Is the controlled-Hadamard gate in the Clifford group?

Question

Is the controlled-Hadamard gate a member of the Clifford group? I understand that Controlled Pauli gates are in the Clifford group. If controlled Hadamard is Clifford member, then is a controlled-SingleClifford also a member of the Clifford group ?

Answer 1

No, the controlled Hadamard isn't a Clifford operation. An operation is Clifford if it conjugates Pauli products into Pauli products. A controlled-Hadamard conjugates an X on its target into something that's not a Pauli product.

$$CH_{c \rightarrow t} \cdot X_t \cdot CH_{c \rightarrow t} = CY_{c \rightarrow t} \cdot X_t \cdot S_c$$

You can also perform two T gates by performing one controlled-Hadamard gate with the right Clifford operations around it. And the T gate isn't Clifford. (Exercise: see if you can find how to get the T gates out. Find Cliffords $U, V$ such that $U \cdot CH \cdot V = T^{\otimes 2}$.)