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Trace-class bounded-observable response

1 · Prerequisites

2 · Summary

This page separates finite-matrix response, trace-class normal-state response, and response on the quasi-local algebra of finite-range quantum spin systems. It proves the bounded-observable Kubo formula with an explicit finite-time remainder, records the precise graph-domain conditions needed for a strong Schrödinger solution, and derives the finite spectral fluctuation–dissipation relation with its zero-frequency qualification.

The infinite-Hilbert-space results use trace-class densities and bounded observables only; they do not use an unbounded trace product or assume finite energy or entropy moments. The infinite-spin result instead uses an invariant algebraic state on the actual finite-range automorphism group, not a global trace Hamiltonian. Neither a spatial finite-range bound nor isolated pure-point dynamics supplies temporal decay or a dc transport coefficient. The two physical postulates are adopted model assumptions, and the prepared-control physics theorem is a conditional prediction rather than an empirical observation.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Banach-space Volterra and Dyson series with a norm remainder

Statement

Let $X$ be a complex Banach space, let $T > 0$ , and let $B (X)$ be the bounded complex-linear operators on $X$ with operator norm $∥ L ∥ : = \sup {∥ L x ∥ : x \in X, ∥ x ∥ \leq 1}$ . Let $K : [0, T] \to B (X)$ be strongly continuous, meaning $t \mapsto K (t) x$ is norm-continuous for every fixed $x \in X$ . Suppose a continuous function $a : [0, T] \to [0, \infty)$ satisfies $∥ K (t) ∥ \leq a (t)$ . Set $c_{T} = \int_{0}^{T} a (s) d s$ , where the scalar integral is the Riemann integral. Define $Δ_{0} (t) = x_{0}$ and recursively $Δ_{n} (t) = \int_{0}^{t} K (s) Δ_{n - 1} (s) d s$ for $n \geq 1$ , using the norm-Riemann integral in $X$ .

For each $ϵ \in C$ , the series $x_{ϵ} (t) = \sum_{n = 0}^{\infty} ϵ^{n} Δ_{n} (t)$ converges uniformly on $[0, T]$ . It is the unique continuously differentiable solution of $x_{ϵ}^{'} (t) = ϵ K (t) x_{ϵ} (t)$ and $x_{ϵ} (0) = x_{0}$ , and

$∥ x_{ϵ} (t) - x_{0} - ϵ \int_{0}^{t} K (s) x_{0} d s ∥ \leq ∥ x_{0} ∥ \frac{(∣ ϵ ∣ c_{T})^{2} e^{∣ ϵ ∣ c_{T}}}{2} .$

If $X$ is a unital complex $C^{*}$ -algebra and $K (t) x = b (t) x$ for a norm-continuous curve $b : [0, T] \to X$ of skew-adjoint elements, then for real $ϵ$ the solution with $x_{0} = 1$ is unitary. If $X = H$ is a complex Hilbert space and each $K (t)$ is a bounded skew-adjoint operator, then for real $ϵ$ the solution operators $W_{ϵ} (t) : x_{0} \mapsto x_{ϵ} (t)$ are unitary even when $K (t)$ is only strongly continuous.

Facts & Assumptions

Given: A complex Banach space $X$ , a strongly continuous bounded-operator family $K (t)$ , a continuous nonnegative majorant $a (t)$ , and $x_{0} \in X$ .

[F1]

Riemann integration of continuous curves in a complete normed space, with the fundamental theorem of calculus and coefficient functionals: a continuous curve in a complete normed real vector space has a norm-Riemann integral; the integral is linear, additive, commutes with bounded linear maps, obeys the norm bound obtained from Riemann sums, and its indefinite integral has derivative equal to the integrand. A complex Banach space is used as its underlying real Banach space.

[F2]

C star algebra: a complex $C^{*}$ -algebra is a complete normed $*$ -algebra with submultiplicative norm and $∥ z^{*} z ∥ = ∥ z ∥^{2}$ ; the unital case has an identity element.

Proof

Given: The Banach space, coefficient family, majorant and initial value in the Statement.

1.1givenF1construct

If $x : [0, T] \to X$ is continuous, then $s \mapsto K (s) x (s)$ is continuous: for $s \to t$ , its difference from $K (t) x (t)$ is bounded by $\sup_{r} ∥ K (r) ∥ ∥ x (s) - x (t) ∥ + ∥ K (s) x (t) - K (t) x (t) ∥$ , and both terms tend to zero. By [F1], the norm-Riemann integral defining $Δ_{n + 1}$ therefore exists whenever $Δ_{n}$ is continuous. Moreover, for a continuous $f : [0, t] \to X$ , the triangle inequality on each tagged Riemann sum and convergence of the scalar Riemann sums give $∥ \int_{0}^{t} f (s) d s ∥ \leq \int_{0}^{t} ∥ f (s) ∥ d s$ .

2.1givenstep 1.1F1algebra

Put $Θ (t) = \int_{0}^{t} a (s) d s$ . By [F1], $Θ$ is continuously differentiable and $Θ^{'} = a$ . Inductively the curves $Δ_{n}$ are continuous and satisfy $∥ Δ_{n} (t) ∥ \leq ∥ x_{0} ∥ Θ (t)^{n} / n!$ : it is equality as a bound for $n = 0$ , and if it holds at $n$ , then $∥ Δ_{n + 1} (t) ∥ \leq \int_{0}^{t} a (s) ∥ Δ_{n} (s) ∥ d s \leq ∥ x_{0} ∥ \int_{0}^{t} a (s) Θ (s)^{n} / n! d s = ∥ x_{0} ∥ Θ (t)^{n + 1} / (n + 1)!$ , where the last equality follows by differentiating $Θ^{n + 1} / (n + 1)!$ .

3.1step 1.1step 2.1F1algebra

Since $Θ (t) \leq c_{T}$ , the bound in step 2.1 gives uniform absolute convergence of $\sum_{n} ϵ^{n} Δ_{n} (t)$ . Uniform convergence, the integral bound in step 1.1, and the recursion show $x_{ϵ} (t) = x_{0} + ϵ \int_{0}^{t} K (s) x_{ϵ} (s) d s$ ; [F1] then gives the differential equation and initial value. If two solutions differ by $z$ , repeated substitution in $z (t) = ϵ \int_{0}^{t} K (s) z (s) d s$ gives $∥ z (t) ∥ \leq \sup_{[0, T]} ∥ z ∥ (∣ ϵ ∣ c_{T})^{n} / n!$ for every $n$ , whose right side tends to zero; hence the solution is unique.

4.1step 2.1step 3.1algebra

With $y = ∣ ϵ ∣ c_{T}$ , the tail after its linear term is at most $∥ x_{0} ∥ \sum_{n \geq 2} y^{n} / n! \leq ∥ x_{0} ∥ y^{2} e^{y} / 2$ , because $n! \geq 2 (n - 2)!$ for every $n \geq 2$ . This is the stated remainder.

4.2givenF2step 3.1algebra

In the $C^{*}$ -algebra case, write $W = x_{ϵ}$ and assume $b (t)^{*} = - b (t)$ and $ϵ \in R$ . The product rule and $W^{'} = ϵ b W$ give $(W^{*} W)^{'} = ϵ W^{*} (b^{*} + b) W = 0$ , so $W^{*} W = 1$ . Also $Z = W W^{*}$ satisfies $Z^{'} = ϵ (b Z - Z b)$ and $Z (0) = 1$ . The uniqueness argument of step 3.1, applied to the bounded operator family $Z \mapsto ϵ (b (t) Z - Z b (t))$ , shows that the constant curve $Z = 1$ is its only solution. Hence $W W^{*} = 1$ as well, so $W$ is unitary in the unital $C^{*}$ -algebra.

5.1givenstep 2.1step 3.1algebra∎

In the Hilbert-space case, the series is linear in $x_{0}$ and step 2.1 gives $∥ W_{ϵ} (t) x_{0} ∥ \leq e^{∣ ϵ ∣ c_{T}} ∥ x_{0} ∥$ , so it defines a bounded operator. Differentiating inner products of two solutions gives $\frac{d}{d t} ⟨ W_{ϵ} (t) x, W_{ϵ} (t) y ⟩ = ϵ ⟨ K (t) W_{ϵ} (t) x, W_{ϵ} (t) y ⟩ + ϵ ⟨ W_{ϵ} (t) x, K (t) W_{ϵ} (t) y ⟩ = 0$ , hence $W_{ϵ} (t)$ is an isometry. For a fixed terminal time $t$ , solve $z^{'} (s) = - ϵ K (t - s) z (s)$ on $[0, t]$ with arbitrary $z (0) = y$ using the first part; then $x (s) = z (t - s)$ solves the forward equation and $W_{ϵ} (t) x (0) = y$ . Thus $W_{ϵ} (t)$ is onto and therefore unitary.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Finite matrix trace-norm ideal inequalities

Statement

Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ . For a matrix $X$ define $∣ X ∣ = (X^{*} X)^{1 / 2}$ by finite Hermitian functional calculus and define its trace norm by $∥ X ∥_{1} = Tr ∣ X ∣$ . This is a complete norm on the finite-dimensional matrix space. Write $∥ Y ∥ : = \sup {∥ Y v ∥ : ∥ v ∥ \leq 1}$ for the operator norm induced by the Hilbert-space norm. For all matrices $X, Y$ ,

∥ Y X ∥_{1} \leq ∥ Y ∥ ∥ X ∥_{1}, ∥ X Y ∥_{1} \leq ∥ Y ∥ ∥ X ∥_{1}, ∣ Tr X ∣ \leq ∥ X ∥_{1} .

For unitary $U$ , $∥ U X U^{*} ∥_{1} = ∥ X ∥_{1}$ .

Facts & Assumptions

Given: A complex inner-product space $H$ of finite dimension $d \geq 1$ and matrices $X, Y$ .

[F1]

Finite matrix spectral and state calculus: every positive Hermitian matrix has a basis-independent orthonormal spectral decomposition and positive square root; the finite trace is basis independent.

Proof

technique · direct

Given: The finite-dimensional Hilbert space and matrices above.

1.1givenF1construct

Diagonalize $X^{*} X$ by [F1], with nonnegative eigenvalues $s_{j}^{2}$ and orthonormal eigenvectors $v_{j}$ . For each $s_{j} > 0$ put $u_{j} = X v_{j} / s_{j}$ ; then $⟨ u_{i}, u_{j} ⟩ = δ_{i j}$ because $⟨ X v_{i}, X v_{j} ⟩ = ⟨ v_{i}, X^{*} X v_{j} ⟩$ . Extending both finite orthonormal families to bases gives the singular-value expansion $X = \sum_{s_{j} > 0} s_{j} R_{u_{j}, v_{j}}$ , where $R_{u, v} x = ⟨ x, v ⟩ u$ . The only nonzero singular value of $R_{u, v}$ is $∥ u ∥ ∥ v ∥$ , so $∥ R_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ and $∥ X ∥_{1} = \sum_{s_{j} > 0} s_{j}$ .

