University Physics V · Angular Momentum in Quantum Mechanics · 9.7

Spherical Harmonics as a Basis on the Sphere

Every central-force problem ends with an angular function to expand, and this is the basis you expand it in. Learn how each Y_ℓ^m is assembled, why the set is orthogonal before you choose anything and orthonormal after one normalisation, how to read ℓ and m off a plot by counting nodes, and exactly what a truncated expansion does and does not promise.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Angular momentum in this unit has so far been mostly an algebra: J₊, J₋ and the commutators forced J² = ħ²j(j+1) and J_z = mħ, and 9.5 and 9.6 supplied the differential operators and the φ factor e^(imφ) without yet writing down a full eigenfunction. Spherical harmonics are what those eigenvectors become when the Hilbert space is L²(S²), square-integrable functions on the unit sphere with inner product ⟨f|g⟩ = ∫ f*g dΩ. Separate L̂² = −ħ²[(1/sin θ)∂_θ(sin θ ∂_θ) + (1/sin²θ)∂_φ²] together with L̂_z = −iħ∂_φ and you get Y_ℓ^m = N_ℓm P_ℓ^m(cos θ) e^(imφ): an azimuthal phase from L̂_z and an associated Legendre function from L̂², with regularity at the two poles as the boundary condition that quantises ℓ to the integers ℓ ≥ |m|.

Orthogonality then costs nothing — both operators are Hermitian under the sin θ dθ dφ measure, so eigenfunctions with different (ℓ, m) are orthogonal automatically, and only the normalisation N_ℓm is a choice. Completeness is the theorem that makes the set a basis: any f in L²(S²) equals Σ c_ℓm Y_ℓ^m with c_ℓm = ⟨ℓ m|f⟩, and for a normalised angular state |c_ℓm|² is the Born probability of finding ℓ and m. Two costs come with it.

The basis is shared by every central potential, so it fixes no energy; E lives in the radial equation. And completeness is convergence in the norm, ‖f − f_L‖ → 0, which lets a truncated sum stay wrong at individual points forever — the step that is 1 on one hemisphere and 0 on the other is met at the midpoint of its jump at every order.

Simple definition: The spherical harmonics Y_ℓ^m(θ, φ) are the simultaneous eigenfunctions of L̂² and L̂_z on the unit sphere, normalised so that ∫ Y_ℓ^m* Y_ℓ′^m′ dΩ = δ_ℓℓ′ δₘₘ′, and together they form a complete orthonormal basis of L²(S²).
Example: Y₁⁰ = √(3/4π) cos θ: ∫|Y₁⁰|² dΩ = (3/4π) × 2π × ∫₋₁¹ u² du = (3/4π)(2π)(2/3) = 1, and it is orthogonal to Y₀⁰ = 1/√(4π) because ∫ cos θ dΩ = 2π ∫₋₁¹ u du = 0.

Construction from associated Legendre functionsY_ℓ^m(θ, φ) = N_ℓm P_ℓ^m(cos θ) e^(imφ), N_ℓm = √[(2ℓ+1)(ℓ−m)! / 4π(ℓ+m)!]

e^(imφ) is the L̂_z eigenfunction from 9.6; P_ℓ^m(cos θ) is the L̂² part, and regularity at both poles is what forces ℓ ≥ |m| to be an integer.

ℓ = 0, 1, 2, …; m = −ℓ … ℓ; Y is dimensionless; Condon–Shortley sign (−1)^m folded into P_ℓ^m

Eigenvalue equations on the sphereL̂² Y_ℓ^m = ħ² ℓ(ℓ+1) Y_ℓ^m, L̂_z Y_ℓ^m = mħ Y_ℓ^m

The same numbers the ladder algebra produced, now recovered by a differential operator — and only for integer ℓ. Single-valuedness is the weak reason (9.6): the ℓ = ½ tower really fails because L̂₋ does not annihilate its would-be bottom rung, so the ladder never closes and the functions it generates are not in L²(S²).

