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Suppose a reversible circuit computes the angle bits $b_l(x)$ in ${\theta }_x={\sum }_l{b}_l(x){\alpha }_l$, where ${\alpha }_l$ are known binary place values. Apply a controlled unitary gate $R_y(2{\alpha }_l)$ to the target for each set angle bit. The rotations about the y-axis commute and their angles add, giving $R_y(2{\theta }_x)$. Finally perform uncomputation of the angle and arithmetic workspace. An $L$-bit angle needs $L$ controlled rotations and twice the reversible angle-computation cost. This preserves coherent superpositions of inputs and underlies the HHL controlled reciprocal rotation.

Put $M=2^m$ and prepare $m$ control qubits in $|0{⟩}^{\otimes m}$ and the target in the supplied eigenstate $|\psi ⟩$. Apply a Hadamard gate to every control qubit. Let $k={\sum }_{j=0}^{m-1}{2}^j{k}_j$ denote the integer encoded in that quantum register.

For every $j$, apply a controlled unitary gate $U^{2^j}$ with control $k_j$ and the same target. The quantum phase kickback yieldsApplying ${\mathrm{QFT}}_M^{-1}$ to the controls therefore produces $|x⟩$ exactly. A computational-basis measurement returns $x$ with certainty, and the algorithm outputsThis is exact quantum phase estimation; exactness uses the finite dyadic phase promise, not a rounding argument.

Only a controlled-U gate is supplied as a primitive. Its $2^j$ repetitions implement controlled-$U^{2^j}$, so the total number of black-box uses is ${\sum }_{j=0}^{m-1}{2}^j=2^m-1$. Access to powers as unit-cost primitives would be a different quantum query complexity model. This distinction is the cost of exact phase estimation on a dyadic spectrum.

Use exact quantum phase estimation with $n$ control qubits initially $|0^n⟩$ and one target quantum register prepared in the given eigenvector $|\psi ⟩$. Apply Hadamard gates to the control qubits. If control bit $x_j$ has significance $2^j$, apply a controlled unitary gate $U^{2^j}$ from that qubit to the target. The target stays in $|\psi ⟩$, while quantum phase kickback producesSince $\varphi =c/2^n$, this is $F_{2^n}|c⟩\otimes |\psi ⟩$. The inverse quantum Fourier transform on the control quantum register therefore gives $|c⟩|\psi ⟩$. The quantum measurement in the computational basis returns $c$ with certainty, soThe quantum circuit below reads left to right. The upper wires are control qubits, with $q_0$ least significant; the vertical ellipsis represents the intervening control wires and powers. The inverse quantum Fourier transform acts on the entire control quantum register, including any bit-order permutations.

A supplied controlled-$U$ can implement controlled-$U^{2^j}$ by $2^j$ repetitions with the same control. Hence this exact quantum circuit uses $2^n-1$ calls to the supplied controlled-$U$, unless controlled powers are additionally available cheaply. The cost of exact phase estimation on a dyadic spectrum is not automatically polynomial in $n$; the question asks for an exact algorithm, not that stronger complexity guarantee. Exactness also uses the promised dyadic eigenphase, ideal gates, and the supplied inverse quantum Fourier transform.

Apply $V^{-1}$ to the data and phase quantum registers, leaving the flag quantum ancilla untouched. This inverse needs no extra oracle assumption: the promised dyadic eigenvalues give $U^{2^n}=I$, hence $U^{-1}=U^{2^n-1}$. The inverse of each controlled unitary gate used in $V$ can therefore be built from repeated uses of the supplied controlled-$U$, and the known Hadamard gates and quantum Fourier transform gates can be reversed. Uncomputation is necessary to erase the eigenvalue label coherently. The resulting quantum state isA quantum measurement in the computational basis of the flag followed by postselection on one yieldsThis requires $A|b⟩\ne 0$. For a normalized input and nonnegative eigenvalues, the success probability of positive quantum spectral filtering satisfies $P_{\mathrm{success}}\ge {\lambda }_{\min }^2$. In the general inequality, equality holds precisely when the input is supported on the minimum-eigenvalue eigenspace. Omitting uncomputation and discarding the phase quantum register would instead leave a mixed state with diagonal weights proportional to $|{\beta }_j{|}^2{\lambda }_j^2$, rather than the desired coherent pure state.

