University Quantum Mechanics II · Rotations and the Addition of Angular Momentum · 5.8
Vector operators and the dipole selection rules derived
Quantum Mechanics I asked you to remember Δl = ±1 and Δm = 0, ±1. Here you earn them: one commutator fixes m, a double commutator and parity fix l, and you learn to say exactly which matrix elements symmetry kills and which ones only a radial integral can settle.
Build the model
Connect the measurement to the mechanism.
Quantum Mechanics I quoted the electric-dipole selection rules; here they become theorems about one operator, r. The starting point is a definition: an operator is a vector when rotating the state turns its expectation value like an arrow, which for small rotations is the statement [Jᵢ, Vⱼ] = iħ εᵢⱼₖ Vₖ. Recombined into spherical components V₀ and V₊₁, V₋₁, each component becomes a ladder that shifts m by exactly q, so one commutator with Lz kills every ⟨n′l′m′|rq|nlm⟩ with m′ ≠ m + q.
The orbital rule needs two facts that belong to r in particular, not to vectors in general: r⋅L = 0, which turns a double commutator with L² into an algebraic condition allowing only l′ = l ± 1 or l = l′ = 0, and odd parity, which demands l + l′ odd and removes the 0 → 0 case. No potential and no radial function enters, so the rules hold in any central potential and, with J in place of L, for an electron with spin. Their reach ends in two places.
They decide which elements vanish, never how large the survivors are: ⟨1 0 0|z|2 1 0⟩ = 0.745 a₀ comes only from Unit 2's radial functions. And they are exact for the dipole operator alone: the full coupling to light adds multipoles weaker in rate by about (ka₀)², which is (3α/8)² ≈ 7.5 × 10⁻⁶ for hydrogen's Lyman-α, and two-photon processes, so a forbidden transition is slow rather than impossible.
- Simple definition
- A vector operator is a set of three operators Vₓ, Vy, Vz obeying [Jᵢ, Vⱼ] = iħ εᵢⱼₖ Vₖ, so that rotating a state turns ⟨V⟩ like an ordinary arrow; position r is one, and its matrix elements obey Δm = 0, ±1 and Δl = ±1.
- Example
- z = r cos θ commutes with Lz, while x + iy = r sin θ e(iφ) obeys [Lz, x + iy] = ħ(x + iy). So ⟨1 0 0|z|2 1 0⟩ is allowed, and Unit 2's radial functions make it 0.745 a₀ = 3.94 × 10⁻¹¹ m, but ⟨1 0 0|z|2 1 1⟩ = 0 exactly because z cannot change m, and ⟨1 0 0|z|2 0 0⟩ = 0 because Δl = 0.
Derived from U†VU = RV for a small rotation U = 1 − iδφ n̂⋅J/ħ. r, p, L, S and d = −er qualify; r² and r⋅p commute with every Jᵢ and are scalars.
J in J s, ħ = 1.055 × 10⁻³⁴ J s; V keeps its own unit (r in m, p in kg m s⁻¹); εᵢⱼₖ is the Levi-Civita symbol
Derived: [Jz, Vq] = qħVq, so JzVq|j m⟩ = (m + q)ħVq|j m⟩. Each component is a ladder that moves m by exactly q.
same unit as V; for position rq = r√(4π/3) Y₁q(θ, φ), the l = 1 harmonics with their Condon-Shortley signs
Derived by sandwiching [Lz, rq] = qħrq, which leaves (m′ − m − q)ħ times the element equal to zero. The three components together allow Δm = 0, ±1.
rq in m; q = 0 for z, +1 for −(x + iy)/√2, −1 for (x − iy)/√2; m and m′ are pure numbers
Derived from [Lᵢ, rⱼ] = iħεᵢⱼₖrₖ, L × L = iħL and r⋅L = 0. It fails for L and S, which commute with L² and so connect only equal l.
