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Finite bounded instruments normalize measurement probabilities

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ carry a normalized finite bounded Kraus instrument datum as defined in Finite bounded Kraus instrument, and let $ρ$ be a density operator on $H$ : a self-adjoint positive trace-class operator with $tr ρ = 1$ (Density operators and scalar spectral probability, Trace class operator, Trace of a trace class operator, Self-adjoint, positive, unitary and normal operators). For each outcome $a$ , put $p_{a} : = tr (J_{a} (ρ))$ . Then each $E_{a}$ is an effect ( $E_{a} \geq 0$ and $I_{H} - E_{a} \geq 0$ ), each $J_{a} (ρ)$ is positive trace class, and

$p_{a} = tr (J_{a} (ρ)) = tr (ρ E_{a}) \in [0, 1], \sum_{a = 1}^{m} p_{a} = 1.$

The nonselective output $\sum_{a} J_{a} (ρ)$ is a density operator. If $p_{a} > 0$ , the normalized conditional state $ρ_{a} : = J_{a} (ρ) / p_{a}$ is a density operator; when $p_{a} = 0$ , this normalized update is undefined. For a unit vector $ψ \in H$ , the rank-one density operator $ρ = T_{ψ, ψ}$ has outcome probability $p_{a} = \sum_{r = 1}^{q_{a}} ∥ M_{a, r} ψ ∥^{2}$ . These are conditional mathematical identities; they do not assert that an apparatus realizes the instrument or supply a physical measurement postulate.

Facts & Assumptions

Given: Countable Choice; a nonzero complex Hilbert space $H$ ; a normalized finite bounded Kraus instrument datum $(M_{a, r})$ ; and a density operator $ρ$ on $H$ .

[A1]

Countable Choice ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) is assumed for the adjoint and trace-class definitions and suppliers, and for the density expansion [F1]. No stronger choice principle is invoked.

[L1]

Finite bounded Kraus instrument: the outcome operators are $E_{a} = \sum_{r} M_{a, r}^{*} M_{a, r}$ , the maps are $J_{a} (X) = \sum_{r} M_{a, r} X M_{a, r}^{*}$ , and normalization means $\sum_{a, r} M_{a, r}^{*} M_{a, r} = I_{H}$ . The adjoint obeys $⟨ M x, y ⟩ = ⟨ x, M^{*} y ⟩$ .

[F1]

Density operators and scalar spectral probability: under Countable Choice, a density operator is self-adjoint, positive and trace class with trace one, and has a trace-norm convergent expansion $ρ = \sum_{j} w_{j} T_{u_{j}, u_{j}}$ with $(u_{j})$ orthonormal, $w_{j} \geq 0$ and $\sum_{j} w_{j} = 1$ . We use statement (a); its separate statement (c) is not used.

[F2]

Nuclear bounded cyclicity for trace-class operators: under Countable Choice, bounded multiplication of a nuclear representation gives trace-norm-convergent series for $B T$ and $T B$ , with $tr (B T) = \sum_{j} ⟨ B u_{j}, v_{j} ⟩$ , $tr (T B) = \sum_{j} ⟨ u_{j}, B^{*} v_{j} ⟩$ , absolute convergence, and $tr (B T) = tr (T B)$ . No unbounded factor is cyclically permuted.

[F3]

Rank-one operator calculus: $T_{u, v} x = ⟨ x, v ⟩ u$ is trace class with $∥ T_{u, v} ∥_{1} = ∥ u ∥ ∥ v ∥$ and $tr (T_{u, v}) = ⟨ u, v ⟩$ ; $T_{x, x}$ is self-adjoint and positive.

[F4]

Rank-one operator calculus: for bounded $B$ , $B T_{u, v} = T_{B u, v}$ and $T_{u, v} B = T_{u, B^{*} v}$ .

[F5]

Trace class operator: a trace-class operator is in $B (H)$ and has finite trace norm. Trace of a trace class operator defines the trace relative to a supplied Hilbert basis. Trace is absolutely convergent and basis independent proves basis independence, linearity and $∣ tr T ∣ \leq ∥ T ∥_{1}$ , hence continuity in trace norm.

[F6]

Self-adjoint, positive, unitary and normal operators: positivity means that the quadratic-form values $⟨ T x, x ⟩$ are real and nonnegative; self-adjointness is the separate identity $T^{*} = T$ .

[F7]

Trace class is a two sided Banach operator ideal: under Countable Choice, if $T$ is trace class and $A, B$ are bounded, then $A T B$ is trace class and $∥ A T B ∥_{1} \leq ∥ A ∥ ∥ T ∥_{1} ∥ B ∥$ .

[F8]

Hilbert-adjoint identities: under ${A C}_{ω}$ , the Hilbert adjoint of every bounded operator $M$ is bounded, $∥ M^{*} ∥ = ∥ M ∥$ , and $M^{* *} = M$ .

