From Classical Mechanics to Quantum Field Theory

Mathematical Foundations of Quantum Mechanics 95

Proposition 2.2.22. Consider an operatorA:D(A)→HwithD(A)dense in
the Hilbert spaceH. The following facts hold.

(a)A†is closed.

(b)Ais closable if and only ifD(A†)is dense and, in this case,A=(A†)†.

Proof. Define the Hermitian scalar product ((x, y)|(x′y′)) :=〈x, x′〉+〈y,y′〉on the
vector space defined as the standard direct sum of vector spaces, we henceforth
denote byH⊕H. It is immediately proven thatH⊕Hbecomes a Hilbert space
when equipped with that scalar product. Next define the operator

τ:H⊕H(x, y)→(−y,x)∈H⊕H.

It is easy to check thatτ∈B(H⊕H) and also (referring the adjoint to the space
H⊕H),

τ†=τ−^1 =−τ. (2.30)

Finally, by direct computation, one sees thatτand⊥(referred toH⊕Hwith the
said scalar product) commute

τ(F⊥)=(τ(F))⊥ ifF⊂H⊕H. (2.31)

Let us pass to prove (a). The following noticeable identity holds true, for an
operatorA:D(A)→HwithD(A)denseinH(so thatA†is defined)

G(A†)=τ(G(A))⊥. (2.32)

Since the right hand side is closed (it being the orthogonal space of a set), the
graph ofA†is closed andA†is therefore closed by definition. To prove (2.32),
observe that in view of the definition ofτ,

τ(G(A))⊥={(y,z)∈H⊕H|((y,z)|(−Ax, x)) = 0,∀x∈D(A)}

that is

τ(G(A))⊥={(y,z)∈H⊕H|〈y,Ax〉=〈z,x〉,∀x∈D(A)}

SinceA†exists, the pairs (y,z)∈τ(G(A))⊥canbewritten(y,A†y) according to
the definition ofA†. Thereforeτ(G(A))⊥=G(A†) proving (a).
(b) From the properties of ⊥we immediately haveG(D(A)) = (G(A)⊥)⊥.
Sinceτand⊥commute by (2.31), andττ=−I(2.30),

G(D(A)) =−τ◦τ((G(A)⊥)⊥)=−τ(τ(G(A))⊥)⊥=τ(τ(G(A))⊥)⊥=τ(G(A†))⊥,