University Physics IV · The Quantum Harmonic Oscillator · 8.9
Thermal Occupation & Vibrational Spectra
Room temperature cannot see most molecular vibrations, and this lesson says why in numbers. Drop the oscillator ladder into a heat bath, sum the Boltzmann weights, and a single ratio ħω/kT falls out that fixes the level populations, the heat capacity, and which infrared bands you will actually record.
Build the model
Connect the measurement to the mechanism.
A vibrational mode is never hot or cold on its own; it is hot or cold relative to its own quantum. Put the harmonic ladder in contact with a bath at temperature T and the only ingredient statistical mechanics needs is the Boltzmann weight e(−Eₙ/kT). Because Eₙ = (n + ½)ħω climbs in equal steps, those weights form a geometric series, the zero-point factor is common to every term and cancels on normalising, and the whole problem closes in one line: Z = 1/[2 sinh(x/2)] and ⟨n⟩ = 1/(ex − 1), with x = ħω/kT.
That single dimensionless ratio then does all the work. When x ≪ 1 the bath cannot tell the ladder from a continuum, ⟨E⟩ → kT, and the mode carries the full classical k of heat capacity. When x ≫ 1 the first rung is unaffordable, the population piles onto n = 0, and the mode contributes nothing to dU/dT — it has frozen out, which is why nitrogen's CV sits at 5R/2 rather than 7R/2 at room temperature.
The cost is the model's rigidity. Giving a whole solid one frequency, as Einstein did, recovers Dulong–Petit and the fall at 300 K but predicts an exponential collapse where experiment shows T³; and a strictly quadratic potential makes every rung radiate at one frequency, hiding the hot-band structure real molecules show.
- Simple definition
- Thermal occupation is the equilibrium Boltzmann distribution over the rungs of a harmonic ladder: a geometric series in e(−ħω/kT) whose mean, ⟨n⟩ = 1/(e(ħω/kT) − 1), is the average number of quanta the mode is carrying.
- Example
- Nitrogen at 300 K has ħω = 292 meV against kT = 25.9 meV, so ħω/kT = 11.3 and ⟨n⟩ = 1.2 × 10⁻⁵ — about one molecule in 82 000 sits above the ground rung, and the mode is frozen.
Equal rung spacing is what makes the populations a geometric series, so one ratio fixes the entire distribution.
x and pₙ are pure numbers. Each rung sits a fixed factor e(−x) below the one beneath it.
One closed sum, and every thermal average — populations, energy, heat capacity — follows by differentiating ln Z.
Dimensionless. It converges for every x > 0 because the weights fall geometrically up an infinite ladder.
The Planck factor: it reads off how far up the ladder the temperature can push before the exponential shuts it down.
A mean count, not a probability. ⟨n⟩ → kT/ħω − ½ for kT ≫ ħω, and ⟨n⟩ → e(−ħω/kT) for kT ≪ ħω.
Splits the energy into a temperature-independent half-quantum and a thermal part; only the thermal part can carry heat.
In eV or J. → kT at high T (equipartition), → ħω/2 at low T, where only the zero-point term is left.
Recovers Dulong–Petit above ΘE and collapses below it — the first time quantisation explained a thermodynamic measurement.
J K⁻¹; per mole 3R = 24.94 J mol⁻¹ K⁻¹. Diamond has ΘE ≈ 1320 K, giving C(300 K) = 6.08 J mol⁻¹ K⁻¹.
The measured low-T law. Einstein's single frequency gives an exponential instead, 17 times too small at T = Θ/10.
Per mole, valid for T ≲ ΘD/50. ΘD = 225 K for silver, 343 K for copper, 428 K for aluminium.
Sum the ladder: it is a geometric series
Put one oscillator in contact with a bath at temperature T and the weight of rung n is e(−Eₙ/kT) with Eₙ = (n + ½)ħω. Factor the exponent: the weight is e(−x/2) × e(−nx), where x = ħω/kT. The zero-point piece e(−x/2) is the same on every rung, so it cancels the moment you normalise — which is the first payoff of doing the sum honestly, because it means zero-point energy can never show up in a population or a heat capacity. What is left is Σ e(−nx), a geometric series with ratio e(−x) < 1, so Z = e(−x/2)/(1 − e(−x)) and pₙ = (1 − e(−x)) e(−nx). The equal spacing of the harmonic ladder is exactly what makes the series geometric; a well whose levels climb as n² gives no such closed form. For iodine vapour at 300 K, e(−x) = 0.357, so the rungs hold 64.3%, 23.0%, 8.2% and 2.9% of the molecules.
One dimensionless ratio decides everything
Temperature alone tells you nothing about a mode; x = ħω/kT does. At 300 K, kT = 25.9 meV, and at 100 K it is 8.62 meV. Set two molecules side by side in the same 300 K room. Nitrogen absorbs at 2359 cm⁻¹, so ħω = 292 meV and x = 11.3; iodine absorbs at 214.5 cm⁻¹, so ħω = 26.6 meV and x = 1.03. The mean occupations are ⟨n⟩ = 1/(ex − 1) = 1.2 × 10⁻⁵ and 0.556 — a factor of 45 000 apart, from a factor of 11 in ħω, because the ratio sits in an exponent. This is why the useful temperature scale is the mode's own: Θ = ħω/k, which is 3394 K for N₂ and 309 K for I₂. Iodine at room temperature sits at 0.97 Θ; nitrogen sits at 0.088 Θ, effectively at absolute zero as far as its bond is concerned.