2.1step 1.1F1algebra

The rank-one trace formula is $Tr (R_{u, v}) = ⟨ u, v ⟩$ , so the expansion in step 1.1 gives $∥ X ∥_{1} = \sup_{U unitary} ∣ Tr (U X) ∣$ . For the upper bound, if $u_{j}, v_{j}$ are unit vectors, then $∥ U u_{j} - ⟨ U u_{j}, v_{j} ⟩ v_{j} ∥^{2} = 1 - ∣ ⟨ U u_{j}, v_{j} ⟩ ∣^{2} \geq 0$ , so $∣ ⟨ U u_{j}, v_{j} ⟩ ∣ \leq 1$ . Equality is attained by a unitary mapping each $u_{j}$ to $v_{j}$ , extended from these finite orthonormal families to bases. The supremum formula implies the triangle inequality $∥ X + Y ∥_{1} \leq ∥ X ∥_{1} + ∥ Y ∥_{1}$ and homogeneity; positivity and definiteness follow from the singular values.

3.1step 1.1step 2.1algebra

Applying step 1.1 to $Y X = \sum_{j} s_{j} R_{Y u_{j}, v_{j}}$ and using the triangle inequality of step 2.1 gives $∥ Y X ∥_{1} \leq \sum_{j} s_{j} ∥ Y u_{j} ∥ ∥ v_{j} ∥ \leq ∥ Y ∥ \sum_{j} s_{j} = ∥ Y ∥ ∥ X ∥_{1}$ . Similarly $X Y = \sum_{j} s_{j} R_{u_{j}, Y^{*} v_{j}}$ gives $∥ X Y ∥_{1} \leq ∥ Y ∥ ∥ X ∥_{1}$ . The rank-one trace formula and step 1.1 also give $∣ Tr X ∣ \leq \sum_{j} s_{j} ∣ ⟨ u_{j}, v_{j} ⟩ ∣ \leq ∥ X ∥_{1}$ .

4.1step 3.1algebra∎

Define the Frobenius norm by $∥ X ∥_{F} = (\sum_{i, j} ∣ X_{i j} ∣^{2})^{1 / 2}$ in an orthonormal basis. The finite singular-value expansion gives $∥ X ∥_{F} = (\sum_{j} s_{j}^{2})^{1 / 2}$ . Since $s_{j} \geq 0$ , $(\sum_{j} s_{j}^{2})^{1 / 2} \leq \sum_{j} s_{j}$ ; also $d \sum_{j} s_{j}^{2} - (\sum_{j} s_{j})^{2} = \sum_{i < j} (s_{i} - s_{j})^{2} \geq 0$ . Thus $∥ X ∥_{F} \leq ∥ X ∥_{1} \leq \sqrt{d} ∥ X ∥_{F}$ . A trace-norm Cauchy sequence is therefore Cauchy in each matrix coordinate and converges to a matrix; the second inequality gives convergence in trace norm. This proves completeness. For unitary $U$ , step 3.1 gives $∥ U X ∥_{1} \leq ∥ X ∥_{1}$ ; applying it to $U^{*}$ and $U X$ gives the reverse inequality. The same argument on the right proves $∥ U X U^{*} ∥_{1} = ∥ X ∥_{1}$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Lorentzian kernels converge to delta and principal-value distributions

Statement

Let $ψ : R \to C$ be continuously differentiable with compact support. For $η > 0$ , define $P_{η} (x) = η / (x^{2} + η^{2})$ and $Q_{η} (x) = x / (x^{2} + η^{2})$ . Define $PV ⁣ \int_{R} \frac{ψ (x)}{x} d x : = \lim_{δ ↓ 0} \int_{∣ x ∣ > δ} \frac{ψ (x)}{x} d x$ when this symmetric limit exists. Then $\lim_{η ↓ 0} \int_{R} P_{η} (x) ψ (x) d x = π ψ (0), \lim_{η ↓ 0} \int_{R} Q_{η} (x) ψ (x) d x = PV ⁣ \int_{R} \frac{ψ (x)}{x} d x .$ Thus $P_{η}$ converges distributionally to $π δ_{0}$ , and $Q_{η}$ converges distributionally to the Cauchy principal-value distribution $PV (1 / x)$ , with the stated test-function convention.

Facts & Assumptions

Given: A compactly supported $C^{1}$ complex test function $ψ$ and the Lebesgue measure on $R$ .

[F1]

Dominated convergence: almost-everywhere convergence under one integrable measurable majorant implies convergence of the integrals.

Proof

Given: The test function and kernels in the Statement.

1.1givenF1algebra

With $x = η y$ , the first integral is $\int_{R} ψ (η y) / (1 + y^{2}) d y$ . Its integrand converges pointwise to $ψ (0) / (1 + y^{2})$ and is bounded in modulus by $∥ ψ ∥_{\infty} / (1 + y^{2})$ , an integrable function. By [F1] the limit is $ψ (0) \int_{R} (1 + y^{2})^{- 1} d y = π ψ (0)$ ; the last integral is $\arctan (y) ∣_{- \infty}^{\infty} = π$ .

1.2givenF1algebra

Let $L = \sup_{x} ∣ ψ^{'} (x) ∣ < \infty$ and set $q (x) = ψ (x) - ψ (0) 1_{[- 1, 1]} (x)$ . The integral of $Q_{η} (x) ψ (0) 1_{[- 1, 1]} (x)$ is zero because its integrand is odd on a symmetric interval. For $0 < ∣ x ∣ \leq 1$ , $∣ Q_{η} (x) q (x) ∣ \leq L x^{2} / (x^{2} + η^{2}) \leq L$ by the mean-value inequality; for $∣ x ∣ > 1$ , $∣ Q_{η} (x) q (x) ∣ \leq ∣ ψ (x) ∣$ . This gives an integrable majorant independent of $η$ , and for $x \neq 0$ the integrand tends to $q (x) / x$ . Applying [F1] yields $\int_{R} q (x) / x d x$ .

2.1step 1.1step 1.2algebra∎

For $0 < δ < 1$ , the constant part $ψ (0) / x$ integrates to zero on ${δ < ∣ x ∣ \leq 1}$ by oddness. Therefore $\int q (x) / x d x = \lim_{δ ↓ 0} \int_{∣ x ∣ > δ} ψ (x) / x d x$ , proving existence of the symmetric principal value and the second limit. The two distributional conclusions are precisely the limits already established in steps 1.1 and 1.2.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Stationary-state covariance identity for time-integrated observables

Statement

Let $A$ be a unital complex C $^{*}$ -algebra, and let $(α_{t})_{t \in R}$ be a group of $*$ -automorphisms of $A$ , so $α_{0} = {id}_{A}$ and $α_{s + t} = α_{s} \circ α_{t}$ . Assume point-norm continuity: for each $X \in A$ , the orbit map $t \mapsto α_{t} (X)$ is norm-continuous. Let $ϕ : A \to C$ be a bounded linear functional with $ϕ (I) = 1$ , $ϕ (X^{*} X) \geq 0$ for every $X \in A$ , and $ϕ \circ α_{t} = ϕ$ for every $t$ . For $m \geq 1$ , take self-adjoint $V_{1}, \dots, V_{m} \in A$ with $ϕ (V_{a}) = 0$ . For $T > 0$ , regard $A$ as its underlying real Banach space with the same norm for these time integrals, and define the norm-Riemann integrals $X_{a, T} = \int_{0}^{T} α_{t} (V_{a}) d t,$ and for $s \geq 0$ define $G_{a b} (s) = \frac{1}{2} (Re ϕ (V_{a} α_{s} (V_{b})) + Re ϕ (V_{b} α_{s} (V_{a}))) .$ Then each $X_{a, T}$ is self-adjoint and $\frac{1}{2 T} Re ϕ (X_{a, T} X_{b, T}) = \int_{0}^{T} (1 - \frac{s}{T}) G_{a b} (s) d s .$ The real symmetric matrix with entries on the left is positive semidefinite.

Facts & Assumptions

Given: The unital C $^{*}$ -algebra, strongly continuous automorphism group, invariant state, self-adjoint centered observables, and $T > 0$ in the Statement.

[F1]

C star algebra: $A$ is complete, multiplication is bounded, and the involution is continuous.

[F2]

Riemann integration of continuous curves in a complete normed space, with the fundamental theorem of calculus and coefficient functionals: continuous curves in a complete normed real vector space have norm-Riemann integrals, with linearity, additive subdivision, bounded-functional interchange, and the norm estimate from Riemann sums; for these integrals, the complex algebra $A$ is viewed as its underlying real Banach space.

Proof

technique · direct double integration over the time square, followed by stationarity and positivity

Given: [F1]–[F2], the bounded normalized positive functional, and the invariant group in the Statement.

1.1givenF1F2algebra

The functional is Hermitian: if $H = H^{*}$ then $ϕ ((I + r H)^{*} (I + r H)) = 1 + 2 r ϕ (H) + r^{2} ϕ (H^{2})$ is real for every real $r$ , so $ϕ (H) \in R$ ; decomposing $X = (X + X^{*}) / 2 + i (X - X^{*}) / (2 i)$ gives $ϕ (X^{*}) = \overline{ϕ (X)}$ . For each $a$ , $t \mapsto α_{t} (V_{a})$ is norm-continuous, so [F2] defines $X_{a, T}$ . Riemann sums of self-adjoint elements are self-adjoint, and continuity of the involution in [F1] passes this property to the limit. The product map $(s, t) \mapsto α_{s} (V_{a}) α_{t} (V_{b})$ is norm-continuous on $[0, T]^{2}$ by continuity of multiplication; approximating each integral by Riemann sums and multiplying the sums shows $X_{a, T} X_{b, T}$ equals the iterated integral of this product.

2.1givenF1F2step 1.1algebra

Invariance of $ϕ$ and the group law give $ϕ (α_{s} (V_{a}) α_{t} (V_{b})) = ϕ (V_{a} α_{t - s} (V_{b}))$ for every $s, t \in [0, T]$ . The scalar integrand is continuous; splitting the square into the triangles $t \geq s$ and $s \geq t$ , and integrating the slices with fixed $u = ∣ t - s ∣$ , whose lengths are $T - u$ , yields $ϕ (X_{a, T} X_{b, T}) = \int_{0}^{T} (T - u) ϕ (V_{a} α_{u} (V_{b})) d u + \int_{0}^{T} (T - u) ϕ (V_{a} α_{- u} (V_{b})) d u .$ The diagonal has zero area and does not affect the Riemann integral. The slice formula follows from the one-variable substitution $u = t - s$ on each triangle and the additivity of the Riemann integral.

3.1step 1.1step 2.1algebra

For $u \geq 0$ , invariance gives $ϕ (V_{a} α_{- u} (V_{b})) = ϕ (α_{u} (V_{a}) V_{b}) = \overline{ϕ (V_{b} α_{u} (V_{a}))}$ , where the last equality uses self-adjointness and step 1.1. Taking real parts in step 2.1 therefore gives the stated identity after dividing by $2 T$ .

4.1givenstep 1.1algebra∎

For any real vector $c = (c_{1}, \dots, c_{m})$ , $X_{c, T} = \sum_{a} c_{a} X_{a, T}$ is self-adjoint and $\sum_{a, b} c_{a} c_{b} Re ϕ (X_{a, T} X_{b, T}) = ϕ (X_{c, T}^{2}) = ϕ (X_{c, T}^{*} X_{c, T}) \geq 0$ by positivity. Hermiticity from step 1.1 gives symmetry of the real matrix, so it is positive semidefinite.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Trace-class bounded-observable response

Statement

Assume the Axiom of Choice. Let $H \neq {0}$ be a separable complex Hilbert space, let $H = H^{*}$ be a densely defined operator with domain $D (H)$ , and let $ℏ > 0$ . Put $U_{0} (t) = e^{- i t H / ℏ}$ . Let $ρ$ be a density operator, meaning a positive trace-class operator with $Tr ρ = 1$ as in Trace-class normal density states, and assume $ρ U_{0} (t) = U_{0} (t) ρ$ for every $t \in R$ . Let $A, B \in B (H)$ be bounded self-adjoint operators, let $T > 0$ be a finite time in s, let $h : [0, T] \to R$ be continuous, and let $ϵ \in R$ be dimensionless. Interpret $H$ in J, $ℏ$ in J s, and $h (t) B$ in J; if $A, B$ carry units $[A], [B]$ , then $R_{A B}$ has units $[A] [B] / (J s)$ .

