96 From Classical Mechanics to Quantum Field Theory. A Tutorial
where we have omitted the minus sign since it appears in front of a subspace that,
by definition, is closed with respect multiplication by scalars and we make use of
(2.32). Now suppose thatD(A†) is dense so that (A†)†exists. In this case, taking
advantage of (2.32) again, we haveG(A)=G((A†)†). Notice that the right-hand
side is the graph of an operator so we have obtained that, ifD(A†)isdense,then
Ais closable. In this case, by definition of closure of an operator, we also have
A=(A†)†.
Vice versa, suppose thatAis closable, so the operatorAexists andG(A)=
G(A). In this caseτ(G(A†))⊥=G(A) is the graph of an operator and thus its
graph cannot include pairs (0,y) withy= 0 by linearity. In other words, if (0,y)∈
τ(G(A†))⊥theny= 0. This is the same as saying that ((0,y)|(−A†x, x)) = 0 for
allx∈D(A†) impliesy= 0. Summing up,〈y|x〉=0forallx∈D(A†) implies
y=0. SinceH=D(A†)⊥⊕(D(A†)⊥)⊥=D(A†)⊥⊕D(A†), we conclude that
D(A†)=H,thatisD(A†)isdense.
An immediate corollary follows.
Corollary 2.2.23. LetA:D(A)→Hbe an operator in the Hilbert spaceH.If
bothD(A)andD(A†)are densely defined then
A†=A†
=A†= (((A†)†)†.The Hilbert space version of theclosed graph theoremholds (e.g., see[ 5 ]).
Theorem 2.2.24(Closed graph theorem).LetA:H→Hbe an operator,H
being a complex Hilbert space.Ais closed if and only ifA∈B(H).
Exercise 2.2.25.Prove that, ifB∈B(H)andAis a closed operator inHsuch
thatRan(B)⊂D(A),thenAB∈B(H).
Solution.ABis well defined by hypothesis andD(AB)=H. Exploiting (c)
in remark 2.2.21 and continuity ofB, one easily finds thatABis closed as well.
Theorem 2.2.24 finally proves thatAB∈B(H).
Definition 2.2.26.An operatorAin the complex Hilbert spaceHis said to be
(0)Hermiteanif〈Ax, y〉=〈x, Ay〉forx, y∈D(A);
(1)symmetricif it is densely defined and Hermitian, which is equivalent to
say thatA⊂A†;
(2)selfadjointif it is symmetric andA=A†;
(3)essentially selfadjointif it is symmetric and(A†)†=A†;