L² in J² s², so each side is in J⁴ s⁴ m; true for p as well, because p⋅L = 0 just as r⋅L = 0
Derived by sandwiching the double commutator. Its roots are l′ = l ± 1 and l = l′ = 0: for l = 2, l′ = 1 gives 16 = 16 and l′ = 3 gives 36 = 36.
l, l′ non-negative integers; the factored form of (λ′ − λ)² = 2(λ′ + λ), λ = l(l + 1), once ħ⁴ is divided out
Derived. It removes the 0 → 0 root the double commutator leaves, and alone it would pass Δl = ±3. Only together do the two give Δl = ±1.
from ΠrΠ† = −r and Π|nlm⟩ = (−1)l|nlm⟩ (Unit 1); holds for p, fails for the even operators L and S
A vector is defined by how it rotates
Three operators do not make a vector; the way they respond to a rotation does. Rotate the state with U = e(−iφ n̂⋅J/ħ), the operator built at the start of this unit, and require every expectation value to turn like an arrow: U†VU = RV, with R the ordinary 3 × 3 rotation matrix. For a small angle U ≈ 1 − iδφ n̂⋅J/ħ and RV ≈ V + δφ n̂ × V, so (i/ħ)[n̂⋅J, V] = n̂ × V, which in components is [Jᵢ, Vⱼ] = iħ εᵢⱼₖ Vₖ. Test it on position: with Lz = xpy − ypₓ, [Lz, x] = −y[pₓ, x] = iħy, exactly what εzxy = +1 demands. So r, p, L, S, J itself and the dipole operator d = −er are vector operators; r² and r⋅p commute with every Jᵢ and are scalars; and xy is neither, since [Lz, xy] = iħ(y² − x²) is not a component of any vector. The word is relative to a particular J: for an electron, r is a vector under L and under J = L + S, but it commutes with S and is a scalar under it.
Spherical components move m by a fixed step
Cartesian components mix whenever the system turns about z; the combinations V₀ = Vz and V₊₁ = −(Vₓ + iVy)/√2, V₋₁ = (Vₓ − iVy)/√2 do not. From [Jz, Vₓ] = iħVy and [Jz, Vy] = −iħVₓ, the combination Vₓ ± iVy obeys [Jz, Vₓ ± iVy] = ±ħ(Vₓ ± iVy), so every spherical component satisfies [Jz, Vq] = qħVq. Act with it on a state of definite m: JzVq|j m⟩ = (VqJz + qħVq)|j m⟩ = (m + q)ħVq|j m⟩. Whatever else Vq does, it shifts m by exactly q, as J₊ shifts it by one; indeed J₊₁ = −J₊/√2. For position the components are familiar functions: r₀ = r cos θ and r₊₁, r₋₁ = ∓r sin θ e(±iφ)/√2, which are r√(4π/3) Y₁q(θ, φ), the l = 1 harmonics of Unit 1 with their Condon-Shortley signs. That is the exact sense in which r carries one unit of orbital angular momentum, and it is why the minus sign on V₊₁ is not optional: drop it and r₊₁ no longer matches Y₁¹.
One commutator gives the Δm rule
Take matrix elements of [Lz, rq] = qħrq between hydrogen states. Lz acting to the right on |nlm⟩ gives mħ, and acting to the left on ⟨n′l′m′| it gives m′ħ, so (m′ − m − q)ħ⟨n′l′m′|rq|nlm⟩ = 0. For z, with q = 0, the element survives only if m′ = m; for x + iy only if m′ = m + 1; for x − iy only if m′ = m − 1. For the vector as a whole, ⟨n′l′m′|r|nlm⟩ vanishes unless Δm = 0 or ±1, and each component fixes its own Δm. The pruning is severe: of the nine elements ⟨1 0 0|rq|2 1 m⟩, only the three with q = −m survive. Light makes the rule visible through the coupling ε⋅r, quoted here from Unit 13: polarisation along z drives Δm = 0, and circular polarisation about z drives Δm = ±1. No potential and no radial function entered. The same steps with Jz in place of Lz give Δmⱼ = q in the coupled basis of the previous lesson, and because r commutes with S, the spin projection mₛ cannot change at all.
