Eigenstates of an observable

Question

Can we use eigenstates of ANY observable as base of the Hilbert space? If we can, is this equal to the statement that those eigenstates are orthogonal to each other and normalizable?

Answer 1

An observable is a self-adjoint operator $\mathcal{O}$ on the Hilbert space of states $\mathcal{H}$.

The spectral theorem tells us that such an operator has an orthonormal basis of eigenvectors in $\mathcal{H}$, if it is compact.

If it is not compact, we have to "enlarge" the Hilbert space to something called rigged Hilbert space or Gelfand tripel. A good discussion of that is in Mathematical surprises and Dirac's formalism in quantum mechanics. Most introductory courses ignore this, however, and also assign "eigenvectors" to the non-compact operators like the position operator. It often works, but one must bear in mind that some non-sensical results coming from that really only occur because one has not been rigorous.

Answer 2

Unfortunately not every operator, which is used to get an observable gives you enough elements to make a base of the Hilbert space. As far as I know every eigenstate is orthogonal to each other. But the “length” could be arbitrary. Mostly used eigenstates in quantum mechanics are normalized for better usage.