Solution of the Schrödinger Equation for Hydrogen

This article stands at a seam, and it is worth drawing the line cleanly before crossing it. The machinery that solves the hydrogen atom — separating the equation in spherical coordinates, reducing it to a radial equation with its effective potential and centrifugal barrier, and forcing a power series to terminate so that the associated Laguerre polynomials fall out and the energy is quantized — is owned by the companion article aq-qm-hydrogen-coulomb-radial-equation, and is recalled here by name rather than re-derived. What this article owns is the other side of that seam: the structure of the solution and everything a spectroscope reads off it. Granting the solution $\psi_{n\ell m} = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi)$ and its spectrum $E_n = -13.6\ \text{eV}/n^2$ , we ask what changed from Bohr — and the answer is not the energies but the states: a probability cloud instead of an orbit, an $\ell = 0$ ground state where Bohr demanded a full unit of angular momentum, a degeneracy of $n^2$ (or $2n^2$ with spin) where Bohr had one orbit per level. Keep the split in mind throughout. The radial equation tells you why the energies are what they are; this article tells you what the atom looks like at those energies and how the pattern reaches a spectrograph — the term diagram, the shell capacities $2, 8, 18$ that seed the periodic table, the reduced-mass shifts that separate hydrogen from deuterium and halve the binding of positronium, and the accidental $\ell$ -degeneracy whose breaking in heavier atoms is the whole reason sodium has its yellow D lines. The Schrödinger solution did not move a single hydrogen level; it replaced a lucky orbit with a complete, computable atom.

Key takeaways

The schrödinger equation for one electron in the coulomb potential $-e^2/4\pi\varepsilon_0 r$ separates into $\psi_{n\ell m} = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi)$ ; the radial-equation machinery that quantizes it is recalled by name from aq-qm-hydrogen-coulomb-radial-equation.
The bound-state energies are exactly Bohr's, $E_n = -13.6\ \text{eV}/n^2$ , depending on the principal quantum number $n$ alone — confirmed here numerically for $(n,\ell) = (2,0),(2,1),(3,0),(3,1),(3,2)$ to better than $0.1\%$ .
Three integers label every state: $n = 1,2,3,\dots$ ; $\ell = 0,\dots,n-1$ ; $m = -\ell,\dots,+\ell$ . The ground state has $\ell = 0$ — no angular momentum, contradicting Bohr's $L = \hbar$ .
The degeneracy is $\sum_{\ell=0}^{n-1}(2\ell+1) = n^2$ , or $2n^2$ with spin: the shell capacities $2, 8, 18$ . The $\ell$ -part is accidental degeneracy, special to the pure $1/r$ potential and lifted by screening in every other atom.
The reduced mass $\mu$ enters the same solution: it splits H from D and, for positronium ( $\mu = m_e/2$ ), halves every energy to $E_1 = -6.8$ eV.
Key idea: the Schrödinger solution reproduces Bohr's energies exactly but replaces his single orbit with a probability cloud $\psi_{n\ell m}$ carrying an $n^2$ -fold degeneracy, an $\ell = 0$ ground state, and the full state-labeling that organizes atomic spectra.

The core insight

Solving the Coulomb Schrödinger equation gives $\psi_{n\ell m} = R_{n\ell}Y_\ell^m$ with energies $E_n = -13.6\ \text{eV}/n^2$ that ignore $\ell$ and $m$ . Same spectrum as Bohr; different atom. What is new is the structure: quantum numbers with fixed ranges, a degeneracy of $n^2$ ( $2n^2$ with spin), an $\ell = 0$ ground state, and — because energy ignores $\ell$ only for pure $1/r$ — an accidental degeneracy whose breaking explains alkali spectra.

Where this lives today

The hydrogen solution is now a benchmark rather than a frontier — precisely because it is exact, every measured deviation from it isolates new physics. The $1S$ – $2S$ interval is known to fifteen digits and, compared with theory, both fixes the Rydberg constant and stress-tests QED; the "proton radius puzzle" of the 2010s was fought out in this comparison, with muonic hydrogen (the electron swapped for a muon) shrinking the orbit until the wavefunction overlaps the proton and reads its size. Antihydrogen trapped at CERN shows the same $1S$ – $2S$ line to a dozen digits — matter and antimatter obeying one radial equation. The structure the solution hands over runs through chemistry and much of technology. The $2n^2$ shell capacities and $s, p, d, f$ labels are the skeleton of the periodic table; the accidental $\ell$ -degeneracy, lifted by core screening, is what makes alkali spectra (and their quantum defect) differ from hydrogen's. And the high- $n$ end is now a resource: Rydberg atoms with $n \sim 60$ are enormous and strongly interacting, the basis of neutral-atom quantum computers running hundreds of qubits.

