- #1
sweet springs
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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.
-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.