The Hydrogen Atom: Coulomb Radial Equation

Pass hydrogen's glow through a prism and you get not a rainbow but a barcode: a red line at 656 nm656\ \mathrm{nm}, a blue-green at 486486, a violet pair, then silence. In 1885 a Swiss schoolteacher, Johann Balmer, noticed the wavelengths obeyed a formula built from whole numbers — and nobody could say why nature counted. The answer had to wait forty years for one differential equation: put the Coulomb potential −e2/4πε0r-e^2/4\pi\varepsilon_0 r into the radial Schrödinger equation, demand that the wavefunction behave at zero and die at infinity, and the integers walk out on their own, carrying the energies En=−13.6 eV/n2E_n = -13.6\ \mathrm{eV}/n^2 and every line of the barcode with them. This is the summit of the chapter — the first real system solved exactly, the calculation that turned the Schrödinger equation from a promising idea into the foundation of atomic physics. The technique deserves the fame as much as the result: peel off the behavior forced at the origin (by the centrifugal barrier) and at infinity (by bound-state decay), and what remains is a polynomial equation whose series must terminate or ruin everything. Termination happens only at special energies. Quantization arrives not as a mysterious rule but as the survival condition of a normalizable wavefunction. Everything the chapter built converges here: separation of variables handed us one function of radius, the effective potential turned it into 1D mechanics, and the boundary conditions became executioners. What walks out is the spectrum, the strict bookkeeping of n,ℓ,mn, \ell, m — and, by one substitution whose story belongs to Hydrogenic Ions and Scaling with Nuclear Charge, every one-electron ion in the universe.

Key takeaways
  • Writing u=rRu = rR collapses hydrogen's 3D problem to one dimension on the half-line, with effective potential −e24πε0r+ℏ2ℓ(ℓ+1)2mr2-\dfrac{e^2}{4\pi\varepsilon_0 r} + \dfrac{\hbar^2\ell(\ell+1)}{2mr^2}.
  • Peeling off the forced asymptotics, u=ρℓ+1e−ρv(ρ)u = \rho^{\ell+1}e^{-\rho}v(\rho), leaves a power series for vv whose termination is what quantizes the energy.
  • The result: En=−13.6 eV/n2E_n = -13.6\ \mathrm{eV}/n^2, with Bohr radius a0=0.529 A˚a_0 = 0.529\ \mathrm{\mathring{A}} as the natural length and the associated Laguerre polynomials Ln−ℓ−12ℓ+1L_{n-\ell-1}^{2\ell+1} as the radial shapes.
  • The bookkeeping: n=N+ℓ+1n = N + \ell + 1, so ℓ≤n−1\ell \leq n - 1 and the state (n,ℓ)(n, \ell) has exactly n−ℓ−1n - \ell - 1 radial nodes.
  • Energy depends on nn alone — an "accidental" degeneracy of n2n^2 states per level, special to the 1/r1/r potential.
  • One substitution (e2→Ze2e^2 \to Ze^2) extends the whole solution to every hydrogenic ion — the story of Hydrogenic Ions and Scaling with Nuclear Charge.
  • Key idea: normalizability forces the series to terminate, and the termination condition is the quantization of energy.
The core insight

Pulling out the asymptotic factors ρℓ+1\rho^{\ell+1} (near the origin) and e−ρe^{-\rho} (at infinity) converts the Coulomb radial equation into a recursion whose termination — required for normalizability — quantizes the energy and produces the associated Laguerre polynomials.

Where this lives today

Hydrogen is now the most precisely measured system in science. The 1S–2S interval is known to fifteen digits — measured against atomic clocks in Garching — and comparing it with theory tests QED and even the proton's size: the "proton radius puzzle" of the 2010s was settled largely inside muonic hydrogen, the electron swapped for a muon 207207 times heavier. The orbit shrinks by the reduced-mass factor of 186186 (the proton is no longer effectively infinite), parking the muon close enough to feel nuclear structure. Antihydrogen, trapped at CERN, shows the same 1S–2S line — matter and antimatter obeying the same radial equation, so far to a dozen digits. The high-nn limit became a technology: Rydberg atoms with n∼60n \sim 60 are a few tenths of a micron across, exaggeratedly polarizable, and interact strongly enough to blockade one another across several microns — the mechanism behind neutral-atom quantum computers now running hundreds of qubits. And the ZZ-scaling runs daily in astrophysics: iron stripped to a hydrogenic ion in accretion disks emits its scaled Lyman lines near 6.9 keV6.9\ \mathrm{keV}, an X-ray barcode read by telescopes to weigh black holes. Balmer's integers, at every scale nature offers.

