University Physics V · The Hydrogen Atom · 11.1

The Central-Field Hamiltonian & Its Commuting Set

Every hydrogen result you will meet rests on three moves made before the algebra starts: strip off the centre of mass, name the operators you intend to diagonalise together, and decide which functions the radial operator is allowed to act on. This lesson makes all three explicit.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A central field is any V that depends on |r| alone, and that one restriction does most of the work in atomic physics. Two bodies come first: the Coulomb Hamiltonian separates exactly into a free plane wave for the centre of mass and one fictitious particle of reduced mass μ = m₁m₂/(m₁+m₂) moving in a fixed V(r), which for hydrogen means μ = 0.999456 mₑ and deuterium lines 0.18 nm to the blue. Rotational invariance then gives [H, L²] = [H, L_z] = 0, so (H, L², L_z) is a commuting set with a shared eigenbasis, and S_z appended on the spin factor completes the labels n, l, mₗ, mₛ.

Separation of variables is that eigenbasis written in the position representation — the spectral theorem guarantees it before any Laguerre polynomial appears. The cost is paid at the origin. Replacing L² by ħ²l(l+1) leaves an operator on the half-line 0 < r < ∞, and a half-line has a boundary: self-adjointness, not formal Hermiticity, requires a domain, and for l = 0 the condition u(0) = 0 is genuine extra input.

Drop it and the free particle acquires a normalisable eigenfunction at every negative energy. The same defect destroys radial momentum outright — pᵣ is symmetric on that domain but its deficiency indices are unequal, so it has no self-adjoint extension and is not an observable at all.

Simple definition: A central-field Hamiltonian is H = p²/2μ + V(r) with V depending only on the distance |r|; because it commutes with L² and L_z, its eigenvectors carry the labels |n l m⟩ and the three-dimensional eigenproblem splits into one half-line radial problem for each l.
Example: For hydrogen μ = mₑmₚ/(mₑ + mₚ) = 0.999456 mₑ, so the n = 1 level sits at −13.5983 eV rather than the infinite-nucleus −13.6057 eV, and the shared eigenbasis is |n l mₗ⟩ ⊗ |mₛ⟩ with l ≤ n − 1.

Centre-of-mass separationH = P²/2M + p²/2μ + V(r), 1/μ = 1/m₁ + 1/m₂

Removes three degrees of freedom with no approximation: the centre of mass is a free plane wave, so the whole spectrum lives in the relative operator.

M = m₁+m₂; R = (m₁r₁+m₂r₂)/M; r = r₁−r₂; μ in kg. Exact for any V(|r₁−r₂|).

Rotational invariance fixes the commuting set[H, L²] = [H, L_z] = [L², L_z] = 0 for every V(r)

(H, L², L_z) is a complete commuting set on the orbital factor, so a shared eigenbasis |E l m⟩ exists: separation of variables is a theorem.

L = r × p, in J s. Needs only that V depends on |r|, never that it goes as 1/r.

Radial–angular split of the kinetic termp² = −ħ²(1/r)∂²ᵣ r + L²/r²

Substituting the L² eigenvalue turns one three-dimensional partial differential equation into one ordinary equation per l. That is all separation does.

Operator identity on L²(ℝ³); L² acts on angles only, with eigenvalue ħ²l(l+1) in J² s².

Reduced radial equation−(ħ²/2μ)u″ + [V(r) + ħ²l(l+1)/2μr²] u = E u, u = rR

Maps the half-line onto a one-dimensional problem with a wall at r = 0, so every 1-D bound-state intuition transfers intact.

u normalised by ∫₀^∞|u|²dr = 1, so u is in m^(−1/2); the bracket is V_eff(r), in J.

The domain that makes Hₗ self-adjointu(0) = 0, ⟨u|Hₗ v⟩ − ⟨Hₗ u|v⟩ = −(ħ²/2μ)[u*v′ − u*′v]₀^∞

Without it H₀ is only symmetric: e^(−r/a) becomes a normalisable eigenstate at E = −ħ²/2μa² for every a, and the spectrum loses its floor.

