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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.

Stationary-state covariance identity for time-integrated observables

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources