The Harmonic Oscillator and Rigid Rotor as Quantum Models
Two exactly solvable systems that between them describe molecular vibration and rotation, and whose results are used constantly in spectroscopy.
Two models, two level patterns
| Harmonic oscillator | Rigid rotor | |
|---|---|---|
| Energy | (v + ½)hν | BJ(J + 1) |
| Lowest level | ½hν — not zero | Zero |
| Spacing | Constant, hν | Increases with J |
| Degeneracy | None in one dimension | 2J + 1 |
| Selection rule | Δv = ±1 | ΔJ = ±1 |
The harmonic oscillator
Modelling a bond as a spring with force constant k and reduced mass μ gives
so a stiffer bond or lighter atoms vibrate faster. Since the levels are evenly spaced and only single-quantum transitions are allowed, a harmonic oscillator absorbs at exactly one frequency regardless of which level it starts from.
Zero-point energy
The lowest level lies at half a quantum above the bottom of the well. A bond therefore vibrates even at absolute zero, which is required by the uncertainty principle — a stationary bond of exactly fixed length would specify both position and momentum precisely.
The practical consequence is the kinetic isotope effect: substituting a heavier isotope raises the reduced mass, lowers the frequency and lowers the zero-point energy, so more energy is needed to break the bond. Bonds to heavier isotopes therefore break more slowly, which is used as evidence that a particular bond is broken in the rate-determining step.
The rigid rotor
Two masses joined by a rigid rod rotate with energies proportional to J(J+1), where the constant B is inversely proportional to the moment of inertia.
Because the energy depends on J only, and there are 2J+1 orientations for each J, every level except the lowest is degenerate. That degeneracy is what makes the intensity distribution in a rotational spectrum non-trivial: population falls with energy through the Boltzmann factor but rises with degeneracy, so the most populated level is at intermediate J rather than at J equal to zero.
Why the lines are equally spaced
The transition from J to J+1 has energy 2B(J+1), so successive transitions differ by exactly 2B. Measuring that spacing gives B, hence the moment of inertia, hence the bond length — the standard chain of inference in microwave spectroscopy.
Where each model fails
| Model | Failure | Correction |
|---|---|---|
| Harmonic oscillator | Predicts infinitely many levels and no dissociation | Anharmonicity; the Morse potential |
| Harmonic oscillator | Forbids overtones | Anharmonicity makes them weakly allowed |
| Rigid rotor | Assumes the bond length is fixed | Centrifugal distortion at high J |
Both corrections are small at low quantum numbers, which is why the simple models remain useful. Recognising when a question is asking about the correction rather than the model is usually signalled by mention of high v or high J.
Frequently asked questions
Why does the harmonic oscillator have zero-point energy?
Because the lowest allowed level is v equal to zero, whose energy is half a quantum. A state of exactly zero energy would violate the uncertainty principle.
Why do rotational lines appear equally spaced when the levels are not?
Because the transition energy is 2B(J+1), which increases in equal steps as J increases. The levels diverge, but the differences between successive transitions are constant.
Why is the most intense rotational line not the lowest?
Because level population depends on both the Boltzmann factor, which falls with J, and the degeneracy 2J+1, which rises. Their product peaks at intermediate J.
Which model does a real molecule follow?
Both approximately and neither exactly. Vibration is nearly harmonic near the bottom of the well; rotation is nearly rigid at low J. Corrections matter at high quantum numbers.
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