The physical picture

Separation of variables has already done the angular work — Separation of Variables in Spherical Coordinates showed that the spherical harmonics YℓmY_\ell^m carry all direction-dependence — and what remains is a single function R(r)R(r) of radius. The masterstroke is the substitution u(r)=rR(r)u(r) = rR(r), built at Effective Potential and the Centrifugal Barrier, which makes the radial equation look exactly like a 1D Schrödinger problem: a particle on the half-line r>0r > 0 in an effective potential — the Coulomb funnel pulling in, plus a centrifugal barrier ℏ2ℓ(ℓ+1)/2mr2\hbar^2\ell(\ell+1)/2mr^2 pushing out for ℓ≥1\ell \geq 1. All your 1D instincts transfer: bound states sit in the pocket, wavefunctions oscillate where allowed and decay where forbidden, and more nodes means more energy.

Figure
The effective potential — a funnel, a wall, and a pocket of bound states
r V(r) V = 0 (ionization) E₁ = −13.6 eV E₂ = −3.4 eV E₃ = −1.5 eV ℓ = 0: pure Coulomb funnel −e²/4πε₀r ℓ = 1: centrifugal wall + ℏ²ℓ(ℓ+1)/2mr² pushes the electron off the nucleus barrier above 0 bound states live in the pocket between wall and funnel rim

The two ends of the half-line dictate everything. Near r=0r = 0 the centrifugal term dominates (1/r21/r^2 beats 1/r1/r), forcing u∼rℓ+1u \sim r^{\ell+1} — the regular root of the indicial analysis that The Radial Equation performed, which also showed why the alternative r−ℓr^{-\ell} is inadmissible. At large rr a bound state must die exponentially, u∼e−κru \sim e^{-\kappa r}, the tail-selection law owned by Boundary Conditions and Normalization of Radial Wavefunctions — whose shooting experiment already bracketed E1=−13.6 eVE_1 = -13.6\ \mathrm{eV} numerically, a number this article now earns with algebra. The solution strategy writes both behaviors in by hand and solves for the interpolation between them. That interpolating function wants to be a polynomial; if its power series ran on forever it would rebuild a growing exponential and destroy the state. Where the series stops — and it can only stop at special energies — is where the spectrum comes from. One habit to carry forward: count nodes. The state (n,ℓ)(n, \ell) has n−ℓ−1n - \ell - 1 radial nodes, so at fixed nn, cranking up ℓ\ell trades radial wiggles for angular motion. The 2p2p state has no radial nodes at all; the 2s2s has one; both sit at the same energy — a coincidence of the pure Coulomb potential that we will have to explain. Here is that habit made visible: the radial probability r2∣Rnℓ∣2r^2|R_{n\ell}|^2 for the first three states. The 1s1s peaks at exactly one Bohr radius a0a_0, the 2s2s pays for its single node with a long second hump, and the 2p2p — maximal ℓ\ell for its level — is nodeless and pushed off the origin by the centrifugal wall.

Figure
Radial probability distributions — nodes count n−ℓ−1
2s node at r = 2a₀ 1s — peak at exactly a₀ 2s — one radial node (n−ℓ−1 = 1) 2p — nodeless: ℓ = n−1 r/a₀ 1 4 8 12 radial probability r²|Rₙℓ|² (each curve scaled to its own peak) more angular momentum at fixed n ⇒ fewer radial wiggles — energy is shared with the centrifugal term

And here is the payoff the algebra is aiming for: the energies the boundary conditions will select stack up as En=−13.6 eV/n2E_n = -13.6\ \mathrm{eV}/n^2, crowding toward the ionization edge, and every jump between rungs is one line of the spectral barcode — Lyman in the ultraviolet, Balmer in the visible, Paschen in the infrared, a taxonomy mapped in full at The Bohr Energy Spectrum and Rydberg Constant. Keep the ladder in view; the whole derivation is a hunt for why the rungs sit exactly there.

