University Physics V · The Hydrogen Atom · 11.2

Angular Wavefunctions & the L², L_z Eigenbasis

The angular half of every central-field problem is solved once and never again. This lesson builds that basis out of the commutator algebra plus a single boundary condition in φ, and the 2l+1 block it produces is the same object whether the potential is the 1/r of hydrogen, the r² of an isotropic trap, or the Woods–Saxon edge of a nucleus.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

L² and L_z contain no r and no ∂ᵣ, so they act on the L²(S²) factor of the state space alone and commute with every function of the radius. That one structural fact makes the angular problem universal: solve it once and it belongs to hydrogen, the isotropic oscillator, a finite spherical well and a Woods–Saxon nucleus alike. The solution takes two separate inputs.

The commutator algebra [Lᵢ, Lⱼ] = iħ εᵢⱼₖ Lₖ on its own, worked with ladder operators, gives L²|l m⟩ = ħ²l(l+1)|l m⟩ and L_z|l m⟩ = mħ|l m⟩ with m running from −l to l in unit steps — and it permits half-integers. The sphere then supplies what the algebra cannot: an amplitude Yₗ^m(θ, φ) = ⟨n̂|l m⟩ must be one function on S², so Φ(φ + 2π) = Φ(φ) forces m to be an integer, and l with it. Half-integer angular momentum survives only where there is no ⟨n̂| to evaluate, which is exactly where spin lives.

What you buy is a complete orthonormal basis of directions, a rotation-irreducible block of dimension 2l + 1 at each l, definite parity (−1)^l, and the free (2l+1)-fold degeneracy of every central field. What you pay is that the answer is blind to dynamics: the angular calculation delivers one number, ħ²l(l+1), into the centrifugal term, and every energy in the problem is settled afterwards, by V(r).

Simple definition: The spherical harmonics Yₗ^m(θ, φ) = ⟨n̂|l m⟩ are the simultaneous eigenfunctions of L² and L_z on the unit sphere, with eigenvalues ħ²l(l+1) and mħ, and they form a complete orthonormal basis for every square-integrable function of direction.
Example: For l = 1 the block is three-dimensional: Y₁⁰ = √(3/4π) cos θ = 0.4886 cos θ has L_z = 0, while Y₁^±1 = ∓√(3/8π) sin θ e^(±iφ) = ∓0.3455 sin θ e^(±iφ) have L_z = ±ħ — and all three share |L| = √2 ħ = 1.414 ħ.

L² and L_z on the sphereL_z = −iħ ∂/∂φ, L² = −ħ²[(1/sinθ) ∂_θ(sinθ ∂_θ) + (1/sin²θ) ∂²_φ]

Neither operator holds r, so both act on the L²(S²) factor alone and commute with every V(r).

θ, φ in rad; L² in J² s². The bracket is the Laplace–Beltrami operator of the unit sphere.

The simultaneous eigenvalue problemL²|l m⟩ = ħ²l(l+1)|l m⟩, L_z|l m⟩ = mħ|l m⟩, Yₗ^m(θ, φ) = ⟨n̂|l m⟩

Two commuting labels: l names the block, m names the state inside it. Neither mentions the potential.

l = 0, 1, 2, …; m = −l, …, +l; Yₗ^m in sr^(−1/2), since ∫|Yₗ^m|² dΩ = 1.

The φ condition that forces integersΦ(φ + 2π) = Φ(φ) ⇒ Φₘ = e^(imφ)/√(2π), m = 0, ±1, ±2, …

Integer m forces integer l, so orbital motion has no l = ½ — an exclusion owed to the sphere, not to the commutators.

The only place integrality enters. The commutators alone allow l and m in half-integer steps.

Ladder construction of the basisL_± = ħe^(±iφ)(±∂_θ + i cotθ ∂_φ), L₊ Yₗ^l = 0 ⇒ Yₗ^l ∝ sin^l θ e^(ilφ)

One first-order ODE plus l − m lowerings replaces the whole Legendre series solution.

Then Yₗ^m ∝ (L_−)^(l−m) Yₗ^l, with L_−|l m⟩ = ħ√(l(l+1) − m(m−1))|l m−1⟩.

Explicit form and orthonormalityYₗ^m = √[(2l+1)(l−m)!/4π(l+m)!] Pₗ^m(cosθ) e^(imφ), ∫Yₗ'^m'* Yₗ^m dΩ = δₗₗ' δₘₘ'

The φ integral delivers δₘₘ' free; Legendre orthogonality at fixed m delivers δₗₗ'.

dΩ = sinθ dθ dφ over 4π sr; Pₗ^m the associated Legendre function; Y₀⁰ = 1/√(4π) = 0.2821.

