University Physics IV · The Hydrogen Atom · 10.4
Solving the Coulomb Radial Equation
This is the one place in the hydrogen problem where a quantum number is manufactured rather than assumed. Follow the series solution honestly and almost every energy produces a wavefunction that explodes; the handful that do not are the spectrum, and l ≤ n − 1 falls out of the same line of algebra.
Build the model
Connect the measurement to the mechanism.
The radial equation for a Coulomb well carries two constants — the energy and the strength of the attraction — so the first move is to make it dimensionless. With κ = √(2μ|E|)/ħ and ρ = κr it becomes d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²]u, all of the energy now hiding in the single number ρ₀ = 2/(κa). Both ends are then legible: near the origin the centrifugal term rules and u goes as ρ(l+1) or ρ(−l); far out the bracket tends to 1 and u goes as e(−ρ) or e(+ρ).
Factor out the two survivors, u = ρ(l+1)e(−ρ)v(ρ), and v is solved by a power series whose coefficients obey one recursion. Here is the whole argument. For a generic ρ₀ that recursion never stops, its large-j ratio tends to 2/j, and those are exactly the coefficients of e(2ρ) — so u recovers the growing branch you thought you had thrown away, and the integral of |u|² diverges.
Demanding a finite norm therefore forces the numerator 2(j+l+1) − ρ₀ to vanish at some integer, that is ρ₀ = 2n with n = jₘₐₓ + l + 1. What it costs is worth naming: n is a truncation index rather than a postulate, l ≤ n − 1 because jₘₐₓ ≥ 0, the radial node count is n − l − 1, and Eₙ = −13.606 eV/n² is a by-product. What it does not buy is the continuum — for E > 0 the runaway branch oscillates instead of growing, nothing needs truncating, and every energy is allowed.
- Simple definition
- Quantisation from normalisability is the demand that the power series in the hydrogen radial solution terminate, because an unterminated series rebuilds the exponentially growing solution and destroys the normalisation integral.
- Example
- For a trial energy of −1.30 eV, ρ₀ = 2√(13.606/1.30) = 6.47, which is not an even integer, so the recursion never stops; only ρ₀ = 6 (E = −1.512 eV) and ρ₀ = 8 (E = −0.850 eV) terminate it.
All of the energy dependence now sits in one number, so "which E are allowed?" becomes "which ρ₀ are allowed?"
ρ = κr, dimensionless; κ = √(2μ|E|)/ħ in m⁻¹; ρ₀ = 2/(κa), also dimensionless
An exact change of unknown, not an approximation: v carries everything the two limits do not already fix.
ρ(l+1) from the centrifugal term as ρ → 0; e(−ρ) from the bracket → 1 as ρ → ∞
One constant generates the whole radial function, which is why no freedom is left over to kill a bad tail.
j = 0, 1, 2, …; every cⱼ is dimensionless and c₀ is fixed only by normalisation
The growing branch you discarded walks back in through the series. This is the step that does the quantising.
e(2ρ) = Σ (2ρ)j/j!, whose consecutive coefficient ratio is 2/(j+1)
n is the index where the series stops, and l ≤ n − 1 restates that same fact rather than adding a second rule.
jₘₐₓ ≥ 0 is the highest surviving power of ρ in v; radial nodes = jₘₐₓ = n − l − 1
The energies are a by-product of the truncation, not an input — nothing was assumed about orbits or angular momentum.
a = 4πε₀ħ²/μe² = 0.0529 nm; a hydrogenic ion gives a/Z and −13.606Z²/n² eV
Make it dimensionless before you try to solve it
With u = rR the radial equation reads −(ħ²/2μ)u″ + [−e²/4πε₀r + ħ²l(l+1)/2μr²]u = Eu. For a bound state E < 0, so κ = √(−2μE)/ħ is real and has units of inverse length; measure r in units of 1/κ by setting ρ = κr. Dividing through by E and tidying gives d²u/dρ² = [1 − ρ₀/ρ + l(l+1)/ρ²]u, where ρ₀ = μe²/(2πε₀ħ²κ) = 2/(κa). Two physical constants have collapsed into one pure number. That is the entire point of the scaling: the energy no longer appears in two places at once, so the question "which E are allowed?" turns into "which ρ₀ are allowed?", and the answer will be a statement about integers.
Peel off what the two ends of the equation already tell you
As ρ → 0 the 1/ρ² term dominates and u″ ≈ l(l+1)u/ρ², whose solutions are ρ(l+1) and ρ(−l). Reject ρ(−l): for l ≥ 1 the norm diverges at the origin. For l = 0 the rejected branch is u → constant, which is square-integrable, so the usual "it is not normalisable" excuse fails — it goes because R = u/r behaves as 1/r, and ∇²(1/r) = −4πδ³(r) puts a δ function at the nucleus that nothing in the equation can balance. As ρ → ∞ the bracket tends to 1 and u″ ≈ u, giving e(±ρ), and e(+ρ) is rejected outright. Now write u = ρ(l+1)e(−ρ)v(ρ). This is not an approximation but an exact change of unknown, valid at every ρ, with the whole solution carried by v.
