Angular wavefunctions and the L², L_z eigenbasis

Write L = r × p in spherical coordinates and two things stand out. L_z = −iħ ∂/∂φ is the generator of rotations about z, and L² = −ħ²[(1/sinθ)∂_θ(sinθ ∂_θ) + (1/sin²θ)∂²_φ] is exactly −ħ² times the Laplace–Beltrami operator of the unit sphere, which is the angular part of the Laplacian: ∇² = (1/r²)∂ᵣ(r²∂ᵣ) − L²/ħ²r². Neither operator contains r or ∂ᵣ. Both therefore commute with every function of the radius, and the angular eigenproblem is settled before the potential has even been named. The quantisation that follows is geometric rather than dynamical. S² is a closed surface of finite area 4π sr, and the Laplace–Beltrami operator on a compact manifold has a purely discrete spectrum, so l takes integer values with no well, no binding and no potential anywhere in sight. Compare a free particle on a line, whose momentum spectrum is continuous because the domain runs to infinity: here the domain has no edge to escape through, and discreteness is the price of closure.

Separate Y = Θ(θ)Φ(φ). The L_z equation is −iħ dΦ/dφ = mħΦ, so Φ ∝ e^(imφ) with no restriction yet on m. Now impose that Y is a single function on the sphere: φ and φ + 2π label the same direction, so Φ(φ + 2π) = Φ(φ) and e^(2πim) = 1, which gives m = 0, ±1, ±2, …. A half-integer m would return e^(iπ) = −1 after one full turn, so one point of space would carry two amplitudes. Ladder termination already forces 2l to be a non-negative integer, and m descends from l in unit steps, so integer m makes l integer too. That is the whole reason orbital l is never ½ — and it is a condition on the representation, not on the algebra, which the misconception below pushes all the way down. Notice what is not excluded. Append a spin factor and J = L + S obeys the very same commutators with j = l ± ½, and those half-integer j are entirely legitimate, because ⟨n̂|j mⱼ⟩ is not a function on the sphere and so answers to no 2π identification. Integrality is a statement about orbital motion, not about angular momentum as such.

The economical construction never touches Legendre's equation. Write L_± = Lₓ ± iL_y = ħe^(±iφ)(±∂_θ + i cotθ ∂_φ) and impose L₊Yₗ^l = 0, which is just the statement that m cannot exceed l. On a trial f(θ)e^(ilφ) the φ derivative brings down il, and the condition collapses to f′ − l cotθ f = 0, a first-order ODE with solution f ∝ sin^l θ. So Yₗ^l ∝ sin^l θ e^(ilφ), normalised on the sphere, and every other member of the block follows from L₋|l m⟩ = ħ√(l(l+1) − m(m−1))|l m−1⟩. Take l = 1, whose top rung is Y₁¹ = −√(3/8π) sinθ e^(iφ). Applying L₋ gives ħe^(−iφ)[√(3/8π) cosθ + cotθ √(3/8π) sinθ]e^(iφ) = 2ħ√(3/8π) cosθ, while the ladder norm says the result is ħ√2 Y₁⁰. Divide: Y₁⁰ = (2/√2)√(3/8π) cosθ = √(3/4π) cosθ = 0.4886 cosθ, the tabulated function, phase and all.

The inner product on directions is ⟨f|h⟩ = ∫f* h dΩ with dΩ = sinθ dθ dφ, and in it ∫Yₗ'^m'* Yₗ^m dΩ = δₗₗ' δₘₘ'. The φ integral supplies δₘₘ' immediately; the θ integral is the orthogonality of associated Legendre functions at fixed m. Completeness follows: any f(θ, φ) with finite ∫|f|² dΩ expands as f = Σₗ Σₘ cₗₘ Yₗ^m with cₗₘ = ∫Yₗ^m* f dΩ, and Σ|cₗₘ|² = ∫|f|² dΩ. This is Fourier analysis on a sphere, and it is what lets you read probabilities of l and of m straight off a state. Expand cos²θ as a check. Since P₂(u) = (3u² − 1)/2, cos²θ = 1/3 + (2/3)P₂(cosθ) = (√(4π)/3) Y₀⁰ + (2/3)√(4π/5) Y₂⁰. Only l = 0 and l = 2 appear, because the function is even under parity, and only m = 0, because it does not depend on φ. Parseval closes it: 4π/9 + (4/9)(4π/5) = 4π/5, and ∫cos⁴θ dΩ = 2π∫ u⁴ du from −1 to 1 = 4π/5.

A rotation acts as U(R)|l m⟩ = Σₘ' D^(l)ₘ'm(R)|l m'⟩: it mixes m and never touches l, because L² is the Casimir invariant of the rotation algebra while L_z is only one generator. Each l therefore spans a (2l+1)-dimensional subspace rotations cannot break up, and three things follow. First, every central field carries an exact (2l+1)-fold degeneracy, since H commutes with L_± as well as with L_z, so the ladder moves between states of one energy. Second, the choice of z is a choice of basis, not of physics: the real orbitals p_z = Y₁⁰, pₓ = (Y₁⁽⁻¹⁾ − Y₁¹)/√2 and p_y = i(Y₁⁽⁻¹⁾ + Y₁¹)/√2 are a unitary rotation inside the same block, and the last two carry no sharp L_z at all. Third, the closure sum Σₘ |Yₗ^m|² = (2l+1)/4π is constant over the sphere — for l = 1, (3/8π)sin²θ + (3/4π)cos²θ + (3/8π)sin²θ = 3/4π = 0.2387 sr⁽⁻¹⁾ — so a filled subshell is exactly spherically symmetric however lobed its individual pictures look.

Every step above used only that L² and L_z act on angles. So the same Yₗ^m are the angular factor of the Coulomb problem, of the three-dimensional isotropic oscillator, of a finite spherical well and of a Woods–Saxon nuclear potential: you solve the sphere once for the whole of atomic and nuclear physics. What the calculation hands the radial equation is a single number, the constant in the centrifugal term ħ²l(l+1)/2μr², and no energy whatever. The spectra it is compatible with could hardly differ more. Hydrogen gives Eₙ = −13.606 eV/n², with 2s and 2p degenerate; the isotropic oscillator gives E = ħω(2nᵣ + l + 3/2), in which 1s lies below 1p lies below the degenerate 2s and 1d. Identical angular functions, unrelated level schemes. And the one degeneracy the angular basis does supply, the 2l + 1 rungs, is also the easiest to destroy: a uniform B along z leaves only rotations about z as symmetries, and the block fans into equally spaced levels μ_B B mₗ — 5.788 × 10⁻⁵ eV per tesla, or 0.467 cm⁻¹ per tesla.