Define $B_{s} = U_{0} (s)^{*} B U_{0} (s)$ , $α_{t} (A) = U_{0} (t)^{*} A U_{0} (t)$ , and

R_{A B} (u) = \frac{i}{ℏ} Tr ⁣ (ρ [α_{u} (A), B]), C_{T} = \frac{2}{ℏ} \int_{0}^{T} ∣ h (s) ∣ ∥ B ∥ d s .

The trace-class interaction-picture solution of $ρ_{I}^{'} (t) = \frac{i ϵ h (t)}{ℏ} [B_{t}, ρ_{I} (t)]$ , $ρ_{I} (0) = ρ$ , satisfies

Tr (ρ_{I} (t) α_{t} (A)) = Tr (ρ α_{t} (A)) + ϵ \int_{0}^{t} h (s) R_{A B} (t - s) d s + r_{ϵ} (t),

where $∣ r_{ϵ} (t) ∣ \leq ∥ A ∥ (∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}} / 2$ for $0 \leq t \leq T$ . Thus the bounded-observable first variation is exact, with an explicit finite-time remainder. No energy or entropy moment of $ρ$ is assumed, and no trace of an unbounded product such as $ρ H A$ is used.

Facts & Assumptions

Given: Choice, the Hilbert space, self-adjoint $H$ , $ℏ > 0$ , trace-class density $ρ$ , bounded self-adjoint $A, B$ , continuous real control $h$ , and finite $T$ .

[A1]

The Axiom of Choice: the full Axiom of Choice is assumed for the unbounded spectral theorem used to define $U_{0} (t)$ .

[F1]

Trace-class normal density states: a density operator is positive, trace class, and has trace one; its state pairing with a bounded operator is $ω_{ρ} (A) = Tr (ρ A)$ .

[F2]

A self-adjoint operator generates a strongly continuous unitary group: under [A1], $U_{0} (t) = e^{- i t H / ℏ}$ is a strongly continuous unitary group, preserves $D (H)$ , and differentiates there; the exact scale change from $e^{i s H}$ is $s = - t / ℏ$ .

[F3]

Trace class is a two sided Banach operator ideal: trace-class operators are complete in trace norm, bounded left/right multiplication is trace-class, and $∥ A X B ∥_{1} \leq ∥ A ∥ ∥ X ∥_{1} ∥ B ∥$ .

[F4]

Cyclicity of the trace: for trace-class $X$ and bounded $Y$ , $Tr (Y X) = Tr (X Y)$ .

[F5]

Trace is absolutely convergent and basis independent: the trace is linear and basis independent, and $∣ Tr X ∣ \leq ∥ X ∥_{1}$ .

[F6]

Banach-space Volterra and Dyson series with a norm remainder: a strongly continuous bounded coefficient on a Banach space has a unique norm-convergent Volterra series solution with the stated quadratic remainder; its Hilbert-space clause gives a unitary solution operator for a strongly continuous skew-adjoint coefficient.

[F7]

Nuclear series characterizes trace norm: each trace-class operator has a nuclear singular-value representation with summable singular values. Its finite-rank partial sums $X_{N}$ satisfy $∥ X - X_{N} ∥_{1} \leq \sum_{j > N} s_{j} (X) \to 0$ : the tail is a nuclear representation, and the supplier's infimum formula bounds its trace norm by that tail sum.

[F8]

Trace of a positive operator is the sum of its eigenvalues: for a positive trace-class operator $ρ$ , $Tr ρ = ∥ ρ ∥_{1}$ .

Proof

technique · direct

Given: [A1], [F1]–[F6], and the stationarity condition $ρ U_{0} (t) = U_{0} (t) ρ$ .

1.1givenA1F2F3F7algebra

For a rank-one operator $R_{u, v} x = ⟨ x, v ⟩ u$ , one has $∥ R_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ , and $∥ R_{u, v} - R_{u^{'}, v^{'}} ∥_{1} \leq ∥ u - u^{'} ∥ ∥ v ∥ + ∥ u^{'} ∥ ∥ v - v^{'} ∥$ . Strong continuity in [F2] therefore makes $t \mapsto U_{0} (t) R_{u, v} U_{0} (t)^{*}$ trace-norm continuous. Finite sums have the same property. By [F7], finite-rank SVD truncations $X_{N}$ of any trace-class $X$ satisfy $∥ X - X_{N} ∥_{1} \to 0$ . The ideal estimate in [F3] and $∥ U_{0} (t) ∥ = ∥ U_{0} (t)^{*} ∥ = 1$ bound the conjugated tails uniformly by $∥ X - X_{N} ∥_{1}$ , so $t \mapsto U_{0} (t) X U_{0} (t)^{*}$ is trace-norm continuous.

1.2givenF2F3F7algebra

For finite-rank $X_{N} = \sum_{j = 1}^{m} R_{u_{j}, v_{j}}$ , $B_{t} X_{N} = \sum_{j} R_{B_{t} u_{j}, v_{j}}$ and $X_{N} B_{t} = \sum_{j} R_{u_{j}, B_{t} v_{j}}$ because $B_{t} = B_{t}^{*}$ . Strong continuity of $B_{t}$ makes both curves trace-norm continuous. For general trace-class $X$ , take the finite-rank approximants from [F7] and use $∥ B_{t} (X - X_{N}) ∥_{1}, ∥ (X - X_{N}) B_{t} ∥_{1} \leq ∥ B ∥ ∥ X - X_{N} ∥_{1}$ uniformly in $t$ . Thus $X \mapsto \frac{i h (t)}{ℏ} [B_{t}, X]$ is strongly continuous on the trace-class Banach space and has norm bound $2 ∣ h (t) ∣ ∥ B ∥ / ℏ$ .

1.3givenF1F3F6F7F8algebra

Apply [F6] on the trace-class Banach space with $K_{t} (X) = \frac{i h (t)}{ℏ} [B_{t}, X]$ and $x_{0} = ρ$ . By [F1] and [F8], $ρ$ is positive, has trace one, and satisfies $∥ ρ ∥_{1} = 1$ ; its Volterra solution has the claimed trace-norm linear approximation and remainder bound with $C_{T} = \frac{2}{ℏ} \int_{0}^{T} ∣ h (s) ∣ ∥ B ∥ d s$ . The Hilbert clause of [F6], applied to $K_{t} = i h (t) B_{t} / ℏ$ on $H$ , gives a unitary solution operator $W (t) = W_{ϵ} (t)$ satisfying $W^{'} (t) u = ϵ i h (t) B_{t} W (t) u / ℏ$ for every $u \in H$ . For a rank-one $R_{u, v}$ , $W (t) R_{u, v} W (t)^{*} = R_{W (t) u, W (t) v}$ is trace-norm differentiable and its derivative is $ϵ K_{t} (W (t) R_{u, v} W (t)^{*})$ , by the vector equation and skew-adjointness. Finite sums give the same integral equation. Approximate $ρ$ by finite-rank $ρ_{N}$ in trace norm using [F7]. Unitary invariance of trace norm follows from the ideal estimate [F3] applied to $W (t)$ and $W (t)^{*}$ , both of norm one; hence $W (t) ρ_{N} W (t)^{*} \to W (t) ρ W (t)^{*}$ uniformly, while the right-hand integral errors are at most $∣ ϵ ∣ C_{T} ∥ ρ - ρ_{N} ∥_{1}$ . Thus $W (t) ρ W (t)^{*}$ solves the trace-class Volterra equation, and uniqueness in [F6] identifies it with $ρ_{I} (t)$ .

2.1givenF2F4step 1.3algebra

The linear term in the expectation is $\frac{i ϵ}{ℏ} \int_{0}^{t} h (s) Tr ([B_{s}, ρ] α_{t} (A)) d s$ . By [F4], the trace integrand equals $Tr (ρ [α_{t} (A), B_{s}])$ . The group law and stationarity give $[α_{t} (A), B_{s}] = α_{s} ([α_{t - s} (A), B])$ and $Tr (ρ α_{s} (C)) = Tr (ρ C)$ for bounded $C$ . Hence the integrand is $Tr (ρ [α_{t - s} (A), B])$ , producing the stated convolution and response kernel.

3.1givenF3F4F5step 1.3step 2.1algebra∎

The trace-norm remainder in step 1.3 changes the expectation by at most $∥ A ∥ ∥ ρ_{I} (t) - ρ - ϵ \int_{0}^{t} K_{s} (ρ) d s ∥_{1}$ by [F3, F5]. This gives the stated bound. Since $A, B$ are self-adjoint and $ρ$ is positive, the commutator trace is purely imaginary by cyclicity; hence $R_{A B} (u)$ is real. No moment of $H$ occurs in any trace.

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Finite unitary dynamics and stationary Gibbs states

Statement

Let $H$ be a complex Hilbert space of finite dimension $d \geq 1$ , let $H = H^{*}$ be a Hamiltonian with energy unit J, and let $ℏ > 0$ have unit J s. A finite density matrix is a positive semidefinite matrix $ρ$ with $Tr ρ = 1$ . For time $t \in R$ measured in s, set $U (t) = e^{- i t H / ℏ}$ and $α_{t} (A) = U (t)^{*} A U (t)$ . Then $U (t)$ is a unitary group, $α_{t}$ is a group of matrix $*$ -automorphisms, and for every finite density matrix $ρ$ and matrix $A$ ,

$Tr (U (t) ρ U (t)^{*} A) = Tr (ρ α_{t} (A)) .$

The von Neumann entropy $S (ρ) = - k_{B} Tr (ρ \log ρ)$ is defined using $\log ρ$ on the positive spectral support, with zero eigenvalues contributing $0$ (the convention $0 \log 0 : = 0$ ), and is finite and unchanged by $U (t) ρ U (t)^{*}$ . For $β > 0$ with unit J $^{- 1}$ , let $Z_{β} = Tr (e^{- β H})$ and $σ_{β} = Z_{β}^{- 1} e^{- β H}$ . Then $Z_{β} > 0$ , $σ_{β}$ is stationary, and the finite Gibbs correlation has the KMS strip identity

$F_{A, B} (t + i β ℏ) = Tr (σ_{β} α_{t} (B) A), F_{A, B} (z) = Tr (σ_{β} A α_{z} (B)),$

where $α_{z} (B) = e^{i z H / ℏ} B e^{- i z H / ℏ}$ . The function $F_{A, B}$ is entire and bounded on $0 \leq Im z \leq β ℏ$ . This finite-dimensional result asserts no approach to equilibrium.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space, $H = H^{*}$ , $ℏ > 0$ , $β > 0$ , matrices $A, B$ , and a positive semidefinite matrix $ρ$ with trace one.

[F1]

Finite matrix spectral and state calculus: every Hermitian matrix has a real spectral decomposition; the finite trace is cyclic; positive matrices have basis-independent spectral functional calculus, including the logarithm on positive support.

[F2]

Finite unitary dynamics and Gibbs KMS identity: for finite-dimensional $H = H^{*}$ and $β > 0$ , $σ_{β}$ is stationary and $F_{A, B}$ is entire, bounded on the strip $0 \leq Im z \leq β ℏ$ , with boundary value $F_{A, B} (t + i β ℏ) = Tr (σ_{β} α_{t} (B) A)$ . Its current draft proof is the inspected finite spectral calculation; no infinite-volume KMS claim is used.

Proof

Given: The finite-dimensional Hamiltonian, density matrix, and constants in the Statement.

1.1givenF1algebra

Diagonalize $H$ using [F1], with eigenvalues $E_{j} \in R$ . The eigenvalues of $U (t)$ are $e^{- i t E_{j} / ℏ}$ , so $U (t)^{*} U (t) = I$ and multiplication of the scalar phases gives $U (t + s) = U (t) U (s)$ . Conjugation therefore makes each $α_{t}$ a unital matrix $*$ -automorphism and gives the group law.

2.1givenF1step 1.1algebra

By cyclicity in [F1], $Tr (U ρ U^{*} A) = Tr (ρ U^{*} A U)$ . Unitary conjugation preserves the eigenvalues of $ρ$ , so it preserves the finite entropy and $Tr ρ = 1$ . Since $e^{- β H}$ is positive definite in finite dimension, its trace is positive; it commutes with $U (t)$ , so $σ_{β}$ is stationary.

3.1givenF1F2step 1.1algebra∎

In the eigenbasis of $H$ , each entry of $α_{z} (B)$ is a constant multiple of $e^{i z (E_{j} - E_{k}) / ℏ}$ , hence is entire. The finite-dimensional trace makes $F_{A, B}$ entire. The KMS boundary identity and boundedness on the specified strip are exactly [F2] under the same hypotheses.

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Finite Cesaro dephasing with degenerate energy blocks

Statement

Under the finite-dimensional unitary dynamics of Finite unitary dynamics and stationary Gibbs states, let $P_{E}$ be the orthogonal spectral projection of $H$ for its distinct energy $E$ . For every density matrix $ρ$ ,

$\frac{1}{T} \int_{0}^{T} U (t) ρ U (t)^{*} d t ⟶ \sum_{E} P_{E} ρ P_{E}$

in trace norm as $T \to \infty$ . The diagonal energy blocks, including all blocks within degenerate eigenspaces, remain. This is Cesaro convergence of the averaged state; pointwise convergence of $U (t) ρ U (t)^{*}$ is not asserted.

Facts & Assumptions

Given: A finite-dimensional $H$ , a Hermitian $H$ , $ℏ > 0$ , and a positive matrix $ρ$ with trace one.

[F1]

Finite unitary dynamics and stationary Gibbs states: $U (t) = e^{- i t H / ℏ}$ is unitary and has the finite-dimensional spectral dynamics of the Statement.

[F2]

Finite matrix trace-norm ideal inequalities: the finite trace norm is a norm on the matrix space, so convergence of finite sums in that norm is well defined.

Proof

Given: The finite-dimensional unitary evolution and density matrix above.

1.1givenF1algebra

Write $H = \sum_{E} E P_{E}$ by finite Hermitian spectral calculus. Then $U (t) ρ U (t)^{*} = \sum_{E, F} e^{- i (E - F) t / ℏ} P_{E} ρ P_{F}$ . There are finitely many ordered pairs $(E, F)$ , so the time average is the corresponding finite sum of averaged coefficients times fixed matrices.

2.1step 1.1F2algebra∎

If $E = F$ , the averaged scalar coefficient is $1$ . If $E \neq F$ , then $∣ T^{- 1} \int_{0}^{T} e^{- i (E - F) t / ℏ} d t ∣ = ∣ \frac{ℏ (e^{- i (E - F) T / ℏ} - 1)}{- i (E - F) T} ∣ \leq \frac{2 ℏ}{T ∣ E - F ∣} \to 0$ . Since the finite trace norm is a norm by [F2], the finite matrix sum converges in that norm to $\sum_{E} P_{E} ρ P_{E}$ . Equal-energy blocks are not removed, so degeneracies are retained.

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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.

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Arbitrarily late finite phase recurrences

Statement

Let $H = H^{*}$ with energy unit J act on a finite-dimensional complex Hilbert space, let $ℏ > 0$ have unit J s, and let $U (t) = e^{- i t H / ℏ}$ for time $t$ measured in s. For every time unit $t_{0} > 0$ measured in s, there is a sequence $t_{n} \to + \infty$ and phases $λ_{n} \in C$ , $∣ λ_{n} ∣ = 1$ , such that $∥ U (t_{n}) - λ_{n} I ∥ \to 0$ , where $∥ T ∥ : = \sup {∥ T v ∥ : ∥ v ∥ \leq 1}$ is the operator norm induced by the Hilbert-space norm. Consequently, for every density matrix $ρ$ and bounded observable $A$ , the expectation $Tr (U (t_{n}) ρ U (t_{n})^{*} A)$ tends to $Tr (ρ A)$ . If $H$ has only one distinct energy, exact recurrences occur at every time.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space, $H = H^{*}$ , $ℏ > 0$ , and a fixed $t_{0} > 0$ .

[F1]

Finite unitary dynamics and stationary Gibbs states: the distinct real eigenvalues of $H$ define the spectral unitary group $U (t) = e^{- i t H / ℏ}$ .

Proof

Given: The finite-dimensional spectrum and unitary group above.

1.1givenF1construct

If the Hilbert space is zero-dimensional, its unique endomorphism is the identity and $U (t) = I$ ; take $t_{n} = n t_{0}$ and $λ_{n} = 1$ , so every operator-norm error is zero. No density operator exists on that space, making the density consequence vacuous there. On a nonzero space, if $H$ has one distinct energy $E_{1}$ , then $U (t) = e^{- i t E_{1} / ℏ} I$ , which proves the exact-recurrence clause. Otherwise list its $r \geq 2$ distinct energies as $E_{1}, \dots, E_{r}$ and put $θ_{j} = (E_{j} - E_{1}) t_{0} / (2 π ℏ)$ for $2 \leq j \leq r$ . For each integer $M \geq 1$ , partition the torus $[0, 1)^{r - 1}$ into $M^{r - 1}$ half-open cubes of side $1 / M$ and consider the $M^{r - 1} + 1$ points $k θ m o d Z^{r - 1}$ for $0 \leq k \leq M^{r - 1}$ . Two lie in the same cube, so their nonzero integer difference $q_{M}$ satisfies $1 \leq q_{M} \leq M^{r - 1}$ and each $q_{M} θ_{j}$ is within $1 / M$ of an integer.

2.1step 1.1algebra

The eigenvalue phase ratios of $U (q_{M} t_{0})$ to its $E_{1}$ phase therefore differ from $1$ by at most $2 π / M$ , since $∣ e^{i x} - 1 ∣ \leq ∣ x ∣$ . The difference $U (q_{M} t_{0}) - λ_{M} I$ is diagonal in the energy basis, so its induced operator norm is the maximum modulus of its diagonal entries and is at most $2 π / M$ , for $λ_{M} = e^{- i q_{M} E_{1} t_{0} / ℏ}$ . If the positive integers $q_{M}$ are unbounded, choose a subsequence tending to infinity; its times and phase errors give the required sequence. If they are bounded, some fixed positive integer $q$ occurs for arbitrarily large $M$ ; then every $q θ_{j}$ is an integer, so $U (q t_{0}) = λ I$ exactly and its positive integer multiples give exact arbitrarily late recurrences.

3.1step 2.1algebra∎

For any density $ρ$ , the expectation difference is bounded by $∥ U (t_{n})^{*} A U (t_{n}) - A ∥ \leq 2 ∥ A ∥ ∥ U (t_{n}) - λ_{n} I ∥$ , because scalar phases cancel in conjugation. The bound tends to zero by step 2.1.

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Bounded-commutator graph-domain Schrödinger response

Statement

Assume the Axiom of Choice. Let $H \neq {0}$ be a separable complex Hilbert space, $H = H^{*}$ a densely defined Hamiltonian on $D (H)$ with unit J, $ℏ > 0$ with unit J s, $ρ$ a positive trace-class density with $Tr ρ = 1$ and $ρ U_{0} (t) = U_{0} (t) ρ$ , and $A = A^{*} \in B (H)$ . Fix a reference energy $E_{0} > 0$ in joules and equip $D (H)$ with the graph norm $∥ ψ ∥_{G} : = ∥ ψ ∥ + ∥ H ψ ∥ / E_{0}$ . Let $B = B^{*} \in B (H)$ with unit $[B]$ . Assume $B (D (H)) \subseteq D (H)$ and that $[H, B] ψ = H B ψ - B H ψ$ on $D (H)$ extends to a bounded operator $C$ on $H$ . Let $h : [0, T] \to R$ be continuous for finite $T > 0$ with $h (t) B$ in J, and let $ϵ \in R$ be dimensionless. Then the bounded-observable response and remainder from Trace-class bounded-observable response hold. In addition, with $U_{0} (t) = e^{- i t H / ℏ}$ , $B_{t} = U_{0} (t)^{*} B U_{0} (t)$ , and $W_{ϵ}^{'} (t) = i ϵ h (t) B_{t} W_{ϵ} (t) / ℏ$ , $W_{ϵ} (0) = I$ , the interaction-picture propagator maps $D (H)$ onto itself, and $Ψ (t) = U_{0} (t) W_{ϵ} (t) ψ$ is a strong solution on $[0, T]$ for each $ψ \in D (H)$ , meaning that $Ψ : [0, T] \to H$ belongs to $C ([0, T]; D (H))$ for the graph norm and to $C^{1} ([0, T]; H)$ , satisfies $Ψ (0) = ψ$ , and satisfies $i ℏ Ψ^{'} (t) = (H - ϵ h (t) B) Ψ (t)$ for every $t \in [0, T]$ (with one-sided derivatives at the endpoints). No domain-preservation claim is made for a general bounded $B$ without the stated commutator hypothesis.

Facts & Assumptions

Given: The Axiom of Choice; self-adjoint $H$ ; bounded self-adjoint $B$ preserving $D (H)$ ; bounded extension $C$ of $[H, B]$ ; continuous real $h$ ; and the trace-class response inputs from Trace-class bounded-observable response.

[A1]

The Axiom of Choice: full Choice is assumed by the unbounded spectral calculus and carried from the normal-state response theorem.

[F1]

Trace-class bounded-observable response: for bounded observables and a trace-class density commuting with $U_{0} (t)$ , the response formula holds with the stated finite-time remainder, without an energy or entropy moment assumption.

[F2]

A self-adjoint operator generates a strongly continuous unitary group: $U_{0} (t)$ is strongly continuous, preserves $D (H)$ , and differentiates there; its parameter is rescaled by $- t / ℏ$ from the supplier's $e^{i s H}$ convention.

[F3]

Symmetric, self-adjoint and essentially self-adjoint operators: a self-adjoint operator is densely defined, closed, and equals its adjoint.

[F4]

Banach-space Volterra and Dyson series with a norm remainder: on any complete normed space, a strongly continuous bounded coefficient has a unique norm-convergent Volterra solution; the same construction applies to the graph norm once that space is shown complete.

Proof

technique · construct the control evolution on the graph Banach space, then differentiate the Schrödinger-picture product

Given: [A1], [F1]–[F4] and the domain/commutator hypotheses in the Statement.

