= Hadamard conjugation of a controlled gate
{c}
{title2=$(I\otimes H)C(V)(I\otimes H)=C(HVH)$}
For a <controlled unitary gate> $C(V)=|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes V$, <Hadamard gates> on the target before and after the gate produce $C(HVH)$. Proof: the control-zero branch applies $HIH=I$, while the control-one branch applies $HVH$. <Linearity> extends the identity to arbitrary <quantum states>. Since $C(H)C(V)C(H)=C(HVH)$ too, a pair of controlled <Hadamard gates> surrounding the same target operation can be replaced by ordinary target <Hadamard gates>. No single controlled <Hadamard gate> is thereby identified with an ordinary one.
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