University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.9
The addition theorem for spherical harmonics
Coulomb's 1/|r − r′| depends on the angle between two position vectors, which ties every direction of one particle to every direction of the other. This identity unties them, so that orthogonality can do the angular integrals. Helium's repulsion, the spherical closed shell and the partial-wave expansion all begin here.
Build the model
Connect the measurement to the mechanism.
The angle γ between two directions n̂ and n̂′ is one number, but it depends on all four angles θ, φ, θ′ and φ′ at once, and Pₗ(cos γ) mixes them inseparably. The addition theorem unmixes it: Pₗ(cos γ) = [4π/(2l + 1)] Σₘ Yₗᵐ*(θ′, φ′) Yₗᵐ(θ, φ), a sum of 2l + 1 products that each hold one direction only. For l = 1 it is nothing more than n̂⋅n̂′ written in spherical coordinates. The general proof takes two moves.
The m-sum over a complete multiplet is unchanged when both directions are rotated together, because a rotation only reshuffles the 2l + 1 states of one l among themselves; that property is quoted here and proved in Unit 5. So the sum can depend on the pair only through γ, and may be evaluated with n̂′ on the z axis, where every Yₗᵐ with m ≠ 0 vanishes and the single m = 0 term is (2l + 1)/(4π) Pₗ(cos γ). The result is exact and convention-free, since any phase on Yₗᵐ cancels against its conjugate. Its limits are just as definite.
Drop one value of m and the rotational invariance is gone, so the theorem needs the whole multiplet. Put into the multipole expansion, it turns 1/|r − r′| into a double sum over l and m whose angular factors separate, while the radial factor rₘᵢₙˡ/rₘₐₓˡ⁺¹ still couples r to r′ through its switch at r = r′.
- Simple definition
- The addition theorem states that the Legendre polynomial of the angle γ between two directions equals 4π/(2l + 1) times the sum, over all 2l + 1 values of m, of Yₗᵐ*(θ′, φ′) Yₗᵐ(θ, φ), a sum of terms that each depend on one direction only.
- Example
- Two equatorial directions, (θ, φ) = (90°, 0°) and (90°, 60°), have cos γ = cos 60° = 0.500. For l = 1 the m = 0 term vanishes and the m = ±1 pair gives (3/4π) × 0.500 = 0.119 sr⁻¹; 4π/3 times that returns P₁(cos γ) = 0.500.
Trades a function of γ, which ties the two directions together, for 2l + 1 products that each hold one direction. Derived here, given one rotation property quoted from Unit 5.
m runs −l … l; γ is the angle between n̂ and n̂′; each Yₗᵐ carries sr(−1/2), so the sum is in sr⁻¹ and Pₗ is a pure number
Derived from the l = 1 harmonics times 4π/3: m = 0 supplies cos θ cos θ′, and the m = ±1 pair adds twice the real part of ½ sin θ sin θ′ e(i(φ − φ′)), the sin θ sin θ′ cos(φ − φ′) term. Every m-sum needs this γ first.
θ, θ′ polar and φ, φ′ azimuthal angles, in rad or degrees; the right side is n̂⋅n̂′ in spherical coordinates
Put n̂′ on the z axis and only m = 0 survives, leaving (2l + 1)/(4π) Pₗ(cos θ) with θ = γ: the whole proof, once rotational invariance is granted.
θ = 0 is the +z pole; Pₗ(1) = 1, and every m ≠ 0 function carries sin(|m|)θ, zero there; in sr(−1/2)
Derived by setting n̂′ = n̂. A filled subshell, two electrons in each of the 2l + 1 values of m, has a spherical angular density, although no single |Yₗᵐ|² is spherical.
in sr⁻¹ and the same in every direction: 0.239 sr⁻¹ for l = 1, 0.398 sr⁻¹ for l = 2
Derived by putting the theorem into the Legendre series of topic 1.4. Directions now sit in separate factors, so angular integrals factorise; radii do not, as rₘᵢₙ and rₘₐₓ swap at r = r′.
