University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.5
Associated Legendre functions, and why |m| cannot exceed l
Quantum Mechanics I stopped the ladder at m = ±l because the algebra said so. Here the polar equation says it too: strip off the pole factor and what remains is a polynomial of degree l − |m|, the |m|-th derivative of Pₗ. You leave able to write any associated Legendre function in two lines and count its nodal cones before plotting it.
Build the model
Connect the measurement to the mechanism.
Topic 1.2 showed that regularity at both poles forces λ = l(l + 1) when m = 0. With m ≠ 0 the polar equation carries the extra term −m²/(1 − x²), and near each pole its two Frobenius behaviours go as (1 − x²)(±μ/2), with μ = |m|. So peel the regular one off: P = (1 − x²)(μ/2) u. The m² term then cancels exactly, and u obeys an equation with polynomial coefficients whose series steps by two with ratio [(k + μ)(k + μ + 1) − λ]/[(k + 1)(k + 2)].
Left infinite, that series diverges at the poles as (1 − x²)(−μ), or as a logarithm for μ = 0, and rebuilds the singular branch, so it must stop, and it stops only if λ = (k + μ)(k + μ + 1) for some k ≥ 0. Call k + μ = l and two results arrive together: λ = l(l + 1), and μ = l − k ≤ l. The bound is a degree that cannot be negative. Differentiating Legendre's equation μ times reproduces the equation for u, which identifies the polynomial as dμ/dxμ Pₗ, and Rolle's theorem then counts its l − μ zeros, the nodal cones.
Nothing here is approximate and no power of α enters: this is the geometry of single-valued functions on the sphere, the same for every central potential, and fine structure later recombines the 2l + 1 values of m with spin without adding a (2l + 2)-th. It does not supply normalisation or phase, which topic 1.6 fixes, or any home for half-integer l, which has no regular single-valued function on the sphere.
- Simple definition
- An associated Legendre function, Pₗμ(x) = (1 − x²)(μ/2) dμ/dxμ Pₗ(x) with x = cos θ and μ = |m|, is, up to a constant, the one solution of the polar equation with λ = l(l + 1) that stays finite at both poles, and it exists only for μ ≤ l.
- Example
- For l = 2, P₂ = (3x² − 1)/2 gives P₂¹ = 3x√(1 − x²) = 3 sin θ cos θ, with one nodal cone at θ = 90°, and P₂² = 3(1 − x²) = 3 sin²θ, with none. The third derivative of a quadratic is zero, so there is no P₂³.
The −m²/(1 − x²) term cancels exactly. The exponents of u at each pole become 0 and −μ, so a regular P means a u that stays finite.
x = cos θ, μ = |m|, λ the eigenvalue of L² in units of ħ². P, u, x and λ are all dimensionless.
An infinite branch sums to u ~ (1 − x²)(−μ), a logarithm at μ = 0, and rebuilds the singular P ~ (1 − x²)(−μ/2), so one branch must stop and the other be set to zero.
Even and odd k decouple. Unbroken, the ratio tends to 1 + 2(μ − 1)/k, so aₖ grows like k(μ − 1).
A degree cannot be negative, so |m| ≤ l follows from the differential equation plus regularity, the bound Quantum Mechanics I took from the ladder.
k₀ is the degree of u. For |m| ≥ l + 1 every numerator is positive, so neither branch can stop.
μ derivatives of Legendre's equation give the equation for u, so u ∝ dμ/dxμ Pₗ. Once μ ≥ l + 1 that derivative, and Pₗμ with it, is identically zero.
μ = 0, 1, …, l and (1 − x²)(μ/2) = sinμ θ. No (−1)ᵐ here: topic 1.6 puts that phase into Yₗᵐ.
u is a constant, so there is no nodal cone: this is the sinˡ θ of the top state Yₗˡ (here reached without any ladder operator).
(2l − 1)!! = 1 × 3 × 5 × … × (2l − 1), which is 1, 3, 15 and 105 for l = 1 to 4.
Count before plotting. For l = 3 the cones number 3, 2, 1 and 0 as μ runs from 0 to 3: at 39.23°, 90° and 140.77° for μ = 0, at 63.43° and 116.57° for μ = 1.
Pₗ has l simple zeros in (−1, 1) (topic 1.3). Zeros of order μ at θ = 0 and π lie on the z axis and are not cones.
Peel off the pole factor first
Topic 1.1 found two Frobenius behaviours at each pole of the polar equation, (1 − x)(±μ/2) at x = 1 and (1 + x)(±μ/2) at x = −1, with μ = |m|. The regular pair multiplies to (1 − x²)(μ/2), so take it out by hand: P = (1 − x²)(μ/2) u. Then (1 − x²)P′ = (1 − x²)(μ/2)[(1 − x²)u′ − μx u], and differentiating once more produces a term +μ²x²u/(1 − x²). Added to the −μ²u/(1 − x²) already in the equation, the pair combines to μ²(x² − 1)u/(1 − x²) = −μ²u, a plain constant times u. What survives is (1 − x²)u″ − 2(μ + 1)x u′ + [λ − μ(μ + 1)]u = 0, whose exponents at x = 1 are 0 and −μ. Regularity of P has become the demand that u stay finite at both poles. Every symbol is a pure number: x = cos θ, and λ is the eigenvalue of L² in units of ħ², so the dimensional check is automatic and the content is all in the powers.
Run the series, and see why it must stop
Put u = Σ aₖ xᵏ. Collecting xᵏ gives aₖ₊₂ = [(k + μ)(k + μ + 1) − λ] aₖ/[(k + 1)(k + 2)], because k(k − 1) + 2(μ + 1)k + μ(μ + 1) = (k + μ)(k + μ + 1). Unbroken, the ratio tends to 1 + 2(μ − 1)/k, so the coefficients of one parity grow like k(μ − 1) and u ~ (1 − x²)(−μ) at the poles, a logarithm when μ = 0. Then P behaves as (1 − x²)(−μ/2), exactly the singular Frobenius branch, and ∫|P|² dx diverges for every μ ≥ 1; topic 1.2 settled μ = 0. No mixture of an even and an odd unbroken branch can cancel at both poles, so a branch must terminate: the numerator must vanish at some k₀ ≥ 0 of its parity, λ = (k₀ + μ)(k₀ + μ + 1), and the other branch, which that λ cannot stop, is set to zero. Name l = k₀ + μ. Then λ = l(l + 1), as topic 1.2 found for m = 0, and |m| = l − k₀ ≤ l. Try λ = 6, so l = 2, with |m| = 3: every numerator is (k + 3)(k + 4) − 6, at least 6, and nothing ever stops.
Identify the polynomial as μ derivatives of Pₗ
Differentiate Legendre's equation (1 − x²)w″ − 2xw′ + l(l + 1)w = 0 μ times and write v = dμ/dxμ w. By Leibniz, (1 − x²)w″ becomes (1 − x²)v″ − 2μx v′ − μ(μ − 1)v, and −2xw′ becomes −2xv′ − 2μv. Collected, (1 − x²)v″ − 2(μ + 1)x v′ + [l(l + 1) − μ(μ + 1)]v = 0: the equation for u, term for term. The regular solution is unique up to a constant, so u ∝ dμ/dxμ Pₗ, of degree l − μ and parity (−1)(l − μ), the branch the recursion kept. Hence Pₗμ(x) = (1 − x²)(μ/2) dμ/dxμ Pₗ(x), and the bound appears a second time: the (l + 1)-th derivative of a degree-l polynomial is identically zero. For l = 3, P₃ = (5x³ − 3x)/2 gives P₃¹ = (3/2)(5x² − 1)√(1 − x²), P₃² = 15x(1 − x²), P₃³ = 15 sin³θ, and P₃⁴ = 0. Both limits check: μ = 0 returns Pₗ itself, and μ = l gives (2l − 1)!! sinˡ θ, the top state that topic 1.7 reaches from L₊Yₗˡ = 0.
Count the nodal cones with Rolle's theorem
Topic 1.3 showed that Pₗ has l simple zeros inside (−1, 1). Between neighbouring zeros of a smooth function its derivative must vanish, which is Rolle's theorem, so Pₗ′ has at least l − 1 zeros there. Being of degree l − 1 it has room for no more, so those are all of its zeros, each simple, one in every gap. Repeat μ times: dμ/dxμ Pₗ has exactly l − μ simple zeros in (−1, 1). The factor sinμ θ vanishes only at θ = 0 and π, on the z axis, so Pₗμ(cos θ) has exactly l − μ nodal cones of constant θ on which |Yₗᵐ|² vanishes, plus zeros of order μ at the poles. For l = 3 the zeros of P₃ sit at θ = 39.23°, 90° and 140.77°. Those of P₃′, at cos θ = ±1/√5, sit at 63.43° and 116.57°, one in each gap, and P₃″ vanishes only at 90°. That gives 3, 2, 1 and 0 cones for μ = 0 to 3, predicted before anything is plotted.
