University Physics IV · The Hydrogen Atom · 10.1
Central Potentials & the 3D Schrödinger Equation
Before hydrogen can have an energy level, it has to become a solvable problem. Two moving bodies collapse to one fictitious particle of reduced mass; three coordinates collapse to one ordinary differential equation plus a table of angular functions you solve once and reuse for the rest of the course. This lesson is that reduction.
Build the model
Connect the measurement to the mechanism.
An atom is two bodies, and a two-body wavefunction lives on six coordinates. Two exact moves cut that down. Because the interaction depends only on r₁ − r₂, the Hamiltonian splits cleanly into a free centre-of-mass piece and a relative piece in which a single fictitious particle of reduced mass μ = m₁m₂/(m₁ + m₂) moves in V(r); the atom's entire internal spectrum belongs to that second piece, and nothing has been approximated to get there.
Then, because V depends on the length of r and not its direction, the angular derivatives in the Laplacian appear only inside L², divided by exactly r² — so multiplying through by r² quarantines them, ψ = R(r)Y(θ, φ) goes through, and the angular half of the problem is solved once for every central potential there will ever be. The cost is worth stating up front. The reduction is exact for two bodies and dies at three, where a mass-polarisation term survives and helium never separates.
Any potential with a direction in it — an applied field, a second electron, a deformed nucleus — destroys the factorisation immediately. And the familiar −e²/4πε₀r is a further idealisation stacked on top, one the separation never asked for: what it buys is not the spherical harmonics but the 1/n² spectrum and the degeneracy in l.
- Simple definition
- A central potential depends only on the distance r between two particles, which lets the two-body problem reduce exactly to one particle of reduced mass μ moving in V(r), with a wavefunction that factors as ψ = R(r)Yₗₘ(θ, φ).
- Example
- For hydrogen μ = mₑ mₚ/(mₑ + mₚ) = mₑ/(1 + 1/1836.15) = 0.9994557 mₑ, which lifts the Rydberg energy from the infinite-mass 13.6057 eV to 13.5983 eV — a 7.4 meV shift that moves Lyman-α by 0.066 nm.
Hydrogen: μ = 0.9994557 mₑ. Positronium: μ = mₑ/2 exactly.
m₁, m₂ and μ in kg, or all as multiples of mₑ; μ is always smaller than either mass.
The R term is a free particle. Every atomic level lives in the r term.
M = m₁ + m₂ in kg; R is the centre of mass and r = r₁ − r₂, both in m.
Every angular derivative sits inside L², divided by exactly r². That is the whole trick.
L² acts on θ and φ only and carries units of ℏ², that is J² s²; r in m.
Two functions of independent variables can only agree by both being one constant.
The left side depends on r only; V and E in J; l(l + 1) is dimensionless.
Solved once. The same Yₗₘ serve every central potential, hydrogen included.
l = 0, 1, 2, … from regularity at the poles; m = −l … +l from 2π-periodicity in φ.
Hydrogen: 0.054% shallower. Muonic hydrogen: 10.1%. Positronium: 50%.
Measured against the M → ∞ values; the fractional shift is exactly 1/(1 + M/m).
Six coordinates, and the move that removes three
An isolated atom is two particles, so its wavefunction lives on six coordinates: Ψ(r₁, r₂), with H = −(ℏ²/2m₁)∇₁² − (ℏ²/2m₂)∇₂² + V(r₁ − r₂). Nothing separates in these variables, because V couples them. Change to the centre of mass R = (m₁r₁ + m₂r₂)/(m₁ + m₂) and the separation r = r₁ − r₂. The kinetic energy — the one piece of algebra worth doing by hand once — becomes −(ℏ²/2M)∇²R − (ℏ²/2μ)∇²ᵣ, with M = m₁ + m₂ and 1/μ = 1/m₁ + 1/m₂, and the cross terms cancel identically. Now Ψ = e(iK⋅R)ψ(r) works: the centre of mass is a free particle carrying kinetic energy ℏ²K²/2M, and every level, every orbital and every spectral line of the atom belongs to the second factor. The only thing used was that V depends on the difference of the positions, which is translation invariance — so this reduction is exact, not a model.
