University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.1
The associated Legendre equation and its poles
This is the course's one review topic, and it earns its place with a single new move. Rewrite the polar equation in x = cos θ and every difficulty lands on the two poles. Reading what the solutions do there, and why only one of each pair is physical, is the skill that quantises λ in the next lesson.
Build the model
Connect the measurement to the mechanism.
Quantum Mechanics I separated ψ = R(r)Θ(θ)e(imφ) with integer m and left one equation for Θ, carrying a separation constant λ that is the eigenvalue of L² in units of ħ². We quote that equation; what is new is reading it. Put x = cos θ and it becomes the associated Legendre equation, whose coefficients are finite at every latitude and fail only at x = ±1, the two halves of the z-axis.
Those are regular singular points, so near each one the solutions go as (1 ∓ x)s, and the lowest power fixes s² = m²/4 whatever λ is. The exponent +|m|/2 gives a solution that vanishes on the axis like sin(|m|)θ; −|m|/2 gives one that diverges like 1/sin(|m|)θ; at m = 0 the two coincide, and Abel's identity, (1 − x²)W = constant, forces the second solution to be a logarithm. Keeping the regular one is a physical demand, not a matter of taste: for |m| ≥ 1 the singular solution cannot be normalised, and for m = 0 it can, but its ⟨L²⟩ is infinite because it hides a source on the axis.
Regularity at both poles is one condition more than a second-order equation with one free ratio can meet at arbitrary λ, which is why λ will be discrete. The argument needs V to depend on r alone and nothing physical on the z-axis: a non-central potential breaks the separation, and a flux line along the axis makes the axis a real place with its own boundary condition.
- Simple definition
- The associated Legendre equation is the polar part of the angular equation written in x = cos θ; its only singular points on the sphere are the poles x = ±1, and at each pole exactly one of its two solutions is regular.
- Example
- For |m| = 2 the two solutions near the north pole go as 1 − x and 1/(1 − x). At θ = 0.01 rad, 1 − x = 5.0 × 10⁻⁵, so one is 5.0 × 10⁻⁵ and the other 2.0 × 10⁴, whatever λ is.
Derived here by the chain rule. The weight sin θ is absorbed into dx, so normalisation over the sphere becomes 2π∫|P|² dx on [−1, 1], with no weight left inside the integral.
x is dimensionless: +1 on the positive z-axis (θ = 0), −1 on the negative z-axis (θ = π), 0 on the equator
Derived here from the separated polar equation. Written as d/dx[(1 − x²)P′] + [λ − m²/(1 − x²)]P = 0 it is in Sturm–Liouville form with unit weight.
P(x) = Θ(θ), dimensionless; m is an integer (quoted from QM I); λ = L²/ħ² is dimensionless and not yet known
Derived here from the lowest power of t. λ does not enter, because λP sits a whole power of t higher, so every λ has the same pole behaviour.
t = 1 − cos θ = 2 sin²(θ/2) ≈ θ²/2 near θ = 0, so tˢ goes as θ(+|m|) or θ(−|m|), with θ in rad
Derived here. W ∝ 1/t at a pole, so the exponents sum to zero, both solutions cannot be regular, and at m = 0, where s = 0 twice, the second must contain ln t.
P₁ and P₂ solve the equation at the same λ and m; the constant vanishes only if they are proportional
Derived here. At m = 0 the singular solution goes as ln t and ∫₀¹ ln²t dt = 2 is finite, so normalisability alone cannot exclude it.
dx = −dt exactly, so this is ∫|Θ|² sin θ dθ; |m| = 1 diverges as ln(1/t₀), |m| = 2 as 1/t₀
Derived from ⟨L²⟩ = Σᵢ‖LᵢY‖² with L = −iħ r × ∇. For m = 0 and P = ln t, |∂Y/∂θ|² = (1 − x²)P′² ≈ 2/t, and ∫dt/t diverges.
Y normalised on the sphere; ħ = 1.055 × 10⁻³⁴ J s, so ⟨L²⟩ is in J² s²; dΩ in sr
One substitution, and where it sends the trouble
Quantum Mechanics I left us, after ψ = R(r)Θ(θ)e(imφ) with integer m, the polar equation (1/sin θ) d/dθ(sin θ dΘ/dθ) + [λ − m²/sin²θ]Θ = 0, with λ the eigenvalue of L² in units of ħ². We quote it, and we do not yet know λ. Put x = cos θ. Then dx = −sin θ dθ, so sin θ d/dθ = −(1 − x²) d/dx, and the first term becomes d/dx[(1 − x²) dΘ/dx]. Writing P(x) = Θ(θ) gives (1 − x²)P″ − 2xP′ + [λ − m²/(1 − x²)]P = 0, the associated Legendre equation. Every trigonometric coefficient is now a ratio of polynomials, which is what a series method needs, and the measure simplifies: ∫|Y|² dΩ = ∫₋₁¹ dx ∫₀(2π) dφ |P|². Check it on a state we know. Y₁⁰ ∝ cos θ is P = x with m = 0, and the equation reads −2x + λx = 0, so λ = 2 and L² = 2ħ², the value the ladder algebra gave. Dimensions: x, P and λ are pure numbers, and λħ² is in J² s², as L² must be.