Differentiate to get the heat capacity of a mode
The mean energy follows from the same distribution: ⟨E⟩ = ħω(½ + 1/(ex − 1)). Differentiate with respect to T, remembering x ∝ 1/T, and the half-quantum drops out because it does not depend on T: C = k x² ex/(ex − 1)². Two limits check it. As x → 0 expand ex = 1 + x + x²/2, and the ratio → 1, so C → k — one whole k per oscillator, not ½k, because equipartition gives ½k for the kinetic term and ½k for the potential term. As x grows, C ≈ k x² e(−x) and dies exponentially. Between them the crossover is sharp: C/k is 0.979 at x = 0.5, 0.921 at x = 1, 0.724 at x = 2, 0.496 at x = 3 and 0.0045 at x = 10. The mode has lost half its classical capacity by x ≈ 3, that is by T ≈ Θ/3.
Freeze-out and the heat capacity of a diatomic gas
Classical equipartition says a diatomic gas should hold 3 translational, 2 rotational and 2 vibrational half-quanta, giving CV = 7R/2 = 29.1 J mol⁻¹ K⁻¹. Nitrogen at 300 K measures 20.8 J mol⁻¹ K⁻¹, which is 5R/2. The quantum count settles it: with x = 11.3 the vibrational term is C/k = x² e(−x) = 1.56 × 10⁻³, so vibration contributes 0.013 J mol⁻¹ K⁻¹ — six parts in ten thousand of the total, and nothing a calorimeter will see. It is not that N₂ lacks a vibrational degree of freedom; it is that the freedom costs 292 meV and the bath has only 25.9 meV to spend. Warm the gas towards Θ/3 ≈ 1130 K and the term wakes up. Iodine, whose Θ is only 309 K, has already woken: its vibration contributes 0.92R = 7.62 J mol⁻¹ K⁻¹ in the same room.
The Einstein solid, and why Debye had to replace it
Einstein's 1907 move was to treat a solid of N atoms as 3N independent oscillators sharing one frequency, so C = 3Nk × the single-mode function. Above ΘE it returns 3R = 24.94 J mol⁻¹ K⁻¹ — Dulong–Petit stops being an empirical rule and becomes a high-temperature limit. Below it, the model explained the anomaly that had defeated classical theory: diamond, with ΘE ≈ 1320 K, gives x = 4.40 and C = 6.08 J mol⁻¹ K⁻¹ at 300 K against 6.11 measured. But at low T the exponential is too severe. At T = Θ/10, Einstein predicts 0.45% of Dulong–Petit while crystals measure about 7.8%, a factor of 17. The repair is not a better single frequency but a spectrum: a real crystal has acoustic modes whose frequency runs to zero with wavevector, so however cold it gets there are always modes with ħω < kT still absorbing heat. Counting them up to a cutoff ΘD gives Debye's C ∝ T³.
Reading the populations off an infrared spectrum
For a harmonic oscillator with a dipole moment linear in displacement, the position matrix element ⟨n+1|x̂|n⟩ = √((n+1)ħ/2mω) is the only non-zero one, so Δn = ±1. Two consequences follow. First, every rung radiates at the same ω, so a strictly harmonic molecule shows exactly one band no matter how hot it is. Second, band strength is set by the populations you just computed: the fundamental 0→1 draws on p₀ − p₁, and the hot bands 1→2 and 2→3 draw on the excited rungs, with an extra factor n+1 from the matrix element. For iodine at 300 K that makes the 1→2 hot band 0.71 as strong as the fundamental. Anharmonicity — the term this model discarded — is what shifts it 2ωₑ xₑ ≈ 1.2 cm⁻¹ to the red so it can be resolved. And the dipole really must change with x: N₂ and O₂ are homonuclear and so have no infrared band at all, which is why the atmosphere's two main gases are not greenhouse gases.
Change one variable at a time
Make the relationship visible.
From 30 meV and 300 K, slide T up: the bars spread up the ladder and the mean-energy dot settles onto the kT line with C/k heading for 1. Now push ħω to 300 meV and everything collapses onto rung 0, C/k falls to zero, and the mode has frozen out.
ħω / kT1.16
MEAN OCCUPATION ⟨n⟩0.456
ON GROUND RUNG n = 068.7 %
HEAT CAPACITY C / k0.895
Live interpretationħω / kT: 1.16. MEAN OCCUPATION ⟨n⟩: 0.456. ON GROUND RUNG n = 0: 68.7 %. HEAT CAPACITY C / k: 0.895
Catch the common trap
Explain before calculating.
