Could someone give me an example of a gate in the Clifford hierarchy which cannot be written as $$ e^{i \theta} V $$ for some unitary $ V $ with entries in terms of $ \zeta_{2^k} $?

If no such example comes to mind, perhaps it is true that every gate in the $ k-1 $ level of the Clifford hierarch be written as a global phase times a determinant $ 1 $ matrix with entries in the cyclotomic field $$ \mathbb{Q}(\zeta_{2^k}) $$ where $ \zeta_{2^k}=e^{2\pi i/2^k} $ is a primitive $ 2^k $ root of unity?

Note that the claim is true for the first level of the hierarchy since the Pauli group is all global phase times determinant $ 1 $ matrices with entries from $ \mathbb{Q}(\zeta_4=i) $.

And the claim is true for second level of the hierarchy because the Clifford group is all global phase times determinant $ 1 $ matrices with entries from $ \mathbb{Q}(\zeta_8=\frac{1+i}{\sqrt{2}}) $. Determinant $ 1 $ Hadamard is $$ \begin{bmatrix} \frac{i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \\ \frac{i}{\sqrt{2}} & \frac{-i}{\sqrt{2}} \end{bmatrix} $$ and determinant 1 phase gate is $$ \begin{bmatrix} \overline{\zeta_8} & 0 \\ 0 & \zeta_8 \end{bmatrix} $$ note that $\overline{\zeta_8}=\zeta_8^7 $ and $ \frac{1}{\sqrt{2}}=\frac{\zeta_8+\zeta_8^7}{2} $ and $ i=\zeta_8^2 $

Maybe someone knows a gate from the third level of the Clifford hierarchy that cannot be written in terms of $ \zeta_{16} $?

  • $\begingroup$ I am confused, isn't that already answered by a combination of your previous questions? $\endgroup$ Nov 6, 2022 at 7:27
  • $\begingroup$ @MarkusHeinrich Ok I think the part I was really confused about was whether all transversal gates of stabilizer codes are in the Clifford hierarchy. I've updated by question to reflect that. $\endgroup$ Nov 17, 2022 at 20:32
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    $\begingroup$ "We use the disjointness to show that all transversal logical operators on stabilizer codes must be in the Clifford hierarchy, as conjectured by Zeng et al." Jochym-O'Connor, Kubica, Yodor: link.aps.org/doi/10.1103/PhysRevX.8.021047 $\endgroup$ Nov 18, 2022 at 8:15
  • $\begingroup$ @MarkusHeinrich Thanks! I thought that fact was known but really started to doubt myself the other day. That's exactly the reference I wanted. Given that that part is established I'm now focusing on the second half of my original question: Is the Clifford hierarchy all defined over $ \zeta_{2^k} $? $\endgroup$ Nov 18, 2022 at 12:50

1 Answer 1


The answer is: Yes (assuming the generalized semi-Clifford conjecture is true)

Conjecture 2 of https://arxiv.org/abs/0712.2084 is that for any number of qubits $ n $ all elements of all levels $ k $ of the Clifford hierarchy are "generalized semi-clifford gates" meaning that they can be expressed as $$ UPDV $$ where $ U,V $ are Clifford , $ P $ is a permutation matrix, and $ D $ is a diagonal gate from the $ k $th level of the Clifford hierarchy.

Recall that the Clifford group is defined over $ \mathbb{Q}(\zeta_8) $, permutation matrices are defined over anything, and the structure of the diagonal gates in the Clifford hierarchy is well known, in particular the diagonal gates in the $ k $th level are generated by all $ C^iZ^{1/2^j} $[https://arxiv.org/abs/2110.11923] gates where $ i+j=k-1 $ and so diagonal gates are defined over $ \mathbb{Q}(\zeta_{2^{k+1}}) $. So all generalized semi Cliffords of level $ k $ are defined over $ \mathbb{Q}(\zeta_{2^{k+1}}) $. Assuming the generalized semi Clifford conjecture then everything in the Clifford hierarchy is generalized semi Clifford and thus is defined over some $ \mathbb{Q}(\zeta_{2^{k+1}}) $.

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    $\begingroup$ I believe this answer is correct but I wanted to point out a slight subtlety. While the structure of diagonal gates in the Clifford Hierarchy is known, it does not (necessarily) imply that the diagonal gates, $D$, in $UPDV$ above must be restricted in exactly the same manner for all generalized semi-Clifford gates. It turns out you can prove that a gate $UPDV$ is in the Clifford Hierarchy iff elements P and D are individually in the Clifford Hierarchy. While expected, I could not find proof of it anywhere. I included a proof in appendix A of this paper. $\endgroup$ Dec 13, 2022 at 12:10

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