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 the directions of momentum. ∫de|e><e|p> = α*|e=p^2/2m> does not give |p>. ∫de|e><e|-p> = β*|e=p^2/2m> does not give |-p>. Your advice on the correct use of the word OBSERVABLE is appreciable. Regards.