L̂² = −ħ²[(1/sin θ)∂_θ(sin θ ∂_θ) + (1/sin²θ)∂_φ²], L̂_z = −iħ∂_φ; ħ = 1.055 × 10⁻³⁴ J s

Orthonormality under the solid-angle inner product⟨ℓ m|ℓ′ m′⟩ = ∫₀²π dφ ∫₀^π sin θ dθ Y_ℓ^m* Y_ℓ′^m′ = δ_ℓℓ′ δₘₘ′

Forced by Hermiticity: different ℓ are separated by L̂², different m by L̂_z. Drop the sin θ and ∫(3cos²θ − 1) dθ = π/2 — Y₂⁰ and Y₀⁰ stop looking orthogonal.

dΩ = sin θ dθ dφ, 4π steradians in total; the sin θ weight is part of the inner product, not decoration

Expansion and Parsevalf(θ, φ) = Σ_ℓ Σₘ c_ℓm Y_ℓ^m, c_ℓm = ∫ Y_ℓ^m* f dΩ, ∫|f|² dΩ = Σ|c_ℓm|²

The coefficients are overlaps and therefore Born amplitudes: for a normalised angular state, |c_ℓm|² is the probability that L̂² and L̂_z return ℓ(ℓ+1)ħ² and mħ.

any f in L²(S²); ℓ from 0 to ∞, m from −ℓ to ℓ; Y is dimensionless, so the c_ℓm carry the dimensions of f and are pure numbers for a normalised angular state

Parity and node countY_ℓ^m(π − θ, φ + π) = (−1)^ℓ Y_ℓ^m(θ, φ)polar nodes = ℓ − |m|

(−1)^(ℓ−m) from P_ℓ^m(−u) times (−1)^m from e^(imφ) gives (−1)^ℓ. Count the circles of latitude where |Y|² vanishes and you have ℓ − |m| with no formula.

r̂ → −r̂ is θ → π − θ, φ → φ + π; the real forms cos mφ, sin mφ add |m| nodal meridian planes, ℓ nodal lines in all

Convergence in the mean‖f − f_L‖² = Σ_(ℓ>L) Σₘ |c_ℓm|² → 0 as L → ∞

What “complete” actually promises: the leftover norm shrinks. At a discontinuity the partial sums sit at the midpoint of the jump at every L and overshoot beside it.

f_L is the sum truncated at ℓ ≤ L; ‖⋅‖ is the L² norm on the sphere; nothing is claimed about any single point

Separate L̂² on the sphere and impose regularity at the poles

Start from the operators, not a table. In spherical coordinates L̂_z = −iħ ∂_φ and L̂² = −ħ²[(1/sin θ)∂_θ(sin θ ∂_θ) + (1/sin²θ)∂_φ²], the angular part of −ħ²r²∇² from 9.5. Try Y = Θ(θ)Φ(φ). Φ must be a function on the circle, Φ(φ + 2π) = Φ(φ), which gives Φ = e^(imφ) with integer m — and 9.6 gave the honest reason no half-integer m can be rescued by a sign convention: its ladder never closes and leaves L²(S²). Θ, written in u = cos θ, obeys the associated Legendre equation (1 − u²)Θ″ − 2uΘ′ + [λ − m²/(1 − u²)]Θ = 0 with λ the eigenvalue of L̂²/ħ². The equation is singular at u = ±1, the two poles, and regularity there is the boundary condition. The solution regular at u = 1 is (1 − u²)^(|m|/2) times a power series in (1 − u); that series diverges at the other pole, u = −1, unless it terminates, and it terminates only for λ = ℓ(ℓ + 1) with ℓ an integer ≥ |m|. The regular solutions are P_ℓ^m(u) = (−1)^m (1 − u²)^(m/2) d^m P_ℓ/du^m for m ≥ 0, the (−1)^m being the Condon–Shortley sign, with negative m fixed by Y_ℓ^(−m) = (−1)^m Y_ℓ^m*, and normalising over the sphere fixes N_ℓm. The first few: Y₀⁰ = 1/√(4π), Y₁⁰ = √(3/4π) cos θ, Y₁^±1 = ∓√(3/8π) sin θ e^(±iφ), Y₂⁰ = √(5/16π)(3cos²θ − 1).

Orthogonality is Hermiticity; only the norm is your choice

The inner product on L²(S²) is ⟨f|g⟩ = ∫ f*g dΩ with dΩ = sin θ dθ dφ, and under exactly that measure L̂_z and L̂² are Hermitian: integrating by parts in φ leaves no boundary term because everything is 2π-periodic, and in θ the boundary terms carry a factor sin θ that vanishes at both poles. Hermitian operators have orthogonal eigenfunctions for distinct eigenvalues, so ⟨ℓ m|ℓ′ m′⟩ = 0 whenever ℓ ≠ ℓ′ (L̂² separates them) or m ≠ m′ (L̂_z does). No integral has to be checked; only the length of each vector is a choice, and N_ℓm = √[(2ℓ+1)(ℓ−m)!/4π(ℓ+m)!] makes it 1. The measure is the whole story. With it, ⟨0 0|2 0⟩ ∝ ∫₋₁¹ (3u² − 1) du = 2 − 2 = 0. Without the sin θ, ∫₀^π (3cos²θ − 1) dθ = 3π/2 − π = π/2, and two functions that must be orthogonal appear not to be.

