In linear variation method using two orthogonal basis functions, the two roots obtained are $\epsilon_0$ and $\epsilon_1$ ($\epsilon_0 < \epsilon_1$). The correct relation of these with exact ground and first excited state energies, $E_0$ and $E_1$, respectively, is
(a)$\epsilon_0 \geq E_0$ and $\epsilon_1 < E_1$
(b)$\epsilon_0 < E_1$ and $\epsilon_1 \geq E_1$
(c)$\epsilon_0 < E_0$ and $\epsilon_1 < E_1$
(d)$\epsilon_0 \geq E_0$ and $\epsilon_1 \geq E_1$
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
By the Hylleraas–Undheim–MacDonald theorem each root of a linear variation calculation is an upper bound to the corresponding exact level: ε₀ ≥ E₀ and ε₁ ≥ E₁.