Read the singular points before solving
Divide by 1 − x² to reach standard form, P″ + p(x)P′ + q(x)P = 0, with p = −2x/(1 − x²) and q = λ/(1 − x²) − m²/(1 − x²)². For every x strictly between −1 and 1 both are finite and smooth, so every latitude is an ordinary point, and a power series about x = 0 converges at least for |x| below 1, the distance to the nearest singular point (Fuchs's theorem, quoted). The trouble sits at x = ±1, where p grows like 1/(1 ∓ x) and q like 1/(1 ∓ x)², and no faster. That is what makes them regular singular points: as x → 1, (x − 1)p → 1 and (x − 1)²q → −m²/4, both finite, and Frobenius's method then promises solutions of the form (1 − x)s times a power series. A third regular singular point lies at x = ∞, off the sphere; it matters only because three of them make this a hypergeometric equation with tabulated properties. Physically, x = ±1 are the positive and negative z-axis, where φ is undefined. The sphere has no special points; our coordinates do.
Frobenius at a pole: the exponents ±|m|/2
Near the north pole let t = 1 − x, so 1 − x² = t(2 − t) and d/dx = −d/dt. Try P = tˢ and keep the lowest power of t. (1 − x²)P″ gives 2s(s − 1)t(s−1), −2xP′ gives +2s t(s−1), and −m²P/(1 − x²) gives −(m²/2)t(s−1), while λP is a whole power of t higher. The coefficient must vanish: 2s² − m²/2 = 0, so s = ±|m|/2, and λ played no part. The equation is unchanged by x → −x, so the south pole has the same pair in t = 1 + x. With t = 2 sin²(θ/2) ≈ θ²/2 the two behaviours are θ(|m|) and θ(−|m|): for |m| = 1 at θ = 0.01 rad, t = 5.0 × 10⁻⁵ and the pair is √t = 7.1 × 10⁻³ against 1/√t = 141. Check it in the flat limit. Near the pole sin θ ≈ θ and the polar equation becomes Θ″ + Θ′/θ + (λ − m²/θ²)Θ = 0, Bessel's equation of order |m| in √λ θ, whose two tabulated solutions go as θ(|m|) and θ(−|m|), the second logarithmic at order 0. A small cap of sphere is a flat disc, where ρ(|m|) e(imφ) = (ξ ± iη)(|m|) is a polynomial in local Cartesian ξ and η, and ρ(−|m|) e(imφ) is not.
Abel's identity: one bad solution per pole, a log at m = 0
Take two solutions P₁ and P₂ at the same λ and m. Multiply the equation for each, in the form d/dx[(1 − x²)P′] + [λ − m²/(1 − x²)]P = 0, by the other solution and subtract: the λ and m² terms cancel, leaving d/dx[(1 − x²)W] = 0 with W = P₁P₂′ − P₁′P₂. So (1 − x²)W is a constant, nonzero for independent solutions, and W grows as 1/t at each pole. Three things follow without a series. Two solutions both of the form t(|m|/2) times a power series have a Wronskian of order t(|m|), never 1/t, so at most one solution per pole is regular. Solutions going as t(s₁) and t(s₂) have W ∝ t(s₁ + s₂ − 1), so s₁ + s₂ = 0, as the indicial equation said. And at m = 0 both exponents are 0: two power series would give a finite W, so the second solution must be P₂ = P₁ ln t + (a power series), whose Wronskian with P₁ is −P₁²/t plus terms that stay finite at the pole. The logarithm is forced, not chosen. It grows slowly: ln t is −9.9 at θ = 0.01 rad and only −28.3 at 10⁻⁶ rad, which is why m = 0 needs a sharper test than the rest.
Why the singular solution is physically excluded
A wavefunction may be infinite at a point and still be acceptable, so the singular solutions must be tested against what a state needs. Since sin θ dθ = −dx = dt, the norm near the north pole is ∫|P|² dt, and a singular solution with |m| ≥ 1 contributes ∫t(−|m|) dt. Cut out a cap of half-angle θ₀ and integrate the leading form θ(−|m|) out to 1 rad: |m| = 1 gives ln(1/θ₀), which is 6.9 at θ₀ = 10⁻³ rad, 13.8 at 10⁻⁶ rad, and has no limit, and |m| = 2 gives (θ₀⁻² − 1)/2 = 5.0 × 10⁵ already at 10⁻³ rad. No normalisation constant rescues that. At m = 0 the same test passes: ∫₀¹ θ ln²θ dθ = 1/4. The decider is ⟨L²⟩ = Σᵢ‖LᵢY‖² = ħ²∫(|∂Y/∂θ|² + |∂Y/∂φ|²/sin²θ) dΩ. With ∂Θ/∂θ ≈ 1/θ the integrand is 1/θ, the same ln(1/θ₀) returns, and ⟨L²⟩ is infinite: a rigid rotor in that state would carry infinite rotational energy ⟨L²⟩/(2I). The pointwise equation L²Θ = λħ²Θ hides this. In the flat patch ln ρ is the potential of a point charge in two dimensions, whose Laplacian is 2π δ²(ρ): the singular m = 0 function solves a different problem, one with a source on the z-axis.
Two poles, one free ratio: why λ will be discrete
A second-order equation has two independent solutions: one overall scale and one free ratio. Demanding regularity at the north pole spends the ratio. Carried to the south pole, the surviving function is some mixture a(λ) × (regular) + b(λ) × (singular) there, and b(λ) = 0 is one equation for one unknown, so only special λ pass, as special energies do in a well. λ = 0 shows it in closed form, derived here. With u = ln tan(θ/2) the polar equation becomes d²Θ/du² = m²Θ, so for |m| = 1 the solution regular at the north pole is tan(θ/2): 1 at the equator, 114.6 at θ = 179°, infinite at θ = π. λ = 0 fails for |m| = 1, while for m = 0 the constant passes. Discreteness is measured. In the rigid-rotor model the axis of ¹²C¹⁶O obeys this equation with E = λħ²/(2I), and the molecule absorbs in sharp lines at 115.271, 230.538 and 345.796 GHz (measured values, quoted), in the ratio 1 : 2.000 : 3.000, not across a band; admit the singular solutions and every λ would be allowed. No α enters: nothing here knows V(r). The model stops when V is not central, and when something real, such as a thin solenoid, sits on the axis.
Change one variable at a time
Make the relationship visible.
Start at |m| = 0 and shrink the cap: the solid norm settles at 0.25, hugging the axis, while the dashed ⟨L²⟩ climbs 2.3 per decade and never stops. Step to |m| = 1 and the norm itself becomes that ramp; at 2 or 3 both are pinned to the top edge, off scale, within the first decade.
CAP HALF-ANGLE θ₀0.0573 °
POLE EXPONENTS s = ±0.0
log₁₀ SINGULAR NORM-0.60
log₁₀ SINGULAR ⟨L²⟩/ħ²0.84
Live interpretationCAP HALF-ANGLE θ₀: 0.0573 °. POLE EXPONENTS s = ±: 0.0. log₁₀ SINGULAR NORM: −0.60. log₁₀ SINGULAR ⟨L²⟩/ħ²: 0.84
Catch the common trap
Explain before calculating.
Near the north pole one solution of the polar equation with m = 0 behaves as ln(1 − cos θ), and one with |m| = 1 behaves as 1/sin θ. Which statement about excluding them is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyThe m = 1 polar function Θ = sin θ cos θ appears in Quantum Mechanics I as part of Y₂¹. Rewrite it as P(x) with x = cos θ, substitute it into the associated Legendre equation to find λ and L², and read off the exponent of 1 − x at the north pole.
- On 0 ≤ θ ≤ π, sin θ = +√(1 − x²), so P(x) = x√(1 − x²).
- Differentiate: P′ = (1 − 2x²)/√(1 − x²) and P″ = x(2x² − 3)/(1 − x²)³⁄².
- Substitute: (1 − x²)P″ − 2xP′ = x(6x² − 5)/√(1 − x²), and −m²P/(1 − x²) = −x/√(1 − x²). Together they give x(6x² − 6)/√(1 − x²) = −6x√(1 − x²) = −6P, so the equation reads (λ − 6)P = 0 and λ = 6.
- So L² = λħ² = 6 × (1.0546 × 10⁻³⁴ J s)² = 6.67 × 10⁻⁶⁸ J² s², and |L| = √6 ħ = 2.58 × 10⁻³⁴ J s. Dimensions: λ is a pure number and ħ² carries J² s².
- Near x = 1 write P = x(1 + x)¹⁄²(1 − x)¹⁄² ≈ √2 (1 − x)¹⁄²: the exponent is +½ = +|m|/2, the regular one. At θ = 0.1 rad, P = 0.0993 while √2 (1 − cos θ)¹⁄² = 0.1000; the 0.6% gap is the factor x√((1 + x)/2) = 0.994 that the leading term drops.
- By the symmetry x → −x, P ≈ −√2 (1 + x)¹⁄² at the south pole: regular there too, as a physical state must be.
Answerλ = 6, so L² = 6ħ² = 6.67 × 10⁻⁶⁸ J² s². Near the north pole P ≈ √2 (1 − x)¹⁄², the exponent +|m|/2 = +½, and P is regular at the south pole as well.