rₘᵢₙ and rₘₐₓ are the smaller and larger of r and r′, in metres; l runs 0 … ∞; every term is in inverse metres
Derived: only l = 0 survives an average over a direction. This is the shell theorem, and it is why a spherical charge cloud interacts with anything through one radial integral.
dΩ′ = sin θ′ dθ′ dφ′ in sr; it follows from ∫ Yₗᵐ dΩ = √(4π) δₗ₀ δₘ₀ applied to the double sum
For l = 1 the theorem is the law of cosines
Take the l = 1 harmonics built earlier in this unit: Y₁⁰ = √(3/4π) cos θ and Y₁(±1) = ∓√(3/8π) sin θ e(±iφ). The m = 0 product is (3/4π) cos θ cos θ′. The m = +1 product is (3/8π) sin θ sin θ′ e(i(φ − φ′)): the two Condon-Shortley minus signs multiply to +1, and conjugating the primed factor turns e(iφ′) into e(−iφ′). The m = −1 product is its complex conjugate, so the pair adds to twice its real part, (3/4π) sin θ sin θ′ cos(φ − φ′). The total is (3/4π)[cos θ cos θ′ + sin θ sin θ′ cos(φ − φ′)] = (3/4π) n̂⋅n̂′, and 4π/3 times it is P₁(cos γ) = cos γ. Two facts hide in this line. Each m term depends on θ and θ′ separately, and the m = ±1 terms are complex, yet the sum is real and depends on the angles only through γ; and any phase attached to Yₗᵐ cancels against its own conjugate, so the convention cannot matter. Redo the sum with the real set √(3/4π) x/r, √(3/4π) y/r, √(3/4π) z/r and it is (3/4π) n̂⋅n̂′ in one step: the sum does not care which orthonormal basis spans the multiplet.
Why the m-sum can depend only on γ
Write the sum in Dirac form. With Yₗᵐ(n̂) = ⟨n̂|l, m⟩, it is Σₘ ⟨n̂|l, m⟩⟨l, m|n̂′⟩ = ⟨n̂|Πₗ|n̂′⟩, where Πₗ = Σₘ |l, m⟩⟨l, m| projects onto the whole l multiplet. Now rotate both directions by the same R. Quoted here and proved in Unit 5: the operator U(R) = exp(−iβ û⋅L/ħ), for a turn through angle β about the unit axis û, carries direction kets rigidly, U(R)|n̂⟩ = |Rn̂⟩. From Quantum Mechanics I, L² commutes with every component of L, so U(R) cannot change l; it only reshuffles the 2l + 1 states of the multiplet among themselves, which is the statement U†ΠₗU = Πₗ. Then ⟨Rn̂|Πₗ|Rn̂′⟩ = ⟨n̂|U†ΠₗU|n̂′⟩ = ⟨n̂|Πₗ|n̂′⟩, and the sum is unchanged. The only thing two directions keep under every common rotation is n̂⋅n̂′ = cos γ, so the sum is a function of γ alone. No single term has this property. The invariance belongs to the complete multiplet, and leaving out even one m destroys it.
Evaluate it with one direction on the z axis
Invariance lets you choose the axes. Turn them so that n̂′ is the +z pole, θ′ = 0; then θ, the polar angle of n̂, is γ itself. At the pole every Yₗᵐ with m ≠ 0 vanishes, because the associated Legendre function Pₗ(|m|)(x) carries the factor (1 − x²)(|m|/2) = sin(|m|)θ, and it has to: φ′ is undefined at the pole, so a continuous function carrying e(imφ′) must be zero there. The m = 0 function is Yₗ⁰ = √((2l + 1)/4π) Pₗ(cos θ), with Pₗ(1) = 1. One term survives, √((2l + 1)/4π) × √((2l + 1)/4π) Pₗ(cos γ) = [(2l + 1)/(4π)] Pₗ(cos γ), which is the theorem. Check the opposite limit as well. With n̂ = −n̂′, γ = π and Pₗ(−1) = (−1)ˡ; parity, Yₗᵐ(−n̂) = (−1)ˡ Yₗᵐ(n̂), turns the sum into (−1)ˡ Σₘ |Yₗᵐ(n̂′)|², which is the same theorem read at γ = 0. Everything in the proof is derived except the rotation property borrowed from Unit 5.