Nitrogen at 300 K has ħω/kT = 11.3, so only about one molecule in 82 000 sits above the ground vibrational rung. What does its vibrational mode contribute to the molar heat capacity, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyCarbon monoxide absorbs its vibrational fundamental at 2143 cm⁻¹. At 300 K, find ħω in meV, the ratio ħω/kT, the fraction of molecules above the ground rung, and the mean occupation ⟨n⟩.
- Convert the wavenumber. One cm⁻¹ is 0.12398 meV, so ħω = 2143 × 0.12398 = 265.7 meV.
- Thermal energy at 300 K: kT = 0.086173 meV K⁻¹ × 300 K = 25.85 meV. So x = ħω/kT = 265.7/25.85 = 10.28.
- The rung-to-rung ratio is e(−x) = e(−10.28) = 3.44 × 10⁻⁵, and since p₀ = 1 − e(−x), the fraction above the ground rung is exactly e(−x) = 3.44 × 10⁻⁵.
- Mean occupation: ⟨n⟩ = 1/(e(10.28) − 1) = 1/29 075 = 3.44 × 10⁻⁵. It matches the excited fraction here because at large x essentially every excited molecule is on rung 1 and no higher.
- Heat-capacity check: C/k = x² e(−x) = 105.6 × 3.44 × 10⁻⁵ = 3.6 × 10⁻³, so the mode adds 0.030 J mol⁻¹ K⁻¹ — nothing a calorimeter will resolve.
Answerħω = 265.7 meV, ħω/kT = 10.28, and both the excited fraction and ⟨n⟩ are 3.4 × 10⁻⁵: about 34 CO molecules in a million are vibrationally excited at 300 K, so the mode is frozen out.
MediumIodine vapour absorbs at 214.5 cm⁻¹. At 300 K, find the populations of the first three rungs, the mean occupation, and the strength of the 1→2 hot band relative to the 0→1 fundamental. Absorption strength goes as (n+1)(pₙ − p_{n+1}).
- ħω = 214.5 × 0.12398 = 26.59 meV, and kT = 25.85 meV, so x = 26.59/25.85 = 1.029 — this mode is barely above the thermal energy.
- Rung-to-rung ratio: r = e(−1.029) = 0.3575.
- Populations: p₀ = 1 − r = 0.6425, p₁ = p₀r = 0.2297, p₂ = p₁r = 0.0821. The fraction above the ground rung is r = 35.7%.
- Mean occupation: ⟨n⟩ = 1/(e(1.029) − 1) = 1/1.798 = 0.556. It exceeds the excited fraction of 0.357 because the excited molecules are not all on rung 1: those higher up add two, three or more quanta to the mean.
- Fundamental 0→1: 1 × (0.6425 − 0.2297) = 0.4129. Hot band 1→2: 2 × (0.2297 − 0.0821) = 0.2952. Ratio = 0.2952/0.4129 = 0.715.
- Physically: over a third of the vapour is already vibrationally excited, so hot bands are strong. The harmonic model puts them at the same wavenumber as the fundamental; real I₂ separates them by 2ωₑ xₑ = 1.23 cm⁻¹.
Answerp₀ = 64.3%, p₁ = 23.0%, p₂ = 8.2%, ⟨n⟩ = 0.556, and the 1→2 hot band is 0.71 as strong as the fundamental — which is why iodine's bands are congested at room temperature.
HardA monatomic solid is modelled as 3N oscillators at one frequency with ΘE = 225 K. Find its molar heat capacity at 300 K and at 22.5 K, compare the low-temperature value with the Debye prediction for the same solid taking ΘD = 225 K, and say which is right.
- Use C = 3R x² ex/(ex − 1)² with x = ΘE/T and 3R = 24.94 J mol⁻¹ K⁻¹.
- At 300 K: x = 225/300 = 0.750, ex = 2.1170, so C/3R = 0.5625 × 2.1170/(1.1170)² = 1.1908/1.2477 = 0.9544 and C = 23.8 J mol⁻¹ K⁻¹ — 95% of Dulong–Petit.
- At 22.5 K: x = 10.0, ex = 22 026, so C/3R = 100 × 22 026/(22 025)² = 4.54 × 10⁻³ and C = 0.113 J mol⁻¹ K⁻¹.
- Debye at the same temperature: C = 1944 (T/ΘD)³ = 1944 × (22.5/225)³ = 1944 × 10⁻³ = 1.94 J mol⁻¹ K⁻¹.
- The two differ by a factor of 1.94/0.113 = 17, and calorimetry follows Debye. Einstein's single frequency has a gap ħω that every mode must pay, so the population — and with it C — dies as e(−Θ/T). A real crystal has acoustic modes with ħω → 0 as the wavevector goes to zero, and those never freeze out; counting them to a cutoff gives the T³ law.
- Note what survives: Einstein is still right at 300 K, and both models agree on the 3R ceiling. The single-frequency assumption fails only where the surviving modes are the soft ones.
AnswerC(300 K) = 23.8 J mol⁻¹ K⁻¹, C(22.5 K) = 0.113 J mol⁻¹ K⁻¹ against Debye's 1.94 J mol⁻¹ K⁻¹ — Einstein is 17 times too small at low T, because a real solid has gapless acoustic modes.