The Lindblad form, the GKSL theorem and a thermal reservoir

Units are SI with ħ and k_B kept explicit; nothing here is relativistic, so the course metric diag(1, −1, −1, −1) never enters. A master equation dρ/dt = ℒ[ρ] is admissible only if ρ(t) stays Hermitian, keeps unit trace and stays positive, and, as topic 14.01 showed with half of a singlet, stays positive while the system is entangled with an untouched partner. The Lindblad form passes the first two tests line by line. Taking the adjoint sends the commutator and the jump term into themselves with ρ replaced by ρ† and swaps the two −½ terms, so (ℒ[ρ])† = ℒ[ρ†] and a Hermitian ρ stays Hermitian. The trace of a commutator vanishes, and cyclicity gives Tr(LₖρLₖ†) = Tr(Lₖ†Lₖρ), which the two −½ terms cancel exactly. Those terms are not optional: drop them from spontaneous emission, L = √Γ σ₋, and d(Tr ρ)/dt = Γpₑ, so an excited atom manufactures probability. Dimensions fix the rest: ρ is a pure number, so every Lₖ carries s^(−1/2).

Positivity needs more than a trace count, and the cleanest proof writes one short step as a quantum operation. Take K₀ = 1 − (i/ħ)H dt − ½ΣₖLₖ†Lₖ dt and Kₖ = √dt Lₖ. Then K₀ρK₀† = ρ − (i/ħ)[H, ρ]dt − ½Σₖ(Lₖ†Lₖρ + ρLₖ†Lₖ)dt + O(dt²) and KₖρKₖ† = LₖρLₖ† dt, so Σ KρK† = ρ + ℒ[ρ]dt + O(dt²): the master equation is this Kraus map to first order. The same expansion gives K₀†K₀ + ΣₖKₖ†Kₖ = 1 + O(dt²). Each step is completely positive because it has Kraus form, and its trace error is O(dt²). Over a finite time t = N dt those errors total O(dt) and vanish as dt → 0; a limit of completely positive maps is completely positive, and the exact identity Tr ℒ[ρ] = 0 restores the trace. So e^(ℒt) is completely positive and trace-preserving for every t ≥ 0. This direction is derived; the converse is not.

Call an evolution Markovian when its maps form a semigroup, ΛₜΛₛ = Λₜ₊ₛ: what happens after time t depends on ρ(t) alone, not on how it got there, and then Λₜ = e^(ℒt). The theorem of Gorini, Kossakowski and Sudarshan for N levels, and of Lindblad for bounded generators, both 1976, says that if every Λₜ is completely positive and trace-preserving, ℒ must have the Lindblad form, with a Hermitian H and at most N² − 1 jump operators. This course quotes it; the proof is a structure theorem about completely positive maps, not physics. Three consequences matter. Every memoryless model, however it was reached, is a choice of H and the Lₖ, and a qubit never needs more than three. The choice is not unique: mixing the Lₖ by a unitary matrix, or shifting one by a constant while adding a compensating term to H, leaves ℒ unchanged. And the theorem is silent on which Lₖ nature picks; that takes a microscopic derivation such as topic 14.03's.

The form inherits every assumption of the semigroup. Put the atom in a lossless cavity mode at resonance with coupling g, starting in |e⟩ with the mode empty, so pₑ = cos²(gt). At gt = π/2 the atom is in |g⟩ whether it started in |e⟩ or in |g⟩, so a semigroup would have to give the two the same future; yet the first returns to |e⟩ at gt = π and the second never leaves |g⟩. No Lindblad equation for the atom alone can do this, because the memory sits in the mode; the remedy is to enlarge the system to atom plus mode. In general the derivation needs three separations of scale: system and bath start uncorrelated and stay nearly so (Born), the bath forgets in a time τ_R short compared with 1/Γ (Markov), and Γ is small compared with ω₀ and with every level splitting (secular). For sodium's 589 nm line in the vacuum, τ_R ≈ 1/ω₀ = 3.1 × 10⁻¹⁶ s and Γ/ω₀ = 1.9 × 10⁻⁸, so all three hold with room to spare.

Topic 14.03 traced out a field in its vacuum. Replace it by the thermal state exp(−H_R/k_B T)/Z, still uncorrelated with the atom at t = 0. Its mode averages are ⟨aₖ†aₖ⟩ = n̄(ωₖ) and ⟨aₖaₖ†⟩ = n̄(ωₖ) + 1, with the Bose-Einstein n̄(ω) = 1/(exp(ħω/k_B T) − 1), and ⟨aₖaₖ⟩ = 0. In the second-order double commutator, terms that lower the atom pair with ⟨aa†⟩ and terms that raise it with ⟨a†a⟩: Unit 12's factors n̄ + 1 and n̄. The Markov step evaluates the mode sum at ω₀ as before, so the vacuum rate Γ is multiplied by n̄(ω₀) + 1 for emission and by n̄(ω₀) for absorption, and the secular step again discards σ₋ρσ₋ and σ₊ρσ₊. With D[L]ρ = LρL† − ½L†Lρ − ½ρL†L the result is dρ/dt = −(i/ħ)[H′, ρ] + Γ(n̄ + 1)D[σ₋]ρ + Γn̄D[σ₊]ρ: Lindblad form, so completely positive. The shift in H′ gains a temperature-dependent part that only moves ω₀. Markov now also needs the thermal spectrum flat across the line, Γ ≪ k_B T/ħ = 3.9 × 10¹³ s⁻¹ at 300 K.

Populations decouple from coherences: dpₑ/dt = −Γ(n̄ + 1)pₑ + Γn̄p_g = −Γ(2n̄ + 1)pₑ + Γn̄. Any deviation from equilibrium decays at the sum of the two rates, 1/T₁ = Γ(2n̄ + 1), towards pₑ = n̄/(2n̄ + 1), so pₑ/p_g = n̄/(n̄ + 1) = exp(−ħω₀/k_B T). Each jump operator removes half its rate from ρ_eg, which therefore decays at Γ(2n̄ + 1)/2, giving T₂ = 2T₁. From any start the atom reaches exp(−H_A/k_B T)/Z, the canonical state Quantum Mechanics II postulated, now with a rate attached. Dimensions: Γ(2n̄ + 1) is in s⁻¹ and ħω₀/k_B T is a pure number. Limits: n̄ → 0 returns topic 14.03, and k_B T ≫ ħω₀ gives pₑ → ½ − ħω₀/(4k_B T) with 1/T₁ → 2Γk_B T/(ħω₀). Measurement: interstellar CN's 113.5 GHz rotational line has n̄ = 0.157 in the 2.7255 K microwave background, and its measured excitation temperature, about 2.73 K toward ζ Ophiuchi, agrees to about 1%, as this steady state requires when radiation outpaces collisions.