University Quantum Mechanics II · The Hydrogen Radial Equation Solved · 2.3

Behaviour at infinity and at the origin, and v(ρ)

This is the step that turns the Coulomb radial equation from something you cannot solve into something you can look up. Read off what u must do at the nucleus and far away, reject the branches physics forbids and say why, then peel the rest off and recognise Kummer's equation in what remains.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The Coulomb radial equation u″ = [1 − ρ₀/ρ + l(l + 1)/ρ²]u cannot be solved by inspection, but each of its ends can. Far from the nucleus the constant term wins, u″ ≈ u, and the solutions are e^(−ρ) and e^(+ρ); close to it the centrifugal term wins and they are ρ^(l+1) and ρ^(−l). Physics removes one branch at each end: e^(+ρ), and ρ^(−l) for l ≥ 1, because they cannot be normalised, and for l = 0 the constant branch because R ∝ 1/r would put a δ-function into the kinetic energy that the Coulomb Hamiltonian cannot match.

Writing u = ρ^(l+1)e^(−ρ)v is then a definition, not an approximation: it moves the known behaviour into prefactors and leaves an exact equation for v, which in the variable 2ρ is Kummer's equation. Its solution regular at the origin, M(l + 1 − ρ₀/2, 2l + 2, 2ρ), is quoted from the tables, and choosing it settles the origin at every ρ₀, so one condition, at infinity, remains for one unknown number, which is why the spectrum will be discrete. The ends are only as good as the potential there.

The 1/r tail dresses e^(−ρ) with a power ρ^(ρ₀/2); screen the charge and the peeling still works but Kummer's equation is lost; and at the nucleus Dirac's equation (quoted) changes the s-state exponent from 1 to √(1 − Z²α²), an order-α² shift of 2.7 × 10⁻⁵ in hydrogen.

Simple definition: The asymptotic factors of the Coulomb problem are ρ^(l+1), which u follows near the nucleus, and e^(−ρ), which sets its decay far from it up to a power of ρ; every bound state shares them, and dividing both out of u defines v(ρ), which then obeys Kummer's equation in 2ρ exactly.
Example: Hydrogen's measured 13.598 eV binding gives 1/κ = 52.9 pm, so at r = 0.529 nm (ρ = 10) the discarded branch e^(+ρ) would outweigh the kept e^(−ρ) by e⁽²⁰⁾ ≈ 4.9 × 10⁸.

Far from the nucleus: two exponentialsρ → ∞: u″ ≈ u, so u ≈ F e^(−ρ) + G e^(+ρ), with G = 0

e^(+ρ) makes ∫|u|² dρ diverge. The survivor sets the decay length 1/κ: 52.9 pm in hydrogen's ground state, which is ħ/(μcα), exactly 1/α reduced Compton wavelengths ħ/(μc).

ρ = κr is dimensionless, κ = √(−2μE)/ħ in m⁻¹; F and G carry the unit of u = rR, m^(−1/2)

Close to the nucleus: two powersρ → 0: u ≈ C ρ^(l+1) + D ρ^(−l), from s(s − 1) = l(l + 1)

For l ≥ 1, ∫ρ^(−2l) dρ diverges at 0, so D = 0. The kept branch gives R ∝ r^l, so only s states reach the nucleus.

Trial u = ρˢ in u″ ≈ l(l + 1)u/ρ²; C and D carry u's unit, m^(−1/2); the Coulomb term is too weak to shift s

Why D = 0 even when l = 0∇²(1/r) = −4π δ³(r)

R ∝ 1/r is square-integrable, but its kinetic term has a δ-function spike at r = 0 that no term of the Coulomb Hamiltonian cancels. So u(0) = 0 for every l.

r in m, δ³(r) in m⁻³; from the divergence theorem, since the flux of ∇(1/r) through any sphere about 0 is −4π

The exact equation for vρv″ + 2(l + 1 − ρ)v′ + [ρ₀ − 2(l + 1)]v = 0

Nothing is approximated: every u gives exactly one v. The constant and centrifugal terms have cancelled, and the unknown energy sits in one coefficient.

v is defined by u = ρ^(l+1) e^(−ρ) v; primes are d/dρ; v, ρ and ρ₀ = 2/(κa) are dimensionless, a the Bohr radius in m

Kummer's equation in z = 2ρz d²v/dz² + (B − z) dv/dz − A v = 0, A = l + 1 − ρ₀/2, B = 2l + 2

Recognising a tabulated equation lets its solutions be quoted rather than rebuilt. Its exponents at z = 0 are 0 and 1 − B = −(2l + 1).

z, A and B are dimensionless; tables write A and B as a and b, and capitals keep a free for the Bohr radius

The solution regular at the origin (quoted)M(A, B, z) = 1 + (A/B) z + [A(A + 1)/(B(B + 1))] z²/2! + …

Taking v = M(A, B, 2ρ) gives u ∝ ρ^(l+1) at every ρ₀. The origin is settled; the condition at infinity is the only one left.

Kummer's function, quoted from standard tables (DLMF §13.2); M(A, B, 0) = 1, and the other solution starts as z^(−2l−1)

Find which term rules each end

Write the radial equation as u″ = W(ρ)u, with W = 1 − ρ₀/ρ + l(l + 1)/ρ², ρ = κr and ρ₀ = μe²/(2πε₀ħ²κ) = 2/(κa) from the previous topic, where a = 4πε₀ħ²/(μe²). Each term of W is a different power of ρ, so each end of the axis has one winner. As ρ → 0 the centrifugal term outgrows the Coulomb term, overtaking it inside ρ = l(l + 1)/ρ₀; as ρ → ∞ both die away and the constant 1 beats the Coulomb term beyond ρ = ρ₀. Inside that interval, provided ρ₀² exceeds 4l(l + 1), lies the classically allowed region, where W is negative and u curves back towards the axis; its edges, the turning points, are the roots of ρ² − ρ₀ρ + l(l + 1) = 0. For hydrogen's 2p state, Quantum Mechanics I's E₂ = −3.40 eV (quoted) gives 1/κ = 2a = 106 pm and ρ₀ = 4. The centrifugal term then rules inside ρ = 0.5 (r = 53 pm), the constant rules beyond ρ = 4 (r = 424 pm), and the turning points are ρ = 2 ± √2, at 62 pm and 362 pm. The asymptotic forms that follow describe the two outer regions only.

At infinity, keep e^(−ρ) and notice the Coulomb power

Where the constant dominates, u″ ≈ u, with solutions e^(−ρ) and e^(+ρ). Any admixture of e^(+ρ) makes ∫|u|² dρ diverge, so it is discarded, and the survivor fixes the decay length 1/κ = ħ/√(2μ|E|). Hydrogen's measured 13.598 eV gives 52.9 pm, and since |E| = ½μc²α² in the ground state (quoted), 1/κ = ħ/(μcα): the reduced Compton wavelength, 386 fm, divided by α. But e^(−ρ) is only the leading behaviour. Try u = ρˢe^(−ρ): then u″ = [1 − 2s/ρ + s(s − 1)/ρ²]u, and matching the −ρ₀/ρ term of W forces s = ρ₀/2. The long-range 1/r tail dresses the decaying exponential with a power, u ≈ ρ^(ρ₀/2)e^(−ρ), and the growing one with ρ^(−ρ₀/2); a short-range potential would leave the bare exponentials. At ρ = 10 in the 2s state (ρ₀ = 4) that power is worth a factor ρ² = 100; the prefactor ρ^(l+1) takes one power of it, and v, here 1 − ρ, must carry the rest.

At the origin, keep ρ^(l+1)

Near ρ = 0 the equation reduces to u″ ≈ l(l + 1)u/ρ². A power u = ρˢ solves it when s(s − 1) = l(l + 1), so s = l + 1 or s = −l, and u ≈ Cρ^(l+1) + Dρ^(−l). The Coulomb term cannot move these exponents, because 1/ρ is one power weaker than 1/ρ²; it only adds higher-order terms. So the exponents depend on l alone, and are the same for a nucleus of finite size, whose potential is finite at the centre. For l ≥ 1 the second branch fails normalisation: |u|² ∝ ρ^(−2l), and ∫ρ^(−2l) dρ diverges at the lower limit. At l = 1, cutting the integral off at ρ = 10⁻⁴ already gives about 10⁴, and each further decade multiplies it by 10. With D = 0, R = u/r ∝ r^l, so the wavefunction vanishes at the nucleus for every l ≥ 1, the more steeply the higher l. That is why only s electrons appreciably sample the proton's interior, whose 0.84 fm radius is some 1.6 × 10⁻⁵ a.

For l = 0, the equation itself forbids R ∝ 1/r

At l = 0 the exponents are 1 and 0, and normalisation no longer helps: u → D stays finite, and R ≈ D/r gives R²r² ≈ D², which integrates happily. What fails is the three-dimensional equation the radial one came from. The flux of ∇(1/r) through a sphere of any radius about the origin is −4π, so by the divergence theorem ∇²(1/r) = −4πδ³(r). A state with ψ ∝ 1/r near the nucleus therefore has a δ-function spike in −(ħ²/2μ)∇²ψ that nothing in −e²/(4πε₀r) can cancel; it would solve Hψ = Eψ only if a point interaction were added to H. It would also carry infinite kinetic energy, since |∇ψ|²r² ∝ 1/r² is not integrable at 0. So u(0) = 0 holds for every l, as Quantum Mechanics I argued for any V(r) that carries no point interaction; the Coulomb 1/r, singular as it is, carries none. Only s states have ψ(0) ≠ 0, and it shows: they alone feel the Fermi contact term behind hydrogen's measured 1420.406 MHz 1s hyperfine line, and the proton's size shifts the 1s level by about 1.1 MHz (both quoted).

Peel both ends off, and derive the equation for v

Define v by u = ρ^(l+1)e^(−ρ)v. The prefactor never vanishes at positive ρ, so every u has exactly one v: the definition assumes nothing. Differentiating, u′ = ρ^l e^(−ρ)[(l + 1 − ρ)v + ρv′] and u″ = ρ^l e^(−ρ)[ρv″ + 2(l + 1 − ρ)v′ + (l(l + 1)/ρ − 2(l + 1) + ρ)v]. The right side of the radial equation is ρ^l e^(−ρ)[ρ − ρ₀ + l(l + 1)/ρ]v. Cancel the common ρ^l e^(−ρ). The ρv and l(l + 1)v/ρ terms, images of the constant and centrifugal terms the prefactors were built from, cancel between the sides, leaving ρv″ + 2(l + 1 − ρ)v′ + [ρ₀ − 2(l + 1)]v = 0. Only the Coulomb strength survives, and the unknown energy now sits in one coefficient. Check it on Quantum Mechanics I's quoted ground state, u ∝ ρe^(−ρ) with ρ₀ = 2: there v = 1, and the equation reduces to (2 − 2) × 1 = 0.

Recognise Kummer's equation and quote its regular solution

Put z = 2ρ, so d/dρ = 2 d/dz, and halve the result: z d²v/dz² + (2l + 2 − z) dv/dz − (l + 1 − ρ₀/2)v = 0. That is Kummer's confluent hypergeometric equation, zw″ + (B − z)w′ − Aw = 0, with A = l + 1 − ρ₀/2 and B = 2l + 2. Its solution regular at z = 0 is quoted from the tables, not derived: M(A, B, z) = Σ (A)ₖzᵏ/[(B)ₖ k!], with (A)ₖ = A(A + 1)…(A + k − 1). The other solution starts as z^(1−B) = z^(−2l−1), and ρ^(l+1) × ρ^(−2l−1) = ρ^(−l), the branch already rejected. So v = M(A, B, 2ρ) imposes the origin condition, completely, at every ρ₀. Check against a tabulated function: for 2s, l = 0 and ρ₀ = 4 give A = −1, B = 2 and M = 1 − ρ, so R ∝ (1 − r/2a)e^(−r/2a), Quantum Mechanics I's R₂₀. What M does not promise is decay: for most ρ₀ it grows fast enough to bring e^(+ρ) back, and that condition at infinity, the only one left, belongs to the next topic.

Change one variable at a time

Make the relationship visible.

Interactive model

Orbital quantum number l1

Energy parameter ρ₀ = 2/(κa)4.6

Hold l = 1 and step ρ₀ through 3.8, 4.0 and 4.2. Near the origin the solid curve follows the dashed ρ² every time, but beyond ρ = ρ₀ it runs off the top, then decays, then runs off the bottom. M has settled the origin at every ρ₀; only the condition at infinity is left to impose.

KUMMER A = l + 1 − ρ₀/2-0.30

HYDROGEN E = −54.39 eV/ρ₀²-2.57 eV

DECAY LENGTH 1/κ = ρ₀a/2121.8 pm

CENTRIFUGAL WINS INSIDE l(l + 1)/ρ₀0.43

Live interpretationKUMMER A = l + 1 − ρ₀/2: −0.30. HYDROGEN E = −54.39 eV/ρ₀²: −2.57 eV. DECAY LENGTH 1/κ = ρ₀a/2: 121.8 pm. CENTRIFUGAL WINS INSIDE l(l + 1)/ρ₀: 0.43

Catch the common trap

Explain before calculating.

