Noncommutative Mathematics for Quantum Systems

76 Noncommutative Mathematics for Quantum Systems

g 1 (Ny)f 1 (Nx), fn(Nx)gn(Ny)···f 1 (Nx)g 1 (Ny), and gn(Ny)fn− 1
(Nx)···f 1 (Nx)g 1 (Ny).

We shall now show that any pair of boolean-independent normal
operators can be reduced to this model.

Theorem 1.7.32 Let X and Y be two normal operators on a Hilbert
space H that are boolean-independent w.r.t. toΩ∈H and letμ=L(X,
Ω),ν=L(Y,Ω).
Then there exists an isometry W:C⊕L^2 (C,μ) 0 ⊕L^2 (C,ν) 0 → H
such that

(1.7.9)

W∗h(X)W





α

ψ 1

ψ 2



=







∫

Ch(x)

(

α+ψ 1 (x)

)

dμ(x)

h(α+ψ 1 )−

∫

Ch(x)

(

α+ψ 1 (x)

)

dμ(x)

h( 0 )ψ 2





,

W∗h(Y)W





α

ψ 1

ψ 2



=







∫

Ch(y)

(

α+ψ 2 (y)

)

dν(y)

h( 0 )ψ 1

h(α+ψ 2 )−

∫

Ch(y)

(

α+ψ 2 (y)

)

dν(y)







for all h∈Cb(C),α∈C,ψ 1 ∈L^2 (C,μ) 0 ,ψ 2 ∈L^2 (C,ν) 0.
We have

W

(

C⊕L^2 (C,μ) 0 ⊕L^2 (C,ν) 0

)

=alg{h(X),h(Y):h∈Cb(C)}Ω.

IfΩ∈H is cyclic for the algebra

alg(X,Y) =alg{h(X),h(Y):h∈Cb(C)}

generated by X and Y, then W is unitary.

Proof For a probability measureμonC, let

Cb(C)μ,0=

{

f∈Cb(C);

∫

C

f(z)dμ(x) = 0

}

,

thenCb(C)μ,0is dense inL^2 (C,μ) 0.
DefineW:C⊕Cb(C)μ,0⊕Cb(C)ν,0→Hby

W





α

f

g



=

(

α+f(X) +g(Y)

)

Ω.