University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.8
Parity, completeness and the recursion for cos θ Yₗᵐ
This is where the Yₗᵐ stop being shapes to plot and become a basis to calculate in. Two short results, that each Yₗᵐ has parity (−1)ˡ and that cos θ moves l up or down by exactly one, decide in advance which dipole matrix elements vanish and turn a whole class of angular integrals into reading off two numbers.
Build the model
Connect the measurement to the mechanism.
Two facts about the Yₗᵐ carry much of atomic physics, and both follow from functions already built in this unit. Inversion r → −r sends θ to π − θ and φ to φ + π; the polar factor Pₗ(|m|) returns (−1)(l + |m|), the azimuthal factor e(imφ) returns (−1)ᵐ, and the product is (−1)ˡ for every m. Multiplying by cos θ = z/r, the simplest odd function on the sphere, must therefore flip the parity, but parity alone would still let l jump by 3 or 5.
The associated Legendre recursion, obtained by differentiating the m = 0 recursion |m| times, is sharper: cos θ Yₗᵐ is exactly A Yₗ₊₁ᵐ + B Yₗ₋₁ᵐ, and orthonormality makes every other coefficient zero. That is the angular content of the dipole rule Δl = ±1, Δm = 0 for light polarised along z, met as algebra before Unit 5 derives it from commutators. Completeness, quoted from Sturm–Liouville theory, licenses the reverse move: any angular function can be expanded on the Yₗᵐ, each coefficient found by one projection.
As mathematics none of this fails; the physics built on it can. An energy eigenstate has definite parity whenever the Hamiltonian is inversion-symmetric and the level is not degenerate with one of opposite parity. A weak electric field breaks the symmetry, and hydrogen's 2s and 2p₁/₂, only 1058 MHz apart, mix strongly in it; the weak interaction breaks parity outright, at about 10⁻¹¹ in atomic states, measured in caesium.
- Simple definition
- The parity of Yₗᵐ is the sign (−1)ˡ it acquires when the direction r̂ is replaced by −r̂, and the cos θ recursion is the exact statement that cos θ Yₗᵐ contains Yₗ₊₁ᵐ and Yₗ₋₁ᵐ and nothing else.
- Example
- cos θ Y₁⁰ = (2/√15) Y₂⁰ + (1/√3) Y₀⁰, coefficients 0.516 and 0.577. Both survivors are even while Y₁⁰ is odd, no Y₃⁰ appears, and 4/15 + 1/3 = 3/5 is ⟨cos²θ⟩ in Y₁⁰.
Parity depends on l alone, so all 2l + 1 states with the same l share it, and an odd operator such as z has zero matrix element between any two of them.
θ, φ in rad; Yₗᵐ in sr(−1/2). Derived: Pₗ(|m|) gives (−1)(l+|m|) and e(imφ) gives (−1)ᵐ
It trades the stray (|m| − 1)-th derivative left behind by Leibniz's rule for |m|-th derivatives of the two neighbours.
x = cos θ, dimensionless; ′ means d/dx. Derived here from (1 − 2xt + t²) ∂g/∂x = t g
At |m| = l the last term is Pₗ₋₁ˡ = 0, so the recursion can never produce a function with |m| greater than l.
x = cos θ; every factor dimensionless; Pₗ(|m|) = (1 − x²)(|m|/2) d(|m|)Pₗ/dx(|m|), with no Condon–Shortley sign
So ⟨l′m′|cos θ|lm⟩ vanishes unless m′ = m and l′ = l ± 1: the angular selection rule for z-polarised dipole transitions, found before any integral is done.
Aₗₘ = √[((l+1)² − m²)/((2l+1)(2l+3))], Bₗₘ = √[(l² − m²)/((2l−1)(2l+1))], pure numbers
It is the squared norm of cos θ Yₗᵐ, so the recursion plus orthonormality replaces an angular integral, and its limits match classical orbits.
