University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.7

The ladder route: solving L₊Yₗˡ = 0 on the sphere

The series route wins each spherical harmonic by solving a second-order equation and discarding a singular solution at each pole. The ladder route needs one first-order equation at the top, then only derivatives: the quickest way to build any Yₗᵐ, the strictest bookkeeper of signs, and the clearest test of why orbital l cannot be ½.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Quantum Mechanics I built the ladder abstractly: L₊ annihilates the top state |l, l⟩, and L₋ steps down with matrix element ħ√((l + m)(l − m + 1)). Written on the sphere, L± = ħe^(±iφ)(±∂/∂θ + i cot θ ∂/∂φ) is a first-order differential operator, so L₊|l, l⟩ = 0 becomes a first-order equation for the polar factor. Up to a constant it has one solution, sinˡθ, and no singular partner to discard: the regularity analysis the series route needed at both poles collapses to one line.

Lowering is then pure differentiation. With x = cos θ, each L₋ replaces F(x) by dF/dx inside e^(imφ) sin⁻ᵐθ F(x), so the whole multiplet is e^(imφ) sin⁻ᵐθ dˡ⁻ᵐ/dxˡ⁻ᵐ(1 − x²)ˡ, and its m = 0 rung is Rodrigues' formula for Pₗ. Positive matrix elements fix every sign relative to the top; the top's own sign is a free choice, and Condon and Shortley's (−1)ˡ is the one that makes Yₗ⁰ positive at the north pole.

Nothing in these equations asks for integer l until the bottom rung, where the algebra demands that L₋ return zero, and that happens only if (1 − x²)ˡ is a polynomial. For l = ½ the top is double-valued in φ and the step past the bottom returns a function that diverges at both poles. The construction is exact, with no expansion in α or anything else, and it builds angular functions only: energies still need the radial equation.

Simple definition: The ladder route builds each spherical harmonic by solving L₊Yₗˡ = 0, a first-order equation on the sphere whose solution is sinˡθ e^(ilφ), then applying L₋ repeatedly and dividing each time by ħ√((l + m)(l − m + 1)).
Example: For l = 1, L₊Y = 0 gives Y₁¹ = −√(3/8π) sin θ e^(iφ). One lowering gives L₋Y₁¹ = 2ħ√(3/8π) cos θ, and dividing by √2 ħ leaves Y₁⁰ = √(3/4π) cos θ, which is +0.489 at the north pole, as Condon–Shortley requires.

Ladder operators as derivatives on the sphereL± = ħe^(±iφ)(±∂/∂θ + i cot θ ∂/∂φ)

Derived here from Quantum Mechanics I's Lₓ and L_y. On g(θ)e^(imφ) it gives ħe^{(imφ ± iφ)}(±g′ − m cot θ g): a step of ±ħ in L_z and one θ-derivative.

ħ = 1.055 × 10⁻³⁴ J s carries every unit; θ and φ are dimensionless angles in rad; cot θ diverges on the z-axis

Top of the ladderL₊Yₗˡ = 0 gives Yₗˡ = Cₗ sinˡθ e^(ilφ)

Derived: g′ = l cot θ g separates to g ∝ sinˡθ. A first-order equation has one solution up to a constant, so unlike the series route there is no singular partner to discard.

Cₗ carries sr⁻¹⁄², usually treated as a pure number; the equation holds for any real l ≥ 0

Modulus of the top-rung constant|Cₗ| = √((2l + 1)!/4π) / (2ˡ l!)

Derived. l = 1, 2, 3 give 0.3455, 0.3863 and 0.4172, the moduli of √(3/8π), √(15/32π) and √(35/64π). The integral fixes the size and leaves the phase free.

from ∫₀^π sin²ˡ⁺¹θ dθ = 2²ˡ⁺¹(l!)²/(2l + 1)!, which follows from Iₙ = [(n − 1)/n]Iₙ₋₂ and I₁ = 2

Condon–Shortley sign of the top rungCₗ = (−1)ˡ |Cₗ|

It is the previous topic's (−1)ᵐ at m = l, and the choice that makes Yₗ⁰ = +√((2l + 1)/4π) at the north pole: 0.4886 for l = 1, 0.6308 for l = 2.

a convention, not a derivation: the ladder fixes every sign relative to the top, never the multiplet's overall phase