Parity and the closure sumYₗ^m(π−θ, φ+π) = (−1)^l Yₗ^m(θ, φ), Σₘ |Yₗ^m|² = (2l+1)/4π

Parity kills Δl = 0 in dipole lines; the closure sum makes every filled subshell spherically symmetric.

r → −r leaves r = |r| untouched, so a central-field state has parity (−1)^l whatever R(r) does.

Two operators that never see the radius

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.

Where the integers come from, and where they do not

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.

Kill the top rung, then lower

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.

Orthonormality, completeness, and a worked expansion

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.

The 2l+1 block is irreducible, so the z axis is a choice

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.

What the angular solution buys, and what it costs

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.

Change one variable at a time

Make the relationship visible.

Interactive model

Angular-momentum label l2

Requested rung m (the ladder sticks at |m| = l)2

Push m past l and it sticks: the ladder stops at |m| = l. At l = 2 the top rung still leans 35.3° off the axis, because √6 ħ = 2.449 ħ beats 2 ħ — L is never parallel to z, and only l = 0, below this slider, makes all three components sharp.

|L| = √(l(l+1))2.449 ħ

L_z = m ħ2 ħ

TILT θ FROM z35.3 °

RUNGS 2l + 15 states

Live interpretation|L| = √(l(l+1)): 2.449 ħ. L_z = m ħ: 2 ħ. TILT θ FROM z: 35.3 °. RUNGS 2l + 1: 5 states

Catch the common trap

Explain before calculating.

Misconception check

The claim that the algebra only allows integer l is backwards. The commutators [Lᵢ, Lⱼ] = iħ εᵢⱼₖ Lₖ, worked with ladder operators and nothing else, give 2l = 0, 1, 2, … with half-integers included, and that is not an oversight — it is the reason spin-½ exists. Nor is the exclusion a matter of normalisability, which is the usual second guess. Build the l = ½ candidate and watch where it actually dies. Solving L₊Y = 0 on f(θ)e^(iφ/2) gives f ∝ sin¹⁄²θ, so the top rung is Y ∝ sin¹⁄²θ e^(iφ/2); the ladder coefficient ħ√(l(l+1) − m(m−1)) is ħ√(3/4 + 1/4) = ħ at l = m = ½, so one lowering hands back the lower rung −cosθ sin^(−1/2)θ e^(−iφ/2) with no extra factor. Both functions are square-integrable on the sphere — as written each returns ∫|Y|² dΩ = π² — so neither is thrown out for blowing up. What fails is closure: L_− on the lower rung gives −ħ[sin¹⁄²θ + cos²θ sin^(−3/2)θ]e^(−3iφ/2), which vanishes nowhere, so the ladder refuses to stop at m = −½, and that next function is no longer square-integrable at all. The pair never closes into a two-dimensional representation inside L²(S²). Spin escapes the whole argument because a spinor is a vector in C² with no amplitude at a direction to be single-valued — which is why a 2π rotation returns −1 on a spin-½ state and +1 on every orbital one.

A three-dimensional isotropic harmonic oscillator, the Coulomb potential and a finite spherical well are each solved by separation of variables. What is true of their angular factors?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn electron in a central field occupies the state with l = 3 and mₗ = −2. Give the eigenvalues of L² and L_z, the magnitude |L| in J s, the angle this state makes with the +z axis, the smallest angle any l = 3 state can reach, and the degeneracy of the level.

L²|l m⟩ = ħ²l(l+1)|l m⟩ with l = 3, so the eigenvalue is 12ħ². L_z|l m⟩ = mₗħ with mₗ = −2, so L_z = −2ħ. Both are sharp at once because [L², L_z] = 0.
|L| = √(l(l+1)) ħ = √12 ħ = 3.4641 ħ = 3.4641 × 1.0546 × 10⁻³⁴ = 3.653 × 10⁻³⁴ J s.
cos θ = L_z/|L| = −2/3.4641 = −0.5774, so θ = 125.3°: the cone opens below the equator because mₗ is negative.
The smallest tilt uses the top rung mₗ = l = 3: cos θ = 3/3.4641 = 0.8660, so θ = 30.0°. L never lies along z, since l < √(l(l+1)) for every l ≥ 1 — the transverse components have zero mean but variance ½(l(l+1) − m²)ħ².
The block holds 2l + 1 = 7 states, degenerate in any central field because H commutes with L_± as well as with L_z.