The math: from funnel to spectrum

The physical picture set the stage; now write it down. Take the general radial equation for a central potential, install the Coulomb attraction −e2/4πε0r-e^2/4\pi\varepsilon_0 r as V(r)V(r), and keep the substitution u=rRu = rR that makes the whole thing read like one-dimensional mechanics on the half-line:

−ℏ22md2udr2+[−e24πε0r+ℏ22mℓ(ℓ+1)r2]u=E u(1)-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[-\frac{e^2}{4\pi\varepsilon_0 r} + \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right]u = E\,u \qquad (1)

where: u=rRu = rR — reduced radial function; mm — electron mass (strictly the reduced mass μ\mu, a refinement quantified at The Bohr Energy Spectrum and Rydberg Constant); ℓ\ell — orbital quantum number; E<0E < 0 for bound states. This is the funnel-plus-wall of the hero figure in symbols: the Coulomb term digs the pocket, the ℓ(ℓ+1)\ell(\ell+1) term walls off the origin, and a bound electron must sit at negative energy inside the pocket. Everything that follows is a hunt for the values of EE this equation actually tolerates.

Derivation APeeling the Coulomb problem down to a recursion

The goal is to reduce equation (1) to something a power series can finish: a clean one-step recursion for polynomial coefficients. The strategy is the standard three-move sequence — go dimensionless so all the physics sits in one number, split off the behavior each end of the half-line forces, and let a series handle the interpolation in between. Nothing here is new machinery: the origin behavior was settled by The Radial Equation and the tail by Boundary Conditions and Normalization of Radial Wavefunctions; we consume both verdicts and watch how much work they do. The thing to track is the parameter ρ0\rho_0: it is the only place the energy survives, so whatever condition eventually falls on ρ0\rho_0 is the quantization of EE.

Step 1. Go dimensionless. With κ=−2mE/ℏ\kappa = \sqrt{-2mE}/\hbar, set ρ=κr\rho = \kappa r and ρ0=me22πε0ℏ2κ\rho_0 = \dfrac{me^2}{2\pi\varepsilon_0\hbar^2\kappa}:

d2udρ2=[1−ρ0ρ+ℓ(ℓ+1)ρ2]u(2)\frac{d^2u}{d\rho^2} = \left[1 - \frac{\rho_0}{\rho} + \frac{\ell(\ell+1)}{\rho^2}\right]u \qquad (2)

Why: all the physics now sits in one number, ρ0∝1/−E\rho_0 \propto 1/\sqrt{-E} — quantizing ρ0\rho_0 will quantize EE.

Step 2. Import the asymptotics. As ρ→∞\rho \to \infty the bracket →1\to 1, so u∼e−ρu \sim e^{-\rho} — the decaying branch, selected because the growing one is not normalizable (Boundary Conditions and Normalization of Radial Wavefunctions owns that execution). As ρ→0\rho \to 0 the centrifugal term dominates, so u∼ρℓ+1u \sim \rho^{\ell+1} — the regular indicial root, the irregular ρ−ℓ\rho^{-\ell} having been ruled out at The Radial Equation. Why: each limit is a two-term balance; solving the extremes first is what makes the middle tractable — and both verdicts arrive pre-proven from earlier in the chapter.

Step 3. Factor out both behaviors. Write u(ρ)=ρℓ+1e−ρv(ρ)u(\rho) = \rho^{\ell+1}e^{-\rho}v(\rho) and substitute:

ρ v′′+2(ℓ+1−ρ) v′+[ρ0−2(ℓ+1)]v=0(3)\rho\,v'' + 2(\ell + 1 - \rho)\,v' + \big[\rho_0 - 2(\ell+1)\big]v = 0 \qquad (3)

Why: the singular and decaying parts are handled; vv only needs to be tame. Equation (3) is the associated Laguerre equation (a confluent-hypergeometric relative) — but pretend we don't know that and solve honestly.

