Mathematical Foundations of Quantum Mechanics 97(4)unitaryifA†A=AA†=I;
(5)normalif it is closed, densely defined andAA†=A†A.Remark 2.2.27.
(a)IfAis unitary thenA, A†∈B(H).FurthermoreA:H→His unitary if
and only if it is surjective and norm preserving. (See the exercises 2.2.31 below).
These operators are nothing but theautomorphismsof the given Hilbert space.
Considering two Hilbert spacesH,H′isomorphismsare linear mapsT:H→H′
which are isometric and surjecitve. Notice thatTalso preserve the scalar products
in view of Proposition 2.2.1.
(b) A selfadjoint operator A does not admit proper symmetric extensions
and essentially selfadjoint operators admit only one self-adjoiunt extension. (See
Proposition 2.2.28 below).
(c)A symmetric operatorAis always closable because A⊂A†andA†is
closed (Proposition 2.2.22), moreover for that operator the following conditions
are equivalent in view of Proposition 2.2.22 and Corollary 2.2.23, as the reader
immediately proves:
(i) (A†)†=A†(Ais essentially selfadjoint);
(ii)A=A†;
(iii)A=(A)†.(d)Unitary and selfadjoint operators are cases of normal operators.The pair of elementary results of (essentially) selfadjoint operators stated in (b)
are worth to be proved.
Proposition 2.2.28. LetA:D(A)→Hbe a densely defined operator in the
Hilbert spaceH. The following facts are true.
(a)IfAis selfadjoint, thenAdoes not admit proper symmetric extensions.
(b)IfAis essentially selfadjoint, thenAadmits a unique selfadjoint exten-
sion, and that this extension isA†.Proof. (a)LetBbe a symmetric extension ofA.A⊂BthenB†⊂A†for (b) in
remark 2.2.17. AsA=A†we haveB†⊂A⊂B.SinceB⊂B†, we conclude that
A=B.
(b)LetBbe a selfadjoint extension of the essentially selfadjoint operatorA,so
thatA⊂B. ThereforeA†⊃B†=Band (A†)†⊂B†=B.SinceAis essentially
selfadjoint, we have foundA†⊂B.HereA†is selfadjoint andBis symmetric