Qutrit metaplectic gates are a subset of Clifford+T

TL;DR

This paper demonstrates that the set of Clifford+R gates for qutrits is a strict subset of Clifford+T gates, establishing the T gate's superior universality and showing R cannot replace T in single-qutrit circuits.

Contribution

It proves that Clifford+T unitaries encompass Clifford+R for multiple qutrits and that R cannot be exactly synthesized by Clifford+T, highlighting T's greater power.

Findings

Clifford+R is a strict subset of Clifford+T for multiple qutrits.

The R gate cannot be exactly synthesized using Clifford+T.

A direct decomposition of R⊗I into Clifford+T circuits is provided.

Abstract

A popular universal gate set for quantum computing with qubits is Clifford+T, as this can be readily implemented on many fault-tolerant architectures. For qutrits, there is an equivalent T gate, that, like its qubit analogue, makes Clifford+T approximately universal, is injectable by a magic state, and supports magic state distillation. However, it was claimed that a better gate set for qutrits might be Clifford+R, where R=diag(1,1,-1) is the metaplectic gate, as certain protocols and gates could more easily be implemented using the R gate than the T gate. In this paper we show that when we have at least two qutrits, the qutrit Clifford+R unitaries form a strict subset of the Clifford+T unitaries, by finding a direct decomposition of $R \otimes \mathbb{I}$ as a Clifford+T circuit and proving that the T gate cannot be exactly synthesized in Clifford+R. This shows that in fact the T gate is at least as powerful as the R gate, up to a constant factor. Moreover, we additionally show that it is impossible to find a single-qutrit Clifford+T decomposition of the R gate, making our result tight.