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Does the two qubit Clifford hierarchy contain all diagonal gates whose entries are $ 2^k $ roots of unity?

In particular, is it true that every $ 4 \times 4 $ diagonal matrix whose diagonal entries are $ 2^k $ roots of unity is in the $ k+1 $ level of the two qubit Clifford hierarchy?

Examples: controlled Z gate is Clifford, controlled P gate is 3rd level, and controlled T gate is 4th level.

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TL;DR: Yes, every $4\times 4$ diagonal gate whose non-zero entries are $2^k$th roots of unity are in the $(k+1)$th level of the two-qubit Clifford hierarchy. This is a consequence of the fact that $Z$ rotations by $\frac{2\pi}{2^k}$ and their controlled variants are all in the $(k+1)$th level of the two-qubit Clifford hierarchy as well as of closure properties that every level has for tensor product and composition of diagonal gates.

Background

Let's denote the $k$th level of the $n$-qubit Clifford hierarchy as $\mathcal{C}_n^{(k)}$. Recall the recursive definition: $\mathcal{C}_n^{(1)}$ is the $n$-qubit Pauli group and $$ \mathcal{C}_n^{(k)}=\{U\in U(2^n)\,\,|\,\, U\mathcal{C}_n^{(1)}U^\dagger\subset\mathcal{C}_n^{(k-1)}\}\tag1 $$ for $k>1$. The $n$-qubit Clifford hierarchy is then $\mathcal{CH}_n:=\bigcup_{k=1}^{\infty}\mathcal{C}_n^{(k)}$. It is well-known that $Z\in\mathcal{C}_1^{(1)}$, $S\in\mathcal{C}_1^{(2)}$ and $T\in\mathcal{C}_1^{(3)}$. This fact can be generalized by defining $$ R_k:=\begin{bmatrix}1&\\&e^{2\pi i/2^k}\end{bmatrix}\tag2 $$ for which it turns out that $R_k\in\mathcal{C}_1^{(k)}$. Similarly, the fact that $CZ\in\mathcal{C}_2^{(2)}$ and $CS\in\mathcal{C}_2^{(3)}$ can be generalized by defining the controlled version of $R_k$ $$ CR_k:=\begin{bmatrix}1&&&\\&1&&\\&&1&\\&&&e^{2\pi i/2^k}\end{bmatrix}\tag3 $$ for which it turns out that $CR_k\in\mathcal{C}_2^{(k+1)}$.

Proof

Now, suppose that $U$ is a $4\times 4$ diagonal unitary whose non-zero entries $2^k$th roots of unity. Since global phase is irrelevant, we can write $U$ as $$ U=\begin{bmatrix}1&&&\\&e^{2\pi ia/2^k}&&\\&&e^{2\pi ib/2^k}&\\&&&e^{2\pi ic/2^k}\end{bmatrix}\tag4 $$ for some $a,b,c\in\mathbb{Z}_{2^k}$, but then $$ U=(R_k^b\otimes R_k^a)\circ CR_{k}^{c-a-b}\tag5 $$ where addition in $c-a-b$ is modulo $2^k$. An argument by induction shows that if $V,W\in\mathcal{C}_1^{(k)}$ then $V\otimes W\in\mathcal{C}_2^{(k)}$. Moreover, even though $\mathcal{C}_n^{(k)}$ is not closed under composition $\circ$, the set of diagonal gates in $\mathcal{C}_n^{(k)}$ is a group. Therefore, $(5)$ implies that $U\in\mathcal{C}_2^{(k+1)}\subset\mathcal{CH}_2$.$\square$

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  • $\begingroup$ Is it true in general that for the $ r $ qubit Clifford hierarchy all diagonal gates defined over $ \zeta_{2^k} $ are in the $ k+r-1 $ level of the hierarchy? $\endgroup$ Commented May 5, 2023 at 0:58

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