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Write the quantum circuit as $U=U_T\cdots U_1$, and let $V_0=I$, $V_t=U_t\cdots U_1$. Use a nonlocal quantum clock with orthonormal states $|0⟩,\dots ,|T⟩$. Let the work space include the quantum witness and the ancilla qubits. The input penalty of a history Hamiltonian is built fromIt annihilates precisely the correctly initialized ancilla qubits, leaving the quantum witness unrestricted. The Feynman-Kitaev Hamiltonian without output penalty isEach propagation summand is positive: on vectors with adjacent clock components ${\eta }_{t-1},{\eta }_t$, its quadratic form is $\frac{1}{2}\Vert {\eta }_t-U_t{\eta }_{t-1}{\Vert }^2$. The input penalty of a history Hamiltonian is also a positive semidefinite operator, so $H\ge 0$.

To verify that this is a stoquastic Hamiltonian, use the work computational basis and the clock basis. Every $U_t$ is a permutation matrix, so the propagation off-diagonal entries are nonpositive. The zero-ancilla projectors are diagonal, while $|-⟩⟨-|=\frac{1}{2}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ also has nonpositive off-diagonal entries. No positive off-diagonal entry is introduced by summing these terms. Thus $H$ is positive semidefinite and stoquastic, with no output penalty.

The construction uses the abstract $T+1$ dimensional clock space. A binary implementation needs diagonal penalties for unused clock labels. The clock transitions are nonlocal; the construction alone does not establish fixed qubit locality of an ordinary local Hamiltonian problem.

For a normalized computational history state, the output projector sees only its final clock component. With $p_{+}(\zeta )$ the quantum circuit's acceptance probability,This proves the hint, including for a computational basis quantum witness. Moreover $H_{\mathrm{out}}=|-⟩⟨-{|}_1\otimes |T⟩⟨T|$ is a stoquastic Hamiltonian term, so adding it with positive coefficient preserves stoquasticity.

There are two substantive problems with the stated promise. First, for any basis quantum witness the initialized state has nonnegative amplitudes, and every permutation matrix preserves them. Write the output as $|0⟩|r_0⟩+|1⟩|r_1⟩$, with both vectors entrywise nonnegative. ThenThe stoquastic acceptance floor rules out the printed one-third soundness condition. There are no NO instances of that literal promise. Second, the ground space includes histories of arbitrary quantum witnesses. It is the maximum over those quantum witnesses, not the maximum over basis quantum witnesses, that determines the lowest output energy. Even the identity quantum circuit accepts each basis input with probability $1/2$, but accepts a $|+⟩$ quantum witness with probability one. Thus a basis-witness soundness bound would not control this Hamiltonian, even if its numerical threshold were repaired.

The requested nontrivial hardness statement therefore needs the standard quantum-witness StoqMA promise, with $1/2\le b<a\le 1$ and inverse-polynomial gap $\gamma =a-b$. The intended reduction can be completed precisely under that corrected promise. Define the quantum witness embedding $J|\zeta ⟩=|\zeta ⟩|a⟩$ and the witness acceptance operatorIt is a positive contraction, and $p_{\max }={\lambda }_{\max }(F)$. The minimum expectation of $H_{\mathrm{out}}$ on the history ground space is $\nu =(1-p_{\max })/(T+1)$. This is also consistent with a nonnegative optimal StoqMA quantum witness: $F$ is entrywise nonnegative, so replacing amplitudes by their absolute values cannot decrease its quadratic form.