The physical picture

Start where the radial equation leaves off. Writing $\psi = R(r)Y_\ell^m$ turns the three-dimensional problem into one dimension in $r$ , with the electron moving in an effective potential — the Coulomb funnel pulling inward plus a centrifugal barrier $\hbar^2\ell(\ell+1)/2\mu r^2$ that, for $\ell \geq 1$ , walls the electron off from the nucleus. This effective potential and its centrifugal barrier are owned by aq-qm-hydrogen-coulomb-radial-equation; what matters here is the picture it paints and one spectroscopic consequence it forces.

Figure

The effective potential for ℓ = 0, 1, 2 (recalled from the radial-equation article). The centrifugal barrier lifts the well bottom as ℓ grows — for ℓ = 1 it sits at −6.8 eV, too shallow to hold the −13.6 eV ground state, so ℓ = 1 needs n ≥ 2.

That last point is the spectroscopic payoff hiding in a recalled figure. The centrifugal barrier for $\ell = 1$ lifts the well bottom to $-6.8$ eV — above the $-13.6$ eV ground state — so no $\ell = 1$ state fits at $n = 1$ ; the lowest $p$ state is $2p$ . For $\ell = 2$ the floor rises to $-2.27$ eV, excluding $n = 1, 2$ . This is exactly the rule $\ell \leq n - 1$ : the geometry of the barrier made into a counting law. Now the counting, which is the punchline. Each level $n$ carries $\ell = 0,\dots,n-1$ , each $\ell$ carries $2\ell+1$ orientations, and doubling for spin gives the chemistry shell capacities $2, 8, 18$ . The energy does not care which box a state sits in — every cell in a row shares the same $E_n$ .

Figure

Punchline — the degeneracy grid. Row n holds ℓ = 0…n−1; each cell carries 2ℓ+1 orbitals; with spin the row totals 2n² — the shell capacities 2, 8, 18, 32. Every state in a row shares one energy E_n.

The core equations

The starting point is the time-independent schrödinger equation for a single electron bound in the attractive coulomb potential of a proton. Its solution — the separation, the radial reduction, the termination that quantizes the energy — is owned by aq-qm-hydrogen-coulomb-radial-equation; here we state the equation and read off its structural output.

\left[-\frac{\hbar^2}{2\mu}\nabla^2 - \frac{e^2}{4\pi\varepsilon_0 r}\right]\psi(\mathbf r) = E\,\psi(\mathbf r) \qquad (1)

where: $\mu = m_e M_p/(m_e + M_p)$ — reduced mass, the same reduced mass that shifts the spectrum; $-e^2/4\pi\varepsilon_0 r$ — the Coulomb attraction; $E < 0$ — a bound-state energy. The spherical symmetry of the potential is what lets the equation separate, and separation is the whole reason the atom is exactly solvable. Because the potential depends only on $r$ , the wavefunction factors into a radial part and an angular part, the latter universal to any central force.

\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)\,Y_\ell^m(\theta,\phi) \qquad (2)

where: $R_{n\ell}$ — radial function (a Laguerre polynomial times a decaying exponential, recalled by name); $Y_\ell^m$ — spherical harmonic, eigenfunction of $\hat L^2$ and $\hat L_z$ ; $n, \ell, m$ — the three quantum numbers that label every state. The radial factor carries the physics of the $1/r$ potential; the angular factor carries only the geometry of rotation. Demanding a normalizable radial function quantizes the energy — the termination condition owned by the radial-equation article — and the result is Bohr's spectrum, now derived rather than postulated, with the reduced mass in place.