Misconception check

“Peeling off ρ^(l+1) and e^(−ρ) has already thrown the bad solutions away.” It has not. The substitution is exact, so the v-equation has every solution the u-equation had, only relabelled. Near the origin its exponents are 0 and −(2l + 1), and v ∝ ρ^(−2l−1) is the rejected ρ^(−l) branch in disguise; at large ρ it admits v growing like e^(2ρ), which is the rejected e^(+ρ). The prefactors say what v should look like if all goes well; they enforce nothing. The origin condition is imposed by choosing Kummer's M rather than the second solution, and the condition at infinity is still open: it is imposed on M's growth in the next topic, and that is what fixes ρ₀. The same slip in another form treats the factorisation as an approximation valid only near the two ends, when it holds at every ρ.

Take l = 1 and ρ₀ = 5. Kummer's first parameter is then l + 1 − ρ₀/2 = −1/2, and the regular solution is u = ρ²e^(−ρ)M(−1/2, 4, 2ρ). Which statement about this u is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe measured ionisation energy of the He⁺ ion is 54.418 eV. Treat its ground state as a bound state of the Coulomb radial equation with Z = 2, reduced mass μc² = 510 929 eV and a = 4πε₀ħ²/(μe²) = 52.925 pm. Find κ, the decay length 1/κ and ρ₀ = 2Z/(κa). Which branch survives at each end for l = 0, and what does the equation for v then say?

κ = √(2μc²|E|)/(ħc) = √(2 × 510 929 eV × 54.418 eV)/(197.327 eV nm) = 7457.0 eV/197.327 eV nm = 37.790 nm⁻¹, so 1/κ = 26.462 pm. The units reduce to nm⁻¹, as a wavenumber should.
With Ze² in place of e², ρ₀ = μZe²/(2πε₀ħ²κ) = 2Z/(κa) = 4 × 26.462 pm/52.925 pm = 2.0000, a pure number. The measured ionisation energy exceeds the Bohr value 54.415 eV by a fraction of about 5 × 10⁻⁵, of order (Zα)²/4 = 5.3 × 10⁻⁵, the relativistic correction this equation leaves out (quoted).
Far end: u ≈ Fe^(−ρ) + Ge^(+ρ), and G = 0 because e^(+ρ) cannot be normalised. The constant term outweighs the Coulomb term 2/ρ beyond ρ = 2, that is r = 2/κ = 52.9 pm.
Origin at l = 0: the exponents are l + 1 = 1 and −l = 0, so u ≈ Cρ + D. R ≈ D/r is square-integrable, so normalisation does not remove D; the δ-function in ∇²(1/r) does. Keep Cρ.
At l = 0 and ρ₀ = 2 the v-equation reads ρv″ + 2(1 − ρ)v′ + (2 − 2)v = 0, solved by v = constant, which is M(0, 2, 2ρ) = 1. So u ∝ ρe^(−ρ) and R ∝ e^(−κr) = e^(−Zr/a): Quantum Mechanics I's 1s with a replaced by a/Z.

Answerκ = 3.779 × 10¹⁰ m⁻¹, 1/κ = 26.46 pm and ρ₀ = 2.0000; keep e^(−ρ) and ρ¹, so v is constant and R ∝ e^{(−r/26.46 pm)}. The measured ionisation energy exceeds the Bohr value by a fraction of about 5 × 10⁻⁵, an order-(Zα)² effect.

MediumCarry the substitution u = ρ²e^(−ρ)v through by hand for l = 1 and obtain the equation for v. Find the values of ρ₀ for which v = 1 and v = 1 + cρ are solutions, compare each with Kummer's M(A, B, 2ρ), and turn the second into a hydrogen energy and node radius, taking a = 52.95 pm and E = −4(13.598 eV)/ρ₀².

Differentiate: u′ = e^(−ρ)[(2ρ − ρ²)v + ρ²v′] and u″ = e^(−ρ)[ρ²v″ + 2(2ρ − ρ²)v′ + (ρ² − 4ρ + 2)v].
The right side of u″ = [1 − ρ₀/ρ + 2/ρ²]u is e^(−ρ)(ρ² − ρ₀ρ + 2)v. Equate and cancel e^(−ρ): the ρ²v and 2v terms cancel, leaving ρ²v″ + 2ρ(2 − ρ)v′ + (ρ₀ − 4)ρv = 0. Divide by ρ: ρv″ + 2(2 − ρ)v′ + (ρ₀ − 4)v = 0, the general result at l = 1.
v = 1 leaves (ρ₀ − 4) × 1 = 0, so ρ₀ = 4. Kummer's A = 2 − ρ₀/2 = 0 and M(0, 4, 2ρ) = 1, in agreement. Then κ = 2/(ρ₀a) = 1/(2a) and E = −54.39 eV/16 = −3.40 eV: the 2p level quoted in Quantum Mechanics I.
v = 1 + cρ gives 2(2 − ρ)c + (ρ₀ − 4)(1 + cρ) = 0 for every ρ. The constant term demands 4c + ρ₀ − 4 = 0 and the ρ term demands c(ρ₀ − 6) = 0, so with c ≠ 0, ρ₀ = 6 and c = −1/2.
Kummer check: A = 2 − 3 = −1 and B = 4, so M(−1, 4, 2ρ) = 1 + (−1/4)(2ρ) = 1 − ρ/2, the same function. ρ, ρ₀ and c are pure numbers, as the equation requires.
With ρ₀ = 6, κ = 1/(3a) and ρ = r/(3a). E = −54.39 eV/36 = −1.511 eV, and v = 0 at ρ = 2 puts a radial node at r = 6a = 318 pm, with R ∝ r(1 − r/6a)e^(−r/3a).

Answerρv″ + 2(2 − ρ)v′ + (ρ₀ − 4)v = 0. v = 1 needs ρ₀ = 4 (E = −3.40 eV, 2p); v = 1 − ρ/2 needs ρ₀ = 6 and equals M(−1, 4, 2ρ), giving E = −1.511 eV and a node at r = 6a ≈ 318 pm (3p).

HardAt every ρ₀ there is a solution that decays at infinity; the question is how it arrives at the origin. Write u = e^(−ρ)f and seek f = ρˢ(d₀ + d₁/ρ + d₂/ρ² + …) at large ρ. Find s and a recursion for the dₖ, run it for l = 0 at ρ₀ = 4 and at ρ₀ = 5, and say what each result means for the two end conditions.

With u = e^(−ρ)f, u″ = e^(−ρ)(f″ − 2f′ + f), and the radial equation becomes f″ − 2f′ + [ρ₀/ρ − l(l + 1)/ρ²]f = 0. The constant term has cancelled, as it must once e^(−ρ) is removed.
Insert f = Σ dₖρ^(s−k). The highest power, ρ^(s−1), collects (ρ₀ − 2s)d₀ = 0, so s = ρ₀/2: the decaying solution is ρ^(ρ₀/2)e^(−ρ) to leading order, not a bare e^(−ρ).
The power ρ^(s−k−1) collects [ρ₀ − 2(s − k)]dₖ + [(s − k + 1)(s − k) − l(l + 1)]dₖ₋₁ = 0. With s = ρ₀/2 the first bracket is 2k, so dₖ = [l(l + 1) − (s − k)(s − k + 1)]dₖ₋₁/(2k).
l = 0, ρ₀ = 4: s = 2, d₁ = −(1 × 2)d₀/2 = −d₀ and d₂ = −(0 × 1)d₁/4 = 0, so every later coefficient vanishes. Then u = −d₀ρ(1 − ρ)e^(−ρ), Quantum Mechanics I's 2s, which vanishes like ρ at the origin: here the decaying solution is the regular one. At ρ = 10 it is 90e⁽⁻¹⁰⁾, so bare e^(−ρ) is low by a factor 90 and ρ²e^(−ρ) high by 11%.
l = 0, ρ₀ = 5: s = 2.5, d₁ = −1.875d₀, d₂ = 0.352d₀, d₃ = 0.0146d₀, d₄ = −0.00137d₀. The factor (s − k)(s − k + 1) is a product of half-odd numbers and never zero, and |dₖ/dₖ₋₁| → k/2, so the series never stops and is only asymptotic. Integrating the equation inward from this tail, numerically from ρ = 25, reaches ρ = 0 with u about 0.22 of its peak: the forbidden l = 0 branch.
So each end admits one acceptable solution at every ρ₀, and a bound state needs them to be the same function: one condition on one number. When ρ₀/2 is an integer greater than l, the tail stops at k = ρ₀/2 − l, and v = uρ^(−l−1)e^(ρ) is a polynomial of degree ρ₀/2 − l − 1.

Answers = ρ₀/2 and dₖ = [l(l + 1) − (s − k)(s − k + 1)]dₖ₋₁/(2k). At ρ₀ = 4 the tail stops after d₁ = −d₀ and is the regular 2s function ρ(1 − ρ)e^(−ρ); at ρ₀ = 5 it never stops and reaches the origin with u(0) ≈ 0.22 of its peak, so it is not a bound state.

Find which term rules each end

At infinity, keep e(−ρ) and notice the Coulomb power

At the origin, keep ρ(l+1)