Orbital angular momentum and spherical harmonics

Classically L = r × p. Quantum mechanically p becomes −iħ∇, and because r × ∇ has no radial component, every component of L acts on angles only. In spherical coordinates the z-component is L_z = −iħ ∂/∂φ, the generator of rotations about z, and L² = −ħ²[(1/sin θ) ∂/∂θ(sin θ ∂/∂θ) + (1/sin²θ) ∂²/∂φ²], the angular part of −ħ²r²∇². The kinetic energy therefore splits into a radial piece plus L²/(2μr²), with μ the reduced mass, so for any potential V(r) the Schrödinger equation separates as ψ = R(r)Y(θ, φ). The angular equation, L²Y = λħ²Y, fixes the constant that the radial equation of the previous topic feels as a centrifugal barrier. Dimensions check: ħ carries J s, the unit of angular momentum, and ħ²/(μr²) is an energy. The Coulomb force never appears, so everything that follows holds for hydrogen, a rotating diatomic molecule and a nucleon in a nucleus alike.

Write Y = Θ(θ)Φ(φ). The φ-dependence obeys d²Φ/dφ² = −m²Φ, so Φ = e^(imφ), and L_zΦ = −iħ(im)Φ = mħΦ: these are the states of sharp L_z. At this stage m could be any real number. The constraint is geometric. The angles φ and φ + 2π label the same point in space, and a wavefunction must take one value there, so e^(2πim) = 1 and m = 0, ±1, ±2, …. A trial m = ½ would give Φ(2π) = −Φ(0), a sign flip on going once round, which no orbital wavefunction may have; half-integers will return only with spin, from experiment. The same line shows what sharp L_z costs: |e^(imφ)|² = 1 for every φ, so the probability density is uniform in azimuth. Knowing L_z exactly means knowing nothing about the angle around the axis, the rotational twin of a plane wave with sharp momentum and no position.

With Φ fixed, the θ-equation in the variable x = cos θ is the associated Legendre equation, with λ as its undetermined constant. Try a power series in x. For m = 0 the ratio of successive coefficients is [k(k+1) − λ]/[(k+1)(k+2)], which tends to 1, so an unending series behaves like Σ xᵏ/k and diverges at x = ±1, the two poles, where a physical wavefunction must stay finite. The only escape is for the series to stop, and it stops only when λ = l(l+1) with l a non-negative integer; for m ≠ 0 the same argument adds l ≥ |m|. The survivors, normalised over the sphere, are the spherical harmonics: Y₀₀ = 1/√(4π), Y₁₀ = √(3/(4π)) cos θ, and the m = ±1 pair ∓√(3/(8π)) sin θ e^(±iφ). Check one: L² acting on cos θ gives −ħ²(1/sin θ) d/dθ(−sin²θ) = 2ħ² cos θ, so l(l+1) = 2 and l = 1. Each l carries 2l + 1 values of m, and under inversion r → −r every Yₗₘ picks up (−1)^l, a parity the selection rules will need.

The components obey [Lₓ, L_y] = iħL_z and its cyclic partners, so no state except l = 0 has two components sharp at once, while L² commutes with each and can be sharp alongside any one; which one is a choice of axis, called z. In the state |l, m⟩ symmetry gives ⟨Lₓ⟩ = ⟨L_y⟩ = 0, and Lₓ² + L_y² = L² − L_z² gives ⟨Lₓ²⟩ = ⟨L_y²⟩ = [l(l+1) − m²]ħ²/2. The Robertson bound is ΔLₓΔL_y ≥ (ħ/2)|⟨L_z⟩| = |m|ħ²/2. At m = l the product is [l(l+1) − l²]ħ²/2 = lħ²/2, exactly the bound: the extra l in l(l+1) is the least room the uncertainty principle allows, which is why |L| can never equal lħ. The vector model draws this as a cone with cos θ = m/√(l(l+1)); for l = 2 the five cones open at 35.3°, 65.9°, 90°, 114.1° and 144.7° from +z. The cone is a picture of what is known, not a precession you could watch: measuring Lₓ returns one of its own eigenvalues, −lħ to +lħ, and leaves L_z unsharp.

|Yₗₘ|² sin θ dθ dφ is the probability that the electron's direction from the nucleus lies in that patch of solid angle, and the integral over all 4π sr is 1. Because Yₗₘ depends on φ only through e^(imφ), every |Yₗₘ|² is symmetric about the z axis; the dumbbells labelled pₓ and p_y in chemistry are real combinations of m = +1 and m = −1, with no sharp L_z. The density can still be strongly directional in θ. For Y₁₀ the probability of lying within 30° of either pole is ∫(3/(4π)) cos²θ dΩ over the two caps, which is 1 − cos³30° = 0.350, against 1 − cos 30° = 0.134 for the isotropic Y₀₀. That is the correspondence picture at work: in the m = 0 state L lies perpendicular to z, so the classical orbit plane contains the z axis and carries the electron over the poles. A filled subshell erases all direction, since Σₘ |Yₗₘ|² = (2l + 1)/(4π) is a constant: closed shells are spherical.

An electric-dipole transition rate is proportional to |⟨f|r|i⟩|², and the components of r are themselves l = 1 harmonics: z = r cos θ goes as rY₁₀, while x ± iy = r sin θ e^(±iφ) carries m = ±1. The φ integral of exp[i(m + q − m′)φ] vanishes unless m′ = m + q with q = 0 or ±1, so Δm = 0 for light polarised along z and ±1 for the circular components. The θ integral is angular-momentum addition: combining l with one unit gives l′ = l − 1, l or l + 1, and because r is odd under inversion the integrand's parity (−1)^{(l + l′ + 1)} must be even, which strikes out l′ = l. So Δl = ±1. The consequences are measurable: hydrogen's 2p state decays to 1s in 1.6 ns, while 2s has no electric-dipole route to 1s and survives 0.12 s, decaying by two photons. The rules bend only through weaker processes, magnetic-dipole emission down by about α² ≈ 5 × 10⁻⁵ and electric-quadrupole emission by about (2πa₀/λ)² ≈ 3 × 10⁻⁷ for visible light.