TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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A zero-probability instrument branch has no conditioned state

Statement

Let $(I_{a})_{a \in Ω}$ be a finite instrument and let $ρ$ be an input density operator. If $p_{a} = tr (I_{a} (ρ)) = 0$ , then $I_{a} (ρ) = 0$ and the conditional value for outcome $a$ is $⊥$ , not a density operator.

Facts & Assumptions

Given: A finite instrument, an input density operator $ρ$ , and an outcome $a$ whose branch weight $p_{a}$ is zero.

[F1]

Finite instruments and conditional states: $p_{a} = tr (I_{a} (ρ))$ , and the conditional value is $⊥$ when $p_{a} = 0$ .

[F2]

Completely positive maps: each finite-ancilla extension of a completely positive branch map preserves nonnegative operators.

[F5]

Density operator: on a nonzero finite-dimensional complex inner-product space, a density operator is nonnegative and has trace one.

[F3]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: a nonnegative operator has an orthonormal basis in which it is diagonal with nonnegative real diagonal entries.

[F4]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: trace is the basis-independent sum of the diagonal entries in any chosen basis.

[F6]

Non-negative and positive operators: a nonnegative operator is self-adjoint and has nonnegative quadratic form on every vector.

Proof

1.1givenF1F2F3F4F5F6

Since $I_{a}$ is completely positive, its $r = 1$ extension sends the nonnegative $ρ$ from [F5] to the nonnegative operator $τ : = I_{a} (ρ)$ by [F2], with nonnegativity as defined in [F6]. By [F3], $τ$ is diagonal with entries $λ_{j} \geq 0$ in an orthonormal basis. Its trace is $\sum_{j} λ_{j} = 0$ by the hypothesis and [F1], [F4]; a finite sum of nonnegative numbers is zero only when every $λ_{j} = 0$ . Hence $τ = 0$ .

2.1givenF1step 1.1∎

The instrument definition [F1] assigns $⊥$ when $p_{a} = 0$ . The algebraic quotient $I_{a} (ρ) / p_{a}$ would have zero denominator, so it is not a normalized conditional density state.

Depends on

Used by

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Sources

John Watrous, The Theory of Quantum Information (2018), §2.3.2 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 3: Foundations II: Measurement and Evolution, §3.2.4 (standard reference, not scraped)