University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.1

Rotation operators and angular momentum as their generator

Quantum Mechanics I built angular momentum out of r × p and then met spin, which has no r × p at all. This week puts things in the right order: start from what a rotation does to a state, call its generator J, and the commutators, the conservation law and the (2j + 1)-fold multiplets follow, for orbit and spin alike.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A rotation R of space must act on quantum states through an operator D(R), and because a turn changes no probability |⟨χ|ψ⟩|², Wigner's theorem (quoted) makes D unitary for every rotation reachable continuously from doing nothing. Near the identity D = 1 − iεJ⋅n̂/ħ, and that line is now the definition of angular momentum: J is whatever generates the turn, Hermitian because D is unitary, and composing many small turns gives D(n̂, φ) = e^{(−iJ⋅n̂φ/ħ)}. Nothing in the definition mentions position. Apply it to a spinless wavefunction, ψ(r) → ψ(R⁻¹r), and J comes out as r × p, so the old angular momentum is one case of the new.

The algebra comes from geometry alone: small turns about x and y, done in opposite orders, differ by a turn through ε² about z, and a D that multiplies the way rotations do must obey [Jᵢ, Jⱼ] = iħεᵢⱼₖJₖ, for an electron's spin or a whole atom exactly as for an orbit. Symmetry then pays out: a Hamiltonian that no rotation changes commutes with J, so J is conserved and J₊, J₋ carry each level through all 2j + 1 values of m. The argument stops in three places. It fixes D only up to a phase, and the next topic finds a 2π turn that returns −1.

It never says which j a system has; that is physics, not geometry. And it conserves only the components whose turns leave H alone.

Simple definition: The angular momentum J of any system is the generator of its rotations: the Hermitian operator for which a turn by a small angle ε about the unit vector n̂ acts on every state as D = 1 − iεJ⋅n̂/ħ.
Example: For a spinless particle a turn by ε about z sends ψ(x, y, z) to ψ(x + εy, y − εx, z) = ψ − (iε/ħ)L_zψ + O(ε²), with L_z = xp_y − ypₓ. Turned by 50°, a pₓ orbital keeps amplitude cos 50° = 0.643 in its old state.

Rotation operator, defined by its generatorD(n̂, φ) = e^{(−iJ⋅n̂φ/ħ)}

Derived: D ≈ 1 − iεJ⋅n̂/ħ defines J, unitarity forces J† = J, and N turns of φ/N compose to the exponential. At φ = 0 it is the identity.

J in J s, like ħ = 1.055 × 10⁻³⁴ J s; φ in rad; n̂ a unit vector; D is dimensionless and unitary

Rotating a spinless wavefunctionψ(R⁻¹r) = ψ(r) − (iε/ħ) n̂⋅(r × p) ψ(r) + O(ε²)

Derived by Taylor expansion. The generator on a scalar wavefunction is L = r × p, so Quantum Mechanics I's angular momentum is one case of J.

r in m; p = −iħ∇ in kg m s⁻¹, so r × p carries J s; ε in rad, the turn about n̂

Two small turns fail to commuteRₓ(ε)R_y(ε) − R_y(ε)Rₓ(ε) = R_z(ε²) − 1 + O(ε³)

Derived by multiplying matrices. At ε = 0.20 rad the exact misfit is a turn of 0.0399 rad, against ε² = 0.0400 rad, about an axis 8° from z.

3 × 3 rotation matrices, ε in rad; the right-hand factor acts first, and every term linear in ε cancels

The commutation relations[Jᵢ, Jⱼ] = iħεᵢⱼₖJₖ

Derived by demanding that the D's multiply as rotations do, so spin, orbit and whole atoms obey them alike. As ħ → 0, divided by iħ, they become the classical Poisson brackets.

εᵢⱼₖ the Levi-Civita symbol; both sides in J² s²; no position or momentum enters

Rotation symmetry means conservation[H, Jᵢ] = 0 ⇒ dJᵢ/dt = (i/ħ)[H, Jᵢ] = 0

Derived: H unchanged by every turn gives [H, Jᵢ] = 0 at first order in ε. Only the components whose turns leave H alone are conserved; a field along z keeps J_z.

