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Spectral first moments and trace products with a density operator

Statement

Assume Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Let $A$ be a self-adjoint operator on a nonzero complex Hilbert space, let $E_{A}$ be its spectral PVM, let $ρ$ be a density operator, and set $μ_{ρ}^{A} (B) : = tr (ρ E_{A} (B))$ . Then:

(a) The finite first mean is defined exactly under the absolute-first-moment condition $\int_{R} ∣ a ∣ d μ_{ρ}^{A} (a) < \infty,$ and its value is $m : = \int a d μ_{ρ}^{A} (a)$ . For every positive-eigenvalue expansion $ρ = \sum_{k} w_{k} T_{u_{k}, u_{k}}$ of Density operators and scalar spectral probability (a), with $w_{k} > 0$ and $\sum_{k} w_{k} = 1$ , one has $\int ∣ a ∣ d μ_{ρ}^{A} = \sum_{k} w_{k} \int ∣ a ∣ d E_{A, u_{k}}, m = \sum_{k} w_{k} \int a d E_{A, u_{k}} .$ The first identity allows infinity; the second is asserted under the stated integrability condition and converges absolutely. This condition does not require $u_{k} \in D (A)$ .

(b) If $A$ is bounded, the condition in (a) holds automatically and $m = tr (A ρ) = tr (ρ A) .$

(c) Suppose the condition in (a) holds and the literal product $A ρ$ is defined on the whole Hilbert space and is trace class; thus $ρ (H) \subset D (A)$ and $T : = A ρ \in S_{1} (H)$ . Then $m = tr (T) .$ Moreover the densely defined product $ρ A$ has bounded closure $T^{*}$ , that closure is trace class, and $tr (\overline{ρ A}) = m$ . This is a conditional trace identity for defined trace-class products. The existence of the mean alone does not assert any of these unbounded-operator product hypotheses.

Facts & Assumptions

Given: The choice hypotheses, $A, E_{A}, ρ, μ_{ρ}^{A}$ , and in (c) the explicitly stated literal-product assumptions.

[F1]

The density eigenexpansion has positive weights and orthonormal vectors, converges in trace norm, and gives $μ_{ρ}^{A} (B) = \sum_{k} w_{k} E_{A, u_{k}} (B)$ . For $u \in D (A)$ , $\int a d E_{A, u} = ⟨ A u, u ⟩$ ; absolute first integrability alone is equivalent to $u \in D (∣ A ∣^{1 / 2})$ (Density operators and scalar spectral probability (a)-(c)).

[F2]

Nonnegative monotone limits commute with integrals, and integrable domination permits limits of signed integrals (Monotone convergence for the integral, Dominated convergence).

[F3]

Bounded Borel spectral integrals agree with their scalar spectral integrals (Bounded borel pvm integral). Bounded multiplication of a density eigenexpansion gives $tr (A ρ) = \sum_{k} w_{k} ⟨ A u_{k}, u_{k} ⟩$ and bounded cyclicity gives $tr (A ρ) = tr (ρ A)$ (Nuclear bounded cyclicity for trace-class operators).

[F4]

A trace-class operator has its trace as an absolutely convergent diagonal in a supplied Hilbert basis; it also has a nuclear representation whose rank-one traces sum to its trace (Trace is absolutely convergent and basis independent, Nuclear series characterizes trace norm). Under AC, order the set of orthonormal families containing the given family by inclusion. This poset is nonempty because it contains the given family. Every nonempty chain has its orthonormal union as an upper bound in this poset; the empty chain has the given family as an upper bound. Thus Zorn's lemma provides a maximal extension. Its orthogonal complement is zero by maximality, hence it is complete by Parseval equivalences for an orthonormal family. Coordinate expansion is Fourier expansion in a Hilbert space. Finite-rank maps and their operator-norm limits are compact (Bounded finite rank operators are compact, Norm limit of compact operators is compact).

Proof

technique · direct; integrate the density eigenexpansion and compare it with correctly defined trace products

Given: The data of the statement and [F1]-[F4].

1.1F1F2algebra

[First-moment expansion] By [F1] the outcome measure is the countable weighted sum of the scalar measures. This gives $\int g d μ_{ρ}^{A} = \sum_{k} w_{k} \int g d E_{A, u_{k}}$ first for nonnegative simple functions by the measure identity and finite linearity, and then for all nonnegative measurable $g$ by increasing simple approximation and [F2]; the nonnegative sums interchange as increasing finite subsums. Taking $g = ∣ a ∣$ gives the first identity in (a), including infinity. When it is finite, both positive and negative parts are integrable, so subtracting their two finite sums gives $m = \sum_{k} w_{k} \int a d E_{A, u_{k}}$ with $\sum_{k} w_{k} ∣ \int a d E_{A, u_{k}} ∣ \leq \sum_{k} w_{k} \int ∣ a ∣ d E_{A, u_{k}} < \infty$ . For each positive $w_{k}$ , only absolute first integrability is forced; [F1] identifies its form domain, without requiring the operator-square moment. This proves (a).

2.1F1F3step 1.1algebra

[Bounded observable] If $A$ is bounded, every $u_{k}$ is in $D (A)$ , and Cauchy-Schwarz for the scalar probability measure gives $\int ∣ a ∣ d E_{A, u_{k}} \leq (\int a^{2} d E_{A, u_{k}})^{1 / 2} = ∥ A u_{k} ∥ \leq ∥ A ∥$ by [F1] (Cauchy-Schwarz inequality for $L^{2}$ ). Step 1.1 therefore bounds the first absolute moment by $∥ A ∥$ . Each $u_{k}$ is in $D (A)$ and [F1] gives $\int a d E_{A, u_{k}} = ⟨ A u_{k}, u_{k} ⟩$ . Thus step 1.1 and the bounded trace formula [F3] give $m = \sum_{k} w_{k} ⟨ A u_{k}, u_{k} ⟩ = tr (A ρ) = tr (ρ A)$ . This proves (b).

2.2F1F4step 1.1choosealgebra

[Literal trace-class left product] Under (c), $ρ u_{k} = w_{k} u_{k} \in D (A)$ and $w_{k} > 0$ , so each $u_{k} \in D (A)$ and $T u_{k} = w_{k} A u_{k}$ . Complete the orthonormal eigenfamily by an orthonormal basis of its perpendicular subspace using the already assumed AC. The density eigenexpansion shows $ρ = 0$ on that perpendicular subspace, so $T = 0$ there. The trace diagonal formula [F4] therefore gives $tr (T) = \sum_{k} ⟨ T u_{k}, u_{k} ⟩ = \sum_{k} w_{k} ⟨ A u_{k}, u_{k} ⟩$ . Step 1.1 and [F1] identify this absolutely convergent scalar sum with $m$ . This proof uses the literal-product hypothesis to obtain $u_{k} \in D (A)$ ; it does not infer that hypothesis from first-moment existence.

3.1F4step 2.2algebra∎

[Closure of the right product] For $x \in D (A)$ and arbitrary $y$ , self-adjointness of $A$ and $ρ$ gives $⟨ ρ A x, y ⟩ = ⟨ A x, ρ y ⟩ = ⟨ x, A ρ y ⟩ = ⟨ x, T y ⟩ = ⟨ T^{*} x, y ⟩$ , so $ρ A$ agrees with the bounded $T^{*}$ on the dense domain $D (A)$ . Its graph closure is all of the graph of $T^{*}$ : approximate any $x$ by $x_{l} \in D (A)$ and use boundedness of $T^{*}$ , while its closed graph contains every graph limit from that restriction. A nuclear representation $T = \sum_{j} T_{v_{j}, z_{j}}$ gives $T^{*} = \sum_{j} T_{z_{j}, v_{j}}$ with the same summable nuclear norms; the operator-norm convergence of these finite-rank partial sums first gives compactness, and then the nuclear criterion gives trace class and [F4] gives $tr (T^{*}) = \overline{tr (T)} = m$ , since the spectral integral $m$ is real. This proves (c), without invoking cyclicity for unbounded multipliers.

Depends on

Dependency tree · 2 levels

Sources

G. Teschl, Mathematical Methods in Quantum Mechanics, Chapters 2-3: density eigenexpansions and spectral integrals; the typed-product comparison is proved locally here