Rotational spectra and the rigid rotor

Born-Oppenheimer lets the slow nuclei move on a potential U(R) set by the fast electrons, and the rigid-rotor approximation then holds the bond at one length, so R enters as a fixed parameter rather than a dynamical variable. The two-body rotation then separates the way every two-body problem does: the centre of mass drifts free, and the internal motion is one fictitious particle of reduced mass μ = m₁m₂/(m₁+m₂) moving on a sphere of radius R. Every detail of how the mass is distributed collapses into a single number, I = μR². For ¹²C¹⁶O, μ = 12 × 15.995/27.995 = 6.856 u = 1.1385 × 10⁻²⁶ kg, and with R = 113.1 pm, I = 1.456 × 10⁻⁴⁶ kg m². Watch how μ behaves when the masses are lopsided: for H³⁵Cl it is 0.9796 u, under 3% below the hydrogen mass alone, because the chlorine barely moves. The light atom does almost all the swinging and the heavy one sits close to the centre of mass — which is also why isotopic substitution at hydrogen shifts a rotational spectrum enormously and substitution at chlorine barely shifts it at all.

Fixing R kills the radial equation and leaves the angular equation you already solved for hydrogen, so the eigenfunctions are the spherical harmonics Y_JM and the eigenvalues follow from L² → J(J+1)ħ²: E_J = J(J+1)ħ²/2I, with 2J+1 values of M_J sharing each energy. Write it as E_J = hcB̃⋅J(J+1). The rungs sit at 0, 2B̃, 6B̃, 12B̃, 20B̃ — visibly spreading apart. But the gaps between neighbouring rungs are 2B̃, 4B̃, 6B̃, 8B̃, and each one exceeds the last by the same 2B̃. Since a photon can only move you one rung, the lines land at ν̃ = 2B̃(J+1) and come out evenly spaced. For CO with B̃ = 1.9225 cm⁻¹ the levels are at 0, 3.845, 11.535 and 23.07 cm⁻¹, while the first three lines are at 3.845, 7.690 and 11.535 cm⁻¹. Levels and lines are two different lists that happen to share their first nonzero entry, and reading one as the other is the classic way to get I wrong by a factor of two.

An electric-dipole transition rate goes as |⟨J′M′|μₑₗ⋅ε̂|JM⟩|². For a diatomic the dipole lies along the bond, so for light polarized along z the operator is μₑₗ cos θ, and cos θ is proportional to Y₁₀; the other two polarizations bring in its M = ±1 partners. An integral over three spherical harmonics survives only when the angular momenta can couple and the parity changes — here ΔJ = ±1 and ΔM_J = 0, ±1, which is the photon's one unit of angular momentum being absorbed by the rotor. Two consequences follow. There are no ΔJ = 2 lines in absorption, so the comb has no extra or missing teeth. And the whole integral carries μₑₗ as a prefactor, so a molecule with no permanent dipole gives a spectrum of exactly zero intensity, not a weak one. This is why CO, present at a mole fraction near 10⁻⁴ in a molecular cloud, is the tracer radio astronomers actually use, while the H₂ that makes up the cloud is invisible in the microwave.

The lines are not all the same height, and two competing factors decide them. The population of level J carries the degeneracy 2J+1, rising linearly, times the Boltzmann factor exp(−hcB̃J(J+1)/kT), falling faster than any power. Their product peaks; setting its derivative to zero gives 2J+1 = √(2kT/hcB̃), so Jₘₐₓ = √(kT/2hcB̃) − ½. Since hc/k = 1.4388 cm K, that is Jₘₐₓ = √(T/2.8776B̃) − ½ with B̃ in cm⁻¹. For H³⁵Cl at 300 K it returns 2.66, so check the integers either side: level 2 has weight 5e^(−0.300) = 3.70 and level 3 has 7e^(−0.601) = 3.84, so J = 3 wins. Cool the gas to 100 K and Jₘₐₓ drops to 1.32. The band envelope is therefore a thermometer. Strictly the absorbed intensity also carries a (J+1)/(2J+1) factor from the transition moment, which drifts slowly, and the line frequency times a stimulated-emission correction, which together grow roughly as (J+1)² while hcν̃ is small next to kT. For HCl at 300 K they move the measured absorption maximum from J = 3 → 4 up to the nearly equal J = 4 → 5 and 5 → 6 lines, so Jₘₐₓ locates the most populated level, and the population model's brightest line, rather than the exact peak of a far-infrared spectrum.

Nothing holds R fixed except a bond of finite stiffness. Balance the centrifugal pull against that spring and the bond extends by x = J(J+1)ħ²/(kμR³); feeding the stretch back, the rotational energy falls by kx² while the spring stores only ½kx², a net drop of ½kx² that gives E_J = hcB̃J(J+1) − hcD̃J²(J+1)², with the Kratzer estimate D̃ ≈ 4B̃³/ω̃² tying the distortion constant to the vibrational wavenumber. Lines move to ν̃ = 2B̃(J+1) − 4D̃(J+1)³. For CO, B̃ = 1.9225 cm⁻¹ and ω̃ = 2170 cm⁻¹ give D̃ = 6.04 × 10⁻⁶ cm⁻¹ against a measured 6.12 × 10⁻⁶, so the correction is six parts per million at the first line and one part in a hundred by J = 39. That cubic growth is why you must never fit a straight line to raw high-J line positions. Divide instead: ν̃/(J+1) = 2B̃ − 4D̃(J+1)², linear in (J+1)² with intercept 2B̃ and slope −4D̃.

Invert the chain: a measured spacing gives B̃ = spacing/2, then I = h/8π²cB̃, then R = √(I/μ). For H³⁵Cl, a 20.88 cm⁻¹ spacing gives B̃ = 10.44 cm⁻¹, I = 2.681 × 10⁻⁴⁷ kg m² and R = 128.4 pm. Two checks are worth running. First, isotopes: Born-Oppenheimer says R is set by the electrons and so is identical for H³⁵Cl and H³⁷Cl, while μ differs by 0.15%, so B̃ must scale as 1/μ — predicting B̃(H³⁷Cl) = 10.424 cm⁻¹, a comb 0.032 cm⁻¹ tighter, with the doublet intensities in chlorine's 3:1 abundance ratio. Second, honesty about what R is. A rotational spectrum of the vibrational ground state measures B̃₀, and hcB̃₀ is an average of ħ²/2μR² over the v = 0 wavefunction rather than the value at the potential minimum, with B̃ᵥ = B̃ₑ − α̃ₑ(v + ½). So HCl's spectroscopic r₀ = 128.4 pm sits about 1 pm above its equilibrium rₑ = 127.5 pm.