A double commutator gives the Δl condition
Lz cannot see l, so the orbital rule must come from L². Start from [Lᵢ, rⱼ] = iħεᵢⱼₖrₖ, which gives [L², r] = iħ(r × L − L × r). Apply L² a second time and simplify the triple cross products with L × L = iħL and r⋅L = L⋅r = 0; the algebra is worth doing once, and it ends at [L², [L², r]] = 2ħ²(rL² + L²r). Test it on the constant function f = 1: L²f = 0 and L²z = 2ħ²z, and both sides give 4ħ⁴z. Now sandwich the identity between ⟨n′l′m′| and |nlm⟩, where L² gives λħ² with λ = l(l + 1): ħ⁴[(λ′ − λ)² − 2(λ′ + λ)]⟨n′l′m′|r|nlm⟩ = 0. A non-zero element needs (λ′ − λ)² = 2(λ′ + λ), which factorises exactly as [(l + l′ + 1)² − 1][(l − l′)² − 1] = 0. For l = 2 the candidates give 16 against 16 at l′ = 1, 36 against 36 at l′ = 3, 0 against 24 at l′ = 2 and 576 against 72 at l′ = 5. The only non-negative roots are l′ = l ± 1, plus the single pair l = l′ = 0.
Parity closes the one loophole
The double commutator leaves l = l′ = 0 open, and parity shuts it. The parity operator sends r to −r, so ΠrΠ† = −r, and Unit 1 gave Π|nlm⟩ = (−1)l|nlm⟩. Inserting Π†Π on both sides of r gives ⟨n′l′m′|r|nlm⟩ = (−1)(l + l′ + 1)⟨n′l′m′|r|nlm⟩, so the element vanishes unless l + l′ is odd. That removes 0 → 0, and with it every s-to-s element of r, 2s → 1s included. Each argument leaks on its own: parity alone would pass Δl = ±3, as in 4f → 1s, and the double commutator alone would pass 3s → 1s. Together they leave Δl = ±1 and nothing else. The rules belong to r and p specifically. L and S are vector operators too, but they are even under parity and commute with L², so they connect only l′ = l. Nothing here assumed hydrogen: in any central potential, and for the valence electron of an alkali atom in the one-electron picture, the same rules hold, which is why sodium's 4s level decays to 3p, near 1.14 μm, and has no dipole route to 3s.
What the rules leave open
The rules sort matrix elements into zero and possibly non-zero; they never compute a survivor. ⟨1 0 0|z|2 1 0⟩ needs Unit 2's functions: the angular factor ⟨Y₀⁰|cos θ|Y₁⁰⟩ = 1/√3 times ∫₀^∞ R₁₀R₂₁r³ dr = 768a₀/(243√6) = 1.290 a₀ gives 128√2a₀/243 = 0.745 a₀, a number no commutator could produce. An allowed element can even vanish: ⟨2 0 0|pz|2 1 0⟩ = 0 although p obeys the same rules, because 2s and 2p share one energy and ⟨f|p|i⟩ = imωfi⟨f|r|i⟩, quoted from Unit 13. The rules are also exact for r alone, and light couples only approximately through ε⋅r: expanding e(ik⋅r) across the atom adds magnetic-dipole and quadrupole terms weaker in rate by about (ka₀)², and for Lyman-α ka₀ = 3α/8 = 2.74 × 10⁻³. So forbidden means slow. The 2s state lives 0.12 s by emitting two photons, a rate of 8.2 s⁻¹ quoted from quantum electrodynamics, while 2p lives 1.60 ns, as measured.
Change one variable at a time
Make the relationship visible.
Set l = 0: the commutator row is empty at l′ = 0 and at l′ = 1, and only the parity bar stops 0 → 0. Then take l = 2 and box l′ = 5: parity passes, but (λ′ − λ)² = 576 against 2(λ′ + λ) = 72.
LEFT SIDE (λ′ − λ)²36 ħ⁴
RIGHT SIDE 2(λ′ + λ)36 ħ⁴
PARITY FACTOR (−1)(l+l′+1)1
Δl = l′ − l1
Live interpretationLEFT SIDE (λ′ − λ)²: 36 ħ⁴. RIGHT SIDE 2(λ′ + λ): 36 ħ⁴. PARITY FACTOR (−1)(l+l′+1): 1. Δl = l′ − l: 1
Catch the common trap
Explain before calculating.
