University Quantum Mechanics II · Deriving the Spherical Harmonics · 1.2

Legendre's equation by series, and why it must terminate

Quantum Mechanics I took l(l + 1) from commutators. Here a differential equation produces it, and the point of the lesson is where exactly it comes from: not from Legendre's equation, which has a convergent series solution for every λ, but from demanding that the solution stay finite at both poles of the sphere.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

For m = 0 the polar equation in x = cos θ is Legendre's equation, (1 − x²)P″ − 2xP′ + λP = 0. Its only singular points at finite x are the poles x = ±1, so a power series about the equator converges for |x| < 1 whatever λ is, and the recursion aₖ₊₂ = [k(k + 1) − λ]aₖ/[(k + 1)(k + 2)] splits it into an even and an odd branch. The recursion never breaks down.

What it does, unless some factor k(k + 1) − λ is exactly zero, is let the coefficients shrink only as 1/k, and Σ xᵏ/k over one parity is a logarithm, infinite at both poles. Mixing the two branches cannot help: expanded about x = 1 the equation has just one solution finite there, and for general λ that one is logarithmically infinite at x = −1. So a solution finite on the whole sphere exists only when one branch terminates, which needs λ = l(l + 1) with l = 0, 1, 2, …, and the other branch, which then never terminates, is given zero weight.

The equation supplies the recursion; the eigenvalue comes from the boundary condition, which Topic 1.1 showed is physical, since the logarithmic solution carries infinite ⟨L²⟩. The argument is about functions on the sphere, so it says nothing about spin, whose half-integers the ladder algebra allows, and for m ≠ 0 it needs the factor (1 − x²)^(|m|/2) removed first, the job of Topic 1.5.

Simple definition: Legendre's equation has a convergent series solution for every separation constant λ, but a solution finite at both poles exists only when that series terminates, which happens exactly when λ = l(l + 1) with l a non-negative integer.
Example: At λ = 6 the recursion gives a₂ = −3a₀ and then a₄ = (6 − 6)a₂/12 = 0, so 1 − 3x² = −2P₂(x) is finite everywhere. At λ = 5, from a₀ = 1, it gives a₂ = −2.5, a₄ = −0.208, a₆ = −0.104 and never stops, with k aₖ settling on −0.393.

Legendre's equation, m = 0(1 − x²)P″ − 2xP′ + λP = 0

Divided by 1 − x², its coefficients are singular only at x = ±1, so a power series about x = 0 converges for |x| < 1 at every λ. The equation alone quantises nothing.

x = cos θ runs from −1 (south pole) to +1 (north pole); λ = L²/ħ² and P are dimensionless.

Two-step recursionaₖ₊₂ = [k(k + 1) − λ] aₖ / [(k + 1)(k + 2)]

It links k to k + 2 only, so each parity has its own fate: a branch stops exactly when k(k + 1) = λ for some k of that parity.

P = Σ aₖ xᵏ with dimensionless aₖ; a₀ launches the even branch and a₁ the odd one, and neither fixes λ.

The tail of an unterminated branch(k + 2) aₖ₊₂ = [1 − λ/(k(k + 1))] k aₖ

Coefficients falling as 1/k sum like Σ xᵏ/k, which is −½ ln(1 − x²) for even k and artanh x for odd k: finite inside, logarithmically infinite at both poles.

The brackets tend to 1 fast enough that k aₖ → C, non-zero unless one bracket vanishes; so aₖ ≈ C/k.

Series about the north pole(n + 1)² cₙ₊₁ = [n(n + 1) − λ] cₙ, where P = Σ cₙ tⁿ and t = (1 − x)/2

Every solution finite at x = 1 is a multiple of this one, and n cₙ also tends to a constant, so it is infinite at x = −1 unless it stops at n = l.

t = 0 at x = 1 and t = 1 at x = −1; c₀ = P(1) = 1. The second solution there carries ln t.

Strength of the far-pole logarithm (quoted)P_ν(x) → (sin νπ / π) ln[(1 + x)/2] + finite, as x → −1

It vanishes only at integer ν, exactly where termination does: λ = 5 gives −0.194, λ = 15/4 gives −1/π, and λ = 6 gives zero.

λ = ν(ν + 1), so ν = √(λ + ¼) − ½. Quoted: the limit of n cₙ is a Γ-function product, evaluated by Euler's reflection formula.

The allowed separation constantsλ = l(l + 1), with l = 0, 1, 2, …

Derived here from finiteness at both poles, not from commutators. The polynomial that survives is Pₗ(x), of degree l and parity (−1)^l.

So L² = ħ²l(l + 1), in J² s², with ħ = 1.055 × 10⁻³⁴ J s. The branch of parity opposite to (−1)^l gets zero weight.

The series exists for every λ

Put P = Σ aₖ xᵏ into (1 − x²)P″ − 2xP′ + λP = 0 and collect the coefficient of xᵏ. P″ supplies (k + 2)(k + 1)aₖ₊₂, while −x²P″ − 2xP′ + λP supply [−k(k − 1) − 2k + λ]aₖ, so (k + 1)(k + 2)aₖ₊₂ = [k(k + 1) − λ]aₖ. Two coefficients stay free, a₀ and a₁, as a second-order equation requires, and they launch an even and an odd branch that never mix. Nothing in this restricts λ. Take λ = 5 and a₀ = 1: a₂ = −2.5, a₄ = −0.208, a₆ = −0.104, and on without end. Because the equation's coefficients are singular only where 1 − x² = 0, both series converge for every |x| < 1 and satisfy the equation exactly there. The oscillator of Quantum Mechanics I behaved the same way: the Hermite recursion accepted any energy until you asked what the series did at large |ξ|. Quantisation is never inside a recursion. It is in what the series does at the edge of its interval.

Coefficients that fall only as 1/k

The ratio aₖ₊₂/aₖ = [k/(k + 2)][1 − λ/(k(k + 1))] tends to 1, so the ratio test is silent at x = ±1. Multiply by k + 2 instead: (k + 2)aₖ₊₂ = [1 − λ/(k(k + 1))] k aₖ. The brackets approach 1 fast enough, since Σ λ/k² converges, that their product settles on a non-zero number unless one bracket is exactly zero. So k aₖ → C and aₖ ≈ C/k. At λ = 0 the odd branch shows it with no approximation at all: aₖ₊₂ = k aₖ/(k + 2) gives aₖ = 1/k, and x + x³/3 + x⁵/5 + … = artanh x = −ln tan(θ/2), which is 4.74 at θ = 1° and unbounded at the pole. In general an even branch ends like −½C ln(1 − x²) and an odd one like C artanh x. The divergence is only logarithmic, far gentler than the oscillator's e^(ξ²), and the function is even square-integrable; but its ⟨L²⟩ is infinite, as Topic 1.1 showed, so it is not an acceptable state.

No mixture of infinite branches survives both poles

Suppose neither branch stops, and try P = A yₑᵥₑₙ + B y_odd with plateaus k aₖ → Cₑᵥₑₙ and C_odd. Near x = 1 both branches end in −½ ln(1 − x), with weights Cₑᵥₑₙ and C_odd; near x = −1 the odd branch changes sign. The logarithms cancel at the north pole if A Cₑᵥₑₙ + B C_odd = 0 and at the south pole if A Cₑᵥₑₙ − B C_odd = 0, and both hold only if A Cₑᵥₑₙ = 0 and B C_odd = 0: each branch present must stop on its own. Expanding about the pole gives the same verdict. With t = (1 − x)/2 the equation becomes t(1 − t)P″ + (1 − 2t)P′ + λP = 0, whose indicial equation has the double root 0, so one solution is a power series in t and the other carries ln t. The finite one obeys (n + 1)²cₙ₊₁ = [n(n + 1) − λ]cₙ, and n cₙ tends to a constant exactly as before, so at t = 1, the south pole, it diverges unless it stops. Every solution finite at the north pole is a multiple of it; nothing is left to mix.

Termination picks λ = l(l + 1) and one parity

The way out is a vanishing factor: if k(k + 1) = λ at k = l, then aₗ₊₂ = 0 and every later coefficient of that parity with it. So λ = l(l + 1) with l a non-negative integer; a negative l only repeats a λ, since l → −l − 1 leaves l(l + 1) unchanged. At λ = 6 the even branch stops at 1 − 3x², which scaled to P(1) = 1 is P₂ = (3x² − 1)/2. The odd branch at λ = 6 cannot stop, because k(k + 1) − 6 = (k − 2)(k + 3) has no positive odd root, and its plateau is k aₖ → −0.500, so its weight must be zero. One branch terminating is necessary; zero weight on the other completes the condition. The survivor Pₗ has degree l and parity (−1)^l, and the series about x = 1 agrees, since its factor n(n + 1) − λ also vanishes at n = l. A half-integer l, which the ladder algebra allowed, gives λ = ¾, 15/4, 35/4, …, never an integer, while every k(k + 1) is one: orbital l is an integer because the wavefunction must be finite on the sphere.

