Past exam of the computer science course of the University of Oxford / 2019 / Quantum information / 3 / b / ii / Solution

Squaring the diagonal gate givesThus the T gate is available up to a global phase. Reading the supplied W columns in order $00,01,10,11$, it sends $|01⟩$ to $|11⟩$, $|11⟩$ to $|01⟩$, and fixes the other two vectors: it is CNOT with the second wire controlling the first. Conjugating both wires by the Hadamard gate reverses that orientation:For instance, write $W=I\otimes |0⟩⟨0|+X\otimes |1⟩⟨1|$. Conjugation uses $HXH=Z$ and $H|0⟩=|+⟩$, $H|1⟩=|-⟩$, yielding $I\otimes |+⟩⟨+|+Z\otimes |-⟩⟨-|$. Expanding the first-wire Z eigenspaces gives $|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes X$, as required.

We have H directly and finite constructions of CNOT and T up to physically irrelevant global phase. By universality transfer by exact gate synthesis, the assumed universality of $\{CNOT,H,T\}$ proves that $\{V,H,W\}$ is universal under the phase-insensitive convention. There is no assertion of arbitrary strict operator-norm phases: on a fixed register these generators have only a finite set of possible determinants, which would obstruct density in the entire unitary group if global phase were treated as an observable target.