Read ℓ and m off a picture: nodes, parity, degeneracy

P_ℓ^m(u) has exactly ℓ − |m| zeros in −1 < u ⟨1, so |Y_ℓ^m|² vanishes on ℓ − |m| circles of latitude. |e^(imφ)|² = 1 adds none, which is why the density of a complex Y_ℓ^m is a figure of revolution about z. The real combinations (Y_ℓ^m ± Y_ℓ^(−m))/√2 ∝ cos mφ, sin mφ — the chemist's pₓ and d_xy — add |m| nodal planes through the poles, bringing the total to ℓ nodal lines. So Y₃¹ shows two polar circles and, in real form, one meridian plane. Parity is the map r̂ → −r̂, that is θ → π − θ and φ → φ + π: P_ℓ^m(−u) = (−1)^(ℓ−m) P_ℓ^m(u) and e^(im(φ+π)) = (−1)^m e^(imφ), so the product is (−1)^ℓ, odd for p, even for d, whatever m is. Degeneracy is the count of m: 2ℓ + 1 functions share L̂² = ℓ(ℓ+1)ħ², seven for ℓ = 3, and any linear combination of them is another eigenfunction with the same ℓ.

Expand a function: overlaps in, probabilities out

Given any angular function, its coordinates in the basis are c_ℓm = ⟨ℓ m|f⟩ = ∫ Y_ℓ^m* f dΩ. Take f = cos²θ. There is no φ-dependence, so only m = 0 survives, and cos²θ = 1/3 + (2/3)(3cos²θ − 1)/2 = (1/3)P₀ + (2/3)P₂. Converting with P₀ = √(4π) Y₀⁰ and P₂ = √(4π/5) Y₂⁰ gives c₀₀ = √(4π)/3 = 1.182 and c₂₀ = (4/3)√(π/5) = 1.057, every other c_ℓm zero. Parseval checks it: c₀₀² + c₂₀² = 4π/9 + 16π/45 = 4π/5 = 2.513, which is ∫cos⁴θ dΩ = 2π × 2/5 computed directly. Read as a state, ψ = √(5/4π) cos²θ is normalised and equals (√5/3)Y₀⁰ + (2/3)Y₂⁰, so a measurement of L̂² returns 0 with probability 5/9 and 6ħ² with probability 4/9, L̂_z returns 0 with certainty, and ⟨L̂²⟩ = 8ħ²/3, a value no single ℓ can give.

Completeness promises the norm, not the points

Completeness, Σ_ℓm Y_ℓ^m(Ω) Y_ℓ^m*(Ω′) = δ(cos θ − cos θ′) δ(φ − φ′), says every f in L²(S²) has an expansion, and it says how the truncation f_L = Σ_(ℓ≤L) fails: by Parseval, ‖f − f_L‖² = Σ_(ℓ>L)|c_ℓm|², the tail of a convergent series, so the error in the norm goes to zero. That is convergence in the mean. Make f the step that is 1 on the northern hemisphere and 0 on the southern. Only m = 0 enters, c_ℓ0 = √((2ℓ+1)π) ∫₀¹ P_ℓ du, and the even ℓ ≥ 2 vanish: c₀₀ = √π = 1.772, c₁₀ = √(3π)/2 = 1.535, c₃₀ = −√(7π)/8 = −0.586, c₅₀ = √(11π)/16 = 0.367. The ℓ ≤ 5 sum holds 5.976 of ‖f‖² = 2π = 6.283, so 95.1% of the norm and rising. At the equator, though, every P_ℓ(0) with odd ℓ is zero, so f_L(π/2) = c₀₀Y₀⁰ = 1/2 at every order: the series lands on the midpoint of the jump and never leaves it, and just beside the jump the partial sums overshoot to about 1.09 — at L = 19 and still at L = 159: the ripple narrows toward the equator as L grows but never flattens. The figure below draws exactly this.

The basis fixes no energy

For any central potential, H = p̂ᵣ²/2μ + L̂²/(2μr²) + V(r) commutes with L̂² and L̂_z, so the stationary states factor as R(r) Y_ℓ^m(θ, φ) with the same Y_ℓ^m whatever V is — hydrogen, the three-dimensional oscillator, a nucleon in a Woods–Saxon well. The angular basis therefore carries no energy of its own; E comes from the radial equation, in which ℓ appears only through the centrifugal term ħ²ℓ(ℓ+1)/(2μr²). Hydrogen shows one extreme: E = −13.6 eV/n² is blind to ℓ, so 2s and 2p are degenerate. The rigid rotor shows the other: freeze r at r₀, the radial equation disappears, and E_ℓ = ħ²ℓ(ℓ+1)/(2I). For HCl, I = μr₀² = 1.627 × 10⁻²⁷ kg × (1.275 × 10⁻¹⁰ m)² = 2.64 × 10⁻⁴⁷ kg m², so ħ²/2I = 2.10 × 10⁻²² J = 10.6 cm⁻¹ and the ladder reads 0, 21.2 and 63.5 cm⁻¹ for ℓ = 0, 1, 2. Same Y_ℓ^m, two utterly different spectra: the harmonics tell you the shape and the quantum numbers, and the radial physics tells you the price.

Change one variable at a time

Make the relationship visible.

Interactive model

Truncation order L3

Probe angle θ75 °

Step L up from 0 to 9 and watch NORM CAPTURED climb while the curve at θ = 90° never leaves ½; then set L = 9 and walk the probe from 60° to 85°: the ripple beside the jump peaks near 1.08 at 75°, and raising L only pushes it toward the equator, never flat.

NORM CAPTURED93.0 %

‖f − f_L‖² TAIL0.442

NON-ZERO TERMS3

EXACT f AT PROBE1.0

Live interpretationNORM CAPTURED: 93.0 %. ‖f − f_L‖² TAIL: 0.442. NON-ZERO TERMS: 3. EXACT f AT PROBE: 1.0

Catch the common trap

Explain before calculating.

Misconception check

The most common error is to test orthogonality with dθ instead of sin θ dθ, as if the sphere were a rectangle in (θ, φ). It is not: a band of width dθ near the equator holds far more solid angle than the same band near a pole, and the inner product ⟨f|g⟩ = ∫ f*g sin θ dθ dφ carries that weight because L̂² is Hermitian only under it. The numbers show the failure. Y₀⁰ and Y₂⁰ satisfy ∫₋₁¹ (3u² − 1) du = 0 in u = cos θ, but ∫₀^π (3cos²θ − 1) dθ = 3π/2 − π = π/2 ≠ 0. A student who drops the weight finds eigenfunctions of a Hermitian operator that are apparently not orthogonal, concludes the table is wrong, and mis-normalises every coefficient that follows. Whenever a spherical integral is written, the sin θ goes in with the dθ.

The angular part of a state is ψ(θ, φ) ∝ cos²θ. A measurement of L̂² is made. What is the probability that it returns 6ħ²?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyY₁¹ = −√(3/8π) sin θ e^(iφ). Show it is normalised on the sphere, show it is orthogonal to Y₁⁰ = √(3/4π) cos θ, and state its parity and its number of polar nodes.

Modulus first: |Y₁¹|² = (3/8π) sin²θ; the phase e^(iφ) and the overall sign both drop out. The measure is dΩ = sin θ dθ dφ.
Integrate: ∫₀²π dφ = 2π, and ∫₀^π sin³θ dθ = ∫₋₁¹ (1 − u²) du = 2 − 2/3 = 4/3 with u = cos θ.
Product: (3/8π) × 2π × 4/3 = 1. Normalised.
Orthogonality to Y₁⁰: the integrand Y₁¹* Y₁⁰ carries e^(−iφ), and ∫₀²π e^(−iφ) dφ = 0. Hermiticity of L̂_z promised this, since the eigenvalues ħ and 0 differ.
Parity: θ → π − θ, φ → φ + π sends sin θ e^(iφ) to sin θ e^(iφ) e^(iπ) = −sin θ e^(iφ), so parity is −1 = (−1)^ℓ. Polar nodes: ℓ − |m| = 0, and |Y₁¹|² ∝ sin²θ indeed vanishes only at the poles, on no circle between them.

