University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.3
Legendre polynomials: Rodrigues, orthogonality and norm
Rodrigues' formula writes every Legendre polynomial as l derivatives of (x² − 1)ˡ, and integration by parts then settles for every l what the recursion could only check one case at a time: orthogonality, the norm 2/(2l + 1) that fixes every m = 0 probability, and why all l nodal cones of Yₗ⁰ lie strictly between the poles.
Build the model
Connect the measurement to the mechanism.
The series route of the previous lesson builds each Pₗ one coefficient at a time: enough to show that the terminating solution exists, useless for proving anything about every l at once. Rodrigues' formula writes the whole polynomial as l derivatives of (x² − 1)ˡ, and that packaging is the point. A function built by differentiation can be integrated by parts, and every boundary term dies because (x² − 1)ˡ and its first l − 1 derivatives vanish at x = ±1.
That one fact makes Pₗ orthogonal to every polynomial of lower degree, fixes the norm at 2/(2l + 1), and, by counting sign changes, puts all l zeros strictly inside (−1, 1). A second proof of orthogonality needs no formula at all, only the Sturm–Liouville form and regularity at the poles: its boundary term carries a factor 1 − x², and it fails for the logarithmic second solution that the series route discarded. The construction, orthogonality, the norm and the zeros are all derived, and they are geometry of the sphere, with no ħ, no α and no potential in them, so they hold exactly for every central force.
It stops at m = 0: the associated functions for m ≠ 0 come later in this unit. Quoted rather than proved: completeness, whether the Pₗ can represent any reasonable function of θ; the large-l asymptotic form used to check the norm; and the NMR and γ-ray correlation results the applications start from.
- Simple definition
- The Legendre polynomial Pₗ(x) is the degree-l solution of Legendre's equation that stays finite at x = ±1, scaled so that Pₗ(1) = 1; the set is orthogonal on −1 ≤ x ≤ 1, with ∫Pₗ² dx = 2/(2l + 1).
- Example
- Rodrigues' formula with l = 2 gives P₂ = (1/8) d²/dx² (x⁴ − 2x² + 1) = (3x² − 1)/2. It equals 1 at x = 1, vanishes at x = ±1/√3 = ±0.577 (θ = 54.7° and 125.3°), and ∫₋₁¹ P₂² dx = 2/5 = 0.40, not 1.
Any Pₗ in l differentiations, in a closed form that integration by parts can work on for every l at once.
x = cos θ and Pₗ are dimensionless; l = 0, 1, 2, …; Pₗ is a polynomial of degree exactly l.
Differentiated l + 1 times by Leibniz's rule it becomes (1 − x²)u″ − 2xu′ + l(l + 1)u = 0 for u = dˡw/dxˡ.
w is dimensionless; the identity holds at every x, not only inside (−1, 1).
Leibniz on (x − 1)ˡ(x + 1)ˡ leaves only l! 2ˡ at x = 1. A convention at one point, not a normalisation: that is the next card.
x = 1 is θ = 0; the prefactor 1/(2ˡ l!) is there to make this true.
Zero for l ≠ k by moving l derivatives across; 2/(2l + 1) = 1/(l + ½) from ∫(1 − x²)ˡ dx. The norms run 2, 2/3, 2/5, 2/7.
δₗₖ is 1 for l = k and 0 otherwise; a pure number, since x and dx = −sin θ dθ are.
At x = 1 the factor 1 − x² kills the right side, so l ≠ k gives zero with no formula for Pₗ. The figure draws both sides.
Primes are d/dx; true for any two solutions regular at x = −1, where both sides start from zero.
Reads off how much of each l a function holds. In an m = 0 state the weight of l is cₗ² × 2/(2l + 1), not cₗ².
For f(x) = Σ cₗPₗ(x); cₗ carries the units of f. Exact and finite when f is a polynomial of degree n (l ≤ n).
Why l derivatives of (x² − 1)ˡ solve Legendre's equation
Start from w = (x² − 1)ˡ. Its slope is w′ = 2lx(x² − 1)(l−1), so (x² − 1)w′ = 2lxw, a first-order equation that w obeys exactly. Differentiate both sides l + 1 times by Leibniz's rule and write u = dˡw/dxˡ. On the left only three terms survive, because x² − 1 has no third derivative: (x² − 1)u″ + 2(l + 1)xu′ + l(l + 1)u. On the right x has no second derivative, leaving 2lxu′ + 2l(l + 1)u. Subtract and change sign: (1 − x²)u″ − 2xu′ + l(l + 1)u = 0, Legendre's equation with λ = l(l + 1). So every multiple of u is a solution, and because u is a polynomial of degree 2l − l = l, finite at both poles, it is the terminating branch the series route found, not the logarithmic one. This is derived, for every non-negative integer l in one stroke.
