Does Controlled-U gate entangle qubits?

Question

I've read C-NOT gate entangles qubits, but is it only for C-NOT or any arbitrary CU gate (apart from the likes if CI) entangles qubits?

Answer 1

For any controlled-$U$, if the input state is $|+\rangle|\phi\rangle$ where $|\phi\rangle$ is not an eigenstate of $U$, then the output state is entangled. This immediately deals with trivial cases such as $U=I$ because in that case all states $|\phi\rangle$ are eigenstates, and so it is not entangling. For any other $U$, there is an input state that is separable that is mapped to an entangled state, and hence controlled=$U$ is entangling for any $U\neq I$.

Proof: Let $|\phi\rangle=\sum_ia_i|\lambda_i\rangle$, where $|\lambda_i\rangle$ are the distinct eigenvectors of $U$. Your gate then evolves $$ |+\rangle|\phi\rangle\mapsto\frac{1}{\sqrt{2}}|0\rangle|\phi\rangle+\frac{1}{\sqrt{2}}|1\rangle\sum_ia_ie^{i\lambda_i}|\lambda_i\rangle. $$ Since $\sum_ia_ie^{i\lambda_i}|\lambda_i\rangle$ is not proportional to $|\phi\rangle$ (under the assumption that the phases $e^{i\lambda_i}$ are distinct and there are at least two non-zero $a_i$), the state is entangled because it is not a product state.