Lowering as differentiation in x = cos θL₋[e^(imφ) sin⁻ᵐθ F(x)] = ħe^{(imφ − iφ)} sin¹⁻ᵐθ dF/dx

Derived from g′ + m cot θ g = sin⁻ᵐθ d(sinᵐθ g)/dθ. Dividing each step by the quoted ħ√((l + m)(l − m + 1)), real and positive, keeps every rung normalised.

x = cos θ is dimensionless; F is any differentiable function; the identity holds for every real m

Every rung from the topYₗᵐ = [(−1)ˡ/(2ˡ l!)] √[(2l + 1)(l + m)!/(4π(l − m)!)] e^(imφ) sin⁻ᵐθ dˡ⁻ᵐ/dxˡ⁻ᵐ(1 − x²)ˡ

Derived. At m = 0 Rodrigues' formula turns it into √((2l + 1)/4π) Pₗ(cos θ); at l = 2, m = 1 it gives −√(15/8π) sin θ cos θ e^(iφ), the series route's answer.

one formula for all 2l + 1 rungs, m = l down to −l; the ladder closes because the (2l + 1)th derivative of (1 − x²)ˡ is zero

Ladder operators become derivatives on the sphere

Quote Quantum Mechanics I's components, Lₓ = iħ(sin φ ∂/∂θ + cot θ cos φ ∂/∂φ) and L_y = iħ(−cos φ ∂/∂θ + cot θ sin φ ∂/∂φ), and combine them. In L± = Lₓ ± iL_y the ∂/∂θ coefficient is iħ(sin φ ∓ i cos φ) = ±ħe^(±iφ), and the ∂/∂φ coefficient is iħ cot θ(cos φ ± i sin φ) = iħ cot θ e^(±iφ). So L± = ħe^(±iφ)(±∂/∂θ + i cot θ ∂/∂φ), derived here. On g(θ)e^(imφ) it gives ħe^{(imφ ± iφ)}(±g′ − m cot θ g): the factor e^(±iφ) is the step of ±ħ in L_z, and what remains is a first-order operator in θ alone. The dimensions check at once, because θ and φ are pure numbers and ħ = 1.055 × 10⁻³⁴ J s carries every unit of an angular momentum. A first test: on sin θ e^(iφ), the angular part of (x + iy)/r, L₊ gives ħe^(2iφ)(cos θ − cot θ sin θ) = 0. That function is already the top of the l = 1 ladder.

The top rung is one first-order equation

The top state has L_z = lħ, so its azimuthal factor is e^(ilφ), as Quantum Mechanics I showed. Then L₊Yₗˡ = 0 reads g′ = l cot θ g, which separates as d(ln g) = l d(ln sin θ), so g = Cₗ sinˡθ. That is the whole derivation. The series route faced a second-order equation with two solutions at each pole and had to discard one; a first-order equation has a single solution up to a constant factor, so there is nothing to discard. Check it against L² without taking a second derivative: the Quantum Mechanics I identity L² = L₋L₊ + L_z² + ħL_z gives 0 + l²ħ² + lħ² = l(l + 1)ħ² on the top rung. Limiting case l = 0: sin⁰θ = 1, the constant Y₀⁰, a ladder of one rung. The shape is a ring about the z-axis: |Yₗˡ|² ∝ sin²ˡθ vanishes at both poles and tightens as l grows, falling to half its equatorial value at 45° and 135° for l = 1 but at 57.2° and 122.8° for l = 2. Notice what the equation did not demand: an integer l. At the top only the single-valuedness of e^(ilφ) asks for that; the bottom of the ladder asks again, as the last section shows.

Normalise the top, and meet the sign no ladder can fix

∫|Yₗˡ|² dΩ = 2π|Cₗ|² Iₙ, with Iₙ = ∫₀^π sinⁿθ dθ and n = 2l + 1. One integration by parts gives Iₙ = [(n − 1)/n]Iₙ₋₂, and I₁ = 2, so I₃ = 4/3, I₅ = 16/15, I₇ = 32/35 and in general I₂ₗ₊₁ = 2²ˡ⁺¹(l!)²/(2l + 1)!. Hence |Cₗ| = √((2l + 1)!/4π)/(2ˡ l!): 0.3455, 0.3863 and 0.4172 for l = 1, 2, 3, the moduli of the tabulated √(3/8π), √(15/32π) and √(35/64π). The integral fixes |Cₗ| and nothing more, and lowering cannot fix the rest: each step divides by a real, positive matrix element, so multiplying the top by any phase e^(iα) multiplies every rung below it by the same e^(iα). One phase per multiplet is therefore a convention. Condon and Shortley take Cₗ ∝ (−1)ˡ, the previous topic's (−1)ᵐ evaluated at m = l, so Y₁¹ = −√(3/8π) sin θ e^(iφ) and Y₂² = +√(15/32π) sin²θ e^(2iφ). What the choice buys shows up at m = 0.

