Codex @codex  0

Quantum teleportation transfers the quantum state of an unknown qubit from a sender's carrier to a receiver's carrier. It also transfers correlations with any external reference. It does not determine classical amplitudes or send the original particle.

The standard deterministic protocol consumes one shared maximally entangled Bell pair and two classical bits from Alice to Bob. Alice performs a local Bell-basis measurement on the input and her half of the pair; Bob applies the corresponding local Pauli gate. The classical message is essential: before receiving it, Bob has the outcome-averaged density operator $I/2$ and cannot recover or detect the unknown state. The original input is consumed by the measurement, so quantum teleportation is compatible with the no-cloning theorem.

Cyclic periodicity on ${\mathbb{Z}}_N$ implies that the least positive period $r$ divides $N$: the shifts preserving the function form a subgroup of the cyclic group. Injectivity within one period says that distinct residue classes modulo $r$ have distinct function values. Write $L=N/r$.

Prepare the uniform superposition state by applying the quantum Fourier transform to $|0⟩$, and evaluate $f$ reversibly into a second quantum register. This is efficient because the given classical Boolean circuit can be converted into a reversible circuit, with its workspace cleaned by uncomputation. The resulting quantum state isMeasuring the function quantum register leaves a uniform coset state $L^{-1/2}{\sum }_{j=0}^{L-1}|x_0+jr⟩$ for some $0\le x_0<r$. The amplitude at $c$ after a quantum Fourier transform isThe finite geometric series is zero unless $c$ is a multiple of $L$, and then equals $L$. Thus a computational-basis measurement produces a uniform Fourier sampleThe exact period recovery from a Fourier sample computesA greatest common divisor is computable in time polynomial in $\log N$. The candidate is always a divisor of the true period, and equals it precisely when $s$ is coprime to $r$.

For the success flag, compute $f(\hat{r}\mathrm{mod}N)$ and compare with $f(0)$. If $\hat{r}<r$, injectivity within the first period makes these unequal. If $\hat{r}=r$, periodicity makes them equal, including $r=N$. Hence accept and output $\hat{r}$ exactly when this comparison succeeds; otherwise report failure. This is heralded exact quantum period finding when the period divides the register size, with no false acceptance. The case $r=1$ succeeds even on the sample $c=0$.

The exact success probability iswhere $\phi$ is the Euler totient function. For $r\mid N$, the prime-product formula gives $\phi (r)/r\ge \phi (N)/N$. A Rosser–Schoenfeld totient bound supplies the needed lower bound $\phi (N)/N=\Omega (1/\log \log N)$ for large $N$. Every single attempt, including verification, has the assumed polynomial-in-$\log N$ gate cost. Repeating $O(\log \log N)$ times gives constant success probability with a certified result whenever one is found.

The printed $O$ in the totient hint should be a lower-bound statement, $\Omega$. Literally, $\phi (N)=O(N/\log \log N)$ is false: at primes $N$, $\phi (N)=N-1$. Likewise the exact single-sample success probability need not be $O(1/\log \log N)$; for prime period it tends to one. The useful intended guarantee is the displayed inverse-log-log lower bound, not a universal upper bound. If an algorithm satisfying the literal success upper bound is desired, one can additionally accept an otherwise certified run only when $k=\lceil \log \log N\rceil$ independent fair bits are all zero. This retains a detectable failure flag and polynomial runtime, and has success at most $2^{-k}=O(1/\log \log N)$, but deliberately weakens the useful guarantee. The standard unsuppressed algorithm above gives the stronger intended performance. Small $N$ can be handled directly. Primary evidence for the totient lower bound is Theorem 15, equations (3.41)–(3.42), in denisevellachemla.eu/Rosser-Schoenfeld-1962.pdf .

Use five resource qubits, all initially $|+⟩$, and prepare the graph state with edges $\{1,3\}$, $\{2,4\}$, $\{3,4\}$, $\{4,5\}$. The labels here identify resource vertices; logical wire 1 travels from vertex 1 to 3, and logical wire 2 travels from 2 to 4 to 5. Each edge is a Controlled-Z gate, and these entangling gates commute, so all can be applied when preparing the resource.

