Independence and L ́evy Processes in Quantum Probability 77This is an isometry, since
〈
Wα 1
f 1
g 1,Wα 2
f 2
g 2〉=〈(
α 1 +f 1 (X) +g 1 (Y))
Ω,(
α 2 +f 2 (X) +g 2 (Y))
Ω〉=α 1 α 2 +∫Cf 1 (x)f 2 (x)dμ(x) +∫Cg 1 (y)g 2 (y)dμ(y),where the mixed terms all vanish because 〈Ω,fi(X)Ω〉 =
〈Ω,gi(Y)Ω〉=0 fori=1, 2. Therefore,Wextends in a unique way
to an isometry onC⊕L^2 (C,μ) 0 ⊕L^2 (C,ν) 0
Let nowh∈Cb(C), then we get
〈
W
α 1
f 1
g 1
,h(X)W
α 2
f 2
g 2
〉=〈(
α 1 +f 1 (X) +g 1 (Y))
Ω,(h(X)−h( 0 ) 1)(
α 2 +f 2 (X) +g 2 (Y))
Ω〉+h( 0 )〈
Wα 1
f 1
g 1,Wα 2
f 2
g 2〉=〈(
α 1 +f 1 (X))
Ω,(
h(X)−h( 0 ) 1)(
α 2 +f 2 (X))
Ω〉+h( 0 )〈
α 1
f 1
g 1,α 2
f 2
g 2〉
,because the boolean independence and〈Ω,gi(Y)Ω〉=0 imply that
all other terms vanish. But since〈Ω,fi(Y)Ω〉=0, this is equal to
〈(
α 1 +f 1 (X))
Ω,h(X)(
α 2 +f 2 (X))
Ω〉+h( 0 )
〈
α 1
f 1
g 1
,
α 2
f 2
g 2
〉
−α 1 α 2 −〈f 1 ,f 2 〉
=〈
α 1
f 1
g 1
,
∫
h(x)(
f 2 (x) +α 2)
dμ(x)
h(f 2 +α 2 )−∫
h(x)(
f 2 (x) +α 2)
dμ(x)
h( 0 )g 2
〉
.