Lowering is differentiation in cos θ

On g(θ)e^(imφ), L₋ gives −ħe^{(imφ − iφ)}(g′ + m cot θ g), and g′ + m cot θ g = sin⁻ᵐθ d(sinᵐθ g)/dθ. With x = cos θ and d/dθ = −sin θ d/dx this collapses to one rule: L₋ turns e^(imφ) sin⁻ᵐθ F(x) into ħe^{(imφ − iφ)} sin¹⁻ᵐθ dF/dx. The top rung already has that form with F = Cₗ(1 − x²)ˡ, since sinˡθ = sin⁻ˡθ(1 − x²)ˡ. So l − m lowerings give dˡ⁻ᵐ/dxˡ⁻ᵐ(1 − x²)ˡ, divided by the product of the quoted matrix elements, √((2l)!(l − m)!/(l + m)!). At m = 0, Rodrigues' formula from earlier in this unit gives dˡ/dxˡ(1 − x²)ˡ = (−1)ˡ 2ˡ l! Pₗ(x), and the constants collapse to Yₗ⁰ = √((2l + 1)/4π) Pₗ(cos θ). The ladder never mentioned Legendre polynomials and produced one. The signs close as well: (−1)ˡ from Condon–Shortley times (−1)ˡ from (1 − x²)ˡ = (−1)ˡ(x² − 1)ˡ is +1, so at the north pole, where Pₗ(1) = 1, Yₗ⁰ = +√((2l + 1)/4π): 0.4886 for l = 1 and 0.6308 for l = 2.

l = 2 by both routes, signs included

Start from Y₂² = √(15/32π) sin²θ e^(2iφ), so g = C₂ sin²θ and m = 2. Then g′ + 2 cot θ g = 4C₂ sin θ cos θ, so L₋Y₂² = −4ħC₂ sin θ cos θ e^(iφ), and dividing by ħ√(4 × 1) = 2ħ gives Y₂¹ = −√(15/8π) sin θ cos θ e^(iφ). Once more, with g = −2C₂ sin θ cos θ and m = 1: g′ + cot θ g = −2C₂(3cos²θ − 1), so L₋Y₂¹ = 2ħC₂(3cos²θ − 1), and dividing by √6 ħ gives Y₂⁰ = +√(5/16π)(3cos²θ − 1). The series route of the previous topic writes Y₂¹ = −√(5/24π) P₂¹(cos θ) e^(iφ) with P₂¹ = 3 sin θ cos θ, the same function, sign included. Two further steps give Y₂⁽⁻¹⁾ = +√(15/8π) sin θ cos θ e^(−iφ) and Y₂⁽⁻²⁾ = √(15/32π) sin²θ e^(−2iφ), so Yₗ^(−m) = (−1)ᵐ(Yₗᵐ)* comes out of the ladder instead of being imposed. Finally, L₋ on Y₂⁽⁻²⁾ needs g′ − 2 cot θ g = 2 sin θ cos θ − 2 sin θ cos θ = 0: the ladder closes by itself, exactly where the matrix element ħ√(0 × 5) vanishes.

