University Physics V · The Quantum Harmonic Oscillator · 7.9

Thermal States & Field Modes

Temperature does not put an oscillator into a state — it puts it into an ensemble. Sum one geometric series for Z, read the Bose–Einstein occupation off it, and learn to tell a hot mode from a driven one by the off-diagonal elements that equilibrium deletes.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Couple one mode to a reservoir at temperature T and quantum mechanics answers with a density matrix, not a ket. Maximising the von Neumann entropy at fixed mean energy — equivalently, tracing a bath out of a global state the mode is entangled with — returns the Gibbs operator ρ̂ = e^(−βĤ)/Z, and since any function of Ĥ is diagonal in Ĥ's own eigenbasis, the Fock basis diagonalises it for free: ρ̂ = Σ pₙ |n⟩⟨n| with pₙ geometric in n. The thermodynamics is then one geometric series, Z = e^(−x/2)/(1 − e^(−x)) with x = ℏω/k_B T, from which n̄ = 1/(e^x − 1) follows by differentiating ln Z.

That single expression is the Planck function, the phonon occupation of a lattice mode and the photon number of a cavity mode at once, because a free field is nothing but one independent oscillator per mode, with â†ₖ adding a quantum of energy ℏωₖ and momentum ℏk. What it costs is phase. Every off-diagonal element is zero, so ⟨â⟩ = 0 and ⟨x̂⟩ = 0 at all times: heating an oscillator never makes it swing, it only spreads probability up the rungs.

And the state is a fiction of the free Hamiltonian, because a strictly quadratic Ĥ cannot thermalise anything — the interaction that brings the mode to equilibrium is exactly the term this model leaves out.

Simple definition: A thermal state of one mode is the mixed state ρ̂ = e^(−βĤ)/Z: an ordinary probability distribution over the Fock rungs, geometric in n, with no coherence at all between them.
Example: At 300 K a 25.9 meV mode has x = ℏω/k_B T = 1, so populations fall by a factor e per rung: p₀ = 1 − e⁽⁻¹⁾ = 0.632, p₁ = 0.233, and n̄ = 1/(e − 1) = 0.582 quanta.

Partition function of one modeZ = Σₙ e^(−βEₙ) = e^(−x/2)/(1 − e^(−x)) = 1/(2 sinh(x/2))

One geometric series carries the rest: F = −k_B T ln Z and ⟨Ĥ⟩ = −∂(ln Z)/∂β.

x = ℏω/k_B T is dimensionless; β = 1/k_B T in J⁻¹; the sum runs n = 0, 1, 2, …

Thermal density matrix in the Fock basisρ̂ = e^(−βĤ)/Z = Σₙ pₙ |n⟩⟨n|pₙ = (1 − e^(−x)) e^{(−n x)}

The half-quantum cancels against Z, so it never touches a single population.

pₙ is a probability, not an amplitude; Σ pₙ = 1; ⟨m|ρ̂|n⟩ = 0 whenever m ≠ n

Bose–Einstein occupation of the moden̄ = ⟨â†â⟩ = 1/(e^x − 1)

Silicon's 520 cm⁻¹ phonon has x = 2.49 at 300 K, so n̄ = 0.090 — nearly frozen out.

A pure number: mean quanta in one mode, not per unit volume or bandwidth

Mean energy and its two limits⟨Ĥ⟩ = ℏω(n̄ + ½) = (ℏω/2) coth(x/2)

Equipartition is the classical limit of the ladder, not a separate law bolted on.

In joules; → k_B T when x ≪ 1, → ℏω/2 when x ≫ 1

Number fluctuations: thermal against coherent(Δn)² = n̄ + n̄², g⁽²⁾(0) = 2coherent: (Δn)² = n̄, g⁽²⁾(0) = 1

Super-Poissonian bunching is how a hot filament is told from a laser at equal power.

Δn is a dimensionless count; one value of n̄ hides two different distributions

Purity and entropy of the thermal modeTr ρ̂² = 1/(2n̄ + 1)S/k_B = (1 + n̄)ln(1 + n̄) − n̄ ln n̄

Both depend on n̄ alone, so one measured occupation fixes the entire state.