Automatic for l ≥ 1 (limit point at r = 0); for l = 0 it selects one of a one-parameter family.

Radial momentum is not an observablepᵣ = −iħ(∂ᵣ + 1/r), [r, pᵣ] = iħ, indices (1, 0)

Unequal deficiency indices mean no self-adjoint extension exists, so pᵣ has no spectral measure — though pᵣ†pᵣ/2μ is a perfectly good kinetic energy.

In the u = rR picture pᵣ = −iħ d/dr; of the pair e^(∓r/a) only e^(−r/a) is square-integrable.

Two bodies collapse to one, exactly

Two masses interacting through V(|r₁ − r₂|) have six coordinates. Change to R = (m₁r₁ + m₂r₂)/M and r = r₁ − r₂ with M = m₁ + m₂, and the kinetic term separates without approximation: H = P²/2M + p²/2μ + V(r), with 1/μ = 1/m₁ + 1/m₂. Nothing couples R to r, so Ψ = e^(iK⋅R)ψ(r) and the centre of mass is a free plane wave carrying no structure. Every energy level belongs to the relative operator p²/2μ + V(r), a genuine one-body problem in three dimensions. The price is that μ, not mₑ, sets the scale. For hydrogen mₚ/mₑ = 1836.153, so μ = 0.9994557 mₑ and every level is 0.0544% shallower than the infinite-nucleus value: −13.5983 eV rather than −13.6057 eV. For deuterium μ = 0.9997276 mₑ, and the ratio of the two reduced masses moves Balmer-α from 656.47 nm to 656.29 nm — the 0.18 nm gap that revealed deuterium in 1931. Positronium, with μ = mₑ/2, halves the entire spectrum.

Why [H, L²] and [H, L_z] vanish for any V(r)

Take L_z = xp_y − ypₓ and a potential depending only on r = |r|. Then [L_z, V] = x[p_y, V] − y[pₓ, V] = −iħ(x ∂V/∂y − y ∂V/∂x), and since ∂V/∂y = V′(r)⋅y/r and ∂V/∂x = V′(r)⋅x/r, the two terms are V′(r)xy/r and V′(r)yx/r, which cancel exactly. The kinetic term goes the same way: [L_z, pₓ] = iħp_y and [L_z, p_y] = −iħpₓ, so [L_z, p²] = iħ(p_ypₓ + pₓₚ_y) − iħ(pₓₚ_y + p_ypₓ) = 0. Nothing singled out z, so [H, Lᵢ] = 0 for all three components and therefore [H, L²] = 0 too. Two consequences follow. First, L is conserved — the quantum form of the classical central-force result. Second, because L_± = Lₓ ± iL_y commute with H, every state in a 2l + 1 multiplet shares one energy: the m-degeneracy is forced by rotational symmetry alone, in any central field. What is not forced is the equality of different l at the same n. That extra Coulomb degeneracy needs a separate symmetry, and it is the first thing screening destroys.

Separation of variables is the spectral theorem

Commuting self-adjoint operators can be diagonalised together; that is the spectral theorem, and it is why ψ = R(r)Yₗ^m(θ, φ) works. The Hilbert space of one spinless particle factorises as L²(ℝ³) ≅ L²((0,∞), r²dr) ⊗ L²(S²), and L²(S²) breaks into rotation-irreducible blocks of dimension 2l + 1 spanned by the Yₗ^m. Because [H, L²] = [H, L_z] = 0, H is block-diagonal in that decomposition: it acts as one radial operator Hₗ on each block, identical for all 2l + 1 values of m. Writing a product function is therefore not an ansatz that happens to succeed; it is the shared eigenbasis expressed in the position representation. Two cautions. The set (H, L², L_z) is complete on the orbital factor — one state per (E, l, m) triple — but a general state is a superposition of products, not a product. And once the electron's spin factor C² is carried along, the space is L²(ℝ³) ⊗ C² and the Schrödinger H acts as H ⊗ I, which cannot fix mₛ at all; S_z must be appended by hand, doubling every hydrogen level to 2n² states.

The half-line has a boundary, so Hₗ needs a domain

Substituting u = rR turns ∫|R|²r²dr into ∫₀^∞|u|²dr and gives −(ħ²/2μ)u″ + V_eff u = Eu — a one-dimensional problem, but on 0 < r < ∞ rather than the whole line, and a half-line has an endpoint. Integrating by parts twice, ⟨u|Hₗ v⟩ − ⟨Hₗ u|v⟩ = −(ħ²/2μ)[u*v′ − u*′v]₀^∞; the piece at infinity dies for normalisable states, and the piece at r = 0 does not die by itself. Requiring it to vanish for every v in the domain is what makes Hₗ self-adjoint rather than merely symmetric, and u(0) = 0 is the condition that does it. For l ≥ 1 the centrifugal term ħ²l(l+1)/2μr² forces the behaviour anyway: the operator is in the limit-point case at the origin, essentially self-adjoint, with the Frobenius roots r^(l+1) and r^(−l) settling the matter. For l = 0 there is a genuine choice — a one-parameter family of self-adjoint extensions exists, and u(0) = 0 selects the Friedrichs member. Physically that is the demand that R = u/r carry no δ function, since ∇²(1/r) = −4πδ³(r) would hide a point source at the nucleus.

Radial momentum: Hermitian, canonical, not an observable

In one dimension p = −iħ d/dx is conjugate to x and a perfectly good observable; on the half-line the analogue fails. The symmetrised radial momentum pᵣ = ½(r̂⋅p + p⋅r̂) = −iħ(∂ᵣ + 1/r) obeys [r, pᵣ] = iħ, and in the u = rR picture it is just −iħ d/dr acting on functions with u(0) = 0, where the boundary term u*v at r = 0 vanishes — so it is symmetric. But symmetric is not self-adjoint. Solve pᵣ†χ = ±i(ħ/a)χ in that picture and χ ∝ e^(∓r/a); only e^(−r/a) is square-integrable on (0, ∞), so the deficiency indices are (1, 0). They are unequal, and von Neumann's theorem then says no self-adjoint extension exists — not several to choose between, none at all. So pᵣ has no real spectrum, no eigenbasis, no projection-valued measure and nothing to measure: a canonical commutator and a formal integration by parts do not make an observable, the domain does. What still works is pᵣ†pᵣ = −ħ²d²/dr² on u(0) = 0, positive and self-adjoint, so radial kinetic energy is fine even though radial momentum is not.

What the commuting set assumes, and what breaks it

Every step above is an assumption you can name. The kinetic term is p²/2μ, so the treatment is non-relativistic; the leading correction is of relative order α² ≈ 5 × 10⁻⁵, the fine structure a good etalon resolves in Balmer-α. The nucleus is a point, which is why the proton's finite radius moves hydrogen levels by parts in 10⁹ but moves muonic hydrogen enormously, μ there being about 186 mₑ. Spin is imported rather than derived: it arrives as a tensor factor because the Schrödinger equation has no place for it. And the commuting set is only as stable as the Hamiltonian. Add the spin-orbit term ξ(r)L⋅S and L_z and S_z stop commuting with H, since L⋅S trades mₗ against mₛ; the surviving set is (H, L², S², J², J_z) and the good labels become n, l, j, mⱼ. Add a uniform field along z and only rotations about z survive, so L_z stays good while L² does not once the field is strong. Naming the commuting set is not decoration — it is the statement of which quantum numbers a given experiment is still entitled to use.

Change one variable at a time

Make the relationship visible.

Interactive model

Decay length a of e^(−r/a)1.0 a₀

Wavenumber k of sin(kr)1.5 1/a₀

Drag a down to 0.3 a₀: the dashed state's energy dives past −150 eV while its norm stays finite at 0.15 a₀, so square-integrability never rejects it. Only u(0) = 0 does — and the solid curve meets the wall at zero for every k you choose.

