University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.7
Normalised radial functions Rₙₗ for n up to 3
Six functions carry most of the hydrogen calculations in the rest of this course, so they are worth building by hand once and trusting afterwards. Take the recursion's polynomials, fix each constant with one factorial integral, and make every result pass the same checks: units, nodes, behaviour at the origin and orthogonality.
Build the model
Connect the measurement to the mechanism.
The recursion of topic 02.04 delivers v(ρ) only up to an overall factor, because the radial equation is linear. Probability supplies the missing number: the electron is somewhere, so ∫₀^∞ R² r² dr = 1, the r² inherited from the volume element because the spherical harmonics carry their own normalisation. With x = r/a every such integral is a sum of factorials, so the six functions with n ≤ 3 come out exact, and each can be read before it is used: a tail e(−r/(na)) that sets the size, n − l − 1 nodes from the polynomial, and an rˡ onset that keeps every l ≥ 1 state off the nucleus.
These six are the course's test bench. Most hydrogen matrix elements from here on are integrals over them, and one number read off them, the contact density 1/(πn³a³), feeds three later corrections. The model stops where its assumptions do.
The nucleus is not a point inside rₚ ≈ 0.84 fm, or 1.6 × 10⁻⁵ a, which is exactly where the contact density is sampled. The Dirac equation, quoted from relativistic quantum mechanics, replaces the finite Rₙ₀(0) by a factor behaving as r(γ − 1) with γ = √(1 − α²), a divergence so weak, an exponent of −2.7 × 10⁻⁵, that at the proton's edge it raises the density by only about 0.06%. Up to corrections of relative order α², the functions derived here are the functions.
- Simple definition
- The normalised radial function Rₙₗ(r) is the r-dependent factor in ψₙₗₘ = Rₙₗ(r) Yₗᵐ(θ, φ), scaled so that ∫₀^∞ Rₙₗ² r² dr = 1; then r²Rₙₗ² dr is the probability that the electron lies between r and r + dr.
- Example
- For the ground state R₁₀ = 2a(−3/2) e(−r/a), the substitution x = r/a turns the condition into 4∫x² e(−2x) dx = 4 × 2!/2³ = 1. For hydrogen a = 52.95 pm, so R₁₀(0) = 5.19 × 10¹⁵ m(−3/2).
Derived from u = rR = ρ(l+1) e(−ρ) v(ρ) with v(0) = 1. The recursion supplies v, and normalisation is left to fix one number, Kₙₗ.
a = 4πε₀ħ²/(μe²) in m, 52.95 pm for hydrogen; ρ, v and Kₙₗ are pure numbers, so R carries m(−3/2).
With x = r/a every term becomes ∫₀^∞ xᵏ e(−bx) dx = k!/b(k+1), so each constant is exact arithmetic rather than a numerical integral.
R in m(−3/2) and r in m, so the integrand is a probability per metre; ∫|Yₗᵐ|² dΩ = 1 is imposed separately.
Quoted from the Laguerre normalisation integral, not derived here. It is trusted only because it reproduces all six constants found by direct integration.
A pure number, for v scaled to v(0) = 1: K₁₀ = 2, K₂₀ = 1/√2, K₂₁ = 1/√6, K₃₂ = 4/(9√30).
Checked by hand for n ≤ 3. In general it rests on the quoted fact that every root of an associated Laguerre polynomial is real, positive and simple.
nᵣ is a count, the degree of v. The nodes: 2s at 2a, 3p at 6a, 3s at (9 ± 3√3)a/2, which is 1.90a and 7.10a.
Derived by setting d(r²R²)/dr = 0. The shell's relative width falls as 1/√(2n + 1), so at large n these states become Bohr's circular orbits.
r in m. For hydrogen a, 4a and 9a are 52.9, 212 and 477 pm, which are Bohr's circular-orbit radii.
Derived for n ≤ 3 from the explicit functions. It is the one number the finite-size shift, the Darwin term and the Fermi contact hyperfine interaction all need.
In m⁻³: 2.14 × 10³⁰ m⁻³ for hydrogen 1s. Replacing a by a/Z multiplies it by Z³; it is zero for every l ≥ 1.
Six polynomials from one recursion
Topics 02.03 to 02.05 left u = rR = ρ(l+1) e(−ρ) v(ρ), with ρ = r/(na) and a series v whose coefficients obey cⱼ₊₁/cⱼ = 2(j + l + 1 − n)/[(j + 1)(j + 2l + 2)]. So R = u/r ∝ ρˡ e(−ρ) v(ρ). Set c₀ = 1 and run the recursion: it stops by itself at j = n − l − 1. For every l = n − 1 state the first ratio already vanishes, so v = 1, giving R₁₀ ∝ e(−r/a), R₂₁ ∝ r e(−r/(2a)) and R₃₂ ∝ r² e(−r/(3a)). For 2s, c₁ = −1 and v = 1 − r/(2a). For 3p, c₁ = −1/2 and v = 1 − r/(6a). For 3s, c₁ = −2 and c₂ = (−2)(−1/3) = 2/3, so v = 1 − 2r/(3a) + 2r²/(27a²). None of this fixes the size of R. The radial equation is linear, so any multiple of a solution is a solution, and a second condition has to supply the number.
One integral fixes the constant
The probability of finding the electron between r and r + dr is R² r² dr multiplied by ∫|Yₗᵐ|² dΩ = 1, so the radial condition is ∫₀^∞ R² r² dr = 1. With x = r/a it reduces to sums of ∫₀^∞ xᵏ e(−bx) dx = k!/b(k+1). For 2s, R = N(1 − x/2) e(−x/2), so N²a³ ∫x²(1 − x + x²/4) e(−x) dx = N²a³(2! − 3! + 4!/4) = 2N²a³, and N = (1/√2) a(−3/2). The same move gives R₁₀ = 2a(−3/2) e(−r/a), R₂₁ = (1/√24) a(−3/2) (r/a) e(−r/(2a)) and R₃₀ = [2/(3√3)] a(−3/2) [1 − 2r/(3a) + 2r²/(27a²)] e(−r/(3a)), where the 3s integral comes to 27/4. The general Kₙₗ is quoted, not derived; it reproduces all six, and that is the check that counts. The overall sign is a convention, since R and −R describe the same state.
Reading each function: tail, nodes and onset
Each R is a product of three factors, and each carries one check. The tail e(−r/(na)) sets the size: the decay length na grows with n, and 1/(na) is the κ of topic 02.05. The polynomial v has degree nᵣ = n − l − 1, and its roots are the radial nodes: 2s at 2a, 3p at 6a, 3s at (9 ± 3√3)a/2, or 1.90a and 7.10a. That each root is real, positive and simple is quoted from the theory of Laguerre polynomials; for n ≤ 3 you have just checked it by hand. The factor ρˡ sets the onset, R ∝ rˡ as r → 0: s states are finite at the nucleus, p states rise linearly and d states quadratically, held out by the centrifugal term ħ²l(l + 1)/(2μr²). Units close the list. ∫R² r² dr is a pure number, so R must carry m(−3/2); a candidate R without a(−3/2) is wrong before a single integral is done.
Orthogonal in n at fixed l, but not across l
At fixed l every Rₙₗ solves one radial eigenvalue problem, −[ħ²/(2μ)]u″ + [−e²/(4πε₀r) + ħ²l(l + 1)/(2μr²)]u = Eu with u(0) = 0, and that operator is Hermitian on such functions. Different energies then force ∫Rₙₗ Rₖₗ r² dr = 0 whenever n ≠ k. Check one: for R₁₀ and R₂₀ the integral is √2 ∫x²(1 − x/2) e(−3x/2) dx = √2[2!/(3/2)³ − (1/2) 3!/(3/2)⁴] = √2(16/27 − 16/27) = 0. Change l and the centrifugal term changes, so the two functions solve different equations and nothing forces a zero: ∫R₁₀ R₂₁ r² dr = (2/√24) 3!/(3/2)⁴ = 0.484. The states are still orthogonal, because ⟨100|21m⟩ is that radial number times ∫Y₀₀* Y₁ᵐ dΩ = 0. Radial integrals across l, with one more power of r, are exactly what survives in the dipole matrix elements of the radiation units.
