Chapter 8, Problem 119FP

(This exercise requires calculus.) In this exercise, use ideas from this chapter to develop the solution to the particle-in-a-box problem. We begin by writing the Schrödinger equation for a particle of mass m moving one dimension:
$-\left(\frac{h^2}{8{\pi }^{2}m}\right)\frac{d^2\phi }{d{x}^2}+V\left(x\right)\phi =E\phi$

The equation above is the one-dimensional version of equation (8.15). For a particle a box, there are no forces acting on the particle (except at the boundaries of box), and so the potential energy, V, of the particle is constant. Without loss of generally, we can assume that the value of V is zero in the bow
a. Show that, for a particle in a box, the equation above can be written in the form $d^2\psi /d{x}^2={\delta }^2{\psi }^2$ , where ${\delta }^2=8{\pi }^{2}mE/h^2$ .
b. Show that $\psi =A\sin \left(ax\right)$ is a solution to the equation $d^2\psi /d{x}^2=-{\delta }^2{\psi }^2$ , by differentiating V twice with respect to x.
c Following the same approach you used (b), show that $\psi =A\cos \left(ax\right)$ is also a solution to the equation $d^2\psi /d{x}^2=-a^2{\psi }^2$
d. For a particle in a box, the probability density, ${\psi }^2$ , must be zero at $x=0$ . To ensure that this is so, we must have $\psi =0$ at $x=0$ . This requirement is called a boundary condition. use this boundary condition to establish that the wave function for a particle in the box must be of the form $\psi =A\sin \left(ax\right)$ , not $\psi =A\cos \left(ax\right)$ .
e. Using the result from (d). show that the boundary condition $\psi =0$ at $x=L$ requires that $aL=n\pi$ so that the wave function may be written as $\psi =A\sin \left(n\pi xL\right)$ . (Hint: $\sin Z=0$ when z is an integer multiple of $\pi .$

f. Using the result $aL=n\pi$ from (e) and the fact that $a^2=8{\pi }^{2}mE/h^2$ , as established in (a), show that $E=n^2{h}^2/\left(8m{L}^2\right).$
g. We know for sure (the probability is 1) that the particle must be somewhere between $x=0$ and $x=L$ . Mathematically, we express this condition as ${\displaystyle {\int }_0^L{\psi }^{2}dx=1}$ . It is caned a normalization condition. using the result $\psi =A\sin \left(n\pi xL\right)$ from (e), show that the normalization condition requires that $A=\sqrt{2/L.}$ (Hint. The integral ${\displaystyle {\int }_0^L{\sin }^2\left(n\pi x/L\right)dx}$ has the value $L/2.$
Working through this problem will walk you through the basic procedure for solving a quantum mechanical problem: Writing down the Schrödinger equation for the system of interest (part a); establishing the general form of the solutions (parts b and c); and using appropriate boundary conditions and a normalization condition to determine not only the specific form of $\psi$ but also the allowed values for E (parts d-g).