Mathematical Foundations of Quantum Mechanics 113
Similarly, from the elementary properties of the scalar product, whenx, y∈D(T)
4 〈x|Ty〉=〈x+y,T(x+y)〉−〈x−y,T(x−y)〉−i〈x+iy, T(x+iy)〉+i〈x−iy, T(x−iy)〉.
It is then obvious that (2.54) implies
〈x, T y〉=
∫
X
f(λ)μ(xyP)(λ) ∀x, y∈Δf,
so that
〈
x,
(
T−
∫
X
f(λ)dP(λ)
)
y
〉
=0 ∀x, y∈Δf
Sincexvaries in a dense set Δf,Ty−
∫
Xf(λ)dP(λ)y=0foreveryy∈Δfwhich
is the thesis.
Example 2.2.53.
(1)∫ Referring to the PVM in (2) of example 2.2.49, directly from the definition of
Xf(λ)dP(λ) or exploiting (3) in exercises 2.2.52 we have that
∫
N
f(λ)dP(λ)z=
∑
n∈N
f(n)Qnz
for everyf:N→C(which is necessarily measurable with our definition of Σ(N)).
Correspondingly, the domain of
∫
Nf(λ)dP(λ) results to be
Δf:=
{
z∈H
∣∣
∣∣
∣
∑
n∈N
|f(n)|^2 ||Qnz||^2 <+∞
}
We stress that we have found a direct generalization of the expansion (2.4) if the
operatorAis now hopefully written as
Az=
∑
n∈N
nQnz.
Weshallseebelowthatitisthecase.
(2)∫ Referring to the PVM in (3) of example 2.2.49, directly from the definition of
Xf(λ)dP(λ) or exploiting (3) in exercises 2.2.52 we have that
(∫
R
f(λ)dP(λ)ψ
)
(x)=f(x)ψ(x),x∈R
Correspondingly, the domain of
∫
Rf(λ)dP(λ) results to be
Δf:=
{
ψ∈L^2 (R,dx)
∣∣
∣∣
∫
R
|f(x)|^2 |ψ(x)|^2 dx <+∞