Spherical coordinates and the angular Laplacian

Start from L_z = x p_y − y pₓ with p = −iħ∇ and change variables: x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ. Holding r and θ fixed, ∂/∂φ = (∂x/∂φ)∂ₓ + (∂y/∂φ)∂_y = −y ∂ₓ + x ∂_y, so −iħ ∂/∂φ = x(−iħ∂_y) − y(−iħ∂ₓ) = L_z. No ∂ᵣ and no ∂_θ survive: L_z acts on the azimuth alone. Its eigenfunctions are e^(imφ), and because φ and φ + 2π label the same point, m must be an integer — the first boundary condition of the topic. The exponential map makes the geometry explicit: exp(−iαL_z/ħ) f(φ) = Σ (−α)ⁿ f⁽ⁿ⁾/n! = f(φ − α), a rotation of the function by α. Lₓ and L_y come out messier because the chart singles out z; the useful combinations are L_± = ħ e^(±iφ)(±∂_θ + i cotθ ∂_φ), and a one-line check that [L_z, L_±] = ±ħ L_± is that L_z acting on the prefactor e^(±iφ) returns ±ħ times it.

Square the vector operator r × ∇ using the identity (r × ∇)⋅(r × ∇) = r²∇² − (r⋅∇)² − r⋅∇. With r⋅∇ = r ∂ᵣ, the last two terms combine into −∂ᵣ(r² ∂ᵣ), so L² = −ħ²(r × ∇)² = −ħ²[ r²∇² − ∂ᵣ(r² ∂ᵣ) ], and rearranging gives the identity the topic turns on: ∇² = (1/r²)∂ᵣ(r² ∂ᵣ) − L²/ħ²r². Written out, L²/ħ² is minus the angular part of the Laplacian, −[(1/sinθ)∂_θ(sinθ ∂_θ) + (1/sin²θ)∂²_φ]. Multiply by −ħ²/2m and the kinetic energy splits: T = pᵣ²/2m + L²/2mr², with pᵣ = −iħ(1/r)∂ᵣ r. Note what pᵣ is not: −iħ ∂ᵣ alone is not Hermitian on the measure r² dr, and its square lacks the (2/r)∂ᵣ term that (1/r²)∂ᵣ r²∂ᵣ carries. Test the split on ψ = z = r cosθ, which is harmonic. Radial part: (1/r²)∂ᵣ(r² cosθ) = 2cosθ/r. Angular part: L²(cosθ) = 2ħ² cosθ, so −L²ψ/ħ²r² = −2cosθ/r. They cancel, ∇²z = 0, and the l = 1 eigenvalue 2ħ² has been read straight off the geometry.

For V = V(r), H = pᵣ²/2m + L²/2mr² + V(r). L contains no ∂ᵣ and commutes with any function of r, so [H, L²] = [H, L_z] = 0 and (H, L², L_z) is a complete commuting set: look for joint eigenfunctions ψ = R(r)Y(θ, φ) with L²Y = λħ²Y and L_zY = mħY. The φ-dependence is e^(imφ), periodicity fixing m ∈ ℤ. With x = cosθ the θ-equation is the associated Legendre equation (1 − x²)P″ − 2xP′ + [λ − m²/(1 − x²)]P = 0, whose singular points x = ±1 are the poles, where the chart's 1/sin²θ blows up. The sphere itself is smooth there, so a physical Y must stay finite at the poles — and for generic λ no solution does: the one regular at x = 1 diverges at x = −1, logarithmically for m = 0 and as (1 + x)^(−|m|/2) otherwise. With the factor (1 − x²)^(|m|/2) taken out, the series terminates, and Y stays finite at both poles, only for λ = l(l+1) with l an integer ≥ |m|. That is the added demand: the differential operator does not know it. Pole regularity says Y is a genuine function on the sphere, the domain on which L² is self-adjoint. The abstract ladder algebra permits half-integer j; its realisation on the sphere does not, and that is a fact about orbital motion only, because a spin-½ state is a C² spinor, not a function of θ and φ.

