Apply the variational method to estimate the ground state energy of a particle confined in a one-dimensional box for which $V=0$ for $-a<x<a$, and $\psi( \pm a)=0$.
(a) First, use an unnormalized trapezoidal trial function which vanishes at $\pm a$ and is symmetric with respect to the center of the well:
$$
\psi_t(x)= \begin{cases}(a-|x|) & b \leq|x| \leq a \\ a-b & |x| \leq b\end{cases}
$$
Try the choice $b=0$ (triangular trial function) and then improve on this by optimizing the parameter $b$.
(b) A more sophisticated trial function is parabolic, again vanishing at the endpoints and even in $x$.
(c) Use a quartic trial function of the form
$$
\psi_i(x)=\left(a^2-x^2\right)\left(\alpha x^2+\beta\right)
$$
where the ratio of the adjustable parameters $\alpha$ and $\beta$ is determined variationally.
(d) Compare the results of the different variational calculations with the exact ground state energy, and, using normalized wave functions, evaluate the meansquare deviation $\int_{-a}^a\left|\psi(x)-\psi_t(x)\right|^2 d x$ for the various cases.
(e) Show that the variational procedure produces, in addition to the approximation to the ground state, an optimal quartic trial function with nodes between the endpoints. Interpret the corresponding stationary energy value. ${ }^7$