A uniform ground-space perturbation bound is needed because the gap shrinks with quantum circuit length. Put $g=(T+1{)}^{-3}$, and choose a positive inverse-polynomial coefficientFor $P$ the history ground projector and any unit vector $\psi =p+q$ with $p=P\psi$, the operator norm bound $\Vert H_{\mathrm{out}}\Vert =1$ givesThe last inequality completes the square in $\Vert q\Vert$ and uses $0\le \nu \le 1$. Testing a ground space minimizing vector gives the upper bound ${\lambda }_{\min }(H^{\prime })\le \delta \nu$. Since $\delta <g/2$, we obtainYES instances of the corrected promise have energy at most $\delta (1-a)/(T+1)$; NO instances have energy at leastThe separation is at least $7\delta \gamma /[8(T+1)]$, an inverse polynomial. The quantum circuit and all these Hamiltonian terms have polynomial-size descriptions. This proves StoqMA hardness for the local Hamiltonian problem variant that permits the nonlocal clock, under the corrected quantum witness and acceptance promise.

The printed perturbation formula also has $H_{\mathrm{out}}$ where a general perturbation $V$ belongs. Its unspecified $O({\delta }^2)$ cannot be treated as uniform in a closing gap. Likewise a circuit-length-independent constant $\delta$ cannot in general satisfy the stated small-perturbation requirement for an arbitrary long quantum circuit. The explicit bound above avoids both issues; it does not claim the defective literal promise defines standard StoqMA.

BQP consists of promise problems decided by a uniform family of polynomial-size quantum circuits. On input $x$, with polynomially many zero-initialized ancilla qubits, a designated output measurement accepts with probability at least $2/3$ on YES inputs and at most $1/3$ on NO inputs. Uniformity means a classical polynomial-time procedure produces the quantum circuit description from the input length, or equivalently produces the verifier quantum circuit with $x$ supplied as input.

For BQP error reduction, run independent copies with freshly initialized registers. If the completeness and soundness thresholds are any constants $a>b$, accept when the fraction of accepting trials exceeds $(a+b)/2$. A Hoeffding inequality bounds either error by $\exp [-r(a-b{)}^2/2]$, with $r$ repetitions. The threshold need not be a simple majority when both $a$ and $b$ lie on the same side of $1/2$.

Choosing $r$ sufficiently large gives the usual $2/3,1/3$ thresholds, or any other fixed separated thresholds. Even an inverse-polynomial gap can be amplified using polynomially many repetitions. Classical threshold evaluation can be incorporated into the uniform computation. Thus fixed separated acceptance probabilities define the same BQP class. This argument concerns unrestricted BQP quantum circuits and does not assume such amplification is available for restricted stoquastic circuits.

The class $C$ uses polynomial-register real-arithmetic computation: its storage restriction counts real registers, without limiting their precision or the running time. Use the usual decision-model interpretation that computed real values may be compared with fixed thresholds, and that the fixed universal gate set's real constants are available. Without a way to test values, arithmetic instructions alone would not specify the intended decision model.

Let a BQP quantum circuit use $q=\mathrm{poly}(n)$ qubits and $T=\mathrm{poly}(n)$ gates. A matrix element can be evaluated by depth-first quantum circuit path summation:Enumerate the intermediate $q$-bit strings recursively. Store the current strings and loop positions, a partial product, and the partial sum at each depth. There are at most $T$ active levels, using $O(qT)$ registers or bits of loop information, not an exponentially long state vector. Each fixed-locality gate matrix element is computed from its few affected bits. Represent a complex number by two real registers; complex multiplication and addition require only real addition, subtraction and multiplication.

Recompute this amplitude separately for every final string $z$ whose output bit is one, accumulating the Born rule probabilityThe outer enumeration needs only another $O(q)$ bits and a real accumulator. Arbitrarily large running time is allowed, so repeated recomputation is harmless. Compare $p_1$ with $1/2$ to distinguish the promised BQP cases. This proves $\mathrm{BQP}\subseteq C$ using polynomial storage.

The printed probability hint omits the squares of the amplitude moduli. Summing their absolute values alone is not the Born rule and can exceed one. The corrected sum above is necessary for the containment proof. The argument is about the stated real-register model, rather than a claim that exponential-precision numbers are free in an ordinary bit-cost model.

