University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.6
The associated Laguerre polynomials as the terminated series
Topic 02.05 stopped the series and left a polynomial known only through its recursion. Here it gets its name: we show it solves the associated Laguerre equation, build it a second way from Rodrigues' formula, and learn to read the three conventions that print one hydrogen state as three different-looking formulas.
Build the model
Connect the measurement to the mechanism.
Once ρ₀ = 2n, the series for v(ρ) stops at degree nᵣ = n − l − 1, and what is left is a polynomial defined by one recursion, cⱼ₊₁/cⱼ = 2(j + l + 1 − n)/[(j + 1)(j + 2l + 2)]. Write its differential equation in x = 2ρ = 2r/(na) and it becomes x v″ + (2l + 2 − x) v′ + (n − l − 1) v = 0, the associated Laguerre equation with upper index p = 2l + 1 and degree k = n − l − 1. The identification is derived, not spotted: the coefficients of Lₖᵖ(x), obtained from Rodrigues' formula and p derivatives, have exactly the recursion's ratio once x is written as 2ρ, for every n and l.
The name imports tabulated facts (the value (n + l)!/[(n − l − 1)!(2l + 1)!] at the origin, zeros that are real, simple and positive, a normalisation integral) and brings a hazard: texts disagree by factorials, a sign and the labelling, and the older physics symbol Lₙ₊ₗ²ˡ⁺¹ is −(n + l)! times the modern Lₙ₋ₗ₋₁²ˡ⁺¹. That hazard is why this unit names its convention at every formula and takes no normalisation constant on trust. The name itself adds no physics: the energies were fixed by termination in 02.05, before the polynomial had a name.
- Simple definition
- The associated Laguerre polynomial Lₖᵖ(x) is the degree-k polynomial that solves x y″ + (p + 1 − x) y′ + k y = 0, fixed in the modern convention by Lₖᵖ(0) = (k + p)!/(k! p!); hydrogen's terminated series is Lₙ₋ₗ₋₁²ˡ⁺¹(2r/(na)).
- Example
- For 2s, n = 2 and l = 0 give L₁¹(x) = 2 − x with x = 2r/(2a) = r/a, so the single radial node sits at r = 2a = 1.06 × 10⁻¹⁰ m. The recursion's c₁ = −c₀ gives v = c₀(1 − ρ), zero at ρ = 1: the same place.
Set beside x y″ + (p + 1 − x) y′ + k y = 0, it reads off p = 2l + 1 and k = n − l − 1 = nᵣ. The name comes from the equation, not from matching a few examples.
x and v dimensionless; r and a = 4πε₀ħ²/(μe²) in m; integers n ≥ 1 and 0 ≤ l ≤ n − 1. Derived from topic 02.03's v equation.
Differentiation alone generates the plain Laguerre polynomials. The Leibniz rule expands it to Σₛ (−1)ˢ k! xˢ/[(k − s)! (s!)²], checked here to solve x y″ + (1 − x) y′ + k y = 0.
x dimensionless, k = 0, 1, 2, …; the 1/k! makes Lₖ(0) = 1, so L₁ = 1 − x and L₂ = 1 − 2x + x²/2
Derived by differentiating Laguerre's equation p times. It gives L₁¹ = −dL₂/dx = 2 − x, L₂¹ = 3 − 3x + x²/2 and L₁³ = −d³L₄/dx³ = 4 − x.
p derivatives lower the degree from k + p to k; the (−1)ᵖ keeps the constant term positive. Hydrogen needs odd p = 2l + 1.
Derived by expansion. In ρ = x/2 successive coefficients have ratio −2(k − s)/[(s + 1)(s + p + 1)]: with k = n − l − 1, p = 2l + 1 it is the recursion factor, for every n and l.
Lₖᵖ(0) = (k + p)!/(k! p!); for hydrogen k + p = n + l, so the 2s, 3s and 3p values are 2, 3 and 4
ρˡ comes from the origin, e(−ρ) from infinity and the polynomial from termination. Take Nₙₗ from your own normalisation integral, never from a table written in another convention.
R and Nₙₗ in m(−3/2); ρ dimensionless; modern convention, Nₙₗ fixed by ∫₀^∞ R² r² dr = 1
Derived from the two definitions. Same zeros, so the same nodes, but a factor −6 for 3s and −5040 for 4f; after direct normalisation only the sign survives, an unobservable phase.
Old: Lₘ = eˣ dᵐ/dxᵐ(e⁻ˣ xᵐ) with no 1/m!, then dᵖLₘ/dxᵖ with no (−1)ᵖ, labelled by m = n + l
Rewrite the v equation in x = 2ρ
Three results arrive from earlier topics and are used, not re-derived: ρ v″ + 2(l + 1 − ρ) v′ + [ρ₀ − 2(l + 1)] v = 0 from 02.03, the recursion from 02.04, and ρ₀ = 2n from 02.05. Put x = 2ρ. Then d/dρ = 2 d/dx, so ρ v″ becomes (x/2)(4 d²v/dx²) = 2x d²v/dx², the middle term becomes 4(l + 1) dv/dx − 2x dv/dx, and the last term is 2(n − l − 1) v. Divide by 2: x v″ + (2l + 2 − x) v′ + (n − l − 1) v = 0. Beside the associated Laguerre equation x y″ + (p + 1 − x) y′ + k y = 0 this reads p = 2l + 1 and k = n − l − 1 = nᵣ. The 2 in 2ρ is not cosmetic: in ρ the middle coefficient is 2l + 2 − 2ρ, and the match fails. Both variables are dimensionless, as a polynomial's argument must be; the scale lives in a = 4πε₀ħ²/(μe²) = ħ/(μcα), which is 5.2947 × 10⁻¹¹ m with hydrogen's reduced mass.
Rodrigues' formula builds the plain polynomials
Lₖ(x) = (eˣ/k!) dᵏ/dxᵏ(e⁻ˣ xᵏ) asks for nothing but differentiation. For k = 2, d²/dx²(x² e⁻ˣ) = (2 − 4x + x²) e⁻ˣ, so L₂ = 1 − 2x + x²/2. For k = 3 the third derivative of x³ e⁻ˣ is (6 − 18x + 9x² − x³) e⁻ˣ, so L₃ = 1 − 3x + 3x²/2 − x³/6. The Leibniz rule gives every case at once: Lₖ(x) = Σₛ (−1)ˢ k! xˢ/[(k − s)! (s!)²]. That this solves Laguerre's equation x y″ + (1 − x) y′ + k y = 0 is checked rather than assumed: a power series Σ aₛ xˢ in that equation gives aₛ₊₁ = (s − k) aₛ/(s + 1)², and with a₀ = 1 this is the same list, stopping at s = k. Two limits guard the formula. Lₖ(0) = 1 for every k, which is what the 1/k! enforces, and the top term is (−1)ᵏ xᵏ/k!, so Lₖ takes the sign of (−1)ᵏ at large x.
Differentiate p times for the associated polynomial
Differentiate Laguerre's equation for y = Lₘ once: x y‴ + (2 − x) y″ + (m − 1) y′ = 0. Each derivative raises the 1 in (1 − x) by one and lowers m by one, so w = dᵖLₘ/dxᵖ obeys x w″ + (p + 1 − x) w′ + (m − p) w = 0. With m = k + p that is the associated equation of degree k, so Lₖᵖ(x) = (−1)ᵖ dᵖ/dxᵖ Lₖ₊ₚ(x) is derived here, the (−1)ᵖ only making the constant term positive. From L₃, L₂¹ = −dL₃/dx = 3 − 3x + x²/2. From L₄ = 1 − 4x + 3x² − 2x³/3 + x⁴/24, L₁³ = −d³L₄/dx³ = 4 − x. And L₀ᵖ = 1 for every p, because the p-th derivative of Lₚ is the constant (−1)ᵖ. Expanding the general case gives the coefficients (−1)ˢ (k + p)!/[(k − s)! (p + s)! s!], and the value at the origin (k + p)!/(k! p!): 2 for 2s, 3 for 3s, 4 for 3p and 120 for 7f.
Check against the recursion for n up to 3
Take c₀ = 1. For n = 2, l = 0 the recursion gives c₁ = −1, so v = 1 − ρ, while L₁¹(2ρ) = 2 − 2ρ = 2(1 − ρ). For n = 3, l = 0 it gives c₁ = −2 and c₂ = 2/3, while L₂¹(2ρ) = 3 − 6ρ + 2ρ² = 3(1 − 2ρ + 2ρ²/3). For n = 3, l = 1 it gives c₁ = −1/2, while L₁³(2ρ) = 4 − 2ρ = 4(1 − ρ/2). The nodeless 1s, 2p and 3d have k = 0 and v = L₀ᵖ = 1. Every case obeys v(ρ) = c₀ Lₙ₋ₗ₋₁²ˡ⁺¹(2ρ)/Lₙ₋ₗ₋₁²ˡ⁺¹(0), and the general proof is one ratio: in ρ, successive Laguerre coefficients have the ratio −2(k − s)/[(s + 1)(s + 2l + 2)], which is the recursion factor. The check earns its keep on the argument. Write L₁¹(ρ) for L₁¹(2ρ) and the 2s node moves from ρ = 1, r = 2a, to ρ = 2, r = 4a; write 2r/a for 2r/(na) and it moves in to r = a. Nodes are not read off a spectrum, but |ψₙ₀₀(0)|² is, through the hyperfine splitting, and it carries L(0); the Hard example below tests it.