Dimensionless. Equals 1/3 at l = 0 and 1/(2l + 3) at |m| = l, and tends to 1/2 for m = 0 as l grows
Parseval, ∫|f|² dΩ = Σ|cₗₘ|², audits any expansion: cos²θ = 1.1816 Y₀⁰ + 1.0569 Y₂⁰ gives 4π/5 = 2.513 on both sides.
Quoted from Sturm–Liouville theory. dΩ = sin θ dθ dφ in sr; cₗₘ has the unit of f times sr¹⁄²
Inversion moves both angles
Replacing r by −r keeps r = |r| and sends the direction to its antipode. In spherical coordinates that is θ → π − θ and φ → φ + π, not θ → −θ: the polar angle must stay between 0 and π, so the azimuth swings half a turn instead. Put x = cos θ. Then x → −x while sin θ = √(1 − x²) is unchanged, so the factor (1 − x²)(|m|/2) is even. Pₗ contains only the powers xˡ, xˡ⁻², …, so Pₗ(−x) = (−1)ˡ Pₗ(x), and each derivative removes one power and one sign: d(|m|)Pₗ/dx(|m|) has parity (−1)(l − |m|) = (−1)(l + |m|). The azimuthal factor gives e(im(φ + π)) = (−1)ᵐ e(imφ). The product is (−1)(l + |m| + m) = (−1)ˡ, because |m| + m is either 0 or 2|m|. Nothing here is quoted; every sign comes from the functions built in topics 1.2 and 1.5. Check it on the two simplest cases: Y₁⁰ ∝ z/r and Y₁¹ ∝ −(x + iy)/r are both visibly odd.
Differentiating the Legendre recursion leaves one stray term
The associated functions inherit their recursion from (2l + 1) x Pₗ = (l + 1) Pₗ₊₁ + l Pₗ₋₁, derived from the generating function in topic 1.4. Differentiate it k = |m| times. Leibniz's rule on x Pₗ gives x Pₗ⁽ᵏ⁾ + k Pₗ⁽ᵏ⁻¹⁾, so (2l + 1)[x Pₗ⁽ᵏ⁾ + k Pₗ⁽ᵏ⁻¹⁾] = (l + 1) Pₗ₊₁⁽ᵏ⁾ + l Pₗ₋₁⁽ᵏ⁾. Every term but one is a k-th derivative, ready to become an associated function once multiplied by (1 − x²)(k/2); the stray k Pₗ⁽ᵏ⁻¹⁾ has one derivative too few. The generating function removes it. From (1 − 2xt + t²) ∂g/∂x = t g, the power tˡ⁺¹ gives P′ₗ₊₁ − 2x P′ₗ + P′ₗ₋₁ = Pₗ, and the x-derivative of the recursion gives (2l + 1)(Pₗ + x P′ₗ) = (l + 1) P′ₗ₊₁ + l P′ₗ₋₁. Eliminating x P′ₗ leaves P′ₗ₊₁ − P′ₗ₋₁ = (2l + 1) Pₗ. At l = 2: P′₃ − P′₁ = (15x² − 3)/2 − 1 = 5(3x² − 1)/2 = 5P₂.
Assembling the recursion for Pₗ(|m|)
Differentiate the identity k − 1 times: (2l + 1) Pₗ⁽ᵏ⁻¹⁾ = Pₗ₊₁⁽ᵏ⁾ − Pₗ₋₁⁽ᵏ⁾. Multiply by k and substitute for the stray term: (2l + 1) x Pₗ⁽ᵏ⁾ = (l + 1 − k) Pₗ₊₁⁽ᵏ⁾ + (l + k) Pₗ₋₁⁽ᵏ⁾. Multiplying through by (1 − x²)(k/2) turns every derivative into an associated function: (2l + 1) x Pₗ(|m|) = (l − |m| + 1) Pₗ₊₁(|m|) + (l + |m|) Pₗ₋₁(|m|). Test it at l = 2, |m| = 1, where P₂¹ = 3x sin θ, P₃¹ = (3/2)(5x² − 1) sin θ and P₁¹ = sin θ. The left side is 5x × 3x sin θ = 15x² sin θ; the right is 2 × (3/2)(5x² − 1) sin θ + 3 sin θ = 15x² sin θ. Two limits check as well. At m = 0 the formula is the Legendre recursion again. At |m| = l the last term holds Pₗ₋₁ˡ, a polynomial of degree l − 1 differentiated l times, which is zero: the recursion itself refuses to produce a function with |m| above l.
