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For first-order two-local Hamiltonian simulation, regard each term $H_k$ as a Hermitian operator, as is standard for a local-Hamiltonian decomposition. The Lie-Trotter product formula for bounded operators statesin operator norm; the reversed factor ordering is just as valid as the displayed one. Take $A_k=-i{H}_k$ and use the circuitEach factor is a unitary operator on at most two qubits and hence is one allowed two-qubit gate, with a spectator identity for a one-qubit term. The factors in the rightmost part of each written product act first. Merely quoting the limit would not establish the requested rate, so we give a quantitative error bound.

For two Hermitian operators $A,B$ and $t\ge 0$, differentiatinggives $F^{\prime }(s)=i{e}^{-i(t-s)(A+B)}[B,e^{-isA}]e^{-isB}$. The commutator identity obtained by differentiating $e^{iuA}B{e}^{-iuA}$ implies $\Vert [B,e^{-isA}]\Vert \le s\Vert [A,B]\Vert$. Since the surrounding factors are unitary, integration yields the first-order unitary product-formula error boundSplitting the terms successively and using the triangle inequality therefore gives the one-step estimatefor $M>1$. Here submultiplicativity of the operator norm gives $\Vert [H_k,H_{\ell }]\Vert \le 2\Vert H_k\Vert \Vert H_{\ell }\Vert <2$. The ordering chosen for the products changes the sign of local commutators, not this norm bound. Applying part (i) to the $r$ steps givesChoose $r=\max (1,\lceil M(M-1)/(2ϵ)\rceil )$. For $M>1$ the error is strictly below $ϵ$; $M=1$ is exact already with $r=1$. The quantum circuit contains $Mr$ two-qubit gates. A sufficient gate count isThus the worst-case polynomial degree in this first-order bound is six. Terms with disjoint supports commute and may improve the count in particular decompositions, but the stated assumptions alone suffice for this bound, without a further sparsity or bounded-degree condition. The requested gate model allows arbitrary two-qubit unitaries, so no additional compilation error is incurred here.

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.

The path graph state is $|{\psi }_3⟩=E_{23}{E}_{12}|+{⟩}_1|+{⟩}_2|+{⟩}_3$. The factor $Z_3^r$ commutes with both the Controlled-Z gates and the quantum measurement of vertex $1$. Treat $Z_1^r|+{⟩}_1$ as the input of one-bit teleportation. The equatorial qubit measurement result $s$ therefore gives the normalized stateHere the subscript on $Z_2^r$ expresses the teleported input factor; it is not an extra operation on the already measured vertex. Use $J(\alpha )Z^r=X^{r}J(\alpha )$ to obtainThe exclusive or exponent is equivalent to the printed sum $r+s$ because $X^2=I$. The conditional probability of $s$ is one half for either $r$, by the one-bit teleportation branch norm. This calculation concerns an equatorial measurement of a graph-state leaf; no third remaining qubit or extra $|-⟩$ factor is present.

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.

Let the ideal two-qubit output before final quantum measurements beThe PDF places $J(\alpha )$ only on the upper wire, followed by the Controlled-Z gate and a quantum measurement in the computational basis on each output wire. Measure vertex $0$ of the square in the computational basis, obtaining $r$, then vertex $1$ of the surviving path in the fixed equatorial qubit measurement basis at angle $\alpha$, obtaining $s$. Put $t=r\oplus s$. The previous part gives $E_{23}{X}_2^{t}J(\alpha {)}_2{Z}_3^r|++⟩$.

Commute its Pauli frame through the Controlled-Z gate. Since $E_{23}{X}_2^t=X_2^t{Z}_3^t{E}_{23}$ and $E_{23}$ commutes with $Z_3^r$, the actual output isFinally measure vertices $2,3$ in the computational basis, with raw results $d_2,d_3$. A Pauli X gate flips a computational basis result, whereas a Pauli Z gate changes only its phase. Thus the purely classical correction isThis four-cycle graph-state simulation of an entangle-and-measure circuit reproduces the full joint output distribution, not just each marginal. All measurement bases are fixed beforehand, and no physical byproduct correction is necessary. As a check, the ideal Controlled-Z gate is diagonal in the output basis, so $P(k,l)=\frac{1}{2}{\cos }^2(\alpha /2)$ for $k=0$ and $\frac{1}{2}{\sin }^2(\alpha /2)$ for $k=1$, independently of $l$. The measurement procedure gives exactly these probabilities after its classical relabelling.

The logical depth of a measurement pattern counts sequential layers of quantum measurements forced by dependence of measurement bases on earlier outcomes. Depth one means every basis can be fixed before measurements start, allowing all quantum measurements on distinct qubits to be performed in parallel. This does not require depth-one resource-state preparation or constant-depth classical parity processing.

Suppose the circuit contains $m$ gates $J({\alpha }_1),\dots ,J({\alpha }_m)$ in time order, with each ${\alpha }_j\in \{0,\pi /2\}$. Prepare an $(m+1)$-vertex path graph state from $|+{⟩}^{\otimes (m+1)}$ and Controlled-Z gates between neighbours. Its first qubit supplies the specified input $|+⟩$. Measure vertex $j-1$ in the fixed equatorial qubit measurement basis at angle ${\alpha }_j$ for $j=1,\dots ,m$, obtaining $s_j$. Measure the final vertex in the computational basis, obtaining $d$.

Track the Pauli frame as $X^{a_j}{Z}^{b_j}$, starting with $a_0=b_0=0$. A hypothetical sequential reading of the same quantum measurements gives the step $X^{s_j}J({\alpha }_j)X^{a_{j-1}}{Z}^{b_{j-1}}$. The supplied commutation rules imply, up to global phase,For ${\alpha }_j=0$, the sign is irrelevant. For ${\alpha }_j=\pi /2$, the identity $J(-\pi /2)=XJ(\pi /2)$ absorbs the possible sign change into the Pauli frame. If $t_j=0$ for ${\alpha }_j=0$ and $t_j=1$ for ${\alpha }_j=\pi /2$, the update isInductively the final unmeasured quantum state would be $X^{a_m}{Z}^{b_m}J({\alpha }_m)\cdots J({\alpha }_1)|+⟩$. The final computational basis outcome is corrected by $a_m$ and is unaffected by $b_m$. The induction on a sequential interpretation proves the joint statistics; projectors on distinct vertices commute, so the identical fixed-basis pattern can actually be measured simultaneously, including its final vertex.

The first printed commutation formula has an incorrect exact scalar phase: with the printed matrix definition, $J(\alpha )X=e^{i\alpha }ZJ(-\alpha )$, rather than the negative-exponent prefactor. For example, at $\alpha =\pi /4$ the two versions differ by a factor $i$. They agree up to global phase, so this error does not affect the Pauli frame recurrence or any quantum measurement probability.

Consequently a fixed-basis path-state pattern of logical depth one simulates every such circuit. Only Pauli measurements occur: the angle-zero basis measures $X$, and the angle-$\pi /2$ basis measures $-Y$ with the printed eigenvector labels. Classical exclusive or processing suffices for all output corrections. These gates are Clifford gates, which explains why angle adaptation can be eliminated. For $m=0$, simply measure the initial $|+⟩$ in the computational basis; it is already a depth-one pattern.

The printed sum contains $Z_n$ although the stated labels end at $n-1$. We use the natural periodic convention $Z_n=Z_0$, with $n\ge 3$. If an open chain was intended, omitting the final term gives the same asymptotic bound.

Set $H_j=X_{j-1}{Z}_j$, with indices modulo $n$, and implement the product-formula Hamiltonian simulationEach factor acts on two qubits, so it is a two-qubit unitary operator. More explicitly, if $a=j-1$, $b=j$, and $C_{a\to b}$ is a controlled-NOT gate,where ${\mathsf{H}}_a$ is a Hadamard gate. This gives a constant number of one-qubit and two-qubit gates per factor; one-qubit gates can also be viewed as two-qubit gates tensored with the identity.

The spectral norm obeys the triangle inequality, submultiplicativity of the operator norm, and invariance under multiplication by unitary operators. In particular, the telescoping bound for products of operators gives $\Vert V^k-W^k\Vert \le k\Vert V-W\Vert$ for unitary $V,W$. The first-order unitary product-formula error bound consequently yieldsOnly neighboring terms can have a nonzero commutator, because all other supports are disjoint. There are $n$ neighboring unordered pairs, and $\Vert [H_j,H_l]\Vert \le 2\Vert H_j\Vert \Vert H_l\Vert =2$. Thus $\Vert U-\tilde{U}\Vert \le n/k$. Choosing $k>n/\epsilon$ givesThe coarser bound that counts all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs also proves the often-used $O(n^3/\epsilon )$ construction. Locality improves that cubic estimate to the quadratic bound above; neither assertion is a lower bound on the best possible circuit.

The spectral norm is the operator norm induced by the Euclidean norm:The Pauli operators $X_1$ and $Z_2$ act on different qubits, and each is a unitary operator. Their product is therefore unitary and preserves the norm of every vector. Hence

For each position $i$, Alice uses $y_i$ to choose a basis and $x_i$ to choose a vector in that basis. Explicitly, she sends the qubitThus $y_i=0$ uses the computational basis $\{|0⟩,|1⟩\}$, while $y_i=1$ uses the Hadamard basis $\{|+⟩,|-⟩\}$.