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?


1 Answer 1


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.

  • $\begingroup$ The first method works ok but the second case is excluding the most interesting cases like $H$ and $CNOT$ $\endgroup$
    – Mauricio
    Nov 23, 2023 at 15:59
  • $\begingroup$ 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. $\endgroup$ Nov 24, 2023 at 5:38
  • $\begingroup$ 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 $\endgroup$
    – Mauricio
    Nov 24, 2023 at 10:55
  • $\begingroup$ 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. $\endgroup$ Nov 24, 2023 at 12:28

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