H in J; Unit 3's Heisenberg equation; J carries no explicit time dependence

The multiplet a symmetric H must haveH J₊|E, j, m⟩ = E J₊|E, j, m⟩, so g = 2j + 1

Derived: J₊ and J₋ commute with H, so they step through m at fixed E. Measured: silver atoms, j = ½, split into 2 Stern–Gerlach beams.

g the number of states one ladder links at a single E, a pure number; a level's full degeneracy is at least g; m runs from −j to j in unit steps (ladder algebra quoted from Quantum Mechanics I)

Define angular momentum by what it does

Quantum Mechanics I defined L = r × p and got its commutators from [x, p] = iħ; spin then arrived with the same commutators and no r × p to explain them. Start instead from the rotation. Turning a system by R must act on its kets through some operator D(R), and since a turn changes no probability |⟨χ|ψ⟩|², Wigner's theorem, quoted here, makes D unitary for every rotation reachable continuously from the identity. A small turn is close to doing nothing, and to first order a turn by ε about n̂ is the sum of turns εnₓ, εn_y and εn_z about the axes, so D(n̂, ε) = 1 − iεJ⋅n̂/ħ + O(ε²). That is the definition: J is whatever operator appears there, with ħ inserted so J carries J s. Unitarity fixes its character, since D†D = 1 + iε(J† − J)⋅n̂/ħ must equal 1, forcing J† = J. For a finite turn about a fixed axis, D(n̂, φ + dφ) = D(n̂, dφ)D(n̂, φ) gives dD/dφ = −(iJ⋅n̂/ħ)D, so D(n̂, φ) = e^{(−iJ⋅n̂φ/ħ)}. Checks: φ = 0 gives the identity, and the exponent is dimensionless.

Rotate a wavefunction and r × p appears

For a particle without spin, rotating the state carries the pattern of ψ to new places: the new wavefunction at r is the old one at the point that was carried there, ψ′(r) = ψ(R⁻¹r). The inverse is what moves the pattern forward. For a turn by ε about z, R⁻¹ sends (x, y, z) to (x + εy, y − εx, z), and a first-order Taylor expansion gives ψ′ = ψ − ε(x ∂ψ/∂y − y ∂ψ/∂x). Matching 1 − iεJ_z/ħ, the generator is J_z = −iħ(x ∂/∂y − y ∂/∂x) = xp_y − ypₓ, the z-component of r × p; every axis gives J = r × p = L. In spherical coordinates the turn only shifts the azimuth, ψ(r, θ, ϕ − φ), and Taylor's series in φ is then exactly e^{(−φ∂/∂ϕ)}, so L_z = −iħ ∂/∂ϕ holds for finite turns. On Yₗᵐ ∝ e^(imϕ) the rotation is the phase e^(−imφ), which is 1 at φ = 2π because m is an integer. An s state is untouched, as a spherical function must be.

Two small rotations fail to commute, by a third

Write the 3 × 3 matrix for a turn by ε about x to second order: its y–z block has 1 − ε²/2 on the diagonal and ∓ε off it, and R_y(ε) has the same pattern in the z–x block. Multiply them in both orders, remembering that the right-hand factor acts first. Everything first order in ε cancels in the difference, and what survives is Rₓ(ε)R_y(ε) − R_y(ε)Rₓ(ε) = R_z(ε²) − 1 + O(ε³): the two orders differ by a turn through ε² about z. Try it on an object. Tip a phone 10° about x, then 10° about y, and a second phone in the other order: they end up turned relative to each other by 1.74° about an axis 7° from z, against ε² = 0.0305 rad = 1.75°. The exact products also hold terms of order ε³, about ε/2 of the leading one, which is where that tilt comes from. Two limits keep the statement honest: the misfit vanishes faster than either turn as ε → 0, and two turns about one axis commute exactly.