What the bound means, and how it is checked
Two more readings agree. Geometrically, sinμ θ e(±iμφ) = [(x ± iy)/r]μ exactly, and the polynomial in cos θ = z/r has degree l − μ and parity (−1)(l − μ), so rˡ Pₗμ(cos θ) e(±iμφ) is (x ± iy)μ times a polynomial in z and r² of degree l − μ. The pole factor is what keeps e(imφ) smooth on the z axis, where φ is undefined, and a polynomial of degree l has room for at most l factors of x ± iy. Algebraically, ⟨Lₓ² + Ly²⟩ = ħ²[l(l + 1) − m²] ≥ 0 from Quantum Mechanics I allows m² ≤ l(l + 1), which for integer m is |m| ≤ l again. No power of α enters: relativistic and QED corrections shift the energies of the 2l + 1 states but cannot add one. Derived here: the u equation, termination, |m| ≤ l, Pₗμ and the node count. Quoted: Rolle's theorem, and the spin factor 2 that turns 2l + 1 into the measured widths of the periodic table's blocks, 2, 6, 10 and 14 columns, the 10 being the d block from scandium to zinc.
Change one variable at a time
Make the relationship visible.
Hold l = 3 and step |m| from 0 to 4: the dashed P₃ keeps its three nodes while the solid curve loses one per step, 3, 2, 1 and 0 dots, is pinned flatter at both poles, and at |m| = 4 lies on the axis, because the fourth derivative of a cubic is zero.
λ = l(l + 1)12
LEAST λ ALLOWING THIS |m|2
NODAL CONES (DOTS)2
TERMS IN u (RECURSION)2
Live interpretationλ = l(l + 1): 12. LEAST λ ALLOWING THIS |m|: 2. NODAL CONES (DOTS): 2. TERMS IN u (RECURSION): 2
Catch the common trap
Explain before calculating.
In the polar equation set λ = 2, so l = 1, and |m| = 2. The series for u then has aₖ₊₂/aₖ = (k + 4)/(k + 2), positive for every k, and (1 − x²) d²/dx² P₁(x) = 0. What do these two facts establish?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn l = 4 state has |m| = 2. Starting from P₄(x) = (35x⁴ − 30x² + 3)/8, build P₄²(cos θ), find the polar angles of its nodal cones, and check the count against l − |m|.
- Differentiate twice: P₄′ = (35x³ − 15x)/2 and P₄″ = (105x² − 15)/2 = (15/2)(7x² − 1), a quadratic, so u has degree l − |m| = 2 as the recursion requires.
- Attach the pole factor (1 − x²)(μ/2) with μ = 2, which is 1 − x² = sin²θ: P₄²(cos θ) = (15/2) sin²θ (7cos²θ − 1).
- Nodes: 7cos²θ = 1 gives cos θ = ±1/√7 = ±0.3780, so θ = 67.79° and 112.21°. The factor sin²θ vanishes only at θ = 0° and 180°, on the z axis, which is not a cone.
- Rolle check: P₄ vanishes at cos θ = ±0.3400 and ±0.8611, P₄′ at 0 and ±√(3/7) = ±0.6547, and the two zeros ±0.3780 of P₄″ fall one in each gap between the zeros of P₄′, as interlacing demands.
AnswerP₄²(cos θ) = (15/2) sin²θ (7cos²θ − 1), with nodal cones at θ = 67.79° and 112.21°: two of them, equal to l − |m| = 4 − 2.
MediumTake |m| = 1 and λ = 12 and run the series recursion for u without looking up any Legendre function. Show which branch terminates and identify it with dP₃/dx, then follow the other branch until its coefficients settle, and say how it behaves at the poles.
- With μ = 1 the numerator is (k + 1)(k + 2) − 12 = (k + 5)(k − 2), so aₖ₊₂/aₖ = (k + 5)(k − 2)/[(k + 1)(k + 2)]. It vanishes only at k = 2, so only the even branch can stop, and λ = 12 = 3 × 4 means l = 3.