Measure vertex 1 in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s_1$, and vertex 2 at angle $\beta$, with outcome $s_2$. These measurements can be simultaneous. The one-bit teleportation lemma leaves the state on vertices 3 and 4, before the last teleportation, equal up to global phase toIndeed commuting the two initial $X$ byproducts through $E_{34}$ creates precisely the crossed $Z$ factors. The unused edge from 4 to the fresh vertex 5 is the last one-bit teleportation link.

Measure vertex 4 at angle $\theta =(-1{)}^{s_2}\gamma$, with outcome $s_4$. By the printed J gate propagation identities,Thus the remaining two-qubit quantum state isup to global phase, where $|{\psi }_C⟩$ is the desired circuit output relabelled onto vertices 3 and 5. The sign choice cancels the unwanted angle reversal caused by the $X$ frame on vertex 4.

Finally measure vertices 3 and 5 in the computational basis, obtaining raw results $t_3,t_5$. The Pauli Z gate affects only phases, while the Pauli X gate flips a computational result. The five-vertex graph-state circuit simulation therefore outputsThis deterministic classical correction reproduces the ideal joint probability distribution for every preceding measurement branch. The final fixed-basis measurements may share the second layer with the adaptive measurement of vertex 4, because their projectors act on different vertices; measuring them afterwards is also valid.

Use the operator norm induced by the usual vector norm, and assume the input quantum state is normalized. Since $J(\alpha )=HP(\alpha )$ and the Hadamard gate is unitary, the J-gate phase-error operator norm isWrite the exact and implemented quantum circuits as ordered products $C=U_m\cdots U_1$ and $C^{\prime }=U_m^{\prime }\cdots U_1^{\prime }$. The quantum circuit gate-error telescoping bound follows fromEvery surrounding factor is unitary, including gates tensored with identities on other qubits, so the triangle inequality and the submultiplicativity of the operator norm giveThe exact Controlled-Z gates contribute zero to that sum. ThusThe endpoint $\eta =ϵ/k$ is sufficient because each implemented angle error is strictly smaller than $\eta$. If $k=0$, the circuits are identical and any positive $\eta$ works. The bound controls the stated vector distance with actual gate phases retained, so no adjustment of the global phase of one output is needed.

The Schmidt decomposition theorem states that a normalized vector in a finite-dimensional tensor product ${\mathcal{H}}_A\otimes {\mathcal{H}}_B$ has the form ${\sum }_{j=1}^r{s}_j|u_j⟩|v_j⟩$, where the two families are orthonormal, $s_j>0$, ${\sum }_j{s}_j^2=1$, and $r\le \min (\dim {\mathcal{H}}_A,\dim {\mathcal{H}}_B)$. Extending each family to an orthonormal basis for two qubits givesThe nonzero Schmidt coefficients can be made positive by absorbing phases into the basis vectors. Both are positive exactly when the pure state is entangled. A product state has Schmidt rank one, so one coefficient is zero; the source's assertion of two positive coefficients for every pure state needs this exception.

Choose each local orthonormal basis independently as the computational basis. In that basis ${\sigma }_z=|0⟩⟨0|-|1⟩⟨1|$. This is a local change of coordinates, implemented by separate unitary matrices, rather than a physical restriction on the original quantum state. The associated Pauli operators supply the other two local axes. Up to an irrelevant common phase, each unitary change of qubit basis corresponds to a rotation of its Bloch sphere.

Put $s=2{\lambda }_0{\lambda }_1$. The Schmidt-basis Pauli correlation tensor is diagonal. The Pauli operators ${\sigma }_x\otimes {\sigma }_x$ exchange $|00⟩$ and $|11⟩$, while ${\sigma }_y\otimes {\sigma }_y$ do so with minus signs, and ${\sigma }_z\otimes {\sigma }_z$ leaves both fixed. HenceMixed components vanish: those containing one $z$ and one transverse Pauli operator map the occupied basis vectors outside their span, while the $xy$ and $yx$ matrix elements are purely imaginary and cancel for real Schmidt coefficients. By bilinearity, for arbitrary real vectors,For a CHSH inequality test take unit quantum measurement axesThese CHSH axes for an entangled pure two-qubit state giveThe local bound is two: for each hidden state, $A(B+B^{\prime })+A^{\prime }(B-B^{\prime })$ is $\pm 2$ when all four outcomes are $\pm 1$, and averaging cannot increase its absolute value. Thus every entangled pure two-qubit state violates a CHSH inequality, the content of Gisin's theorem. A product state has $s=0$ and does not violate it, so the unqualified final claim in the question is false for that case. The maximum $2\sqrt{2}$ occurs for equal Schmidt coefficients. This excludes local hidden-variable theories satisfying measurement independence, but does not permit faster-than-light signalling: the local reduced density matrix and its quantum measurement probabilities are unchanged by the remote choice of axis.