QMA is the class of promise problems for which a uniform polynomial-size quantum circuit checks a polynomial-size quantum witness. A YES instance has some quantum witness accepted with probability at least $2/3$; on a NO instance every quantum witness is accepted with probability at most $1/3$. The verifier initializes its own ancilla qubits to zero. Mixed quantum witnesses do not improve the maximum because acceptance is linear in their density operator.

For a verifier $U_x$ and quantum witness embedding $J|w⟩=|w⟩|0\cdots 0⟩$, define the witness acceptance operatorThe Rayleigh quotient gives ${\max }_{\Vert w\Vert =1}⟨w|F|w⟩={\lambda }_{\max }(F)$. Every entry of $F$ can be recomputed by the polynomial-storage path summation of part (b); no full quantum witness matrix needs to be stored.

Let the quantum witness have $m$ qubits, so its dimension is $D=2^m$. For this positive operator, trace-power witness optimization usesThis is the useful exponentiated form of the supplied logarithmic inequality. Positivity is required; such a logarithmic statement is not true for an arbitrary operator. No logarithms or roots need to be calculated.

Choose $d=2m+2$. On a NO instance,since $2^m(2/3{)}^{2m+2}=(4/9)(8/9{)}^m<1$. On a YES instance, $\mathrm{Tr}(F^d)\ge (2/3{)}^d>(1/2{)}^d$. Thus comparison of $2^d\mathrm{Tr}(F^d)$ with one distinguishes the cases using only multiplication and a final comparison.

Evaluate the trace byThere are $d=O(m)$ active quantum witness indices, each requiring $m$ bits. Accumulate one product at a time and recompute its matrix entries as required. Together with the verifier path workspace, this uses polynomially many real registers. The potentially enormous time and the numerical precision are unrestricted in $C$. Therefore $\mathrm{QMA}\subseteq C$, without solving an exponentially large eigenvalue problem by storing its matrix.

Because each term in the history-subspace propagation Hamiltonian is positive semidefinite, the uniform history vector has zero energy and is a ground state:Indeed every edge term annihilates the uniform coefficients. Conversely, a zero-energy vector must have equal coefficients across every edge of the connected path, so this computational history state is the unique ground state in $\mathcal{L}$. It coherently records every intermediate state of the quantum circuit.

A promise problem $(L_{\mathrm{yes}},L_{\mathrm{no}})$ belongs to QMA if there are polynomials $p,q$ and a uniform family of quantum circuits $V_x$ of size at most $q(|x|)$, taking a $p(|x|)$-qubit witness and clean ancillas, with one output qubit interpreted as acceptance, such thatNo condition is imposed outside the promise. The witness may be an arbitrary density operator, and the verifier is efficient but need not know how to prepare an honest witness. Pure witnesses suffice for completeness because the acceptance probability is linear in the witness. QMA is quantum verification with polynomially many witness qubits and a constant completeness-soundness gap.

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.

First require $x\ne 0$. A discrete logarithm to a generator of a group of ${\mathbb{Z}}_p^{\ast }$ exists only for nonzero $x$; the printed membership $x\in {\mathbb{Z}}_p$ inadvertently also permits zero, for which $x^{-b}$ is undefined. Zero can be rejected before any quantum circuit.

Put $M=p-1$ and write $x=g^y$. Since $g$ is a generator of a group of order $M$, the map $k\mapsto g^k$ is a bijection from ${\mathbb{Z}}_M$ to the multiplicative group of a finite field. Hence every nonzero $c$ has a unique $k\in {\mathbb{Z}}_M$ with $c=g^k$. Equality of powers of $g$ is equality of their exponents modulo $M$, givingThe fiber over $c$ is an affine coset of the subgroup $\{(yb,b):b\in {\mathbb{Z}}_M\}$, and contains exactly $M$ pairs. This identifies the hidden subgroup problem underlying discrete-logarithm Fourier sampling; knowing $k$ numerically is not needed to construct the coset state.

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.

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.

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.