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Thermal balance makes Gibbs stationary

Statement

Let $d \geq 1$ , let $H = H^{*} \in M_{d} (C)$ have entries in joules and a finite spectral decomposition $H = \sum_{E} E P_{E}$ with pairwise orthogonal projections summing to $I_{d}$ , and let $S_{1}, \dots, S_{m}$ be a finite family of dimensionless Hermitian matrices, where $m \geq 0$ . Set $Ω_{H} = {(E^{'} - E) / ℏ : E, E^{'} \in spec (H)}$ and, for each $ω \in Ω_{H}$ , define $S_{a} (ω) = \sum_{E^{'} - E = ℏ ω} P_{E} S_{a} P_{E^{'}}$ . Suppose that the rate matrices $Γ (ω) = (γ_{a b} (ω))$ , $ω \in Ω_{H}$ , have entries in seconds $^{- 1}$ , are positive semidefinite in the explicit sense that $Γ (ω) = Γ (ω)^{*}$ and $v^{*} Γ (ω) v \geq 0$ for every $v \in C^{m}$ , and satisfy $γ_{a b} (- ω) = e^{- β ℏ ω} γ_{b a} (ω)$ , with $β \in R$ in joules $^{- 1}$ and $ℏ > 0$ in joule-seconds. Let $H_{L S} = H_{L S}^{*}$ have entries in joules and commute with $H$ , and let $L$ be the finite generator

$L (X) = - \frac{i}{ℏ} [H + H_{L S}, X] + \sum_{ω \in Ω_{H}} \sum_{a, b = 1}^{m} γ_{a b} (ω) (S_{b} (ω) X S_{a} (ω)^{*} - \frac{1}{2} {S_{a} (ω)^{*} S_{b} (ω), X}) .$

For $ρ \in M_{d} (C)$ , being a density matrix means $ρ = ρ^{*}$ , $v^{*} ρ v \geq 0$ for every $v \in C^{d}$ , and $Tr ρ = 1$ ; it is faithful when it is positive definite. For $Z_{β} : = Tr (e^{- β H}) > 0$ , define the faithful Gibbs density $σ_{β} : = e^{- β H} / Z_{β}$ . For time $t \geq 0$ measured in seconds, $T_{t} : = e^{t L}$ is completely positive and trace-preserving, and $T_{t} (σ_{β}) = σ_{β}$ . This statement asserts stationarity only; it asserts neither uniqueness of the stationary density nor convergence to $σ_{β}$ .

Statement references

Finite energy-resolved GKSL thermal model

Finite completely positive trace-preserving maps

Finite GKSL generators and units

Finite GKSL generates CPTP semigroups

Finite matrix spectral and state calculus

Facts & Assumptions

Given: The finite-dimensional system, energy-resolved couplings, rate matrices, balance relation, and commuting Lamb-shift matrix in the Statement. Use the ordered finite spectral projections and SI units there.

[A1]

Finite energy-resolved GKSL thermal model: for each $ω$ , $S_{a} (ω) = \sum_{E^{'} - E = ℏ ω} P_{E} S_{a} P_{E^{'}}$ ; $Γ (ω) = Γ (ω)^{*}$ and $v^{*} Γ (ω) v \geq 0$ for every $v \in C^{m}$ ; and the matrices obey $γ_{a b} (- ω) = e^{- β ℏ ω} γ_{b a} (ω)$ . Also, $H_{L S}$ is Hermitian and commutes with $H$ .

[F1]

Finite completely positive trace-preserving maps: complete positivity means positivity is preserved by every finite ancilla amplification, and trace preservation means traces agree on every matrix input; a finite channel is a complex-linear map with both properties.

[F2]

Finite GKSL generators and units: a finite GKSL generator has the form $- i [H, X] / ℏ + \sum_{j} (J_{j} X J_{j}^{*} - \frac{1}{2} {J_{j}^{*} J_{j}, X})$ for Hermitian $H$ and finite jump matrices $J_{j}$ .

[F3]

Finite GKSL generates CPTP semigroups: the finite GKSL generator of [F2] has an exponential semigroup of finite channels for all $t \geq 0$ .

[F4]

Finite matrix spectral and state calculus: a finite Hermitian matrix has orthogonal spectral projections and basis-independent spectral calculus.

Proof

1.1givenA1F1F2F3F4algebraconstruct

For each $ω$ , positive semidefiniteness of the finite matrix $Γ (ω)$ gives a finite factorization $γ_{a b} (ω) = \sum_{r} \overline{c_{r a} (ω)} c_{r b} (ω)$ , with $c_{r a} (ω)$ in seconds $^{- 1 / 2}$ . Define $J_{r, ω} : = \sum_{b} c_{r b} (ω) S_{b} (ω)$ . Expanding yields $\sum_{r} J_{r, ω} X J_{r, ω}^{*} = \sum_{a, b} γ_{a b} (ω) S_{b} (ω) X S_{a} (ω)^{*}$ and $\sum_{r} J_{r, ω}^{*} J_{r, ω} = \sum_{a, b} γ_{a b} (ω) S_{a} (ω)^{*} S_{b} (ω)$ . Thus the displayed generator has the finite GKSL form of [F2] with Hamiltonian $H + H_{L S}$ and finite jumps $J_{r, ω}$ , so [F3] gives a finite channel semigroup; [F1] states its complete-positivity and trace-preservation content.

1.2givenA1F4algebra

From the spectral definition, $[H, S_{a} (ω)] = - ℏ ω S_{a} (ω)$ and $S_{a} (ω)^{*} = S_{a} (- ω)$ . Therefore $S_{a} (ω) e^{- β H} = e^{- β ℏ ω} e^{- β H} S_{a} (ω)$ . Every $S_{a} (- ω) S_{b} (ω)$ commutes with $H$ because its two energy shifts cancel, hence it commutes with each spectral projection of $H$ and with $σ_{β}$ . Also $[H_{L S}, H] = 0$ implies $H_{L S}$ preserves every eigenspace of $H$ : if $H x = E x$ , then $H (H_{L S} x) = H_{L S} (H x) = E (H_{L S} x)$ . Thus $H_{L S}$ commutes with every spectral projection of $H$ and with $σ_{β}$ . Finally, the exponential $e^{- β H}$ is positive definite with positive finite trace, making $σ_{β}$ a faithful density.

2.1givenA1step 1.2algebra

Fix $ω > 0$ and put $q = e^{- β ℏ ω}$ . The $ω$ contribution is $D_{ω} (σ_{β}) = σ_{β} \sum_{a, b} γ_{a b} (ω) (q S_{b} (ω) S_{a} (- ω) - S_{a} (- ω) S_{b} (ω))$ , using $S_{b} (ω) σ_{β} = q σ_{β} S_{b} (ω)$ and commutation of the anticommutator product with $σ_{β}$ . For the negative frequency, substitute $γ_{a b} (- ω) = q γ_{b a} (ω)$ and $S_{a} (- ω)^{*} = S_{a} (ω)$ , then reindex $a, b$ to obtain $D_{- ω} (σ_{β}) = σ_{β} \sum_{a, b} γ_{a b} (ω) (S_{a} (- ω) S_{b} (ω) - q S_{b} (ω) S_{a} (- ω)) = - D_{ω} (σ_{β})$ . Thus every nonzero frequency pair cancels.

3.1givenA1step 1.2step 2.1algebra∎

At $ω = 0$ , the balance relation gives $γ_{a b} (0) = γ_{b a} (0)$ . Each $S_{a} (0)$ is Hermitian and commutes with $H$ , hence with $σ_{β}$ ; consequently the zero-frequency contribution is $σ_{β} \sum_{a, b} γ_{a b} (0) (S_{b} (0) S_{a} (0) - S_{a} (0) S_{b} (0)) = 0$ , after swapping $a, b$ in the first finite sum. The Hamiltonian contribution vanishes because both $H$ and $H_{L S}$ commute with $σ_{β}$ . Together with step 2.1 this gives $L (σ_{β}) = 0$ , and the exponential series then gives $T_{t} (σ_{β}) = σ_{β}$ for all $t \geq 0$ .

Thermal balance makes Gibbs stationary

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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Sources