According to this article, any Clifford gate, acting on $n$ qubits, can be generated by Hadamard, CNOT, and S gates.

What are the set of generators for qutrit Cliffords?


1 Answer 1


From the paper Normal form for single-qutrit Clifford+T operators and synthesis of single-qutrit gates, the Clifford group in $p>2$ dimensions acting on a sigle qudit is generated by $S$ and $H$ given by:

$$ \begin{gather} S=\sum_{j=0}^{p-1}\omega^{j(j+1)2^{-1}}|j\rangle\langle j| \\ H = \frac{1}{\sqrt{p}}\sum_{j=0}^{p-1}\sum_{k=0}^{p-1}\omega^{jk}|j\rangle\langle k| \end{gather} $$

where $j$ and $k$ denote elements of the finite field $\mathbb{Z}_p$ with multiplication and addition defined modulo $p$, $2^{-1}$ is an element of the same field, and $\omega$ denotes a $p$th rooth of unity $\omega^p=1$. You can set $p=3$ for your case.

Now, for the equivalent of the $CNOT$ gate you want to look into this answer.

  • $\begingroup$ Thanks! An additional question, just out of curiosity. Is there something like a Solovay Kitaev theorem for the Clifford group and its generators? If we have a set of generators, is there any guarantee on how many gates we need from that generating set to construct a Clifford gate? I think the answer is no, I just thought of doing a sanity check. $\endgroup$
    – BlackHat18
    Commented Sep 7, 2021 at 19:50
  • $\begingroup$ As a follow up question, let's say we have two different generating sets for the qubit/qutrit/qudit Clifford group: if we care about efficiency, because of the absence of any Solovay-Kitaev type theorem, the set of Clifford gates efficiently generated by these two generating sets may be very different --- is that correct? $\endgroup$
    – BlackHat18
    Commented Sep 7, 2021 at 19:57
  • 1
    $\begingroup$ @BlackHat18 yes I think that's right. As you state, the absence of any equivalent of the Solovay-Kitaev theorem leaves the efficiency to depend on the selected set of generators. In this paper the author gives a generator and relations presentation of the Clifford group. AFAIK, different relations (and/or different sets of generators) may yield different efficiencies on generating an arbitrary operator on the group. So this would make your statement true. Don't take my word for it since I'm not an expert on this, but the statement seems true. $\endgroup$
    – epelaez
    Commented Sep 7, 2021 at 20:20
  • 3
    $\begingroup$ @BlackHat18 for the "standard" set of generators, the number of non-local generators needed is $O(n^2/log(n))$ and this is asymptotically optimal. For other sets of generators, this can be vastly different of course. Just by increasing your set of generators, you will get better counts. For instance, if you include all CZ circuits and all CX circuits, the gate count will be $O(n)$. Include any Clifford unitary then it's 1. Thus, if you want to compare generating sets, you should consider minimal generating sets (similar to a vector space basis). But then, the mentioned scaling is optimal. $\endgroup$ Commented Sep 8, 2021 at 11:29
  • $\begingroup$ Might you provide a reference for the $O(n^{2}/\log n)$ scaling? Also, what's a "non-local generator"? $\endgroup$
    – BlackHat18
    Commented Sep 8, 2021 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.