Independence and L ́evy Processes in Quantum Probability 71after some tedious, but straightforward, computation.
Remark 1.7.22 Ifν=δ 0 , thenMy=0 onL^2 (R+×R+,μ⊗ν), and
therefore
√
SxMy√
Sx=0. This is of course a positive operator, and
its distribution isδ 0.
Theorem 1.7.23 Let X and Y be two positive self-adjoint operators on
a Hilbert space H such that X− 1 and Y are monotonically independent
w.r.t. to a unit vectorΩ ∈ H. Assume, furthermore, thatΩis cyclic,
that is,
alg{h(X),h(Y);h∈Cb(R+)}Ω=H.Then
√
XY√
X is essentially self-adjoint and the distribution w.r.t.Ω
of its closure is equal to the multiplicative monotone convolution of the
distributions of X and Y w.r.t.Ω, that is,
L(√
XY√
X,Ω)
=L(X,Ω)mL(Y,Ω).Proof Letμ=L(X,Ω),ν=L(Y,Ω).
By Theorem 1.7.13 it is sufficient to consider the caseX=Sxand
Y = My. In this case Proposition 1.7.21 shows thatz−
√
XY√
X
has a bounded inverse for all√ z∈C\R. This implies that Ran(z−
XY
√
X)is dense for allz∈C\Rand that√
XY√
Xis essentially
self-adjoint,cf.[RS80, Theorem VIII.3].
Using Equation (1.7.6), we can compute the Cauchy transform
of the distribution of the closure of
√
XY√
X. Letz∈C+, then we
have
G√XY√X(z) =〈
Ω,(
z−√
XY√
X)− 1
Ω〉=〈
1 ,(
z−√
SxMy√
Sx)− 1
1〉=〈
1 ,1 +g 1
z−y+h 1〉where
g 1 (x) =√
x−x+ (x− 1 )zGν(x)
( 1 −x)zGν(z) +x=√
x
( 1 −x)zGν(z) +x−1,h 1 (x) =( 1 −√
x)Gν(z)
( 1 −x)zGν(z) +x.