University Physics IV · The Hydrogen Atom · 10.2
The Angular Equation & Spherical Harmonics
Do this part once and you never do it again. The angles in a central-force problem know nothing about the force, so the same Yₗm close the sphere for hydrogen, for a nuclear well, for a quark pair. What you buy here is the pair of boundary conditions that turn two arbitrary separation constants into integers.
Build the model
Connect the measurement to the mechanism.
Write ψ = R(r)Y(θ, φ) in a potential that depends on r alone and the Schrödinger equation splits, because the angular part of the Laplacian is nothing but −L²/(ħ²r²). What is left on the sphere contains no V, no E and no mass: it is the eigenvalue problem L²Y = λħ²Y, so its solution is common property, computed once for every central potential there will ever be. Solving it costs two boundary conditions and nothing else.
Demanding that Φ(φ) = e(imφ) come back to itself after a full turn forces m to be a whole number, which is the statement Lz = mħ. Demanding that the θ solution stay finite at the poles — where the associated Legendre equation is singular — truncates its series and forces λ = l(l+1) with l a non-negative integer and l ≥ |m|. Neither integer was postulated; both are the price of a wavefunction you can normalise.
The cost is that only one component of L can be sharp at a time, since [Lₓ, Ly] = iħLz, so a state carries √(l(l+1)) ħ of total angular momentum but only mħ of it along z, and the rest is genuinely undetermined rather than merely unmeasured. The payoff is that a three-dimensional partial differential equation is now one ordinary radial equation, and the labels n, l, m are exactly the eigenvalues of the three commuting operators H, L² and Lz.
- Simple definition
- The angular equation is what is left of a central-potential Schrödinger equation once the radial part is separated off; its normalisable solutions are the spherical harmonics Yₗm(θ, φ), eigenfunctions of L² with eigenvalue l(l+1)ħ² and of Lz with eigenvalue mħ.
- Example
- For l = 2, m = 1: L² returns 2(3)ħ² = 6ħ², so |L| = √6 ħ = 2.449ħ while Lz = 1ħ, and the vector sits at arccos(1/√6) = 65.9° from the z axis. Even the top rung, m = 2, only closes to 35.3°.
Separation works because the entire angular dependence sits in one operator that commutes with any V(r).
L² = −ħ²[(1/sinθ) ∂θ(sinθ ∂θ) + (1/sin²θ) ∂²φ]; L² has units J² s²
No V, no E and no mass appear, so one solution set serves every central potential ever written down.
λ is the dimensionless separation constant; l = 0, 1, 2, … labels it
Integer m is a boundary condition, not a postulate: the seam mismatch 2|sin πm| must close.
Lz = −iħ ∂φ, so LzΦ = mħΦ; m is a pure number, Lz is in J s
Solutions finite at both poles exist only for λ = l(l+1) with integer l ≥ |m| — that is where l comes from.
x = cos θ runs over [−1, 1]; x = ±1 are the two poles, where the equation is singular
Y₀⁰ = 1/√(4π) = 0.2821 is flat; Y₁⁰ = 0.4886 cos θ already has a nodal cone.
orthonormal on the sphere: ∫ Y*ₗ'm' Yₗₘ dΩ = δₗ'l δₘ'm, with dΩ = sinθ dθ dφ
|L| always exceeds the best projection lħ, so L never lies along z: that gap is [Lₓ, Ly] ≠ 0.
−l ≤ m ≤ l, so one l carries 2l+1 states of identical energy
Separation needs only that V depends on r
In spherical polars the Laplacian splits into a radial block and an angular block, and the angular block is not a new object: ∇² = (1/r²)∂ᵣ(r²∂ᵣ) − L²/(ħ²r²), with L² = −ħ²[(1/sinθ)∂θ(sinθ ∂θ) + (1/sin²θ)∂²φ]. Put ψ = R(r)Y(θ, φ) into −(ħ²/2μ)∇²ψ + V(r)ψ = Eψ, multiply through by 2μr²/(ħ²RY), and every surviving term depends either on r alone or on the angles alone. Two functions of independent variables can agree only if both equal the same constant; call it λ, so that the angular side reads L²Y = λħ²Y. That is the whole trick, and notice what it did not require. It did not require a point charge, an inverse-square force, or a clamped nucleus. It required exactly one thing: that V is a function of r and of nothing else, so that V never appears on the angular side of the equation. Break that — apply a field along z, or add a second electron — and the split fails immediately.
Round the axis once: single-valuedness fixes m
Separate again, Y(θ, φ) = Θ(θ)Φ(φ). The azimuth enters only through ∂²φ, so Φ″ = −m²Φ and Φ(φ) = e(imφ), where at this stage m is any complex number the equation will accept. Physics now imposes something the differential equation cannot see: φ and φ + 2π label the same direction in space, so a single-valued ψ demands e(2πim) = 1. Only whole numbers pass. The failure is quantitative — the mismatch is |e(2πim) − 1| = 2|sin πm|, which is 1.414 at m = 1/4 and 2 at m = 1/2, and falls to zero exactly on the integers. Since Lz = −iħ ∂φ, the same line reads LzΦ = mħΦ: the projection of angular momentum on a chosen axis comes in whole units of ħ, derived rather than assumed. Half-integer m cannot appear here, because no single-valued function on a sphere carries it; spin gets its half-integers from the commutation algebra, which has no position representation to be single-valued in.