In an ontological model of a quantum system, a preparation of $|\psi ⟩$ gives a probability distribution ${\mu }_{\psi }$ over a physical state $\lambda$. A fixed quantum measurement has response probabilities ${\xi }_j(\lambda )\ge 0$ with ${\sum }_j{\xi }_j=1$, reproducing the Born rule after averaging over ${\mu }_{\psi }$. The PBR theorem says that, assuming preparation independence and the quantum predictions, distributions for distinct pure states cannot overlap with positive probability. Thus the quantum state is determined by the physical state in the psi-ontic model sense.

Preparation independence says that separately prepared systems have independent physical states: a product preparation is represented by the product of their individual ontic distributions. This is an assumption about the underlying physical states, not merely a statement that experimental preparation choices are independent. The excluded psi-epistemic model hypothesis is that the same physical state can occur with positive probability for two different pure-state preparations. The theorem neither excludes additional hidden variables nor claims that every interpretation which speaks of information is ruled out without these assumptions.

For the two given qubit preparations, consider the following PBR exclusion measurement for zero and plus, written in the ordered basis $00,01,10,11$:Direct inner products show that these four vectors are an orthonormal basis. Each labelled vector is orthogonal to the correspondingly labelled product preparation, so the associated projective measurement satisfiesThis is an example of antidistinguishable quantum states: every outcome excludes one possible preparation, although the preparations cannot be perfectly distinguished.

To prove the contradiction without requiring deterministic quantum measurement responses, choose a common dominating measure for ${\mu }_0,{\mu }_{+}$ and write their densities $m_0,m_{+}$. If they overlap, $\nu (\lambda )=\min (m_0,m_{+})$ has mass $\epsilon >0$. By preparation independence, every one of the four product densities dominates $\nu ({\lambda }_1)\nu ({\lambda }_2)$. Its total mass is ${\epsilon }^2$. The zero Born rule probability for outcome $xy$ implies that its nonnegative response ${\xi }_{xy}({\lambda }_1,{\lambda }_2)$ vanishes almost everywhere for its matching preparation, hence also under this common product measure. All four responses would then vanish on a set of positive measure, contradicting their sum being one. Therefore ${\mu }_0$ and ${\mu }_{+}$ are mutually singular.

For the second pair, group the $2n$ independent preparations into two blocks of $n$. Define $|{\Psi }_i⟩=|{\psi }_i{⟩}^{\otimes n}$. The tensor-power reduction of PBR overlap givesEach block therefore has an effective two-dimensional Hilbert space. Explicitly,are orthonormal and $|{\Psi }_2⟩=(|e_0⟩+|e_1⟩)/\sqrt{2}$. Embed the four-vector exclusion basis above into the tensor product of these two block spaces. To obtain a complete quantum measurement on all $2n$ qubits, add the orthogonal complement of that four-dimensional subspace to one of its four projectors. All four allowed preparations lie in the subspace, so the four forbidden probabilities remain zero.

If the single-copy preparation distributions overlap with common mass $\epsilon >0$, their $2n$-fold products all dominate the common measure ${\nu }^{\otimes 2n}$ of mass ${\epsilon }^{2n}>0$. The same zero-response contradiction now applies to the four block preparations. ThusThe argument is exact for the ideal devices specified in the question; no finite experimental resolution or noisy overlap bound is assumed.

The GRW model takes a normalized wave function in $L^2({\mathbb{R}}^{3N})$. Between jumps it obeys the Schrodinger equation, $i\hslash {\partial }_t\Psi =H\Psi$, with the usual many-particle Hamiltonian operator, for example $H={\sum }_i[-{\hslash }^2{\nabla }_i^2/(2m_i)]+V({\mathbf{x}}_1,\dots ,{\mathbf{x}}_N)$. Each particle has an independent Poisson process of collapse times with rate $1/\tau$. Thus the total jump rate is $N/\tau$, the mean wait is $\tau /N$, and the jumping label is uniform on $\{1,\dots ,N\}$.

At a jump of particle $i$, use the GRW localization operatorIts centre has probability density function $p_i(\mathbf{c})=\Vert L_i(\mathbf{c})\Psi {\Vert }^2$. The post-jump quantum state is $L_i(\mathbf{c})\Psi /\Vert L_i(\mathbf{c})\Psi \Vert$. The normalization is consistent because $\int d^3\mathbf{c}{L}_i(\mathbf{c}{)}^2=I$. The collapse times and labels have these state-independent rates; the centres depend on the wave function through the Born rule-like squared norm.

For precision about the stated spatial event, let $B=Y_{10a}=\{\mathbf{c}:\mathrm{dist}(\mathbf{c},Y)\le 10a\}$. With $H=0$, the GRW collapse-centre event probabilities are exactlyThe last expression is defined only for $p_1(B)>0$. It includes the possibility that the next jump hits the same particle.

Write $f_i^Y(\mathbf{x})=|\varphi (\mathbf{x}-{\mathbf{y}}_i){|}^2$, and similarly $f_i^Z$. Putwith the same definitions for $Z$. The macroscopic separation and negligible packet tails make the two branches essentially orthogonal, so their interference contribution can be neglected. The first-event probability is thereforeThe $Z$ contribution is extremely small: its actual positions are within $r$ of $Z$, whereas centres in $B$ are within $10a$ of $Y$, leaving a distance much larger than $a$ in the Gaussian. Packet-tail errors add to this Gaussian-tail error.

For the next-event calculation, two distinct labels have independent position densities within either product branch, giving factors ${\eta }_i{\eta }_j$. A repeated label instead gives $R_i$, since both localization Gaussians multiply the same position variable. ConsequentlyThese formulas, with $p_{2\mid 1}=p_{12}/p_1$, answer the spatial questions without an extra containment assumption. If the first event is dominated by the $Y$ branch, they simplify toNeglecting the $Z$ term in this conditional expression requires it to be small relative to the retained first-event probability, not merely small in absolute value. The exact operator formulas above cover rare-event exceptions as well.

The intended GRW branch persistence versus geometric containment approximation is obtained if $Y$ and $Z$ contain essentially all the position mass of their corresponding branches. Then positions in $Y$ have a Gaussian-centre margin of $10a$ inside $B$, so ${\eta }_i^Y\simeq R_i^Y\simeq 1$, while ${\eta }_i^Z,R_i^Z\simeq 0$. This givesFor the Gaussian convention used here, the three-dimensional probability of a displacement exceeding $10a$ is $\mathrm{erfc}(10)+20e^{-100}/\sqrt{\pi }$, an extremely small number. A first centre near $Y$ also suppresses the $Z$ branch relative to the $Y$ branch by the very small ratio $|\beta {|}^2{q}_i^Z(\mathbf{c})/(|\alpha {|}^2{q}_i^Y(\mathbf{c}))$ when the latter branch has appreciable weight. All particles are correlated with the same pointer alternative, so one localization selects the macroscopic branch even when the next particle label differs.

The literal PDF, however, only places the packet centres ${\mathbf{y}}_i$ in $Y$; it does not say that the packets themselves fit inside $Y$. Since it permits $r>a$, the two numerical approximations just displayed do not follow for every allowed packet and region. For example, take all ${\mathbf{y}}_i=0$, let $Y$ be a very small ball, and let $|\varphi {|}^2$ be uniform on a ball of radius $r\gg 10a$. Translate the other branch to a similarly small region $Z$ at distance much larger than $100r$. ThenIf these equal packet probabilities are denoted by $\eta$ and their repeated-label integral by $R$, the next probability iswhich approaches $\eta$, not one, for large $N$. Thus packet containment is needed for the stated near-region event to have the usual branch probabilities. The literal general answer is given by the integrals above.

