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

## 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.

• Self-dual CSS codes do not necessarily implement all logical Clifford gates transversally, they are only guaranteed to have transversal Paulis (because stabilizer) transversal CNOT (because CSS) and transversal $H$ (because self-dual). You need to add the assumption that the Self-dual CSS code is also doubly-even to guarantee transversal phase gate and thus all of transversal Clifford. See quantumcomputing.stackexchange.com/questions/26701/… for an example of a $[[25,1,5]]$ self-dual CSS code that is not doubly even. Mar 4 at 13:41

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.)
• Thanks for mentioning this paper! Indeed, I also came up with the same "cheating" idea. However, I realized the gate $V$ is generally not simultaneously transversal. In another word, it just transferred the non-transversality to another part and I am not really satisfied with this solution. Dec 6, 2023 at 9:18