At γ = 0: Unsöld's rule and spherical closed shells
Put n̂′ = n̂. Then γ = 0, Pₗ(1) = 1, and the theorem says Σₘ |Yₗᵐ(θ, φ)|² = (2l + 1)/(4π) in every direction: 3/(4π) = 0.239 sr⁻¹ for l = 1 and 5/(4π) = 0.398 sr⁻¹ for l = 2. Check l = 1 by hand: (3/4π) cos²θ + 2 × (3/8π) sin²θ = 3/(4π), with no θ left. Each |Yₗᵐ|² on its own is lopsided, a dumbbell along z for m = 0 and a ring round the equator for m = ±1, yet the complete set sums to a sphere. Physically, a filled subshell, two electrons in each of the 2l + 1 values of m, has a spherically symmetric angular density, and that is derived here. It is why a closed core such as Na⁺ presents its valence electron with a central potential, the approximation Unit 7 builds on, and it is consistent with the measured ground terms of helium, neon and argon, all ¹S₀ in the spectroscopic tables. That closed shells carry L = 0 and S = 0 is quoted here; it follows from the exclusion principle in Unit 7.
The multipole expansion as a double sum over l and m
Topic 1.4 gave 1/|r − r′| = Σₗ (rₘᵢₙˡ/rₘₐₓˡ⁺¹) Pₗ(cos γ), with rₘᵢₙ and rₘₐₓ the smaller and larger of r and r′. Replace each Pₗ(cos γ) by its m-sum and 1/|r − r′| = Σₗ Σₘ [4π/(2l + 1)] (rₘᵢₙˡ/rₘₐₓˡ⁺¹) Yₗᵐ*(θ′, φ′) Yₗᵐ(θ, φ). Nothing was approximated, and each term is still an inverse length, since 4π/(2l + 1) is in sr and the two harmonics together in sr⁻¹. What changed is that the direction of r and the direction of r′ now sit in different factors. An interaction ∫∫ ρ₁(r) ρ₂(r′)/|r − r′| d³r d³r′ therefore splits, term by term and at each pair of radii r and r′, into the projection of ρ₂ onto Yₗᵐ times the conjugate projection of ρ₁, and orthogonality decides which (l, m) survive. If ρ₂ is spherically symmetric, its projection is √(4π) δₗ₀ δₘ₀ times its radial profile, so every term but l = 0 dies whatever ρ₁ looks like. The average of 1/|r − r′| over either direction is 1/rₘₐₓ: Newton's shell theorem, recovered. As r′ → 0 the factor rₘᵢₙˡ kills every l ≥ 1 and the sum returns 1/r.
The radii stay coupled, and the order in α
The double sum separates directions, never radii. rₘᵢₙ and rₘₐₓ trade places at r = r′, so every radial integral must be split there, one piece with r′ inside r and one with r′ outside. For helium's 1s² ground state the double sum does the angular half of the repulsion integral, leaving only l = 0, and Unit 10 finishes the radial half: ⟨1/r₁₂⟩ = 5Z/(8a₀), a repulsion of (5/8) × 2 × 27.21 eV = 34.0 eV. That moves the independent-electron energy from −108.8 eV to −74.8 eV, against a measured −79.0 eV, the sum of helium's two ionisation energies, 24.6 eV and 54.4 eV, taken from the tables. The remaining 4.2 eV is a failure of the unscreened wavefunction, not of the expansion. The expansion is exact and carries no power of the fine-structure constant α. What is approximate is the Coulomb law it expands: magnetic and retardation corrections to e²/(4πε₀r₁₂), the Breit interaction, enter at relative order (Zα)², about 2.1 × 10⁻⁴ for helium, a result quoted from relativistic quantum mechanics.
Change one variable at a time
Make the relationship visible.
Set θ′ = 0: both cycling curves flatten and the bold sum becomes the m = 0 line, the proof's frame. Then compare θ = 60°, θ′ = 60°, Δφ = 180° with θ = 0°, θ′ = 120°: every piece changes, the sum stays at −0.125, because γ = 120° both times.
cos γ-0.140
ANGLE γ98.1 °
(4π/5) × SUM OVER m (DOT)-0.471
P₂(cos γ) DIRECT (RING)-0.471
Live interpretationcos γ: −0.140. ANGLE γ: 98.1 °. (4π/5) × SUM OVER m (DOT): −0.471. P₂(cos γ) DIRECT (RING): −0.471
Catch the common trap
Explain before calculating.
Two directions have (θ, φ) = (60°, 0°) and (θ′, φ′) = (60°, 180°). What is Σₘ Y₂ᵐ*(θ′, φ′) Y₂ᵐ(θ, φ), summed over m = −2 … 2?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA filled 3d subshell holds two electrons, one of each spin, in every m from −2 to 2. Use the addition theorem at γ = 0 to find Σₘ |Y₂ᵐ(θ, φ)|² and the number of electrons per steradian, then check the sum term by term at θ = 30°.
- Put n̂′ = n̂, so γ = 0 and P₂(1) = 1. The theorem reads 1 = (4π/5) Σₘ |Y₂ᵐ|², so Σₘ |Y₂ᵐ(θ, φ)|² = 5/(4π) = 0.398 sr⁻¹, with no θ or φ left.
- Two electrons per m give 2 × 5/(4π) = 10/(4π) = 0.796 electrons per steradian in every direction; over the full 4π sr that returns all 10 electrons.
- Check at θ = 30°, where cos²θ = 0.750 and sin²θ = 0.250: |Y₂⁰|² = (5/16π)(3 × 0.750 − 1)² = 0.09947 × 1.5625 = 0.1554 sr⁻¹.
- |Y₂(±1)|² = (15/8π) sin²θ cos²θ = 0.5968 × 0.1875 = 0.1119 sr⁻¹ each, and |Y₂(±2)|² = (15/32π) sin⁴θ = 0.14921 × 0.0625 = 0.00933 sr⁻¹ each.
- Total: 0.1554 + 2 × 0.1119 + 2 × 0.00933 = 0.3979 sr⁻¹ = 5/(4π). No single m is isotropic: the m = 0 share is 39% at θ = 30° and 25% at θ = 90°, and the other four make up the difference exactly.
AnswerΣₘ |Y₂ᵐ|² = 5/(4π) = 0.398 sr⁻¹ in every direction, so the filled 3d subshell carries 0.796 electrons per steradian everywhere: its angular density is spherical.
MediumTwo charges sit at r = 1.00 a₀ in the direction (θ, φ) = (40°, 10°) and at r′ = 3.00 a₀ in the direction (75°, 115°). Evaluate the l = 2 term of 1/|r − r′| through the m-sum, check it against P₂(cos γ), and compare the multipole series through l = 2 with the exact 1/|r − r′|.
- cos γ = cos 40° cos 75° + sin 40° sin 75° cos(−105°) = 0.1983 − 0.1607 = 0.0376, so γ = 87.8°: the two directions are nearly perpendicular.
- The m-sum times 4π/5, pair by pair. m = 0: ¼(3cos²θ − 1)(3cos²θ′ − 1) = ¼(0.7605)(−0.7990) = −0.1519. m = ±1: 3 sin θ cos θ sin θ′ cos θ′ cos Δφ = 3(0.4924)(0.2500)(−0.2588) = −0.0956.
- m = ±2: ¾ sin²θ sin²θ′ cos 2Δφ = 0.75(0.4132)(0.9330)(−0.8660) = −0.2504. The sum is −0.4979; directly, P₂(0.0376) = (3 × 0.00141 − 1)/2 = −0.4979. The m = 0 term alone would have given −0.152.