The nodeless states and the circular-orbit limit
For l = n − 1, v = 1 and the radial density is P(r) = r²R² ∝ r(2n) e(−2r/(na)). Setting dP/dr = 0 gives 2n/r = 2/(na), so the most probable radius is n²a: a, 4a and 9a, or 52.9, 212 and 477 pm in hydrogen, which are Bohr's orbit radii. The same factorial integrals give the whole shell. Because P has the form rk e(−r/θ) with k = 2n and θ = na/2, ⟨r⟩ = (2n + 1)na/2 = n²a + na/2 and the width is σ = (na/2)√(2n + 1), so σ/⟨r⟩ = 1/√(2n + 1): 58% for 1s, 45% for 2p, 38% for 3d. The shell sharpens slowly, but it sharpens. A circular Rydberg state with n = 50 is a ring of radius 2500a ≈ 0.13 μm with a 10% width, the classical circular orbit recovered as the correspondence principle requires. Topic 02.08 reproduces ⟨r⟩ from Kramers' relation without an integral. States with l below n − 1 are the eccentric orbits: their extra lobes reach inside n²a and their outer lobe beyond it.
The contact density, recorded for later units
Only s states survive at r = 0, where |ψₙ₀₀(0)|² = Rₙ₀(0)² |Y₀₀|² = Rₙ₀(0)²/(4π). The explicit functions give Rₙ₀(0)² = 4/a³, 1/(2a³) and 4/(27a³) for n = 1, 2, 3, so |ψₙ₀₀(0)|² = 1/(πa³), 1/(8πa³) and 1/(27πa³): the law 1/(πn³a³), which the quoted constant extends to every n, since Rₙ₀(0) = 2(na)(−3/2). For hydrogen 1s it is 2.14 × 10³⁰ m⁻³; a → a/Z scales it by Z³, and since a = ħ/(αμc) it is third order in α. Three later effects are proportional to it: the finite-size shift of Unit 6, the Darwin term of Unit 8 and the Fermi contact hyperfine interaction of Unit 9. The n⁻³ law is measured. Hydrogen's hyperfine splitting is 1420.406 MHz in 1s and 177.557 MHz in 2s, a ratio of 7.9997 against the predicted 8. The 3.4 × 10⁻⁵ shortfall is of order α² = 5.3 × 10⁻⁵, the size of the relativistic correction to ψ near the nucleus, quoted from relativistic quantum mechanics.
Change one variable at a time
Make the relationship visible.
Hold l = n − 1 and step n from 1 to 3: the nodeless peak moves out to a, 4a and 9a and spreads, so it sinks even drawn n times taller. At n = 3, lower l: nodes appear, inner lobes reach towards the nucleus, and only l = 0 gives a nonzero contact density.
RADIAL NODES n − l − 12
NORMALISATION Kₙₗ0.3849
CONTACT πa³|ψ(0)|²0.037
l = n − 1 PEAK n²a477 pm
Live interpretationRADIAL NODES n − l − 1: 2. NORMALISATION Kₙₗ: 0.3849. CONTACT πa³|ψ(0)|²: 0.037. l = n − 1 PEAK n²a: 477 pm
Catch the common trap
Explain before calculating.
A student normalises R₁₀ = 2a(−3/2) e(−r/a) and R₂₁ = (1/√24) a(−3/2)(r/a) e(−r/(2a)) correctly, then finds ∫₀^∞ R₁₀ R₂₁ r² dr = 0.484 rather than 0. What should she conclude?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUse the recursion to build the 3d radial function R₃₂, normalise it, and find the radius at which a 3d electron in hydrogen, with a = 52.95 pm, is most likely to be found.
- For n = 3, l = 2 the first ratio is c₁/c₀ = 2(0 + 2 + 1 − 3)/[(1)(6)] = 0, so v = 1 and R₃₂ = N r² e(−r/(3a)): no radial node, as n − l − 1 = 0 requires, and an r² onset, as l = 2 requires.
- Normalise with ∫₀^∞ rᵏ e(−br) dr = k!/b(k+1), taking k = 6 and b = 2/(3a): N² × 6! × (3a/2)⁷ = 1. With 6! = 720 and (3/2)⁷ = 17.086, the integral is 1.2302 × 10⁴ a⁷.
- So N = (1.2302 × 10⁴)(−1/2) a(−7/2) = 9.016 × 10⁻³ a(−7/2). Units: a(−7/2) × r² is m(−3/2), as a radial function needs.
- In the tabulated form R₃₂ = [4/(81√30)] a(−3/2) (r/a)² e(−r/(3a)). Check: (81√30/4)² = 196 830/16 = 12 301.875, exactly the integral.
- The radial density r²R² ∝ r⁶ e(−2r/(3a)) peaks where 6/r = 2/(3a), so r = 9a = n²a, which is 9 × 52.95 pm ≈ 477 pm.