1.1givenF3construct

Use the graph norm on $D (H)$ fixed in the Statement. If $(ψ_{n})$ is Cauchy in this norm, then $ψ_{n} \to ψ$ and $H ψ_{n} \to ξ$ in $H$ ; closedness of $H$ from [F3] gives $ψ \in D (H)$ and $H ψ = ξ$ , so the graph norm is complete.

2.1givenF2step 1.1algebra

For $B_{t} = U_{0} (t)^{*} B U_{0} (t)$ and $ψ \in D (H)$ , invariance and commutation in [F2] give $H B_{t} ψ = B_{t} H ψ + U_{0} (t)^{*} C U_{0} (t) ψ$ . Hence $B_{t}$ is uniformly bounded on the graph space by a constant depending only on $∥ B ∥, ∥ C ∥, E_{0}$ . To check graph-norm continuity at $t_{0}$ , the Hilbert-norm part follows from strong continuity of $U_{0}$ ; for the $H$ -part use the displayed identity and continuity of $t \mapsto B_{t} H ψ$ and $t \mapsto U_{0} (t)^{*} C U_{0} (t) ψ$ . The latter follows from $∥ U_{0} (t)^{*} C U_{0} (t) ψ - U_{0} (t_{0})^{*} C U_{0} (t_{0}) ψ ∥ \leq ∥ C ∥ ∥ U_{0} (t) ψ - U_{0} (t_{0}) ψ ∥ + ∥ (U_{0} (t)^{*} - U_{0} (t_{0})^{*}) C U_{0} (t_{0}) ψ ∥,$ whose two terms tend to zero by strong continuity of $U_{0}$ and the group law.

3.1givenA1F2F4step 1.1step 2.1algebra

Apply [F4] on the graph Banach space with coefficient $K_{t} = i h (t) B_{t} / ℏ$ ; its Volterra series defines $W_{ϵ} (t)$ on $D (H)$ . For any $y \in D (H)$ , solve the reversed-time graph equation $z^{'} (s) = - ϵ K_{t - s} z (s)$ with $z (0) = y$ on $[0, t]$ ; then $x (s) = z (t - s)$ solves the forward equation and $W_{ϵ} (t) x (0) = y$ . Hence $W_{ϵ} (t) D (H) = D (H)$ . On the Hilbert space the same coefficient is skew-adjoint, and the Hilbert clause of [F4] gives the unitary propagator used in the Statement; uniqueness identifies both constructions. The domain-preservation conclusion therefore uses the full-Choice spectral group hypothesis [A1] together with [F2], [F4] and the commutator condition.

4.1F2F3step 3.1algebra

For $ψ \in D (H)$ , the graph-space Volterra solution makes $W_{ϵ} (t) ψ$ continuous and differentiable in graph norm. Also $U_{0} (t)$ acts continuously on $D (H)$ in graph norm because it is strongly continuous on both $ψ$ and $H ψ$ and commutes with $H$ there. Thus $Ψ \in C ([0, T]; D (H))$ in graph norm, and the product rule gives $Ψ \in C^{1} ([0, T]; H)$ with $Ψ (0) = ψ$ . Using $U_{0} (t) B_{t} = B U_{0} (t)$ and $i ℏ W_{ϵ}^{'} = - ϵ h (t) B_{t} W_{ϵ}$ , we obtain $i ℏ Ψ^{'} (t) = (H - ϵ h (t) B) Ψ (t)$ for every $t \in [0, T]$ , with one-sided derivatives at the endpoints. For each fixed $t$ , $H - ϵ h (t) B$ is self-adjoint on $D (H)$ : it is symmetric, and if $y \in D ((H - ϵ h (t) B)^{*})$ , subtracting the bounded $B$ term in the adjoint functional shows $y \in D (H^{*}) = D (H)$ , proving equality of the adjoint domain.

5.1F1step 4.1∎

The trace-class response assertion is exactly [F1] under its stated density, stationarity, bounded-observable and control hypotheses; step 4.1 adds the strong Schrödinger realization only for the graph-domain class in this Statement. The two conclusions therefore retain their separate hypotheses.

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Adopted isolated prepared unitary model

Physical scope

A specified finite-dimensional, isolated, prepared nonrelativistic quantum subsystem with a fixed Hermitian Hamiltonian and bounded-observable Born expectations; no bath limit, relaxation claim, apparatus model, or empirical realization is asserted.

Postulate

For a specified finite-dimensional complex Hilbert space $H$ , $H = H^{*}$ in J, and adopted $ℏ > 0$ in J s, take a prepared finite density matrix $ρ_{0}$ (positive semidefinite with trace one). For time $t$ measured in s, adopt isolated evolution $U (t) = e^{- i t H / ℏ}, ρ_{t} = U (t) ρ_{0} U (t)^{*},$ and the Born expectation $E_{t} [A] = Tr (ρ_{t} A)$ for each bounded self-adjoint matrix observable $A$ . This adopts the finite subsystem's preparation and unitary time evolution as a model assumption. It inherits the full Axiom of Choice convention, stated in The Axiom of Choice, of Normal-state quantum measurement interpretation for the wider Hilbert-space spectral-measure framework, although this finite bounded-observable clause uses no unbounded spectral measurement. No actual apparatus, measured value, thermalization, or infinite-volume Hamiltonian is asserted.

The finite unitary and Gibbs conventions are stated in Finite unitary dynamics and stationary Gibbs states as formulation context; they do not prove that this postulate describes nature.

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Prepared control and expectation convention

Physical scope

An idealized finite-dimensional subsystem with a prescribed continuous bounded control on a finite time interval and a fixed prepared initial state; no apparatus calibration, measured response, re-preparation during control, or universal response law is asserted.

Postulate

On a finite-dimensional subsystem with unperturbed $H = H^{*}$ in J, bounded self-adjoint control observable $B$ , $ℏ > 0$ in J s, and finite interval $[0, T]$ with $T > 0$ in s, adopt the idealized controlled Hamiltonian $H_{ϵ} (t) = H - ϵ h (t) B,$ where $h : [0, T] \to R$ is continuous, $ϵ \in R$ is dimensionless, and $h (t) B$ has unit J. The initial prepared state is held fixed at $t = 0$ as the Gibbs matrix $σ = Z_{β}^{- 1} e^{- β H}$ , $β > 0$ in J $^{- 1}$ ; the model does not re-prepare a Gibbs state for $H_{ϵ} (t)$ . Expectations use the adopted normal-state Born convention of Normal-state quantum measurement interpretation. The full Axiom of Choice convention, stated in The Axiom of Choice, of that wider measurement framework is retained. This is an adopted control model; it makes no claim about a particular device or data.

For the controlled propagator $U_{ϵ}$ satisfying $i ℏ U_{ϵ}^{'} (t) = H_{ϵ} (t) U_{ϵ} (t)$ and $U_{ϵ} (0) = I$ , define $ρ_{ϵ} (t) = U_{ϵ} (t) σ U_{ϵ} (t)^{*}$ and $E_{ϵ, t} [A] = Tr (ρ_{ϵ} (t) A)$ for bounded self-adjoint matrices $A$ . Existence and uniqueness of this finite-dimensional propagator are supplied by Finite-time Kubo formula with norm remainder.

The finite-time response theorem Finite-time Kubo formula with norm remainder supplies the exact mathematical response interface used to formulate this postulate. It does not establish the physical applicability of the control model.

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Finite spectral balance and distributional fluctuation response

Statement

Let $H$ be finite-dimensional, $H = H^{*}$ with unit J, $ℏ > 0$ with unit J s, $β > 0$ with unit J $^{- 1}$ , and $σ = Z_{β}^{- 1} e^{- β H}$ as in Finite unitary dynamics and stationary Gibbs states. Let $A = A^{*}$ be a matrix observable, write $H u_{i} = E_{i} u_{i}$ in an orthonormal eigenbasis, $r_{i} = e^{- β E_{i}} / Z_{β}$ , $A_{i j} = ⟨ A u_{j}, u_{i} ⟩$ , and $ω_{j i} = (E_{j} - E_{i}) / ℏ$ . Define the finite positive spectral measure $S_{A A} = 2 π \sum_{i, j} r_{i} ∣ A_{i j} ∣^{2} δ_{ω_{j i}},$ where $δ_{ν}$ is unit point mass at frequency $ν$ and $S_{A A} (- \cdot)$ denotes pushforward under $ω \mapsto - ω$ ; $ω_{j i}$ has unit s $^{- 1}$ . With observable unit $[A]$ , the atomic masses of the finite measure $S_{A A}$ have units $[A]^{2}$ . For every smooth compactly supported complex-valued function $φ : R \to C$ , the distribution represented by the measure $S_{A A}$ acts as $⟨ S_{A A}, φ ⟩ = \int_{R} φ (ω) d S_{A A} (ω)$ . In the Fourier convention below, the generalized density $S_{A A} (ω)$ relative to $d ω$ has units $[A]^{2}$ multiplied by seconds; it is a finite sum of Dirac distributions, not an ordinary function density. For time $t \in R$ measured in s, the equilibrium correlation is $C_{A A} (t) = Tr (σ α_{t} (A) A)$ ; its Fourier variable $ω$ is measured in s $^{- 1}$ , and its Fourier convention is $S_{A A} (ω) = \int_{R} e^{i ω t} C_{A A} (t) d t$ in the distributional sense.

For $t \geq 0$ , set $R_{A A} (t) = \frac{i}{ℏ} Tr (σ [α_{t} (A), A])$ . For $ω \in R$ and regulator $η > 0$ , both measured in s $^{- 1}$ , define the retarded transform $χ_{η} (ω) = \int_{0}^{\infty} e^{(i ω - η) t} R_{A A} (t) d t$ . With observable unit $[A]$ , $R_{A A}$ has units $[A]^{2} / (J s)$ and $χ_{η}$ has units $[A]^{2} / J$ . Then $χ_{η} (ω) = \sum_{i, j} \frac{(r_{i} - r_{j}) ∣ A_{i j} ∣^{2}}{ℏ (ω_{j i} - ω - i η)} .$ As $η ↓ 0$ , its boundary value is the corresponding sum of principal-value real parts and Lorentzian imaginary terms weighted by $r_{i} - r_{j}$ , and distributionally $Im χ = \frac{1}{2 ℏ} (1 - e^{- β ℏ ω}) S_{A A} (ω) .$ Detailed balance is $S_{A A} (- d ω) = e^{- β ℏ ω} S_{A A} (d ω)$ . Thus, on $R ∖ {0}$ , $S_{s y m} (d ω) = ℏ \coth ⁣ (\frac{β ℏ ω}{2}) Im χ (d ω), S_{s y m} = \frac{1}{2} (S_{A A} + S_{A A} (- \cdot)) .$ Any zero-frequency mass of $S_{A A}$ is retained separately: the commutator factor vanishes there, so the formula does not determine that atom or define a value for the singular product with $\coth$ at $ω = 0$ .

Facts & Assumptions

Given: The finite-dimensional Hamiltonian, Gibbs matrix, bounded self-adjoint observable, Fourier convention, and positive regulator $η$ above.

[F1]

Finite unitary dynamics and stationary Gibbs states: the finite Hermitian spectral decomposition gives $U (t)$ and the stationary Gibbs matrix $σ$ .

[F2]

Finite matrix spectral and state calculus: finite Hermitian matrices admit basis-independent real spectral decompositions and the finite trace is cyclic.

[F3]

Finite-time Kubo formula with norm remainder: under the fixed Gibbs preparation and control $H - ϵ h (t) B$ , the retarded commutator kernel has the sign $+ i / ℏ$ and is the finite-time first variation.

[F4]

Lorentzian kernels converge to delta and principal-value distributions: $η / (x^{2} + η^{2})$ converges to $π δ_{0}$ and $x / (x^{2} + η^{2})$ to $PV (1 / x)$ on compactly supported $C^{1}$ test functions, with full dominated-convergence proofs.

Proof

technique · finite spectral expansion, exact retarded transform, and the Lorentzian distribution limits

Given: [F1]–[F4], finite $d \geq 1$ , $β > 0$ , and $A = A^{*}$ .

1.1givenF1F2constructalgebra

In the eigenbasis of $H$ , $α_{t} (A)_{i j} = e^{- i ω_{j i} t} A_{i j}$ , so $C_{A A} (t) = \sum_{i, j} r_{i} ∣ A_{i j} ∣^{2} e^{- i ω_{j i} t}$ and the displayed $S_{A A}$ is its Fourier measure under the stated convention. Swapping $i, j$ in the finite sum changes $S_{A A} (- d ω)$ to coefficients $r_{j} ∣ A_{i j} ∣^{2}$ at $ω_{j i}$ ; since $r_{j} / r_{i} = e^{- β ℏ ω_{j i}}$ , this proves detailed balance. The coefficients of $S_{A A}$ are nonnegative.

2.1F1F2F3step 1.1algebra

Expanding the commutator and swapping indices in its second trace term gives $R_{A A} (t) = \frac{i}{ℏ} \sum_{i, j} (r_{i} - r_{j}) ∣ A_{i j} ∣^{2} e^{- i ω_{j i} t}$ . For $η > 0$ , each term is integrable on $[0, \infty)$ and $\int_{0}^{\infty} e^{(i (ω - ω_{j i}) - η) t} d t = - i / (ω_{j i} - ω - i η)$ ; multiplying by $i / ℏ$ yields the stated rational expression.

3.1F1F4step 2.1algebra

For each summand, write $x = ω_{j i} - ω$ and $1 / (x - i η) = x / (x^{2} + η^{2}) + i η / (x^{2} + η^{2})$ . Applying [F4] after the affine change from $ω$ to $x$ gives the principal-value real part and $π δ (ω - ω_{j i})$ imaginary part. On each atom $ω = ω_{j i}$ , detailed balance gives $r_{i} - r_{j} = r_{i} (1 - e^{- β ℏ ω})$ , so summing the finite terms proves the distributional commutator relation.

4.1F1step 1.1step 3.1algebra∎

Detailed balance gives $S_{s y m} = \frac{1}{2} (1 + e^{- β ℏ ω}) S_{A A}$ away from zero, while step 3.1 gives $Im χ = \frac{1}{2 ℏ} (1 - e^{- β ℏ ω}) S_{A A}$ . Dividing on $R ∖ {0}$ yields the stated $ℏ \coth (β ℏ ω / 2)$ factor. At $ω = 0$ , the detailed-balance pair has equal Gibbs weights and the commutator coefficient is zero, whereas the positive correlation measure may have a nonzero atom; therefore that atom must be retained separately.

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Absolute convergence of trace-Gibbs bounded correlation series

Statement

Assume the Axiom of Choice. Let $H \neq {0}$ be separable, $K = K^{*}$ a densely defined Hamiltonian bounded below with energy unit J, and $β$ lie in an open interval $I \subset (0, \infty)$ measured in J $^{- 1}$ on which $e^{- s K}$ is trace class for every $s \in I$ . Set $Z_{β} = Tr (e^{- β K})$ , $σ_{β} = Z_{β}^{- 1} e^{- β K}$ , and $U (t) = e^{- i t K / ℏ}$ with $ℏ > 0$ in J s and $t$ in s. By Self-adjoint trace-Gibbs normalization and form-energy optimization, $K$ has a complete orthonormal energy basis $(u_{i})$ , $K u_{i} = E_{i} u_{i}$ , with Gibbs weights $r_{i} = e^{- β E_{i}} / Z_{β} > 0$ and $\sum_{i} r_{i} = 1$ . For bounded $A, B$ , put $A_{i j} = ⟨ A u_{j}, u_{i} ⟩$ , $B_{i j} = ⟨ B u_{j}, u_{i} ⟩$ , and $ω_{j i} = (E_{j} - E_{i}) / ℏ$ . Then $C_{A B} (t) : = Tr (σ_{β} U (t)^{*} A U (t) B) = \sum_{i, j} r_{i} A_{i j} B_{j i} e^{- i ω_{j i} t}$ converges absolutely and uniformly on $R$ , with $\sum_{i, j} r_{i} ∣ A_{i j} ∣ ∣ B_{j i} ∣ \leq ∥ A ∥ ∥ B ∥$ . Consequently $S_{A B} : = 2 π \sum_{i, j} r_{i} A_{i j} B_{j i} δ_{ω_{j i}}$ is a finite complex Borel measure with total variation at most $2 π ∥ A ∥ ∥ B ∥$ , and $C_{A B} (t) = (2 π)^{- 1} \int e^{- i ω t} d S_{A B} (ω)$ . A Dirac measure $δ_{ν}$ assigns mass one to the point $ν$ . When $A, B$ carry units $[A], [B]$ , the atomic masses of the finite measure $S_{A B}$ have units $[A] [B]$ . For every smooth compactly supported complex-valued function $φ : R \to C$ , the distribution represented by $S_{A B}$ acts as $⟨ S_{A B}, φ ⟩ = \int_{R} φ (ω) d S_{A B} (ω)$ . In the Fourier convention, the generalized density $S_{A B} (ω)$ relative to $d ω$ has units $[A] [B]$ multiplied by seconds; it is a sum of Dirac distributions, not an ordinary function density.

Facts & Assumptions

Given: The Axiom of Choice; the trace-class Gibbs hypotheses in the Statement; bounded operators $A, B$ ; and the specified $ℏ$ and $β$ .

[A1]

The Axiom of Choice: assume the full Axiom of Choice for the unbounded self-adjoint spectral theorem.

[F1]

Self-adjoint trace-Gibbs normalization and form-energy optimization: under [A1], self-adjoint $K$ bounded below with $e^{- s K}$ trace class on an open positive interval has $0 < Z_{β} < \infty$ , a faithful normal Gibbs density, and a complete energy eigenbasis with eigenvalues $e^{- β E_{i}} / Z_{β}$ .

[F2]

Trace-state cyclicity and spectral probability calculus: for a trace-class density $σ$ and bounded $X$ , $Tr (σ X) = \sum_{i} r_{i} ⟨ X u_{i}, u_{i} ⟩$ with absolute convergence; bounded cyclicity holds only for the trace-class factor and bounded factor.

[F3]

Parseval equivalences for an orthonormal family: for a complete orthonormal basis $(u_{j})$ , $\sum_{j} ∣ ⟨ v, u_{j} ⟩ ∣^{2} = ∥ v ∥^{2}$ .

[F4]

Hilbert-adjoint identities and The operator norm as the least bound and as the unit-sphere or unit-ball supremum: $∥ A^{*} ∥ = ∥ A ∥$ and $∥ A x ∥ \leq ∥ A ∥ ∥ x ∥$ for bounded operators.

[F5]

Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs: for square-summable coefficient sequences, the absolute inner-product sum is bounded by the product of their $ℓ^{2}$ norms.

Proof

technique · direct spectral expansion with an absolute-summability estimate

Given: [A1], [F1]–[F3], and the bounded matrices of the Statement.

1.1givenA1F1F3F4F5algebra

The eigenbasis and weights in the Statement are provided by [F1], and $\sum_{i} r_{i} = 1$ follows from $Tr σ_{β} = 1$ . For each fixed $i$ , Parseval [F3] and [F4] give $\sum_{j} ∣ A_{i j} ∣^{2} = ∥ A^{*} u_{i} ∥^{2} \leq ∥ A ∥^{2}$ and $\sum_{j} ∣ B_{j i} ∣^{2} = ∥ B u_{i} ∥^{2} \leq ∥ B ∥^{2}$ . By [F5] and summing the nonnegative bounds over $i$ , $\sum_{i, j} r_{i} ∣ A_{i j} ∣ ∣ B_{j i} ∣ \leq ∥ A ∥ ∥ B ∥$ .

2.1givenF1F2step 1.1algebra

Apply [F2] to $X = U (t)^{*} A U (t) B$ . Expanding the diagonal matrix element in the energy basis gives $⟨ U (t)^{*} A U (t) B u_{i}, u_{i} ⟩ = \sum_{j} e^{- i (E_{j} - E_{i}) t / ℏ} A_{i j} B_{j i}$ . The absolute bound from step 1.1 justifies the double sum and is independent of $t$ ; hence the series converges uniformly by the uniform tail bound and equals the stated trace correlation.

3.1givenstep 1.1step 2.1algebra∎

For each Borel set $E \subseteq R$ , define $S_{A B} (E) = 2 π \sum_{i, j : ω_{j i} \in E} r_{i} A_{i j} B_{j i}$ . Absolute convergence from step 1.1 makes this finite and makes the sum countably additive: for disjoint $E_{n}$ , each atom belongs to exactly one $E_{n}$ , and the absolutely convergent series may be regrouped. The same bound gives total variation at most $2 π ∥ A ∥ ∥ B ∥$ . Integrating the bounded function $e^{- i ω t}$ termwise, justified by that absolute sum, gives the stated Fourier relation.

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Trace-Gibbs bounded correlation measure and balance

Statement

Assume the Axiom of Choice. Let $H \neq {0}$ be separable, $K = K^{*}$ densely defined and bounded below with energy unit J, and let $e^{- s K}$ be trace class for every $s$ in an open interval $I \subset (0, \infty)$ measured in J $^{- 1}$ . Fix $β \in I$ , $Z_{β} = Tr (e^{- β K})$ , $σ_{β} = Z_{β}^{- 1} e^{- β K}$ , and $U (t) = e^{- i t K / ℏ}$ for $ℏ > 0$ in J s and $t$ in s. For bounded $A, B$ , let $C_{A B}$ and $S_{A B}$ be the correlation function and finite spectral measure defined in Absolute convergence of trace-Gibbs bounded correlation series. Write $S_{B A} (- \cdot)$ for the pushforward under $ω \mapsto - ω$ . Then for every bounded Borel test function $f$ with compact support, $\int f (ω) S_{B A} (- d ω) = \int f (ω) e^{- β ℏ ω} S_{A B} (d ω) .$ The weighted measure on the right extends to a finite complex measure: the reindexed coefficient family is absolutely summable and has the same total-variation bound as $S_{B A}$ . For Hermitian $A = B$ , $S_{A A}$ is a positive finite measure. This is a discrete-spectrum equilibrium correlation identity; it asserts no temporal decay, continuum limit, or ordinary zero-frequency transport coefficient.

Facts & Assumptions

Given: The Axiom of Choice, the self-adjoint trace-Gibbs hypotheses, the bounded observables, and the correlation measure from the Statement.

[A1]

The Axiom of Choice: full Choice is the exact assumption of the unbounded spectral theorem and trace-Gibbs supplier.

[F1]

Self-adjoint trace-Gibbs normalization and form-energy optimization: the stated open-interval trace-class hypothesis yields $0 < Z_{β} < \infty$ , a faithful normal Gibbs density, and a complete discrete energy basis.

[F2]

Absolute convergence of trace-Gibbs bounded correlation series: the bounded correlation has an absolutely summable atomic spectral measure with coefficients $r_{i} A_{i j} B_{j i}$ at $ω_{j i} = (E_{j} - E_{i}) / ℏ$ .

