Solution
= Solution
Write the normalized trial state as $\sqrt{1-\epsilon^2}|0\rangle+\epsilon|\eta\rangle$ with $\eta\perp0$. The cross term vanishes because $H|0\rangle=E_0|0\rangle$, so the energy error is $\epsilon^2(\langle\eta|H|\eta\rangle-E_0)=O(\epsilon^2)$.