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Let $U_a|x⟩|b⟩=|x⟩|b\oplus (a\cdot x)⟩$ be the Boolean quantum oracle for a linear Boolean function, and put $W=H^{\otimes (n+1)}$. The CNOT gates $C_{j\to y}^{a_j}$ implement $U_a$, where the data wires control the ancilla qubit. CNOT control reversal in the Hadamard basis and $W^2=I$ imply the exact unitary operator identityHere $\oplus$ on strings is bitwise exclusive or. To conjugate a product, insert $W^2=I$ between its factors; each factor reverses its control and target, and the unused Hadamard gates cancel. The global phase is exactly $1$. In particular, the identity holds on arbitrary quantum superpositions, not only on the initialized inputs of the Bernstein-Vazirani algorithm.

Reverse communication is also possible. Prepare Alice in $|+{⟩}_A$. Bob encodes $m=0$ as $|+{⟩}_B$ and $m=1$ as $|-{⟩}_B$. Since $X|\pm ⟩=\pm |\pm ⟩$, phase kickback giveswhere $|(-{)}^0⟩=|+⟩$ and $|(-{)}^1⟩=|-⟩$. Alice measures in the Hadamard basis: $+$ means zero and $-$ means one, deterministically.

The target's Pauli X gate eigenvalue becomes a relative phase on the control. This is the operational content of CNOT control reversal in the Hadamard basis; the computational basis control arrow alone does not determine every possible direction of communication.

The required identity is CNOT control reversal in the Hadamard basis. To derive it using standard quantum circuit identities, let $\mathrm{CZ}$ be the Controlled-Z gate, which multiplies $|11⟩$ by $-1$ and fixes the other computational-basis states. Since $HXH=Z$,The Controlled-Z gate is symmetric in its two wires. Combining this symmetry with $H^2=I$ givesThis is an exact operator identity, including its global phase, and applies to arbitrary inputs. One can also check it directly from the orthogonal projectors $P_b=|b⟩⟨b|$: conjugating $C_{1\to 2}=P_0\otimes I+P_1\otimes X$ by $H\otimes H$ gives

For the three-wire quantum circuit, put $W=H^{\otimes 3}$. Conjugation distributes over products because $W^2=I$: $WABW=(WAW)(WBW)$. Each CNOT gate in $U_a$ therefore reverses direction, while the Hadamard gates on untouched wires cancel. The Hadamard-conjugated linear oracle isThese are the simplified quantum circuits in the figure: for $a=00$ there are no gates; for $a=01$ there is $C_{y\to 2}$; for $a=10$ there is $C_{y\to 1}$; and for $a=11$ there are both. The ancilla qubit becomes their common control. On every computational-basis state, the entire simplified circuit acts asStarting the data qubits at $|00⟩$, the prescribed computational basis preparation of the ancilla qubit must therefore be $|1⟩$: it activates the required flips and gives $|a⟩|1⟩$. Starting it at $|0⟩$ instead gives $|00⟩|0⟩$ for every $a$, so no information is obtained.

More generally, an initial ancilla qubit state $\alpha |0⟩+\beta |1⟩$ in this same fixed quantum circuit givesFor nonzero $a$, the desired data output has probability $|\beta {|}^2$, and exact identification for every $a$ requires $\alpha =0$. This condition refers to the displayed ancilla preparation for the Bernstein-Vazirani algorithm: the state entering the function oracle must be $|-⟩$, up to global phase. Other quantum state preparation procedures that deliver this same state are equivalent; $|1⟩$ is the prescribed input before the first Hadamard gate.

For arbitrary $n$, linearity of the linear Boolean function suppliesThis decomposition is for analysis: the Bernstein-Vazirani algorithm calls the unknown function oracle as one supplied operation. The common control starts at $1$, so the simplified quantum circuit takes $|0{⟩}^{\otimes n}|1⟩$ to $|a⟩|1⟩$. Equivalently, phase kickback gives a product statewhich the final Hadamard transform decodes to $|a⟩$. Thus one quantum oracle call suffices for every $n\ge 1$, compared with $n$ classical calls. This counts function oracle uses; the surrounding Hadamard gates and final computational-basis measurement still act on $n+1$ and $n$ qubits, respectively.

In the Hadamard basis, $X_D$ is diagonal with projectors $Q_{\pm }=(I\pm X_D)/2$. Rewrite the same CNOT gate in the original tensor-factor order $S\otimes D$ asExpanding $P_0\otimes I+P_1\otimes X_D$ verifies this identity directly. Now $Z_S|+⟩=|-⟩$ and $Z_S|-⟩=|+⟩$, so $Z_S$ is ${\tilde{X}}_S$ in the complementary basis. If $D$ is plus, the identity acts on $S$; if $D$ is minus, its sign-basis label flips. Hence $D$ is the control and $S$ the target in this basis.

Equivalently,This CNOT control reversal in the Hadamard basis concerns the matrix in changed local bases, rather than a physical exchange of the two carriers.