An electron in a central potential is in the state |n, l = 2, m = 0⟩. For which final states |n′, l′, m′⟩ can the matrix element ⟨n′ l′ m′|(x + iy)|n 2 0⟩ be non-zero?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen atom is in the 3d state |n l m⟩ = |3 2 1⟩. Of the 15 matrix elements ⟨n′l′m′|rq|3 2 1⟩ with n′ ≤ 2 (the states 1s, 2s and the three 2p) and q = −1, 0, +1, which can be non-zero, and what photon does the surviving decay emit?
- Δl first: l′ must be 2 ± 1, so 1 or 3. No l = 3 level lies below n = 3, and 1s and 2s have l′ = 0, a jump of two. Only the three 2p states remain, so 9 of the 15 elements are still in play.
- Δm next: m′ = m + q = 1 + q. For q = 0, the z component, m′ = 1, which 2p has; for q = −1, m′ = 0, which 2p has; for q = +1, m′ = 2, which no l′ = 1 state has.
- Survivors: ⟨2 1 1|z|3 2 1⟩ and ⟨2 1 0|r₋₁|3 2 1⟩, with r₋₁ = (x − iy)/√2. The other 13 elements vanish exactly, whatever the radial functions are.
- The photon: E = 13.61 eV × (1/4 − 1/9) = 1.890 eV, so λ = hc/E = 1239.8 eV nm ÷ 1.890 eV = 656 nm, the red Balmer-α line. Units: eV nm over eV leaves nm.
- Check against measurement: including the reduced mass moves it to 656.5 nm in vacuum, and Hα is measured at 656.3 nm in air. No electric-dipole 3d → 1s line exists; Lyman-β at 102.6 nm comes from 3p alone.
AnswerOnly 2 of the 15 survive: ⟨2 1 1|z|3 2 1⟩ and ⟨2 1 0|(x − iy)/√2|3 2 1⟩. The decay emits Balmer-α at λ ≈ 656 nm; 3d has no electric-dipole route to 1s or 2s.
MediumThe previous lesson built |2p₃/₂, mⱼ = ½⟩ = √(2/3)|mₗ = 0, ↑⟩ + √(1/3)|mₗ = 1, ↓⟩ and |2p₁/₂, mⱼ = ½⟩ = −√(1/3)|mₗ = 0, ↑⟩ + √(2/3)|mₗ = 1, ↓⟩. Given ⟨1 0 0|z|2 1 0⟩ = 0.745 a₀, find which components rq connect each state to 1s₁/₂ with mⱼ = ±½, and evaluate every z element.
- r acts on position only and commutes with S, so ⟨1s, mₛ′|rq|2p, mₗ, mₛ⟩ is zero unless mₛ′ = mₛ, and then equals ⟨1 0 0|rq|2 1 mₗ⟩, which the Δm rule under L allows only for q = −mₗ.
- Final 1s₁/₂, mⱼ = +½, which is |0, ↑⟩: only the |mₗ = 0, ↑⟩ piece matches its spin, and it needs q = 0. So z is the only component, and ⟨1s₁/₂, +½|z|2p₃/₂, +½⟩ = √(2/3) × 0.745 a₀ = 0.608 a₀.
- Final 1s₁/₂, mⱼ = −½, which is |0, ↓⟩: only the |mₗ = 1, ↓⟩ piece matches, and it needs q = −1, so only r₋₁ = (x − iy)/√2 connects; z and x + iy give zero. Both cases have Δmⱼ = q, as [Jz, rq] = qħrq requires, because r is a vector under J = L + S.
- For 2p₁/₂ the same pieces appear with the other weights: ⟨1s₁/₂, +½|z|2p₁/₂, +½⟩ = −√(1/3) × 0.745 a₀ = −0.430 a₀, and again only r₋₁ reaches mⱼ = −½.
