Noncommutative Mathematics for Quantum Systems

80 Noncommutative Mathematics for Quantum Systems

Gν(z), it can not vanish forzwith Imz<0 either. The functionsz−^1 x
andz−xxare bounded onRforz∈C\R, therefore, Equation (1.7.10)
defines a bounded operator.
Let

φ 1 =

ψ 1 +βx−cx

z−x

and φ 2 =

ψ 2 +βy−cy

z−y

,

then

(z−Nx−Ny)





β

φ 1

φ 2



 =







zβ+dx+dy

(z−x)φ 1 −βx−dx

(z−y)φ 2 −βy−dy







=







zβ+dx+dy

ψ 1 −cx−dx

ψ 2 −cy−dy







where

dx=

∫

x

(

φ 1 (x) +β

)

dμ(x), dy=

∫

y

(

φ 2 (y) +β

)

dν(y).

Sinceψ 1 ∈L^2 (R,μ) 0 ,ψ 2 ∈L^2 (R,ν) 0 , integrating over the second
and third component in the formula above givescx = −dxand
cy=−dy. Therefore,

(z−Nx−Ny)





β

φ 1

φ 2



=







zβ−cx−cy

ψ 1

ψ 2







We have to show that the first component is equal toα. We get

zβ−cx−cy=zβ−

∫ ψ 1 (x)

z−xdμ(x) +β

(

zGμ(z)− 1

)

Gμ(z)

−

∫ ψ 1 (x)

z−xdμ(x) +β

(

zGν(z)− 1

)

Gν(z)

=β

Gμ(z) +Gν(z)−zGμ(z)Gν(z)

Gμ(z)Gν(z)

−

1

Gμ(z)

∫ ψ

1 (x)

z−x

dμ(x)

−

1

Gν(z)

∫ ψ

2 (y)

z−y

dν(y)