# Is every Clifford gate conjugate to a diagonal Clifford gate?

Let $$C$$ be a Clifford gate. Let $$D$$ be the diagonalization of $$C$$. In other words $$D$$ is a diagonal gate and $$C=VDV^{-1}$$ for some $$V$$. Is $$D$$ also a Clifford gate?

Update: Filling in the details of my comment below.

Claim: There exists a Clifford gate whose diagonalization is not in any level of the Clifford hierarchy.

Proof: Let $$H$$ be the hadarmard and $$P=diag(1,i)$$ the phase gate. Then $$M:=(H*P)^8$$ is a Clifford gate of order 3. Recall that (up to global phase) a diagonal matrix in the $$k$$ level of the Clifford hierarchy has all its entries $$2^k$$ roots of unity. See https://arxiv.org/abs/1608.06596 . So every diagonal matrix in the Clifford hierarchy is a global phase times a matrix with order $$2^k$$. Thus a diagonal matrix which is in the Clifford hierarchy but has order relatively prime to $$2^k$$ must be a global phase times the identity (a unitary scalar matrix).

Suppose for the sake of contradiction that the diagonalization of $$M$$ is in the Clifford hierarchy. Since $$M$$ has order 3 and is not a scalar matrix then $$D$$ must also have order 3 and also not be a scalar matrix. So this non scalar diagonal matrix $$D$$ of order prime to $$2^k$$ contradicts the fact above.

No, $$D$$ isn't guaranteed to be Clifford. That would require all entries on the diagonal to differ by multiples of 90 degrees (all be 1, $$i$$, $$-i$$, or -1 up to global phase). But that would imply $$D$$ had period 1, 2, or 4 up to global phase. But there are Clifford gates that have periods other than 1, 2, or 4 up to global phase. For example, the 120 degree rotation around X+Y+Z is Clifford but has period 3.
• Ah you're right of course. Indeed $D$ is not even in any level of the Clifford hierarchy. Diagonal gates in the $k$ level of the Clifford hierarchy have order $2^k$ up to global phase. Since a gate and its diagonalization have the same order then there is no way for a Clifford gate with order divisible by 3 to have a diagonalization in the Clifford hierarchy. Hadamard $H$ times phase gate $P$ is a good example of a Clifford gate with order divisible by 3. Commented Feb 10, 2023 at 1:28