Why people believe this
CNOT is the standard two-qubit entangling gate. It is used to create Bell states. So it must create entanglement whenever it is applied.
The correction
CNOT does not always create entanglement. If the control qubit is in a computational basis state (|0> or |1>) rather than superposition, CNOT acts as a classical conditional flip — no entanglement is created. Entanglement requires a superposition on the control qubit before the CNOT. The standard Bell state preparation works because H creates superposition first, then CNOT correlates the two qubits. A CNOT on |00> simply gives |00> — no entanglement.
Try it in the simulator
What to do
First run the Bell state (H then CNOT) — you get entanglement, 50% |00> and 50% |11>. Now remove the H gate and run CNOT alone on |00>. You get |00> with 100% probability — no entanglement. The H gate is what creates the superposition that CNOT then entangles.
Research notes
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