- Even branch: a₂ = (5)(−2)/(1 × 2) a₀ = −5a₀ and a₄ = 0, so u = a₀(1 − 5x²). Since dP₃/dx = (15x² − 3)/2 = −(3/2)(1 − 5x²), the choice a₀ = −3/2 gives P₃¹ = (3/2)(5cos²θ − 1) sin θ.
- Its nodal cones sit at cos θ = ±1/√5 = ±0.4472, that is θ = 63.43° and 116.57°: two, as l − |m| = 2 requires.
- Odd branch: a₃ = (6)(−1)/(2 × 3) a₁ = −a₁, a₅ = (8)(1)/(4 × 5) a₃ = −0.4000a₁, a₇ = (10)(3)/(6 × 7) a₅ = −0.2857a₁ and a₉ = −0.2381a₁. In closed form a₂ⱼ₊₁ = −a₁(j + 1)(j + 2)/[2(2j − 1)(2j + 1)] for j ≥ 1, which tends to −a₁/8, not to zero.
- Coefficients that level off make u ≈ −(a₁/8)x/(1 − x²) near the poles, so P = sin θ × u grows like |a₁|/(8 sin θ). Then |P|² sin θ ≈ a₁²/(64θ) and the norm diverges logarithmically: this is the singular Frobenius branch, exponent −1/2, and it is discarded.
AnswerOnly the even branch stops: u = a₀(1 − 5x²) ∝ dP₃/dx, giving P₃¹ = (3/2)(5cos²θ − 1) sin θ with cones at 63.43° and 116.57°. The odd coefficients tend to −a₁/8, so that branch grows like 1/sin θ at the poles and is rejected.
HardTry to build an l = 2 state with |m| = 3. Show that the recursion for u never terminates, sum its even branch in closed form, verify that the resulting P solves the polar equation, and decide from the norm whether it can be a state. Then find the smallest λ that does admit |m| = 3.
- With μ = 3 and λ = 6 the numerator is (k + 3)(k + 4) − 6 = (k + 1)(k + 6), so aₖ₊₂/aₖ = (k + 6)/(k + 2): positive for every k ≥ 0, and neither branch can stop.
- Even branch, k = 2j: a₂ⱼ₊₂/a₂ⱼ = (j + 3)/(j + 1), so a₂ⱼ = a₀(j + 1)(j + 2)/2, the triangular numbers 1, 3, 6, 10, 15 times a₀. Their generating function is Σ (j + 1)(j + 2)yʲ/2 = (1 − y)⁻³, so with y = x² the branch sums to u = a₀/(1 − x²)³.
- So P = (1 − x²)³⁄² u = a₀/sin³θ. Check with q = 1 − x²: (1 − x²)P″ = a₀(3q(−3/2) + 15x²q(−5/2)), −2xP′ = −6a₀x²q(−5/2), and (6 − 9/q)P = a₀(6q(−3/2) − 9q(−5/2)). The sum is 9a₀q(−5/2)(q + x² − 1) = 0: an exact solution.
- Norm: |P|² sin θ = a₀²/sin⁵θ ≈ a₀²θ⁻⁵ near θ = 0, and the integral of θ⁻⁵ from ε upward grows as 1/(4ε⁴). The odd branch, a₂ⱼ₊₁ = a₁(2j + 3)(2j + 5)/15, grows as fast, and no mixture cancels both poles, because the even part is symmetric under x → −x and the odd part antisymmetric.
- Cross-check: ⟨Lₓ² + Ly²⟩ = ⟨L²⟩ − ⟨Lz²⟩ would be ħ²(6 − 9) = −3ħ², negative for a sum of squares of Hermitian operators. The smallest λ admitting μ = 3 makes the numerator vanish at k = 0: λ = 3 × 4 = 12, so l = 3, u is a constant and P₃³ = 15 sin³θ, whose pole factor is sin³θ where the λ = 6 solution had 1/sin³θ.
AnswerNo such state: with λ = 6 and |m| = 3 the even solution is exactly a₀/sin³θ, whose norm diverges like the integral of θ⁻⁵, and the odd one fails the same way. |m| = 3 first becomes possible at λ = 12 (l = 3), with P₃³ = 15 sin³θ.