AnswerL² = 12ħ² and L_z = −2ħ, so |L| = 3.4641 ħ = 3.653 × 10⁻³⁴ J s at θ = 125.3° from +z. The closest any l = 3 state comes to the axis is 30.0°, and the level is 7-fold degenerate.

MediumAn electron's angular state is ψ(θ, φ) = N cos²θ. Normalise it on the sphere, expand it in spherical harmonics, give the probability of each outcome of a measurement of L² and of L_z, and find ⟨L²⟩.

Normalise: ∫|ψ|² dΩ = N² · 2π ∫ u⁴ du from −1 to 1 = N² · 2π · (2/5) = 4πN²/5 = 1, so N = √(5/4π) = 0.6308.
Rewrite in Legendre polynomials: P₀ = 1 and P₂(u) = (3u² − 1)/2, so cos²θ = 1/3 + (2/3)P₂(cosθ). In harmonics, 1 = √(4π) Y₀⁰ and P₂(cosθ) = √(4π/5) Y₂⁰.
Hence ψ = N[(√(4π)/3) Y₀⁰ + (2/3)√(4π/5) Y₂⁰] = (√5/3) Y₀⁰ + (2/3) Y₂⁰, and (√5/3)² + (2/3)² = 5/9 + 4/9 = 1 confirms the normalisation independently.
Measuring L²: P(0) = 5/9 = 0.556 and P(6ħ²) = 4/9 = 0.444, and no other l appears. Odd l are absent because ψ is even under parity, and nothing above l = 2 survives because ψ is quadratic in cos θ.
Measuring L_z: both terms carry m = 0, so L_z = 0 with certainty. The state is a superposition in l and an eigenstate of L_z at the same time.
⟨L²⟩ = (5/9)(0) + (4/9)(6ħ²) = 24ħ²/9 = 2.667 ħ².

AnswerN = √(5/4π) = 0.6308 and ψ = (√5/3) Y₀⁰ + (2/3) Y₂⁰. L² returns 0 with probability 5/9 and 6ħ² with probability 4/9; L_z returns 0 with certainty; ⟨L²⟩ = 24ħ²/9 = 2.667 ħ².

HardBuild the l = 2 harmonics from the algebra alone. Solve L₊Y₂² = 0 for Y₂² and normalise it, lower once with L_− to obtain Y₂¹, then verify by direct integration that the result is normalised. Use L_± = ħe^(±iφ)(±∂_θ + i cotθ ∂_φ).

Top rung: put Y₂² = f(θ)e^(2iφ) into L₊Y = 0. The φ derivative brings down 2i, so the bracket is f′ + i cotθ (2i) f = f′ − 2cotθ f = 0, giving f ∝ sin²θ. So Y₂² = A sin²θ e^(2iφ).
Normalise: ∫|Y₂²|² dΩ = |A|² · 2π ∫ sin⁵θ dθ from 0 to π = |A|² · 2π · (16/15) = 32π|A|²/15 = 1, so A = √(15/32π) = 0.3863.
Lower once: L_−(A sin²θ e^(2iφ)) = ħe^(−iφ)[−2A sinθ cosθ + i cotθ (2i) A sin²θ]e^(2iφ) = −4ħA sinθ cosθ e^(iφ), the two terms being equal because cotθ sin²θ = sinθ cosθ.
The ladder norm gives L_−|2 2⟩ = ħ√(l(l+1) − m(m−1))|2 1⟩ = ħ√(6 − 2)|2 1⟩ = 2ħ|2 1⟩, so Y₂¹ = (−4A/2) sinθ cosθ e^(iφ) = −2√(15/32π) sinθ cosθ e^(iφ) = −√(15/8π) sinθ cosθ e^(iφ) = −0.7725 sinθ cosθ e^(iφ).
Verify: ∫|Y₂¹|² dΩ = (15/8π) · 2π ∫ sin³θ cos²θ dθ = (15/4) ∫ (1 − u²)u² du from −1 to 1 = (15/4)(4/15) = 1. The ladder coefficient normalises the new state automatically.
The minus sign is the Condon–Shortley convention, a factor (−1)^m on positive m. It cancels in |Y₂¹|², but it must be carried consistently into Clebsch–Gordan tables and dipole matrix elements.

AnswerY₂² = √(15/32π) sin²θ e^(2iφ) = 0.3863 sin²θ e^(2iφ) and Y₂¹ = −√(15/8π) sinθ cosθ e^(iφ) = −0.7725 sinθ cosθ e^(iφ), the second normalised by direct integration to ∫|Y₂¹|² dΩ = 1.