Reduced mass is what an exact reduction leaves behind
μ = m₁m₂/(m₁ + m₂) is smaller than either mass and is dominated by the lighter one: for hydrogen μ = mₑ × 1836.15/1837.15 = 0.9994557 mₑ. Let the nucleus grow without limit and μ → mₑ, which is the clamped-nucleus picture — so nailing the proton down is the approximation, and μ is the exact answer, not the other way round. The correction is small but never invisible. Every level of a hydrogenic system carries one factor of μ and every length one factor of 1/μ, so hydrogen's Rydberg energy is 13.5983 eV against the infinite-mass 13.6057 eV, and Lyman-α sits at 121.568 nm rather than 121.502 nm. Deuterium, with a nucleus twice as heavy, has μ = 0.9997276 mₑ, and the 0.18 nm gap between the two Balmer-α lines near 656 nm is how Urey identified deuterium in 1932 — a new isotope found in a mass ratio. Positronium, where the two masses are equal, gives μ = mₑ/2 exactly: half the binding, twice the radius.
Why r-dependence alone lets the angles leave
In spherical coordinates ∇² = (1/r²)∂ᵣ(r²∂ᵣ) + (1/r²sinθ)∂θ(sinθ ∂θ) + (1/r²sin²θ)∂²φ. The three angular terms are not a mess to be endured: together they are exactly −L²/ℏ², an operator built only from θ and φ, and they sit divided by exactly r². So H = −(ℏ²/2μ)(1/r²)∂ᵣ(r²∂ᵣ) + L²/(2μr²) + V(r). If V depends on r alone, then L² commutes with every term — it cannot see V(r), and the radial derivatives cannot see θ or φ. Put ψ = R(r)Y(θ, φ), divide by RY and multiply by −2μr²/ℏ². The r² cancels the 1/r² carried by the angular block, and what is left is a function of r alone on one side and a function of θ and φ alone on the other. Two such functions can be equal only if both are the same constant, and that constant is written l(l + 1). That step, and only that step, is where separability comes from.
One boundary condition per separation constant
Separation manufactures constants; boundary conditions choose their values, and each of the three comes from a different demand. Take φ first: d²Φ/dφ² = −m²Φ gives Φ = e(imφ), and going once round the axis must return the same wavefunction, Φ(φ + 2π) = Φ(φ), so m is an integer. The θ equation then has solutions that diverge at cosθ = ±1 unless its series terminates; demanding a finite ψ on the polar axis forces l to be a non-negative integer with l ≥ |m|, so each l carries 2l + 1 values of m. Only the third constant, E, comes from the radial equation, where R must stay finite at the origin and normalisable at infinity. Notice what that ordering means: two of the three quantum numbers are fixed by geometry before the potential has even been named, and they come out the same for a nuclear well as for hydrogen. E is the only one that ever knows what V is.
The angular half is universal; only the radial half knows V
The saving is large. Yₗₘ(θ, φ) is the answer for every central potential there is: the three-dimensional isotropic oscillator, the finite spherical well, the Woods-Saxon potential of a nucleus, the Yukawa and Thomas-Fermi screened forms, the confinement of a spherical quantum dot, and the Coulomb 1/r. That is why the s, p, d labels and the 2l + 1 counting carry across from atoms to nuclei to nanocrystals with no rederivation. Rotational symmetry also hands you a degeneracy for free: because H contains no preferred direction, the 2l + 1 states of a given l share one energy in any V(r), which is exactly why something with a direction in it — an applied field — is what splits them. What the potential controls is the radial equation, and through it E(n, l) and the node count. The extra l-degeneracy of hydrogen is not generic: move the exponent off −1 and states of different l separate at once, while every Yₗₘ stays untouched.
Where the reduction stops
Three limits are worth naming in advance. First, two bodies only. Remove the centre of mass from a three-body atom and you are left not merely with reduced masses but with a mass-polarisation term, −(ℏ²/M)∇₁⋅∇₂, coupling the electrons through the recoiling nucleus; helium has no exact one-particle reduction, and that is the first reason it needs approximation methods. Second, central means central. An external electric or magnetic field, the r₁₂ repulsion between two electrons, a quadrupole-deformed nucleus — each puts a direction into V and kills the factorisation on the spot. Third, spherical symmetry is a far weaker requirement than a point charge. A proton of radius 0.84 fm gives a potential that is not −e²/4πε₀r inside the nucleus, yet it is still a function of r alone, so ψ = R Yₗₘ survives intact and only the radial equation changes. Keep the two ideas apart: separation is a statement about symmetry, and 1/r is a statement about the source.
Change one variable at a time
Make the relationship visible.
Set the ratio to 0 and the two bodies are twins: equal orbits, μ exactly m/2, positronium. Drag right to 3.3 and the heavy body's orbit collapses to 0.0005 of the separation while μ climbs to 0.9995 m — the clamped nucleus is the limit M → ∞, not an assumption you get for free.
