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The square-root Lüders maps form an instrument

Statement

Let (Ea)a∈Ω be a finite POVM on H, and let Ma=Ea1/2 be the nonnegative square root of each effect. The maps

Ia(X):=MaXMa,X∈L(H),

form an instrument from H to itself.

Facts & Assumptions

Given: A nonzero finite-dimensional complex-inner-product space H, a finite POVM (Ea)a∈Ω, and Ma=Ea1/2.

[F1]

Finite POVMs and their outcome law: each effect Ea is nonnegative and ∑aEa=IH.

[F2]

Finite instruments and conditional states: a family of completely positive outcome maps whose output traces sum to the input trace on every operator is an instrument.

[F3]

Completely positive maps: a branch map is completely positive when all its finite-ancilla extensions preserve nonnegative operators.

[F4]

Finite ancilla amplification is a well-defined block map: for an extension input with blocks Xij, the amplification of X↦MaXMa has blocks MaXijMa.

[F5]

A non-negative operator has a unique non-negative square root: Ma is nonnegative and Ma2=Ea.

[F6]

The space L(V,W) of linear maps with pointwise addition and scalar multiplication: L(H) has pointwise addition and scalar multiplication.

[F7]

L(V,W) is a vector space over the common scalar field: those pointwise operations make L(H) a vector space.

[F8]

Identity maps and composites of linear maps are linear: a composition of linear maps is linear.

[F10]

Trace is a linear functional on Mn(F): trace is linear on finite-dimensional matrix spaces.

[F11]

Rectangular matrix multiplication and the identity matrix In, including zero-sized shapes: compatible block matrices multiply by row-by-column finite sums.

[F12]

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

[F13]

Ordered tensor registers, their inner product, and wire permutations: after specifying ordered orthonormal bases for the factors, their tensor register carries the product inner product, linear in the first argument.

[F14]

Finite-dimensional vector space, and its dimension dim⁡FV; infinite-dimensional means having no finite basis: finite-dimensional H has a finite basis; since H is nonzero, this basis is nonempty.

[F15]

Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans: Gram–Schmidt turns a finite linearly independent list, in particular a basis, into an orthonormal list with the same span.

[F16]

The product basis is orthonormal and tensor dimensions multiply: product vectors from specified orthonormal bases of the factors form an orthonormal basis of the tensor register.

Proof

1.1givenF6F7F8

For each a, X↦MaXMa takes L(H) to L(H) by composition of linear maps [F8]. For X,Y∈L(H), scalars s,t∈C, and v∈H, evaluate at v to obtain Ma((sX+tY)(Mav))=sMaXMav+tMaYMav. By the pointwise operations [F6]–[F7], this is sIa(X)v+tIa(Y)v, so Ia is complex-linear.

1.2givenF3F4F5F11F12F13F14F15F16

Fix any finite r and a nonnegative X on H⊗Cr in the sense of [F12]. By [F14] choose a finite basis of H, which is linearly independent by the basis definition, and apply [F15] to obtain an orthonormal basis; together with the standard basis of Cr, [F16] gives the product basis of H⊗Cr. By [F4], the extended branch map sends the blocks of X to MaXijMa. Let Sa be the block-diagonal operator with Ma on every diagonal block and zero off the diagonal. By the block product formula [F11], (SaXSa)ij=∑k,ℓ(Sa)ikXkℓ(Sa)ℓj=MaXijMa, so the extension is SaXSa. Write z=∑jzj⊗ej and w=∑jwj⊗ej using that product basis. The product inner product [F13] gives ⟨Saz,w⟩=∑j⟨Mazj,wj⟩=∑j⟨zj,Mawj⟩=⟨z,Saw⟩, since Ma is self-adjoint by [F5]; hence Sa is self-adjoint. For all z,w, self-adjointness of Sa and X gives ⟨SaXSaz,w⟩=⟨XSaz,Saw⟩=⟨Saz,XSaw⟩=⟨z,SaXSaw⟩, so SaXSa is self-adjoint. Therefore ⟨SaXSaz,z⟩=⟨XSaz,Saz⟩≥0 by [F12]. Thus the extension is nonnegative for every r, so [F3] gives complete positivity of each Ia.

2.1givenF1F2F5F9F10∎

For any X∈L(H), rectangular trace cyclicity [F9] and Ma2=Ea give tr⁡(Ia(X))=tr⁡(MaXMa)=tr⁡(XMa2)=tr⁡(XEa). By [F10] and completeness in [F1], ∑atr⁡(Ia(X))=tr⁡(X∑aEa)=tr⁡(X). Thus the family satisfies the instrument trace condition [F2].

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