# Does Controlled-U gate entangle qubits?

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?

• Definitely not any arbitrary $U$. Consider $U=I$. – nippon Apr 7 '20 at 9:34
• That's a fair answer. But I'm interested in let's say U = Ry or something like that – Ajay Rao Apr 7 '20 at 9:37

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.