From Classical Mechanics to Quantum Field Theory

Mathematical Foundations of Quantum Mechanics 99

U†U=I=UU†, is equivalent toUA†U†UAU†=UAU†UA†U†=U†U=I,that
is (UA†U†)UAU†=(UAU†)UA†U†=Iand thusUAU†is unitary as well. IfAis
normal,UAU†is normal too, with the same reasoning as in the unitary case.

Exercise 2.2.31.
(1)Prove that ifAis unitary thenA, A†∈B(H).
Solution. It holdsD(A)=D(A†)=D(I)=Hand||Ax||^2 =〈Ax, Ax〉=
〈x, A†Ax〉=||x||^2 ifx∈H,sothat||A||= 1. Due to (c) in remark 2.2.17,
A†∈B(H).
(2)Prove thatA:H→His unitary if and only if is surjective and norm pre-
serving.
Solution.IfAis unitary ((3) Def 2.2.26), it is evidently bijective, moreover as
D(A†)=H,||Ax||^2 =〈Ax, Ax〉=〈x, A†Ax〉=〈x, x〉=||x||^2 ,soAis isometric. If
A:H→His isometric its norm is 1 and thusA∈B(H). ThereforeA†∈B(H).
The condition||Ax||^2 =||x||^2 can be re-written〈Ax, Ax〉=〈x, A†Ax〉=〈x, x〉
and thus〈x,(A†A−I)x〉=0forx∈H.Usingx=y±zandx=y±iz, the found
indentity implies〈z,(A†A−I)y〉=0forally,z∈H. Takingz=(A†A−I)y,we
finally have||(A†A−I)y||=0forally∈Hand thusA†A=I.Inparticular,Ais
injective as it admits the left inverseA†.SinceAis also surjective, it is bijective
and thus its left inverse (A†) is also a right inverse, that isAA†=I.
(3)Prove that, ifA:H→Hsatisfies〈x, Ax〉∈Rfor all x∈H(and in
particular ifA≥ 0 , which means〈x, Ax〉≥ 0 for allx∈H), thenA†=Aand
A∈B(H).
Solution.We have〈x, Ax〉=〈x, Ax〉∗=〈Ax, x〉=〈x, A†x〉where, asD(A)=
H, the adjointA†is well defined everywhere onH.Thus〈x,(A−A†)x〉=0for
everyx∈H.Usingx=y±zandx=y±iz,weobtain〈y,(A−A†)z〉=0for
ally,z∈H. Choosingy=(A−A†)z, we conclude thatA=A†. Theorem 2.2.29
concludes the proof.

Example 2.2.32.TheFourier transform,F:S(Rn)→S(Rn), defined as^6

(Ff)(k):=

1

(2π)n/^2

∫

Rn

e−ik·xf(x)dnx (2.33)

(k·xbeing the standardRnscalar product ofkandx) is a bijective linear map
with inverseF−:S(Rn)→S(Rn),

(F−g)(x):=

1

(2π)n/^2

∫

Rn

eik·xg(k)dnk. (2.34)

(^6) In QM, adopting units with=1,k·xhas to be replaced fork·xand (2π)n/ (^2) for (2π)n/ (^2).