U = U_{T} \dots U_{1}

, and let

V_{0} = I

V_{t} = U_{t} \dots 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 from

Q = \sum_{a \in A_{0}} ∣1 ⟩ ⟨ 1 ∣_{a} + \sum_{a \in A_{+}} ∣ - ⟩ ⟨ - ∣_{a} .

(1)

It annihilates precisely the correctly initialized ancilla qubits, leaving the quantum witness unrestricted. The Feynman-Kitaev Hamiltonian without output penalty is

H = H_{in} + H_{prop}, H_{in} = Q \otimes ∣0 ⟩ ⟨ 0∣,

(2)

H_{prop} = \frac{1}{2} \sum_{t = 1}^{T} [I \otimes (∣ t ⟩ ⟨ t ∣ + ∣ t - 1 ⟩ ⟨ t - 1∣) - U_{t} \otimes ∣ t ⟩ ⟨ t - 1∣ - U_{t}^{†} \otimes ∣ t - 1 ⟩ ⟨ t ∣] .

(3)

Each propagation summand is positive: on vectors with adjacent clock components

η_{t - 1}, η_{t}

, its quadratic form is

\frac{1}{2} ∥ η_{t} - U_{t} η_{t - 1} ∥^{2}

. The input penalty of a history Hamiltonian is also a positive semidefinite operator, so

H \geq 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} (1 - 1 - 1 1)

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.