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Positive-probability instrument branches are density operators

Statement

Let ρ be an input density operator for a finite instrument (Ia)a∈Ω, and let pa=tr⁡(Ia(ρ)). If pa>0, then the conditional operator ρa=Ia(ρ)/pa is a density operator on the instrument's output space.

Facts & Assumptions

Given: A finite instrument, an input density operator ρ, and an outcome a with pa>0.

[F1]

Finite instruments and conditional states: each branch map is completely positive, pa=tr⁡(Ia(ρ)), and the conditional operator is Ia(ρ)/pa when pa>0.

[F2]

Completely positive maps: at ancilla dimension r=1, a completely positive 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]

Instrument outcome weights normalize: the branch weights satisfy pa≥0 and sum to one.

[F4]

Trace is a linear functional on Mn(F): trace commutes with scalar multiplication.

[F6]

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

Proof

1.1givenF3

The outcome-law theorem [F3] shows that pa is the probability of branch a; the hypothesis pa>0 selects a positive-probability branch.

2.1givenF1F2F5F6step 1.1

Put τa=Ia(ρ). By [F1]–[F2] and [F5]–[F6], τa is nonnegative. Since pa>0 is real and τa is self-adjoint, for all u,v one has ⟨(τa/pa)u,v⟩=pa−1⟨τau,v⟩=pa−1⟨u,τav⟩=⟨u,(τa/pa)v⟩, so τa/pa is self-adjoint. Division by pa>0 also preserves nonnegativity of every quadratic form. Thus ρa=τa/pa is nonnegative by [F6].

3.1givenF1F4F5step 2.1∎

By the definition in [F1], tr⁡(τa)=pa. Linearity [F4] gives tr⁡(ρa)=pa−1tr⁡(τa)=1. Thus ρa is a density operator by [F5]. The hypothesis pa>0 also forces the output space to be nonzero, since its zero space has only trace-zero endomorphisms.

Depends on

Used by

Cited to discharge well-definedness by Finite instruments and conditional states.

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