Proof

1.1givenL1F6F8algebra

For each $a, r$ , $M_{a, r}^{*} M_{a, r}$ is bounded by [F8]. Also [F8] gives $M_{a, r}^{* *} = M_{a, r}$ , so $⟨ M_{a, r}^{*} M_{a, r} x, y ⟩ = ⟨ M_{a, r} x, M_{a, r}^{* *} y ⟩ = ⟨ M_{a, r} x, M_{a, r} y ⟩ = ⟨ x, M_{a, r}^{*} M_{a, r} y ⟩$ ; hence it is self-adjoint. Its quadratic form is $∥ M_{a, r} x ∥^{2} \geq 0$ . Hence $E_{a} \geq 0$ . Normalization gives $I_{H} - E_{a} = \sum_{b \neq a} \sum_{r} M_{b, r}^{*} M_{b, r} \geq 0$ , including empty sums, so $E_{a}$ is an effect.

1.2givenA1L1F1F6F7F8algebra

By [F8], each $M_{a, r}^{*}$ is bounded, so [F7] makes $M_{a, r} ρ M_{a, r}^{*}$ trace class; the finite sum $J_{a} (ρ)$ is trace class. By [F8], $M_{a, r}^{* *} = M_{a, r}$ , and for all $x, y \in H$ , $⟨ M_{a, r} ρ M_{a, r}^{*} x, y ⟩ = ⟨ ρ M_{a, r}^{*} x, M_{a, r}^{*} y ⟩ = ⟨ M_{a, r}^{*} x, ρ M_{a, r}^{*} y ⟩ = ⟨ x, M_{a, r}^{* *} ρ M_{a, r}^{*} y ⟩ = ⟨ x, M_{a, r} ρ M_{a, r}^{*} y ⟩$ , where the first equality uses the adjoint identity for $M_{a, r}$ , the second uses $ρ = ρ^{*}$ , and the fourth uses the adjoint identity for $M_{a, r}^{*}$ . Also $⟨ M_{a, r} ρ M_{a, r}^{*} x, x ⟩ = ⟨ ρ M_{a, r}^{*} x, M_{a, r}^{*} x ⟩ \geq 0$ . Finite summation shows $J_{a} (ρ)$ is self-adjoint and positive.

2.1givenA1L1F1F2F3F4F5F7F8step 1.2algebra

By [F1], write $ρ = \sum_{j} w_{j} T_{u_{j}, u_{j}}$ in trace norm. For its partial sums $ρ_{N} = \sum_{j \leq N} w_{j} T_{u_{j}, u_{j}}$ , [F8] gives $∥ M_{a, r}^{*} ∥ = ∥ M_{a, r} ∥ < \infty$ , so [F7] gives $∥ M_{a, r} (ρ - ρ_{N}) M_{a, r}^{*} ∥_{1} \leq ∥ M_{a, r} ∥ ∥ ρ - ρ_{N} ∥_{1} ∥ M_{a, r}^{*} ∥ \to 0$ . Thus [F4] gives $J_{a} (ρ) = \sum_{r, j} w_{j} T_{M_{a, r} u_{j}, M_{a, r} u_{j}}$ in trace norm; the trace-norm sum is finite since $\sum_{j} w_{j} ∥ M_{a, r} u_{j} ∥^{2} \leq ∥ M_{a, r} ∥^{2} \sum_{j} w_{j} = ∥ M_{a, r} ∥^{2}$ for each $r$ . Trace continuity [F5] and [F3] yield $p_{a} = \sum_{r, j} w_{j} ∥ M_{a, r} u_{j} ∥^{2} \geq 0$ . Cyclicity [F2] gives $p_{a} = \sum_{r} tr (ρ M_{a, r}^{*} M_{a, r}) = tr (ρ E_{a})$ .

3.1givenA1L1F1step 2.1algebra

Because the outcome and Kraus index sets are finite, [F1] and step 2.1 give $\sum_{a} p_{a} = \sum_{j} w_{j} ⟨ (\sum_{a, r} M_{a, r}^{*} M_{a, r}) u_{j}, u_{j} ⟩ = \sum_{j} w_{j} ∥ u_{j} ∥^{2} = 1$ by normalization. Since every $p_{a} \geq 0$ , this also gives $p_{a} \leq 1$ for every $a$ .

4.1givenA1L1step 1.2step 2.1step 3.1F1F5algebra

If $p_{a} > 0$ , step 1.2 gives a positive trace-class numerator $J_{a} (ρ)$ and step 2.1 gives its trace $p_{a}$ , so $ρ_{a} = J_{a} (ρ) / p_{a}$ is positive trace class with trace one, hence a density operator by [F1]. The finite sum $\sum_{a} J_{a} (ρ)$ is likewise positive trace class, and its trace is $\sum_{a} p_{a} = 1$ by step 3.1, so it is a density operator. If $p_{a} = 0$ , division by $p_{a}$ is undefined and no normalized update is asserted.

5.1givenL1F1F3F4step 2.1algebra∎

For a unit vector $ψ$ , [F3] gives that $T_{ψ, ψ}$ is self-adjoint, positive and trace class with trace $∥ ψ ∥^{2} = 1$ , hence a density operator by [F1]. Also [F4] gives $M_{a, r} T_{ψ, ψ} M_{a, r}^{*} = T_{M_{a, r} ψ, M_{a, r} ψ}$ . Taking traces and summing over the finite Kraus list yields $p_{a} = \sum_{r} ∥ M_{a, r} ψ ∥^{2}$ , as claimed.

Finite bounded instruments normalize measurement probabilities

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources