For arbitrary single-qubit unitary gate, does there exist a code such that the given gate is transversal?

Question

Background

Transversal gates are considered to be fault-tolerant logical operation as they won't lead to the propagation of local errors. Here are some examples:

1. CSS code has transversal Pauli and CNOTs gates. Particularly, self-dual CSS code has transversal Clifford gates.

2. $[[5, 1, 3]]$ has transversal Pauli and Phase-Hadamard $SH$ gate.

Motivation

I am recently interested in QEC codes with non-Clifford transversal gates. Particularly, I find the family of quantum Reed-Muller code that allows the transversal Pauli rotation $Z(\frac{\pi}{2^n})$ gate. Although the allowed rotation angle is discrete, I may argue to myself that it's approximately continuous and for any $Z(\theta)$ gate, we can find a code which allows it to be transversal gate.

As Pauli bases are symmetric, I believe this intuition also holds for $X(\theta)$ and $Y(\theta)$.

Problem

Pauli rotation gates are very special and I wanna generalize this intuition to arbitrary single-qubit unitary $U(\theta)$: For arbitrary single-qubit unitary gate, does there exist a code such that the given gate is transversal?

Thought by far

A common practice is to decompose $U(\theta)$ into

$$ U(\theta) = X(\beta_0)Z(\beta_1)X(\beta_2). $$

Maybe we should find out a code both allows small-angle transversal X and Z rotations? But I highly doubt the presence of such code.

Answer 1

You might be interested in the following paper, which is part of a sequence that helps to classify the possible forms of transversal gates: https://arxiv.org/abs/1409.8320 (up to some local equivalence). Pretty much, it's the type of diagonal gates you've mentioned + Clifford transformations.

On the other hand, I might cheat a little bit (OK, quite a lot!). Imagine you want to implement a $U=VZ(\theta)V^\dagger$ where $Z(\theta)$ is one of the cases for which you already have a code that implements it. Let $|\tilde 0\rangle_L$ and $|\tilde 1\rangle_L$ be the logical states of this code. Now, note that the choice of basis within a subspace is arbitrary. So, I can just happen to choose to define my logicals instead as $|0\rangle_L=V^\dagger|\tilde 0\rangle_L$. Now if I apply what was $Z(\theta)$, its effective action on my new basis is just $U$. So you're done! (The cheat here is that you no longer have nice descriptions of logical X and Z, which you probably, implicitly, want to keep.)