Past exam of the laser science and quantum information processing course of the University of Oxford / 2006 / 5 (DPHY 2752, DPHP 2752)

Source extract: users.physics.ox.ac.uk/~Foot/C2/pdf/C22006Q5.pdf

The year is attributed from the preserved Oxford support-page record; it is not printed on the held page.

TODO: The remainder of this historical paper is not held. Only the supplied question is covered.

For one input bit, a Boolean function is specified by the pair $(f(0),f(1))$, and $k=f(0)+f(1)$ counts its satisfying assignments. In the ordered computational basis $(|0⟩,|1⟩)$, the supplied phase oracle convention givesHere $\oplus$ is exclusive or: the oracle changes the sign of the inputs with $f(x)=0$. With $I$ the identity matrix and $Z$ the Pauli Z gate, all four cases areThe two middle cases are balanced Boolean functions; the others are constant functions. The specified zero-state sign change and the Hadamard gate are

Use the top qubit as the first tensor product factor. Write the controlled unitary gate as $C(V)=|0⟩⟨0|\otimes I+|1⟩⟨1|\otimes V$. Reading the bracketed quantum circuit from left to right, its four operations are $C(U_f)$, $C(H)$, $C(U_0)$, $C(H)$. Thus the target operation in its control-one branch iswhere the rightmost matrix acts first. The control-zero branch is the identity operator. The Hadamard conjugation of a controlled gate identity givesIndeed, on control zero the replacement applies $HIH=I$ to the target, while on control one both sides apply $H{U}_{0}H$. By linearity, the identity holds on every quantum superposition and therefore in every repetition of the bracket. This replaces the two internal controlled Hadamard gates as a pair.

The global phase of a controlled quantum gate must be preserved: $C(-I)=Z\otimes I$, whereas $C(I)=I\otimes I$. Consequently the minus signs in $U_0=-Z$ and the special phase oracles $U_f=-I,-Z$ cannot be discarded as an unobservable global phase of the complete quantum circuit.

Define the Hadamard basis states $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$. The initial two Hadamard gates prepare $|+⟩|+⟩$. For one bracket, the two control branches giveUsing the Pauli X gate $X=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, unitary conjugation by the Hadamard gate gives $H{U}_{0}H=-X$, so $G=-X{U}_f$. Keeping every sign, the target states along the control-one branch, in the exact gate order, areThe last column also includes the final target Hadamard gate, which lies outside the bracket. Writing $|{\chi }_f⟩=HG|+⟩$, applying the final top Hadamard gate gives the exact normalized two-qubit quantum stateThus the complete outputs, in top-then-bottom computational basis order, areFor the first qubit, the Pauli matrix ${\sigma }_z=Z$ has expectation value $P(0)-P(1)$ by the Born rule. The two $k=1$ cases each have $P(0)=P(1)=1/2$, while the $k=0$ and $k=2$ cases are deterministic. Equivalently, this Hadamard test measuresThis explains the approximate quantum counting interpretation: repeated preparations estimate the expectation value, and hence $k$. A single top-qubit computational-basis measurement need not identify $k$.

For $r$ brackets the control-zero branch still does nothing, and the control-one branch applies $G^r$ to the target. Immediately before the final two Hadamard gates, the quantum state isFor the constant function with $k=0$, the phase oracle is $U_f=-I$, so $G=(-X)(-I)=X$. Since $|+⟩$ is an eigenstate of the Pauli X gate with eigenvalue $+1$, $G^r|+⟩=|+⟩$ for every integer $r$. The final two Hadamard gates therefore giveIn particular, the first qubit always ends in $|0⟩$.

For the constant function with $k=2$, $U_f=I$ and $G=-X$, so $G^r|+⟩=(-1{)}^r|+⟩$. Thus the target remains in $|+⟩$, and the top qubit before its final Hadamard gate is $(|0⟩+(-1{)}^r|1⟩)/\sqrt{2}$. The sign is a relative quantum phase between the control branches. The Hadamard gate converts it into the integer parity of $r$:The first qubit is therefore $|0⟩$ for even $r$ and $|1⟩$ for odd $r$, including all requested $r>1$.