(a) As shown in Figure 4.16, the wave function in the region x < 0 is zero, \psi (x)=0. In the region x > 0 the Schrödinger equation for the bound state solutions, -V_{0}<E<0, is given by
\left(\frac{d^{2} }{dx^{2} }+k^{2}_{1} \right) \psi _{1} (x)=0 (0 < x < a), (4.304)
\left(\frac{d^{2} }{dx^{2} }-k^{2}_{2} \right) \psi _{2} (x)=0 (x > a), (4.305)
where k^{2}_{1}=2m(V_{0}+E)/\hbar ^{2} and k^{2}_{2}=-2mE/\hbar ^{2}. On one hand, the solution of (4.304) is oscillatory,\psi _{1} (x)=A\sin k_{1}x+B\cos k_{1}x , but since \psi _{1} (0)=0 we must have B = 0. On
the other hand, eliminating the physically unacceptable solutions which grow exponentially for large values of x, the solution of (4.305) is \psi _{2} (x)=Ce^{-k_{2}x}. Thus, the wave function is given by
\psi (x)=\begin{cases} 0, & x<0, \\ \psi _{1} (x)=A\sin k_{1}x, & 0<x<0,\\ \psi _{2} (x)=Ce^{-k_{2}x}, & x>a.\end{cases} (4.306)
(b) To determine the eigenvalues, we need to use the boundary conditions at x = a. The condition \psi _{1}(a)=\psi _{2}(a) yields
A\sin k_{1}a=Ce^{-k_{2}a}, (4.307)
while the continuity of the first derivative, \psi _{1}^{\prime } (a)=\psi _{2}^{\prime } (a), leads to
Ak_{1}\cos k_{1}a=-Ck_{2}e^{-k_{2}a}. (4.308)
Dividing (4.308) by (4.307) we obtain
k_{1}a\cot k_{1}a=-k_{2}a. (4.309)
Since k^{2}_{1}=2m(V_{0}+E)/\hbar ^{2} and k^{2}_{2}=-2mE/\hbar ^{2}, we have
(k_{1}a)^{2} +(k_{2}a)^{2}=\gamma ^{2} , (4.310)
where \gamma =\sqrt{2mV_{0}} a/\hbar .
The transcendental equations (4.309) and (4.310) can be solved graphically. As shown in Figure 4.16, the energy levels are given by the intersection of the circular curve (k_{1}a)^{2} +(k_{2}a)^{2} = \gamma ^{2} with k_{1}a\cot k_{1}a=-k_{2}a.
(c) If \pi /2<\gamma <3\pi /2 there will be only one bound state, the ground state n 1, for there is only one crossing between the curves (k_{1}a)^{2} +(k_{2}a)^{2}=\gamma ^{2} and k_{1}a\cot k_{1}a=-k_{2}a. The lowest value of V_{0} that yields a single bound state is given by the relation \gamma =\pi /2, which leads to 2ma^{2} V_{0}/\hbar ^{2}=\pi ^{2}/4 or to
V_{0}=\frac{\pi ^{2}\hbar ^{2}}{8ma^{2} }. (4.311)
Similarly, if 3\pi /2<\gamma <5\pi /2 there will be two crossings between (k_{1}a)^{2} +(k_{2}a)^{2}=\gamma ^{2} and k_{1}a\cot k_{1}a=-k_{2}a. Thus, there will be two bound states: the ground state, n = 1, and the first excited state, n = 2. The lowest value of V_{0} that yields two bound states corresponds to 2ma^{2} V_{0}/\hbar ^{2}=9\pi ^{2}/4 or to
V_{0}=\frac{9\pi ^{2}\hbar ^{2}}{8ma^{2} }. (4.312)
We may thus infer the following general result. If n\pi -\pi /2< \gamma <n\pi +\pi /2, there will be n crossings and hence n bound states:
n\pi -\frac{\pi }{2} <\frac{\sqrt{2mV_{0}} }{\hbar } a<n\pi +\frac{\pi }{2}\Longrightarrow there are n bound states. (4.313)
The lowest value of V_{0} giving n bound states is
V_{0}=\frac{\pi ^{2}\hbar ^{2}}{8ma^{2}} (2n-1)^{2}. (4.314)
