Independence and L ́evy Processes in Quantum Probability 75Proof Under the identification C⊕L^2 (C,μ) 0 ⊕L^2 (C,ν) 0 ∼=
L^2 (C,μ)⊕L^2 (C,ν) 0 , where
α
ψ 1
ψ 2
∼=(
ψ 1 +α
ψ 2)
,the operator Nx becomes multiplication by the variable x on
L^2 (C,μ). It is clearly normal and we have
h(Nx)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
and〈ω,h(Nx)ω〉=
∫
Chdμfor allh∈Cb(C), that is,L(Nx,ω) =μ.
Similarly,
h(Ny)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 allh∈Cb(C), andL(Ny,ω) =ν.
Letf 1 ,... ,fn,g 1 ,... ,gn∈ Cb(C), withf 1 ( 0 ) =··· = fn( 0 ) =
g 1 ( 0 ) =···=gn( 0 ) =0. Thenfn(Nx)gn− 1 (Ny)···g 1 (Ny)f 1 (Nx)ω
=
∏nk= 1∫
Cfkdμ∏n− 1
`= 1∫
Cg`dν
∏nk=−^11∫
Cfkdμ∏n− 1
`= 1∫
Cg`dν(
fn−∫
Cfndμ)0
and therefore
〈ω,fn(Nx)gn− 1 (Ny)···g 1 (Ny)f 1 (Nx)ω〉=n
∏
k= 1∫Cfkdμn− 1
∏
`= 1∫Cg`dν=n
∏
k= 1〈ω,fk(Nx)ω〉n− 1
∏
`= 1〈ω,g`(Ny)ω〉that is, the condition for boolean independence is satisfied in this
case. Similarly, one checks the expectation ofgn(Ny)fn(Nx)···