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Let $\lambda =e^{2\pi i\varphi }$. In terms of the Pauli X gate, the given unitary operator is $U_{\varphi }=aI+bX$, with $a=(1+\lambda )/2$ and $b=(1-\lambda )/2$. Since $X|+⟩=|+⟩$ and $X|-⟩=-|-⟩$,Thus the normalized eigenvectors are $|\pm ⟩=(|0⟩\pm |1⟩)/\sqrt{2}$, with eigenvalues $1$ and $e^{2\pi i\varphi }$ respectively. At $\varphi =0$ they coincide as eigenvalues, and the whole two-dimensional Hilbert space is the eigenspace; the same chosen eigenbasis remains valid.

Prepare $|-⟩$ independently of $\varphi$ and use the exact quantum phase estimation circuit of the preceding parts on the controlled-$U_{\varphi }$ black box. It returns $\varphi$ exactly usingThe chosen eigenstate contains all the unknown phase information, unlike $|+⟩$, whose eigenvalue is independent of $\varphi$.

Under the usual reversible-circuit interpretation, there is a genuine obstruction to uniform probability lowering by a unitary in the printed premise. The universal assertion cannot hold for arbitrary $p$: it fails already at the search density $p=k/N$.

An explicit counterexample uses $N=4$, one good basis state $|0⟩$, $p=1/4$ and $p^{\prime }=1/8$. The six pure statesall have good probability $1/4$. Their equally weighted density operator is $I_4/4$, a maximally mixed state. Every unitary operator $C$ leaves this mixture unchanged, so its average output good probability remains $1/4$. The printed assertion would instead make all six output probabilities $1/8$, a contradiction. More generally, averaging states $\sqrt{k/N}|g⟩+e^{i\alpha }\sqrt{1-k/N}|b⟩$ uniformly over good and bad basis vectors and two opposite phases gives $I_N/N$, proving the same obstruction at the actual search density $p=k/N$.

The intended conditional construction needs only a supplied reversible preparation that lowers the success probability of the particular starting state, not of every state. Here is the complete argument under that weaker resource assumption. Count $C$ and $C^{†}$ as supplied, query-free operations, as required for the claimed input-oracle bound. Let $\theta =\arcsin \sqrt{k/N}$ and chooseIf equality holds, use the original preparation. Otherwise use the stipulated preparation on $|{\psi }_0⟩$ to obtain $|{\psi }_{\ast }⟩=C|{\psi }_0⟩$ with good probability $p_{\ast }$. Its reflection operator is implementable byThe good and bad components may be nonuniform, but amplitude amplification applies to this same two-dimensional decomposition. After $j$ iterations of $-I_{{\psi }_{\ast }}{I}_f$, its good probability isThus the stated exact-query conclusion follows from accessible preparation of one known-overlap state and its inverse. If the supplied preparation uses oracle queries, those costs cannot be omitted.

A physically realizable known-state success dilution for exact amplitude amplification is available in the standard enlarged marking-oracle model. Append a flag in $\sqrt{1-q}|0⟩+\sqrt{q}|1⟩$, where $q=p_{\ast }/p$, and call a joint state good only when $f(x)=1$ and the flag is one. This prepared state's success probability is exactly $pq=p_{\ast }$. Reflect about this known product preparation and mark the joint good subspace; $j$ ordinary iterations then succeed with certainty. A supplied controlled phase oracle gives one marking query per iteration, or a Boolean bit oracle computes $f$, applies the joint phase and uncomputes $f$ in two queries. Measuring the data therefore returns a good $x$ with certainty in $O(\sqrt{N/k})$ queries. This changes the state space and marking test, and does not assert the impossible universal $n$-qubit circuit. Controlled access is an additional resource in a bare phase-oracle model, so it is made explicit rather than silently assumed.

This is the Deutsch-Jozsa test with an arbitrary uniform-state unitary, and does not require $N$ to be a power of two. Choose a known unitary operator $F$ such that $F|0⟩=|s⟩=N^{-1/2}{\sum }_i|i⟩$. Prepare the target qubit in $|-⟩=(|0⟩-|1⟩)/\sqrt{2}$. One Boolean-oracle call gives quantum phase kickback:Apply $F^{†}$ to the index register. The amplitude on $|0⟩$ isIt equals $+1$ or $-1$ for the two constant strings, and equals zero for a balanced string. Consequently a single query decides the promised problem exactly: measure the index register and report constant for outcome zero, balanced for any other outcome. This is the Deutsch-Jozsa algorithm with the available exact state-preparation operation.

For $n\ge 2$, the Pauli X gate and Pauli Z gate act on different qubits in $X_0{Z}_1$, so they commute. Their product is a Hermitian matrix and squares to the identity operator. Its eigenvalues have modulus one; equivalently it is a unitary operator. Therefore its spectral norm isThe $n\ge 2$ qualification is necessary because the qubit labelled one does not exist for $n=1$.

For a unitary operator $W$, $W{W}^{†}=I$. Thus for every nonnegative integer $j$,Insert this into the convergent matrix exponential series in the finite-dimensional qubit setting:Consequently unitary conjugation commutes with the exponential:For an unbounded self-adjoint Hamiltonian, the same identity follows from the spectral theorem for normal operators, with the domain transported by $W$; no unbounded power-series manipulation is needed.

Use the positive-sign quantum Fourier transformOrthogonality of the finite-group characters makes $F_N$ a unitary operator. Here $r$ is the least positive period. The injectivity within one period implies $r\mid N$: write $N=qr+s$, $0\le s<r$. Periodicity gives $f(s)=f(N)=f(0)$, and injectivity forces $s=0$. Put $L=N/r$.

Start with two available computational basis states $|0⟩|0⟩$. Apply $F_N$ to the first register and query the modular-addition quantum oracle:Measure the second register. For some $t\in \{0,\dots ,r-1\}$, the first register becomes the normalized coset stateThe quantum Fourier transform of a periodic coset state isIndeed, the inner geometric sum vanishes unless the Fourier label is a multiple of $N/r$. Measuring that label gives $j=sN/r$ for a uniformly random $s$. Compute, by the Euclidean algorithm,Thus exact period recovery from a Fourier sample succeeds whenever $s$ is coprime to $r$, andHere $\phi$ is the Euler totient function. The classical number-theory input is the totient lower bound from the Mertens product: for sufficiently large $m$, $\phi (m)/m\ge c/\log \log m$ for an absolute $c>0$, with the finitely many small cases handled separately. For $r=1$ recovery is certain; for $r=2$ its probability is $1/2$. The useful asymptotic probability statement is a lower bound, hence $\Omega$ notation; it can be much larger for particular periods, for example a prime period.

There is one oracle query, two Fourier transforms and two measurements. The final greatest common divisor and division use polynomial time in $\log N$. This meets the specified primitive-operation model without assuming that an arbitrary-$N$ Fourier gate is free in another gate model. If a constant success probability is wanted, repeat independently $O(\log \log N)$ times and return the least common multiple of the candidates: each candidate divides $r$, and one successful sample makes that least common multiple exactly $r$.