Step 4. Solve by power series. Insert v=∑jcjρjv = \sum_j c_j\rho^j and match powers of ρ\rho:

cj+1=2(j+ℓ+1)−ρ0(j+1)(j+2ℓ+2) cj(4)c_{j+1} = \frac{2(j + \ell + 1) - \rho_0}{(j+1)(j + 2\ell + 2)}\,c_j \qquad (4)

Why: a one-step recursion — each coefficient built from the last, with ρ0\rho_0 sitting in the numerator, waiting. *Verified: substituting u=ρℓ+1e−ρvu = \rho^{\ell+1}e^{-\rho}v into (2) reproduces (3) symbolically (sympy), and the recursion (4) checked by direct series substitution; matches Griffiths §4.2.*

Go deeper · Derivation B · 1 minTermination quantizes the energy
Derivation BTermination quantizes the energy

The goal is the spectrum: the discrete list of energies at which equation (1) admits a normalizable solution. The strategy is to interrogate the recursion (4) about its large-jj behavior — series that run forever have a growth rate, and this one's is fatal — then demand the only escape: termination. Watch how a global requirement (normalizability, a statement about r→∞r \to \infty) reaches back and imposes an arithmetic condition on ρ0\rho_0; that conversion of analysis into integers is the whole magic of bound-state quantum mechanics, and it is the same mechanism that terminated the oscillator's Hermite series.

Step 1. Face the catastrophe. For large jj, the recursion gives cj+1/cj→2/jc_{j+1}/c_j \to 2/j — precisely the growth pattern of e2ρe^{2\rho}. An infinite series rebuilds u∼ρℓ+1e+ρu \sim \rho^{\ell+1}e^{+\rho}: non-normalizable, no particle. Why: generic solutions of (3) really do blow up; boundary behavior at both ends can only be reconciled at exceptional energies.

Step 2. Demand termination. The series stops at j=Nj = N only if the numerator of (4) vanishes there: ρ0=2(N+ℓ+1)≡2n\rho_0 = 2(N + \ell + 1) \equiv 2n. Unwinding ρ0=2n\rho_0 = 2n through the definitions gives κ=1/na0\kappa = 1/na_0 and

En=−[m2ℏ2(e24πε0)2]1n2=−13.6 eVn2,a0=4πε0ℏ2me2=0.529 A˚(5)E_n = -\left[\frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\varepsilon_0}\right)^2\right]\frac{1}{n^2} = -\frac{13.6\ \mathrm{eV}}{n^2}, \qquad a_0 = \frac{4\pi\varepsilon_0\hbar^2}{me^2} = 0.529\ \mathrm{\mathring{A}} \qquad (5)

Why: this is the Bohr spectrum, now derived rather than postulated — and Balmer's integers are the possible stopping points of a power series.

Step 3. Name the survivors. The terminated series are the associated Laguerre polynomials, v=Ln−ℓ−12ℓ+1(2ρ)v = L_{n-\ell-1}^{2\ell+1}(2\rho). One concrete instance to see it happen: for the 2s2s state (n=2n = 2, ℓ=0\ell = 0, so ρ0=4\rho_0 = 4), the recursion gives c1/c0=(2−4)/2=−1c_1/c_0 = (2 - 4)/2 = -1 and then c2=0c_2 = 0 — the series is v∝1−ρv \propto 1 - \rho, which is exactly L11(2ρ)L^1_1(2\rho) up to normalization. Its single zero at ρ=1\rho = 1, i.e. r=2a0r = 2a_0, is the 2s2s node in the figure above. Why: the polynomial family, its conventions (which differ between textbooks), and the normalization integrals are the machinery of "Hydrogenic Wavefunctions and Quantum Numbers" — this article only needs to know the series stops and where. *Verified: the prefactor me4/2(4πε0)2ℏ2=2.18×10−18 J=13.6057 eVme^4/2(4\pi\varepsilon_0)^2\hbar^2 = 2.18\times10^{-18}\ \mathrm{J} = 13.6057\ \mathrm{eV} recomputed from CODATA constants; c1/c0=−1c_1/c_0 = -1 and v∝1−ρ=L11(2ρ)/2v \propto 1-\rho = L^1_1(2\rho)/2 checked against sympy's assoc_laguerre; κa0=1/n\kappa a_0 = 1/n confirmed numerically; matches Griffiths §4.2.*

