Noncommutative Mathematics for Quantum Systems

62 Noncommutative Mathematics for Quantum Systems

Definition 1.7.11 Let XandY be two normal operators on a
Hilbert spaceH, not necessarily bounded. We say thatXandYare
monotonically independentw.r.t. Ω, if the ∗-algebras alg 0 (X) =
{h(X);h ∈Cb(C),h( 0 ) = 0 }and alg 0 (Y) = {h(Y);h ∈ Cb(C),
h( 0 ) = 0 }are monotonically independent w.r.t.Ω.

Recall that for monotone independence we have to work with
non-unital sub-algebras,cf.Remark 1.7.10.(b). This is the reason for
the Conditionh( 0 ) =0 in the definition of the algebras alg 0 (X)and
alg 0 (Y).
Let us now introduce the model we shall use for calculations
with monotonically independent operators.

Proposition 1.7.12 Letμ,νbe two probability measures onCand
define normal operators X and Y on L^2 (C×C,μ⊗ν)by

DomX=

{

ψ∈L^2 (C×C,μ⊗ν);

∫

C

∣

∣∣

∣x

∫

C

ψ(x,y)dν(y)

∣

∣∣

∣

2

dμ(x)<∞

}

,

DomY=

{

ψ∈L^2 (C×C,μ⊗ν);

∫

C×C

|yψ(x,y)|^2 dμ⊗ν(x,y)<∞

}

,

(Xψ)(x,y) = x

∫

C

ψ(x,y′)dν(y′),

(Yψ)(x,y) = yψ(x,y).

ThenL(X, 1 ) = μ,L(Y, 1 ) = ν, and X and Y are monotonically
independent w.r.t. the constant function 1.

Proof Denote by P 2 the orthogonal projection on the space of
functions inL^2 (C×C,μ⊗ν)that do not depend on the second
variable, and by Mx multiplication by the first variable, then
X=MxP 2. This operator is normal, we have

h(X)ψ(x,y) =

(

h(x)−h( 0 )

)∫

C

ψ(x,y)dν(y) +h( 0 )ψ(x,y)

and 〈 1 ,h(X) 1 〉 =

∫
Ch(x)dμ(x) for all h ∈ Cb(C), that is,
L(X, 1 ) = μ. The operator Y is multiplication by the second
variable; it is clearly normal. We have

h(Y)ψ(x,y) =h(y)ψ(x,y)