CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Outcome probabilities do not determine the state update

Statement refuted

For finite quantum instruments on $C^{2}$ , equality of all outcome effects forces equality of the outcome update maps on every input matrix.

Statement references

Finite instrument dilation with an outcome register

Finite effects and instruments: normalization and non-determination of the update

Facts & Assumptions

Given: The standard basis $∣ 0 ⟩, ∣ 1 ⟩$ of $C^{2}$ , the associated projections $P_{a} = ∣ a ⟩ ⟨ a ∣$ , and $∣ + ⟩ = (∣ 0 ⟩ + ∣ 1 ⟩) / \sqrt{2}$ .

[A1]

The state $ρ = P_{0}$ and the Kraus families are exactly those specified in the Counterexample.

[F1]

Finite effects and instruments: normalization and non-determination of the update: Kraus operators $M_{a, r}$ define effects $E_{a} = \sum_{r} M_{a, r}^{*} M_{a, r}$ and outcome maps $I_{a} (ρ) = \sum_{r} M_{a, r} ρ M_{a, r}^{*}$ ; the family is normalized when all effects sum to $I$ .

[F2]

Finite instrument dilation with an outcome register: a finite family of completely positive outcome maps whose sum is trace-preserving admits the finite orthogonal outcome-register dilation proved on the A page.

[F3]

Complete finite Choi-Kraus equivalence: a finite Kraus representation defines a completely positive map, and the map is trace-preserving when the Kraus adjoint products sum to identity.

Counterexample

Consider the two-outcome sharp measurement $P_{0}, P_{1}$ . The Lüders instrument has one Kraus operator $M_{a}^{L} = P_{a}$ per outcome. The measure-and-prepare instrument has $M_{a}^{R} = ∣ + ⟩ ⟨ a ∣$ per outcome. Both have effects $E_{0} = P_{0}, E_{1} = P_{1}$ , but their update maps differ on the input $ρ = P_{0}$ .

1.1givenA1F1algebra

Since $P_{a}^{*} = P_{a} = P_{a}^{2}$ , the Lüders Kraus operators have $M_{a}^{L *} M_{a}^{L} = P_{a}$ . For the other family, $M_{a}^{R *} M_{a}^{R} = (∣ a ⟩ ⟨ + ∣) (∣ + ⟩ ⟨ a ∣) = ⟨ + ∣ + ⟩ P_{a} = P_{a}$ , since the displayed normalized vector satisfies $⟨ + ∣ + ⟩ = 1$ . Thus both families are normalized, with the same outcome effects, because $P_{0} + P_{1} = I$ .

2.1A1F2F3step 1.1algebra

By [F3], the one-operator Kraus forms give completely positive maps, and their sum is trace-preserving because the combined adjoint products sum to identity by step 1.1. Hence both are finite instruments in the sense of the A-page theorem, and each admits its stated outcome-register dilation by [F2].

3.1A1F1step 1.1step 2.1algebra∎

On $ρ = P_{0}$ , the Lüders update for outcome $0$ is $I_{0}^{L} (P_{0}) = P_{0}$ , whereas the other update is $I_{0}^{R} (P_{0}) = ∣ + ⟩ ⟨ + ∣$ . These matrices differ: $P_{0} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $∣ + ⟩ ⟨ + ∣ = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ . The outcome probabilities are nevertheless identical, $Tr (P_{0} P_{0}) = 1$ and $Tr (P_{0} P_{1}) = 0$ . This refutes the statement.

Outcome probabilities do not determine the state update

Statement refuted

Statement references

Facts & Assumptions

Counterexample

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