Three conventions, one set of zeros
Texts differ in three places: whether Rodrigues' formula carries 1/k!, whether the p-fold derivative carries (−1)ᵖ, and whether the lower index is the degree k = n − l − 1 or k + p = n + l. This unit uses the modern convention of the NIST Digital Library of Mathematical Functions, which scipy.special.genlaguerre also follows. Drop only the 1/k! and every hydrogen polynomial is multiplied by (n + l)!. The older physics convention drops both and labels by n + l, so its Lₙ₊ₗ²ˡ⁺¹ is −(n + l)! times the modern Lₙ₋ₗ₋₁²ˡ⁺¹: for 3s its L₃¹(x) = −18 + 18x − 3x², which is −6 times 3 − 3x + x²/2. A constant factor moves no zero, so every convention agrees on the nodes, and a directly normalised Rₙₗ absorbs the factorial and keeps at most a sign. What fails is mixing: an old-convention polynomial under a modern normalisation constant makes ∫R² r² dr equal to [(n + l)!]², 36 for 3s and 2.5 × 10⁷ for 4f.
What the name buys, and where it stops
Knowing the polynomial is Lₖᵖ imports results proved once for the whole family, which this lesson quotes. Its k zeros are real, simple and positive, so the nᵣ radial nodes are genuine sign changes and never mere touching points. The normalisation integral ∫₀^∞ x(p + 1) e⁻ˣ [Lₖᵖ(x)]² dx = (k + p)!(2k + p + 1)/k! comes from tables; for 2s it is 2! × 4 = 8, which you can confirm in one line, and topic 02.07 turns it into the general Nₙₗ. None of this is new physics. The energies came from termination, and the polynomial is exact only for a non-relativistic, spinless electron bound to a point charge. The Dirac equation, quoted from relativistic quantum mechanics, replaces the integer power ρˡ by powers set by γ = √(κ² − (Zα)²), with energy shifts of relative order (Zα)², 5.3 × 10⁻⁵ for hydrogen. In sodium the valence electron's effective n is not an integer, so its series never terminates and there is no polynomial to name.
Change one variable at a time
Make the relationship visible.
Hold nᵣ = 2 and raise l from 0 to 3: there are still two nodes, but they move out from 1.90 a and 7.10 a to 18.0 a and 36.0 a, because p = 2l + 1 grows and the argument 2r/(na) stretches with n. Then set l = 0 and step nᵣ up: each step adds one Laguerre zero and one node.
PRINCIPAL n = nᵣ + l + 13
L AT ORIGIN (n + l)!/(nᵣ! p!)3
OLD ÷ MODERN = −(n + l)!-6
OUTER NODE (0 = NONE)7.10 a
Live interpretationPRINCIPAL n = nᵣ + l + 1: 3. L AT ORIGIN (n + l)!/(nᵣ! p!): 3. OLD ÷ MODERN = −(n + l)!: −6. OUTER NODE (0 = NONE): 7.10 a
Catch the common trap
Explain before calculating.
Hydrogen's 3p state has n = 3 and l = 1. In this lesson's modern convention, which polynomial factor belongs in R₃₁(r), and where is its radial node?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyFor hydrogen 3s (n = 3, l = 0), build L₂¹(x) from Rodrigues' formula and one derivative, confirm it against the recursion with c₀ = 1, and locate both radial nodes in units of a and in metres, taking a = 5.29 × 10⁻¹¹ m.
- Indices: k = n − l − 1 = 2 and p = 2l + 1 = 1, so L₂¹ = −dL₃/dx, and L₃ comes first.
- Rodrigues: d³/dx³(x³ e⁻ˣ) = (6 − 18x + 9x² − x³) e⁻ˣ, so L₃ = (6 − 18x + 9x² − x³)/3! = 1 − 3x + 3x²/2 − x³/6. Then L₂¹ = −(−3 + 3x − x²/2) = 3 − 3x + x²/2.
- Recursion with n = 3, l = 0: c₁ = 2(0 + 1 − 3)/(1 × 2) = −2, c₂ = 2(1 + 1 − 3)/(2 × 3) × (−2) = 2/3, and c₃ = 0. In x = 2ρ, L₂¹ = 3 − 6ρ + 2ρ² = 3(1 − 2ρ + 2ρ²/3): the same polynomial, with L₂¹(0) = 3.
- Zeros: x² − 6x + 6 = 0 gives x = 3 ∓ √3 = 1.268 and 4.732. With x = 2r/(3a), r = 1.5 a × x = 1.902 a and 7.098 a.
- In metres: 1.902 × 5.29 × 10⁻¹¹ m = 1.01 × 10⁻¹⁰ m and 7.098 × 5.29 × 10⁻¹¹ m = 3.75 × 10⁻¹⁰ m. Checks: two nodes, as nᵣ = 2 requires, both inside the classical turning point 2n²a = 18a.
AnswerL₂¹(x) = 3 − 3x + x²/2 = 3(1 − 2ρ + 2ρ²/3), as the recursion gives; nodes at r = 1.90 a and 7.10 a, that is 1.01 × 10⁻¹⁰ m and 3.75 × 10⁻¹⁰ m.
MediumAn older table gives the 4p (n = 4, l = 1) polynomial as L₅³(x) = −60x² + 600x − 1200. Convert it to this lesson's convention, check it two independent ways, and find the radial nodes of the 4p state of He⁺ (Z = 2) in picometres, taking a = 52.9 pm.
- Here n + l = 5, so the modern polynomial has k = n − l − 1 = 2 and p = 3, and the conversion factor is −(n + l)! = −120: L₂³(x) = (−60x² + 600x − 1200)/(−120) = 10 − 5x + x²/2.
- Explicit coefficients, (k + p)!/[(k − s)! (p + s)! s!] with alternating signs: 5!/(2! 3!) = 10, 5!/(1! 4! 1!) = 5 and 5!/(0! 5! 2!) = 1/2. It agrees.
- Recursion with n = 4, l = 1, c₀ = 1: c₁ = 2(0 + 2 − 4)/(1 × 4) = −1, c₂ = 2(1 + 2 − 4)/(2 × 5) × (−1) = 1/5, c₃ = 0. In x = 2ρ, L₂³ = 10 − 10ρ + 2ρ² = 10(1 − ρ + ρ²/5). The same polynomial again.
- Zeros: x² − 10x + 20 = 0 gives x = 5 ∓ √5 = 2.764 and 7.236. For one electron and nuclear charge Ze, a becomes a/Z, so x = 2Zr/(na) = r/a for Z = 2, n = 4.
- Nodes: r = 2.764 × 52.9 pm = 146 pm and 7.236 × 52.9 pm = 383 pm. Using a = 52.9 pm ignores He⁺'s reduced-mass correction, 1.4 parts in 10⁴, far below three figures.
AnswerL₂³(x) = 10 − 5x + x²/2 (the table's entry divided by −5! = −120); the 4p nodes of He⁺ lie at r = (5 ∓ √5) a = 146 pm and 383 pm.
HardStart from an old-convention entry for 2s, L₂¹(x) = 2x − 4 with x = r/a. Normalise R₂₀ directly, repeat with the modern L₁¹(x) = 2 − x, and compare |ψ₂₀₀(0)|² with |ψ₁₀₀(0)|² = 1/(πa³). Hyperfine intervals are proportional to |ψ(0)|² (quoted from Unit 9); measured, they are 1420.405752 MHz for 1s and 177.556834 MHz for 2s. Does the ratio agree, and at what order in α does it fail?
- Old convention, with s = r/a: R = N e(−s/2)(2s − 4), and ∫R² r² dr = N² a³ ∫(4s⁴ − 16s³ + 16s²) e⁻ˢ ds = N² a³ (4 × 24 − 16 × 6 + 16 × 2) = 32 N² a³, so N = (32a³)(−1/2) and R(0) = −4N = −(2a³)(−1/2).
- Modern: ∫(2 − s)² s² e⁻ˢ ds = 4 × 2 − 4 × 6 + 24 = 8, so N = (8a³)(−1/2) and R(0) = +(2a³)(−1/2). The factor −(n + l)! = −2 has vanished except for an overall sign.
- Dimensions: R(0)² = 1/(2a³) is in m⁻³, as a density must be. With Y₀⁰ = 1/√(4π), |ψ₂₀₀(0)|² = 1/(8πa³), one-eighth of the 1s value: the predicted frequency ratio is 8, with every reduced-mass factor cancelling.
- Measured: 1420.405752/177.556834 = 7.99972, so 8f(2s)/f(1s) − 1 = 3.44 × 10⁻⁵.
- Order in α: α² = 5.33 × 10⁻⁵. The leading relativistic correction to 8f(2s)/f(1s), quoted from Breit's treatment, is (5/8)α² = 3.33 × 10⁻⁵; the remaining 0.12 × 10⁻⁵ belongs to QED and nuclear-recoil corrections beyond this course.
AnswerBoth conventions give R₂₀(0)² = 1/(2a³) and |ψ₂₀₀(0)|² = 1/(8πa³), so the ratio should be 8; the measured 7.99972 falls short by 3.4 × 10⁻⁵, an order-α² effect, mostly the quoted relativistic (5/8)α² = 3.3 × 10⁻⁵.