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 $U \cdot CH \cdot V = T^{\otimes 2}$.)

Answer 2

Adding a control always increases the level of the Clifford hierarchy by $ 1 $. Since Hadamard is in the 2nd level of the Clifford hierarchy (i.e. the Clifford group) then controlled Hadamard is in the 3rd level of the Clifford hierarchy not the 2nd level (i.e. it is not in the Clifford group).

Another easy way to see that $ CH $ is in the 3rd level is to show what Craig suggested that $ CH $ can be multiplied by Cliffords to obtain $ T^{\otimes 2} $. It is well known that $ T $ and $ T^{\otimes 2} $ are in the third level of the hierarchy. Clifford multiplication preserves the level of the hierarchy so this is an independent proof that $ CH $ is in the third level.