With the spectrum won, reassemble the wavefunctions from the three pieces the derivation peeled apart — the ρℓ+1\rho^{\ell+1} behavior forced at the origin, the exponential tail forced at infinity, and the terminated polynomial interpolating between them. Restoring physical units through κ=1/na0\kappa = 1/na_0 gives the full radial eigenfunctions:

Rnℓ(r)=Nnℓ(2rna0)ℓe−r/na0 Ln−ℓ−12ℓ+1 ⁣(2rna0)(6)R_{n\ell}(r) = N_{n\ell}\left(\frac{2r}{na_0}\right)^{\ell} e^{-r/na_0}\, L_{n-\ell-1}^{2\ell+1}\!\left(\frac{2r}{na_0}\right) \qquad (6)

where: NnℓN_{n\ell} — normalization fixed by ∫0∞∣Rnℓ∣2r2 dr=1\int_0^\infty |R_{n\ell}|^2 r^2\,dr = 1; the decay length is na0na_0 — bigger orbits for higher levels; the polynomial degree n−ℓ−1n - \ell - 1 counts the radial nodes. The explicit table of these functions, the Laguerre conventions, and the closed-form NnℓN_{n\ell} are owned by Hydrogenic Wavefunctions and Quantum Numbers.

Go deeper · Derivation C · 1 minCounting the degeneracy of level n
Derivation CCounting the degeneracy of level n

The goal is the size of each energy level: how many independent states share EnE_n. The strategy is pure bookkeeping on the termination condition, plus one imported theorem — the mm-count is not ours to prove. What deserves attention is that the final n2n^2 has two different explanations stacked together: a symmetry guarantee for the mm-direction and an unexplained coincidence in the ℓ\ell-direction. Keeping those two strands separate is the difference between understanding hydrogen and merely memorizing it.

Step 1. Bound ℓ\ell. Since n=N+ℓ+1n = N + \ell + 1 with N≥0N \geq 0, we get ℓ≤n−1\ell \leq n - 1: level nn hosts ℓ=0,1,…,n−1\ell = 0, 1, \dots, n-1. Why: the termination index NN (the node count) cannot be negative.

Step 2. Count orientations. Each ℓ\ell carries 2ℓ+12\ell + 1 values of the magnetic quantum number mm, all at the same energy — the theorem proved at Degeneracy and the Role of the Magnetic Quantum Number: for any central potential, rotational symmetry forbids the energy from knowing mm. Why: this factor is imported, not re-derived — it is a symmetry guarantee, and it would survive any deformation of V(r)V(r) that keeps it central.

Step 3. Sum.

∑ℓ=0n−1(2ℓ+1)=n2(7)\sum_{\ell=0}^{n-1}(2\ell + 1) = n^2 \qquad (7)

Why: the ℓ\ell-independence is NOT guaranteed by rotational symmetry — a generic central potential splits different ℓ\ell at the same nn. The pure 1/r1/r potential hides an extra conserved quantity, and the story of how it enlarges SO(3) to SO(4) and forces the coincidence is owned by "Accidental Degeneracy and the SO(4) Symmetry". Spin doubles the count to 2n22n^2 — the shell capacities 2,8,18,…2, 8, 18, \dots of the periodic table. *Verified: ∑ℓ=0n−1(2ℓ+1)=n2\sum_{\ell=0}^{n-1}(2\ell+1) = n^2 proven symbolically and checked n=1..5n = 1..5 (1,4,9,16,251, 4, 9, 16, 25); n=2n = 2: one ss state + three pp states =4=22= 4 = 2^2; matches Sakurai §3.7.*

One line completes the family: replace e2→Ze2e^2 \to Ze^2 and the identical solution covers He+\mathrm{He}^+, Li2+\mathrm{Li}^{2+}, and every one-electron ion — energies ×Z2\times Z^2, sizes ×1/Z\times 1/Z — the single substitution worked through, with its stellar detective story, at Hydrogenic Ions and Scaling with Nuclear Charge.

