University Physics V · Angular Momentum in Quantum Mechanics · 9.5
Spherical Coordinates & the Angular Laplacian
Before hydrogen, learn to read ∇² in spherical coordinates as a radial operator minus L²/ħ²r². That single identity turns every central-potential problem into a one-dimensional radial equation with a centrifugal barrier — and shows exactly where the integer l comes from, and where the method stops applying.
Build the model
Connect the measurement to the mechanism.
Unit 9 so far built angular momentum from commutators alone. This topic gives L a concrete home: on functions of position L = −iħ r × ∇, and because r × ∇ contains no ∂/∂r, every component acts on the angles only. Lz becomes −iħ ∂/∂φ, since a rotation about z is nothing but a shift in φ, and L² turns out to be −ħ² times the Laplacian on the unit sphere.
That identity is the whole point: ∇² = (1/r²)∂ᵣ(r²∂ᵣ) − L²/ħ²r², or p² = pᵣ² + L²/r², kinetic energy split into a radial part and a rotational part with no cross term. For a central potential H then commutes with L² and Lz, so ψ = R(r)Y(θ, φ) separates into an angular eigenproblem solved once for all potentials and a radial equation carrying the barrier ħ²l(l+1)/2mr². The price is paid in boundary conditions and in scope.
The spherical chart is singular at the poles and at the origin, and demanding that ψ stay regular there is an added condition, not a consequence of the operator: it is what selects integer l ≥ |m| and forces u = rR to vanish at r = 0. The differential form describes orbital motion only — a spin-½ state is not a function on the sphere, so nothing here constrains it — and the separation belongs to the potential, not the coordinates: a non-central V still has a Laplacian that splits, but no longer a wavefunction that factorises.
- Simple definition
- In spherical coordinates the Laplacian splits into a radial derivative and an angular operator equal to −L²/ħ²r², so for any central potential the Schrödinger equation separates into an angular eigenproblem for L² and Lz and a radial equation with a centrifugal term.
- Example
- For an electron in the hydrogen Coulomb field with l = 1, the centrifugal term ħ²l(l+1)/2mr² at r = a₀ is 27.2 eV and cancels the −27.2 eV Coulomb energy exactly; the effective potential bottoms out at r = 2a₀ at −6.8 eV.
A rotation about z is a shift in φ, so the z-component is the φ-derivative and its eigenfunctions are pure phases with integer winding.
φ is the azimuth in radians and ħ = 1.055 × 10⁻³⁴ J s; ψ(φ + 2π) = ψ(φ) forces m to be an integer
No ∂/∂r appears, so L acts on the angles alone and its eigenproblem can be solved once and reused for every central potential.
θ ∈ [0, π] is the polar angle, φ ∈ [0, 2π); the bracket is ∇² on the unit sphere; L² carries J² s²
Kinetic energy is radial plus rotational with no cross term — the operator version of the classical p² = pᵣ² + L²/r².
pᵣ = −iħ (1/r) ∂ᵣ r = −iħ(∂ᵣ + 1/r), in kg m s⁻¹; Hermitian on r² dr only if rR → 0 as r → 0
The angular factor is universal; only R remembers the potential, and it feels the angular motion through l alone.
l = 0, 1, 2, … and m = −l, …, l; valid because [H, L²] = [H, Lz] = 0 when V = V(r)
A one-dimensional problem on the half-line: each l gets its own effective potential and its own ladder of levels.
u(0) = 0; ħ²/2mₑ = 0.0381 eV nm²; for an electron ħ² l(l+1)/(2mₑ r²) = 13.6 l(l+1) (a₀/r)² eV
The barrier keeps higher-l states away from the origin — and it is a boundary condition you impose, not something the equation supplies.
holds for V finite or Coulomb-like at r = 0; R = u/r ∝ rl, so |ψ|² near the nucleus scales as r(2l)
Rotation about z is a shift in φ
Start from Lz = x py − 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) = Lz. No ∂ᵣ and no ∂θ survive: Lz 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(−iαLz/ħ) f(φ) = Σ (−α)ⁿ f⁽ⁿ⁾/n! = f(φ − α), a rotation of the function by α. Lₓ and Ly 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 [Lz, L_±] = ±ħ L_± is that Lz acting on the prefactor e(±iφ) returns ±ħ times it.
L² is the Laplacian on the unit sphere
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)∂²φ]. 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.
Separate, and name every boundary condition
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, Lz] = 0 and (H, L², Lz) is a complete commuting set: look for joint eigenfunctions ψ = R(r)Y(θ, φ) with L²Y = λħ²Y and LzY = 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 the Frobenius series diverges logarithmically at x = ±1. It terminates 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.
The radial equation and the centrifugal barrier
Put ψ = R(r)Yₗm 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.
What the commuting set buys: 2l + 1 for free
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 |m| nodal planes through the z axis. 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 Lz still commutes with H, so m survives, but z ∝ rY₁⁰ couples Yₗm 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.
Diagonalise the radial Hamiltonian in NumPy
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 rN = rₘₐₓ, 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.
Change one variable at a time
Make the relationship visible.
Step l from 0 to 3 at E = −3.4 eV: the inner turning point moves out from 0 to 1.17 a₀ at l = 1, and at l = 2 the level lies below the −2.3 eV well bottom, so the allowed shell collapses. Then set l = 1 and lower E to −6.8 eV to watch the shell shrink to the single point r = 2 a₀.
COULOMB −27.2 a₀/r-13.6 eV
BARRIER 13.6 l(l+1)/r²6.8 eV
Veff AT THE PROBE-6.8 eV
MINIMUM r = l(l+1) a₀2 a₀
Live interpretationCOULOMB −27.2 a₀/r: −13.6 eV. BARRIER 13.6 l(l+1)/r²: 6.8 eV. Veff AT THE PROBE: −6.8 eV. MINIMUM r = l(l+1) a₀: 2 a₀
Catch the common trap
Explain before calculating.