- Checks: the squares 0.370 a₀² and 0.185 a₀² add to 0.555 a₀² = (0.745 a₀)², the whole z strength of |mₗ = 0, ↑⟩ shared between j = 3/2 and j = ½ because the two coupled states span the same pair as the uncoupled ones. In SI, e × 0.608 a₀ = 1.602 × 10⁻¹⁹ C × 3.22 × 10⁻¹¹ m = 5.16 × 10⁻³⁰ C m.
Answerz connects each state only to 1s₁/₂, mⱼ = +½: 0.608 a₀ from 2p₃/₂ and −0.430 a₀ from 2p₁/₂ (|dz| = 5.16 × 10⁻³⁰ and 3.65 × 10⁻³⁰ C m). Only r₋₁ reaches mⱼ = −½, and x + iy reaches neither.
HardTest the derived rules against a brute-force integral. For hydrogen's 3d state |3 2 0⟩ and the component z = r cos θ, say which elements ⟨n′l′m′|z|3 2 0⟩ can be non-zero. Then show directly from the angular functions that every element the rules forbid is zero, and evaluate ⟨2 1 0|z|3 2 0⟩ with Unit 2's R₂₁ = a₀(−3/2)(r/a₀)e(−r/2a₀)/(2√6) and R₃₂ = 4a₀(−3/2)(r/a₀)²e(−r/3a₀)/(81√30).
- Rules first: z is r₀, so q = 0 and m′ = 0; the double commutator and parity then allow only l′ = 1 or 3. The candidates are |n′ 1 0⟩ with n′ ≥ 2 and |n′ 3 0⟩ with n′ ≥ 4, so below n = 3 only |2 1 0⟩ remains.
- Brute force, angular part: the table identity cos θ Y₂⁰ = √(9/35) Y₃⁰ + √(4/15) Y₁⁰, quoted with Condon-Shortley phases, contains exactly the l′ and m′ the rules allow. Completeness shows nothing else hides in it: with Y₂⁰ = √(5/16π)(3cos²θ − 1) and x = cos θ, ∫|cos θ Y₂⁰|² dΩ = (5/8)∫₋₁¹ x²(3x² − 1)² dx = (5/8)(88/105) = 11/21, while 9/35 + 4/15 = 27/105 + 28/105 = 55/105 = 11/21.
- The two coefficients use up the whole norm, so the angular element ⟨Y|cos θ|Y₂⁰⟩ is exactly zero for every other harmonic Y, whatever its l′ or m′: the commutator and parity result is confirmed by an integral that knows nothing about commutators.
- Radial part: with u = r/a₀, R₂₁R₃₂r³ dr = a₀u⁶e(−5u/6) du/(243√5), so ∫₀^∞ R₂₁R₃₂r³ dr = a₀ × 6! × (6/5)⁷/(243√5) = 2579.9 a₀/543.4 = 4.748 a₀. Then ⟨2 1 0|z|3 2 0⟩ = √(4/15) × 4.748 a₀ = 0.5164 × 4.748 a₀ = 2.452 a₀ = 1.297 × 10⁻¹⁰ m; the angular factor is a pure number and the radial integral carries a₀, so the element is a length.
- Check against measurement: this element drives a component of Balmer-α, measured at 656.5 nm in vacuum. With k = 2π/λ, k × 2.452 a₀ = 2π × 1.297 × 10⁻¹⁰ m ÷ 6.565 × 10⁻⁷ m = 1.24 × 10⁻³ ≪ 1, so the atom is small against the wavelength, the dipole operator r is the right one, and the quadrupole correction is of order (1.24 × 10⁻³)² ≈ 1.5 × 10⁻⁶ in rate.
AnswerOnly m′ = 0 with l′ = 1 or 3 can survive, which below n = 3 leaves |2 1 0⟩ alone. cos θ Y₂⁰ = √(9/35) Y₃⁰ + √(4/15) Y₁⁰ uses up the full norm 11/21, so every forbidden element is exactly zero, and ⟨2 1 0|z|3 2 0⟩ = √(4/15) × 4.748 a₀ = 2.452 a₀ ≈ 1.30 × 10⁻¹⁰ m.