When one studies quantum mechanics under a more rigorous point of view, the very first postulate states that the underlying Hilbert space $\mathscr{H}$ is separable. This means that $\mathscr{H}
$ has a countable orthonormal basis.
I remember when I took a first course in quantum mechanics the first part of the course was dedicated to solving the Schrödinger equation $H\psi = E\psi$ to find the eigenvalues of the Hamiltonian. When the system was bounded, the Hamiltonian had only a countable set of eigenvalues $\{E_{n}\}_{n\in \mathbb{N}}$ and its corresponding eigenvectors defined a basis for the Hilbert space. This was done, e.g. to study a particle in a square well, the harmonic oscillator or even the Hydrogen atom.
At a first sight, I thought the separability condition was reasonable because one was trying to find a basis of eigenvectors of the time-independent Hamiltonian $H$. This was for me the basic machinery of quantum mechanics, as the aforementioned examples would suggest.
However, when I started studying the mathematics of quantum mechanics more deeply, one finds out that the Hamiltonian is supposed to be a self-adjoint operator on some densely-defined domain on $\mathscr{H}$ and, of course, it is not always the case it has a countable set of eigenvectors.
It is clear to me that the separability condition is necessary. What is not clear is what it means when it is not connected with states of well-defined energy. For instance, suppose we have a many-particle quantum system described by a Fock space $\mathcal{F}(\mathscr{H})$. To study bosonic or fermionic Fock spaces one usually uses a countable basis for $\mathscr{H}$ to find a countable basis for $\mathcal{F}(\mathscr{H})$. If $\mathscr{H}$ has a countable basis of eigenvectors of the Hamiltonian of a single particle state, each associated to an energy $E_{n}$, it is clear that a state $|n_{1},n_{2},…\rangle$ of the Fock space can be thought as a state with $n_{1}$ particles with energy $E_{1}$, $n_{2}$ particles of energy $E_{2}$ and so on. If, on the other hand, $\mathscr{H}$ has a countable basis which has nothing to do with the Hamiltonian, then the states $|n_{1},n_{2},…\rangle$ mean nothing to me because I don't know what this basis represents.
So, what is the point of having a basis for $\mathscr{H}$ which has nothing to do with the Hamiltonian? Is it physically relevant? If so, what it represents?
Best Answer
Separability is not a necessary physical requirement, at least in modern approaches. First of all, all mathematical technology, as the spectral theory, is valid both for separable or non-separable Hilbert spaces.
The Hilbert space of a quantum system turns out to be separable under some quite standard circumstances however. In particular, when it is the representation space of a strongly continuous unitary irreducible representation of a (finite dim) Lie group. This is the case for every elementary system. The group is the Poincaré one, the Weyl-Heisenberg one or more complicated groups including some also non-abelian internal symmetries. The Fock space constructed upon that space is separable as well by construction.
An apparent physical consequence of non separability is that, a bounded below Hamiltonian, even if equipped with pure point spectrum only, cannot produce thermal mixed states of the usual form $e^{-\beta H}/Z_\beta$, since these operators cannot be trace class. However there is the way out of the algebraic formalism to describe thermal states in that case, using KMS algebraic states.
The structure of Fock space is independent of the notion of eigenvector of a Hamiltonian. For instance, the symmetric Fock space is nothing but the direct orthogonal sum of all possible symmetrized tensor products of the one particle space. It does not matter if there is a basis made of eigenvectors of one or another Hamiltonian indicated with the popular notation.
ADDENDUM Separability plays however a role in motivating the structure of QM from more basic principles. It takes place in the hypotheses of the Soler theorem and the Gleason theorem in particular. But QM as a whole does not need that hypothesis to be stated.