Hydrogen is placed in a uniform electric field ℰ along z, so V = −e²/4πε₀r + eℰz. You still write ∇² in spherical coordinates. Which quantum numbers of the separated form ψ = R(r)Yₗm survive as good labels of the energy eigenstates?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyShow by direct differentiation that f(θ, φ) = cosθ is a joint eigenfunction of Lz = −iħ ∂/∂φ and of L² = −ħ²[(1/sinθ)∂θ(sinθ ∂θ) + (1/sin²θ)∂²φ], read off l and m, and normalise it over the sphere.
- Lz first: cosθ has no φ-dependence, so −iħ ∂φ cosθ = 0. The eigenvalue mħ is zero, hence m = 0.
- L²: the ∂²φ term vanishes. In the θ term, sinθ ∂θ cosθ = −sin²θ, and (1/sinθ)∂θ(−sin²θ) = (1/sinθ)(−2 sinθ cosθ) = −2cosθ.
- So L² cosθ = −ħ²(−2cosθ) = 2ħ² cosθ. Matching ħ²l(l+1) = 2ħ² gives l(l+1) = 2, so l = 1; the other root, l = −2, is not a quantum number.
- Normalise: ∫|N cosθ|² dΩ = N² · 2π ∫₋₁¹ x² dx = N² · 2π · 2/3 = 4πN²/3 = 1, so N = √(3/4π) = 0.489 — the standard Y₁⁰.
Answerl = 1, m = 0: cosθ is Y₁⁰ up to the factor √(3/4π) ≈ 0.489, with L² eigenvalue 2ħ² and Lz eigenvalue 0.
MediumAn electron moves in the hydrogen Coulomb potential V = −e²/4πε₀r with l = 1. Using e²/4πε₀ = 27.2 eV⋅a₀ and ħ²/mₑ = 27.2 eV⋅a₀², find where the effective potential Veff = V + ħ²l(l+1)/2mₑ r² is deepest, its value there, and the classical turning points of the n = 2 level, E = −3.4 eV.
- Write s = a₀/r. The barrier is ħ²l(l+1)/2mₑ r² = (27.2/2) · l(l+1) · s² eV = 27.2 s² eV for l = 1, and the Coulomb term is −27.2 s eV, so Veff = 27.2(s² − s) eV.
- dVeff/ds = 27.2(2s − 1) = 0 at s = ½, i.e. r = 2a₀, where Veff = 27.2(¼ − ½) = −6.8 eV. In general rₘᵢₙ = l(l+1)a₀ and Vₘᵢₙ = −13.6/l(l+1) eV.
- Check the pieces at r = a₀ (s = 1): Coulomb −27.2 eV, barrier +27.2 eV, so Veff = 0 — the l = 1 barrier has cancelled the whole Coulomb attraction at the Bohr radius.
- Turning points solve 27.2(s² − s) = −3.4, i.e. s² − s + 0.125 = 0: s = [1 ± √(1 − 0.5)]/2 = (1 ± 0.7071)/2, so s = 0.8536 or 0.1464.
- r = a₀/s = 1.17a₀ and 6.83a₀, exactly (4 ∓ 2√2)a₀. The 2p electron is classically confined to that shell; an l = 0 state at the same energy would reach in to r = 0 and out to 27.2/3.4 = 8a₀.
Answerrₘᵢₙ = 2a₀ with Vₘᵢₙ = −6.8 eV; at E = −3.4 eV the l = 1 shell runs from 1.17a₀ to 6.83a₀.
HardA particle of mass m is trapped inside an impenetrable sphere of radius a (V = 0 for r < a). With u = rR, find the l = 0 levels, then the lowest l = 1 level, given u ∝ sin(kr)/(kr) − cos(kr) for l = 1 and that tan x = x first at x = 4.4934. Then list the four lowest distinct energies, with degeneracies, for an electron in a = 1.00 nm, taking ħ²/2mₑ = 0.0381 eV nm² and the first zero of j₂ at 5.7635.
- Inside the sphere V = 0, so the radial equation is −(ħ²/2m)u″ + [ħ²l(l+1)/2mr²]u = Eu with u(0) = 0 and u(a) = 0. Set k² = 2mE/ħ².
- l = 0: u″ = −k²u gives u = A sin kr, the cos kr branch failing u(0) = 0. The wall demands ka = nπ, so E_(n0) = n²π²ħ²/2ma².
- l = 1: u = r j₁(kr) ∝ sin(kr)/(kr) − cos(kr) behaves as kr²/3 ∝ r(l+1) near the origin, as regularity requires. u(a) = 0 gives sin(ka)/(ka) = cos(ka), i.e. tan(ka) = ka, first root ka = 4.4934.
- So E₁₁ = 4.4934² ħ²/2ma² = 20.19 ħ²/2ma². Likewise E₁₂ = 5.7635² ħ²/2ma² = 33.22 ħ²/2ma² for l = 2, while the second l = 0 level is 4π² = 39.48 in the same unit.
- For an electron with a = 1.00 nm, ħ²/2ma² = 0.0381 eV: E₁₀ = 9.870 × 0.0381 = 0.376 eV, E₁₁ = 0.769 eV, E₁₂ = 1.266 eV, E₂₀ = 1.504 eV, with degeneracies 2l + 1 = 1, 3, 5 and 1.
AnswerE_(n0) = n²π²ħ²/2ma²; the lowest l = 1 level has ka = 4.4934. For a = 1.00 nm: 0.376 eV (×1), 0.769 eV (×3), 1.266 eV (×5), 1.504 eV (×1).