Bell-basis conversion circuit ($U=\mathrm{CNOT}(H\otimes I)$)

Apply a Hadamard gate to the first qubit, then a controlled-NOT gate from the first to the second. This maps a computational-basis state $|ij⟩$ to the Bell state with phase bit $i$ and parity bit $j$. Its inverse uses the same gates in reverse order because each gate is self-inverse but they do not commute. Applying the forward sequence twice is not generally a decoding operation.