Just a remark from the physical side.
If a physical system is described in a non-separable Hilbert space whatever Hamiltonian operator one chooses, thermal (Gibbs canonical or grand canonical) states cannot be defined as density matrices (mixed states) in the given Hilbert space.
So, if one wants to describe thermodynamics of that system he/she necessarily has to deal with either the micro canonical description or the algebraic formalism that, in this case, makes useless the initial Hilbert space approach.
Proof. If ${\cal H}$ is a generic Hilbert space and $\rho$ is a non-negative self-adjoint trace class operator in it, its non-vanishing eigenvalues, taking their multiplicity into account, must be countable at most, otherwise:
$$tr(\rho) = \sum_{\lambda \in \sigma(\rho)} m_\lambda\lambda$$
diverges ($*$). Above $m_\lambda\geq 1$ is the (always finite for $\lambda >0$), multiplicity of $\lambda$.
A mixed state describing a Gibbs canonical or grand canonical ensemble is, by construction a non-negative self-adjoint trace class operator with strictly positive eigenvalues and there is a Hilbertian basis of eigenvectors of $\rho$. If $\cal H$ is non-separable this basis must be uncountable and this, in turn, implies that the associated (non-vanishing) eigenvalues, counted with their multiplicity, form an uncountable set.
footnotes
($*$) if $M= \sup \sigma(\rho) = ||\rho|| <+\infty$, then $$(0, M)=\cup_{n=1}^{+\infty} (M/(n+1), M/n]\:.$$ If each interval $(M/(n+1), M/n]$ contained a finite number of eigenvalues $\lambda >0$ (taking their multiplicity into account), their number would be countable. So, if they are uncountably many, at least some interval, say $(M/(n_0+1), M/n_0]$, has to include and infinite number of them and thus $\sum_{\lambda \in \sigma(\rho)} m_\lambda \lambda$ diverges because $$\sum_{\lambda \in \sigma(\rho)} m_\lambda\lambda \geq \infty M/(n_0+1)\:.$$