The physical GRW amplification mechanism remains intact: with properly chosen macroscopic branch regions, spontaneous localization selects the alternatives with approximately the squared branch amplitudes and stabilizes the selected record, on a time scale $\tau /N$. Gaussian tails are not exactly eliminated, so the persistence is approximate. The containment qualification concerns how the region is defined, rather than whether the dynamics selects a macroscopically separated branch.

Assume a complete relativistically normalized basis of scalar momentum eigenstates transforms asThis states both reversal of spatial momentum and preservation of the normalization of basis quantum states. Antilinearity alone would not suffice: multiplying a conjugation operator by two gives an antilinear operator that multiplies squared norms by four.

Expand $|\varphi ⟩=\int d{\Pi }_p\varphi (\mathbf{p})|\mathbf{p}⟩$ and similarly for $|\psi ⟩$, with $d{\Pi }_p=d^{3}p/((2\pi {)}^{3}2E_p)$. This Lorentz-invariant phase-space measure is unchanged under $\mathbf{p}\mapsto -\mathbf{p}$. The antilinearity of the quantum time-reversal operator gives conjugated expansion coefficients. Orthogonality of the momentum eigenstates and cancellation of their unit-modulus phases yieldSince momentum reversal is a bijection of the complete basis, $\hat{T}$ is also onto. ThusThis proves antiunitarity, with the normalization hypothesis explicitly included. For scalar multiparticle quantum states the same argument uses the complete occupation-state basis and reverses all momenta; it is not restricted to a single-particle wave packet.

Work first in units $G=c=\hslash =k_B=1$. Black-hole thermodynamics starts with the laws of black-hole mechanics. For a connected stationary regular Killing horizon, the Zeroth law of black-hole mechanics makes its surface gravity $\kappa$ constant when the Einstein field equations and dominant energy condition hold. This parallels uniform equilibrium temperature. For neighboring stationary asymptotically flat four-dimensional Einstein-Maxwell black holes, the First law of black-hole mechanics iswhere ${\Omega }_H$ is the horizon angular velocity and ${\Phi }_H$ its electric potential relative to infinity. The last terms are rotational and electromagnetic work, analogous to the work terms in $\delta E=T\delta S+\text{work}$. The second law of black-hole mechanics, expressed by Hawking's area theorem, makes the total future horizon area nondecreasing for classical matter obeying the null energy condition and global assumptions such as strong asymptotic predictability. The third law of black-hole mechanics is an unattainability statement: subject to the usual regularity assumptions, the weak energy condition, and a bounded stress-energy tensor, no finite physical process reduces the surface gravity of a regular horizon to zero. It does not assert that extremal black holes have zero area or zero entropy.

Classical geometry alone fixes an analogy, not a nonzero physical temperature. Quantum field theory supplies Hawking temperature $T_H=\kappa /(2\pi )$. Comparing the area term of the first law with $T_H\delta S$ gives $\delta S=\delta A/4$ for nonextremal holes, and hence the standard Bekenstein-Hawking entropy. Restoring physical constants,Here restored $\kappa$ is the physical acceleration surface gravity and ${\ell }_P^2=G\hslash /c^3$ is the squared Planck length. The first-law comparison determines the area coefficient, leaving an additive entropy constant unspecified; the displayed formula is the usual convention. Its area scaling, rather than ordinary volume scaling, signals that the available thermodynamic degrees of freedom of a gravitating system are constrained by the horizon geometry.

Particle production explains the quantum input. For a real scalar field obeying the massless covariant wave equation, the Klein-Gordon inner product on solutions isThe wave equation makes its current divergence-free, so the product is independent of a Cauchy hypersurface when the boundary flux vanishes. Use suitably normalized wave packets or the appropriate continuum distributions. Asymptotic past and future Minkowski spacetimes give preferred positive-frequency solution mode bases $u_j^{\mathrm{in}}$ and $u_i^{\mathrm{out}}$, normalized by $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Between those regions, a nonstationary geometry generally has no preferred positive-frequency splitting.

The mode bases are related by a Bogoliubov transformation,The last identities follow from conservation of the Klein-Gordon inner product and are the canonical identities for a bosonic Bogoliubov transformation. Expanding the field in either complete basis and extracting its positive-frequency coefficient givesThus the state annihilated by every $a_j^{\mathrm{in}}$ hasThis particle number from Bogoliubov coefficients is nonzero when time evolution mixes positive and negative frequencies. It describes one state as vacuum in the early particle basis and populated in the late basis; it does not require an arbitrary choice of a vacuum at each intermediate time. For a finite implementable transformation the state remains a squeezed pure quantum state. In infinitely many modes, a common unitary bosonic Fock space implementation requires additional conditions, such as the beta map being a Hilbert-Schmidt operator; finite wave packet observables need not share every global divergence of an ideal continuum calculation.

For gravitational collapse, the future is not globally Minkowski: it contains a black hole. The relevant late out-basis includes modes reaching future null infinity together with modes entering the future event horizon. Modes on infinity alone are not a complete Cauchy basis. Nevertheless the same mode-mixing calculation determines the outgoing particle flux. Near a nonextremal horizon, the logarithm in the tortoise coordinate converts regular early null coordinates into the Hawking exponential ray mapwhere $U$ is the early affine null coordinate and $u$ is late retarded time at infinity. Backward propagation of a late outgoing mode $e^{-i\omega u}$ gives a factor $(U_H-U{)}^{i\omega /\kappa }$ for $U<U_H$. Its positive- and negative-frequency Fourier integrals differ by analytic continuation around the logarithmic branch. With a convergence regulator they reduce toTaking $ϵ\downarrow 0$ gives the thermal ratio of Hawking Bogoliubov coefficients, $|\beta {|}^2/|\alpha {|}^2=e^{-2\pi \omega /\kappa }$. Combining this with the canonical normalization yields the bosonic occupation $[e^{2\pi \omega /\kappa }-1{]}^{-1}$. Late-time wave packets turn formal continuum coefficients into a finite number flux.

A Schwarzschild black hole has $\kappa =1/(4M)$, so this spectrum has $T_H=1/(8\pi M)$. The exterior curvature potential partly reflects the outgoing modes. Its transmission probabilities are the greybody factors, giving, for one massless scalar species in the stationary late-time approximation,Thus the horizon temperature is universal, while the spectrum received at infinity is not a perfect featureless blackbody radiation spectrum. The derivation assumes the near-horizon quantum state inherited from a regular collapse vacuum and ignores rapid backreaction over the timescale of the packets. Exponentially blueshifted precursor frequencies indicate the usual short-distance assumption in the semiclassical calculation, not an independently demonstrated quantum-gravity description of the endpoint.

The outgoing positive energy flux drives black-hole evaporation. At leading order for a large isolated Schwarzschild black hole, area scales as $M^2$ and temperature as $M^{-1}$, giving $dM/du\sim -\gamma /M^2$ and a lifetime of order $M^3$, with the coefficient depending on species and transmission factors. Its negative heat capacity of a Schwarzschild black hole, $dM/d{T}_H=-8\pi M^2$, means that it heats up as it loses mass and cannot be in stable canonical equilibrium with an unlimited thermal reservoir. The approximation fails when quantum-gravitational scales are reached and does not fix whether the endpoint is complete evaporation, a remnant, or something else.

Area loss does not contradict the classical area theorem, because the quantum stress-energy tensor need not obey the classical null energy condition; negative horizon energy accompanies the positive outgoing flux. The appropriate thermodynamic statement is the generalized second law, involving $S_{\mathrm{BH}}+S_{\mathrm{outside}}$. Finally, Hawking radiation raises the black hole information paradox: if a pure quantum state completely evaporates into an exactly thermal mixed quantum state and its partners disappear, the result conflicts with ordinary unitary time evolution. Early outgoing radiation can be mixed simply because of its entanglement with interior modes while the complete state remains a pure quantum state; that fact alone is not information loss. Resolving the fate of all correlations through the evaporation endpoint requires more than the leading semiclassical flux calculation.

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.

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.