Proof

technique · reindex the absolutely convergent spectral measure

Given: [A1], [F1]–[F2] and bounded Borel $f$ with compact support.

1.1givenA1F1F2algebra

Under the unbounded spectral convention [A1], apply [F2]: $\int f (ω) S_{B A} (- d ω) = 2 π \sum_{i, j} r_{i} B_{i j} A_{j i} f (- ω_{j i}) .$ The coefficient family is absolutely summable after multiplication by $∥ f ∥_{\infty}$ , so interchanging indices $i$ and $j$ is permitted.

2.1givenF1F2step 1.1algebra

Reindexing the sum in step 1.1 gives $2 π \sum_{i, j} r_{j} A_{i j} B_{j i} f (ω_{j i})$ . Since $r_{j} / r_{i} = e^{- β (E_{j} - E_{i})} = e^{- β ℏ ω_{j i}}$ , this is exactly $\int f (ω) e^{- β ℏ ω} S_{A B} (d ω)$ . The absolute sum of the reindexed coefficients equals that defining $S_{B A}$ and is finite by [F2]; compact support also makes the test-function weight bounded. Thus the weighted measure is finite and equals the reflected measure.

3.1F1F2algebra∎

If $A = B = A^{*}$ , then $B_{j i} = \overline{A_{i j}}$ , so each atom coefficient $r_{i} ∣ A_{i j} ∣^{2}$ of $S_{A A}$ is nonnegative. The measure is finite by [F2], proving positivity.

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Conditional prepared finite quantum control response

Physical scope

Exact prediction of the specified finite-dimensional isolated-control model under its adopted preparation, unitary-evolution, and Born-expectation postulates; it is not an observation, an apparatus validation, or evidence that the model describes nature.

Statement

Assume the adopted finite isolated model and fixed-preparation control convention of Adopted isolated prepared unitary model and Prepared control and expectation convention. Let $T > 0$ be finite in s, let $A = A^{*}$ be a bounded matrix observable, and set $α_{t} (A) = e^{i t H / ℏ} A e^{- i t H / ℏ}$ for time $t$ measured in s. 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 for $0 \leq t \leq T$ the model expectation obeys $E_{ϵ, t} [A] = Tr (σ α_{t} (A)) + ϵ \int_{0}^{t} h (s) R_{A B} (t - s) d s + r_{ϵ} (t), ∣ r_{ϵ} (t) ∣ \leq ∥ A ∥ \frac{(∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}}}{2} .$ Here $H$ and $h B$ have unit J, $ℏ$ has unit J s, $t$ has unit s, $β$ has unit J $^{- 1}$ , and $ϵ, C_{T}$ are dimensionless. This is a conditional model prediction for the adopted preparation and dynamics, not a claim that an observed finite dataset must equal the prediction.

Facts & Assumptions

Given: The Axiom of Choice, the two adopted physical postulates, finite $T$ , the matrix observable $A$ , and the stated control $h, B$ .

[A1]

Adopted isolated prepared unitary model: the specified finite prepared state evolves as $ρ_{t} = U (t) ρ_{0} U (t)^{*}$ and the model expectation of bounded $A$ is $Tr (ρ_{t} A)$ .

[A2]

Prepared control and expectation convention: the controlled generator is $H - ϵ h (t) B$ on $[0, T]$ while the initial Gibbs matrix $σ$ remains fixed.

[F1]

Finite-time Kubo formula with norm remainder: for this fixed initial Gibbs density, the Heisenberg expectation has the displayed retarded commutator response and quadratic finite-time remainder.

[A3]

The Axiom of Choice: the full Choice convention is carried from the adopted normal-state measurement framework; no step below uses it to derive the finite matrix formula.

Proof

technique · conditional substitution of the adopted model inputs into the finite-time mathematical response theorem

Given: [A1]–[A3], [F1], and the exact finite-dimensional matrices and controls in the Statement.

1.1givenA1A2F1algebra

By [A1] and [A2], the initial state, controlled Hamiltonian, and model expectation are exactly the inputs used by [F1]; the control does not differentiate or replace the initial Gibbs matrix. The controlled density satisfies $Tr (ρ_{ϵ} (t) A) = Tr (ρ_{I} (t) α_{t} (A))$ by the finite trace identity in [F1]. Applying [F1] with this $A, B, h, ϵ, T$ gives the stated convolution and remainder bound.

2.1givenA1A2A3step 1.1∎

The equation is therefore a deduction within the adopted finite model, conditional on [A1] and [A2]. The full-Choice convention [A3] is inherited from the wider measurement framework, but the finite-matrix calculation in step 1.1 does not need it. The remainder is a mathematical bound; it supplies no measured response, apparatus reliability claim, or unconditional assertion about nature.

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Bounded C-star Dyson response for finite-range spin dynamics

Statement

Use $Γ = Z^{d}$ , $d \geq 1$ , finite on-site dimension $q \geq 1$ , and a uniformly bounded finite-range interaction $Φ$ with parameters $M \geq 1$ , $R < \infty$ , $κ < \infty$ , and $ℏ > 0$ as in Uniformly bounded finite-range quantum spin interaction. Assume Countable Choice and the exact finite-matrix C $^{*}$ -identity condition supplied in Finite matrix C-star algebra and density-state duality. Let $A$ be the norm-completed quasi-local spin algebra of Norm-completed quasi-local quantum spin algebra, and let $α_{t}$ be the strongly continuous automorphism group of Strongly continuous thermodynamic spin automorphism group. Let ${ρ_{Λ}}$ be consistent positive trace-one local density matrices; write $ϕ$ for their unique state extension supplied by Consistent local densities extend to an algebraic state, and additionally assume $ϕ \circ α_{t} = ϕ$ for every $t \in R$ . This is an algebraic state on $A$ , not a trace-class density on an infinite tensor-product Hilbert space.

For $A \in A$ , $B = B^{*} \in A$ , continuous real $h : [0, T] \to R$ , finite $T > 0$ , and dimensionless $ϵ \in R$ , put $B_{t} = α_{t} (B)$ and let $W_{ϵ}$ solve $W_{ϵ}^{'} (t) = \frac{i ϵ h (t)}{ℏ} B_{t} W_{ϵ} (t), W_{ϵ} (0) = I .$ The C $^{*}$ -algebra Dyson solution is unitary. Define $α_{t}^{ϵ} (A) = W_{ϵ} (t)^{*} α_{t} (A) W_{ϵ} (t)$ , $R_{A B} (u) = \frac{i}{ℏ} ϕ ([α_{u} (A), B]), C_{T} = \frac{2}{ℏ} \int_{0}^{T} ∣ h (s) ∣ ∥ B ∥ d s .$ Then $ϕ (α_{t}^{ϵ} (A)) = ϕ (α_{t} (A)) + ϵ \int_{0}^{t} h (s) R_{A B} (t - s) d s + r_{ϵ} (t), ∣ r_{ϵ} (t) ∣ \leq ∥ A ∥ \frac{(∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}}}{2} .$ When energy is the interpretation of $H_{Λ}$ and $h B$ , their units are J, $t$ is s, $ℏ$ is J s, and $C_{T}$ is dimensionless. No global trace Hamiltonian, spatial-to-temporal decay implication, or dc coefficient follows.

Facts & Assumptions

Given: Countable Choice, the finite local spin system and interaction, the finite-matrix identity, the completed quasi-local algebra and group, consistent local densities, an invariant extension $ϕ$ , bounded $A, B$ , continuous $h$ , and finite $T$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is used for the norm-completion construction; no full Axiom of Choice is assumed here.

[F1]

Finite matrix C-star algebra and density-state duality: every nonzero finite local matrix algebra with its induced norm satisfies $∥ T^{*} T ∥ = ∥ T ∥^{2}$ .

[F2]

Uniformly bounded finite-range quantum spin interaction: $Φ (Z)$ is self-adjoint, vanishes for $∣ Z ∣ > M$ or diameter above $R$ , has per-site norm sum $κ$ , and defines the free-boundary finite-volume Hamiltonians.

[F3]

Norm-completed quasi-local quantum spin algebra: $A$ is the norm completion of the directed union of finite local matrix algebras, with no selected global tensor-product Hilbert space.

[F4]

Consistent local densities extend to an algebraic state: consistent positive trace-one local densities extend uniquely to a bounded normalized positive functional $ϕ$ on $A$ .

[F5]

Strongly continuous thermodynamic spin automorphism group: under [A1] and [F1]–[F3], the finite-volume dynamics converge to a strongly continuous one-parameter group $α_{t}$ of isometric $*$ -automorphisms.

[F6]

Banach-space Volterra and Dyson series with a norm remainder: on a unital C $^{*}$ -algebra, a norm-continuous skew-adjoint coefficient has a unique unitary Dyson solution; on a Banach space, the Volterra series has the stated norm remainder.

Proof

technique · apply the Volterra series to the dual Banach space of bounded functionals and identify its solution with unitary conjugation

Given: [A1], [F1]–[F6], and the invariant-state condition in the Statement.

1.1givenA1F1F2F3F4F5algebra

The finite-range group application uses Countable Choice [A1], the local matrix C $^{*}$ identity [F1], the finite-range interaction [F2], the quasi-local completion [F3], and the exact group conclusion [F5]. The extension $ϕ$ exists by [F4]. Since $t \mapsto α_{t} (B)$ is norm-continuous by [F5], $B_{t}$ is norm-continuous and $K_{t}^{*} (ψ) (X) : = \frac{i h (t)}{ℏ} ψ ([X, B_{t}])$ is a norm-continuous bounded operator on the continuous dual $A^{*}$ . Its norm is at most $2 ∣ h (t) ∣ ∥ B ∥ / ℏ$ , because automorphisms are isometric and $∥ [X, B_{t}] ∥ \leq 2 ∥ X ∥ ∥ B ∥$ . The dual $A^{*}$ is complete: a Cauchy sequence of bounded functionals has pointwise scalar limits, and the uniform Cauchy bound passes to that limit in the dual norm.

2.1givenF4F6step 1.1algebra

By [F6], the initial functional $ϕ$ has a unique Volterra solution $ψ_{ϵ} (t)$ in $A^{*}$ with first-order term $ϵ \int_{0}^{t} K_{s}^{*} (ϕ) d s$ and remainder dual norm at most $(∣ ϵ ∣ C_{T})^{2} e^{∣ ϵ ∣ C_{T}} / 2$ , since $∥ ϕ ∥ = 1$ . The C $^{*}$ -algebra clause of [F6] gives a norm-differentiable unitary $W_{ϵ}$ . Differentiating $X \mapsto ϕ (W_{ϵ} (t)^{*} X W_{ϵ} (t))$ shows this functional solves the same dual Volterra equation, so uniqueness gives $ψ_{ϵ} (t) (X) = ϕ (W_{ϵ} (t)^{*} X W_{ϵ} (t))$ .

3.1givenF5step 2.1algebra

Evaluate the first-order functional term at $X = α_{t} (A)$ . It is $\frac{i ϵ}{ℏ} \int_{0}^{t} h (s) ϕ ([α_{t} (A), B_{s}]) d s$ . The group law gives $[α_{t} (A), B_{s}] = α_{s} ([α_{t - s} (A), B])$ , and invariance of $ϕ$ removes $α_{s}$ . This is the stated convolution.

4.1givenF4step 2.1step 3.1algebra∎

Since $∥ α_{t} (A) ∥ = ∥ A ∥$ , evaluating the dual-norm remainder from step 2.1 at $α_{t} (A)$ gives the bound in the Statement. If $A = A^{*}$ as well as $B = B^{*}$ , positivity and Hermiticity of $ϕ$ make $R_{A B} (u)$ real.

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Nonzero pure-point equilibrium covariance is not integrable

Statement

Assume the Axiom of Choice and the trace-class Gibbs hypotheses of Trace-Gibbs bounded correlation measure and balance for a self-adjoint Hamiltonian $K$ in J with a complete discrete energy basis and faithful Gibbs density $σ_{β}$ at $β > 0$ in J $^{- 1}$ . Let $ℏ > 0$ be in J s, $t$ in s, and let $A = A^{*}$ be a nonzero bounded observable with $Tr (σ_{β} A) = 0$ . Define $C_{A A} (t) = Tr (σ_{β} U (t)^{*} A U (t) A), G_{A A} (t) = Re C_{A A} (t), U (t) = e^{- i t K / ℏ} .$ Then neither $C_{A A}$ nor $G_{A A}$ belongs to $L^{1} ([0, \infty))$ . In particular, the ordinary absolutely convergent isolated-system correlation integral used in a dc Green–Kubo formula does not exist under these hypotheses. This claim concerns discrete-spectrum trace-Gibbs dynamics only; it asserts no failure of temporal decay or transport limits in an infinite-volume or open-system model.

Facts & Assumptions

Given: The Axiom of Choice, a faithful trace-class Gibbs state with a complete energy eigenbasis, a nonzero bounded self-adjoint centered $A$ , and $ℏ > 0$ .

[A1]

The Axiom of Choice: full Choice is assumed by the unbounded self-adjoint spectral theorem and trace-Gibbs supplier.

[F1]

Trace-Gibbs bounded correlation measure and balance: for the Gibbs energy basis, $C_{A A} (t)$ has an absolutely summable atomic expansion with nonnegative weights $r_{i} ∣ A_{i j} ∣^{2}$ , whose total is $C_{A A} (0) = \sum_{i} r_{i} ∥ A u_{i} ∥^{2} \leq ∥ A ∥^{2}$ ; $σ_{β}$ is faithful.

[F2]

Dominated convergence: on any measure space, pointwise convergence under an integrable majorant permits passage of limits through integrals.

Proof

technique · grouped pure-point frequencies have a nonzero mean-square limit, which contradicts absolute integrability

Given: [A1], [F1], [F2], and the centered nonzero observable in the Statement.

1.1givenA1F1algebra

Group the nonnegative coefficients in [F1] by equal frequency, writing $C_{A A} (t) = \sum_{ω \in Ω} w_{ω} e^{- i ω t}$ with $w_{ω} \geq 0$ and $\sum_{ω} w_{ω} = C_{A A} (0) = Tr (σ_{β} A^{2})$ . This sum is strictly positive: in the complete energy basis, $Tr (σ_{β} A^{2}) = \sum_{i} r_{i} ∥ A u_{i} ∥^{2}$ , every $r_{i} > 0$ , and $A \neq 0$ prevents all $A u_{i}$ from vanishing. It is bounded by $∥ A ∥^{2}$ because each $∥ A u_{i} ∥^{2} \leq ∥ A ∥^{2}$ and $\sum_{i} r_{i} = 1$ .

2.1givenF1F2step 1.1algebra

Absolute summability gives $\sum_{ω, ν} w_{ω} w_{ν} < \infty$ and allows termwise integration of $∣ C_{A A} (t) ∣^{2}$ . For each pair, $T^{- 1} \int_{0}^{T} e^{- i (ω - ν) t} d t$ tends to $1$ if $ω = ν$ and $0$ otherwise, and its modulus is at most $1$ . Apply [F2] to the countable pair set with summable majorant $w_{ω} w_{ν}$ ; then $\lim_{T \to \infty} \frac{1}{T} \int_{0}^{T} ∣ C_{A A} (t) ∣^{2} d t = \sum_{ω} w_{ω}^{2} > 0.$ If $C_{A A} \in L^{1} ([0, \infty))$ , the left side is at most $∥ A ∥^{2} T^{- 1} \int_{0}^{T} ∣ C_{A A} (t) ∣ d t \to 0$ , because step 1.1 gives $∣ C_{A A} (t) ∣ \leq \sum_{ω} w_{ω} \leq ∥ A ∥^{2}$ . This contradiction proves $C_{A A} \notin L^{1}$ .

3.1F1F2step 1.1step 2.1algebra∎

The real part has grouped nonnegative weights $g_{ω} = (w_{ω} + w_{- ω}) / 2$ , with absent frequencies assigned weight zero. They satisfy $\sum_{ω} g_{ω} = \sum_{ω} w_{ω} > 0$ and $G_{A A} (t) = \sum_{ω} g_{ω} e^{- i ω t}$ . Repeating step 2.1 with $g_{ω}$ gives $\lim_{T \to \infty} T^{- 1} \int_{0}^{T} G_{A A} (t)^{2} d t = \sum_{ω} g_{ω}^{2} > 0$ . If $G_{A A} \in L^{1}$ , then this average is at most $∥ A ∥^{2} T^{- 1} \int_{0}^{T} ∣ G_{A A} (t) ∣ d t \to 0$ , again a contradiction.

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Integrability-qualified quantum variance Green–Kubo limit

Statement

Assume Countable Choice and the finite-matrix C $^{*}$ -identity hypothesis used by Bounded C-star Dyson response for finite-range spin dynamics. Use its finite-range spin algebra $A$ , strongly continuous automorphism group $α_{t}$ , and invariant state $ϕ$ obtained from consistent local densities. Let $V = V^{*} \in A$ satisfy $ϕ (V) = 0$ , and define $X_{T} = \int_{0}^{T} α_{t} (V) d t, G (t) = Re ϕ (V α_{t} (V)) .$ If $G \in L^{1} ([0, \infty))$ , then $\lim_{T \to \infty} \frac{ϕ (X_{T}^{2})}{2 T} = D : = \int_{0}^{\infty} G (t) d t \geq 0.$ For a finite list of centered self-adjoint observables $V_{1}, \dots, V_{m}$ , if each symmetrized covariance $G_{a b} (t) = \frac{1}{2} (Re ϕ (V_{a} α_{t} (V_{b})) + Re ϕ (V_{b} α_{t} (V_{a})))$ is absolutely integrable on $[0, \infty)$ , then $D_{a b} = \int_{0}^{\infty} G_{a b} (t) d t$ is a real positive semidefinite matrix. For a velocity observable measured in m s $^{- 1}$ , $D$ has units m $^{2}$ s $^{- 1}$ ; for general $V$ , its units are $[V]^{2}$ s. The result is a variance of a time-integrated observable, not a joint classical trajectory or a universal transport law.

Facts & Assumptions

Given: Countable Choice and the finite-matrix C $^{*}$ -identity hypothesis of the finite-range group construction; the invariant state and group of the algebraic response theorem; centered self-adjoint observables; and absolute integrability of the stated covariance functions.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): Countable Choice is the exact choice assumption for the local-algebra completion; no full Axiom of Choice is used.

[F1]

Finite matrix C-star algebra and density-state duality: every finite local matrix algebra has its induced-norm C $^{*}$ -identity.

[F2]

Bounded C-star Dyson response for finite-range spin dynamics: under [A1], [F1], and the stated finite-range interaction assumptions, $α_{t}$ is a strongly continuous automorphism group on the quasi-local algebra and $ϕ$ is an invariant bounded state.

[F3]

Stationary-state covariance identity for time-integrated observables: for centered self-adjoint observables, the time-integrated symmetrized covariance equals the finite-time symmetrized variance and its matrix is positive semidefinite.

[F4]

Dominated convergence: pointwise convergence under an integrable majorant permits passage of limits through Lebesgue integrals.

Proof

technique · apply the stationary covariance identity and pass to the infinite-time limit by dominated convergence

Given: [A1], [F1]–[F4], the centered observables, and the integrability hypotheses.

1.1givenA1F1F2F3F4algebra

The algebraic group/state hypotheses follow from [A1], [F1], and the exact interaction and state assumptions carried by [F2]; applying the stationary covariance identity [F3] to the scalar observable gives $\frac{ϕ (X_{T}^{2})}{2 T} = \int_{0}^{T} (1 - \frac{t}{T}) G (t) d t .$ Extend the integrand by zero for $t > T$ . For each fixed $t \geq 0$ it converges to $G (t)$ as $T \to \infty$ , and its absolute value is at most $∣ G (t) ∣$ , which is integrable by hypothesis. Applying [F4] on $[0, \infty)$ yields the limit $\int_{0}^{\infty} G (t) d t$ .

2.1givenF2F3step 1.1algebra

For finite $T$ , $X_{T} = X_{T}^{*}$ by [F3], so $ϕ (X_{T}^{2}) = ϕ (X_{T}^{*} X_{T}) \geq 0$ . The limit in step 1.1 is therefore nonnegative.

3.1F2F3F4step 1.1algebra∎

For $m$ observables, [F3] gives the finite-time real symmetric covariance matrix with entries $D_{a b} (T) = \int_{0}^{T} (1 - t / T) G_{a b} (t) d t$ ; for every real vector $c$ , $c^{T} D (T) c = ϕ ((\sum_{a} c_{a} X_{a, T})^{2}) / (2 T) \geq 0$ . Each entry converges to $D_{a b} = \int_{0}^{\infty} G_{a b} (t) d t$ by the same dominated-convergence argument as step 1.1. For any $c$ , $c^{T} D c = \lim_{T} c^{T} D (T) c \geq 0$ , so $D$ is positive semidefinite.

5 · Examples, experiments and counterexamples

None yet.

Trace-class bounded-observable response

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

Banach-space Volterra and Dyson series with a norm remainder

Statement

Facts & Assumptions

Proof

Finite matrix trace-norm ideal inequalities

Statement

Facts & Assumptions

Proof

Lorentzian kernels converge to delta and principal-value distributions

Statement

Facts & Assumptions

Proof

Stationary-state covariance identity for time-integrated observables

Statement

Facts & Assumptions

Proof

Trace-class bounded-observable response

Statement

Facts & Assumptions

Proof

Finite unitary dynamics and stationary Gibbs states

Statement

Facts & Assumptions

Proof

Finite Cesaro dephasing with degenerate energy blocks

Statement

Facts & Assumptions

Proof

Finite-time Kubo formula with norm remainder

Statement

Facts & Assumptions

Proof

Arbitrarily late finite phase recurrences

Statement

Facts & Assumptions

Proof

Bounded-commutator graph-domain Schrödinger response

Statement

Facts & Assumptions

Proof

Adopted isolated prepared unitary model

Postulate

Prepared control and expectation convention

Postulate

Finite spectral balance and distributional fluctuation response

Statement

Facts & Assumptions

Proof

Absolute convergence of trace-Gibbs bounded correlation series

Statement

Facts & Assumptions

Proof

Trace-Gibbs bounded correlation measure and balance

Statement

Facts & Assumptions

Proof

Conditional prepared finite quantum control response

Statement

Facts & Assumptions

Proof

Bounded C-star Dyson response for finite-range spin dynamics

Statement

Facts & Assumptions

Proof

Nonzero pure-point equilibrium covariance is not integrable

Statement

Facts & Assumptions

Proof

Integrability-qualified quantum variance Green–Kubo limit

Statement

Facts & Assumptions

Proof

5 · Examples, experiments and counterexamples

Sources