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Every finite instrument induces a unique POVM

Statement

Let (Ia)a∈Ω be a finite instrument from a nonzero finite-dimensional complex inner-product space H to a finite-dimensional complex inner-product space K. There is a unique finite POVM (Ea)a∈Ω on H such that tr⁡K(Ia(X))=tr⁡H(XEa)(X∈L(H), a∈Ω). Consequently, for every density operator ρ on H, the instrument probability tr⁡K(Ia(ρ)) equals the POVM weight tr⁡H(ρEa).

Facts & Assumptions

Given: The finite instrument and its finite nonempty outcome set Ω.

[F1]

Finite instruments and conditional states: every Ia:L(H)→L(K) is completely positive and ∑atr⁡K(Ia(X))=tr⁡H(X) for every X.

[F2]

Kraus and Choi characterizations of complete positivity: every completely positive finite-dimensional map between full operator spaces has a finite Kraus representation.

[F3]

Finite Kraus representation: a finite Kraus family (Mj) represents Φ when Φ(X)=∑jMjXMj† for every input operator X.

[F4]

Instrument branch Kraus families satisfy total normalization: for finite branch Kraus families of an instrument, ∑a,jMa,j†Ma,j=IH.

[F6]

Non-negative and positive operators: non-negativity means self-adjointness and a non-negative quadratic form.

[F7]

Finite positive effect family and Finite POVMs and their outcome law: a finite POVM is a nonempty finite family of non-negative operators summing to identity, and its outcome weights are tr⁡(ρEa).

[F8]

Trace is a linear functional on Mn(F) and For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA): trace is linear and tr⁡K(AB)=tr⁡H(BA) for compatible rectangular maps A:H→K, B:K→H.

[F9]

The trace pairing on a finite endomorphism space is nondegenerate: if tr⁡H(AX)=0 for every X∈L(H), then A=0.

Proof

1.1givenF1F2F3choose

For each a∈Ω, [F1] makes Ia completely positive, so [F2] supplies a finite family (Ma,j)j representing it as in [F3]. If the supplied family is empty, append one zero operator; this preserves the formula and makes each index list nonempty. Since Ω is finite, the lists can be selected one at a time by finite induction; no arbitrary-index choice is used.

2.1givenstep 1.1F1F3F4algebra

Define Φ=∑a∈ΩIa. Concatenating the finite lists in step 1.1 gives a Kraus representation of Φ. These are exactly the finite branch Kraus families of the instrument in [F1], so [F4] gives ∑a,jMa,j†Ma,j=IH.

3.1givenstep 2.1F5F6F7algebra

For each outcome define Ea=∑jMa,j†Ma,j. By [F5], each summand is self-adjoint and ⟨Ma,j†Ma,jx,x⟩=⟨Ma,jx,Ma,jx⟩≥0. Thus each Ea is non-negative by [F6]. Step 2.1 gives ∑aEa=IH, so (Ea) is a finite POVM by [F7].

4.1givenstep 1.1step 3.1F3F8algebra

For every X∈L(H), the Kraus formula [F3], trace linearity, and rectangular cyclicity [F8] give tr⁡K(Ia(X))=∑jtr⁡K(Ma,jXMa,j†)=∑jtr⁡H(XMa,j†Ma,j)=tr⁡H(XEa). This proves the displayed outcome-law identity for every density operator as well.

5.1givenstep 4.1F8F9algebra∎

If another family Ea′ has the same identity for every X, then tr⁡H(X(Ea−Ea′))=0 for all X. By [F8], this equals tr⁡H((Ea−Ea′)X); [F9] implies Ea−Ea′=0. This holds for each a, so the induced POVM is unique.

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