Independence and L ́evy Processes in Quantum Probability 67
then we can check that this equals
ψ(x,y) +
x
∫ψ(x,y′)
z−y′ dν(y
′)
1 −xGν(z)
−x
∫
R
ψ(x,y′)
z−y′
dν(y′)
−x
∫
R
x
∫
R
ψ(x,y′′)
z−y′′ dν(y
′′)
(z−y′)
(
1 −xGν(z)
)dν(y′)
=ψ(x,y)
(
(z−y)−(z−y)
(
1 −xGν(z)
)
−xGν(z)(z−y)
)
x
∫
R
ψ(x,y′)
z−y′ dν(y
′)
(z−y)
(
1 −xGν(z)
)
=ψ(x,y),
as it should.
Theorem 1.7.18 Let X and Y be two self-adjoint operators on a Hilbert
space H that are monotonically independent w.r.t. to a unit vectorΩ∈H.
Assume, furthermore, thatΩis cyclic, that is, that
alg{h(X),h(Y);h∈Cb(R)}Ω=H.
Then X+Y is essentially self-adjoint and the distribution w.r.t.Ωof its
closure is equal to the additive monotone convolution of the distributions
of X and Y w.r.t. toΩ, that is,
L(X+Y,Ω) =L(X,Ω).L(Y,Ω).
Proof Letμ=L(X,Ω),ν=L(Y,Ω).
By Theorem 1.7.13 and Lemma 1.7.8 it is sufficient to consider
the case where X and Y are given by Proposition 1.7.12.
Proposition 1.7.17 shows thatz−X−Yadmits a bounded inverse
and therefore that Ran(z−X−Y)is dense forz∈C\R. By [RS80,
Theorem VIII.3] this is equivalent to X+Y being essentially
self-adjoint.
Using Equation (1.7.4), we can compute the Cauchy transform of
the distribution of the closure ofX+Y. Letz∈C+, then we have
