For a single qubit, the Clifford hierarchy is defined to be $$ \mathcal C^{(k)} = \Bigl\{ U \in \mathbf U(2) \mathrel{\Big\vert} \forall P \in \mathcal C^{(1)} : U P U^\dagger \in \mathcal C^{(k-1)} \Bigr\} $$ Is it true that all elements of $ \mathcal C^{(k)} $ must have entries in the field generated by $ \zeta_{2^k} $, a primitive $ 2^k $th root of unity? In particular, is it the case that a matrix like $$ \Phi= \frac{1}{2} \begin{bmatrix} \varphi+ i \varphi^{-1} & 1\\ -1 & \varphi - i \varphi^{-1} \end{bmatrix} $$must not be in the Clifford hierarchy? Here $ \varphi:= \frac{1+\sqrt{5}}{2} $ is the golden ratio and $ \varphi^{-1}= \frac{-1+\sqrt{5}}{2} $ is its multiplicative inverse.


1 Answer 1


I assume that you mean entries in $\mathbb Q[\zeta_{2^k}]$, up to a global phase, right? This is certainly true for $k=1,2$: Clearly, any element of the Pauli group has, up to a global phase, entries in $\mathbb Q$. It is also known that the minimal matrix group which is projectively isomorphic to the Clifford group has entries in $\mathbb Q[i]$.

In the more general case, this may be true, but it is far from being clear. The form of diagonal unitaries in the Clifford hierarchy was characterized (see Cui-Gottesman-Krishna), and it is known that they involve powers of $2^m$-th roots of unity for $m\leq k$. The full Clifford hierarchy is still not completely understood. There is a subset called semi-Clifford unitaries (Ref) which are exactly those unitaries which can be written as $C_1 D C_2$ where $C_1$,$C_2$ are Clifford and $D$ is a diagonal matrix. For this subset, your claim certainly still holds. Beyond this, we get into unsettled territory.

To address your example, it is not in the Clifford hierarchy. This is because it is known that for $n=1,2$ qubits, all gates in the Clifford hierarchy are semi-Clifford.

If you want to dive into this, here are some more recent takes on the Clifford hierarchy:

  1. "Unifying the Clifford Hierarchy via Symmetric Matrices over Rings" (make sure you have v3) http://arxiv.org/abs/1902.04022
  2. "Un-Weyl-ing the Clifford Hierarchy" http://arxiv.org/abs/2006.14040

Final note: This holds only for qubits. In prime dimensions $p>2$, the interplay between $p$, the order of the root of unity, and the level of the hierarchy is more involved (e.g. Cliffords can be written using a $p$-th root of unity). See the paper by Cui et al.

Very final note: On second thought, you might not need the full machinery. Maybe it's possible to prove this by induction using the definition alone. But I'm not sure how to argue in this case.

  • $\begingroup$ Ah I see since clifford is defined over $ \zeta_{2^k} $ and by the other paper you mention all diagonal Clifford hierarchy gates are defined over some $ \zeta_{2^k} $ then all semi-clifford are defined over some $ \zeta_{2^k} $ And so for $ n=1,2 $ qubits it is know that all the Clifford hierarchy is semi-clifford (what's a good reference for that?) and thus all Clifford hierarchy is defined over some $ \zeta_{2^k} $ for $ n=1,2 $ qubits. $\endgroup$ Nov 4, 2022 at 21:42
  • $\begingroup$ @IanGershonTeixeira indeed. The reference is already in the cited paper by Zheng et al. $\endgroup$ Nov 6, 2022 at 6:28
  • $\begingroup$ Given a diagonal gate $ D$ in the Clifford hierarchy is it always the case that the semi Clifford gate $ V_1 D V_2 $ is also in the Clifford hierarchy for any Clifford gates $ V_1,V_2$ ? $\endgroup$ Nov 8, 2022 at 18:29
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    $\begingroup$ @IanGershonTeixeira That's proposition 3 in the Zheng et al. paper: If $R$ is in the $k$-th level, then so is $L_1 R L_2$ for all Cliffords $L_1$ and $L_2$. (here $R$ does not need to be diagonal) $\endgroup$ Nov 9, 2022 at 6:51
  • $\begingroup$ The $ \Phi $ gate in this question plays a prominent role in my paper with Eric Kubischta arxiv.org/abs/2305.07023. In our 3rd revision (just uploaded 2 days ago) we decided to include in our appendix a proof that $ \Phi $ is an exotic gate (not in the Clifford hierarchy). I forgot that you are the one who first told me about this proof nine months ago! We'll make sure to include you in the acknowledgments in the next revision! $\endgroup$ Jul 26, 2023 at 21:37

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