-?? (Section 7.8) Consider the problem of a particle in finite square well of depth $U_{0}$, as described in Section 7.8. To solve this problem, use a coordinate system centered on the box, with the left edge of the well at $x=-a / 2$ and the right edge at $x=+a / 2 .$ From the symmetry of the potential, one can argue that the stationary states should have symmetric probability distributions, that is, $|\psi(x)|^{2}=|\psi(-x)|^{2}$. This implies that solutions are either symmetric and satisfy $\psi(x)=\psi(-x)$ or are antisymmetric and satisfy $\psi(x)=-\psi(-x)$. Thus, with our choice of coordinates, the solutions within the well are either of the form $\cos (k x)$ (for $n=1,3,5, \ldots)$ or of the form $\sin (k x)$
(for $n=2,4,6, \ldots)$. (a) For the case of the symmetric solutions, derive an equation relating $k$ and $\alpha$. [Hint:
Using the boundary conditions that both $\psi(x)$ and $\psi^{\prime}(x)$ are continuous at $x=-a / 2$, you can produce two equations that relate $k, \alpha$, and the arbitrary coefficients $A$ and $G$ in Section 7.8. By dividing these equations, you can eliminate the coefficients. The final equation you produce is a transcendental equation relating $E$ and $U_{0}$; it cannot be solved for $E$ using elementary methods.] (b) Show that the equation derived in (a) produces the correct stationary-state energies in the limit $U_{0} \rightarrow \infty$. (c) Using numerical techniques, find the ground-state energy $E_{1}$ of a square well of depth $U_{0}=3 E_{1} .[$ Hint $:$ A simple trialand-error search for the solution works surprisingly well.]