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Fixed-dissipator finite response and integrable kernel

Statement

Let $L_{0}$ be the finite Schrödinger generator trace-dual to $G$ in Specified finite GNS detailed balance, and assume the gap hypotheses of Faithful scalar-kernel detailed balance gives a relaxation gap with $d > 1$ . Let $A \in M_{d} (C)$ , $B = B^{*} \in M_{d} (C)$ be dimensionless observables, and let $h : [0, T] \to R$ be continuous with units joules. For dimensionless $ϵ \in R$ , perturb only the Hamiltonian by $H_{ϵ} (t) = H_{s} - ϵ h (t) B$ ; hold the dissipator fixed. Let $ρ_{t}^{ϵ}$ be the unique state solution with $ρ_{0}^{ϵ} = σ$ , supplied by Continuous finite GKSL channel propagators.

Define $A_{0} : = A - Tr (σ A) I$ and, for $τ \geq 0$ ,

$R_{A B} (τ) : = \frac{i}{ℏ} Tr ⁣ (σ [e^{τ G} A_{0}, B]) .$

Then, on every finite interval $[0, T]$ , the derivative at $ϵ = 0$ exists and

${\frac{d}{d ϵ} Tr (ρ_{t}^{ϵ} A) ∣}_{ϵ = 0} = \int_{0}^{t} h (s) R_{A B} (t - s) d s .$

The kernel is absolutely integrable on $[0, \infty)$ . If $μ$ is the smallest eigenvalue of $σ$ , then

$∣ R_{A B} (τ) ∣ \leq \frac{2 ∥ B ∥_{\infty}}{ℏ \sqrt{μ}} e^{- γ τ} ∥ A_{0} ∥_{σ} .$

This is an open-system model response with a fixed dissipator. It does not assert that the driven stationary state is the Gibbs state of $H_{s} - ϵ h B$ .

Facts & Assumptions

Given: The finite GKSL dynamics of $L_{0}$ , its trace-dual $G$ , faithful stationary density $σ$ , positive gap $γ$ , Hermitian $B$ , continuous field $h$ , and initial preparation $σ$ .

[F1]

Specified finite GNS detailed balance: $G$ is the trace-dual observable generator $i [H_{s}, \cdot] / ℏ + D$ , with $D (I) = 0$ and $[H_{s}, σ] = 0$ .

[F2]

Continuous finite GKSL channel propagators: a continuous finite GKSL family has a unique CPTP propagator on every compact time interval.

[F3]

Faithful scalar-kernel detailed balance gives a relaxation gap: $σ$ is stationary, centered observables decay in $∥ \cdot ∥_{σ}$ at least as $e^{- γ t}$ , and $σ$ is faithful.

[F4]

Finite matrix trace cyclicity and nondegeneracy: the finite trace is cyclic.

Proof

1.1givenF2F3algebra

Put $V (X) : = i [B, X] / ℏ$ . The perturbed Schrödinger generator is $L_{ϵ} (t) = L_{0} + ϵ h (t) V$ . Since $L_{0}$ leaves $σ$ stationary by [F3], variation of constants gives

ρ_{t}^{ϵ} = σ + ϵ \int_{0}^{t} h (s) e^{(t - s) L_{0}} V (σ) d s + ϵ^{2} \int_{0 \leq u \leq s \leq t} h (s) h (u) e^{(t - s) L_{0}} V e^{(s - u) L_{0}} V (ρ_{u}^{ϵ}) d u d s .

The second integral is $O (ϵ^{2})$ uniformly on $[0, T]$ : $h$ is bounded, $V$ is a bounded linear map on the finite matrix space, $\sup_{0 \leq s \leq t \leq T} ∥ e^{(t - s) L_{0}} ∥_{1} < \infty$ , and $∥ ρ_{u}^{ϵ} ∥_{1} = 1$ because [F2] gives a channel propagator. Thus the first derivative exists and equals $\int_{0}^{t} h (s) e^{(t - s) L_{0}} V (σ) d s$ .

2.1F1F4step 1.1algebra

By the generator duality in [F1], the exponential power series gives $Tr (A e^{τ L_{0}} X) = Tr (e^{τ G} A X)$ for every finite matrix $X$ . Using [F4] and $V (σ) = i [B, σ] / ℏ$ ,

Tr ⁣ (A e^{τ L_{0}} V (σ)) = \frac{i}{ℏ} Tr ⁣ (σ [e^{τ G} A, B]) .

The scalar part of $A$ commutes with $B$ , so $A$ may be replaced by $A_{0}$ . Substituting this identity into step 1.1 gives the stated response formula.

3.1F3step 2.1algebra∎

Let $C = e^{τ G} A_{0}$ . The gap theorem gives $∥ C ∥_{σ} \leq e^{- γ τ} ∥ A_{0} ∥_{σ}$ . If $μ > 0$ is the smallest eigenvalue of $σ$ , then

∥ C ∥_{\infty} \leq \frac{∥ C σ^{1 / 2} ∥_{2}}{\sqrt{μ}} = \frac{∥ C ∥_{σ}}{\sqrt{μ}} .

The finite density trace satisfies $∣ Tr (σ M) ∣ \leq ∥ M ∥_{\infty}$ : diagonalize $σ$ , write its nonnegative eigenvalues summing to one, and bound each diagonal matrix element of $M$ by $∥ M ∥_{\infty}$ . Therefore $∣ Tr (σ [C, B]) ∣ \leq 2 ∥ B ∥_{\infty} ∥ C ∥_{\infty}$ , which proves the exponential bound and absolute integrability.

Remarks

If the bath rates or jump operators depended on the perturbing field, a derivative of the dissipator would appear in step 1.1. It is absent here by the explicit fixed-dissipator hypothesis. The formula is not a universal fluctuation-dissipation theorem and is not an empirical observation.

Fixed-dissipator finite response and integrable kernel

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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