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For an arbitrary vector $|\Psi ⟩={\sum }_{t=0}^T|{\eta }_t⟩|t⟩$, the preceding positive quadratic forms show that zero energy requiresThus all its clock components are generated by one correctly initialized input. Ifthen $\ker Q$ consists of $|\zeta ⟩\otimes |a⟩$, with unrestricted quantum witness $\zeta$. Normalization gives the computational history stateConversely, every such computational history state annihilates every propagation term and the input term, hence has zero energy. These states are exactly the ground space, not merely examples of its vectors. In particular, histories of computational basis quantum witnesses span the ground space; their coherent superpositions are also zero-energy states.

Another way to see the propagation constraint is to conjugate by $W={\sum }_t{V}_t\otimes |t⟩⟨t|$. Then $W^{†}{H}_{\mathrm{prop}}W=I\otimes E$, whereThe spectrum of a path propagation Hamiltonian has a unique uniform clock zero mode. This is the path with $T+1$ vertices. The upper summation index in the printed auxiliary formula must be $T-1$, rather than $T$, to agree with its stated dimension and spectrum; using $T$ would introduce an extra vertex.

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.

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

In the computational basis ordered as $00,01,10,11$, the Hermitian matrix isIts trace is one, and its eigenvalues are $(1+\alpha )/2$, $(1-\alpha )/2$, $0$, and $0$. Thus the remaining density operator condition, positivity, holds exactly when both nonzero candidates are nonnegative:This is the admissible parameter interval for the dephased Bell-state mixture.

The coefficient matrix in the computational basis isFor a bipartite pure state, the partial traces give ${\psi }_A=M{M}^{†}$ and ${\psi }_B=M^T\overline{M}$. Therefore the requested reduced density matrices areThe columns of $M$ are orthogonal. Their norms are $\sqrt{2/3}$ and $\sqrt{1/3}$, so normalizing them gives the orthonormal setWith $|v_0⟩=|0⟩$ and $|v_1⟩=|1⟩$, a Schmidt decomposition isExpanding this expression reproduces the minus sign in the $|01⟩$ coefficient. The two reduced density matrices both have eigenvalues $2/3$ and $1/3$, consistent with these Schmidt coefficients.

Choose $P_X$ uniform on ${\mathcal{A}}_M$ and zero outside it, so the marginal density operator of $Q$ is ${\bar{\rho }}_Q=k^{-1}{\sum }_m\rho (m{)}_Q$. Define the binary test for quantum decoding success byEach $E(m)$ is a positive contraction. The computational basis blocks of $F(0)$ are these effects on ${\mathcal{A}}_M$ and zero elsewhere. Thus $0\le F(0)\le I$, making $F$ a POVM on the whole space, including labels outside ${\mathcal{A}}_M$.

On the correlated classical-quantum state, its acceptance probability isOn the product of the marginal density operators,since ${\sum }_{m}E(m)=I_Q$. The two requested probabilities are therefore $1-ϵ$ and $1/k$. In the notation for binary output density operators, the measurement channel sends the joint input to $\omega [ϵ]$ and the product input to $\omega [1-1/k]$; their parameters describe outcome one, not outcome zero.

Interpret a function collision as two distinct inputs with equal outputs. Without distinctness the request is vacuous, since an input paired with itself always qualifies. Set $m=\lceil N^{1/3}\rceil$ (handling small $N$ by direct queries), choose a known set $A$ of size $m$, and query all its inputs. Store a classical table of their outputs and corresponding inputs. If this table contains a function collision, return it immediately.

Otherwise the table contains $m$ distinct outputs. Because $g$ is exactly two-to-one, each $a\in A$ has exactly one partner in $B$, the known complement of $A$. None lies in $A$, and different table outputs have different partners. Thus exactly $m$ inputs in $B$, of size $L=N-m$, satisfy the known-table membership predicateBuild a marked-state phase oracle for this predicate by a compute-phase-uncompute construction: use $U_g$ on a clean answer quantum register, reversibly compare its output against the classical table, apply the conditional minus sign, undo the comparison, and apply $U_g$ again. Since the answer operation is bitwise exclusive or, $U_g^{-1}=U_g$. All workspace returns to zero, and each phase query costs two calls to $U_g$.