The poles are singular, and regularity truncates the series
What remains is (1/sinθ) d/dθ(sinθ dΘ/dθ) + [λ − m²/sin²θ]Θ = 0. Substitute x = cos θ and it becomes the associated Legendre equation, d/dx[(1−x²)dP/dx] + [λ − m²/(1−x²)]P = 0. The leading coefficient (1−x²) vanishes at x = ±1 — the north and south poles — so those are regular singular points. A power series built about x = 0 converges for |x| < 1 but generically diverges like ln(1−x²) at the ends, and a wavefunction that blows up along the polar axis cannot be normalised over the sphere. So the series must terminate. Pushing the two-term recursion through shows that it terminates only when λ = l(l+1) with l a non-negative integer obeying l ≥ |m|. One boundary condition, two results: the L² eigenvalue is l(l+1)ħ² rather than l²ħ², and the number of m values riding on a given l is capped at 2l+1. For l = 2 that means λ = 6 and five states.
Reading a harmonic: nodes, parity, orthonormality
The normalised product is Yₗm(θ, φ) = N Pₗ^|m|(cos θ) e(imφ) with N = √[(2l+1)(l−|m|)!/(4π(l+|m|)!)]. Count the nodes and the two labels become visible. Pₗ^|m|(cos θ) has l − |m| zeros strictly between the poles, so the density carries l − |m| nodal cones, while the e(imφ) factor supplies |m| nodal planes once you take the real combinations that give the familiar pₓ, py and dxy pictures. Y₁⁰ = 0.4886 cos θ has its single cone flattened into the xy plane at θ = 90°; Y₂⁰ ∝ 3cos²θ − 1 has two, at cos θ = ±1/√3, that is θ = 54.7° and 125.3°. Parity comes free: inversion sends (θ, φ) to (π−θ, φ+π) and Yₗm to (−1)l Yₗm, which is the seed of the Δl = ±1 dipole rule two topics from here. And the set is orthonormal over the sphere, so any angular function whatever can be expanded in it.
Three commuting operators, three labels
For a central potential [H, L²] = 0, [H, Lz] = 0 and [L², Lz] = 0, so all three share one complete set of eigenfunctions and a state may carry all three eigenvalues at once. You cannot do better than three: [Lₓ, Ly] = iħLz is not zero, so a second component may not join the set, and the z axis has to be chosen by hand. Three commuting observables for a three-dimensional problem is exactly the right count, and it is why a hydrogen state carries the labels n, l, m and no others until spin adds a fourth. One consequence is immediate. The angular equation never mentions the energy, so all 2l+1 values of m must give identical E — the m-degeneracy shared by every central potential, which breaks only when a field or a neighbouring atom singles out a direction. Unsöld's theorem makes it concrete: summing |Yₗm|² over all m gives (2l+1)/4π, a constant, so a filled subshell is exactly spherical.
One sphere, every central force
Nothing in L²Y = l(l+1)ħ²Y mentions the potential, the mass or the energy. The same Yₗm therefore solve the angular half of the hydrogen atom, the isotropic three-dimensional oscillator, a finite spherical well, the deuteron and a charmonium quark pair; only R(r) and the energies differ, and the labour of this topic is spent once for all of them. State that as a cost as well as a benefit. Because the angular solution is fixed before any dynamics is specified, it can say nothing about the spectrum: the whole physics of the force law lives in the radial equation, which is the next topic. And the moment the potential stops being central the construction fails. An external electric or magnetic field, a molecular framework, a second electron — each mixes different m, and usually different l too, so Yₗm survives as a basis to expand in rather than as a solution.
Change one variable at a time
Make the relationship visible.
Set l = 2 and step m from −2 to +2: the arrow tilts but never stands upright, because √6 ħ always beats 2ħ. Then push m past l — try l = 1, m = 3 — and the arrow shoots above the dashed ceiling while l(l+1) − m² turns negative. That is the state regularity at the poles refuses to allow.
|L| = √(l(l+1)) ħ2.449 ħ
Lz = m ħ1 ħ
L⊥² = l(l+1) − m²5 ħ²
ORIENTATIONS 2l + 15 states
Live interpretation|L| = √(l(l+1)) ħ: 2.449 ħ. Lz = m ħ: 1 ħ. L⊥² = l(l+1) − m²: 5 ħ². ORIENTATIONS 2l + 1: 5 states
Catch the common trap
Explain before calculating.
