Two trial wavefunctions $\phi = c_1 x(a-x)$ and $\phi_2 = c_1 x(a-x) + c_2 x^{2}(a-x)^{2}$ give ground state energies $E_1$ and $E_2$, respectively, for the microscopic particle in a 1-D box by using the variation method. If the exact ground state energy is $E_0$, the correct relationship between $E_0$, $E_1$ and $E_2$ is:
(a)$E_0 = E_1 = E_2$
(b)$E_0 < E_1 < E_2$
(c)$E_0 < E_2 < E_1$
(d)$E_0 > E_2 = E_1$
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
The variation theorem bounds every trial energy from above; the two-parameter function φ₂ contains φ₁ as a special case, so E₀ < E₂ < E₁ (c).