Perform known-subset Grover search in the span of the computational basis states labelled by $B$, starting in their uniform superposition state. Reflection about this known state uses no $U_g$ queries; choosing $A$ as an initial interval makes $B$ a known interval that can also be encoded by its own $L$ labels. Its marked fraction is $m/L$, so the previous Grover rotation angle analysis uses $O(\sqrt{L/m})$ phase queries. On measuring $b$, one further $U_g$ query checks the output and retrieves its partner $a$ from the table. The inputs are distinct because the two sets are disjoint.

Consequently the quantum query complexity isFor large $N$, rounding the optimal Grover search algorithm iteration count gives success at least $1-m/(N-m)>1/2$. An internal table function collision already gives certain success. The balance $m\asymp \sqrt{N/m}$ explains the cube-root choice. This is the Brassard–Høyer–Tapp collision algorithm: its table uses $O(m\log N)$ classical bits. The bound counts calls to $U_g$, as requested; it does not charge table lookup and its reversible implementation as constant-time gates, or claim an overall polynomial time in $\log N$.

Label the measured vertex of the four-cycle by $0$, its two neighbours by $1,3$, and the opposite vertex by $2$. The edges are $01,12,23,30$. By the definition of a graph state,All Controlled-Z gates commute. Conditioning on the computational basis value $r$ of vertex $0$, the factors on edges $01$ and $03$ act on their other endpoints as $Z_1^r$ and $Z_3^r$, since $E|r,a⟩=(-1{)}^{ra}|r,a⟩$. The remaining edges form the three-vertex path $1-2-3$:The Born rule gives probability one half for either result. Normalizing provesThis is computational-basis measurement of a graph-state vertex: remove the measured vertex and apply a known Pauli Z gate byproduct to each neighbour. The chosen labels specify precisely which factors occur; any vertex of the square can serve as vertex $0$ by relabelling.

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 map $x\mapsto -x(\mathrm{mod}K)$ is a bijection and an involution, so its permutation matrix satisfiesIn particular $S^{†}S=I$, which proves that $S$ is a unitary operator. Equivalently, it permutes the computational basis and therefore preserves every inner product.

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 $\{|+⟩,|-⟩\}$.

This is superdense coding. To send the two classical bits $(z,x)$, Alice applies $Z^z{X}^x$ to her half of the shared Bell state $|{\beta }_{00}⟩$, producing $|{\beta }_{zx}⟩$, and sends that one qubit to Bob. Bob applies $\mathrm{CX}$ followed by $H$ on the first qubit, as in part (i), and measures both qubits in the computational basis. The outcome is $zx$ with certainty, so one transmitted qubit communicates two classical bits using the previously shared quantum entanglement.

Starting from the $n$-qubit computational basis state $|0{⟩}^{\otimes n}$, apply one Hadamard gate to every qubit. Since $H|0⟩=(|0⟩+|1⟩)/\sqrt{2}$,Thus parallel Hadamard gates prepare the uniform quantum superposition.

The Hadamard gate conjugates the Pauli X gate into the Pauli Z gate:When the control qubit is $|0⟩$, all three circuits act as the identity on the target. When it is $|1⟩$, the right-hand circuit acts on the target as $HXH=Z$. It therefore agrees with the Controlled-Z gate on every computational basis state, provingConjugating both sides by $I\otimes H$ and using $H^2=I$ immediately gives the reverse identity

The three controlled-NOT gates have directions $1\to 2$, $2\to 1$, and $1\to 2$. On computational basis bits they act asBy linearity, the whole quantum circuit is the SWAP gate. Thereforeso $|\alpha ⟩=|\varphi ⟩$ and $|\beta ⟩=|\psi ⟩$.