 Solution

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.