 Solution

Normalize the trial quantum state as $(|{\varphi }_0⟩+c|{\varphi }_1⟩)/\sqrt{1+c^2}$. With the phases used above, the off-diagonal position matrix element is $x_{01}=-32a/(9{\pi }^2)$. DefineThe diagonal perturbation matrix elements vanish by parity, so the Rayleigh quotient isThe stationarity equation is $W+\Delta c-W{c}^2=0$. The minimum continuously connected to the unperturbed ground energy eigenstate hasEquivalently, diagonalize the Hamiltonian operator matrix in this two-dimensional trial space. Its lower eigenvalue isThereforeThis retains exactly the $k=1$ term of the full ground-state second-order nondegenerate perturbation theory sum. Every omitted term is negative, so the full quadratic correction is more negative than this restricted one. This is consistent with the Rayleigh-Ritz variational principle: minimizing over a two-state subspace gives an upper bound on the exact ground energy. The bound applies to the unexpanded two-state minimum; a truncated series is to be compared order by order.