟩| is built from a commutator and has no classical object to inherit. Second, the algebra is more permissive than its classical origin. Realised as r × p acting on functions of position, it supports the integer towers only; the abstract algebra [Jᵢ, Jⱼ] = iħ εᵢⱼₖ Jₖ also admits half-integer j, which no r × p can produce. Classical mechanics therefore hands over the commutation relations and the Casimir L², but neither the list of allowed j nor the fact that electrons carry one. Those come from the ladder argument and from experiment.

Change one variable at a time

Make the relationship visible.

Interactive model

Turn about x, α30 °

Turn about y, β30 °

Set both angles to 5° and watch GAP ÷ αβ sit at 0.998 — the commutator of two small turns is a third turn of size αβ about z. Push both to 90° and the two orders end a full 90° apart, which is why only the infinitesimal statement [Tₓ, T_y] = T_z is exact.

GAP BETWEEN TIPS0.259

SMALL-ANGLE αβ0.274 rad

GAP ÷ αβ0.944

ANGLE BETWEEN ORDERS14.87 °

Live interpretationGAP BETWEEN TIPS: 0.259. SMALL-ANGLE αβ: 0.274 rad. GAP ÷ αβ: 0.944. ANGLE BETWEEN ORDERS: 14.87 °

Catch the common trap

Explain before calculating.

A classical particle moves in a potential V(x, y, z) with no symmetry whatever. Its angular-momentum components still satisfy {Lₓ, L_y} = L_z. What does that non-zero bracket assert?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 0.20 kg puck sits at r = (0.30, −0.40, 0) m and moves with v = (0, 5.0, 2.0) m s⁻¹. Find L about the origin, then about the point a = (0, 0, 0.50) m, and express the first magnitude in units of ħ = 1.055 × 10⁻³⁴ J s.

Momentum first: p = mv = 0.20 × (0, 5.0, 2.0) = (0, 1.0, 0.40) kg m s⁻¹.
Now L = r × p, component by component. Lₓ = y p_z − z p_y = (−0.40)(0.40) − (0)(1.0) = −0.16. L_y = z pₓ − x p_z = (0)(0) − (0.30)(0.40) = −0.12. L_z = x p_y − y pₓ = (0.30)(1.0) − (−0.40)(0) = 0.30, all in kg m² s⁻¹.
About a, use r′ = r − a = (0.30, −0.40, −0.50): L′ₓ = (−0.40)(0.40) − (−0.50)(1.0) = 0.34, L′_y = (−0.50)(0) − (0.30)(0.40) = −0.12, L′_z = 0.30.
Check the shift: L′ − L = (0.50, 0, 0), and −a × p = −(0, 0, 0.50) × (0, 1.0, 0.40) = −(−0.50, 0, 0) = (0.50, 0, 0). The origin is part of the definition, not a detail.
Magnitude about O: |L| = √(0.16² + 0.12² + 0.30²) = √(0.0256 + 0.0144 + 0.0900) = √0.1300 = 0.3606 kg m² s⁻¹.
In units of ħ: 0.3606 ÷ (1.055 × 10⁻³⁴) = 3.4 × 10³³. A step of one in that quantum number is 33 orders of magnitude below the value itself — which is what calling the motion classical means.

AnswerL = (−0.16, −0.12, 0.30) kg m² s⁻¹ about O and (0.34, −0.12, 0.30) kg m² s⁻¹ about a, differing by −a × p = (0.50, 0, 0). |L| = 0.36 kg m² s⁻¹ ≈ 3.4 × 10³³ ħ.

MediumUsing only {xᵢ, pⱼ} = δᵢⱼ, antisymmetry, and the Leibniz rule {AB, C} = A{B, C} + {A, C}B, evaluate {Lₓ, L_y} for a single particle, then show (L², L_z) = 0 and say what each result becomes after quantisation.

