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For a Boolean quantum oracle whose function is $g(x,y)=a_y\cdot x$, place its target in $|-⟩$. Conjugating the oracle by a Walsh-Hadamard transform on the $x$ register translates that register by $a_y$. This is Bernstein-Vazirani phase kickback without measuring its output, so the index $y$ may remain in an arbitrary quantum superposition. A second application undoes the translation. A phase evaluation between the two calls therefore transfers a hidden-label-dependent phase to the index register while uncomputation returns the decoding register to zero.

The linear decoding must be done coherently, so its output is available inside the next Boolean quantum oracle call. Use registers $Z,X$ of $n$ qubits and a shared phase ancilla qubit in $|-⟩$. Define the Bernstein-Vazirani decoding controlled by a quantum registerThe ancilla qubit is implicit. For every computational index $x$, the Walsh-Hadamard transform calculation givesIn particular $W_g^2=I$ on these states. Initialize $Z$ to $|0⟩$ and $X$ to $H^{\otimes n}|0⟩$. The three query stages areThe middle equality uses the matching index $z=a_x$, not a classical guess of that string. The second $W_g$ performs uncomputation, removing the hidden-string register without losing its phase on $X$. Apply $H^{\otimes n}$ to $X$ to obtainThis requires precisely two queries to $U_g$ and one to $U_f$, and only $O(n)$ additional fixed quantum gates. No measurement of $a_x$ is made: such a measurement would spoil the required coherence. Preparing all registers and the phase ancilla qubit uses only the initially available zero states.

Prepare an additional qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$ by applying $X$ and then the Hadamard gate to $|0⟩$. Apply $H^{\otimes n}$ to the data register. The Boolean quantum oracle then produces quantum phase kickback:A final Walsh-Hadamard transform gives an amplitude for $z$ equal toTo see this identity, the sum factors over the bits; any position at which $a$ and $z$ differ contributes $1-1=0$. This is Bernstein-Vazirani phase kickback, and its output isThere is exactly one oracle query, $O(n)$ fixed quantum gates, and no probabilistic intermediate step. The ancilla qubit can be left in $|A⟩=|-⟩$ or reset to $|0⟩$ using its known inverse preparation. The construction includes the case $a=0$.

Assume first $0<k<N$. Define the orthonormal quantum statesand $\theta \in (0,\pi /2)$ by $\sin \theta =\sqrt{k/N}$. The relevant invariant subspace is the two-dimensional span of these quantum states, not the entire Hilbert space. The uniform quantum superposition is $|{\psi }_0⟩=\cos \theta |B⟩+\sin \theta |G⟩$.

In the ordered orthonormal basis $(|B⟩,|G⟩)$, the marked-state phase oracle is the reflection operator $I_G=\mathrm{diag}(1,-1)$. Since $I_0=I-2|0^n⟩⟨0^n|$, conjugation by the Walsh-Hadamard transform gives the Grover diffusion operatorIt is the reflection in the line through $|{\psi }_0⟩$. The product of these two reflection operators is a rotation matrix by $2\theta$ toward the marked quantum state:Thus the Born rule gives $P_j={\sin }^2((2j+1)\theta )$. Choose a nonnegative integer $j_{\ast }$ nearest to $\pi /(4\theta )-1/2$. Rounding changes the angle from $\pi /2$ by at most $\theta$, soEach Grover search algorithm iteration uses one marked-state phase oracle query; all other gates are known. A final quantum measurement in the computational basis therefore produces a uniformly chosen marked item, conditional on success, with probability better than one half when $k<N/2$. This uses the supplied marked count $k$ to select the stopping time; an unknown $k$ needs an additional search schedule or counting procedure. If $k=N$, no oracle query is needed; if $k=0$, no marked item exists, so a successful search is impossible.

If $f(b_1{b}_2)=b_1\oplus b_2$, thenso

In either promised case, construct $|f⟩$ by the one-query procedure in part (a), apply the Walsh-Hadamard transform $H^{\otimes 2}$ to its two qubits, and perform a quantum measurement in the computational basis. The outcome is certainly $00$ for case (i) and certainly $11$ for case (ii). ThereforeThis is a two-bit instance of Bernstein-Vazirani phase kickback, and also a promised Deutsch-Jozsa test.

Starting from $|0^n⟩|0^n⟩$, the first layer of Hadamard gates givesThe Boolean quantum oracle then givesUsing the Walsh-Hadamard transform identitythe final state is