For some one-electron system with $l=0$ and $m=0$, the functions $N_0 e^{-\sigma}$ and $N_1(2-\sigma)e^{-\sigma}$ refer respectively to the ground $(E_0)$ and first excited $(E_1)$ energy levels. If a variational wave function $N_2(3-\sigma)e^{-\sigma}$ yields an average energy $\bar{E}$, it will satisfy
(a)$\bar{E} \geq 0$
(b)$0 \leq \bar{E} \leq E_0$
(c)$\bar{E} \geq E_1$
(d)$E_o \leq \bar{E} \leq E_1$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed C; on checking, D is correct — see the explanation.
Printed C; a mixture of two eigenfunctions cannot lie above the higher of the two energies.
Explanation
The trial function can be written as (3 − σ)e^(−σ) = (2 − σ)e^(−σ) + e^(−σ), i.e. a linear combination of the ground-state eigenfunction (energy E₀) and the first-excited eigenfunction (energy E₁). For ψ = c₀ψ₀ + c₁ψ₁ (orthogonal eigenfunctions), Ē = (|c₀|²E₀ + |c₁|²E₁)/(|c₀|² + |c₁|²), a weighted average of E₀ and E₁. Hence E₀ ≤ Ē ≤ E₁ (d).