Purity reaches 1 only at n̄ = 0; S is the von Neumann entropy in nats

Temperature hands you a density matrix, not a ket

A ket is maximal information about a system, and a mode in contact with a bath does not have it: the two are entangled, so tracing the bath out leaves a mixed ρ̂. Fix the mean energy, maximise S = −k_B Tr(ρ̂ ln ρ̂), and the stationary point is the Gibbs operator ρ̂ = e^(−βĤ)/Z. Now use the structure of Ĥ = ℏω(N̂ + ½): the exponential of an operator is diagonal wherever that operator is, so ⟨m|ρ̂|n⟩ = pₙ δₘₙ in the Fock basis with no calculation at all. The zero-point term then drops out of every population, because each Boltzmann factor splits into a common e^(−x/2) times e^{(−n x)}, and that common factor cancels against Z — the half-quantum shifts the energy, never the statistics.

Sum one geometric series and the thermodynamics is done

Write r = e^(−x) with x = ℏω/k_B T. The spectrum Eₙ = ℏω(n + ½) makes every Boltzmann factor a power of r: Z = Σₙ e^(−βEₙ) = e^(−x/2) Σₙ rⁿ = e^(−x/2)/(1 − r), which converges for any T > 0 because the spectrum is bounded below and evenly spaced — the two facts the ladder argument delivered. Symmetrising gives Z = 1/(2 sinh(x/2)). Differentiate: ⟨Ĥ⟩ = −∂(ln Z)/∂β = (ℏω/2) coth(x/2) = ℏω(n̄ + ½), and the same derivative returns n̄ = 1/(e^x − 1). At x = 1 that is Z = 0.960, n̄ = 0.582, ⟨Ĥ⟩ = 1.08 ℏω. One derivative more gives the Einstein heat capacity C = k_B x² e^x/(e^x − 1)², equal to 0.921 k_B at x = 1 and dying like x² e^(−x) beyond it. That exponential freeze-out is what a classical solid cannot produce.

Read the exponent first: x = ℏω/k_B T decides everything

Two conversions do most of the work: k_B T = 25.9 meV = 208.5 cm⁻¹ at 300 K, and hf/k_B = 48.0 mK per GHz. Then read off the regime. Silicon's 520 cm⁻¹ optical phonon at 300 K has x = 2.49, n̄ = 0.090, and 91.7% of the ensemble on the ground rung — cold, in a warm room. A 6.0 GHz circuit mode, ℏω/k_B = 0.288 K, has n̄ = 13.4 at the 4 K stage but 5.6 × 10⁻⁷ at 20 mK, which is why superconducting qubits live in dilution refrigerators. A 100 MHz LC mode at 300 K has x = 1.6 × 10⁻⁵ and n̄ = 6.3 × 10⁴, where n̄ ≈ k_B T/ℏω − ½ to five figures: classical. Visible light is the far end — a 500 nm mode at 300 K has x = 95.9 and n̄ ≈ 2 × 10⁻⁴², which is why a warm room is dark.

Mixed is not superposed: apply the off-diagonal test

Take n̄ = 3 and compare the thermal ρ̂ with the pure state |χ⟩ = Σₙ √pₙ |n⟩ built from the same populations. Measure n̂ and the two are indistinguishable: identical histograms, p₀ = 0.250 for both. Everything else separates them. Tr ρ̂² = 1/(2n̄ + 1) = 0.143 against 1. ⟨â⟩ = 0 against ⟨â⟩ = 1.60, so ⟨x̂⟩ stays at zero for the mixture while |χ⟩ swings at ω. And ρ̂ commutes with Ĥ, so it is stationary — a thermal state has no dynamics to watch. The fluctuations differ too: (Δn)² = n̄ + n̄² = 12 for the thermal mode against 3 for a coherent state of the same mean, and g⁽²⁾(0) = 2 against 1. That bunching is the Hanbury Brown–Twiss signature of thermal light.

