Given a non-Clifford quantum circuit $U$, is it possible to construct a commuting Clifford circuit $C$?

Question

Let a non-Clifford circuit $U$, say $$U = \prod_{i=1}^k e^{i \theta_i P_i} $$ for $P_i \in \{ I,X,Y,Z\}^{\otimes n} $ with $\theta_i \in \mathbb{R}$ be given. Is it possible to construct a non-trivial Clifford circuit $C$ with commutes with $U$?

Answer 1

Generically, no.

Consider the single-qubit unitary $U=HT$. We want to find a $C$ such that $$ CUC^\dagger=U $$ but since $C$ is Clifford, it transforms Paulis into Paulis. So, decompose $U$ in terms of Paulis. If you work though the maths, you'll find that you need $CYC^\dagger=Y$. It wouldn't matter if $CXC^\dagger=X$ or $Z$ so long as $CZC^\dagger$ is the other. Except that $CXC^\dagger=Z$ and $CZC^\dagger=X$ is unphysical in the way that the transpose operation is unphysical (it's the same as Hadamard followed by transpose). That means that the only solution is the trivial one.

You can extend this argument to the generic case. Any $U$ has a Pauli decomposition $$ \sum_{x\in\{0,1,2,3\}^n}\alpha_x\sigma_x. $$ If you require $CUC^\dagger=U$ and $C$ Clifford, then for any unique values of $\alpha_x$, $C\sigma_x C^\dagger=\sigma_x$. In the generic case, they're all different, and so the only possible solution is that $C$ is the identity operation.