From Classical Mechanics to Quantum Field Theory

Mathematical Foundations of Quantum Mechanics 107

The operatorsA−λIandA−λ∗Iare injective, and||(A−λI)−^1 ||≤|ν|−^1 ,where
(A−λI)−^1 :Ran(A−λI)→D(A). Notice that, from (2.44),

Ran(A−λI)

⊥

=[Ran(A−λI)]⊥=Ker(A†−λ∗I)=Ker(A−λ∗I)={ 0 },

where the last equality makes use of the injectivity ofA−λ∗I. Summarising:A−λI

in injective, (A−λI)−^1 bounded andRan(A−λI)

⊥
={ 0 }, i.e. Ran(A−λI)is
dense inH; thereforeλ∈ρ(A), by definition of resolvent set. Let us pass to
(ii). Supposeλ∈σ(A), butλ∈σp(A). ThenA−λImust be one-to-one and
Ker(A−λI)={ 0 }.SinceA=A†andλ∈Rby (i), we haveKer(A†−λ∗I)={ 0 },
so [Ran(A−λI)]⊥=Ker(A†−λ∗I)={ 0 }andRan(A−λI)=H.Consequently
λ∈σc(A). Proving (iii) is easy: ifλ=μandAu=λu,Av=μv,then

(λ−μ)〈u, v〉=〈Au, v〉−〈u, Av〉=〈u, Av〉−〈u, Av〉=0;

fromλ, μ∈RandA=A†.Butλ−μ=0,so〈u, v〉=0.

(2)LetA:D(A)→HbeaclosedoperatorinH(in particularA∈B(H)). Prove
thatλ∈ρ(A)if and only ifA−λIadmits an inverse which belongs toB(H).

Solution.If (A−λI)−^1 ∈B(H), it must beRan(A−λI)=Ran(A−λI)=
H and (A−λI)−^1 is bounded, so thatλ∈ρ(A) by definition. Let us prove
the converse. Suppose thatλ∈ρ(A), we know that (A−λI)−^1 is defined on
the dense domainRan(A−λI) and is bounded. To conclude, it is therefore
enough proving thaty∈Himpliesy∈Ran(A−λI). To this end, notice that if
y∈H=Ran(A−λI), theny= limn→+∞(A−λI)xnfor somexn∈D(A−λI).
The sequence ofxnconverges. IndeedHis complete and{xn}n∈Nis Cauchy as
(1)xn=(A−λI)−^1 yn,(2)||xn−xm|| ≤ ||(A−λI)−^1 || ||yn−ym||,and(3)
yn→y. To end the proof, we observe that,A−λIis closed sinceAis such ((b) in
remark 2.2.21). It must consequently be ((c) in remark 2.2.21)x= limn→+∞xn∈
D(A−λI)andy=(A−λI)x∈Ran(A−λI).

Example 2.2.44.Them-axis position operatorXminL^2 (Rn,dnx) introduced
in (1) of example 2.2.39 satisfies

σ(Xm)=σc(Xm)=R. (2.45)

The proof can be obtained as follows. First observe thatσ(Xm)⊂Rsince the
operator is selfadjoint. Howeverσp(Xm)=∅as observed in the first section and
σr(Xm)=∅becauseXmis selfadjoint ((1) in exercise 2.2.43). Suppose that, for
somer∈R,(Xm−rI)−^1 is bounded. Ifψ∈D(Xm−rI)=D(Xm) with||ψ||=1
we have||ψ||=||(Xm−rI)−^1 (Xm−rI)ψ||and thus||ψ||≤||(Xm−rI)−^1 ||||(Xm−