Write the pieces: Lₓ = y p_z − z p_y and L_y = z pₓ − x p_z. Bilinearity splits the bracket into four: {y p_z, z pₓ} − {y p_z, x p_z} − {z p_y, z pₓ} + {z p_y, x p_z}.
Two die at once. {y p_z, x p_z} needs {y, x}, {y, p_z}, {p_z, x} and {p_z, p_z}, every one of them zero; the same audit kills {z p_y, z pₓ}.
The survivors keep a single δ each. {y p_z, z pₓ} = y {p_z, z} pₓ = −y pₓ, since {p_z, z} = −{z, p_z} = −1. And {z p_y, x p_z} = x {z, p_z} p_y = +x p_y.
Add them: {Lₓ, L_y} = −y pₓ + x p_y = x p_y − y pₓ = L_z. Cyclic relabelling gives {L_y, L_z} = Lₓ and {L_z, Lₓ} = L_y, that is {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ.
For the Casimir, apply Leibniz to L² = Lₓ² + L_y² + L_z²: (L², L_z) = 2Lₓ{Lₓ, L_z} + 2L_y{L_y, L_z} + 2L_z{L_z, L_z} = 2Lₓ(−L_y) + 2L_y(Lₓ) + 0 = 0.
Under {A, B} → [Â, B̂]/iħ these become [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ and [L̂², L̂_z] = 0 — the first forbidding a shared eigenbasis for two components, the second permitting one for L̂² and L̂_z.

Answer{Lₓ, L_y} = L_z, and cyclically {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ; (L², L_z) = 0. Quantised: [L̂ᵢ, L̂ⱼ] = iħ εᵢⱼₖ L̂ₖ and [L̂², L̂_z] = 0, which is why (H, L̂², L̂_z) can be a complete commuting set.

HardTake the right-handed rotation matrices Rₓ(α) and R_y(β). (a) Obtain Tₓ and T_y by differentiating at zero and verify [Tₓ, T_y] = T_z. (b) Apply both orders to v₀ = x̂ with α = β = 0.20 rad and find the angle between the two results. (c) Compare that angle with αβ, and name the direction the discrepancy points.

Rₓ(α) has rows (1, 0, 0), (0, cos α, −sin α), (0, sin α, cos α); differentiating at α = 0 leaves Tₓ with only (Tₓ)₂₃ = −1 and (Tₓ)₃₂ = +1. Likewise (T_y)₁₃ = +1 and (T_y)₃₁ = −1. Both match (Tᵢ)ⱼₖ = −εᵢⱼₖ.
(TₓT_y)₂₁ = (Tₓ)₂₃(T_y)₃₁ = (−1)(−1) = +1 is the only non-zero entry of TₓT_y; (T_yTₓ)₁₂ = (T_y)₁₃(Tₓ)₃₂ = (+1)(+1) = +1 is the only one of T_yTₓ. Subtracting leaves −1 in slot 12 and +1 in slot 21, which is T_z exactly. So [Tᵢ, Tⱼ] = εᵢⱼₖ Tₖ, the same constants as {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ.
x̂ is fixed by Rₓ, so the x-turn first then the y-turn gives v_A = R_y(β)x̂ = (cos β, 0, −sin β). The other order gives v_B = Rₓ(α)R_y(β)x̂ = (cos β, sin α sin β, −cos α sin β).
With α = β = 0.20 rad: cos = 0.98007 and sin = 0.19867, so v_A = (0.98007, 0, −0.19867) and v_B = (0.98007, 0.03947, −0.19471).
Both are unit vectors, so cos θ = v_A⋅v_B = cos²β + cos α sin²β = 0.96053 + 0.03868 = 0.99921, giving θ = arccos(0.99921) = 0.03967 rad = 2.273°.
Compare: αβ = 0.0400 rad, so the true mismatch falls 0.8% short of the leading-order value, and the shortfall shrinks with the angles. The difference v_B − v_A = (0, 0.03947, 0.00396) points essentially along +ŷ, which is the reading ε_xyz = +1 in [Tₓ, T_y] = T_z.

Answer[Tₓ, T_y] = T_z, matching {Lᵢ, Lⱼ} = εᵢⱼₖ Lₖ. At α = β = 0.20 rad the two orders leave x̂ 0.03967 rad (2.273°) apart, against the prediction αβ = 0.0400 rad — agreeing to 0.8% — and the gap lies along +ŷ.