Quantum system A has a basis $\{a_1, a_2\}$. System B has a basis $\{b_1, b_2, b_3\}$. A and B evolve according to their own Hamiltonian and do not interact at all. If I consider A and B as one large system, is the dimension of the vector space 5 or 6?
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The Hilbert space is the tensor product of the two, i.e. $\mathcal{H} = \mathcal{H}_A \otimes \mathcal{H}_B$, and its dimension is the product of the two dimensions: $\mathrm{dim}(\mathcal{H}) = \mathrm{dim}(\mathcal{H}_A) \mathrm{dim}(\mathcal{H}_B)$. In your case $\mathrm{dim}(\mathcal{H}) = 2 \cdot 3 = 6$. In simple words, when $A$ is in state $a_1$, $B$ can be in one of the three states $b_1$, $b_2$, $b_3$ and the same happens when $A$ is in state $a_2$. By the way this happens also when the two subsystems are interacting.
Matteo
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As per usual in quantum mechanics lets say that A is described by states in a space $\mathcal{E}_A$ and B by $\mathcal{E}_B$. These systems are fully determined by the basis $\{a_1, a_2\}$ and $\{b_1, b_2, b_3\}$.
If the quantum systems do not interact, then the space containing the states of the full system A+B is \begin{equation} \mathcal{E}=\mathcal{E}_A \otimes \mathcal{E}_B \end{equation} whose basis vectors can be written always as \begin{equation} |a_n b_m \rangle := |a_n\rangle \otimes |b_m\rangle. \end{equation}
So the real question is: how many different combinations of $a_n$ and $b_m$ can you make? If $a_1 \neq a_2$ and $b_1 \neq b_2 \neq b_3$, which they must be, the answer is 6.
