Energy levels, scales and the continuum

Series truncation gave n = nᵣ + l + 1 and an energy; the useful form strips the SI clutter away. Write Eₙ = −½ μc²α²Z²/n², with α = e²/4πε₀ħc = 1/137.036. Everything on the right is either a rest energy or a pure number, so the arithmetic is one line: ½ × 510 999 eV × (1/137.036)² = ½ × 27.211 eV = 13.606 eV. That number, the Rydberg energy E_R, is the whole atomic energy scale, and the formula says where it comes from — a rest energy demoted by α²/2 ≈ 2.66 × 10⁻⁵. Two consequences follow at once. By the virial theorem ⟨T⟩ = −E₁ = 13.6 eV, so ⟨v²⟩^½/c = √(2 × 13.598/511 000) = 0.00730 ≈ α, and the non-relativistic p²/2μ we assumed is self-consistent rather than merely convenient. And the leading relativistic correction, one further factor of α² down, lands near 10⁻⁴ eV — invisible without an etalon, and the business of the next unit.

The energy scale fixes the length scale. Combining a₀ = ħ/(mₑ c α) with the reduced Compton wavelength ƛ_C = ħ/(mₑ c) = 386.16 fm gives a₀ = ƛ_C/α = 386.16 fm × 137.036 = 52 918 fm = 0.052918 nm, and E_R = ħ²/(2mₑ a₀²) closes the loop between them. Notice which mass sits where: the atomic length goes as 1/μ while Eₙ goes as μ, so the reduced mass moves length and energy in opposite directions and leaves the product E_R a₀ = ½ αħc = 0.7200 eV⋅nm alone — the product E_R a₀², by contrast, is ħ²/2μ and does depend on the mass. The n and Z scaling is what to carry, and it belongs to an expectation value rather than to an orbit: ⟨1/r⟩ = Z/(n²a₀) exactly, for every l, so n²a₀/Z is the scale 1/⟨1/r⟩, while the mean radius itself is l-dependent, ⟨r⟩ = (a₀/2Z)[3n² − l(l+1)], and only the circular state l = n − 1 has its radial density peaking at n²a₀/Z. The virial theorem turns the same scale into a speed, ⟨v²⟩^½/c = Zα/n. So He⁺ in n = 3 has 1/⟨1/r⟩ = 0.238 nm and E = −6.05 eV, while a hydrogen Rydberg state at n = 50 sits on a scale of 2500a₀ = 132 nm and binds by only 5.44 meV — a fifth of room-temperature kT, which is why such atoms survive only in cold, dilute beams.

The Coulomb problem is two-body, and reducing it to one body replaces mₑ by μ = mₑ M/(mₑ + M). For hydrogen M/mₑ = 1836.15, so μ/mₑ = 1836.15/1837.15 = 0.999456: every level is shallower by 5.44 parts in 10⁴, and E₁ is −13.598 eV, not the infinite-mass −13.6057 eV. That is not a rounding detail. Deuterium has μ/mₑ = 3670.48/3671.48 = 0.999728, larger by 2.72 × 10⁻⁴, so every deuterium line sits that fraction to the blue: Balmer-α at 656.46 nm in hydrogen appears at 656.28 nm in deuterium, a 0.179 nm doublet and the signature Urey used in 1931. Push μ further and the whole atom rescales. Positronium has μ = mₑ/2 exactly, so E₁ = −6.80 eV and its length scale is 2a₀. Muonic hydrogen has μ = 185.8 mₑ, giving E₁ = −2.53 keV and a scale of 285 fm — small enough that the proton's finite size shifts the levels measurably.

The levels do not run out; they crowd. E_{n+1} − Eₙ = E_R(1/n² − 1/(n+1)²) ≈ 2E_R/n³, so the 1→2 gap is 10.20 eV while the 50→51 gap is 0.211 meV, and the density of levels, |dn/dE| = n³/(2E_R), diverges as E → 0⁻. There are therefore infinitely many bound states, and they accumulate at threshold rather than stopping at some deepest-but-one. The reason is the range of the potential, not the depth of the well: −Ze²/4πε₀r falls off too slowly for the centrifugal term to win at large r, so there is always another shallow state further out. Replace the tail by a screened Yukawa form e^(−r/λ)/r and the count becomes finite — for l = 0 the last bound state vanishes once λ drops below about 0.84 a₀. That is why H⁻, whose departing electron sees a neutral atom and only a short-range polarisation tail, has exactly one bound state while neutral hydrogen has an endless series.

At E = 0 the character of the spectrum changes. For E > 0 every value is an eigenvalue: write E = ħ²k²/2μ with k continuous, and the radial solutions oscillate all the way out, so they are not square-integrable and cannot be normalised to one. They are normalised to a delta instead, ⟨E|E′⟩ = δ(E − E′), whose argument carries units of energy, so the kets carry energy^(−1/2) — the same move made for plane waves in the momentum basis. What is not the same is the phase. The 1/r tail never stops acting, so the asymptotic form is sin(kr − lπ/2 − η ln 2kr + σₗ), with the Sommerfeld parameter η = −Z/ka₀ supplying a logarithmic distortion no finite-range potential produces. Physically this is photoionisation. A 20.00 eV photon on ground-state hydrogen leaves E = −13.598 + 20.00 = +6.40 eV of kinetic energy, and because the eigenvalues are continuous, so is the absorption: the Lyman lines converge on 91.18 nm and are followed by a continuous edge, not by nothing.

The resolution of the identity on a fixed (l, m) subspace is Σₙ |n l m⟩⟨n l m| + ∫₀^∞ dE |E l m⟩⟨E l m| = Î — bound sum plus continuum integral — and the continuum half is not a small correction. Change the nuclear charge suddenly, as the beta decay of tritium does, and the overlap of the old ground state with the new bound states does not reach one; the missing weight is real ionisation, called shake-off. A numerical diagonalisation on a grid of radius R never sees the integral. It returns discrete positive eigenvalues instead, spaced as the box modes kⱼ ≈ jπ/R allow. Those pseudo-states are not states of the atom, and the test is to double R: a true bound level moves by a fraction of a per cent, a pseudo-state moves a long way and multiplies in number. They still earn their place, though — as a quadrature rule for the continuum integral they restore the sum rule Σ|c|² = 1 that the bound states alone can never reach.