The printed universal nonzero-success request needs a nonkernel-input hypothesis. An $n$-qubit Hermitian operator has $2^n$ eigenvalues, counted with multiplicity. All are distinct, and the printed dyadic grid contains exactly $2^n$ possible values. They therefore occupy the entire grid, including zero: this is a multiplicity-free complete dyadic spectrum, and ${\lambda }_{\min }=0$. Taking $n=1$, $A=\mathrm{diag}(0,1/2)$, and $|b⟩=|0⟩$ satisfies every printed spectral promise but gives $A|b⟩=0$. No normalized output vector exists, so no algorithm can deliver it with nonzero probability.

For every input outside the kernel, the procedure above has $P_{\mathrm{success}}>0={\lambda }_{\min }^2$, giving the requested strict bound on the meaningful domain. The corrected result is therefore exact quantum spectral filtering conditional on $A|b⟩\ne 0$, with the boxed probability. This does not require knowing the input probability amplitudes or making extra copies of an unknown quantum state. Arbitrarily small nonzero support outside the kernel gives arbitrarily small success probability, and a failed flag measurement disturbs the input; fresh independent trials cannot be assumed when only one unknown physical input is supplied.

Use a reversible circuit $T$ to compute an $L=O(n)$-bit representation of the angle into a workspace register, with all additional work bits retained:Write that representation as ${\theta }_x={\sum }_{l=1}^L{b}_l(x){\alpha }_l$, with known binary place values ${\alpha }_l$ (including the fixed scale and any integer bit). On the target, apply a controlled unitary gate $R_y(2{\alpha }_l)$ controlled by each angle bit $b_l$. Rotations about the same axis add their angles, so their product isFinally apply $T^{†}$ to perform uncomputation. The workspace returns to zero while the control register and rotated target remain unchanged; the construction works coherently on every superposition of $x$.

A polynomial time classical calculation has a polynomial-size reversible circuit with Toffoli gates, and each Toffoli gate has a constant-size decomposition into one-qubit and two-qubit gates. The $L$ controlled rotations are themselves two-qubit gates. Thus the total quantum circuit size is $\mathrm{poly}(n)$. This is the binary-angle implementation of a quantum variable rotation; the stipulated angle precision is understood.

After the first Hadamard gate and the controlled unitary gate, the state isThe final Hadamard gate givesHence the Hadamard test returns one with probabilityIf $U|\psi ⟩=|\psi ⟩$, then $\theta =0$ and the outcome is zero with certainty. If $U|\psi ⟩=-|\psi ⟩$, then $\theta =\pi$ modulo $2\pi$ and the outcome is one with certainty. The single measurement therefore distinguishes the two promised cases exactly.

First construct a controlled unitary gate from the uncontrolled oracle. Keep the supplied $|v_0⟩$ in a reference register. Controlled on an extra qubit, swap the data and reference registers, query $U$ on the register that contains $|v_0⟩$ in one branch, and swap back. Because $U|v_0⟩=|v_0⟩$, that branch is unchanged, while the other branch acquires $U$ on the data. Reversing the control convention with Pauli X gates gives controlled-$U$. The reference state is returned unchanged, and each use costs one query to $U$ and $O(n)$ controlled-SWAP gates.

Expand $|b⟩={\sum }_j{b}_j|v_j⟩$ in an eigenbasis of $U$. Exact quantum phase estimation with an $m$-qubit phase register producesApply the available phase gates, controlled by the corresponding bits of $c_j$, to multiply branch $j$ byThen reverse phase estimation. Although no $U^{†}$ oracle was supplied, every eigenvalue obeys ${\lambda }_j^{2^m}=1$, so $U^{-q}=U^{2^m-q}$. The inverse controlled powers can therefore be implemented with forward calls to $U$. Since $2^m=\mathrm{poly}(n)$, phase estimation and its inverse use only $\mathrm{poly}(n)$ queries. The phase register returns to $|0^m⟩$, while the data register isAs a direct check, the same spectral promise implies $U^{†}=U^{2^m-1}$.