 Solution

Let $U(t)=e^{-itH}$. Define the second initial state by $|{\tilde{\psi }}_0⟩=U(t{)}^{†}|{\psi }_2⟩$, while $|{\psi }_0⟩=U(t{)}^{†}|{\psi }_1⟩$. The unitary time evolution preserves their inner product:Both initial states are normalized and evolve to the chosen final pair. This supplies the orthogonal partner needed for topological quantum order. The remaining requirement is local indistinguishability, which is proved in the next condition's solution.