Separating the Schrödinger equation for a central potential gives the azimuthal factor Φ(φ) = e(imφ). What forces m to be an integer?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA particle moves in a central potential in a state with l = 3. Find the magnitude of its orbital angular momentum, list the allowed values of m and count them, and find the smallest angle the vector L can make with the z axis.
- The angular equation gives L²Y = l(l+1)ħ²Y, so with l = 3 the eigenvalue is 3(4)ħ² = 12ħ² and |L| = √12 ħ = 2√3 ħ = 3.464ħ.
- Regularity at the poles allows only l ≥ |m|, so m = −3, −2, −1, 0, +1, +2, +3 — that is 2l + 1 = 7 states, all with the same energy because the angular equation never mentions E.
- Lz = mħ, so the largest projection available is 3ħ. Note that 3ħ < 3.464ħ: the projection cannot equal the magnitude.
- cos θₘᵢₙ = mₘₐₓ/√(l(l+1)) = 3/3.4641 = 0.86603 = √3/2, so θₘᵢₙ = arccos(√3/2) = 30.0° exactly.
Answer|L| = √12 ħ = 3.464ħ; seven states, m = −3 … +3; the closest L ever comes to the z axis is 30.0°.
MediumShow that Y₁⁰(θ, φ) = √(3/4π) cos θ is normalised over the sphere and orthogonal to Y₀⁰ = 1/√(4π). Then find ⟨cos²θ⟩ in the state Y₁⁰ and compare it with the value for Y₀0.
- The solid angle element is dΩ = sinθ dθ dφ. With x = cos θ, sinθ dθ = −dx, so every angular integral becomes ∫₀²π dφ ∫₋₁+1 … dx.
- Normalisation: ∫|Y₁⁰|² dΩ = (3/4π)(2π)∫₋₁+1 x² dx = (3/4π)(2π)(2/3) = 1. The factor √(3/4π) exists precisely to make this come out at 1.
- Orthogonality: ∫ Y₀⁰* Y₁⁰ dΩ = (1/√(4π))√(3/4π)(2π)∫₋₁+1 x dx = 0, because x is odd on a symmetric interval. Different l cannot overlap.
- Expectation value: ⟨cos²θ⟩ = (3/4π)(2π)∫₋₁+1 x⁴ dx = (3/2)(2/5) = 3/5 = 0.600.
- For the isotropic Y₀⁰ the same integral gives (1/4π)(2π)(2/3) = 1/3 = 0.333. The pz density is pulled towards the axis, exactly as its single nodal cone in the xy plane suggests.
AnswerNormalised (the integral is 1) and orthogonal to Y₀⁰ (∫x dx = 0 over [−1, 1]); ⟨cos²θ⟩ = 3/5 = 0.600 for Y₁⁰, against 1/3 = 0.333 for the isotropic s state.
HardCarbon monoxide can be modelled as a rigid rotor: a C and an O atom held at a fixed separation r = 0.1128 nm, free to rotate, with V = 0. Find the frequency and wavelength of the l = 0 → 1 microwave absorption line, and say how many states the upper level holds. Take mC = 12.000 u, mO = 15.995 u.
- Freezing r removes the radial equation entirely, so Ĥ = L²/(2I) and the eigenfunctions are precisely the spherical harmonics, with Eₗ = ħ²l(l+1)/(2I). The angular equation has already solved the whole problem.
- Reduced mass: μ = mC mO/(mC + mO) = (12.000 × 15.995)/27.995 u = 6.856 u = 6.856 × 1.6605 × 10⁻²⁷ = 1.1385 × 10⁻²⁶ kg.
- Moment of inertia: I = μr² = 1.1385 × 10⁻²⁶ × (1.128 × 10⁻¹⁰)² = 1.1385 × 10⁻²⁶ × 1.2724 × 10⁻²⁰ = 1.4486 × 10⁻⁴⁶ kg m².
- ΔE = E₁ − E₀ = ħ²[1(2) − 0(1)]/(2I) = ħ²/I = 1.1121 × 10⁻⁶⁸ / 1.4486 × 10⁻⁴⁶ = 7.677 × 10⁻²³ J, or 0.479 meV.
- ν = ΔE/h = 7.677 × 10⁻²³ / 6.626 × 10⁻³⁴ = 1.159 × 10¹¹ Hz = 115.9 GHz, so λ = c/ν = 2.998 × 10⁸ / 1.159 × 10¹¹ = 2.59 mm.
- Degeneracy: l = 0 holds 2(0)+1 = 1 state and l = 1 holds 2(1)+1 = 3, with m = −1, 0, +1. Those three stay degenerate because nothing in the angular equation picks out an axis; a magnetic field would split them.
AnswerΔE = ħ²/I = 7.68 × 10⁻²³ J, giving ν = 116 GHz and λ = 2.59 mm — against the observed CO line at 115.271 GHz, the 0.5% gap being vibrational averaging of r. The upper level holds 2l+1 = 3 degenerate m states.