In this problem, we will consider a particle in a finite potential well. whose
mathematical form is $$\begin{aligned}
V_{0} & & x<-a \\
V(x)=0 & &-aa
\end{aligned}$$
Note that this potential describes what we have called a "particle in a box"
if \(V_{0} \rightarrow \infty\) Show that if \(0a
\end{array}$$
where
$$k_{1}=\left(\frac{2 m\left(V_{0}-E\right)}{\hbar^{2}}\right)^{1 / 2} \quad
\text { and } \quad \alpha=\left(\frac{2 m E}{\hbar^{2}}\right)^{1 / 2}$$
Now apply the conditions that \(\psi(x)\) and \(d \psi / d x\) must be continuous
at \(x=-a\) and \(x=a\) to obtain
$$\begin{array}{c}
A e^{-k_{1} a}=-B \sin \alpha a+C \cos \alpha a \\
D e^{-k_{1} a}=B \sin \alpha a+C \cos \alpha a \\
k_{1} A e^{-k_{1} a}=\alpha B \cos \alpha a+\alpha C \sin \alpha a
\end{array}$$
and $$-k_{1} D e^{-k_{1} a}=\alpha B \cos \alpha a-\alpha C \sin \alpha a$$
Add and subtract Equations 4 and 5 and add and subtract Equations 6 and 7 to
obtain $$\begin{array}{c}
2 C \cos \alpha a=(A+D) e^{-k_{1} a} \\
2 B \sin \alpha a=(D-A) e^{-k_{1} a} \\
2 \alpha C \sin \alpha a=k_{1}(A+D) e^{-k_{1} a}
\end{array}$$ and $$2 \alpha B \cos \alpha a=-k_{1}(D-A) e^{-k_{1} a}$$ Now
divide Equation 10 by Equation 8 to get $$\frac{\alpha \sin \alpha a}{\cos
\alpha a}=\alpha \tan \alpha a=k_{1} \quad(D \neq-A \text { and } C \neq 0)$$
and then divide Equation 11 by Equation 9 to get $$\frac{\alpha \cos \alpha
a}{\sin \alpha a}=\alpha \cot \alpha a=-k_{1} \quad \text { and } \quad(D \neq
A \text { and } B \neq 0)$$ Referring back to Equation \(3,\) note that
Equations 12 and 13 give the allowed values of \(E\) in terms of \(V_{0} .\) It
turns out that these two equations cannot be solved simultaneously, so we have
two sets of equations $$\alpha \tan \alpha a=k_{1}$$ and $$\alpha \cot \alpha
a=-k_{1}$$ Let's consider Equation 14 first. Multiply both sides by \(a\) and
use the definitions of \(\alpha\) and \(k_{1}\) to get $$\left(\frac{2 m a^{2}
E}{\hbar^{2}}\right)^{1 / 2} \tan \left(\frac{2 m a^{2}
E}{\hbar^{2}}\right)^{1 / 2}=\left[\frac{2 m
a^{2}}{\hbar^{2}}\left(V_{0}-E\right)\right]^{1 / 2}$$ Show that this equation
simplifies to $$\varepsilon^{1 / 2} \tan \varepsilon^{1 /
2}=\left(v_{0}-\varepsilon\right)^{1 / 2}$$ where \(\left.\varepsilon=2 m a^{2}
E / \hbar^{2} \text { and } v_{0}=2 m a^{2} V_{0} / \hbar^{2} . \text { Thus,
if we fix } v_{0} \text { (actually } 2 m a^{2} V_{0} / \hbar^{2}\right)\) then
we can use Equation 17 to solve for the allowed values of \(\varepsilon\)
(actually \(2 m a^{2} E / \hbar^{2}\) ). Equation 17 cannot be solved
analytically, but if we plot both \(\varepsilon^{1 / 2} \tan \varepsilon^{1 /
2}\) and \(\left(v_{0}-\varepsilon\right)^{1 / 2}\) versus \(\varepsilon\) on the
same graph, then the solutions are given by the intersections of the two
curves. Show that the intersections occur at \(\varepsilon=2 m a^{2} E /
\hbar^{2}=1.47\) and 11.37 for \(v_{0}=12\) The other value(s) of \(\varepsilon\)
are given by the solutions to Equation \(15,\) which are obtained by finding the
intersection of \(-\varepsilon^{1 / 2} \cot \varepsilon^{1 / 2}\) and
\(\left(v_{0}-\varepsilon\right)^{1 / 2}\) plotted against
\(\varepsilon .\) Show that \(\varepsilon=2 m a^{2} E / \hbar^{2}=5.68\) for
\(v_{0}=12 .\) Thus, we see there are only three bound states for a well of
depth \(V_{0}=12 \hbar^{2} / 2 m a^{2} .\) The important point here is not the
numerical values of \(E\) but the fact that there is only a finite number of
bound states. Show that there are only two bound states for \(v_{0}=2 m a^{2}
V_{0} / \hbar^{2}=4\).
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