Mathematical Foundations of Quantum Mechanics 101Exercise 2.2.35.Prove that a symmetric operator that admits a unique selfad-
joint extension is necessarily essentially selfadjoint.
Solution.By (b) of Proposition 2.2.33,n+=n−.Ifn±=0therearemany
selfadjoint extension. The only possibility for the uniqueness of the selfadjoint
extension isn±= 0. (a) of Proposition 2.2.33 implies thatAis essentially selfad-
joint.
A very useful criterion to establish the essentially selfadjointness of a symmetric
operator is due to Nelson. It relies upon an important definition.
Definition 2.2.36.LetAbe an operator in the complex Hilbert spaceH.If
ψ∈∩n∈ND(An) satisfies
+∑∞n=0tn
n!||Anψ||<+∞ for somet>0,thenψis said to be ananalytic vectorofA.
We can state Nelson’s criterion here[ 5 ].
Theorem 2.2.37(Nelson’s essentially selfadjointness criterium).LetAbe a sym-
metric operator in the complex Hilbert spaceH,Ais essentially selfadjoint ifD(A)
contains a dense setDof analytic vectors (or – which is equivalent – a setDof
analytic vectors whose finite span dense inH).
The above equivalence is due to the fact that a finite linear combination of analytic
vector is an analytic vector as well, the proof being elementary. We have the
following evident corollary.
Corollary 2.2.38. IfAis a symmetric operator admitting a Hilbertian basis of
eigenvectors inD(A),thenAis essentially selfadjoint.
Example 2.2.39.
(1)Form∈{ 1 , 2 ,...,n}, consider the operatorsXm′ andXm′′inL^2 (Rn,dnx) with
dense domainsD(X′m)=C 0 ∞(Rn;C),D(Xm′′)=S(Rn)forx∈Rnand, forψ,φ
in the respective domains,
(Xm′ψ)(x):=xmψ(x), (X′′mφ)(x):=xmφ(x),wherexmis them-th component ofx∈Rn. Both operators are symmetric but
not selfadjoint. They admit selfadjoint extensions because they commute with the
standard complex conjugation of functions (see remark 2.2.34). It is furthermore
possible to prove that both operators are essentially selfadjoint as follows. First