- With rₘᵢₙ = 1.00 a₀ and rₘₐₓ = 3.00 a₀, the l = 2 term is (rₘᵢₙ²/rₘₐₓ³) P₂(cos γ) = (1/27)(−0.4979) a₀⁻¹ = −0.0184 a₀⁻¹.
- Series: l = 0 gives 1/3 = 0.3333 a₀⁻¹; l = 1 adds (1/9)(0.0376) = 0.0042; l = 2 adds −0.0184, for 0.3191 a₀⁻¹. Exact: 1/√(1 + 9 − 6 × 0.0376) = 1/√9.774 = 0.3199 a₀⁻¹. With γ near 90° the dipole term is small and the quadrupole carries the correction.
AnswerThe l = 2 term is −0.0184 a₀⁻¹, with P₂(cos γ) = −0.498 from both routes; the series through l = 2 gives 0.3191 a₀⁻¹ against the exact 0.3199 a₀⁻¹, within 0.3%.
HardA hydrogen atom is in the 2p state with m = 0, and a proton sits at distance R = 20 a₀ from its nucleus, in the direction with polar angle Θ. Use the double sum to find which multipoles of the electron cloud the proton feels, then the electron-proton energy for Θ = 0 and Θ = 90°. Take ⟨r²⟩ = 30 a₀² for 2p and e²/(4πε₀a₀) = 27.21 eV.
- Only 1.7 × 10⁻⁵ of the 2p probability lies beyond 20 a₀, so take rₘᵢₙ = r and rₘₐₓ = R throughout: ⟨1/|r − R|⟩ = Σₗ Σₘ [4π/(2l + 1)] (⟨rˡ⟩/Rˡ⁺¹) Yₗᵐ*(Θ, Φ) ∫ |Y₁⁰|² Yₗᵐ dΩ.
- Angular selection: |Y₁⁰|² = (3/4π) cos²θ = (1/4π)[1 + 2P₂(cos θ)] has no φ in it and only l = 0 and l = 2 parts, so orthogonality keeps just (l, m) = (0, 0) and (2, 0). The dipole term is absent, as parity requires of |ψ|².
- The (0, 0) term is 1/R. The (2, 0) term is (4π/5)(⟨r²⟩/R³) Y₂⁰(Θ) ∫ |Y₁⁰|² Y₂⁰ dΩ = (⟨r²⟩/R³) P₂(cos Θ) ⟨P₂(cos θ)⟩, and ⟨cos²θ⟩ = 3/5 gives ⟨P₂⟩ = 2/5. So ⟨1/|r − R|⟩ = 1/R + 12 a₀² P₂(cos Θ)/R³, each term in inverse length.
- U = −27.21 eV × a₀⟨1/|r − R|⟩: the monopole is −27.21/20 = −1.3605 eV, and the quadrupole −27.21 × 12/8000 × P₂(cos Θ) = −0.0408 P₂(cos Θ) eV, 3% of the monopole. On the z axis P₂ = 1, so U = −1.3605 − 0.0408 = −1.401 eV; in the xy plane P₂ = −½, so U = −1.3605 + 0.0204 = −1.340 eV.
- Unsöld check: for m = ±1, ⟨cos²θ⟩ = 1/5 and ⟨P₂⟩ = −1/5, so a filled 2p subshell has 2/5 − 1/5 − 1/5 = 0 and no quadrupole, as Σₘ |Y₁ᵐ|² = 3/(4π) requires. These are averages over the 2p₀ cloud held fixed: 2s is degenerate with 2p, so the proton's l = 1 term would mix them, the linear Stark effect of topic 9.6, which this calculation leaves out.
AnswerOnly the monopole and the m = 0 quadrupole survive: U = −1.401 eV with the proton on the z axis and −1.340 eV in the xy plane, the axial proton meeting the elongated 2p₀ cloud; averaging over m = −1, 0, 1 leaves the monopole, −1.3605 eV, at every angle.