E_n = -\frac{\mu e^4}{2(4\pi\varepsilon_0)^2\hbar^2}\frac{1}{n^2} = -\frac{13.6\ \text{eV}}{n^2}, \qquad n = 1, 2, 3, \dots \qquad (3)

where: $n$ — the principal quantum number, and the only quantum number the energy depends on; $13.6$ eV — the Rydberg energy. This $\ell$ - and $m$ -independence is the central surprise of the solution, and everything below is a consequence of it. The termination also fixes the allowed ranges of the three integers, and hence how many states share each energy.

\ell = 0, 1, \dots, n-1; \quad m = -\ell, \dots, +\ell; \qquad g_n = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2 \quad(2n^2 \text{ with spin}) \qquad (4)

where: $\ell$ — orbital quantum number, capped at $n-1$ ; $m$ — magnetic quantum number, $2\ell+1$ values per $\ell$ ; $g_n$ — degeneracy of level $n$ . The $m$ -degeneracy is guaranteed by rotational symmetry, but the $\ell$ -degeneracy is not — it is the accidental degeneracy peculiar to the $1/r$ Coulomb law.

Derivation AThe separated structure and the (n, ℓ, m) ladder

The goal is to turn the solution's existence into a complete labeling scheme: given that normalizability forces the radial series to terminate — the verdict owned by aq-qm-hydrogen-coulomb-radial-equation — pin down exactly which triples $(n, \ell, m)$ are allowed, and read the ground-state contradiction with Bohr straight off the ranges. We do not re-derive the termination; we consume it and do the bookkeeping that spectroscopy actually uses.

Step 1. Take the separation as given: $\psi = R_{n\ell}Y_\ell^m$ , with the angular factor supplying $\hat L^2 Y_\ell^m = \hbar^2\ell(\ell+1)Y_\ell^m$ and $\hat L_z Y_\ell^m = \hbar m\,Y_\ell^m$ . Why: the central potential commutes with $\hat L^2$ and $\hat L_z$ , so energy, $|\mathbf L|$ , and $L_z$ are simultaneously sharp — three good quantum numbers, one per degree of freedom.

Step 2. Import the termination condition. Writing the radial solution as $r^{\ell+1}$ near the origin times $e^{-r/na_0}$ times a polynomial of degree $n_r$ , normalizability requires the polynomial to stop, fixing $n = n_r + \ell + 1$ with $n_r \geq 0$ . Why: this is the recalled result — an infinite series would regrow the divergent exponential — and it is the single equation that couples the radial and angular labels.

Step 3. Read the ranges. Since $n_r \geq 0$ , $\ell \leq n - 1$ ; and since $\hat L_z$ has $2\ell+1$ eigenvalues, $m = -\ell,\dots,+\ell$ . So $n = 1$ admits only $\ell = 0$ ; $n = 2$ admits $\ell = 0, 1$ ; and so on. Why: the cap $\ell \leq n-1$ is the same barrier-geometry rule the effective-potential figure showed — no $\ell = 1$ state fits at $n = 1$ .

Step 4. Confront Bohr. The ground state $(1,0)$ has $\ell = 0$ : zero orbital angular momentum, a spherically symmetric cloud with a peak — not an orbit — at $a_0$ . Bohr's model demanded $L = 1\cdot\hbar$ for the same state. Why: this is the sharpest break between the two pictures — identical energy, opposite angular momentum, and only experiment (the s-state hyperfine contact term) settles it in Schrödinger's favor. *Verified: the allowed-triple counts (n=1→1, n=2→4, n=3→9 orbitals) and the $\ell = 0$ ground state are checked in verify_aq-optics-schrodinger-hydrogen-solution.py.*

Go deeper · Derivation B · 1 minThe energy ignores ℓ — a numerical verdict

Derivation BThe energy ignores ℓ — a numerical verdict

The goal is to establish, beyond the algebra, that hydrogen's energy truly depends on $n$ alone — the accidental degeneracy — by solving the radial equation numerically for several $(n, \ell)$ and watching the energies collapse onto $-13.6\ \text{eV}/n^2$ . Where the owner article proves quantization symbolically, this is the independent check a skeptic would run: integrate the actual differential equation and see whether $2s$ and $2p$ really land at the same energy.

