# The Clifford hierarchy and $e^{2 \pi i/2^k}$

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 $$\zeta_{16}$$?

• I am confused, isn't that already answered by a combination of your previous questions? Nov 6, 2022 at 7:27
• @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. Nov 17, 2022 at 20:32
• "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 Nov 18, 2022 at 8:15
• @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 $\zeta_{2^k}$? Nov 18, 2022 at 12:50

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 $$\mathbb{Q}(\zeta_{2^{k+1}})$$.
• 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. Dec 13, 2022 at 12:10