Limiting cases

CaseResultMeaning
ground state (1,0)(1, 0)R10=2a03/2e−r/a0R_{10} = \dfrac{2}{a_0^{3/2}}e^{-r/a_0}nodeless exponential; ⟨r⟩=32a0\langle r \rangle = \tfrac32 a_0, peak probability at exactly a0a_0
large rru∼e−r/na0u \sim e^{-r/na_0}every level dies exponentially, with reach na0na_0 growing with excitation
small rrRnℓ∼rℓR_{n\ell} \sim r^{\ell}the centrifugal barrier expels high-ℓ\ell states; only ss-states touch the nucleus
large nnEn→0−E_n \to 0^-, ⟨r⟩∼n2a0\langle r \rangle \sim n^2a_0Rydberg atoms: fragile, nearly classical orbits reaching thousands of a0a_0
nuclear charge ZZEn×Z2E_n \times Z^2, sizes × 1/Z\times\, 1/Zone substitution, no new equation — owned by Hydrogenic Ions and Scaling with Nuclear Charge

Worked examples

1 · The red line of hydrogen (confidence builder)

Compute the wavelength of the n=3→2n = 3 \to 2 transition — the Hα\alpha line that colors emission nebulae red — and the Balmer-series limit.

Step 1. Transition energy from (5): ΔE=13.6(14−19)=13.6×536=1.89 eV\Delta E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 13.6 \times \dfrac{5}{36} = 1.89\ \mathrm{eV}.

Step 2. Wavelength: λ=hcΔE=1240 eV⋅nm1.89 eV=656 nm\lambda = \dfrac{hc}{\Delta E} = \dfrac{1240\ \mathrm{eV\cdot nm}}{1.89\ \mathrm{eV}} = \boxed{656\ \mathrm{nm}} — Balmer's red line, to three figures, from a boundary condition. (At the next digit the reduced mass matters: the vacuum wavelength is 656.47 nm656.47\ \mathrm{nm} once m→μm \to \mu — the refinement quantified at The Bohr Energy Spectrum and Rydberg Constant.)

Step 3. The series limit is ionization from n=2n = 2: ΔE=3.4 eV\Delta E = 3.4\ \mathrm{eV}, so λ=1240/3.4=365 nm\lambda = 1240/3.4 = \boxed{365\ \mathrm{nm}}. All Balmer lines crowd between 656656 and 365 nm365\ \mathrm{nm} — the barcode's whole visible span, fixed by two numbers.

2 · One state by hand: the 2s from the recursion (exam level)

Use the recursion (4) to construct the complete 2s2s radial function, locate its node, and contrast it with the 2p2p.

Step 1. Set the dials. For n=2n = 2: ρ0=2n=4\rho_0 = 2n = 4 and κ=1/2a0\kappa = 1/2a_0, so ρ=r/2a0\rho = r/2a_0. For the 2s2s, ℓ=0\ell = 0, and the series must terminate at N=n−ℓ−1=1N = n - \ell - 1 = 1: a degree-one polynomial.

Step 2. Run the recursion once. With j=0j = 0, ℓ=0\ell = 0: c1=2(0+0+1)−4(1)(2)c0=−c0c_1 = \dfrac{2(0+0+1) - 4}{(1)(2)}c_0 = -c_0. With j=1j = 1: the numerator 2(1+0+1)−4=02(1+0+1) - 4 = 0 kills c2c_2. So v(ρ)=c0(1−ρ)v(\rho) = c_0(1 - \rho) — the series terminated exactly where the bookkeeping promised.

Step 3. Assemble the state: u=ρ e−ρv∝ρ(1−ρ)e−ρu = \rho\,e^{-\rho}v \propto \rho(1-\rho)e^{-\rho}, i.e.

R20(r)∝(1−r2a0)e−r/2a0.R_{20}(r) \propto \left(1 - \frac{r}{2a_0}\right)e^{-r/2a_0}.

The node sits where vv vanishes: ρ=1\rho = 1, i.e. r=2a0=1.06 A˚\boxed{r = 2a_0 = 1.06\ \mathrm{\mathring{A}}} — the red dot in the probability figure. The decay length is 2a02a_0: the n=2n = 2 shell reaches twice as far as the ground state.

Step 4. Now the 2p2p (ℓ=1\ell = 1): termination at N=0N = 0 means v=c0v = c_0, a constant — the recursion numerator 2(0+1+1)−4=02(0 + 1 + 1) - 4 = 0 vanishes before a second term is ever born. So R21∝r e−r/2a0R_{21} \propto r\,e^{-r/2a_0}: nodeless, pushed off the origin, peaking (in shell probability) at 4a04a_0. Same energy, opposite anatomy — and only the 1/r1/r potential lets them tie.