Step 1. Discretize the radial equation $-\tfrac12 u'' + [\ell(\ell+1)/2r^2 - 1/r]u = E u$ (atomic units, $u = rR$ ) on a grid and diagonalize the resulting tridiagonal matrix. Why: turning the ODE into a matrix eigenvalue problem lets a solver find the bound energies directly, with no assumed form for the answer.

Step 2. Extract, for each $\ell$ , the $(n-\ell-1)$ -th eigenvalue — the state $n$ for that subshell. Run $(2,0), (2,1), (3,0), (3,1), (3,2)$ . Why: these are the smallest cases where $\ell$ -degeneracy can be tested — two states at $n=2$ , three at $n=3$ .

Step 3. Compare. The solver returns $E(2,0) = E(2,1) = -3.401$ eV and $E(3,0) = E(3,1) = E(3,2) = -1.512$ eV, each matching $-13.6\ \text{eV}/n^2$ to under $0.1\%$ . Why: the energies are numerically independent of $\ell$ — the accidental degeneracy is real, not an artifact of a slick derivation.

Step 4. Name what it means and what breaks it. The $\ell$ -independence reflects a hidden symmetry of the pure $1/r$ potential (the Runge–Lenz vector, SO(4)); its microscopic story lives in the QM track. Any departure from $1/r$ — relativity, screening — lifts it. Why: this is the hinge to the rest of atomic physics: hydrogen is degenerate by accident, and every other atom is not. *Verified: the scipy eigensolver reproduces $E_{n\ell}$ for all five $(n,\ell)$ to $<0.1\%$ of $-13.6\ \text{eV}/n^2$ in verify_aq-optics-schrodinger-hydrogen-solution.py.*

Go deeper · Derivation C · 1 minCounting the states: Σ(2ℓ+1) = n² and the 2n² shells

Derivation CCounting the states: Σ(2ℓ+1) = n² and the 2n² shells

The goal is the number every chemist knows — the shell capacities $2, 8, 18, 32$ — derived from the quantum-number ranges as a clean closed form, and connected to why these numbers, not others, structure the periodic table. This is pure bookkeeping on the ranges of Derivation A, but it is the bookkeeping spectroscopy and chemistry are built on, so it earns its own card.

Step 1. Fix a level $n$ . It contains subshells $\ell = 0, 1, \dots, n-1$ , and subshell $\ell$ holds $2\ell + 1$ orbitals (its $m$ values). Why: the orbital count per subshell is the $2\ell+1$ orientation multiplicity, guaranteed by rotational symmetry.

Step 2. Sum the arithmetic series: $\sum_{\ell=0}^{n-1}(2\ell+1) = 1 + 3 + 5 + \dots + (2n-1) = n^2$ . Why: the sum of the first $n$ odd numbers is $n^2$ — an identity, so the spatial degeneracy is exactly $n^2$ with no approximation.

Step 3. Double for spin: each orbital holds two electrons ( $m_s = \pm\tfrac12$ ), giving $2n^2 = 2, 8, 18, 32$ for $n = 1, 2, 3, 4$ . Why: spin is an extra two-valued label the spatial solution does not know about; multiplying by two gives the full state count.

Step 4. Connect to the table. These capacities set the periods' lengths and, once screening lifts the $\ell$ -degeneracy so subshells fill in energy order, the whole aufbau sequence. Why: hydrogen hands the labels and counts to every atom; the energies of those labels are rearranged by electron–electron repulsion — the subject of multi-electron atoms. *Verified: $\sum_{\ell=0}^{n-1}(2\ell+1) = n^2$ proved symbolically (sympy) and the capacities $2, 8, 18, 32, 50$ checked for $n = 1$ – $5$ in verify_aq-optics-schrodinger-hydrogen-solution.py.*

Limiting cases

Case	Result	Meaning
ground state $(1,0)$	$\ell = 0$ , cloud peaks at $a_0$ , $\langle r\rangle = 1.5a_0$	spherically symmetric, zero angular momentum — Bohr's orbit is gone
energy vs $\ell$	$E_{n\ell} = -13.6\ \text{eV}/n^2$ , no $\ell$	the accidental degeneracy; numerically confirmed to $<0.1\%$
infinite nuclear mass	$\mu \to m_e$	drops the $0.054\%$ reduced-mass shift; using $m_e$ overstates $E$ by 1 part in 1836
positronium, $\mu = m_e/2$	$E_1 = -6.8$ eV, size $2a_0$	the reduced-mass showcase — every energy halved, every length doubled
hydrogenic ion, charge $Z$	$E_n = -Z^2\,13.6\ \text{eV}/n^2$ , $a_0 \to a_0/Z$	one substitution $e^2 \to Ze^2$ covers He⁺, Li²⁺, …
relativistic $\alpha^2$ terms	fine structure splits by $j$	the $\ell$ -degeneracy breaks — beyond the plain Schrödinger equation

