University Physics IV · Angular Momentum and Spin · 9.3
Spherical Harmonics & Integer l
Every central-force problem in this course hands you the same angular factor, so it is worth solving once and keeping. Here we separate it, learn to read nodes and parity straight off it, and settle the question textbooks usually fudge: why orbital l comes out an integer when spin one-half plainly does not.
Build the model
Connect the measurement to the mechanism.
Write the Schrödinger equation for any potential that depends on r alone and the angles fall out on their own: what survives of the Laplacian after the radial derivatives is exactly −L̂²/ħ²r², so the angular problem is L̂²Y = ħ²l(l+1)Y and it never sees V(r). Solve it once and you own the angular half of hydrogen, of the isotropic oscillator, of the finite spherical well and of the shell-model nucleus. Two further separations do it. The azimuth gives Φ = e(imφ) because L̂z = −iħ∂/∂φ; the polar factor obeys the associated Legendre equation, whose power series must terminate or blow up at the poles, and termination at l − |m| forces integer l ≥ |m|.
But that last step is only as strong as the claim that m is an integer, and the usual justification — that the wavefunction must return to itself as φ advances by 2π — is not sound, because an overall sign is unobservable and spinors flip theirs routinely. The honest argument is algebraic: build the l = ½ tower explicitly and it refuses to close. L̂₋ does not annihilate the m = −½ state it produces, and one more lowering yields a function whose norm on the sphere is infinite. The cost is worth naming.
This excludes half-integers only for angular momentum represented by functions of position. Spin has no such function to break, which is exactly why s = ½ survives.
- Simple definition
- The spherical harmonics Yₗm(θ, φ) are the simultaneous eigenfunctions of L̂² and L̂z on the unit sphere, and they exist only for integer l ≥ 0, with m running in integer steps from −l to +l.
- Example
- Y₂⁰ = √(5/16π)(3cos²θ − 1) has its two nodal cones where 3cos²θ = 1, that is at θ = 54.74° and 125.26°; it is one of the 2l + 1 = 5 states of the l = 2 level, all of parity (−1)² = +1.
The separation constant is everything the radial equation learns about the angles; it arrives as the centrifugal term ħ²l(l+1)/2mr².
θ polar, φ azimuthal, both dimensionless; L̂² carries units of J² s², with ħ = 1.055 × 10⁻³⁴ J s.
One separation constant per coordinate: m is fixed by the azimuth before the polar equation is even written down.
m is a pure number; the measured projection is mħ, in J s. Φ is normalised over 0 ≤ φ < 2π.
The Frobenius ratio tends to 1, so the series must stop at k = l − |m| or P blows up at the poles — termination, not a wall, quantises l.
x = cosθ runs over [−1, 1]. Solutions regular at both poles exist only for integer l ≥ |m|.
Orthonormality on the sphere is what lets any angular function be expanded in harmonics, and what makes selection rules computable.
Dimensionless in θ, φ; ∫Y*_(l′m′) Y_(lm) dΩ = δ_(ll′)δ_(mm′) over 4π sr. Y₀⁰ = 1/√(4π) = 0.2821.
The tool that kills half-integer l: it turns the closure condition into a first-order equation you can solve on sight.
On Θ(θ)e(imφ): L̂₊ → ħ(Θ′ − m cotθ Θ)e(i(m+1)φ); L̂₋ → −ħ(Θ′ + m cotθ Θ)e(i(m−1)φ).
Parity (−1)l is what forces Δl = ±1 on electric-dipole transitions; the 2l + 1 count is exact for every central potential.
Node counts are angular surfaces, pure numbers. The |m| planes appear in the real combinations, not in |Yₗm|².
The angles come off any central potential
In spherical coordinates the Laplacian splits cleanly: ∇² = (1/r²)∂ᵣ(r²∂ᵣ) − L̂²/(ħ²r²), where L̂² is exactly the angular operator −ħ²[(1/sinθ)∂θ(sinθ∂θ) + (1/sin²θ)∂²φ]. Put ψ = R(r)Y(θ, φ) into −(ħ²/2m)∇²ψ + V(r)ψ = Eψ and the angular part detaches as L̂²Y = ħ²l(l+1)Y, with l(l+1) the separation constant. Notice what is missing from that equation. There is no V, no E, no particle mass — nothing that could carry an energy. The angular problem is pure geometry on the unit sphere, so its solutions are identical for hydrogen, for the isotropic oscillator, for a finite spherical well and for the shell-model nucleus. All the angular half ever exports to the radial equation is one number, entering as the centrifugal term ħ²l(l+1)/2mr². Solve it once and you have solved it for every central force there will ever be.
Two more separations: the azimuth, then the pole
Write Y(θ, φ) = Θ(θ)Φ(φ). Because L̂z = −iħ∂/∂φ commutes with L̂², we may demand L̂zY = mħY as well, and that fixes Φ = e(imφ)/√(2π) at once. What remains is the associated Legendre equation, which in x = cosθ reads d/dx[(1 − x²)dP/dx] + [l(l+1) − m²/(1 − x²)]P = 0. Pull out the factor (1 − x²)(|m|/2) that the m² term demands and expand the rest as a power series; the recursion ratio a_(k+2)/aₖ tends to 1 for large k, and a series behaving that way diverges at x = ±1 — the two poles — unless it terminates. It terminates only when l = |m| + k for some non-negative integer k, so given integer m, integer l follows and l ≥ |m| comes free. The first few results are Y₀⁰ = 1/√(4π) = 0.2821, Y₁⁰ = √(3/4π)cosθ = 0.4886 cosθ, and Y₂⁰ = √(5/16π)(3cos²θ − 1).
The single-valuedness argument is not the argument
Everything above hangs on m being an integer, and the reason usually given is that ψ must return to itself when φ advances by 2π. That is a postulate dressed as a derivation, and a suspect one. Half-integer m gives Φ(φ + 2π) = −Φ(φ), an overall sign, and an overall phase changes no probability, no expectation value and no interference pattern within a single tower. Worse for the argument, nature uses exactly that behaviour: a spin-½ spinor picks up a minus sign under a 2π rotation and needs 4π to come home, and that sign has been seen directly in neutron interferometry. Nor can the half-integer candidate be dismissed as too singular to normalise. The l = ½ top rung works out to f = (1/π)√(sinθ)e(iφ/2), and ∫|f|²dΩ = (1/π²)(2π)∫₀π sin²θ dθ = (1/π²)(2π)(π/2) = 1. Finite, smooth away from the poles, and perfectly respectable-looking. If half-integer l is to be excluded, something else must do the excluding.
Kill the l = ½ tower with the ladder itself
Use L̂_± = ±ħe(±iφ)(∂θ ± i cotθ ∂φ). Acting on Θ(θ)e(imφ), L̂₊ returns ħ(Θ′ − m cotθ Θ)e(i(m+1)φ) and L̂₋ returns −ħ(Θ′ + m cotθ Θ)e(i(m−1)φ). Demand L̂₊f = 0 at the top of a j = ½ tower and you get Θ′/Θ = ½cotθ, so Θ ∝ √(sinθ) — the normalisable state just quoted. Lower it once: L̂₋f = −ħ(1/π)cosθ(sinθ)(−1/2)e(−iφ/2), whose norm is exactly ħ, matching the ħ√[j(j+1) − m(m−1)] = ħ√(¾ + ¼) = ħ that the algebra demands. Still consistent. Now the tower must stop, because m = −½ is the bottom rung and L̂₋ has to annihilate it. It does not: the bracket comes out −(sinθ)¹⁄² − cos²θ(sinθ)(−3/2), which vanishes nowhere on 0 < θ < π. And that leftover is not even on the sphere — its norm integrand is sin²θ + 2cos²θ + cos⁴θ/sin²θ, whose last term integrates like ∫dθ/θ² at θ → 0. The ladder neither closes nor stays normalisable. That is the exclusion.