Origins · 1 minNiels Bohr and how this was discovered
NB
Niels Bohr
1885–1962 · Copenhagen

Bohr said that the moment he saw Balmer's formula, "everything became clear." His 1913 trilogy quantized the hydrogen atom by decree — stationary orbits, quantum jumps — and derived the Rydberg constant from ee, mem_e, ℏ\hbar: the first time a spectroscopic number came out of fundamental constants. The Pickering-line episode was his model's trial by fire, and the reduced-mass verdict made his reputation. Yet Bohr held the theory lightly, knowing decree is not derivation; he spent the 1920s assembling the people — Heisenberg, Pauli, Dirac among them — who would replace his orbits with honest mechanics, and his institute in Copenhagen became quantum theory's capital. (How he named the non-radiating states, and how his correspondence principle steered the new mechanics toward its classical limit, are stories told at their own articles; this band is about the constant he made calculable.) The radial equation of this article is the repayment: his −13.6 eV/n2-13.6\ \mathrm{eV}/n^2, re-derived with nothing postulated but the Schrödinger equation. He argued with Einstein about what it all meant for thirty years, gently, and never lost.

History
1885
— Balmer, a Basel schoolteacher, fits hydrogen's four visible lines with integers: λ∝n2/(n2−4)\lambda \propto n^2/(n^2 - 4).
1913
— Bohr's model derives the Rydberg constant from fundamental constants and settles the Pickering lines as He+\mathrm{He}^+ — the reduced-mass shift convinces Einstein.
1926
— Schrödinger and, independently by matrix methods, Pauli solve hydrogen exactly; quantization becomes a boundary-value statement.
1928
— Dirac's relativistic equation adds fine structure: the n2n^2 degeneracy splits by jj.
1947
— Lamb measures a tiny splitting Dirac forbids; explaining the Lamb shift launches quantum electrodynamics.

Going further

Texts. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §4.2 · Sakurai & Napolitano, Modern Quantum Mechanics, §3.7 · Shankar, Principles of Quantum Mechanics, Ch. 13 · Cohen-Tannoudji, Diu & Laloë, Vol. 1, Ch. VII · Bethe & Salpeter, Quantum Mechanics of One- and Two-Electron Atoms (the classic reference).

Provenance. Equations and substitutions follow Griffiths §4.2 (the u=ρℓ+1e−ρvu=\rho^{\ell+1}e^{-\rho}v method and recursion) and Sakurai §3.7. A verifier should recompute the energy prefactor me4/(2(4πε0)2ℏ2)=13.6 eVme^4/(2(4\pi\varepsilon_0)^2\hbar^2)=13.6\,\mathrm{eV}, the Bohr radius a0=0.529 A˚a_0=0.529\,\text{Å}, the Rydberg constant R∞=1.097×107 m−1R_\infty=1.097\times10^7\,\mathrm{m^{-1}}, the degeneracy sum ∑ℓ=0n−1(2ℓ+1)=n2\sum_{\ell=0}^{n-1}(2\ell+1)=n^2, and ⟨r⟩=a02[3n2−ℓ(ℓ+1)]\langle r\rangle=\tfrac{a_0}{2}[3n^2-\ell(\ell+1)], all standard textbook values (Griffiths, Shankar Ch. 13, Cohen-Tannoudji Vol. 1 Ch. VII). Edition note (2026-07-08). This article folds the approved v3 pilot into the track as its narrative spine, with the AllQFT base grafted for depth; every spine number was machine re-verified before shipping (41 checks — all clean, with two precision tightenings: muonic hydrogen's orbit shrinks by the reduced-mass factor 186186, not the bare mass ratio 207207; and n∼60n \sim 60 Rydberg atoms are tenths of a micron across, with blockade radii of microns). The spine's ZZ-scaling derivation ("one substitution, every hydrogenic ion") and its Pickering-lines worked example are relocated to "Hydrogenic Ions and Scaling with Nuclear Charge" (QUANT.HYDR.ZSCAL.0945), which owns the one-electron family; this article keeps the hydrogen (Z=1Z=1) solve and defers by name. Derivations are recast to consume the chapter's earlier serials (indicial roots, tail selection, the (2ℓ+1)(2\ell+1) theorem) rather than re-derive them.