Common trap“The Schrödinger equation changed hydrogen's energy levels”

It did not. To the accuracy of the gross structure, the Schrödinger energies are identical to Bohr's: $E_n = -13.6\ \text{eV}/n^2$ , the same Rydberg formula, the same Balmer lines. What changed is everything around the energies. Bohr had one circular orbit per level with $L = n\hbar$ ; Schrödinger has $n^2$ degenerate states (with $\ell = 0,\dots,n-1$ and $2\ell+1$ orientations each), a ground state with zero orbital angular momentum, probability clouds rather than trajectories, and — crucially — transition amplitudes you can actually compute, which give intensities and selection rules Bohr could never supply. The energies move only when you go beyond Schrödinger: fine structure ( $\alpha^2$ ), the Lamb shift (QED), hyperfine structure (the proton's moment). Say "Schrödinger re-derived Bohr's spectrum and replaced his orbits with states," never "Schrödinger corrected the energies."

Worked examples

1 · Enumerating the n = 3 level (confidence builder)

List every state of the $n = 3$ level of hydrogen, count them, and give the mean radius of the most and least penetrating subshells.

Step 1. Allowed $\ell$ : from $\ell \leq n - 1$ , the level has $\ell = 0$ ( $3s$ ), $\ell = 1$ ( $3p$ ), $\ell = 2$ ( $3d$ ). Orbital counts $2\ell+1$ : one $3s$ , three $3p$ , five $3d$ .

Step 2. Total orbitals: $1 + 3 + 5 = 9 = n^2$ . With spin ( $\times 2$ ), $\mathbf{18 = 2n^2}$ distinct states, all sharing $E_3 = -13.6/9 = -1.51$ eV.

Step 3. Mean radii from $\langle r\rangle_{n\ell} = \tfrac{a_0}{2}[3n^2 - \ell(\ell+1)]$ : the $3s$ ( $\ell=0$ ) reaches $\langle r\rangle = \tfrac{a_0}{2}(27) = 13.5\,a_0$ , while the $3d$ ( $\ell=2$ ) sits at $\tfrac{a_0}{2}(27-6) = 10.5\,a_0$ . The low- $\ell$ state reaches farther on average even though it feels less centrifugal repulsion — a foretaste of penetration.

Step 4. Why it matters. In hydrogen all nine orbitals are degenerate, so this is just labeling. In sodium, the $3s$ electron's deep penetration toward the core lowers its energy well below $3p$ — lifting the accidental degeneracy and producing the sodium D lines. The $n=3$ level: 9 orbitals, 18 states, all at $-1.51$ eV in hydrogen.

2 · Positronium: the reduced mass on full display (exam level)

Positronium is an electron bound to a positron. Find its ground-state energy and size, and say what the reduced mass does to the whole spectrum.

Step 1. Reduced mass: with two equal masses, $\mu = m_e m_e/(m_e + m_e) = m_e/2$ — exactly half the electron mass, the largest reduced-mass effect any hydrogenic system shows.

Step 2. Energy scales linearly with $\mu$ : $E_1 = -13.6\ \text{eV}\times(\mu/m_e) = -13.6/2 = \mathbf{-6.80\ eV}$ . Every level halves; the whole spectrum, Lyman to Paschen, shifts to half the hydrogen wavenumbers.

Step 3. Size scales as $1/\mu$ : the Bohr radius doubles, $a_{0,\text{Ps}} = 2a_0 = 1.06$ Å, so positronium is a puffed-up, loosely bound twin of hydrogen. It is also unstable — the two particles annihilate — but lives long enough for its $1S$ – $2S$ interval to be measured and compared with QED.