Normalised, cos θ moves l by one step and no further
Multiply by e(imφ) and by the norms of topic 1.6, Nₗₘ² = [(2l + 1)/4π](l − |m|)!/(l + |m|)!. The ratios of neighbouring norms turn the integer coefficients into cos θ Yₗᵐ = Aₗₘ Yₗ₊₁ᵐ + Bₗₘ Yₗ₋₁ᵐ, and the Condon–Shortley sign, the same for every l at fixed m, cancels, so A and B are positive. Orthonormality then gives ⟨l′m′|cos θ|lm⟩ = A when l′ = l + 1, B when l′ = l − 1, and zero otherwise, always with m′ = m because cos θ holds no φ. Parity alone only demands that l′ − l be odd and would allow l → l + 3; the recursion forbids it. The checks: Y carries sr(−1/2) on both sides, so A and B are pure numbers. At l = 0, A = 1/√3 and ⟨cos²θ⟩ = A² + B² = 1/3, the isotropic average. At |m| = l, B = 0 and A² = 1/(2l + 3), falling towards zero like a classical orbit in the equatorial plane. At m = 0 and large l, A² + B² → 1/2, the average of cos² round an orbit whose plane contains z.
Completeness, and cos²θ by projection
Completeness is quoted, not derived. For each m the Pₗ(|m|) with l ≥ |m| are the eigenfunctions of a singular Sturm–Liouville problem on −1 ≤ x ≤ 1, which that theory guarantees complete, and the e(imφ) are complete over one turn of φ by Fourier's theorem. So any square-integrable f(θ, φ) equals Σ cₗₘ Yₗᵐ, converging in the mean, with cₗₘ = ∫ Yₗᵐ* f dΩ by orthonormality. Take f = cos²θ. It has no φ, so only m = 0 survives; it is even under inversion, so only even l; it is a quadratic in cos θ, orthogonal to every Pₗ with l above 2. That leaves Y₀⁰ = 1/√(4π) and Y₂⁰ = √(5/16π)(3cos²θ − 1). Projecting gives c₀₀ = (1/√(4π))(4π/3) = 2√π/3 = 1.1816 and c₂₀ = (4/3)√(π/5) = 1.0569, in sr¹⁄². Parseval checks both at once: 1.1816² + 1.0569² = 2.513 = 4π/5 = ∫cos⁴θ dΩ. In Legendre form, cos²θ = 1/3 + (2/3)P₂(cos θ).
What the rule forbids, measured in the laboratory
In hydrogen cos θ Y₀⁰ = Y₁⁰/√3 has no Y₀⁰ in it, so ⟨1s|z|2s⟩ = 0 exactly, and x and y vanish by parity. The 2s level has no one-photon electric-dipole route to 1s; 2p has one. The recursion is geometry with no α in it, but the rates it switches on and off carry powers of α, quoted from Unit 13 and quantum electrodynamics: 2p → 1s runs at (2/3)⁸ α⁵ mc²/ħ = 6.27 × 10⁸ s⁻¹, a lifetime of 1.60 ns that matches the measured 1.60 ns, while 2s must emit two photons, an α⁸ process at 8.2 s⁻¹, a lifetime of 0.12 s. Lamb and Retherford built their 1947 experiment on exactly this, a beam of metastable 2s atoms. The same algebra aims photoelectrons. Light polarised along z couples through z (quoted from Unit 13), so an ejected s electron leaves as cos θ Y₀⁰ ∝ Y₁⁰ and its distribution goes as cos²θ = (1/3)(1 + 2P₂): an asymmetry parameter β = 2, and helium 1s photoelectrons are measured close to 2. The rule governs one-photon electric-dipole processes only; magnetic-dipole, quadrupole and two-photon channels, each down by further powers of α, carry what it forbids.