Every mode of a free field is exactly one oscillator

Expand a free field in normal modes and each wavevector-polarisation pair carries its own quadratic Hamiltonian ℏωₖ(â†ₖ âₖ + ½), with ωₖ = c|k| for light. The operator â†ₖ adds one quantum carrying energy ℏωₖ and momentum ℏk, so the excitation is a photon or a phonon rather than a rung on an abstract ladder. Because the modes are independent, each thermalises separately, and n̄ = 1/(e^x − 1) evaluated at x = ℏω/k_B T is the Planck function itself: multiply by the mode density 8πν²/c³ and by ℏω and the blackbody spectrum falls out with nothing else added. A 160 GHz mode of the cosmic microwave background at 2.725 K has x = 2.82 and n̄ = 0.064. The zero-point sum Σ ℏωₖ/2 diverges and is normal-ordered away; only differences in it, Casimir and Lamb, are measurable.

Where the free-mode model runs out

The thermal state is the fixed point of dynamics this Hamiltonian does not contain. Under a strictly quadratic Ĥ every eigenstate keeps its population forever, a coherent state stays coherent, and nothing relaxes — so equilibrium has to be imported from a coupling the model dropped: anharmonic phonon–phonon scattering, a cavity mode leaking into lossy walls, electron–photon scattering in a filament. That same coupling sets the linewidth, so a mode narrow enough to call sharp is a mode slow to thermalise; and once the coupling is not weak, the mode's own temperature stops being well defined, because ρ̂ for the mode alone is then no longer the Gibbs state of its own Ĥ. Mode independence carries the same warning: modes decouple only while the field is free.

Change one variable at a time

Make the relationship visible.

Interactive model

Mode quantum ℏω25 meV

Temperature T300 K

Hold ℏω at 25 meV and drag T from 30 K to 370 K: the ladder fills from the bottom, but the n = 0 bar stays the tallest at every temperature. A thermal ladder never grows a bump at ⟨n⟩ — that is what makes it geometric rather than Poissonian.

x = ℏω/k_B T0.97

MEAN OCCUPATION ⟨n⟩0.613 quanta

GROUND RUNG p₀0.620

ENERGY ⟨H⟩/ℏω1.113

Live interpretationx = ℏω/k_B T: 0.97. MEAN OCCUPATION ⟨n⟩: 0.613 quanta. GROUND RUNG p₀: 0.620. ENERGY ⟨H⟩/ℏω: 1.113

Catch the common trap

Explain before calculating.

Misconception check

The commonest error is to read the pₙ as amplitudes and write the thermal state as |ψ⟩ = Σ √pₙ |n⟩ — a hot oscillator pictured as a superposition sloshing harder as T rises. It is not a superposition at all. Every off-diagonal element of ρ̂ = e^(−βĤ)/Z is zero, so ⟨â⟩ = 0, ⟨x̂⟩ = ⟨p̂⟩ = 0 at every instant, and ρ̂ commutes with Ĥ, which makes the state stationary. The two objects have identical number histograms and differ in everything else: at n̄ = 3 the purity is 1/(2n̄ + 1) = 0.143 against 1, and ⟨â⟩ is 0 against 1.60. Heating an oscillator spreads probability up the ladder while destroying every relative phase. The second habit worth breaking is treating n̄ as the energy: n̄ is a mean quantum number, and ⟨Ĥ⟩ = ℏω(n̄ + ½), with the half surviving at T = 0.

One mode is in equilibrium at the temperature where ℏω = k_B T exactly, so x = 1. Which statement about ρ̂ = e^(−βĤ)/Z is correct?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySilicon's Raman-active optical phonon sits at 520 cm⁻¹. At 300 K find x = ℏω/k_B T, the mean occupation n̄, the fraction of the ensemble on the ground rung, and the anti-Stokes to Stokes intensity ratio, ignoring the ω⁴ scattering factor. Take k_B T = 25.85 meV at 300 K and 1 cm⁻¹ = 0.1240 meV.