One recursion generates every coefficient
Substituting gives ρv″ + 2(l+1−ρ)v′ + [ρ₀ − 2(l+1)]v = 0. Put v = Σ cⱼ ρj, collect equal powers of ρ, and each power returns the same relation: c_(j+1) = [2(j+l+1) − ρ₀] cⱼ / [(j+1)(j+2l+2)]. Two features matter. First, this fixes the entire function from c₀ alone — the second constant of a second-order equation was already spent rejecting ρ(−l) at the origin, and c₀ is nothing but overall scale, so there is no third knob to turn. Second, ρ₀ sits in the numerator and not in the denominator, which is where the energy will make its move. Try l = 0 with ρ₀ = 6: c₁ = (2−6)c₀/(1⋅2) = −2c₀, then c₂ = (4−6)c₁/(2⋅3) = (2/3)c₀, then c₃ = (6−6)c₂/(3⋅4) = 0. The series has stopped of its own accord.
An endless series is the growing solution in disguise
At large j the recursion tends to c_(j+1)/cⱼ ≈ 2j/j² = 2/j. Those are exactly the coefficients of e(2ρ) = Σ (2ρ)j/j!, whose consecutive ratio is 2/(j+1). So an unterminated v behaves as c₀e(2ρ) at large ρ, and u = ρ(l+1)e(−ρ)v behaves as c₀ρ(l+1)e(+ρ). The branch discarded during the asymptotic step has walked back in through the series, and ∫|u|²dr diverges. This is the whole quantisation argument, and notice what it does not say: not that the growth is ugly, or unphysical, or hard to normalise. It says the object has no finite norm, so it is not a state, and the energy that produced it is not in the spectrum.
Termination hands you n, and l ≤ n − 1 in the same breath
The only escape is that the numerator vanish: 2(jₘₐₓ + l + 1) = ρ₀ for some non-negative integer jₘₐₓ, after which c_(jₘₐₓ+1) and everything beyond it is zero and v is a polynomial of degree jₘₐₓ. Define n ≡ jₘₐₓ + l + 1, so ρ₀ = 2n with n a positive integer. Because jₘₐₓ ≥ 0, l ≤ n − 1 follows at once, and the number of radial nodes is jₘₐₓ = n − l − 1. Nothing was postulated anywhere in that sentence. Take n = 3, l = 0: jₘₐₓ = 2, and the recursion above gave v = c₀(1 − 2ρ + 2ρ²/3), whose roots are ρ = (3 ± √3)/2 = 0.634 and 2.366. Since ρ = r/(na), those sit at r = 1.90a and 7.10a — the two nodes of the 3s radial function, read off a quadratic.
Read the spectrum off, and what the argument never touched
ρ₀ = 2/(κa) = 2n gives κ = 1/(na), so E = −ħ²κ²/2μ = −ħ²/(2μa²n²) = −13.606 eV/n², with a = 4πε₀ħ²/μe² = 0.0529 nm. Swap e² for Ze² and the same algebra returns a/Z and −13.606Z²/n² eV. In practice the useful rearrangement is ρ₀ = 2√(ER/|E|) with ER = 13.606 eV, which tests any trial energy for integer-ness in one line. But every step assumed E < 0. For E > 0, κ is imaginary, the runaway solution oscillates instead of growing, no truncation is needed, and every energy above the ionisation limit is allowed — which is why a discrete series has an edge with a continuum beyond it. The treatment is also non-relativistic, spinless, and point-nucleus, so fine structure, the Lamb shift and hyperfine splitting all sit below its resolution.
Change one variable at a time
Make the relationship visible.
Leave l = 0 and drag the trial n from 2.50 up to 3.00: the escaping tail collapses and the 3s curve appears with its two nodes. Then hold n = 3 and raise l to 3, where the series never stops - normalisability alone forbidding l > n - 1.
TRIAL ρ0 = 2n5.00
TRIAL ENERGY-2.177 eV
SERIES STOPS AT j1.50
FIRST RATIO c1/c0-1.500
Live interpretationTRIAL ρ0 = 2n: 5.00. TRIAL ENERGY: −2.177 eV. SERIES STOPS AT j: 1.50. FIRST RATIO c1/c0: −1.500
Catch the common trap
Explain before calculating.
