Past exam of the computer science course of the University of Oxford / 2020 / Quantum information / 1 / e

Directly applying the Hadamard gate giveswhileHence the two states are interchanged up to the displayed global phases. Their relative factors $-i$ and $+i$ are retained. They are the two Pauli eigenstates of the Pauli Y gate, and the Hadamard interchange of Y eigenstates is consistent with $HYH=-Y$.

Compute the parity into the second wire, phase that wire with the T gate, then uncompute with the same controlled-NOT gate. The answer isFor computational bits $x,y$, the successive states are $|x,y⟩\mapsto |x,x\oplus y⟩\mapsto e^{i\pi (x\oplus y)/4}|x,x\oplus y⟩\mapsto e^{i\pi (x\oplus y)/4}|x,y⟩$. The parity $x\oplus y$ is one precisely on the two odd-parity inputs. Thus, in the ordered computational basis $00,01,10,11$, the matrix is exactly $\mathrm{diag}(1,e^{i\pi /4},e^{i\pi /4},1)$, with no omitted overall phase. This is a two-qubit parity phase gate.

Let $R=\mathrm{diag}(1,e^{i\pi /12})$. Since the Hadamard gate maps the standard basis to $|+⟩,|-⟩$, the given gate is $Q=HRH$. Because $H^2=I$,Next, use $HZH=X$ on the target wire of the Controlled-Z gate:The control-zero block remains the identity, and the control-one block becomes $HZH=X$, proving the identity with the stated control convention. We have therefore constructed each gate in the assumed universal quantum gate set $\{\mathrm{CNOT},H,T\}$ using only $CZ,Q,H$.

Replacing every gate of a circuit by its displayed finite construction reproduces that circuit exactly. Thus $\{CZ,Q,H\}$ is universal under the usual arbitrary-accuracy convention: it can perform every approximation obtainable from the given universal set. The claim concerns approximation of arbitrary unitary operators; it does not assert that finitely many circuits exactly represent an uncountable continuum of gates. This is universality transfer by exact gate synthesis.