Change one variable at a time
Make the relationship visible.
At l = 3, |m| = 1 the polar part alone is even, so the left half mirrors itself about +x, yet the right half is its negative: e(imφ) adds (−1)ᵐ. Step l − |m| and the chord flips between level and crossing the axis; the bars stand only at l ± 1, and at l − |m| = 0 the left one vanishes.
PARITY (−1)ˡ-1
POLAR PART ALONE1
ANTIPODE θ, AT φ = 180°140 °
⟨cos²θ⟩ = A² + B²0.467
Live interpretationPARITY (−1)ˡ: −1. POLAR PART ALONE: 1. ANTIPODE θ, AT φ = 180°: 140 °. ⟨cos²θ⟩ = A² + B²: 0.467
Catch the common trap
Explain before calculating.
Using only results derived in this lesson, which spherical harmonics does the product cos θ Y₃¹ contain?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyY₂¹ = −√(15/8π) sin θ cos θ e(iφ). Evaluate Re Y₂¹ at θ = 30°, φ = 20° and at the antipodal direction, and say what sign the polar factor alone would have predicted.
- The antipode of (30°, 20°) is θ = 180° − 30° = 150°, φ = 20° + 180° = 200°.
- Polar factor: sin θ cos θ = 0.5 × 0.8660 = 0.4330 at 30°, and 0.5 × (−0.8660) = −0.4330 at 150°. It flips, as (−1)(l + |m|) = (−1)³ predicts.
- Azimuthal factor: Re e(iφ) = cos φ is 0.9397 at 20° and −0.9397 at 200°. It flips too, as (−1)ᵐ = −1 predicts.
- With √(15/8π) = 0.7725: Re Y₂¹(30°, 20°) = −0.7725 × 0.4330 × 0.9397 = −0.3143, and at the antipode −0.7725 × (−0.4330) × (−0.9397) = −0.3143. The two flips cancel.
- Im Y₂¹ uses sin φ, which is 0.3420 at 20° and −0.3420 at 200°, so it is also equal at both points: −0.7725 × 0.4330 × 0.3420 = −0.1144.
AnswerRe Y₂¹ = −0.314 sr(−1/2) at both points, and Im Y₂¹ = −0.114 sr(−1/2) at both: parity (−1)² = +1. The polar factor alone would have called Y₂¹ odd.
MediumExpand cos³θ on the spherical harmonics without evaluating a single angular integral: decide which Yₗᵐ can appear, find their coefficients from the cos θ recursion, and check the result against Parseval's theorem.
- Selection first. cos³θ has no φ, so only m = 0; it is odd under inversion, so only odd l; it is a cubic in cos θ, orthogonal to every Pₗ with l above 3. That leaves Y₁⁰ and Y₃⁰.
- Start from the projected result cos²θ = c₀₀ Y₀⁰ + c₂₀ Y₂⁰, with c₀₀ = 2√π/3 = 1.18164 and c₂₀ = (4/3)√(π/5) = 1.05689, and multiply by cos θ.
- Recursion coefficients: cos θ Y₀⁰ = A₀₀ Y₁⁰ with A₀₀ = 1/√3 = 0.57735; cos θ Y₂⁰ = A₂₀ Y₃⁰ + B₂₀ Y₁⁰ with A₂₀ = 3/√35 = 0.50709 and B₂₀ = 2/√15 = 0.51640.
- Collect: c₁₀ = 1.18164 × 0.57735 + 1.05689 × 0.51640 = 0.68222 + 0.54578 = 1.22800, and c₃₀ = 1.05689 × 0.50709 = 0.53594.
- Parseval: 1.22800² + 0.53594² = 1.50798 + 0.28723 = 1.79521, while ∫cos⁶θ dΩ = 2π × 2/7 = 4π/7 = 1.79520. In Legendre form the result is cos³θ = (3/5)P₁ + (2/5)P₃, which gives 1 at θ = 0, as it must.