Step 4. Generalize. For ordinary hydrogen $\mu/m_e = 0.99946$ , a $0.054\%$ shift; for deuterium the heavier nucleus pushes $\mu$ up slightly, blue-shifting the lines and betraying the isotope. The same $\mu$ in equation (1) does all of it. Positronium: $E_1 = -6.80$ eV, radius $2a_0$ — the reduced mass halving the atom.

Physics of the portrait

The equation that dissolved Bohr's orbits into probability clouds was written in a burst of work over the winter of 1925–26, and its author was, by the standards of the revolution he joined, an old man of thirty-eight.

Origins · 1 minErwin Schrödinger and how this was discovered

Erwin Schrödinger

1887–1961 · Zürich · Berlin · Dublin

Erwin Schrödinger was a Viennese physicist of wide culture — fluent in four languages, steeped in philosophy and poetry — who found his great result late and almost by accident. Prompted by de Broglie's matter waves and a pointed question from Debye about what equation they obeyed, he retreated to a villa at Arosa over Christmas 1925 and returned with a wave equation and its first triumph: the exact solution of hydrogen. His four 1926 papers, "Quantisierung als Eigenwertproblem," ran from January to June and recast quantization as a boundary-value problem — the visualizable alternative to Heisenberg's matrices, which he soon proved equivalent to his own. He shared the 1933 Nobel Prize with Dirac, left Germany in disgust at the Nazi regime, and settled at the Dublin Institute for Advanced Studies, where his 1944 What Is Life? turned a generation of physicists toward molecular biology. He is equally remembered for the 1935 thought experiment — the cat both alive and dead — devised not to celebrate quantum superposition but to protest how absurd he found it.

History

1913

— Bohr reproduces the hydrogen spectrum by decree, with quantized orbits but no mechanism for the quantization

1925

— matrix mechanics is built by Heisenberg, Born and Jordan; Uhlenbeck and Goudsmit propose electron spin

1926

— Pauli solves the hydrogen atom by matrix mechanics (submitted 17 January), days before Schrödinger's first wave-mechanics paper; Schrödinger's four papers run from January to June, the first solving hydrogen exactly; Born reads |ψ|² as a probability density

1928

— Dirac's relativistic equation adds spin and fine structure, splitting the n² degeneracy by j

1947

— Lamb and Retherford measure a splitting Dirac's theory forbids; explaining the Lamb shift launches quantum electrodynamics

Going further

Texts. Foot, Atomic Physics (2005), ch. 2 · Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., ch. 4 · Bransden & Joachain, Physics of Atoms and Molecules, ch. 3 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, vol. 1, ch. VII · Demtröder, Laser Spectroscopy, vol. 1, ch. 2.

Provenance. Equations are the standard Coulomb Schrödinger separation; a verifier should confirm $\mathrm{Ry} = \mu e^4/[2(4\pi\varepsilon_0)^2\hbar^2] = 13.6\,\text{eV}$ , $a_0 = 4\pi\varepsilon_0\hbar^2/(\mu e^2) = 0.529\,\text{Å}$ , $\alpha \approx 1/137$ , the degeneracy $\sum_{\ell=0}^{n-1}(2\ell+1) = n^2$ , and the recurrence/termination giving $n = n_r + \ell + 1$ — all standard textbook results, cross-checked against Griffiths (ch. 4), Bransden & Joachain (ch. 3), Foot (ch. 2), and Cohen-Tannoudji vol. 1 (ch. VII). The numerical $\ell$ -independence of $E_{n\ell}$ , the positronium $E_1 = -6.8$ eV, and $\langle r\rangle_{n\ell} = \tfrac{a_0}{2}[3n^2 - \ell(\ell+1)]$ are recomputed in verify_aq-optics-schrodinger-hydrogen-solution.py. The separation, effective potential, centrifugal barrier and associated-Laguerre termination are owned by, and recalled by name from, aq-qm-hydrogen-coulomb-radial-equation.

Prerequisites. The Bohr Model of Hydrogen · Quantum Numbers and Orbital Angular Momentum · The Hydrogen Spectrum and Spectral Series. Leads to. Hydrogenic Wavefunctions and Radial Distributions · Fine Structure and Spin-Orbit Coupling · The Lamb Shift · Multi-electron Atoms and the Central-Field Approximation.

