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

It is an open conjecture, see Controlled Gates in the Clifford Hierarchy, that if $ U $ is a gate in the Clifford hierarchy with order a power of $ 2 $ then 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 should be in the 3rd level of the Clifford hierarchy not the 2nd level (i.e. it is not in the Clifford group). Note that in general adding a control does not have such a nice effect. For example, the facet gate is a single qubit Clifford but adding a control produces a new gate which is neither in the Clifford group, nor even in any level of the Clifford hierarchy. Indeed the Facet gate has order divisible by 3 and it is proven in Controlled Gates in the Clifford Hierarchy that for any gate of order not a power of 2, then the controlled version of the gate is not in the Clifford hierarchy.

A direct way to see that the conjecture holds in this case, i.e. 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 (and not in the 2nd level since $ T^{\otimes 2} $ is not in the 2nd level.