Is Hamiltonian of a particle in free space H=P^2/2m OBSERVABLE ?
-Yes, we can surely observe energy in some manner.
-No, ∫de|e><e| is not identical operator I, thus |e>s does not form a complete set.
As an example, energy eigenstate |e> degenerates, as |e=p^2/2m> = α|p> + β|-p>, according to...