Prerequisites. The radial equation · separation of variables in spherical coordinates · effective potential and the centrifugal barrier · boundary conditions and normalization of radial wavefunctions. Leads to. The Bohr energy spectrum and Rydberg constant · hydrogenic wavefunctions and quantum numbers · radial probability distribution and expectation values · the Bohr radius and characteristic atomic scales · accidental degeneracy and SO(4) · hydrogenic ions and scaling with nuclear charge.

How do I actually see that the terminated series is an associated Laguerre polynomial?

Run the recursion for a small case and compare. For the 2s2s (ρ0=4\rho_0 = 4, ℓ=0\ell = 0) you get v∝1−ρv \propto 1 - \rho; sympy's `assoc_laguerre(1, 1, x)` returns 2−x2 - x, which at x=2ρx = 2\rho is 2(1−ρ)2(1-\rho) — the same polynomial up to the overall constant that normalization absorbs. In general, equation (3) is the associated Laguerre ODE with argument 2ρ2\rho, so its polynomial solutions are Ln−ℓ−12ℓ+1(2ρ)L^{2\ell+1}_{n-\ell-1}(2\rho) by definition. Beware: two normalization conventions for LqkL^k_q circulate in textbooks — the dictionary is kept at Hydrogenic Wavefunctions and Quantum Numbers.

Energy is independent of m and of ℓ. Is that one fact or two?

Two, with different pedigrees. The mm-independence is a theorem: any central potential commutes with all three components of L\mathbf{L}, and the ladder argument at Degeneracy and the Role of the Magnetic Quantum Number forces all 2ℓ+12\ell+1 orientations to share an energy. The ℓ\ell-independence is an accident of 1/r1/r: nothing about rotational symmetry demands that 2s2s tie with 2p2p, and in any real multi-electron atom they don't. The hidden conservation law that explains the accident is the business of Accidental Degeneracy and the SO(4) Symmetry.

Bohr's model got exactly the same energies. Was it right after all?

It got one thing right — the spectrum — for a reason that only becomes clear here: both calculations are controlled by the same combination of constants, and the 1/r1/r potential is special enough that Bohr's circular-orbit shortcut lands on the exact eigenvalues. Everything else in the model is wrong. The electron is not on an orbit; the ground state has zero orbital angular momentum (ℓ=0\ell = 0), where Bohr's model demands ℏ\hbar; and the state is a standing probability distribution peaking at a0a_0, not a planet at radius a0a_0. The model's own story — and why it still earned the Nobel — belongs to the optics track's Bohr-model article.

What happened to the positive-energy solutions?

They exist and they matter — they just aren't normalizable to 1. For E>0E > 0, κ\kappa is imaginary, the tail oscillates instead of decaying, and no termination condition arises: every positive energy is allowed. These are the scattering states — an ionized electron flying past a proton — normalized as a continuum. They complete the basis (bound states alone are not complete), and they are the arena of scattering theory later in the track.

Why quantize ρ₀ instead of just solving for E directly?

Because ρ0\rho_0 is where the equation stores the energy after nondimensionalization — fixing ρ0=2n\rho_0 = 2n is fixing EE. The move pays twice: the recursion only ever sees ρ0\rho_0, so the termination condition reads off cleanly; and it exposes the structure — one dimensionless number controls the whole problem, which is why one substitution rescales the entire family at Hydrogenic Ions and Scaling with Nuclear Charge.

When do I need the reduced mass instead of the electron mass?

The moment you care about the fourth significant figure — or about any comparison between atoms. The proton moves too: the honest two-body problem separates into center-of-mass motion plus a one-body problem with μ=meM/(me+M)\mu = m_e M/(m_e + M), which for hydrogen shifts everything by about one part in 1836. That is invisible in 13.6 eV13.6\ \mathrm{eV} but glaring in spectroscopy — it is how deuterium was discovered, and for positronium (μ=me/2\mu = m_e/2) it halves every energy. The machinery lives at The Bohr Energy Spectrum and Rydberg Constant.

Before this

Where this leads