SPURIOUS E −ħ²/2μa²-13.6 eV

NORM ∫|uₐ|² dr0.50 a₀

ALLOWED E +ħ²k²/2μ30.6 eV

FIRST NODE π/k2.09 a₀

Live interpretationSPURIOUS E −ħ²/2μa²: −13.6 eV. NORM ∫|uₐ|² dr: 0.50 a₀. ALLOWED E +ħ²k²/2μ: 30.6 eV. FIRST NODE π/k: 2.09 a₀

Catch the common trap

Explain before calculating.

Misconception check

The u(0) = 0 condition is usually explained as "otherwise the wavefunction blows up at the origin". That is not the reason. For l = 0 the rejected solution R = e^(−r/a)/r is perfectly square-integrable: ∫₀^∞|R|²r²dr = ∫₀^∞ e^(−2r/a) dr = a/2, finite for every a, so normalisability does not exclude it. What excludes it is the domain. Without u(0) = 0 the radial operator is symmetric but not self-adjoint, and it then owns a normalisable eigenfunction at E = −ħ²/2μa² for every a > 0 — a free particle with no lowest state. Equivalently, ∇²(1/r) = −4πδ³(r), so a nonzero u(0) means the Schrödinger equation is violated at the origin by a hidden point source. The general lesson: "Hermitian", as usually written, is a claim about an integration by parts and says nothing about the boundary; self-adjointness is a claim about a domain, and the boundary is what decides it.

The l = 0 radial equation for a free particle, −(ħ²/2μ)u″ = Eu on 0 < r < ∞, is solved by u(r) = e^(−r/a) at E = −ħ²/2μa², and ∫₀^∞|u|²dr = a/2 is finite for every a > 0. Why does a free particle nevertheless not possess this infinite tower of bound states?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTake mₚ/mₑ = 1836.153 and m_d/mₑ = 3670.483. Find the reduced mass of hydrogen and of deuterium in units of mₑ, give both ground-state energies from E₁ = −13.6057 eV × (μ/mₑ), and find the wavelength gap between the two Balmer-α lines given that hydrogen's sits at 656.47 nm.

1/μ = 1/mₑ + 1/M gives μ = mₑ/(1 + mₑ/M), so the reduced mass always sits just below mₑ and the correction is set by the mass ratio alone.
Hydrogen: μ_H/mₑ = 1836.153/1837.153 = 0.9994557. Deuterium: μ_D/mₑ = 3670.483/3671.483 = 0.9997276.
Energies scale linearly with μ: E₁(H) = −13.6057 × 0.9994557 = −13.5983 eV and E₁(D) = −13.6057 × 0.9997276 = −13.6020 eV, a gap of 3.7 meV.
Every transition energy carries the same factor, so λ ∝ 1/μ and λ_D/λ_H = μ_H/μ_D = 0.9994557/0.9997276 = 0.9997280.
λ_D = 656.47 nm × 0.9997280 = 656.29 nm, so the deuterium line lies 0.18 nm to the blue of hydrogen's.

Answerμ_H = 0.9994557 mₑ and μ_D = 0.9997276 mₑ; E₁ = −13.5983 eV and −13.6020 eV, 3.7 meV apart; D-α at 656.29 nm sits 0.18 nm blue of H-α at 656.47 nm — the shift that revealed deuterium.

MediumShow explicitly that [L_z, V(r)] = 0 and [L_z, p²] = 0 for any V depending only on r, then count the states of the hydrogen n = 3 level and say which part of that count rotational symmetry actually explains.

Potential: [L_z, V] = [xp_y − ypₓ, V] = x[p_y, V] − y[pₓ, V] = −iħ(x ∂V/∂y − y ∂V/∂x).
With V a function of r alone, ∂V/∂y = V′(r)⋅y/r and ∂V/∂x = V′(r)⋅x/r, so the bracket is −iħV′(r)(xy/r − yx/r) = 0. This used only that V depends on |r|, never that it goes as 1/r.
Kinetic term: [L_z, pₓ] = [x, pₓ]p_y = iħp_y and [L_z, p_y] = −[y, p_y]pₓ = −iħpₓ, with [L_z, p_z] = 0. Hence [L_z, p²] = iħ(p_ypₓ + pₓₚ_y) − iħ(pₓₚ_y + p_ypₓ) = 0.
So [H, L_z] = 0, and since z was not special, [H, Lₓ] = [H, L_y] = 0 and therefore [H, L²] = 0. Because L_± commute with H, the 2l + 1 states of each multiplet are degenerate.
n = 3 allows l = 0, 1, 2, giving 1 + 3 + 5 = 9 orbital states; the spin factor doubles this to 18 = 2n². Rotational symmetry explains the 1-, 3- and 5-fold blocks; the spin doubling is separate, because the Schrödinger Hamiltonian never touches spin.
What rotation does not explain is why l = 0, 1 and 2 share one energy. That equality holds only for the 1/r potential — screening in any other central field splits it while leaving every 2l + 1 block intact.

Answer[L_z, V(r)] = [L_z, p²] = 0 for every central V, so [H, L²] = [H, L_z] = 0. The n = 3 level holds 9 orbital states and 18 with spin; rotation explains the 1, 3 and 5 blocks, not their equality.

HardFor the l = 0 free radial operator H₀ = −(ħ²/2μ)d²/dr² on 0 < r < ∞: (a) find the boundary term in ⟨u|H₀v⟩ − ⟨H₀u|v⟩; (b) show that dropping u(0) = 0 admits uₐ = e^(−r/a), and evaluate its energy for a = a₀ = 0.0529 nm and for a = a₀/10, taking μ = mₑ; (c) say what the same defect does to pᵣ.

Integrating by parts twice, ⟨u|H₀v⟩ − ⟨H₀u|v⟩ = −(ħ²/2μ)[u*v′ − u*′v]₀^∞. Normalisable states kill the upper limit, leaving +(ħ²/2μ)(u*(0)v′(0) − u*′(0)v(0)), which vanishes for every v in the domain only if u(0) = 0.
Drop that condition: uₐ = e^(−r/a) satisfies uₐ″ = uₐ/a², so H₀uₐ = −(ħ²/2μa²)uₐ, and ∫₀^∞|uₐ|²dr = a/2 is finite. It is a genuinely normalisable eigenfunction at negative energy.
Numbers: ħ²/2mₑa₀² is exactly one Rydberg, 13.6057 eV. So a = a₀ gives E = −13.61 eV — a free particle sitting at hydrogen's ground-state energy — and a = a₀/10 gives E = −13.6057 × 100 = −1360.6 eV. As a → 0, E → −∞.
The spectrum is therefore unbounded below and labelled by a continuous parameter, with no lowest state: H₀ without the boundary condition is symmetric but not self-adjoint. Its deficiency indices are (1, 1), so extensions do exist, and u(0) = 0 picks the Friedrichs one, restoring the physical spectrum E ≥ 0 with generalised eigenfunctions sin(kr).
For radial momentum the algebra is the same and the outcome is worse. In the u picture pᵣ = −iħ d/dr, and pᵣ†χ = ±i(ħ/a)χ gives χ ∝ e^(∓r/a), of which only e^(−r/a) is square-integrable on (0, ∞). The indices are (1, 0): unequal, so no self-adjoint extension exists at all and pᵣ is not an observable, while pᵣ†pᵣ/2μ still is.

AnswerThe boundary term vanishes only on u(0) = 0. Without it e^(−r/a) is a normalisable eigenstate at −13.61 eV for a = a₀ and −1360.6 eV for a₀/10, so H₀ has no floor; pᵣ fares worse, with indices (1, 0) and no self-adjoint extension at all.