# Is every diagonal gate whose non-zero entries are $2^k$th roots of unity in the two qubit Clifford hierarchy?

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.

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.
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 $$CR_k\in\mathcal{C}_2^{(k+1)}$$.
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 $$U\in\mathcal{C}_2^{(k+1)}\subset\mathcal{CH}_2$$.$$\square$$
• 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? May 5, 2023 at 0:58