Why the ladder refuses l = ½, and what spectra say

Nothing so far used integer l, so try l = ½. L₊Y = 0 gives Y = (1/π)e^(iφ/2)√sin θ, normalisable because 2π∫sin²θ dθ = π². But e^{(iφ/2 + iπ)} = −e^(iφ/2): the function changes sign once round the z-axis, so it is not a single-valued function of position. Lower it anyway, with matrix element ħ√(1 × 1) = ħ: the m = −½ rung is −(1/π)e^(−iφ/2) cos θ/√sin θ, which diverges at both poles yet has norm exactly 1. It is the would-be bottom rung, so the algebra demands that L₋ annihilate it. Instead L₋ returns −(ħ/π)e^(−3iφ/2) sin⁻³⁄²θ, neither zero nor normalisable. In derivative language the reason is one line: the ladder closes only if the (2l + 1)th derivative of (1 − x²)ˡ vanishes, which needs (1 − x²)ˡ to be a polynomial, so l = 0, 1, 2, …. Spectra agree. ¹²C¹⁶O has a ¹Σ⁺ ground state, with no electronic spin and no electronic orbital angular momentum about its axis, and spinless nuclei, so its J is the orbital angular momentum of the rotating nuclei; its rotational lines at 115.271, 230.538 and 345.796 GHz (quoted from spectroscopic tables) stand in the ratio 1 : 2.000 : 3.000 that E = BJ(J + 1) gives for integer J. Half-integer J would give 3 : 5 : 7.

Change one variable at a time

Make the relationship visible.

Interactive model

Angular momentum quantum number l2.0

Lowerings applied k2

At l = 2 step k from 0 to 3: sin²θ gains a node, becomes 3cos²θ − 1, positive at the north pole, and k = 3 mirrors k = 1. At l = 1, k = 3 steps past the bottom and gives flat zero; at l = 0.5 that step is k = 2, and the curve blows up at both poles as sin⁻³⁄²θ, too fast to normalise.

RUNG m = l − k0.0

L² ON EVERY RUNG6.00 ħ²

NEXT LOWERING FACTOR2.449 ħ

PRODUCT SO FAR / ħᵏ4.899

Live interpretationRUNG m = l − k: 0.0. L² ON EVERY RUNG: 6.00 ħ². NEXT LOWERING FACTOR: 2.449 ħ. PRODUCT SO FAR / ħᵏ: 4.899

Catch the common trap

Explain before calculating.

Start from the top rung Y₃³ = −√(35/64π) sin³θ e^(3iφ). Apply L₋ = −ħe^(−iφ)(∂/∂θ − i cot θ ∂/∂φ) once and divide by the Quantum Mechanics I matrix element for this step. Which function is Y₃²?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyUse L₊ = ħe^(iφ)(∂/∂θ + i cot θ ∂/∂φ) to find the polar factor of Y₄⁴, normalise it with the Condon–Shortley sign, and find the polar angles at which |Y₄⁴|² falls to half its value on the equator.

Y₄⁴ has L_z = 4ħ, so write it as g(θ)e^(4iφ). Then L₊Y₄⁴ = ħe^(5iφ)(g′ − 4 cot θ g), and setting this to zero gives g′/g = 4 cos θ/sin θ, so g = C₄ sin⁴θ.
Normalise: ∫|Y₄⁴|² dΩ = 2π|C₄|² I₉, and the reduction formula gives I₉ = ∫₀^π sin⁹θ dθ = (8/9)(6/7)(4/5)(2/3) × 2 = 256/315 = 0.8127.
So |C₄|² = 315/(512π) = 0.1958 and |C₄| = 0.4425. The Condon–Shortley sign is (−1)⁴ = +1, so C₄ = +0.4425, the tabulated (3/16)√(35/2π). Units: |C₄|² is per steradian, so ∫|Y|² dΩ is a pure number, as a probability must be.
Half maximum: sin⁸θ = ½ gives sin θ = 2⁻¹⁄⁸ = 0.9170, so θ = 66.5° and 113.5°, a band 47.0° wide. For Y₁¹, sin²θ = ½ gives 45° to 135°, 90° wide: the top-rung ring tightens round the equator as l grows.

AnswerY₄⁴ = +(3/16)√(35/2π) sin⁴θ e^(4iφ) ≈ 0.443 sin⁴θ e^(4iφ); |Y₄⁴|² stays above half its equatorial value between θ = 66.5° and 113.5°, a band 47.0° wide.

MediumLower Y₄⁴ = (3/16)√(35/2π) sin⁴θ e^(4iφ) twice, with L₋ = −ħe^(−iφ)(∂/∂θ − i cot θ ∂/∂φ), to reach Y₄². Give its constant and sign, locate its polar nodes, and confirm the constant against the product of matrix elements.

