From the book Quantum Mechanics by Cohen-Tannoudji it seems that the only requirement for an Operator to be an Observable is to form an orthonormal basis in the state space (finite or infinite dimensional)

"By definition, the Hermitian operator $A$ is an observable if this orthonormal system of vectors forms a basis in the state space."

In the rest of the book, the Hamiltonian $H$ is always referred to as an Operator (e.g. when talking about the Schrödinger picture). However I fail to see why in some cases the Hamiltonian cannot fulfil the definition of Observable.

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    $\begingroup$ The Hamiltonian is always an observable. $\endgroup$ – Jahan Claes Nov 9 '20 at 23:27
  • $\begingroup$ Aren't observables time independent in the Schrodinger picture? Does this mean that a time dependent $\mathcal H$ needs to be manipulated and strip of its time dependency to express it in the Schrodinger picture? $\endgroup$ – Alberto Gomez Saiz Nov 9 '20 at 23:49
  • $\begingroup$ No, a time dependent Hamiltonian describes the energy observable at that instance in time. $\endgroup$ – Jahan Claes Nov 10 '20 at 13:30
  • $\begingroup$ Thanks, I was getting confused with the fact that the Hamiltonian is time dependent and the statement that operators are time independent in the Schrodinger picture. My understanding is that only unitary operators need to be time independent in the S picture and hence the Hamiltonian being non unitary can have a time dependency. Source: physics.stackexchange.com/questions/441656/… $\endgroup$ – Alberto Gomez Saiz Nov 15 '20 at 12:55

Eigenvalues of $H$ are the energy states of the system, so in order to have a real value for the energy $H$ has to be hermitian and so an observable. Also the hamiltonian operator is based on the momentum operator that is hermitian too. You can try to write the integral form of $\langle \psi|H|\psi\rangle $ and $\langle \psi|H^\dagger|\psi\rangle $ for a generic $H$ and you would get the same result.

  • $\begingroup$ Hermiticity is a necessary but no sufficient condition for an operator to be an observable. As stated in the question you need it's eigenvectors to form an orthonormal basis of the state space $\endgroup$ – Alberto Gomez Saiz Nov 15 '20 at 12:51

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