Coupled and uncoupled qubits - Hilbert space representation Suppose I have two situations: one where two qubits, $q_A$ and $q_B$, exist independently (on separate sides of the quantum chip, maybe), and one where they exist with some coupling between them. And consider operations on the qubits in these two situations. For the independent qubits, only single-qubit gates are possible, and for the coupled qubits, entangling operations would then become possible.
Both of these situations can be described by using the Hilbert space $\mathcal H = \mathcal H_A \otimes \mathcal H_B$, right? But I want to be able to express mathematically the idea that without coupling between the qubits, that is, only using single-qubit operations, there is an inability to reach certain points in this Hilbert space. Can you help push me in the right direction?
 A: This isn't my realm of expertise, so my terminology & notation may be unorthodox and possibly off-base.  But since this question hasn't garnered any attention so far, I'll give it a go.
An "unentangled state" is known as a separable state.  It is effectively any state that can be written in the form
$$
|\Psi\rangle = |\psi_A \rangle \otimes |\psi_B \rangle,
$$
where $|\psi_A\rangle \in \mathcal{H}_A$ and $|\psi_B \rangle \in \mathcal{H}_B$.  Single-bit operations correspond to operators that are of the form
$$
\mathcal{O}_\text{single} = \mathcal{O}_A \otimes \mathbb{1}_B \quad \text{or}  \quad \mathbb{1}_A \otimes \mathcal{O}_B
$$
(i.e., they act on "one part" of the Hilbert space and leave the other part alone.)  It is not too difficult to see that the successive application of any single-bit operations is equivalent to applying an operator of the form $\mathcal{O} = \mathcal{O}_A \otimes \mathcal{O}_B$.  We can think of operators of this form as "separable operators" in the sense that they map separable states $|\Psi\rangle$ to other separable states, since
$$
\mathcal{O} |\Psi \rangle = \left( \mathcal{O}_A |\psi_A \rangle \right) \otimes \left( \mathcal{O}_B |\psi_B \rangle \right).
$$
So regardless of whether there is any coupling between the qubits, single-bit operations will only take separable states to separable states;  they cannot take an "unentangled" state to an "entangled" state.
Where the coupling comes in is how the states themselves evolve in time.  If the qubits are uncoupled, then the Hamiltonian is of the form
$$
H = H_A \otimes \mathbb{1}_B + \mathbb{1}_A \otimes H_B
$$
meaning that the time-evolution operator is
\begin{align}
\mathcal{O}(t) &= e^{i (H_A \otimes \mathbb{1}_B + \mathbb{1}_A \otimes H_B) t/\hbar} \\
&= e^{i (H_A \otimes \mathbb{1}_B) t/\hbar} e^{i (\mathbb{1}_A \otimes H_B) t/\hbar} \\
&= \left(e^{i H_A t/\hbar} \otimes \mathbb{1}_B \right)\left( \mathbb{1}_A \otimes e^{i H_B t/\hbar} \right) \\
&= e^{i H_A t/\hbar} \otimes e^{i H_B t/\hbar}.
\end{align}
In the second step, we have used the Baker-Campbell-Hausdorff formula, which relies on the two parts of the Hamiltonian commuting with each other;  for the third step, see this question.  This is a "separable operator" in the same way that the product of single-qubit operators is separable:  it maps separable states to separable states, and so an unentangled system will remain unentangled.
However, if the qubits are coupled, then there will be other pieces of the Hamiltonian that are not of the form $H_A \otimes \mathbb{1}_B$ or $\mathbb{1}_A \otimes H_B$.  It can be shown in this case that the time-evolution operator is no longer separable, and so a separable state can evolve to a non-separable (entangled) state.