Convert the wavenumber to an energy: ℏω = 520 cm⁻¹ × 0.1240 meV/cm⁻¹ = 64.5 meV.
Form the exponent, the only thing the statistics depend on: x = ℏω/k_B T = 64.5/25.85 = 2.494, so e^(−x) = 0.0826.
Occupation: n̄ = 1/(e^x − 1) = 1/(12.11 − 1) = 0.0900 quanta in the mode.
Ground rung: p₀ = 1 − e^(−x) = 0.917, so 91.7% of the ensemble carries no phonon at all, and P(n ≥ 1) = e^(−x) = 0.083.
Anti-Stokes must destroy a phonon and scales as n̄, Stokes creates one and scales as n̄ + 1, so the ratio is n̄/(n̄ + 1) = e^(−x) = 0.0826, about 1 : 12. That ratio is the standard Raman thermometer; restoring the ω⁴ factor for a 532 nm laser raises it to 0.103.

Answerx = 2.49, n̄ = 0.0900, p₀ = 0.917, and I(anti-Stokes)/I(Stokes) = 0.083.

MediumA superconducting resonator mode at 6.0 GHz is cooled in a dilution refrigerator. (a) Express its quantum as a temperature ℏω/k_B. (b) Find n̄ at the 4.0 K pulse-tube stage and at the 20 mK mixing chamber. (c) Find the temperature at which n̄ falls to 0.010. Use h/k_B = 48.0 mK per GHz.

Energy as a temperature: ℏω/k_B = hf/k_B = 6.0 GHz × 48.0 mK/GHz = 0.288 K.
At 4.0 K: x = 0.288/4.0 = 0.0720, so n̄ = 1/(e^x − 1) = 1/0.0746 = 13.4. Since x ≪ 1 the classical estimate k_B T/ℏω − ½ = 13.4 agrees to three figures.
At 20 mK: x = 0.288/0.020 = 14.4, so e^x = 1.79 × 10⁶ and n̄ = 5.6 × 10⁻⁷ — the mode is in |0⟩ to better than a part in a million.
Invert for a target occupation: n̄ = 0.010 needs e^x = 1 + 1/n̄ = 101, so x = ln 101 = 4.615 and T = 0.288/4.615 = 0.0624 K, or 62 mK.
Read the physics off the exponent: the 4 K stage leaves 13 thermal photons in the mode, enough to dephase any qubit coupled to it, while 20 mK leaves essentially none. Occupation is exponential in x, not linear in T, so a factor of 200 in temperature buys seven orders of magnitude in n̄.

Answerℏω/k_B = 0.288 K; n̄ = 13.4 at 4.0 K and 5.6 × 10⁻⁷ at 20 mK; and n̄ = 0.010 at T = 62 mK.

HardA 7.5 GHz mode is held at 1.25 K. (a) Find n̄. (b) Find p₀ and P(n ≥ 3). (c) Compare Δn with that of a coherent state of the same mean. (d) Give the purity Tr ρ̂² and the entropy S/k_B. Use h/k_B = 48.0 mK per GHz.

Exponent: ℏω/k_B = 7.5 × 0.0480 K = 0.360 K, so x = 0.360/1.25 = 0.288 and r = e^(−x) = 0.750.
Occupation: pₙ = (1 − r)rⁿ is geometric, so n̄ = r/(1 − r) = 0.750/0.250 = 3.00 quanta, matching 1/(e^x − 1).
Populations: p₀ = 1 − r = 0.250, and the tail sums exactly because it is geometric, P(n ≥ 3) = r³ = 0.422. The most likely single outcome is still n = 0, even though the mean is three.
Fluctuations: (Δn)² = n̄ + n̄² = 3 + 9 = 12, so Δn = 3.46. A coherent state of the same n̄ = 3 is Poissonian, Δn = √3 = 1.73 — exactly half, and g⁽²⁾(0) = 2 against 1.
Purity: Tr ρ̂² = Σ pₙ² = (1 − r)/(1 + r) = 1/(2n̄ + 1) = 1/7 = 0.143, nowhere near the 1 of any pure state.
Entropy: S/k_B = (1 + n̄)ln(1 + n̄) − n̄ ln n̄ = 4 ln 4 − 3 ln 3 = 5.545 − 3.296 = 2.249 nats, about 3.24 bits of ignorance about which rung the mode is on.

Answern̄ = 3.00, p₀ = 0.250, P(n ≥ 3) = 0.422, Δn = 3.46 against 1.73 for a coherent state, Tr ρ̂² = 0.143 and S = 2.25 k_B.