AnswerR₃₂ = [4/(81√30)] a(−3/2)(r/a)² e(−r/(3a)), with 4/(81√30) = 9.016 × 10⁻³; the radial density peaks at 9a ≈ 0.477 nm.
MediumBuild R₃₁ from the recursion, locate its node, normalise it, and show that it is orthogonal to R₂₁ = (1/√24) a(−3/2)(r/a) e(−r/(2a)).
- With n = 3, l = 1 and ρ = r/(3a): c₁/c₀ = 2(0 + 1 + 1 − 3)/[(1)(4)] = −1/2 and c₂/c₁ = 2(1 + 1 + 1 − 3)/[(2)(5)] = 0, so v = 1 − ρ/2 = 1 − r/(6a) and R₃₁ = N r[1 − r/(6a)] e(−r/(3a)).
- One node, as n − l − 1 = 1 requires, at r = 6a, which is 318 pm in hydrogen. Near the origin R₃₁ ∝ r, as l = 1 requires.
- Normalise with x = r/a: N²a⁵ ∫x⁴(1 − x/3 + x²/36) e(−2x/3) dx = 1. Term by term, 4!(3/2)⁵ = 182.25, −(1/3) 5!(3/2)⁶ = −455.625 and (1/36) 6!(3/2)⁷ = 341.719, summing to 68.344.
- So N = 68.344(−1/2) a(−5/2) = 0.1210 a(−5/2), which is the tabulated 8/(27√6) = 0.1210, giving R₃₁ = [8/(27√6)] a(−3/2)(r/a)[1 − r/(6a)] e(−r/(3a)).
- Overlap with R₂₁: the exponentials combine to e(−5x/6), and ∫x⁴(1 − x/6) e(−5x/6) dx = 4!(6/5)⁵ − (1/6) 5!(6/5)⁶ = (6/5)⁵(24 − 20 × 1.2) = 0. Equal l and different n require the zero, and the recursion delivers it exactly.
AnswerR₃₁ = [8/(27√6)] a(−3/2)(r/a)[1 − r/(6a)] e(−r/(3a)), with 8/(27√6) = 0.1210; one node at 6a ≈ 0.318 nm; ∫R₂₁R₃₁ r² dr = 0 exactly.
HardPositronium is an electron bound to a positron. Its singlet ground state annihilates into two photons at a rate that quantum electrodynamics gives as Γ = 4πrₑ²c|ψ(0)|², with rₑ = 2.818 fm the classical electron radius. Use this topic's contact density to predict the lifetime, check it by dimensions and by its order in α, and compare with the measured 125.1 ps.
- The explicit s functions give Rₙ₀(0)² = 4/a³, 1/(2a³) and 4/(27a³) for n = 1, 2, 3. Dividing by 4π gives |ψₙ₀₀(0)|² = 1/(πn³a³), the n⁻³ law confirmed term by term.
- Reduced mass: μ = mₑmₑ/(2mₑ) = mₑ/2, and a ∝ 1/μ, so aPs = 2a₀ = 2 × 52.918 pm = 105.84 pm.
- Contact density: |ψ₁₀₀(0)|² = 1/(π aPs³) = 1/[π(1.0584 × 10⁻¹⁰ m)³] = 2.685 × 10²⁹ m⁻³, one eighth of the value for an infinitely heavy nucleus, because a has doubled.
- Dimensions of the quoted rate: m² × m s⁻¹ × m⁻³ = s⁻¹. Numerically Γ = 4π(2.818 × 10⁻¹⁵ m)²(2.998 × 10⁸ m s⁻¹)(2.685 × 10²⁹ m⁻³) = 8.03 × 10⁹ s⁻¹, so τ = 1/Γ = 124.5 ps.
- Order in α: with rₑ = α²a₀ and a₀ = ħ/(αmₑc), the rate becomes Γ = α⁵mₑc²/(2ħ), five powers of α times the electron's rest-energy frequency, and it evaluates to the same 8.03 × 10⁹ s⁻¹.
- The measured 125.1 ps is 0.5% longer. That is the size of the first radiative correction, of relative order α/π ≈ 0.0023 times a coefficient near 2.5, quoted from QED. The 2¹S₀ state, with one eighth of the contact density, is predicted to live 8 × 124.5 ps ≈ 1.0 ns.
Answerτ ≈ 124.5 ps, from Γ = α⁵mₑc²/(2ħ) = 8.03 × 10⁹ s⁻¹, within 0.5% of the measured 125.1 ps; the n = 2 singlet would live about 1.0 ns.