Two trial wavefunction $\phi=\mathrm{c}_{1} \mathrm{x}(\mathrm{a}-\mathrm{x})$ and $\phi_{2}=\mathrm{c}_{1} \mathrm{x}(\mathrm{a}-\mathrm{x})+\mathrm{c}_{2} \mathrm{x}^{2}(\mathrm{a}-\mathrm{x})^{2}$ give ground state energies $\mathrm{E}_{1}$ and $\mathrm{E}_{2}$, respectively, for the microscopic particle in a 1-D box by using the variation method. If the exact ground state energy is $\mathrm{E}_{0}$, the correct relationship between $\mathrm{E}_{0}, \mathrm{E}_{1}$ and $\mathrm{E}_{2}$ is:
(a)$\mathrm{E}_{0}=\mathrm{E}_{1}=\mathrm{E}_{2}$
(b)$\mathrm{E}_{0}<\mathrm{E}_{1}<\mathrm{E}_{2}$
(c)$\mathrm{E}_{0}<\mathrm{E}_{2}<\mathrm{E}_{1}$
(d)$\mathrm{E}_{0}>\mathrm{E}_{2}=\mathrm{E}_{1}$
Answer
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