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Finite-time Kubo formula with norm remainder

Statement

Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ , $H = H^{*}$ with energy unit J, $ℏ > 0$ with unit J s, $β > 0$ with unit J $^{- 1}$ , and let $σ = Z_{β}^{- 1} e^{- β H}$ be the stationary Gibbs density from Finite unitary dynamics and stationary Gibbs states. Let $A = A^{*}$ and $B = B^{*}$ be matrices, let $T > 0$ be finite in s, let $h : [0, T] \to R$ be continuous, and let $ϵ \in R$ be dimensionless; $h (t) B$ has unit J. Put $U_{0} (t) = e^{- i t H / ℏ}$ for time $t$ in s, $α_{t} (A) = U_{0} (t)^{*} A U_{0} (t)$ , and $B_{t} = α_{t} (B)$ . Let $W_{ϵ}$ solve $W_{ϵ}^{'} (t) = \frac{i ϵ h (t)}{ℏ} B_{t} W_{ϵ} (t), W_{ϵ} (0) = I,$ and set $U_{ϵ} (t) = U_{0} (t) W_{ϵ} (t)$ and $ρ_{I} (t) = W_{ϵ} (t) σ W_{ϵ} (t)^{*}$ . Then $U_{ϵ}$ solves $i ℏ U_{ϵ}^{'} (t) = (H - ϵ h (t) B) U_{ϵ} (t)$ , $U_{ϵ} (0) = I$ , and $ρ_{ϵ} (t) : = U_{ϵ} (t) σ U_{ϵ} (t)^{*}$ satisfies $Tr (ρ_{ϵ} (t) A) = Tr (ρ_{I} (t) α_{t} (A))$ . Define $R_{A B} (u) = \frac{i}{ℏ} Tr ⁣ (σ [α_{u} (A), B]), C_{T} = \frac{2}{ℏ} \int_{0}^{T} ∣ h (s) ∣ ∥ B ∥ d s .$ Then $W_{ϵ} (t)$ is unitary and, for $0 \leq t \leq T$ , $Tr (ρ_{I} (t) α_{t} (A)) = Tr (σ α_{t} (A)) + ϵ \int_{0}^{t} h (s) R_{A B} (t - s) d s + r_{ϵ} (t), ∣ r_{ϵ} (t) ∣ \leq ∥ A ∥ \frac{(∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}}}{2} .$ The initial Gibbs matrix is fixed as $ϵ$ varies. The result is an exact finite-time first variation with a controlled remainder, not a static derivative of a re-prepared Gibbs family. $C_{T}$ and $ϵ$ are dimensionless, and $R_{A B}$ has units $[A] [B] / (J s)$ .

Facts & Assumptions

Given: The finite-dimensional Hamiltonian and Gibbs density, bounded self-adjoint matrices $A, B$ , continuous real $h$ , $ϵ \in R$ , and $T > 0$ .

[F1]

Finite unitary dynamics and stationary Gibbs states: $U_{0}$ is a unitary group, $σ$ is a positive trace-one matrix commuting with $U_{0} (t)$ , and finite trace expectation is cyclically represented in the Heisenberg picture.

[F2]

Finite matrix spectral and state calculus: finite Hermitian matrices have basis-independent spectral functional calculus and the finite trace is cyclic.

[F3]

Finite matrix trace-norm ideal inequalities: $∥ Y X ∥_{1}, ∥ X Y ∥_{1} \leq ∥ Y ∥ ∥ X ∥_{1}$ , $∣ Tr X ∣ \leq ∥ X ∥_{1}$ , and the finite trace-norm matrix space is complete.

[F4]

Banach-space Volterra and Dyson series with a norm remainder: a strongly continuous bounded coefficient on a Banach space yields a unique Volterra solution and the explicit quadratic remainder; a strongly continuous skew-adjoint Hilbert-space coefficient yields a unitary solution.

Proof

technique · interaction-picture Volterra series and finite trace cyclicity

Given: [F1]–[F4] and the control/observable hypotheses in the Statement.

1.1givenF1F3algebra

The matrix curve $t \mapsto B_{t}$ is norm-continuous by the finite matrix exponential in [F1]. On the Hilbert space, $K_{t} (ψ) = i h (t) B_{t} ψ / ℏ$ is a strongly continuous skew-adjoint bounded coefficient. On the trace-norm matrix space, $L_{t} (X) = i h (t) [B_{t}, X] / ℏ$ is strongly continuous and satisfies $∥ L_{t} (X) ∥_{1} \leq 2 ∣ h (t) ∣ ∥ B ∥ ∥ X ∥_{1} / ℏ$ by [F3].

2.1givenF1F3F4step 1.1algebra

Apply [F4] to $K_{t}$ with initial vector $ψ$ to obtain a unitary $W_{ϵ} (t)$ , and apply [F4] to $L_{t}$ with initial matrix $σ$ . The trace-norm solution has linear term $ϵ (i / ℏ) \int_{0}^{t} h (s) [B_{s}, σ] d s$ and remainder norm at most $(∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}} / 2$ , since $∥ σ ∥_{1} = 1$ . Differentiating the finite matrix product $W_{ϵ} (t) σ W_{ϵ} (t)^{*}$ shows it solves the same commutator equation; uniqueness in [F4] identifies it with that trace-norm solution.

3.1givenF1F2step 2.1algebra

The product rule and $U_{0} (t) B_{t} = B U_{0} (t)$ give $i ℏ (U_{0} W_{ϵ})^{'} = H U_{0} W_{ϵ} - ϵ h (t) B U_{0} W_{ϵ}$ , so $U_{ϵ}$ solves the displayed controlled Schrödinger equation. Since $U_{ϵ} = U_{0} W_{ϵ}$ , cyclicity of the finite trace gives $Tr (U_{ϵ} σ U_{ϵ}^{*} A) = Tr (W_{ϵ} σ W_{ϵ}^{*} α_{t} (A))$ .

3.2givenF1F2step 2.1algebra

Applying $Tr (\cdot α_{t} (A))$ to the linear term, cyclicity [F2] gives $Tr ([B_{s}, σ] α_{t} (A)) = Tr (σ [α_{t} (A), B_{s}])$ . Since $B_{s} = α_{s} (B)$ and $α_{t} (A) = α_{s} (α_{t - s} (A))$ , stationarity of $σ$ from [F1] turns this into $Tr (σ [α_{t - s} (A), B])$ . This is exactly the convolution in the Statement.

4.1F1F2F3step 2.1step 3.2algebra∎

The trace-norm remainder from step 2.1 changes the expectation by at most $∥ A ∥$ times its trace norm, by [F3] and $∥ α_{t} (A) ∥ = ∥ A ∥$ . This gives the displayed bound. For self-adjoint $A, B$ , $Tr (σ [α_{u} (A), B])$ is purely imaginary by cyclicity and adjoints, so $R_{A B} (u)$ is real.

Finite-time Kubo formula with norm remainder

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources