sweet springs said:
Hi. SpectraCat. Let me write my understanding of QM.
Any real physical variable forming complete set is Hermitian, but not all the Hermitian form complete set.
Energy is a real physical variable ... so your example falls into case 1 (I guess), but your language is *very* imprecise here .. I have corrected it below (my text in red).
The eigenstates of the momentum operator p form a complete set and p is Hermitian.
This means that the momentum eigenvalues are real, and the momentum eigenstates satisfy the relations:
\hat{\textbf{p}}\left|\vec{p}\:\right\rangle=p\left|\vec{p}\:\right\rangle (eigenvalue/eigenvector relation)
\left\langle\vec{p}\:'|\vec{p}\:\right\rangle = \delta\left(\vec{p}\:'-\vec{p}\:\right) (orthonormality)
\int|d\vec{p}\:\left|\vec{p}\:\right\rangle\left\langle\vec{p}\:'\right| = 1 (completeness, or closure)
\left\langle\vec{p}\:'\right|\hat{\textbf{p}}\left|\vec{p}\:\right\rangle = \left\langle\vec{p}\:\right|\hat{\textbf{p}}\left|\vec{p}\:'\right\rangle^{*} (Hermicity of operator)
where \delta is the Dirac delta function, and I have used vector notation to help distinguish eigenstates and eigenvalues.
If we apply the Hamiltonian operator for the free particle, \hat{\textbf{p}}^{2}/2m, to one of these states, it is easy to see they are also eigenstates of that operator (i.e. they are energy eigenstates). Of course the orthonormality and completeness relations still hold, since we are talking about the same states.
Energy of free particle P^2 is Hermitian but does not form complete set.
Here completeness means projection operator ∫de or Σe |e><e|=I where e is eigenvalue of operator.
No, as I told you many times, even if you corrected the language to state what you are trying to say correctly, the content is wrong. I gave you explicit justification above why it is wrong. Yet you keep claiming that it is true without any justification or proof. Why do you think the eigenstates of the free-particle energy operator do not form a complete set? Show a detailed proof.
PS1
As for Identity operator I, there's only one eigenvalue, 1, thus one eigenstate |1> so if it formed complete set, projection should be |1><1|. Projection |1><1|any vector>∝|1> thus vector does not come back to itself but fall into |1>. It means that identity operator is of course Hermitian but does not form complete set, the case of ultimate degeneracy.
Identity operator I is expressed by help of any observable E as ∫de or Σe |e><e| where e is eigenvalue of E. Similarly by held of |p>s, P^2 is expressed as ∫dp |p>p^2<p|. These examples show that I or P^2 does not form a complete set but a complete set helps representation of I or P^2.
As far as I can tell, everything you wrote above is nonsense, and indicates that you don't have a clear idea what the terms eigenstate, projection operator, observable, Hermetian operator, closure relation, or identity operator mean. I suggest that you go back to the textbook, read carefully, and work some problems so that you understand better what you are talking about. Then you should be able to phrase your question better, if you haven't already realized the answer for yourself.