In the series solution u = ρ(l+1) e(−ρ) v(ρ) of the hydrogen radial equation, with v(ρ) = Σ cⱼ ρj, why must the series for v terminate?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA hydrogen bound state has ρ₀ = 8. Using the termination condition ρ₀ = 2n, state n, list every orbital quantum number l the truncation permits, give the number of radial nodes for each, and compute the energy in eV. Take the Rydberg energy as 13.606 eV.
- ρ₀ = 2n gives n = 8/2 = 4.
- Termination needs jₘₐₓ = n − l − 1 to be a non-negative integer, so l ≤ n − 1 = 3, and the permitted values are l = 0, 1, 2, 3.
- Radial nodes = jₘₐₓ = n − l − 1: three for l = 0, two for l = 1, one for l = 2, none for l = 3. The angular nodes make up the difference, so all four states carry n − 1 = 3 nodes in total.
- E₄ = −13.606/n² = −13.606/16 = −0.8504 eV. As a check on the counting, Σ(2l+1) from l = 0 to 3 is 1 + 3 + 5 + 7 = 16 = n² orbital states at that one energy.
Answern = 4; l = 0, 1, 2, 3 carrying 3, 2, 1 and 0 radial nodes; E₄ = −0.850 eV, shared by 16 orbital states.
MediumFor n = 3 and l = 0, run the recursion c_(j+1) = [2(j+l+1) − ρ₀] cⱼ / [(j+1)(j+2l+2)] from c₀ = 1 until it stops, then locate the radial nodes in units of the Bohr radius a.
- ρ₀ = 2n = 6 and l = 0, so the recursion reads c_(j+1) = [2(j+1) − 6] cⱼ / [(j+1)(j+2)].
- j = 0: c₁ = (2 − 6)(1)/(1×2) = −2. j = 1: c₂ = (4 − 6)(−2)/(2×3) = (−1/3)(−2) = 2/3.
- j = 2: the numerator is 2(3) − 6 = 0, so c₃ = 0 and every coefficient after it. v(ρ) = 1 − 2ρ + (2/3)ρ², a polynomial of degree jₘₐₓ = n − l − 1 = 2, exactly as the termination condition promised.
- Nodes are the roots of v. Multiply by 3/2: ρ² − 3ρ + 3/2 = 0, so ρ = (3 ± √3)/2 = 0.6340 and 2.3660.
- ρ = κr = r/(na) = r/(3a), so r = 3aρ = 1.902a and 7.098a. Neither the origin nor infinity counts as a node: u vanishes there for every state.
Answerv(ρ) = 1 − 2ρ + (2/3)ρ², terminating at jₘₐₓ = 2, and the two 3s radial nodes lie at r = 1.90a and 7.10a.
HardA numerical integration of the radial equation is started for l = 1 at the trial energy E = −1.30 eV. Use ρ₀ = 2√(ER/|E|) with ER = 13.606 eV to show that this energy cannot be an eigenvalue, follow the first four coefficient ratios to see what goes wrong, then find the nearest energy that does work and the number of radial nodes it carries.
- ρ₀ = 2√(13.606/1.30) = 2√10.4662 = 2(3.2352) = 6.4703.
- Termination for l = 1 needs ρ₀ = 2(jₘₐₓ + 2), that is ρ₀ ∈ (4, 6, 8, …). 6.4703 is not in that set, so nothing in the recursion can stop it.
- Ratios c_(j+1)/cⱼ = [2(j+2) − 6.4703]/[(j+1)(j+4)]: j = 0 gives −2.4703/4 = −0.6176; j = 1 gives −0.4703/10 = −0.0470; j = 2 gives +1.5297/18 = +0.0850; j = 3 gives +3.5297/28 = +0.1261. The numerator has turned positive for good, so every later coefficient shares one sign and the ratio decays like 2/j — the signature of e(2ρ).
- So u behaves as ρ²e(+ρ) far out and the norm diverges. Legal values need ρ₀ = 2n: ρ₀ = 6 gives n = 3, ρ₀ = 8 gives n = 4.
- Energies from |E| = 4ER/ρ₀²: n = 3 gives 4(13.606)/36 = 1.5118 eV and n = 4 gives 4(13.606)/64 = 0.8504 eV. The trial sits 0.212 eV from the first and 0.450 eV from the second, so the nearest eigenvalue is E₃ = −1.512 eV.
- At n = 3 the value l = 1 is permitted, since l ≤ n − 1 = 2, and jₘₐₓ = n − l − 1 = 1. The recursion gives c₁ = (2(0+1+1) − 6)/(1×4) = −1/2, so v = 1 − ρ/2 with its single root at ρ = 2, that is r = 3a(2) = 6a.
Answerρ₀ = 6.470 is not an even integer, so the series runs away as e(2ρ); the nearest eigenvalue is n = 3, E₃ = −1.512 eV, whose l = 1 radial function has exactly one node, at r = 6a.