Why focus on T gates and not some other single qubit rotation R making Clifford + R universal?

Background: In many error correction codes in particular the surface code, the Clifford operations generated by the S,H and CNOT are transversal for quantum computation (meaning that these logical gates can be implemented in easily with very low circuit depth).

However, the Eastin-Knill theorem tells us that one can never get a universal gate set transversally. Whenever the Clifford gates are transversal, such as for the important example of the surface code, the gate the is considered to make the gate set universal is usually the T gate

$$T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i \frac{\pi}{4}} \end{pmatrix},$$ which is a single qubit rotation. The $$T$$ gates are then usually implemented using T-states and T-state distillation

QuestionThe T-gate and counting T states is so common that we have the Clifford+T paradigm. But why the focus on $$T$$ gates? Most rotation gates would complete the gate set to a universal set. Is it because it is "furthest" away from Clifford?

• I think a decent amount of applications in chemistry (including phase estimation) require you rotate by small Rz angles. I don't think there is a way to doing these rotations transverally so I think they're typically approximated by strings of {H, S, T} gates. Commented Jun 19 at 14:00
• Yes, but if you are building a quantum computer for chemistry applications why not do the distillation of these small Rz angles instead of making them out of T-gates? Commented Jun 19 at 14:20

The $$n$$-th level of the Clifford hierarchy, $$C_n$$, is defined recursively as those gates which preserve $$C_{n-1}$$ under conjugation. That is, $$g \in C_n$$, if for all $$h \in C_{n-1}$$, $$ghg^{-1} \in C_{n-1}$$ with $$C_1$$ the Pauli group.

Gates in $$C_n$$ also have the property that they can be teleported up to an operation in $$C_{n-1}$$. In this sense, a gate in $$C_3$$ is the most "simple:" It can be teleported with a Clifford correction applied.

Unfortunately, $$C_n$$ with $$n \ge 3$$ does not form a group. However, the set of diagonal gates in $$C_n$$ does form a group. Such gates have now been fully characterized. As far as I am aware, this means that $$T$$ is essentially the only single qubit diagonal unitary in $$C_3 \setminus C_2$$.

In this sense $$T$$ is relatively special: It is a diagonal $$C_3$$ gate.

That said, we are not required to make our "easy" operations the Clifford group. For example the binary icosahedral group $$2I$$ and the $$\tau_{60}$$ gate

• The authors of the last reference are quite active on this stack exchange, so any questions would likely be answered Commented Jun 20 at 4:38

The reason I focus on T states is because I want to do billions of Toffoli gates, and you can get Toffoli states of sufficient fidelity by performing one round of 15-to-1 T state distillation followed by 8T-to-1TOF distillation. I'm not aware of any cheaper way to get Toffoli states with that kind of error rate. If someone can do that with a different intermediate state, I'd switch immediately.

I'm not an expert on all the ways a magic states can be used to implement a $$T$$ gate on the circuit, but lets just assume we do a teleportation circuit to teleport the $$T$$ onto the state $$|\psi\rangle$$. The result of the teleportation would be $$TX^xZ^x |\psi \rangle$$ where $$x,z$$ are measurement outcomes of the teleportation. We can manipulate this to $$TX^xZ^xT^\dagger T |\psi \rangle = X^x S^x Z^z T| \psi \rangle$$ (up to a phase or possible extra $$Z$$).

This makes us happy as $$X,Z,S$$ are all Clifford, so we can implement their logical operations and obtain $$T|\psi \rangle$$. This went well for us as $$T X^x Z^z T^\dagger \in \text{Clifford}$$ because $$T$$ is in the second level of the Clifford hierarchy.

This is probably the nice property of $$T$$ we wanted to choose when picking an additional gate.