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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?

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  • $\begingroup$ Definitely not any arbitrary $U$. Consider $U=I$. $\endgroup$ – nippon Apr 7 at 9:34
  • $\begingroup$ That's a fair answer. But I'm interested in let's say U = Ry or something like that $\endgroup$ – Ajay Rao Apr 7 at 9:37
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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.

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