 Solution

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.