Alphabeta Math
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POVM outcome weights are probabilities

Statement

Let ρ be a density operator on a nonzero finite-dimensional complex inner-product space H, and let E=(Ea)a∈Ω be a finite POVM on H. Then

pρ(a)=tr⁡(ρEa)∈[0,1]for each a∈Ω,∑a∈Ωpρ(a)=1.

Facts & Assumptions

Given: H, the density operator ρ, the finite outcome set Ω, and the POVM E stated above.

[F1]

Finite POVMs and their outcome law: Ea is nonnegative, ∑aEa=IH, and pρ(a)=tr⁡(ρEa).

[F7]

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

[F8]

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

[F2]

A non-negative operator has a unique non-negative square root: every nonnegative operator on a finite-dimensional real or complex inner-product space has a unique nonnegative square root R with R2=T.

[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 its matrix is diagonal with nonnegative real diagonal entries.

[F4]

For A∈Mm×n(F) and B∈Mn×m(F), tr⁡(AB)=tr⁡(BA): for compatible rectangular matrices A,B, tr⁡(AB)=tr⁡(BA), including zero-sized cases.

[F5]

Trace is a linear functional on Mn(F): trace preserves finite sums and scalar multiples.

[F6]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: endomorphism trace is the matrix trace in any basis and is basis-independent.

Proof

1.1givenF1F2F3F4F6F7F8

Fix a∈Ω. Let Ma=Ea1/2 be the nonnegative square root supplied by [F2]. Since Ma and ρ are self-adjoint by [F2] and [F7]–[F8], for every x,y∈H one has ⟨MaρMax,y⟩=⟨ρMax,May⟩=⟨Max,ρMay⟩=⟨x,MaρMay⟩, so MaρMa is self-adjoint. For every x∈H, ⟨MaρMax,x⟩=⟨ρMax,Max⟩≥0 by [F7]–[F8], so MaρMa is nonnegative. By [F3] it is diagonal with nonnegative entries in an orthonormal basis, hence its trace is nonnegative by [F6]. By [F4] and Ma2=Ea, pρ(a)=tr⁡(ρEa)=tr⁡(MaρMa)≥0.

1.2givenF1F5F6F7

Linearity of trace [F5] and the completeness relation in [F1] give ∑a∈Ωpρ(a)=tr⁡(ρ∑aEa)=tr⁡(ρIH)=tr⁡(ρ)=1 by [F7] and [F6].

2.1step 1.1algebra∎

For fixed a, 1−pρ(a)=∑b∈Ω, b≠apρ(b)≥0 by step 1.1; if Ω={a} this is the empty sum 0. Thus pρ(a)≤1, and together with step 1.1 each weight lies in [0,1].

Depends on

Used by

Cited to discharge well-definedness by Finite POVMs and their outcome law.

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources