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Write $N=2^n$ and $|s⟩=N^{-1/2}{\sum }_{x=0}^{N-1}|x⟩$. The initial Hadamard gates prepare $|s⟩$ in the search register and $|-⟩$ in the final quantum ancilla. Since the Pauli X gate satisfies $X|-⟩=-|-⟩$, the Boolean quantum oracle acts by quantum phase kickback:For a single marked entry this is the marked-state phase oracle $O_a=I-2|a⟩⟨a|$ on the search register. The ancilla stays separate and unchanged throughout.

The Grover diffusion operator is $V=2|s⟩⟨s|-I$, because $H^{\otimes n}|0^n⟩=|s⟩$. The PDF circuit places one diffusion after each oracle call, so the state before measurement isIn the supplied amplitude amplification theorem, take $|\psi ⟩=|s⟩$ and $|\varphi ⟩=|a⟩$. Their overlap has modulus $1/\sqrt{N}$, so the Grover rotation angle is $\theta =\arcsin (N^{-1/2})$. A quantum measurement in the computational basis therefore givesThe inverse sine here is applied to the overlap of the two different states. The converted TeX's self-overlap in the angle definition is a transcription error; the original PDF has $|⟨\varphi |\psi ⟩|$.

No measurement on the supplied system alone distinguishes ensembles with the same density operator. A completely general measurement in quantum mechanics has POVM effects $E_r\ge 0$ with ${\sum }_r{E}_r=I$. The outcome probability for a quantum ensemble isIf both ensembles have the same $\rho$, every outcome probability is identical. This proves the claim for all positive operator-valued measures, and includes projective measurements as a special case. An quantum ancilla in a fixed independent state does not help: the input density operators remain identical after tensoring with that ancilla, and the same argument applies to any joint measurement.

The premise concerns access only to the prepared state. An accessible preparation label or a correlated purification of a density operator is additional information outside that premise and can distinguish global preparations. Without such side information, different ensemble decompositions do not define distinguishable states.

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.

Let $V$ denote the coherent exact quantum phase estimation unitary operator from the preceding part, with the data quantum register written first. For every eigenvector of $A$,because $e^{2\pi iA}|u_j⟩=e^{2\pi i{\lambda }_j}|u_j⟩$ and ${\lambda }_j=c_j/2^n$ lies in $[0,1)$, with no phase-aliasing ambiguity. Apply $V$ coherently to the input quantum state and a clean quantum register; do not measure the eigenvalue label. By linearity it produces ${\sum }_j{\beta }_j|u_j⟩|c_j⟩$.

Append a quantum ancilla in $|0⟩$, and use the provided quantum variable rotation with ${\theta }_{c_j}=\arcsin {\lambda }_j$, choosing the nonnegative cosine. The resulting normalized quantum state isThe sum covers both terms; this fixes the potentially ambiguous sum placement in the abbreviated display. The orthonormal basis of eigenvectors and the unit length of each rotated quantum ancilla show that its norm is one. This is quantum spectral filtering by multiplication, not the reciprocal rotation used in the HHL algorithm.

The basic one-bit teleportation primitive uses the input as qubit 1 and a fresh $|+⟩$ quantum ancilla as qubit 2. Apply the Controlled-Z gate $E_{12}$ and measure qubit 1 in the equatorial qubit measurement basis with angle $\alpha$. If $s\in \{0,1\}$ is the result, the normalized post-measurement state on qubit 2 isBoth results have probability one half. For example, writing $|\psi ⟩=a|0⟩+b|1⟩$, the unnormalized branch is $(a|+⟩+(-1{)}^s{e}^{i\alpha }b|-⟩)/\sqrt{2}=X^{s}J(\alpha )|\psi ⟩/\sqrt{2}$. Thus $J(\alpha )$ is applied deterministically as a logical gate with known Pauli frame $X^s$; a permitted conditional Pauli X gate would remove this byproduct physically.

The literal resource restriction does not need an unstated direct Pauli X gate. A Pauli Z gate can be enacted by applying $E$ to the data and a fixed $|1⟩$ quantum ancilla. Two consecutive angle-zero one-bit teleportations, with results $p,q$, map an arbitrary current state $|\chi ⟩$ to $X^{q}H{X}^{p}H|\chi ⟩=X^q{Z}^p|\chi ⟩$. Apply the available $Z^p$ to remove the latter factor, up to an irrelevant global phase. This implements a heralded $X^q$: if $q=0$, the input is unchanged; if $q=1$, the required Pauli X gate has been applied.

If the original result is $s=1$ and a physically corrected output is required, repeat this heralded Pauli X correction using controlled-Z and measurements until $q=1$. Each attempt succeeds with probability one half independent of the input, so it terminates with probability one, using two attempts on average. Only $E$, fixed ancillary quantum states and single-qubit quantum measurements are used. This exact physical correction has no finite worst-case measurement bound; the standard finite deterministic realization is the logical Pauli frame version, which suffices for the later output-simulation parts.

Since $A|x⟩=(-1{)}^{f(x)}|x⟩$, the spectral decomposition givesReplace the middle gate of the previous compute-phase-uncompute construction bya one-qubit rotation about the z-axis. ThenThe circuit uses the gates of $C$, their inverses, and one fixed rotation; it returns the quantum ancilla to $|0⟩$. The same construction implements $e^{itA}$ with middle gate $e^{itZ}$, an instance of simulation of a computable diagonal Hamiltonian.

Use a clean quantum ancilla and apply the quantum circuit $C$, then the Pauli Z gate on the ancilla, then the inverse quantum circuit $C^{†}$:This compute-phase-uncompute construction implements $A\otimes I$ on every input with the ancilla initially $|0⟩$, and by linearity on any superposition of those inputs. The action on ancilla-$|1⟩$ inputs need not coincide with $A\otimes I$; the question only requires the clean-ancilla subspace.

Write the normalized input as $|b⟩={\sum }_j{\beta }_j|u_j⟩$, where $A|u_j⟩={\lambda }_j|u_j⟩$ and ${\lambda }_j>0$. Use a data register, an $n$-qubit eigenvalue register, and a flag quantum ancilla. Begin with $|b⟩|0^n⟩|0⟩$.

Apply exact quantum phase estimation to $e^{2\pi iA}$. Its controlled evolutions and inverse quantum Fourier transform produceApply the HHL controlled reciprocal rotation, with $0<c\le {\lambda }_{\min }$:The function of ${\ell }_j$ is computed by quantum arithmetic; the rotation is a quantum variable rotation. On a zero eigenvalue label with no input amplitude, define any unitary action, for example no rotation.

Run the inverse phase-estimation circuit. Because each amplitude multiplier depends only on ${\lambda }_j$, the eigenvalue register is reset to $|0^n⟩$ on both flag branches. Measuring the flag and conditioning on outcome one leavesThe successful unnormalized branch is $c{A}^{-1}|b⟩$, giving the probability in the preceding part. The operations independent of $A,b$ are the Hadamard gates preparing the phase register, the inverse quantum Fourier transform, the reversible reciprocal/angle arithmetic, the controlled fixed-angle rotations, and the final flag measurement. Uncomputation removes all arithmetic workspace; the $A$-dependent controlled evolutions and the $b$-dependent state-preparation circuit supply the input-specific operations.

Introduce a quantum ancilla with basis $|i⟩$ and defineThe completeness relation gives $V^{†}V=I$, so $V$ is an isometry and extends to a unitary operator on a sufficiently large system-plus-ancilla space. Prepare the ancilla in a fixed state, apply that unitary, and perform the projective measurement $\{I\otimes |i⟩⟨i|\}$. Outcome $i$ has probability $\Vert M_i|\psi ⟩{\Vert }^2$ and leaves the system in the normalized state $M_i|\psi ⟩$. By linearity the same holds for mixed states, implementing the generalized measurement.

Attach a quantum ancilla prepared with amplitudeon $|1⟩$; this is possible because $p>1/4$. Mark a state as good only when both the original success qubit and this ancilla equal one. The enlarged state's good probability is $p{c}^2=1/4$, so its good amplitude is $1/2=\sin (\pi /6)$. One amplitude amplification iteration rotates the angle from $\pi /6$ to $3\pi /6=\pi /2$. It therefore prepares $|{\psi }_1⟩$ exactly, after which the flag and ancillary qubits may be discarded. This is an instance of exact amplitude amplification.

Implement the assumed efficient classical algorithm for $x\mapsto {\theta }_x$ as a reversible circuit. On input $|x⟩|0^s⟩$, it computes an $O(m)$-bit binary expansion of the angle in a work register using $\mathrm{poly}(m)$ Toffoli gates and elementary reversible gates. Write the computed angle as a sum of binary-weighted angles. For each angle bit, apply the corresponding controlled single-qubit $R_y$ rotation to the target. These rotations have the same axis, so their angles add and produceFinally apply uncomputation to erase the work register. Each Toffoli gate and controlled rotation has a constant-size decomposition into one- and two-qubit gates when arbitrary one-qubit rotations are available. Ignoring the stipulated precision costs, the resulting circuit has size $\mathrm{poly}(m)$ and returns every quantum ancilla to zero.

For $p=(d+1)/(2d)$, the weighted difference of the two Werner–Holevo channels isThus the map is the transposition map divided by $d$. The diamond norm of the transposition map and invariance of the trace norm under matrix transpose giveSubstitution into the optimal-error formulas shows that an entangled quantum ancilla permits perfect discrimination, whereas every ancilla-free strategy has

For a fixed input, the Holevo–Helstrom theorem gives the optimal error for the two output states. Optimizing the input and quantum ancilla therefore givesThe stabilization in the diamond norm is exactly the optimization over ancillary systems. Without an ancilla the same argument instead giveswhere $\Vert \cdot {\Vert }_1$ is the induced trace norm.

In quantum channel discrimination, prepare a density operator ${\rho }_{HR}$ on the channel input $H$ and an optional quantum ancilla $R$. Under hypothesis $j\in \{1,2\}$ the output isUse a two-outcome quantum measurement $\{Q,I-Q\}$ and decide for $T_1$ on outcome $Q$. The conditional Type I error and Type II error areWith prior probabilities $p$ and $1-p$, symmetric Bayesian discrimination minimizes the average error $p\alpha +(1-p)\beta$ over the input and measurement.

A positive operator-valued measure is a family of positive operators $E_i$ satisfying ${\sum }_i{E}_i=I$. Each $E_i$ is called an effect and determines a probability $p_i=\mathrm{Tr}(E_i\rho )$ for input density operator $\rho$. The effects need not be orthogonal projections or commute. This permits measurements with more outcomes than the dimension of the Hilbert space, such as the trine qubit POVM. Coupling to a quantum ancilla and measuring the larger system supplies a physical route to generalized measurements. ^[ref].

For an elementary unsharp $Z$ example on a qubit, takeOn the eigenstate $|0⟩$, $p_{\pm }=(1\pm \eta )/2$; at $\eta =1/2$ they are $3/4$ and $1/4$. At $\eta =0$ the outcome is a fair coin independent of the input, and at $\eta =1$ these effects specify a sharp $Z$ projective measurement. Choosing $K_{\pm }=\sqrt{E_{\pm }}$ specifies one quantum instrument, but $U_{\pm }\sqrt{E_{\pm }}$ for outcome-dependent unitary operators has the same effects and different state updates. Thus POVM does not determine the post-measurement state.

Quantum channel discrimination feeds a common input into one of several candidate quantum channels and measures the output. An entangled quantum ancilla permits inputs ${\rho }_{HR}$ and compares $(T_j\otimes {\mathrm{id}}_R)({\rho }_{HR})$; optimal binary discrimination is governed by the diamond norm of the weighted channel difference.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.