Reading a harmonic: nodes, parity, degeneracy
Pₗ^|m|(cosθ) factorises as (sinθ)^|m| times a polynomial of degree l − |m| in cosθ, so |Yₗm|² has l − |m| nodal cones at fixed polar angles. For Y₂⁰ they sit where 3cos²θ = 1, at θ = 54.74° and 125.26° — the magic angle that NMR spinning exploits. The azimuthal nodes need care. |Yₗm|² is independent of φ, so the complex harmonic has none at all; take the real combinations that chemists draw, carrying cos(mφ) or sin(mφ), and |m| nodal planes containing the z axis appear. Either way the total comes to l angular nodal surfaces. Parity is a one-line calculation: r → −r means θ → π − θ and φ → φ + π, giving (−1)(l+|m|) from the Legendre factor and (−1)m from the exponential, hence (−1)l overall. Because the electric-dipole operator is odd, that parity is precisely what forces Δl = ±1. And every l carries 2l + 1 states, an exact degeneracy for any central potential; hydrogen's extra n²-fold degeneracy is a peculiarity of the 1/r potential, not a general rule.
What this settles, and what it deliberately does not
The argument excludes half-integer l for angular momentum built from position and momentum — anything, that is, whose eigenstates must be functions on the sphere. It says nothing about angular momentum that has no such representation. Spin obeys identical commutators, so the ladder derivation of j(j+1)ħ² and 2j+1 projections applies to it unchanged, but there is no Θ(θ) for the l = ½ calculation to break, and s = ½ survives untouched. That is the honest statement of where the sphere's authority ends. Two further limits are worth naming. The angular equation returns no energy at all, so nothing here predicts a spectrum: the 2l + 1 degeneracy within a level is exact, and everything else waits on V(r) and the radial equation. And that degeneracy is a statement about a rotationally symmetric Hamiltonian — switch on a magnetic field and it lifts, which is the entire content of the Zeeman effect coming next.
Change one variable at a time
Make the relationship visible.
Set j = 1: rung 3 lies flat on the axis, which is the lowering operator annihilating the bottom rung and the ladder closing. Now set j = 0.5, where rung 2 should be that same flat zero — instead it dives off the frame at both poles, and dragging θ₀ to 3° sends its readout to −19.
STATES NEEDED 2j + 12
LOWERINGS TO BOTTOM 2j1
RUNG 2 AT θ₀-9.57
RUNG 3 AT θ₀273.07
Live interpretationSTATES NEEDED 2j + 1: 2. LOWERINGS TO BOTTOM 2j: 1. RUNG 2 AT θ₀: −9.57. RUNG 3 AT θ₀: 273.07
Catch the common trap
Explain before calculating.
A student argues that orbital angular momentum could have l = ½, since L̂² would then have the perfectly respectable eigenvalue ¾ħ². What actually rules l = ½ out?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe l = 1, m = 0 spherical harmonic has the form Y₁⁰ = N cosθ. Normalise it over the sphere, then find the probability that a measurement of the polar angle returns θ < 60°.
- Normalisation runs over the full solid angle, dΩ = sinθ dθ dφ: ∫|Y₁⁰|² dΩ = N² ∫₀(2π) dφ ∫₀π cos²θ sinθ dθ.
- Substitute u = cosθ, du = −sinθ dθ: ∫₀π cos²θ sinθ dθ = ∫₋₁¹ u² du = 2/3. So N²(2π)(2/3) = 4πN²/3 = 1, giving N = √(3/4π) = 0.4886.
- Cut the same integral at θ = 60°, where u runs from 1 down to cos 60° = 1/2: ∫₀(π/3) cos²θ sinθ dθ = ∫_(1/2)¹ u² du = (1 − 1/8)/3 = 7/24 = 0.2917.
- So P(θ < 60°) = N²(2π)(7/24) = (3/4π)(2π)(7/24) = (3/2)(7/24) = 7/16 = 0.4375.
AnswerN = √(3/4π) = 0.4886, and P(θ < 60°) = 7/16 = 0.4375. That is 87.5% of the upper hemisphere's probability packed within 60° of the +z axis — the honest content of the claim that a pz orbital points along z.
