Can every unitary be approximated by gates from the Clifford Hierarchy?

Question

For $k > 1$, we recursively define $\mathcal C^{(k)}(n)$ as $$ \mathcal C^{(k)}(n) = \Bigl\{ U \in \mathbf U(2^n) \mathrel{\Big\vert} \forall P \in \mathcal C^{(1)}(n) : U P U^\dagger \in \mathcal C^{(k-1)}(n) \Bigr\} $$ That is, the set of unitaries that, when used to conjugate a Pauli operator, yield an operator at one level lower in the hierarchy. $\mathcal C^{(1)}(n) $ is the Pauli group. Define $$ \mathcal C^{(\infty)}(n):= \bigcup_{k=1}^\infty \mathcal C^{(k)}(n) $$ Can every unitary in $ \mathbf U(2^n) $ be approximated by elements of $ \mathcal C^{(\infty)}(n) $? In other words, is the set $ \mathcal C^{(\infty)}(n) $ dense in $ \mathbf U(2^n) $? If not, what is the closure of the set $ \mathcal C^{(\infty)}(n) $ in $ \mathbf U(2^n) $.

To be clear, I mean "closure" and "closed" in a geometric/topological sense. (The totally different algebraic question of closure under products is addressed here Is there a closure property for the entire Clifford hierarchy? )

Thoughts so far: Using this definition it is clear that all $\mathcal C^{(k)}(n)$ contain the multiplies of the identity $ e^{i \theta} I $. So the closure is at least one dimensional containing the scalar matrices $ e^{i \theta} I $.

Specializing to the case $ n=1 $ for ease, it seems that the diagonal matrix $ \begin{bmatrix} \zeta_{2^{k+1}} & 0 \\ 0 & \overline{\zeta_{2^{k+1}}} \end{bmatrix} $ is in $\mathcal C^{(k)}(n)$. So the closure in $ \mathbf U(2) $ of $ \mathcal C^{(\infty)}(n) $ should include all matrices of the form $ \begin{bmatrix} e^{i \alpha} & 0 \\ 0 & e^{-i \alpha} \end{bmatrix} $.

Together with the previous claim, that implies all diagonal matrices in $ \mathbf U(2) $ should be in the closure of $ \mathcal C^{(\infty)}(n) $.

Edit: I added the notation $ (n) $ to emphasize that the Clifford hierarchy is defined for a fixed number of qubits $ n $.

Answer 1

This is too long for a comment but not an answer.

First we must define some notion of closeness of unitaries. Other metrics should be fine, but operator norm seems like a good one. Now I think your question becomes for all $U \in U(2^n)$ and $\epsilon$ does there exist a $V \in \mathcal{C}^{(\infty)}$ such that $||U-V||_\infty < \epsilon$.

From Cui et al. Diagonal matrices in $U(2^n)$ with $2^k$ root-of-unity entries are all in $\mathcal{C}^{(\infty)}$ at some level. This should get you `close' to any diagonal gate in $U(2^n)$.

I'll try to give a sketchy proof that $U$ exist which are not close to any element in $\mathcal{C}^{(\infty)}$. We know that all $U(2)$ gates in the Clifford Hierarchy are semi-Clifford and can therefore be written as $C e^{i\theta P}$. $C$ is a one-qubit Clifford gate and $P$ is a non-trivial Pauli gate. $\theta$ is restricted, but even if we allow arbitrary $\theta$ each $V_{C,P}(\theta)=C e^{i\theta P}$ is parameterized by one real variable, $\theta$. There are 24 elements (up to phase) in the one-qubit Clifford group. So there are $3\times 24$ different $V_{C,P}$. (Note I think there are actually $3\times 6$ since these aren't all independent, but that doesn't matter for the following argument.) Each $V_{C,P}$ is a rotation about some axis and the set of unitaries $V_{C,P}(\theta)$ is a circle on the Bloch sphere (or in the space of $U(2)$ matrices). Since we have only a finite (at most 72) number of these circles on the sphere there must be some 'gaps' where for some $U$, $||U-V||_\infty > \epsilon$ for all $V$.