Demand the same product law of D, and the commutators follow

A representation must multiply as the rotations do, D(R₁)D(R₂) = D(R₁R₂), so the misfit must reappear among the D's. Expand each factor to second order, D(Rₓ(ε)) = 1 − iεJₓ/ħ − ε²Jₓ²/2ħ². In D(Rₓ)D(R_y) − D(R_y)D(Rₓ) the zeroth-order, first-order and squared terms cancel, leaving −(ε²/ħ²)(JₓJ_y − J_yJₓ), while the right side is D(R_z(ε²)) − 1 = −iε²J_z/ħ. So [Jₓ, J_y] = iħJ_z, and cycling x → y → z gives [Jᵢ, Jⱼ] = iħεᵢⱼₖJₖ. No r, p or wavefunction was used, so spin obeys the same relations, and they admit the half-integer j that r × p cannot supply. Checks: both sides carry J² s², and as ħ → 0 the commutator divided by iħ becomes the classical Poisson bracket. No power of α enters; this is exact geometry. One assumption was hidden: states are rays, so D(R₁)D(R₂) may carry an extra phase. Near the identity that phase only adds a constant to each commutator, which shifting each Jₖ by a constant removes; a sign at 2π can survive, and the next topic finds it.

Rotation-invariant H conserves J and fills multiplets

If no turn changes the energy operator, D(R)†HD(R) = H for every R, and at first order in ε that is [H, Jᵢ] = 0 for all three components. Unit 3's Heisenberg equation, dJᵢ/dt = (i/ħ)[H, Jᵢ], then makes each component a constant of motion. J² and J_z label the energy eigenstates |E, j, m⟩, and since J₊ commutes with H, H(J₊|E, j, m⟩) = E(J₊|E, j, m⟩): the ladder steps through m without changing the energy, stopping only at m = ±j. So every level of an isotropic system is at least (2j + 1)-fold degenerate. The measured record agrees. Silver atoms, j = ½, split into exactly 2 beams in Stern and Gerlach's field. Carbon monoxide rotates freely with H = L²/2I and only integer J, as the ψ(R⁻¹r) route demands; its lines at 115.271, 230.538 and 345.796 GHz, taken from a spectroscopic table, stand as 1 : 2.000 : 3.000, the spacings of J(J + 1), and the 2J + 1 states per level show up in the band's intensities.

Partial symmetry buys partial conservation

The argument conserves only the generators whose turns leave H alone, and that is how later units decide what a good quantum number is. In a uniform magnetic field along z, H is unchanged by turns about z but not about x, so J_z is conserved while Jₓ and J_y precess, and without J₊ the 2j + 1 states of a level spread into 2j + 1 Zeeman components. Spin–orbit coupling, proportional to L⋅S, is unchanged when orbit and spin are turned together but not when either is turned alone. L and S separately generate those partial turns, so neither is conserved, while J = L + S is. That is why fine structure labels states by j rather than by mₗ and mₛ; it is an effect of order α², and hydrogen's measured 2p splitting of 10.97 GHz sits close to α²/4 of the 3.40 eV binding energy, 10.95 GHz (quoted from the fine-structure unit). A degeneracy larger than 2j + 1 signals extra symmetry, as hydrogen's l-degeneracy does; its source, the Runge–Lenz vector, is quoted here, not derived.

Change one variable at a time

Make the relationship visible.

Interactive model

Rotation angle φ about z60 °

Azimuthal number m of cos mϕ1

First-order steps N4

Start at N = 1, one first-order step ψ − φ ∂ψ/∂ϕ: it slides the curve the right way but swells it. Raise N and the dashed curve closes on the solid one, the exponential of L_z. Then set m = 2 and φ = 45°, which turns cos 2ϕ into sin 2ϕ, and m = 0, which no turn can move.

OVERLAP ⟨ψ|D|ψ⟩0.500

EIGENPHASE mφ60 °

N-STEP NORM1.142

N-STEP PHASE SHORTFALL1.32 °

Live interpretationOVERLAP ⟨ψ|D|ψ⟩: 0.500. EIGENPHASE mφ: 60 °. N-STEP NORM: 1.142. N-STEP PHASE SHORTFALL: 1.32 °

Catch the common trap

Explain before calculating.