MediumFor l = 3, locate every angular node of Y₃⁰ and of the real l = 3, |m| = 1 harmonic, then state the parity and the degeneracy of the level. Use P₃(cosθ) ∝ cosθ(5cos²θ − 3) and P₃¹(cosθ) ∝ sinθ(5cos²θ − 1).
- Y₃⁰ vanishes where cosθ(5cos²θ − 3) = 0. One root is cosθ = 0, giving θ = 90.00°.
- The others satisfy cos²θ = 3/5, so cosθ = ±0.7746 and θ = 39.23° and 140.77°. That is l − |m| = 3 − 0 = 3 nodal cones, and with |m| = 0 there are no nodal planes: three angular nodal surfaces in all.
- For |m| = 1 the sinθ prefactor vanishes only on the z axis, which is a line and not a cone, so the cones come from 5cos²θ = 1: cosθ = ±1/√5 = ±0.4472, giving θ = 63.43° and 116.57° — two cones, exactly as l − |m| = 2 requires.
- The complex Y₃(±1) has |Y|² independent of φ, so its |m| = 1 nodal plane shows up only in the real combination ∝ sinθ(5cos²θ − 1)cosφ, which vanishes where cosφ = 0, that is on the single plane φ = 90° and 270°. Two cones plus one plane is again l = 3 surfaces.
- Parity is (−1)l = (−1)³ = −1, and the degeneracy of the l = 3 level is 2l + 1 = 7, exact for any central potential.
Answerm = 0: three nodal cones, at θ = 39.23°, 90.00° and 140.77°. |m| = 1: two cones at 63.43° and 116.57°, plus one nodal plane through the z axis in the real combination. Either way l = 3 angular nodal surfaces, parity −1, and 2l + 1 = 7 degenerate states.
HardUsing L̂_± = ±ħe(±iφ)(∂θ ± i cotθ ∂φ), try to build an orbital tower with j = ½. Find the top rung from L̂₊f = 0, normalise it, lower it once and check the norm against the algebra, then test whether the ladder closes at m = −½.
- The top rung has m = +½, so write f = Θ(θ)e(iφ/2). Then L̂₊f = ħ(Θ′ − ½cotθ Θ)e(3iφ/2), and setting it to zero gives Θ′/Θ = ½cotθ, so Θ = A√(sinθ).
- Normalise: ∫|f|² dΩ = A² ∫₀(2π) dφ ∫₀π sinθ · sinθ dθ = A²(2π)(π/2) = A²π², so A = 1/π. The candidate is square-integrable — nothing has gone wrong yet, and the single-valuedness objection is not available to us.
- Lower once. For general m, L̂₋(Θe(imφ)) = −ħ(Θ′ + m cotθ Θ)e(i(m−1)φ); with m = ½ and Θ = A√(sinθ) the bracket is A cosθ(sinθ)(−1/2), so L̂₋f = −ħA cosθ(sinθ)(−1/2) e(−iφ/2).
- Check its norm: ħ²A²(2π)∫₀π (cos²θ/sinθ) sinθ dθ = ħ²(1/π²)(2π)(π/2) = ħ². The algebra demanded ħ√[j(j+1) − m(m−1)] = ħ√(¾ + ¼) = ħ. Still perfectly consistent.
- Now close the ladder. m = −½ is the bottom rung, so L̂₋ must annihilate it. With Θ₂ = cosθ(sinθ)(−1/2) and m = −½ the bracket is Θ₂′ − ½cotθ Θ₂ = −(sinθ)¹⁄² − cos²θ(sinθ)(−3/2), which is nowhere zero on 0 < θ < π.
- That leftover is not even a function on the sphere: its norm integrand is sin²θ + 2cos²θ + cos⁴θ/sin²θ, and the last term behaves as 1/θ² near θ = 0, so the integral diverges at both poles.
AnswerThe j = ½ tower fails at closure: L̂₋ does not annihilate the m = −½ rung, and the state it produces has infinite norm on the sphere. Half-integer l is excluded for orbital motion — nothing here touches spin, which has no wavefunction on the sphere to break.