MediumFor |m| = 2, a classmate proposes that one solution of the polar equation behaves near the south pole as (1 + x)¹⁄². Find the indicial equation at x = −1 and test the claim. Then, keeping only the two leading forms with unit coefficients, compare the norm each puts in the band from 0.5° to 5° off the negative z-axis, and find what happens to the singular one when the inner edge moves to 0.25°.
- Put t = 1 + x, so 1 − x² = t(2 − t) and d/dx = d/dt. For P = tˢ the lowest power is t(s−1), with coefficient 2s(s − 1) + 2s − m²/2 = 2s² − m²/2. Setting it to zero gives s = ±|m|/2 = ±1.
- The proposed exponent ½ is not a root; it belongs to |m| = 1, so the claim is wrong. The allowed behaviours are (1 + x)¹ and (1 + x)⁻¹.
- With β = π − θ, t = 1 + cos θ = 1 − cos β = 2 sin²(β/2) ≈ β²/2, so the pair is about β²/2, vanishing on the axis, and 2/β², diverging. At the band edges t₁ = 1 − cos 0.5° = 3.808 × 10⁻⁵ and t₂ = 1 − cos 5° = 3.805 × 10⁻³.
- Here sin θ dθ = −dt, and t falls from t₂ to t₁ as θ crosses the band towards the axis, so the norm per 2π of these forms is exactly ∫|P|² dt from t₁ to t₂. Regular: (t₂³ − t₁³)/3 = 1.84 × 10⁻⁸. Singular: 1/t₁ − 1/t₂ = 26263 − 263 = 2.600 × 10⁴.
- Move the inner edge to 0.25°: t₁ = 9.52 × 10⁻⁶ and the singular norm becomes 1.048 × 10⁵, 4.03 times larger, because 1/t₁ scales as 1/β₁². Each halving roughly quadruples it, so it has no limit as the band reaches the axis, while the regular part is unchanged to three significant figures.
Answers = ±1 at the south pole, so (1 + x)¹⁄² is impossible for |m| = 2. In the band the regular form holds 1.84 × 10⁻⁸ and the singular one 2.600 × 10⁴ (per 2π, unit coefficients); moving the inner edge to 0.25° multiplies the singular share by 4.03.
HardAt m = 1 and λ = 2 the polar equation has the regular solution P₁ = √(1 − x²) = sin θ. Use Abel's identity to build the second solution, find the power of 1 − x it starts with at the north pole and where a logarithm enters, and follow its norm, ∫|P₂|² sin θ dθ from θ₀ to π/2, as θ₀ shrinks from 10⁻² rad to 10⁻⁴ rad.
- Check P₁: (1 − x²)P₁″ − 2xP₁′ − P₁/(1 − x²) = −2P₁, so it solves the equation at λ = 2, and it is regular at both poles.
- Abel's identity with the constant set to 1: W = P₁²(P₂/P₁)′ = 1/(1 − x²), so (P₂/P₁)′ = 1/[(1 − x²)P₁²] = 1/(1 − x²)².
- Integrate with ∫dx/(1 − x²)² = x/[2(1 − x²)] + ¼ ln[(1 + x)/(1 − x)], which differentiates back to 1/(1 − x²)². Hence P₂ = x/[2√(1 − x²)] + ¼√(1 − x²) ln[(1 + x)/(1 − x)] = ½ cot θ + ½ sin θ ln cot(θ/2).
- Near θ = 0, ½ cot θ ≈ 1/(2θ) = 1/[2√2 (1 − x)¹⁄²]: the exponent is −½ = −|m|/2. The logarithm arrives only as P₁ times a log, of order (1 − x)¹⁄² ln(1 − x), a whole power of 1 − x below the leading term. At θ = 0.01 rad the two terms are 49.998 and 0.026.
- Integrating |P₂|² sin θ from θ₀ to π/2 numerically gives 1.378 at θ₀ = 10⁻² rad, 1.954 at 10⁻³ rad and 2.530 at 10⁻⁴ rad. Each decade adds 0.576 = ¼ ln 10, exactly what the leading integrand 1/(4θ) produces, so the norm grows without limit.
- Checks: (1 − x²)W built from P₁ and P₂ is 1.000 at every x tried, and P₂ is odd under x → −x, so it is singular at the south pole too. At the allowed λ = 2 one solution is regular at both poles and its partner at neither, as Abel's identity requires.
AnswerP₂ = ½ cot θ + ½ sin θ ln cot(θ/2). It starts as (1 − x)(−1/2), with the logarithm entering only at order (1 − x)¹⁄² ln(1 − x), and its norm grows by ¼ ln 10 = 0.576 per decade of θ₀ (1.378, 1.954, 2.530), so it cannot be normalised.