An atom sits in a uniform magnetic field along z, so its Hamiltonian is unchanged by every rotation about z but not by rotations about x or y. What does the generator argument let you conclude?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA hydrogen atom is prepared in the 2p state ψ = N x e^(−r/2a₀), a pₓ orbital with its lobes along x. It is rotated by φ = 50° about the z-axis. Write the rotated wavefunction, find the probability that the rotated atom is found in the original state, and check the result at φ = 90° and φ = 360°.

The rotated state is ψ′(r) = ψ(R⁻¹r). For a turn by φ about z, R⁻¹ carries (x, y, z) to (x cos φ + y sin φ, −x sin φ + y cos φ, z) and leaves r unchanged, so ψ′ = N(x cos φ + y sin φ)e^(−r/2a₀).
With cos 50° = 0.643 and sin 50° = 0.766, ψ′ = N(0.643x + 0.766y)e^(−r/2a₀): the same orbital with its lobe turned 50° from x towards y, the forward direction for an active rotation.
Overlap: ⟨ψ|ψ′⟩ = cos φ ⟨ψ|ψ⟩ + sin φ N² ∫ xy e^(−r/a₀) d³r. The integral is odd in x and vanishes, so ⟨ψ|ψ′⟩ = cos 50° = 0.643 and the probability is cos²50° = 0.413.
The generator gives the same number. With the Condon–Shortley phase, pₓ = (−|1, 1⟩ + |1, −1⟩)/√2, and D(ẑ, φ) = e^{(−iφL_z/ħ)} multiplies |1, ±1⟩ by e^(∓iφ), so ⟨ψ|D|ψ⟩ = (e^(−iφ) + e^(iφ))/2 = cos φ, the average of the two eigenphases.
Checks: at φ = 90° the state is N y e^(−r/2a₀), a p_y orbital, and the probability is 0; at φ = 360°, cos²(2π) = 1 because m is an integer. ψ′ keeps the units of ψ, m^(−3/2), and the probability is a pure number.

Answerψ′ = N(0.643x + 0.766y)e^(−r/2a₀), the pₓ lobe turned 50° towards y; the probability of finding the original state is cos²50° = 0.413.

MediumTurns of ε = 0.20 rad about x and about y are applied to a rigid body in both orders. Find the exact difference Rₓ(ε)R_y(ε) − R_y(ε)Rₓ(ε), compare it with R_z(ε²) − 1, and use the comparison to fix the commutator [Jₓ, J_y] that any quantum description of the body must obey. Check the result on a spin-½.

With c = cos ε = 0.98007 and s = sin ε = 0.19867, multiplying the matrices gives Rₓ(ε)R_y(ε) with rows (c, 0, s), (s², c, −sc), (−sc, s, c²), and R_y(ε)Rₓ(ε) with rows (c, s², sc), (0, c, −s), (−s, sc, c²).
The difference has +s² = 0.03947 in row 2, column 1 and −0.03947 in row 1, column 2; the four entries that touch z all equal s(1 − c) = 0.00396, and the diagonal cancels exactly.
R_z(ε²) − 1 has sin ε² = 0.03999 and −0.03999 in the same two places and only cos ε² − 1 = −0.0008 on its diagonal. The ε² parts agree to 1.3%; the leftover s(1 − c) ≈ ε³/2 = 0.0040 is the O(ε³) remainder, ε/2 = 10% of the leading term. The net turn is 0.0399 rad about an axis 8.1° from z.
Any D must repeat this: D(Rₓ)D(R_y) − D(R_y)D(Rₓ) = D(R_z(ε²)) − 1. To second order the left side is −(ε²/ħ²)[Jₓ, J_y] and the right side is −iε²J_z/ħ, so [Jₓ, J_y] = iħJ_z.
Spin-½, J = ħσ/2: [Jₓ, J_y] = (ħ²/4)(σₓσ_y − σ_yσₓ) = (ħ²/4)(2iσ_z) = iħ(ħσ_z/2) = iħJ_z, using σₓσ_y = iσ_z from Quantum Mechanics I. Both sides carry J² s².

