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Projective measurement outcome weights are probabilities

Statement

Let ρ be a density operator on a nonzero finite-dimensional complex inner-product space H, and let (Pa)a∈Ω be a finite projective measurement on H. Then pρ(a)=tr⁡(ρPa) lies in [0,1] for every a, and ∑a∈Ωpρ(a)=1.

Facts & Assumptions

Given: The density operator ρ and the finite projective measurement (Pa)a∈Ω stated above.

[F1]

Finite projective measurements: Pa∗=Pa=Pa2, PaPb=0 for a≠b, and ∑aPa=IH; zero projections are allowed.

[F6]

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

[F7]

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

[F2]

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.

[F3]

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

[F4]

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

[F5]

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

1.1givenF1F2F3F5F6F7

Fix a. Set τa=PaρPa. Self-adjointness of Pa and ρ gives ⟨τax,y⟩=⟨ρPax,Pay⟩=⟨x,τay⟩ for all x,y, so τa is self-adjoint. It is nonnegative, since ⟨τax,x⟩=⟨ρPax,Pax⟩≥0 for every x by [F6]–[F7]. By [F2], τa has nonnegative diagonal entries in an orthonormal basis, so [F5] gives tr⁡(τa)≥0. By [F3] and Pa2=Pa, pρ(a)=tr⁡(ρPa)=tr⁡(PaρPa)=tr⁡(τa)≥0.

1.2givenF1F4F5F6

By [F4], the resolution of the identity in [F1], and [F6], ∑apρ(a)=tr⁡(ρ∑aPa)=tr⁡(ρIH)=tr⁡(ρ)=1.

2.1step 1.1algebra∎

For each a, 1−pρ(a)=∑b≠apρ(b)≥0 by step 1.1, with the empty sum interpreted as 0 when Ω={a}. Therefore every weight is at most one and, by step 1.1, belongs to [0,1].

Depends on

Used by

…and 1 more result.

Cited to discharge well-definedness by Finite projective measurements.

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