Fixing the scale, and checking P₀ to P₃
The prefactor 1/(2ˡ l!) is chosen so that Pₗ(1) = 1. Split (x² − 1)ˡ = (x − 1)ˡ(x + 1)ˡ and apply Leibniz again: any term that leaves a factor x − 1 undifferentiated vanishes at x = 1, and the single term that spends all l derivatives on (x − 1)ˡ gives l!(x + 1)ˡ = l! 2ˡ there. And since (x² − 1)ˡ is even and each derivative flips parity, Pₗ(−x) = (−1)ˡPₗ(x). Now expand: P₀ = 1, P₁ = (1/2) d/dx (x² − 1) = x, P₂ = (1/8)(12x² − 4) = (3x² − 1)/2, and P₃ = (1/48) d³/dx³ (x⁶ − 3x⁴ + 3x² − 1) = (120x³ − 72x)/48 = (5x³ − 3x)/2. The terminated recursion aₖ₊₂/aₖ = [k(k + 1) − l(l + 1)]/[(k + 1)(k + 2)] agrees: it gives −3 for l = 2, matching (3/2)/(−1/2), and −5/3 for l = 3, matching (5/2)/(−3/2). The leading coefficient is (2l)!/(2ˡ(l!)²) in general: 3/2, 5/2 and 35/8 for l = 2, 3 and 4.
Orthogonality and the norm by moving derivatives across
Integrate a smooth f against Pₗ and move the derivatives off (x² − 1)ˡ one at a time. Each step leaves a boundary term containing a derivative of (x² − 1)ˡ of order l − 1 or less, which still carries a factor x² − 1 and so vanishes at ±1. After l steps, ∫₋₁¹ fPₗ dx = [(−1)ˡ/(2ˡ l!)] ∫₋₁¹ (dˡf/dxˡ)(x² − 1)ˡ dx. If f is any polynomial of degree below l, dˡf/dxˡ = 0: Pₗ is orthogonal to all of them, every Pₖ with k below l included. For f = Pₗ, dˡPₗ/dxˡ is the constant (2l)!/(2ˡ l!), leaving ∫Pₗ² dx = [(2l)!/(2(2l)(l!)²)] Iₗ with Iₗ = ∫₋₁¹ (1 − x²)ˡ dx. One more integration by parts gives Iₗ = [2l/(2l + 1)] Iₗ₋₁ from I₀ = 2, and the product collapses to 2/(2l + 1). For l = 2: (24/64)(16/15) = 0.40. Limits: l = 0 gives 2, the length of the interval, and at large l the quoted asymptotic form Pₗ(cos θ) ≈ √(2/(π(l + ½) sin θ)) cos((l + ½)θ − π/4) gives 1/(l + ½) once cos² is averaged to ½, which is the exact norm written another way.
Orthogonality again, from the boundary term alone
Legendre's equation is already in Sturm–Liouville form, d/dx[(1 − x²)Pₗ′] + l(l + 1)Pₗ = 0, with p(x) = 1 − x² and weight 1. Multiply the l equation by Pₖ and the k equation by Pₗ, subtract, and the derivative terms combine into one total derivative. Integrating from −1 to x gives [l(l + 1) − k(k + 1)] ∫₋₁ˣ PₗPₖ dx′ = (1 − x²)(PₗPₖ′ − PₖPₗ′). At x = 1 the right side vanishes because 1 − x² does while the polynomials and their slopes stay finite, so for l ≠ k the full integral is zero. This proof never used Rodrigues' formula, only regularity at both poles, and it explains what parity cannot: P₀P₂ is even, yet its integral vanishes, because the positive caps beyond |x| = 0.577 exactly cancel the negative band around the equator. It also shows its limits. At l = k the left side is zero times the integral, so the norm must come from the previous section, and for the singular solution Q₀, with (1 − x²)Q₀′ = 1, the boundary term survives.
Why all l zeros sit strictly between the poles
Orthogonality to every lower-degree polynomial forces the zeros inside. Suppose Pₗ changed sign in (−1, 1) at only n points x₁, …, xₙ, with n smaller than l. The polynomial q(x) = (x − x₁)⋯(x − xₙ) has degree n and changes sign at exactly those points, so Pₗq keeps one sign across the whole interval and cannot integrate to zero, contradicting orthogonality. So Pₗ changes sign at least l times inside, and a degree-l polynomial has at most l roots: exactly l simple zeros, none at the poles, where Pₗ = ±1. In θ they are the l nodal cones of Yₗ⁰. P₃ vanishes at x = 0 and ±√(3/5) = ±0.7746, so θ = 90°, 39.23° and 140.77°. P₂ vanishes at x = ±1/√3, θ = 54.74°, the magic angle of solid-state NMR: the dipolar coupling of a spinning sample is scaled by P₂(cos θR), θR the angle between rotor axis and field (quoted), and spectrometers are tuned experimentally, by sharpening the ⁷⁹Br signal of KBr, to spin at 54.74°.
What the norm is for: projections and probabilities
If f(x) = Σ cₗPₗ(x), multiply by Pₖ and integrate: orthogonality leaves one term, so cₖ = [(2k + 1)/2] ∫₋₁¹ fPₖ dx. For a polynomial of degree n this is exact and finite, since P₀ to Pₙ have degrees 0 to n and so span every such polynomial (derived); for an arbitrary function the infinite series needs completeness, which this unit quotes. Take x³: c₁ = (3/2)(2/5) = 3/5 and c₃ = (7/2)(4/35) = 2/5, and the pieces rebuild ∫x⁶ dx = 2/7 as (3/5)²(2/3) + (2/5)²(2/7) = 6/25 + 8/175. For an angular state ψ ∝ cos³θ those two pieces over 2/7 are the probabilities of l = 1 and l = 3: 0.84 and 0.16. The same norm fixes the m = 0 harmonic. With dΩ = sin θ dθ dφ = −dx dφ, ∫|Yₗ⁰|² dΩ = [(2l + 1)/4π] × 2π × 2/(2l + 1) = 1 for Yₗ⁰ = √((2l + 1)/4π) Pₗ(cos θ), which carries units of sr(−1/2).
Change one variable at a time
Make the relationship visible.
At l = 2, k = 0 the product is even, so parity cannot kill it, yet the solid curve rises, dips below zero and lands on exactly zero at x = 1, where 1 − x² vanishes. Then set k = l: it ends on 2/(2l + 1) instead, falling from 2 at l = 0 to 0.18 at l = 5.
∫₋₁¹ PₗPₖ dx0.000
NORM OF Pₗ 2/(2l + 1)0.400
GAP l(l + 1) − k(k + 1)6
PARITY OF PₗPₖ (−1)(l+k)1
Live interpretation∫₋₁¹ PₗPₖ dx: 0.000. NORM OF Pₗ 2/(2l + 1): 0.400. GAP l(l + 1) − k(k + 1): 6. PARITY OF PₗPₖ (−1)(l+k): 1
Catch the common trap
Explain before calculating.
A particle's angular state is ψ ∝ cos²θ, with no φ dependence. With x = cos θ, x² = (1/3)P₀(x) + (2/3)P₂(x). What is the probability that a measurement of L² returns 0?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUse Rodrigues' formula to build P₄(x). Check P₄(±1) and its leading coefficient, then find the polar angles θ of the four nodal cones of Y₄⁰ ∝ P₄(cos θ).
- Expand w = (x² − 1)⁴ = x⁸ − 4x⁶ + 6x⁴ − 4x² + 1. The prefactor is 1/(2⁴ × 4!) = 1/384.
- Differentiate four times: x⁸ gives 8 × 7 × 6 × 5 x⁴ = 1680x⁴, −4x⁶ gives −4 × 360x² = −1440x², 6x⁴ gives 6 × 24 = 144, and −4x² and 1 vanish. So P₄ = (1680x⁴ − 1440x² + 144)/384 = (35x⁴ − 30x² + 3)/8.
- Checks: P₄(1) = (35 − 30 + 3)/8 = 1; P₄ is even, so P₄(−1) = +1 = (−1)⁴; the leading coefficient 35/8 = 4.375 equals (2l)!/(2ˡ(l!)²) = 40320/(16 × 576) = 4.375.
- Zeros: with y = x², 35y² − 30y + 3 = 0 gives y = (30 ± √480)/70 = 0.74156 or 0.11559, so x = ±0.8611 and ±0.3400. Four simple zeros, all strictly inside (−1, 1), as orthogonality demands.
- Angles: θ = arccos x = 30.56° and 70.12°, and by the evenness of P₄ their mirror images 109.88° and 149.44°. The equator is not a node, since P₄(0) = 3/8.
AnswerP₄ = (35x⁴ − 30x² + 3)/8, with P₄(±1) = 1 and leading coefficient 35/8. Nodal cones at θ = 30.56°, 70.12°, 109.88° and 149.44°, from x = ±0.8611 and ±0.3400.
MediumQ₀(x) = (1/2) ln[(1 + x)/(1 − x)] also solves Legendre's equation with l = 0, but it diverges at x = ±1: it is the branch the series route discarded. Predict ∫₋₁¹ Q₀P₁ dx from the Sturm–Liouville boundary term, confirm it by direct integration, and explain why the same argument gives zero for P₀ and P₁ but not here.