Put ψ = R(r)Yₗ^m into Hψ = Eψ with L² replaced by ħ²l(l+1), then set u = rR. The radial operator simplifies because (1/r²)∂ᵣ(r²∂ᵣ)(u/r) = u″/r, leaving −(ħ²/2m)u″ + [V(r) + ħ²l(l+1)/2mr²]u = Eu on the half-line r > 0, with the condition u(0) = 0 that keeps R finite and pᵣ Hermitian. Near r = 0 with V finite, u″ ≈ l(l+1)u/r² has the two solutions r^(l+1) and r^(−l). The second is rejected: for l ≥ 1 its |R|²r² dr ∝ r^(−2l) dr is not integrable at the origin, and for l = 0 it gives R ∝ 1/r, which is not a solution at all since ∇²(1/r) = −4πδ³(r). So u ∝ r^(l+1) and |ψ|² ∝ r^(2l) near the origin: every l > 0 state is pushed outward. Numbers for an electron: ħ²/2mₑ = 0.0381 eV nm², and in the hydrogen Coulomb field, with ħ²/mₑ a₀² = 27.2 eV, the barrier is 13.6 l(l+1)(a₀/r)² eV against a Coulomb term of −27.2 (a₀/r) eV. Their sum bottoms out at r = l(l+1)a₀ with Vₘᵢₙ = −13.6/l(l+1) eV: −6.8 eV at 2a₀ for l = 1, −2.27 eV at 6a₀ for l = 2. At r = a₀ the l = 1 barrier cancels the Coulomb attraction exactly.

The radial equation contains l but not m, so once R(r) is found for a given l its energy is shared by all 2l + 1 values of m: the multiplet is exactly the set of states L_± reach from one another, and [H, L_±] = 0 guarantees they sit at one energy. Nothing more is degenerate for a generic central potential; hydrogen's extra n-only degeneracy is a Coulomb accident, not a consequence of this separation. Label a state by the eigenvalues of the commuting set — a radial quantum number nᵣ counting the nodes of u for r > 0, plus l and m — and the node count is complete: nᵣ radial nodes, l − |m| nodal cones in θ, and, in the real combinations of e^(±imφ), |m| nodal planes through the z axis; e^(imφ) itself never vanishes, so the complex Yₗ^m has none. Now test the scope. The identity ∇² = (1/r²)∂ᵣ(r²∂ᵣ) − L²/ħ²r² is kinematics and holds for any potential whatsoever; the separation is dynamics and needs [H, L²] = 0. Add eℰz to hydrogen and L_z still commutes with H, so m survives, but z ∝ rY₁⁰ couples Yₗ^m to Y_(l±1)^m and l is lost — the eigenstates become sums over l. Spherical coordinates did not do the separating; the spherically symmetric potential did.

When V(r) has no closed-form solution, the separated problem is a tridiagonal eigenproblem. Take a grid rᵢ = ih for i = 1, …, N with h = rₘₐₓ/(N + 1), and build H = −(ħ²/2m)D₂ + diag(V(rᵢ) + ħ²l(l+1)/2m rᵢ²), where D₂ is the three-point second-difference matrix with 1/h² off the diagonal and −2/h² on it. Leaving out r = 0 and r = rₘₐₓ is the boundary condition: the matrix already assumes u₀ = u_(N+1) = 0, which is exactly u(0) = 0 at the origin and an impenetrable wall at rₘₐₓ. Then np.linalg.eigh(H) returns the levels and the columns u(rᵢ); the centrifugal term at the first grid point is large but finite and simply enforces u ∝ r^(l+1). Benchmark it. For an impenetrable sphere of radius a the exact levels are E = ħ²x²/2ma² with x a zero of the spherical Bessel function jₗ: x = π and 2π for l = 0, 4.4934 for l = 1, 5.7635 for l = 2. With V = 0 and rₘₐₓ = a, an electron in a = 1.00 nm should return 0.376, 0.769, 1.266 and 1.504 eV in that order, with an error falling as h². That check, not a pretty plot, is what lets you trust the code on a potential with no known answer.