On g(θ)e^(imφ), L₋ gives −ħe^{(imφ − iφ)}(g′ + m cot θ g). For g = C₄ sin⁴θ and m = 4, g′ + 4 cot θ g = 8C₄ sin³θ cos θ, so L₋Y₄⁴ = −8ħC₄ sin³θ cos θ e^(3iφ).
Divide by ħ√((4 + 4)(4 − 4 + 1)) = √8 ħ: Y₄³ = −2√2 C₄ sin³θ cos θ e^(3iφ) = −1.2517 sin³θ cos θ e^(3iφ).
Lower again with g = D sin³θ cos θ, D = −2√2 C₄ and m = 3: g′ + 3 cot θ g = D(3 sin²θ cos²θ − sin⁴θ + 3 sin²θ cos²θ) = D sin²θ(7cos²θ − 1). So L₋Y₄³ = −ħD sin²θ(7cos²θ − 1)e^(2iφ), to be divided by ħ√((4 + 3)(4 − 3 + 1)) = √14 ħ.
Y₄² = (2/√7)C₄ sin²θ(7cos²θ − 1)e^(2iφ) = 0.3345 sin²θ(7cos²θ − 1)e^(2iφ), which is √(45/128π) and positive: the minus signs of the two L₋ steps cancel. Check the constant another way: √8 × √14 = √112, and √((2l)!(l − m)!/(l + m)!) = √(40320 × 2/720) = √112.
Nodes: 7cos²θ = 1 gives cos θ = ±0.3780, so θ = 67.8° and 112.2°, two nodes as l − |m| = 2 requires. Parity: sin²θ(7cos²θ − 1) and e^(2iφ) are both unchanged by θ → π − θ, φ → φ + π, so the state is even, (−1)⁴ = +1.

AnswerY₄² = +√(45/128π) sin²θ(7cos²θ − 1)e^(2iφ) ≈ 0.335 sin²θ(7cos²θ − 1)e^(2iφ), with polar nodes at θ = 67.8° and 112.2°.

HardNothing in L₊Y = 0 forces l to be an integer. Run the ladder for l = 3/2: find and normalise the top rung, lower it three times with the Quantum Mechanics I matrix elements, and find the first rung that is not square-integrable on the sphere.

L₊Y = 0 with Y = g e^(3iφ/2) gives g = C sin³⁄²θ. Normalising, 2π|C|²∫sin⁴θ dθ = 2π|C|²(3π/8) = 1, so |C| = 2/(√3 π) = 0.3676. It is normalisable, but e^{(3iφ/2 + 3iπ)} = −e^(3iφ/2): the top rung is already double-valued.
Write the top as e^(3iφ/2) sin⁻³⁄²θ F(x) with F = C(1 − x²)³⁄², so each lowering is one x-derivative. First step: F′ = −3Cx(1 − x²)¹⁄² and the matrix element is √3 ħ, giving the m = ½ rung −√3 C cos θ √sin θ e^(iφ/2), regular at both poles, with norm 3|C|² × 2π × π/8 = 1.
Second step: F″ = 3C(2x² − 1)(1 − x²)⁻¹⁄² and the matrix element is 2ħ, giving the m = −½ rung (√3/2)C cos 2θ e^(−iφ/2)/√sin θ. It diverges at both poles, yet its norm is (3/4)|C|² × 2π × π/2 = 1: singular but square-integrable.
Third step, to the would-be bottom m = −3/2: F‴ = 3Cx(3 − 2x²)(1 − x²)⁻³⁄² and the matrix element is √3 ħ, giving the m = −3/2 rung (C/2) cos θ(3 − 2cos²θ)e^(−3iφ/2)/sin³⁄²θ. Within a small angle ε of either pole, |Y|² sin θ ≈ |C|²/(4ε²), and ∫dε/ε² diverges: this rung is not a state.
Two cross-checks. On the m = −½ rung ψ the algebra predicts ‖L₋ψ‖² = ħ²[l(l + 1) − m(m − 1)] = ħ²(15/4 − 3/4) = 3ħ², finite, but the function produced has infinite norm, so L₋ has stopped being the adjoint of L₊. And solving L₋Y = 0 directly at m = −3/2 gives sin³⁄²θ e^(−3iφ/2), not the ladder's function. Generally a half-integer rung goes as sinᵐθ near a pole, integrable only if m > −1.

AnswerThe top and m = ½ rungs are regular and the m = −½ rung is singular but normalised; the bottom rung, ∝ cos θ(3 − 2cos²θ)/sin³⁄²θ, has a divergent norm, so no l = 3/2 multiplet of orbital functions exists.