If $H_A$ and $H_B$ are Hilbert spaces corresponding to two systems $A$ and $B$, then the Hilbert space for the combined system is given by the tensor product of $H_A$ and $H_B$. Does this hold irrespective of whether $A$ and $B$ are interacting or not?
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Yes, the combined space of two systems is always the tensor product $\mathcal{H}_A\otimes \mathcal{H}_B$ of the individual state spaces $\mathcal{H}_A$ and $\mathcal{H}_B$, see also this question and its answers.
However, for non-interacting systems, the Hamiltonian is the sum of the two individual Hamiltonians $H_A$ and $H_B$, $H = H_A \otimes \mathbf{1}_B + \mathbf{1}_A\otimes H_B$.
This means that if you start with a non-entangled state $\psi = \psi_A\otimes \psi_B\in \mathcal{H}_A\otimes \mathcal{H}_B$, the time evolution of this state will forever remain non-entangled as $$\mathrm{e}^{\mathrm{i}Ht}\psi = \left(\mathrm{e}^{\mathrm{i}H_At}\psi_A\right)\otimes\left(\mathrm{e}^{\mathrm{i}H_Bt}\psi_B\right),$$ so effectively you can describe the state of the system by pairs $(\psi_A,\psi_B)$ from the direct sum $\mathcal{H}_A\oplus\mathcal{H}_B$ since they can never become entangled.
However, there's rarely a reason to do this: If the two systems don't interact, why would you want to model them as subsystem of a combined system and not just as two separate system?
answered Jul 1, 2023 at 19:56
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The Hilbert space of a composite system is a closed subspace of the Hilbert tensor product of the Hilbert spaces of the constituents. If the system consists of identical particles then it is either the symmetric subspace (bosons) or the anti-symmetric subspace (fermions). If none of the particles are identical then it is simply the whole Hilbert tensor product.
The Hilbert tensor product is the completion of the tensor product with respect to the norm induced by the natural scalar product on the tensor product and not just the algebraic tensor product, because the algebraic tensor product is in general not a Hilbert space.
The only thing that changes if interactions are allowed is the form of the Hamiltonian. The Hamiltonian is then not just the tensor product of the individual Hamiltonians, but can have extra terms that model the interaction of the subsystems.
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If the states you want to describe are either states of $A$ or states of $B$, its Hilbert space will be the direct sum $H_A\oplus H_B$. This is the case when considering non-interacting particles, e.g. spinless fermions
$$H_A = \sum_{i,j} h_{ij}^{A} c_i^\dagger c_j^{\phantom{\dagger}},$$ $$H_B = \sum_{i,j} h_{ij}^{B} d_i^\dagger d_j^{\phantom{\dagger}},$$ $$H_{AB} = \sum_{i,j} h_{ij}^{AB} c_i^\dagger d_j^{\phantom{\dagger}}.$$
The Hamiltonians $H_A,H_B$ and $H_{AB}$ are quadratic in creation and annihilation operators, and thus the systems $A$, $B$, and also the combined one $AB$, are non-interacting (even though $A$ and $B$ are in "contact" with each other through $H_{AB}$). States of the combined system can be constructed by taking direct sums of states in the corresponding subsystems $\vert i_A\rangle \oplus \vert j_B\rangle$.
If we have a physical system whose states are states of $A$ and states of $B$, its Hilbert space will be the tensor product $H_A \otimes H_B$. For example, taking $H_A$ and $H_B$ as in the example above, but including interactions between particles of $A$ and $B$
$$H_{AB} = \sum_{i,j} h_{ij}^{AB} c_i^\dagger c_i^{\phantom{\dagger}} d_j^{\dagger} d_j^{\phantom{\dagger}}$$
which would correspond to a Hubbard-type interaction. In this particular case, one cannot describe the system $AB$ by states of $A$ or $B$ alone, hence the states of the combined system are constructed by taking the tensor product of states of the corresponding subsystem, $\vert i_A\rangle \otimes \vert j_B\rangle$.
