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A Feynman-Kitaev Hamiltonian penalizes incorrect input initialization, disagreement between successive quantum circuit steps and clock labels, and optionally a rejecting output. Its propagation quadratic form is a sum of $\frac{1}{2}\Vert {\eta }_t-U_t{\eta }_{t-1}{\Vert }^2$. Without output penalty, its zero-energy space consists of correctly initialized computational history states. A nonlocal quantum clock makes the propagation formula simple but does not supply fixed qubit locality by itself.

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.

Use the unitary change of basis from part (b). It reduces the Feynman-Kitaev Hamiltonian to $A+B$, whereThe positive eigenvalues of $Q$ are positive integers, since its commuting ancilla projectors act on different qubits. The positive spectral gap of $B$ is $1-\cos (\pi /(T+1))\ge 2/(T+1{)}^2$. Thus both positive spectra are bounded below by $\mu \ge 2/(T+1{)}^2$, for $T\ge 1$.

Let $|u⟩=(T+1{)}^{-1/2}{\sum }_t|t⟩$, and split the work space into $G=\ker Q$ and $G^{\perp }$. The common ground space is $K=G\otimes |u⟩$. A unit vector in $\ker B$ orthogonal to $K$ has the form $|z⟩|u⟩$, where $z\in G^{\perp }$. Its orthogonal projection onto $\ker A$ simply removes its time-zero component, so the projected norm is $\sqrt{T/(T+1)}$. Consequently the smallest angle between two subspaces, after removing their common intersection, satisfiesThe Kitaev geometrical lemma now givesHere $1-\sqrt{1-x}\ge x/2$ supplies the penultimate step. HenceIf there are no input constraints, $Q=0$ and the propagation gap is already $\Omega (T^{-2})$, which is stronger.

The printed geometric-lemma notation needs a correction: the maximum overlap defines $\cos \vartheta$, not $\vartheta$, and is taken over normalized vectors in the two kernels, with the common ground space removed. The ground space restriction is essential when many quantum witnesses are allowed.