AnswerThe two orders differ by 0.0395 in the z-turn entries, against ε² = 0.0400, with an O(ε³) remainder of 0.0040; this forces [Jₓ, J_y] = iħJ_z, which the spin-½ matrices satisfy.

HardThe ground state of the negatively charged nitrogen-vacancy centre in diamond is a spin-1 whose zero-field Hamiltonian is H = (Δ/ħ²)S_z², with z along the centre's symmetry axis and Δ/h = 2.870 GHz measured at room temperature. Decide which components of S are conserved, explain why the three m states form a single level and a pair rather than one triplet or three separate levels, and follow ⟨S⟩ for a spin prepared pointing along x.

Test each generator against H. A turn about z leaves S_z² alone, and [H, S_z] = 0, so S_z is conserved and m stays good. A turn about x does not: [Sₓ, S_z²] = S_z[Sₓ, S_z] + [Sₓ, S_z]S_z = −iħ(S_zS_y + S_yS_z), which is not zero, so Sₓ and S_y are not conserved. S² = 2ħ² is fixed because s = 1.
Levels: H|1, m⟩ = m²Δ|1, m⟩, so m = 0 lies at 0 and m = ±1 at Δ. S₊ no longer commutes with H, so nothing holds all 2j + 1 = 3 states together. One discrete turn survives: the commutators give D(x̂, θ)†S_zD(x̂, θ) = S_z cos θ + S_y sin θ, which is −S_z at θ = π, so a half-turn about x leaves S_z² and H unchanged and carries the eigenstate |1, 1⟩ into |1, −1⟩ up to a phase. The pairing needs both symmetries: strain that spoils the turns about z splits the pair.
Measured check: in zero field the centre shows one magnetic-resonance line, from m = 0 to the m = ±1 pair, at 2.870 GHz, split by only a few MHz in strained crystals. Δ is quoted from that measurement, not derived; it comes from the magnetic coupling of the centre's two unpaired electrons.
Spin along x. With the spin-1 matrices quoted from Quantum Mechanics I, Sₓ = (ħ/√2) times the matrix with rows (0, 1, 0), (1, 0, 1), (0, 1, 0), whose Sₓ = +ħ eigenvector is (|1, 1⟩ + √2|1, 0⟩ + |1, −1⟩)/2. At time t the m = ±1 amplitudes become c₁ = c₋₁ = ½e^(−iΔt/ħ) while c₀ = 1/√2 stays, and S₊ raises m with matrix elements √2ħ, so ⟨S₊⟩ = √2ħ[(½e^(iΔt/ħ))(1/√2) + (1/√2)(½e^(−iΔt/ħ))] = ħ cos(Δt/ħ). It is real, so ⟨Sₓ⟩ = ħ cos(Δt/ħ) and ⟨S_y⟩ = 0, with ⟨S_z⟩ = 0 throughout. The spin does not precess: ⟨S⟩ shrinks to zero at t = h/4Δ = 0.0871 ns, points along −x at h/2Δ = 0.1742 ns and is back after h/Δ = 0.3484 ns.
Checks: ⟨S_z⟩ = ħ(¼ − ¼) = 0 at t = 0 and stays 0, as its conservation requires; as Δ → 0 every turn leaves H alone and ⟨Sₓ⟩ = ħ stays put; Δt/ħ is dimensionless. A spin-½ could not do this, since each of its pure states has |⟨S⟩| = ħ/2, while a spin-1 can carry ⟨S⟩ = 0.

AnswerS_z and S² are conserved, Sₓ and S_y are not; the triplet splits into m = 0 and an m = ±1 pair 2.870 GHz above it, held together by the half-turn about x; a spin prepared along x has ⟨Sₓ⟩ = ħ cos(Δt/ħ) and ⟨S_y⟩ = ⟨S_z⟩ = 0, so ⟨Sₓ⟩ reaches −ħ after 0.174 ns instead of precessing.