Past exam of the computer science course of the University of Oxford / 2020 / Quantum information / 1 / f / iii / Solution

Now $T^2=S=\mathrm{diag}(1,i)$. The gates $H,S$ and CNOT are Clifford gates: conjugating a Pauli matrix or a tensor product of them gives another signed Pauli tensor. On one wire,For CNOT the generator images are $X\otimes I\mapsto X\otimes X$, $I\otimes X\mapsto I\otimes X$, $Z\otimes I\mapsto Z\otimes I$, and $I\otimes Z\mapsto Z\otimes Z$. These identities can be checked on the four computational vectors and generate the preservation property for all Pauli tensors.^[ref]

For a fixed number $n$ of wires there are only finitely many possible images of the $2n$ generators $X_j,Z_j$ in the Pauli group. Their images determine a unitary up to global phase. To see this, if two unitary operators induce identical images, their quotient commutes with every $Z_j$, so is diagonal in the computational basis; commuting with every $X_j$ then forces all diagonal entries to be equal. The quotient is a scalar phase. Thus the generated gates form a finite set modulo global phase, which cannot be dense in the continuous space of all unitary operators. This is finiteness of the projective Clifford group.

A small visible consequence is that from $|0⟩$, one-qubit Clifford circuits reach only the six states $|0⟩,|1⟩,|+⟩,|-⟩,|+i⟩,|-i⟩$ up to global phase. For instance, $T|+⟩$ is outside this finite set. The fixed-$n$ Pauli argument also covers circuits using CNOT on several wires. Therefore $\{\mathrm{CNOT},H,T^2\}$ is not universal.