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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.

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.

StoqMA is a restricted quantum-verifier class using stoquastic circuits and polynomial-size quantum witnesses. A YES instance has a quantum witness accepted with probability at least $a$; on a NO instance every quantum witness has acceptance at most $b$, with an inverse-polynomial gap $a-b$. An optimal quantum witness can be chosen with nonnegative amplitudes, since the witness acceptance operator is entrywise nonnegative. The quantum witness is generally a superposition, rather than a computational basis vector. The stoquastic acceptance floor explains the lower limit $b\ge 1/2$ for a nontrivial soundness promise.

For a positive witness acceptance operator on dimension $D$, large trace powers approximate its largest eigenvalue. With an $m$-qubit QMA quantum witness and $d=2m+2$, the usual NO instances give $\mathrm{Tr}(F^d)<2^{-d}$ and YES instances give a value above $2^{-d}$. A closed product-index expansion computes the trace by recomputation with polynomial storage, avoiding logarithms and roots in a real-arithmetic implementation.

For quantum witness embedding $J$, verifier unitary $U$ and accepting projector $\Pi$, the witness acceptance operator is the positive contraction $F=J^{†}{U}^{†}\Pi UJ$. The maximum acceptance over normalized quantum witnesses is ${\lambda }_{\max }(F)$ by the Rayleigh quotient. Restricting to basis quantum witnesses only tests its diagonal entries and need not find that maximum.