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Abstract trace-class instruments: effects and Heisenberg duality

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a nonzero complex Hilbert space (Hilbert space) and let $(X, Σ)$ be a measurable space. In this item, a positive trace-class operator means a self-adjoint trace-class operator $T$ with $⟨ T x, x ⟩ \geq 0$ for every $x \in H$ ; a bounded effect means a self-adjoint $F \in B (H)$ with $0 \leq ⟨ F x, x ⟩ \leq ∥ x ∥^{2}$ for every $x \in H$ .

By an abstract instrument on $(X, Σ)$ we mean a map $J : Σ \to B (S_{1} (H))$ into bounded linear maps of the trace-class operators such that (i) each $J_{B}$ is completely positive (every finite amplification $I_{C^{n}} \otimes J_{B}$ preserves positivity on positive trace-class operators); (ii) for pairwise disjoint $B_{n}$ with union $B$ , $J_{B} (ρ) = \sum_{n} J_{B_{n}} (ρ)$ in trace norm for every trace-class $ρ$ ; (iii) $J_{X}$ is trace preserving.

Then:

(a) for every $B \in Σ$ there is a unique effect $F_{B} \in B (H)$ , $0 \leq F_{B} \leq I$ (Self-adjoint, positive, unitary and normal operators), with $tr (J_{B} (T)) = tr (T F_{B})$ for every trace-class $T$ ;

(b) $F_{X} = I$ , and $B \mapsto F_{B}$ is weakly countably additive: for pairwise disjoint $B_{n}$ with union $B$ and all $x, y \in H$ , $⟨ F_{B} x, y ⟩ = \sum_{n} ⟨ F_{B_{n}} x, y ⟩$ ;

(c) for every $B$ and every bounded $A \in B (H)$ there is a unique bounded operator $J_{B}^{*} (A)$ with $tr (J_{B} (T) A) = tr (T J_{B}^{*} (A))$ for every trace-class $T$ ; the map $A \mapsto J_{B}^{*} (A)$ is linear and bounded with $J_{B}^{*} (A) \geq 0$ for $A \geq 0$ and $J_{X}^{*} (I) = I$ ;

(d) for a density operator $ρ$ and $B \in Σ$ the number $p (B) = tr (J_{B} (ρ)) = tr (ρ F_{B})$ lies in $[0, 1]$ , and $p$ is countably additive. Whether an experimental procedure realizes such an instrument is a separate model assumption; the conditional update $J_{B} (ρ) / p (B)$ is defined only when $p (B) > 0$ .

No universal representation theorem for arbitrary instruments is asserted.

Facts & Assumptions

Given: A nonzero complex Hilbert space $H$ ; an abstract instrument $J$ as defined; bounded $A$ ; vectors $x, y, u, v$ .

[F1]

$T_{x, y}$ is trace class with $tr (T_{x, y}) = ⟨ x, y ⟩$ , $∥ T_{x, y} ∥_{1} = ∥ x ∥ ∥ y ∥$ ; $T_{x, y}^{*} = T_{y, x}$ ; $tr (S T_{x, y}) = ⟨ S x, y ⟩$ for trace-class $S$ ; and $∣ tr T ∣ \leq ∥ T ∥_{1}$ (Rank-one operator calculus, Nuclear bounded cyclicity for trace-class operators, Trace is absolutely convergent and basis independent).

[F2]

Trace-class operators form a Banach space in trace norm, and the trace is linear and trace-norm continuous; for bounded $B$ , $∥ B T ∥_{1} \leq ∥ B ∥ ∥ T ∥_{1}$ (Trace class is a two sided Banach operator ideal).

[F3]

A bounded linear functional $f$ on $H$ has a unique Riesz representative $y$ with $f (x) = ⟨ x, y ⟩$ (Riesz representation for Hilbert spaces). Applying this to the conjugate of each bounded conjugate-linear section of a sesquilinear form represents that form by a unique bounded operator. If the diagonal values of the form are real, polarization makes it Hermitian, and the representing operator is self-adjoint.

[F4]

Complete positivity implies positivity of $J_{B}$ by the definition in this Statement. A self-adjoint positive trace-class operator has an orthonormal rank-one expansion with nonnegative weights summing to its trace, and its trace is the sum of its positive eigenvalues (Density operators and scalar spectral probability, Trace of a positive operator is the sum of its eigenvalues).

Proof

1.1givenF1F2algebra

[The functional and its sesquilinear form] Fix $B \in Σ$ . The map $f_{B} (T) : = tr (J_{B} (T))$ is a bounded linear functional on $S_{1} (H)$ with $∣ f_{B} (T) ∣ \leq ∥ J_{B} ∥ ∥ T ∥_{1}$ by [F2]. Define $b_{B} (u, v) : = f_{B} (T_{u, v})$ on rank-one operators. Because $T_{u, v}$ is linear in $u$ and conjugate-linear in $v$ and $f_{B}$ is linear, $b_{B}$ is sesquilinear; by [F1], $∣ b_{B} (u, v) ∣ \leq ∥ J_{B} ∥ ∥ u ∥ ∥ v ∥$ .

2.1step 1.1F1F2F3F4algebra

[The effect] For fixed $u$ , Riesz representation of $v \mapsto \overline{b_{B} (u, v)}$ gives a unique vector $F_{B} u$ with $b_{B} (u, v) = ⟨ F_{B} u, v ⟩$ ; the sesquilinear bound in step 1.1 makes $u \mapsto F_{B} u$ a bounded linear operator. By [F4], $J_{B} (T_{u, u})$ is self-adjoint positive trace class and has nonnegative trace, so $b_{B} (u, u) \geq 0$ . The complement relation $J_{X} = J_{B} + J_{X ∖ B}$ and positivity of $J_{X ∖ B}$ give $b_{B} (u, u) \leq b_{X} (u, u) = tr (T_{u, u}) = ∥ u ∥^{2}$ . Thus the diagonal form values are real, so polarization makes $b_{B}$ Hermitian and $F_{B}$ self-adjoint; the two inequalities show it is an effect in the stated sense. Linearity and trace-norm continuity [F1, F2] extend $tr (J_{B} (T_{u, v})) = ⟨ F_{B} u, v ⟩ = tr (T_{u, v} F_{B})$ to every trace-class $T$ . Uniqueness follows because all pairings $⟨ F_{B} u, v ⟩$ are fixed. This is (a).

3.1step 2.1F1F2algebra

[Normalization and countable additivity] For $T = T_{u, v}$ and $B = X$ the trace-preservation clause of the instrument gives $⟨ F_{X} u, v ⟩ = tr (J_{X} (T_{u, v})) = tr (T_{u, v}) = ⟨ u, v ⟩$ , so $F_{X} = I$ . For pairwise disjoint $B_{n}$ with union $B$ , the instrument's countable additivity says $\sum_{n \leq N} J_{B_{n}} (T) \to J_{B} (T)$ in trace norm; by linearity and trace-norm continuity of the functional $tr (\cdot F_{B})$ -pairing [F1, F2], $tr (T_{u, v} F_{B}) = tr (J_{B} (T_{u, v})) = \lim_{N} \sum_{n \leq N} tr (J_{B_{n}} (T_{u, v})) = \sum_{n} ⟨ F_{B_{n}} u, v ⟩$ by step 2.1. As $u, v$ are arbitrary, this is $⟨ F_{B} u, v ⟩ = \sum_{n} ⟨ F_{B_{n}} u, v ⟩$ , which is (b).

4.1step 2.1step 3.1F1F2F3F4algebra

[Heisenberg dual] Fix $B$ and bounded $A$ . Define $c_{A} (u, v) : = tr (J_{B} (T_{u, v}) A)$ . This is sesquilinear in $(u, v)$ with $∣ c_{A} (u, v) ∣ \leq ∥ J_{B} ∥ ∥ A ∥ ∥ u ∥ ∥ v ∥$ by [F1, F2]; by the Riesz construction in [F3] there is a unique bounded $J_{B}^{*} (A)$ with $c_{A} (u, v) = ⟨ J_{B}^{*} (A) u, v ⟩$ and $∥ J_{B}^{*} (A) ∥ \leq ∥ J_{B} ∥ ∥ A ∥$ . Then for rank-one $T_{u, v}$ , $tr (J_{B} (T_{u, v}) A) = ⟨ J_{B}^{*} (A) u, v ⟩ = tr (T_{u, v} J_{B}^{*} (A))$ by [F1]; both sides are linear and trace-norm continuous in $T$ [F2], so the identity holds for all trace-class $T$ . Linearity of $A \mapsto J_{B}^{*} (A)$ follows by uniqueness of the representing operator. If $A \geq 0$ and $x \in H$ , then $⟨ J_{B}^{*} (A) x, x ⟩ = tr (J_{B} (T_{x, x}) A) \geq 0$ : by [F4], $J_{B} (T_{x, x}) = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ with $w_{n} \geq 0$ , and the trace is $\sum_{n} w_{n} ⟨ A u_{n}, u_{n} ⟩$ , a convergent nonnegative sum. Its diagonal values are real, so polarization also gives $J_{B}^{*} (A)^{*} = J_{B}^{*} (A)$ . Finally $J_{X}^{*} (I) = I$ by $tr (J_{X} (T) I) = tr T = tr (T I)$ and uniqueness. This is (c).

5.1step 2.1step 3.1F1F2F4algebra∎

[Probabilities] For a density operator $ρ = \sum_{n} w_{n} T_{u_{n}, u_{n}}$ , step 2.1 gives $p (B) = tr (J_{B} (ρ)) = tr (ρ F_{B}) = \sum_{n} w_{n} ⟨ F_{B} u_{n}, u_{n} ⟩$ . Since $0 \leq F_{B} \leq I$ , each diagonal term lies in $[0, 1]$ and $\sum_{n} w_{n} = 1$ , so $p (B) \in [0, 1]$ . Trace-norm additivity of $J$ and trace continuity imply countable additivity of $p$ , and $p (X) = 1$ by trace preservation. Positivity and normalization of $J_{B} (ρ) / p (B)$ hold precisely when $p (B) > 0$ . Whether the prescribed instrument is experimentally realized remains a separate model assumption.

Abstract trace-class instruments: effects and Heisenberg duality

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources