TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Finite instrument dilation with an outcome register

Statement

Let $d, e \geq 1$ and let $A$ be a finite nonempty outcome set. A finite completely positive instrument from $C^{d}$ to $C^{e}$ is a family of completely positive complex-linear maps $I_{a} : M_{d} (C) \to M_{e} (C)$ , indexed by $a \in A$ , whose sum is trace-preserving. Then there are an integer $q \geq 1$ , an isometry $V : C^{d} \to C^{e} \otimes C^{q}$ , and pairwise orthogonal projections $Q_{a}$ on $C^{q}$ with $\sum_{a} Q_{a} = I_{q}$ such that, for every $X \in M_{d} (C)$ ,

I_{a} (X) = {Tr}_{E} ⁣ [(I_{e} \otimes Q_{a}) V X V^{*} (I_{e} \otimes Q_{a})] .

Here $E = C^{q}$ is the finite register space with its displayed orthonormal basis, and the partial trace is defined on its matrix units by ${Tr}_{E} (Y \otimes ∣ e_{a, r} ⟩ ⟨ e_{b, s} ∣) = δ_{(a, r), (b, s)} Y$ , extended linearly. The register basis is indexed by the Kraus labels $(a, r)$ ; a zero outcome map has no labels and corresponds to $Q_{a} = 0$ .

Statement references

Finite completely positive trace-preserving maps

Complete finite Choi-Kraus equivalence

Facts & Assumptions

Given: $d, e \geq 1$ , a finite nonempty set $A$ , and completely positive maps $I_{a} : M_{d} (C) \to M_{e} (C)$ such that $\sum_{a} I_{a}$ is trace-preserving.

[A1]

The outcome set $A$ is finite and nonempty, each $I_{a}$ is completely positive, and their sum is trace-preserving, exactly as in the Statement.

[F1]

Finite completely positive trace-preserving maps: complete positivity means every finite ancilla amplification ${id}_{r} \otimes Φ$ preserves positive semidefinite matrices; trace preservation means $Tr Φ (X) = Tr X$ for every input matrix.

[F2]

Complete finite Choi-Kraus equivalence: every finite completely positive map $Φ : M_{d} (C) \to M_{e} (C)$ has a finite Kraus representation $Φ (X) = \sum_{r} K_{r} X K_{r}^{*}$ ; for any such representation, $Φ$ is trace-preserving exactly when $\sum_{r} K_{r}^{*} K_{r} = I_{d}$ .

Proof

1.1givenA1F1F2construct

By [F1], each $I_{a}$ is completely positive in the finite matrix sense. Apply [F2] to obtain a finite Kraus family $(K_{a, r})_{r}$ for it, allowing the empty family for a zero map. Since $A$ is finite, these finitely many choices require only finite choice. Put $q = \sum_{a} q_{a}$ , where $q_{a}$ is the number of Kraus operators for outcome $a$ .

2.1givenA1F1F2step 1.1algebra

The combined map $\sum_{a} I_{a}$ has Kraus family $(K_{a, r})_{a, r}$ and is trace-preserving by [A1] and [F1]. Thus [F2] gives $\sum_{a, r} K_{a, r}^{*} K_{a, r} = I_{d}$ . Since $d \geq 1$ , this identity implies $q \geq 1$ .

3.1A1step 2.1constructalgebra

Let $(e_{a, r})_{a, r}$ be the standard orthonormal basis of $C^{q}$ indexed by these labels. Define $V x = \sum_{a, r} K_{a, r} x \otimes e_{a, r}$ and $Q_{a} = \sum_{r} ∣ e_{a, r} ⟩ ⟨ e_{a, r} ∣$ . Then $V^{*} V = \sum_{a, r} K_{a, r}^{*} K_{a, r} = I_{d}$ , so $V$ is an isometry. Disjoint label sets make the $Q_{a}$ pairwise orthogonal and their sum $I_{q}$ .

4.1A1step 1.1step 3.1algebra∎

Expanding the sandwich gives $\sum_{r, s} K_{a, r} X K_{a, s}^{*} \otimes ∣ e_{a, r} ⟩ ⟨ e_{a, s} ∣$ with both labels restricted to outcome $a$ . The partial trace sends each register matrix unit $∣ e_{a, r} ⟩ ⟨ e_{a, s} ∣$ to $δ_{r s}$ , leaving $\sum_{r} K_{a, r} X K_{a, r}^{*} = I_{a} (X)$ by step 1.1. This proves the claimed formula.

Remarks

The finite Kraus family in this proof carries both the outcome label and its internal Kraus label. The projection $Q_{a}$ groups all labels that yield the same outcome; tracing out the register after that projection removes cross terms. The identity is a finite-dimensional algebraic dilation statement and makes no claim about a laboratory apparatus.

The following diagram records the factorization of the displayed map. Its arrows name the maps in the formula; the proof is the matrix expansion in step 4.1 and does not rely on the diagram.

diagram failed to render:
M_d(\mathbb C)
  \arrow[r, "X\mapsto VXV^*"]
  \arrow[dr, "I_a"']
&
M_e(\mathbb C)\otimes M_q(\mathbb C)
  \arrow[r, "Z\mapsto(I_e\otimes Q_a)Z(I_e\otimes Q_a)"]
&
M_e(\mathbb C)\otimes M_q(\mathbb C)
  \arrow[dl, "\operatorname{Tr}_E"]
\\
& M_e(\mathbb C)

Finite instrument dilation with an outcome register

Statement

Statement references

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources