# When is the square root of a Clifford gate a Clifford gate?

Is there any condition for the square root or any rational power of a Clifford gate to be a Clifford gate (generated by $$S$$, Hadamard $$H$$, and CNOT)? It can be easily shown that $$\sqrt{X}$$ is Clifford (because $$S=\sqrt{Z}$$ is a generator, and powers of $$Z$$ can be changed into power of $$X$$ using $$H$$), but $$\sqrt{H}$$ and $$\sqrt{\mathrm{CNOT}}$$ are not-so-obviously non-Clifford. Is there an easy way to tell?

TL;DR A simple sufficient condition is: if the order $$r$$ of a given Clifford $$U$$ and the root's degree $$n$$ are relatively prime, then $$U$$ has an$$^1$$ $$n$$th root.

Let $$r$$ be the order of $$U$$, i.e. the smallest positive integer such that $$U^r=I$$. If $$\gcd(n,r)=1$$, then there exists a Clifford $$V$$ such that $$V^n=U$$. In fact, we can choose $$V=U^b$$ where $$b$$ is the multiplicative inverse of $$n$$ modulo $$r$$. For example, this criterion shows that $$HS$$ has a Clifford square root.

In the case $$\gcd(n,r)>1$$, a Clifford $$n$$th root may or may not exist. For example, if $$U$$ is a Pauli operator, then $$U=V^2$$ where $$V=\sqrt{\frac{i}{2}}U+\sqrt{-\frac{i}{2}}I$$ is Clifford. However, if $$U$$ is the Hadamard, then there is no suitable $$V$$, because Hadamard induces an odd permutation on the axes of the Bloch sphere.

$$^1$$ Note that the $$n$$th root is only well-defined for positive semidefinite operators. Instead of literal interpretation of the question I assume that we would like a criterion that, for a given positive integer $$n$$ and Clifford $$U$$, ensures that there exists a Clifford $$V$$ such that $$V^n=U$$. Such $$V$$ will not be unique.

• The first method works ok but the second case is excluding the most interesting cases like $H$ and $CNOT$ Nov 23, 2023 at 15:59
• This gives two sufficient conditions: one for general $n$ (namely: $\gcd(n, r)=1$) and one for $n=2$ (namely: $U$ is a Pauli). It also shows an example demonstrating a necessary condition for $n=2$ (namely: $U$ must induce an even permutation on every set acted on by Cliffords). This actually handles one of the interesting cases (namely: $H$). However, I don't know a general condition that is both necessary and sufficient. Nov 24, 2023 at 5:38
• Wouldn't being Pauli or having gdc$(n,r)=1$ be a necessary and sufficient condtions for $n=2$? Or there is something escaping these two Nov 24, 2023 at 10:55
• Arrange twelve qubits around the face of a clock. Let $S_k$ denote the clockwise shift by $k$ hours, constructible as a circuit out of $\text{SWAP}$ gates. Clearly, $S_2$ isn't a Pauli and has an even order, yet it has a square root. Nov 24, 2023 at 12:28