Answercos³θ = 1.228 Y₁⁰ + 0.536 Y₃⁰, coefficients in sr¹⁄², exactly (3/5)√(4π/3) and (2/5)√(4π/7); both sides of Parseval give 4π/7 = 1.795 sr.
HardA linear molecule's axis has rotor states Yₗᵐ with energies ε l(l + 1), ε = ħ²/(2I). In rotational Raman scattering with z-polarised light in and out, the part of the amplitude that can change l is proportional to ⟨l′m′|cos²θ|lm⟩ (quoted from molecular physics). Apply the recursion twice to find every l′ reachable from l and the amplitudes from l = 0, then predict the Stokes shifts and test them on N₂ lines measured at 11.9, 19.9, 27.9 and 35.8 cm⁻¹.
- Selection first. cos²θ holds no φ, so m′ = m; it is even under inversion, so l′ − l is even. Parity alone would therefore allow l′ = l, l ± 2, l ± 4 and so on.
- Apply the recursion twice, with Aₗ and Bₗ short for Aₗₘ and Bₗₘ at fixed m: cos²θ Yₗᵐ = AₗAₗ₊₁ Yₗ₊₂ᵐ + (AₗBₗ₊₁ + BₗAₗ₋₁) Yₗᵐ + BₗBₗ₋₁ Yₗ₋₂ᵐ. Nothing else appears, so Δl = 0 or ±2, never ±4. Since Bₗ₊₁ = Aₗ and Aₗ₋₁ = Bₗ, the middle coefficient is Aₗ² + Bₗ² = ⟨cos²θ⟩, as it must be.
- From l = 0, with A₀ = 1/√3, A₁ = 2/√15 and B₀ = 0: cos²θ Y₀⁰ = (2/(3√5)) Y₂⁰ + (1/3) Y₀⁰ = 0.2981 Y₂⁰ + 0.3333 Y₀⁰. There is no Y₄⁰, though parity would admit one, and no l − 2 term. This is cos²θ = 1/3 + (2/3)P₂ times Y₀⁰, and Parseval checks it: 0.0889 + 0.1111 = 0.2000 = 1/5, the mean of cos⁴θ over the sphere.
- Shifts. The Stokes lines l → l + 2 sit at ε[(l + 2)(l + 3) − l(l + 1)] = (4l + 6)ε: 6ε, 10ε, 14ε, 18ε, in the ratios 3 : 5 : 7 : 9 whatever ε is. An l → l + 4 line would sit at (8l + 20)ε = 20ε, 28ε, …, a multiple of 4ε and so midway between two allowed lines. Δl = 0 leaves the rotor alone: the unshifted Rayleigh line.
- Test. The measured ratios are 3 × 19.9/11.9 = 5.0, 3 × 27.9/11.9 = 7.0 and 3 × 35.8/11.9 = 9.0. The spacing (35.8 − 11.9)/3 = 7.97 cm⁻¹ is 4ε/hc, so ε/hc = 1.99 cm⁻¹ and 6ε/hc = 11.94 cm⁻¹ places the first line. Nothing appears at 20ε/hc = 39.8 cm⁻¹ or 28ε/hc = 55.7 cm⁻¹.
- Checks. Each shift is ε/hc, a wavenumber, times a pure number. No α enters: the coefficients are geometry, and ε is set by the nuclear masses and the bond length. Classical limit: (4l + 6)ε/ħ is twice (2l + 3)ε/ħ, the rotation rate L/I at the midway L = ħ(l + 3/2), because cos²θ runs through two cycles per turn.
AnswerOnly l′ = l and l ± 2: cos²θ Y₀⁰ = 0.298 Y₂⁰ + 0.333 Y₀⁰, and the Stokes lines fall at (4l + 6)ε = 6ε, 10ε, 14ε, 18ε, matching N₂ with ε/hc = 1.99 cm⁻¹. The l → l + 4 lines that parity alone would allow, at 39.8 and 55.7 cm⁻¹, are absent.