MASS RATIO M/m8
REDUCED MASS μ/m0.8882
HEAVY-BODY ORBIT0.1118 of r
SHIFT FROM INFINITE MASS11.182 %
Live interpretationMASS RATIO M/m: 8. REDUCED MASS μ/m: 0.8882. HEAVY-BODY ORBIT: 0.1118 of r. SHIFT FROM INFINITE MASS: 11.182 %
Catch the common trap
Explain before calculating.
An electron is bound to a nucleus by the screened potential V(r) = −(A/r)e(−r/d), with A and d fixed constants. What happens when you try to separate its three-dimensional Schrödinger equation?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTake mₚ/mₑ = 1836.15 and the infinite-nuclear-mass Rydberg energy as 13.6057 eV. Find the reduced mass of hydrogen as a multiple of mₑ, then the ground-state energy and the size of the correction. Finally quote that correction as a wavelength shift on the Lyman-α line, whose infinite-mass value is 121.502 nm.
- 1/μ = 1/mₑ + 1/mₚ, so μ = mₑ mₚ/(mₑ + mₚ) = mₑ × 1836.15/1837.15.
- μ/mₑ = 1 − 1/1837.15 = 0.9994557. The reduced mass sits below the electron mass, by 1 part in 1837.
- Every hydrogenic energy carries one factor of μ, so E₁ = −13.6057 eV × 0.9994557 = −13.5983 eV — the level is 7.4 meV shallower than the infinite-mass value.
- Wavelengths carry one factor of 1/μ: λ = 121.502 nm ÷ 0.9994557 = 121.568 nm, a shift of 0.066 nm. A 0.01 nm spectrograph resolves that several times over, so the reduced mass is not optional bookkeeping.
Answerμ = 0.999456 mₑ, E₁ = −13.5983 eV, which is 7.4 meV above the infinite-mass value, and Lyman-α moves from 121.502 nm to 121.568 nm.
MediumA deuterium nucleus has mass 3670.48 mₑ. Hydrogen's Balmer-α line (n = 3 → 2) lies at 656.28 nm. Predict deuterium's Balmer-α wavelength and the splitting between the two lines, and state the resolving power a spectrograph needs to separate them.
- Reduced masses: μH/mₑ = 1836.15/1837.15 = 0.9994557 and μD/mₑ = 3670.48/3671.48 = 0.9997276.
- Transition energies carry one factor of μ and wavelengths one factor of 1/μ, so λD/λH = μH/μD = 0.9994557/0.9997276 = 0.9997280.
- λD = 656.28 nm × 0.9997280 = 656.28 nm − 0.179 nm = 656.10 nm. The heavier nucleus recoils less, so it binds more tightly and emits at the shorter wavelength.
- Δλ = 0.18 nm, so the resolving power needed is λ/Δλ = 656.28/0.179 = 3.7 × 10³ — modest for a grating, which is why Urey could find deuterium this way in 1932.
AnswerλD = 656.10 nm, a splitting of 0.18 nm with deuterium on the short-wavelength side; a resolving power of about 3.7 × 10³ separates the pair.
HardMuonic hydrogen replaces the electron with a muon, mμ = 206.77 mₑ, bound to a proton of 1836.15 mₑ. Find the reduced mass, the ground-state energy and the Bohr radius of the system, taking a₀ = 52918 fm and the Rydberg energy as 13.6057 eV. Then repeat with mμ used unreduced in place of mₑ, and say how large the difference is and why it dwarfs the one in ordinary hydrogen.
- μ = mμ mₚ/(mμ + mₚ) = mₑ × (206.77 × 1836.15)/(206.77 + 1836.15) = mₑ × 379658/2042.92 = 185.84 mₑ.
- Energies scale as μ: E₁ = −13.6057 eV × 185.84 = −2528 eV = −2.53 keV.
- Lengths scale as 1/μ: a = a₀ mₑ/μ = 52918 fm ÷ 185.84 = 285 fm, so the muon orbits some 186 times closer in than an electron would.
- Now skip the reduction and put 206.77 mₑ straight in: E₁ = −13.6057 eV × 206.77 = −2813 eV and a = 52918 ÷ 206.77 = 256 fm.
- The unreduced answers are 11.3% too deep and 10.1% too small. The fractional correction is 1/(1 + mₚ/mμ) = 1/9.88 = 10.1%, against 1/1837 = 0.054% in hydrogen — 186 times larger, because the muon is 1/8.88 of the proton mass rather than 1/1836, so the proton's recoil is no longer a footnote.
Answerμ = 185.84 mₑ, E₁ = −2.53 keV, a = 285 fm. Using the muon mass unreduced gives −2.81 keV and 256 fm: 11% too deep and 10% too small, because the recoil correction here is 1/9.9 rather than 1/1837.