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The Bernstein-Vazirani algorithm identifies the coefficient string of a linear Boolean function with one Boolean quantum oracle call. It contrasts with the $n$ calls required by the classical query complexity of linear Boolean function identification. Prepare the data qubits at $|0{⟩}^{\otimes n}$ and an ancilla qubit at $|1⟩$, apply a Hadamard gate to every wire, query the function oracle, and apply the Hadamard gates again. The full output is $|a⟩|1⟩$.

The mechanism is Bernstein-Vazirani phase kickback. The Pauli X gate has eigenvector $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$ with eigenvalue $-1$. Thus the target flips become relative quantum phases on the data, givingThe final Hadamard transform maps this product to $|a⟩$, so computational-basis measurement recovers every coefficient with certainty. The ancilla preparation for the Bernstein-Vazirani algorithm ensures the required minus sign, and the Hadamard-conjugated linear oracle gives an exact quantum circuit interpretation. One query is optimal for $n\ge 1$, since without a query the output cannot depend on the unknown string.

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.

Use the Bernstein-Vazirani algorithm: initialize n qubits in $|0^n⟩$, apply $H^{\otimes n}$, query the supplied phase oracle once, apply $H^{\otimes n}$ again, and measure all n computational bits. Immediately after the query the state iswhere the dot product in the exponent is taken modulo two. The Walsh-Hadamard transform gives amplitudeEvery factor is one if $y_j=a_j$ and zero otherwise. Thus the final state is exactly $|a⟩$, and the measurement outputs a with certainty after one oracle query. This is the character cancellation behind Bernstein-Vazirani phase kickback; because access here is already to the reduced phase oracle, no extra target ancilla or conversion from a bit-flip oracle is required.

The Walsh-Hadamard transform is the tensor product of one Hadamard gate on each qubit. Indeed,The network therefore consists of $n$ parallel Hadamard gates, with no interaction between wires. Applied to an all-zero quantum register, it produces the uniform quantum superposition of all $2^n$ input strings. This lets a Boolean quantum oracle act coherently on all those inputs; a subsequent operation can turn their relative phases into useful interference. It does not permit reading all oracle values from a single measurement in quantum mechanics.

For the supplied linear-phase input, the amplitude on output $y$ isEach factor is obtained by summing over the independent bit $x_j$. This directly proves the relevant orthogonality of the binary Fourier phases, and yields the output state $|a⟩$ exactly.

For exact classical identification, query the Boolean function at the $n$ standard unit vectors: $f(e_j)=a_j$. Hence $n$ oracle calls suffice. They are also necessary in the worst case, even with adaptive queries: every answer supplies one binary linear equation in the unknown string. With fewer than $n$ queries, their coefficient matrix has matrix rank at most the number of queries, so its null space contains a nonzero string $h$. The strings $a$ and $a\oplus h$ give identical answers along that adaptive transcript and cannot both be identified correctly. Equivalently, a depth-$t$ binary decision tree has at most $2^t$ leaves and must distinguish $2^n$ possibilities. Thus the exact classical query count is $n$. This concerns exact recovery, rather than allowing an arbitrary error probability.

The quantum network in the lower panel starts its data quantum register in $|0{⟩}^{\otimes n}$ and its answer ancilla qubit in $|-⟩$, which can be prepared as $H|1⟩$. Apply Hadamard gates to the data wires, make one Boolean quantum oracle call, then apply Hadamard gates again to all data wires. Quantum phase kickback producesA computational-basis measurement of the data wires gives all bits of $a$ with certainty. This is Bernstein-Vazirani phase kickback: one quantum query, $2n$ data Hadamard gates, and one answer-state preparation suffice. The answer ancilla qubit stays separate from the data at the output.

Let $\omega =e^{2\pi i/3}$. The positive-exponent quantum Fourier transform gives $|\xi ⟩=3^{-1/2}{\sum }_{y=0}^2{\omega }^{2y}|y⟩$. Reindexing the cyclic shift operator givesThus $|\xi ⟩$ is an eigenstate with eigenvalue $\omega$, with this sign fixed by the printed quantum Fourier transform convention.

Prepare the first two qutrits in $({\mathrm{QFT}}_3|0⟩{)}^{\otimes 2}$ and the answer qutrit in $|\xi ⟩$. The modular-addition quantum oracle applies $S^{f(x_1,x_2)}$ to the answer. Its eigenvalue produces quantum phase kickback, leavingApply ${\mathrm{QFT}}_3^{-1}$ to each input qutrit, then measure them in the computational basis. The result isThe preparation, inverse transforms and measurements are independent of $f$, and there is exactly one use of $U_f$. This is qutrit linear-function identification, the ternary version of Bernstein-Vazirani phase kickback.

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 $a=0$.

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.

Let $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. After the Hadamards and the oracle, Bernstein-Vazirani phase kickback gives