If the energies are identical to Bohr's, what did the Schrödinger solution actually accomplish?

It replaced a lucky guess with a theory. Bohr postulated quantized circular orbits and got the right energies, but could not say what the atom looked like, could not compute a line's intensity, gave the ground state a unit of angular momentum it does not have, and failed for helium. The Schrödinger solution derives the same energies from one principle — normalizability — and hands you the full states $\psi_{n\ell m}$ , the correct $\ell = 0$ ground state, the $n^2$ degeneracy, and the dipole matrix elements behind selection rules and transition rates. The energies were the one thing Bohr got right; almost everything else needed the wave equation.

Why does energy depend on n alone, when rotational symmetry only seems to fix the m-degeneracy?

Because two different facts are stacked. The $m$ -independence is a theorem of rotational symmetry: any central potential commutes with all three components of $\mathbf L$ , forcing the $2\ell+1$ orientations of a given $\ell$ to share an energy. The $\ell$ -independence is not — a generic central potential splits $2s$ from $2p$ . It is special to the exact $1/r$ Coulomb law, which hides an extra conserved quantity (the quantum Runge–Lenz vector, enlarging the symmetry to SO(4)), and it is called "accidental" for exactly that reason. The microscopic proof is owned by the QM track; the spectroscopic point is that this accident is fragile — relativity, finite nuclear size, and electron screening all break it, which is why real multi-electron spectra look nothing like hydrogen's.

What sets the ranges ℓ ≤ n − 1 and |m| ≤ ℓ — are they separate rules?

They come from two places, bundled by the termination condition. The cap $|m| \leq \ell$ is angular — a property of the spherical harmonics $Y_\ell^m$ , the allowed projections of an angular momentum $\sqrt{\ell(\ell+1)}\,\hbar$ . The cap $\ell \leq n - 1$ is radial: normalizability forces the polynomial to terminate at degree $n_r = n - \ell - 1 \geq 0$ , so a larger $\ell$ eats into the radial nodes and eventually leaves none. The effective-potential picture makes it physical — the centrifugal barrier lifts the well bottom above the lower $E_n$ , so those states have no home. A " $1p$ " would need $n_r = -1$ : it does not exist.

Positronium halves the energies, but why does deuterium only shift them slightly?

It is all in the reduced mass $\mu = m_e M/(m_e + M)$ . For positronium the "nucleus" is a positron of mass $m_e$ , so $\mu = m_e/2$ — halving every energy. For deuterium the deuteron is about $3670$ electron masses, so $\mu$ differs from hydrogen's by only $\sim 2.7\times10^{-4}$ — yet enough to shift the Balmer lines by fractions of a nanometre, which is how deuterium was found. The effect scales as $m_e/M$ : enormous when the partner is light (positronium, muonium), a whisper for a heavy nucleus.

The solution uses the reduced mass, but the electron and proton are in three dimensions — how does the two-body problem become one-body?

Exactly as in classical mechanics. For two particles interacting through a potential depending only on their separation, change to the center-of-mass and relative coordinates $\mathbf r = \mathbf r_e - \mathbf r_p$ . The center of mass moves freely (uninteresting); the relative motion obeys a one-body Schrödinger equation with the reduced mass $\mu$ in place of $m_e$ . So equation (1) is not an approximation that "fixes the proton" — it is the exact relative-motion equation, and the only trace of the proton's finite mass is $m_e \to \mu$ . Fixing the proton would give $\mu = m_e$ ; the finite-mass correction restores the real $\mu$ .

Where do the intensities and selection rules come from, if not from the energies?

From the wavefunctions, which Bohr never had. Given $\psi_{n\ell m}$ , an electric-dipole transition rate is set by $\langle \psi_{n'\ell'm'}|\mathbf r|\psi_{n\ell m}\rangle$ ; its angular part vanishes unless $\Delta\ell = \pm 1$ and $\Delta m = 0, \pm 1$ — the dipole selection rules — and its radial part fixes the oscillator strength, hence the line's brightness and the level's lifetime. This is what the Schrödinger solution adds on top of the unchanged energies, developed further in the selection-rule and transition-rate articles. The energies tell you where the lines are; the wavefunctions tell you which appear and how strong.