Answer∫|Y₁¹|² dΩ = 1; ⟨1 0|1 1⟩ = 0 by the φ integral; parity −1; zero polar nodes.

MediumA particle's angular wavefunction is ψ(θ, φ) = A cos²θ. Normalise it, expand it in spherical harmonics, and find the probabilities for each outcome of a measurement of L̂² and of L̂_z, together with ⟨L̂²⟩.

Normalise: ∫cos⁴θ dΩ = 2π ∫₋₁¹ u⁴ du = 2π × 2/5 = 4π/5, so A = √(5/4π) = 0.631.
No φ-dependence means only m = 0 harmonics can appear. In Legendre polynomials, with P₂ = (3u² − 1)/2, cos²θ = (1/3)P₀ + (2/3)P₂.
Convert using P₀ = √(4π) Y₀⁰ and P₂ = √(4π/5) Y₂⁰: ψ = √(5/4π)[(1/3)√(4π) Y₀⁰ + (2/3)√(4π/5) Y₂⁰] = (√5/3) Y₀⁰ + (2/3) Y₂⁰.
Born rule: P(ℓ = 0) = 5/9 = 0.556 and P(ℓ = 2) = 4/9 = 0.444; they sum to 1, which is Parseval for this state. L̂_z returns 0 with probability 1.
⟨L̂²⟩ = (5/9)(0) + (4/9)(6ħ²) = 8ħ²/3 = 2.67ħ². Direct check: L̂² cos²θ = 6ħ²cos²θ − 2ħ², so ⟨L̂²⟩ = (5/4π)ħ²[6 × 4π/5 − 2 × 4π/3] = ħ²(6 − 10/3) = 8ħ²/3.

Answerψ = (√5/3)Y₀⁰ + (2/3)Y₂⁰; P(ℓ = 0) = 5/9, P(ℓ = 2) = 4/9; m = 0 with certainty; ⟨L̂²⟩ = 8ħ²/3.

HardLet f(θ, φ) = 1 on the northern hemisphere (θ < π/2) and 0 on the southern. Find c_ℓm for ℓ ≤ 5, the fraction of ‖f‖² captured by the truncation at ℓ = 5, and the value every partial sum takes at the equator. Use ∫₀¹ P_ℓ du = [P_(ℓ−1)(0) − P_(ℓ+1)(0)]/(2ℓ+1) for ℓ ≥ 1.

f is φ-independent, so c_ℓm = 0 for m ≠ 0, and c_ℓ0 = ∫ Y_ℓ⁰ f dΩ = √((2ℓ+1)/4π) × 2π × ∫₀¹ P_ℓ(u) du = √((2ℓ+1)π) ∫₀¹ P_ℓ du.
Legendre values at 0: P₀ = 1, P₂ = −1/2, P₄ = 3/8, P₆ = −5/16, and every odd P_ℓ(0) = 0. So ∫₀¹P₀ = 1, ∫₀¹P₁ = (1 + 1/2)/3 = 1/2, ∫₀¹P₃ = (−1/2 − 3/8)/7 = −1/8, ∫₀¹P₅ = (3/8 + 5/16)/11 = 1/16, while ∫₀¹P₂ = ∫₀¹P₄ = 0 because an even P_ℓ with ℓ ≥ 2 integrates to zero over the whole interval.
Coefficients: c₀₀ = √π = 1.772, c₁₀ = √(3π)/2 = 1.535, c₂₀ = 0, c₃₀ = −√(7π)/8 = −0.586, c₄₀ = 0, c₅₀ = √(11π)/16 = 0.367.
‖f‖² = ∫|f|² dΩ = 2π = 6.283. The truncation holds π + 3π/4 + 7π/64 + 11π/256 = 1.9023π = 5.976, so 95.1% of the norm; the tail past ℓ = 5 is the mean-square error, 0.307, about 4.9%, and it keeps shrinking as L grows.
At the equator u = 0 every odd P_ℓ vanishes and the even coefficients beyond ℓ = 0 are zero, so f_L(π/2) = c₀₀ Y₀⁰ = √π/√(4π) = 1/2 for every L. The series converges to the midpoint of the jump and never to either side of it: the norm converges, the point does not.

Answerc₀₀ = 1.772, c₁₀ = 1.535, c₂₀ = 0, c₃₀ = −0.586, c₄₀ = 0, c₅₀ = 0.367; ℓ ≤ 5 captures 95.1% of ‖f‖² = 2π; every partial sum equals 1/2 at the equator.