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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?

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    $\begingroup$ Use \langle and \rangle instead of < and >. Can you elaborate your question: What is the problem in having a basis not being the eigenbasis of a Hamiltonian on $\mathscr H$? $\endgroup$ Commented Sep 14, 2022 at 18:54
  • $\begingroup$ @JasonFunderberker I edited the symbols with your suggestions. Thanks. About the question, I just want to know what it means having a basis not connected to the Hamiltonian. In a many-particle system, would it mean anything at all? I mean, $|n_{1},n_{2},...\rangle$ would represent a system of $n_{1}$ particles with some well-defined property? $\endgroup$
    – MathMath
    Commented Sep 14, 2022 at 18:58
  • $\begingroup$ Yes: If you have a single-particle basis $\{\varphi_n\}_{n\in \mathbb N}$ in $\mathscr H$, then a state $|n_1,n_2,\ldots\rangle$ means $n_1$ particles in $\varphi_1$ etc. $\endgroup$ Commented Sep 14, 2022 at 19:17

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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.

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  • $\begingroup$ Valter, thanks for the answer. It seems that it confirms my impression that the basis of the Hilbert space in príncipe has nothing to do with its physical properties. However, how to understand this basis physically? On a Fock space, a creation operator creates a one particle with state given by such vectors. How to interpret these vectors then? $\endgroup$
    – MathMath
    Commented Sep 14, 2022 at 19:24
  • $\begingroup$ A state in a Fock space with n particles is nothing but a state for that system of identical particles and thus it is suitably symmetric or antisymmetric under swap of particles. As the space is however still a Hilbert space, you can always decompose these vectors under bais. Natural basis are constructed by symmetrising thensor products of elements of one-particle basis. $\endgroup$ Commented Sep 14, 2022 at 19:28
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    $\begingroup$ There is no physics involved in that procedure to construct Fock bases from a one-particle basis. It us just a consequence of the mathematical definition. $\endgroup$ Commented Sep 14, 2022 at 19:30
  • $\begingroup$ In some cases, but this fact is not general, the Fock basis has a physical sense. When constructing the Fock basis out of a one particle basis of eigenvectors of a Hamiltonian H, you find the basis of eigenvectors of the Hamiltonian of the system of particles viewed as independent of each other. The big Hamiltonian is here the sum of the Hamiltonians of the single particles. But it is by no means obvious that this us the true Hamiltonian of the system: maybe there are further interactions between the particles. $\endgroup$ Commented Sep 14, 2022 at 19:34
  • $\begingroup$ Valter, I get your point. You are basically saying that if a Hilbert space for a quantum system is separable, every vector (state) can be written in terms of this basis, regardless if this basis has some physical meaning or not. But from the physics perspective, this still bothers me because it does not seem helpful for physicists to know that the Hilbert space has a countable basis if they don't know what this basis of vectors is or represent. Isn't the whole point of the theory to find a known physically relevant basis, from which all possible states can be obtained? $\endgroup$
    – MathMath
    Commented Sep 14, 2022 at 20:49

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