- Q₀′ = (1/2)[1/(1 + x) + 1/(1 − x)] = 1/(1 − x²), so (1 − x²)Q₀′ = 1 at every x and d/dx[(1 − x²)Q₀′] = 0: an l = 0 solution, like P₀ = 1.
- The identity [l(l + 1) − k(k + 1)] ∫uₗuₖ dx = [(1 − x²)(uₗuₖ′ − uₖuₗ′)] between the limits holds for any two solutions. With uₗ = Q₀ (l = 0) and uₖ = P₁ = x (k = 1), the gap is 0 − 2 = −2 and the bracket is (1 − x²)Q₀ − x.
- Near either pole |Q₀| grows only like (1/2)|ln(1 − x²)|, so (1 − x²)Q₀ tends to 0. So the bracket tends to −1 at x = 1 and to +1 at x = −1, and the boundary term is −1 − (+1) = −2. Then −2 ∫Q₀P₁ dx = −2 predicts ∫₋₁¹ Q₀P₁ dx = 1.
- Direct check: the integrand is even, so the integral is ∫₀¹ x ln[(1 + x)/(1 − x)] dx. Integrate by parts with v = (x² − 1)/2, chosen to vanish at x = 1: the boundary term is zero at both ends, and what remains is −∫₀¹ [(x² − 1)/2][2/(1 − x²)] dx = ∫₀¹ 1 dx = 1. A two-million-point midpoint sum agrees to better than 10⁻⁶.
- Why it fails here: for a regular solution uₗ′ is finite, so (1 − x²)uₗ′ tends to 0 at the poles and the bracket dies. For Q₀ the zero of 1 − x² is cancelled exactly by the pole of Q₀′, leaving (1 − x²)Q₀′ = 1. Orthogonality belongs to the solutions regular at both poles.
Answer∫₋₁¹ Q₀P₁ dx = 1 exactly: the boundary term −2 divided by the eigenvalue gap −2. Q₀ and P₁ belong to different l yet are not orthogonal, because (1 − x²)Q₀′ = 1 does not vanish at the poles.
HardThe two γ rays of the 4 → 2 → 0 cascade in ⁶⁰Ni, fed by ⁶⁰Co decay, are correlated in direction. Angular-correlation theory (quoted, not derived) gives the coincidence rate at angle θ between them as W(θ) ∝ 1 + (1/8)cos²θ + (1/24)cos⁴θ. Use orthogonality and the norm to write W = c₀[1 + A₂P₂(cos θ) + A₄P₄(cos θ)], find A₂ and A₄, and check them against the 90° to 180° anisotropy.
- Set x = cos θ. W is even, so every odd cₗ vanishes by parity; W has degree 4, so every cₗ with l ≥ 5 vanishes because Pₗ is orthogonal to all lower-degree polynomials. Only c₀, c₂ and c₄ survive, each from cₗ = [(2l + 1)/2] ∫₋₁¹ WPₗ dx.
- Moments needed: ∫x² dx = 2/3, ∫x⁴ dx = 2/5, ∫x²P₂ dx = 4/15, ∫x⁴P₂ dx = 8/35, and ∫x⁴P₄ dx = [35(2/9) − 30(2/7) + 3(2/5)]/8 = 16/315.
- c₀ = (1/2)[2 + (1/8)(2/3) + (1/24)(2/5)] = 1 + 1/24 + 1/120 = 21/20. c₂ = (5/2)[(1/8)(4/15) + (1/24)(8/35)] = (5/2)(1/30 + 1/105) = 3/28. c₄ = (9/2)(1/24)(16/315) = 1/105.
- A₂ = c₂/c₀ = (3/28)(20/21) = 5/49 = 0.1020, and A₄ = c₄/c₀ = (1/105)(20/21) = 4/441 = 0.00907.
- Checks: at θ = 90°, P₂ = −1/2 and P₄ = 3/8, so W = (21/20)(1 − 0.05102 + 0.00340) = 1.0000, matching 1 + 0 + 0; at θ = 180°, W = (21/20)(1.11111) = 1.1667 = 7/6, matching 1 + 1/8 + 1/24. The anisotropy W(180°)/W(90°) − 1 is 1/6 = 0.167.
- Against measurement: coincidence counts on ⁶⁰Co, corrected for the finite solid angle of the detectors, give A₂ ≈ 0.10 and A₄ ≈ 0.01, which is why ⁶⁰Co is the usual test source for an angular-correlation set-up.
AnswerA₂ = 5/49 ≈ 0.102 and A₄ = 4/441 ≈ 0.0091, with c₀ = 1.05. Then W(180°)/W(90°) = 7/6, an anisotropy of 0.167.