Eigenvalues are the zeros of a far-pole divergence

Launch the solution finite at the north pole, unique up to a factor and scaled to P(1) = 1, and follow it to the south pole. There it behaves as S ln[(1 + x)/2] with S = sin(νπ)/π, a strength quoted from the connection formulae but checked here by summing the series. S is a smooth function of λ, and the eigenvalues are its zeros: 0, 2, 6, 12 and on. Near a zero it is linear, S ≈ (−1)^l (ν − l) with ν − l ≈ [λ − l(l + 1)]/(2l + 1), so λ = 5.75 gives S ≈ −0.050, against the exact −0.0503. That is the logic of a shooting method, and it hides a trap, because a logarithm announces itself late. At λ = 5.75 the solution is 1.315 at 1° from the south pole, 1.547 at 0.1° and 1.779 at 0.01°, climbing by only 2|S| ln 10 = 0.232 for each tenfold approach: on any plot it looks finite. The reliable test is analytic, not graphical: n cₙ must go to zero, and it does so only when the series stops.

Checks: dimensions, limits and a measured spectrum

x = cos θ is a pure number, so λ is too, and L² = ħ²l(l + 1) carries the J² s² of ħ², as it must. No charge and no c enter, so there is no order in α to count: the result is geometry, the same for an electron, a nucleus or a molecule. For large l, l(l + 1) = (l + ½)² − ¼, so |L| → (l + ½)ħ, the semiclassical value; at l = 10 the two differ by 0.012ħ. The measurement is a rotational spectrum. A rigid rotor has E = ħ²λ/2I and absorbs on l → l + 1, so λ = l(l + 1) puts the lines in the ratio 1 : 2 : 3. The lowest three of ¹²C¹⁶O, at 115.2712, 230.5380 and 345.7960 GHz, stand at 1 : 1.99996 : 2.99985, the shortfall being centrifugal stretching (quoted). The classical-looking λ = l² predicts 1 : 3 : 5. A spectrum cannot tell l(l + 1) from (l + ½)², though: the ¼ shifts every level alike and cancels from every line. Fixing that constant is the series argument's job, and the centrifugal term of the hydrogen radial equation, Unit 2, depends on it.

Change one variable at a time

Make the relationship visible.

Interactive model

Separation constant λ5.00

Start at λ = 5 and watch each curve escape at its far pole; drag to 6 and both close onto P₂ = (3x² − 1)/2, ending at +1 at both poles. At λ = 2 the dashed curve is the solid one flipped, −x against x: parity (−1)^l.

ν = √(λ + ¼) − ½1.791

FAR-POLE LOG STRENGTH-0.194

NEAREST INTEGER l2

RECURSION FACTOR aₗ₊₂/aₗ0.083

Live interpretationν = √(λ + ¼) − ½: 1.791. FAR-POLE LOG STRENGTH: −0.194. NEAREST INTEGER l: 2. RECURSION FACTOR aₗ₊₂/aₗ: 0.083

Catch the common trap

Explain before calculating.

At λ = 2 the recursion stops the odd branch of Legendre's series at a₁x, while the even branch runs on as 1 − x² − x⁴/3 − x⁶/5 − …. Which solutions are finite at both poles?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyRun the recursion aₖ₊₂ = [k(k + 1) − λ]aₖ/[(k + 1)(k + 2)] at λ = 12, once from a₁ = 1 and once from a₀ = 1. Which branch stops? Scale that one to P(1) = 1, and use the other branch's first coefficients to decide whether it may be kept.

A branch stops where k(k + 1) = 12, which is k = 3. That k is odd, so only the odd branch can terminate.
Odd branch: a₃ = (2 − 12)/6 = −5/3, then a₅ = (12 − 12)/20 × a₃ = 0, and every later odd coefficient vanishes. The solution is x − (5/3)x³.
At x = 1 it equals 1 − 5/3 = −2/3, so multiply by −3/2: P₃(x) = (5x³ − 3x)/2. Check: P₃(1) = 1, P₃(−1) = −1, odd parity (−1)³, zeros at x = 0 and ±√(3/5) = ±0.775.
Even branch: a₂ = (0 − 12)/2 = −6, a₄ = (6 − 12)/12 × (−6) = 3, a₆ = (20 − 12)/30 × 3 = 0.800, a₈ = (42 − 12)/56 × 0.800 = 0.429, a₁₀ = (72 − 12)/90 × 0.429 = 0.286.
So k aₖ runs 12, 4.80, 3.43, 2.86. No non-negative even k has k(k + 1) = 12, so no zero factor lies ahead, and iterated far out k aₖ settles on 1.50. The branch ends like −0.75 ln(1 − x²), infinite at both poles.

AnswerThe odd branch stops, giving P₃(x) = (5x³ − 3x)/2. The even branch never stops (k aₖ → 1.50), diverges at both poles, and must be given zero weight.

MediumThe three lowest rotational absorption lines of ¹²C¹⁶O are at 115.2712, 230.5380 and 345.7960 GHz. A rigid rotor has E = ħ²λ/2I and absorbs on l → l + 1. Decide whether the lines support λ = l(l + 1) or the classical-looking λ = l², then find the moment of inertia and the bond length, given μ = 6.8562 u and 1 u = 1.6605 × 10⁻²⁷ kg.

A line has hf = ħ²[λ(l + 1) − λ(l)]/2I. For λ = l(l + 1) the bracket is 2(l + 1), so the lines stand in the ratio 1 : 2 : 3; for λ = l² it is 2l + 1, giving 1 : 3 : 5.
Measured: 230.5380/115.2712 = 1.99996 and 345.7960/115.2712 = 2.99985. The l² rule would put the second line at 3 × 115.2712 = 345.8 GHz and has nothing near 230.5 GHz: it fails by 50%.
The small shortfalls, 4 × 10⁻⁵ and 1.5 × 10⁻⁴, grow with l. They are centrifugal stretching of the bond, a quoted correction E → E − D[l(l + 1)]² with D/h = 183 kHz, not a defect in l(l + 1).
Lowest line, l = 0 → 1: hf₁ = ħ²(2 − 0)/2I = ħ²/I, so I = h/(4π²f₁) = 6.626 × 10⁻³⁴ J s ÷ (39.48 × 1.152712 × 10¹¹ s⁻¹) = 1.456 × 10⁻⁴⁶ kg m².
μ = 6.8562 × 1.6605 × 10⁻²⁷ kg = 1.1385 × 10⁻²⁶ kg, so r = √(I/μ) = √(1.2789 × 10⁻²⁰ m²) = 1.131 × 10⁻¹⁰ m. Dimensions: kg m² ÷ kg = m², so the root is a length.
Limit of the test: (l + ½)² has the same differences 2(l + 1), since the extra ¼ shifts every level by ħ²/8I, h × 14.4 GHz here, and cancels from every line. The spectrum fixes the l(l + 1) structure; the series fixes the constant.

AnswerThe lines stand at 1 : 2.0000 : 2.9998, as λ = l(l + 1) requires (l² would give 1 : 3 : 5). I = 1.456 × 10⁻⁴⁶ kg m² and r = 113.1 pm.

HardTake λ = 15/4, halfway in ν between the eigenvalues 2 and 6. Show that no branch stops, find the plateaus Cₑᵥₑₙ and C_odd of the two branches about x = 0, use the series about the north pole to build the one solution finite at x = 1, and find the strength of its logarithm at x = −1 in two independent ways.

λ = ν(ν + 1) gives ν = 3/2, and k(k + 1) − 15/4 = (k − 3/2)(k + 5/2) has no integer root, so no branch of either series stops.
Iterating aₖ₊₂ = [k(k + 1) − 15/4]aₖ/[(k + 1)(k + 2)] to k ≈ 10⁶ gives k aₖ → Cₑᵥₑₙ = −0.8090 from a₀ = 1 and C_odd = −0.3934 from a₁ = 1. The early even values, −3.75, −1.41, −1.14, close in on the limit only as λ/k.
About x = 1 with c₀ = 1: c₁ = −15/4, c₂ = (2 − 15/4)/4 × c₁ = 1.6406, c₃ = (6 − 15/4)/9 × c₂ = 0.4102. At the equator t = ½ it converges geometrically, giving P(0) = Σ cₙ/2ⁿ = −0.3934 and P′(0) = −Σ n cₙ/2ⁿ = 0.8090.
So the north-finite solution is P = P(0) yₑᵥₑₙ + P′(0) y_odd. Its logarithm at x = 1 has coefficient −½[P(0)Cₑᵥₑₙ + P′(0)C_odd] = −½[(−0.3934)(−0.8090) + (0.8090)(−0.3934)] = 0, as it must.
At x = −1 the odd branch changes sign, so the coefficient of ln(1 + x) is −½[P(0)Cₑᵥₑₙ − P′(0)C_odd] = −½(0.3183 + 0.3183) = −0.3183.
Second route: iterating the north-pole series gives n cₙ → 0.3183, so near the south pole P ≈ −0.3183 ln(1 − t) = −0.3183 ln[(1 + x)/2]. Both routes give −0.3183 = −1/π, the quoted sin(νπ)/π at ν = 3/2, and P rises to +∞ there.

AnswerNo branch stops (Cₑᵥₑₙ = −0.8090, C_odd = −0.3934). The unique solution finite at x = 1 carries −0.318 ln(1 + x) at x = −1, equal to sin(νπ)/π = −1/π by both routes, so at λ = 15/4 no solution is finite at both poles.