From Classical Mechanics to Quantum Field Theory

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 <+∞

}

.