There are mathematical solutions to the Schrödinger equation for the finite well for any energy, and in fact. They can be made smooth everywhere. Guided by A Closer Look: Solving the Finite Well. Show this as follows:

(a) Don't throw out any mathematical solutions. That is in region Il $\left(x<0\right)$, assume that $\left(C{e}^{+ax}+D{e}^{-ax}\right)$, and in region III $\left(x>L\right)$, assume that$\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathbf{\text{Fe}}}^{\mathbf{+}\mathbf{a}\mathbf{x}}\mathbf{+}{\mathbf{\text{Ge}}}^{\mathbf{-}\mathbf{a}\mathbf{x}}$. Write the smoothness conditions.

(b) In Section 5.6. the smoothness conditions were combined to eliminate $\mathit{A}\mathbf{,}\mathbf{}\mathit{B}\mathbf{}\mathbf{\text{and G}}$in favor of $\mathit{C}$. In the remaining equation. $\mathit{C}$canceled. leaving an equation involving only $\mathit{k}$and $\mathit{\alpha }$, solvable for only certain values of $\mathit{E}$. Why can't this be done here?

(c) Our solution is smooth. What is still wrong with it physically?

(d) Show that

localid="1660137122940" $$\begin{gathered} \mathit{D}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}\mathbf{B}\mathbf{-}\frac{\mathbf{k}}{\mathbf{\alpha }}\mathbf{A}\mathbf{)}\mathbf{}\mathbf{\text{and}} \\ \mathit{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{\alpha }\mathbf{L}}\mathbf{[}\left(A-B\frac{k}{\alpha }\right)\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{k}\mathbf{L}\mathbf{\right)}\mathbf{+}\left(A\frac{k}{\alpha }+B\right)\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{k}\mathbf{L}\mathbf{\right)}\mathbf{]} \end{gathered}$$

and that setting these offending coefficients to 0 reproduces quantization condition (5-22).

The finite well has two boundaries, dividing the space into three regions. The necessary condition for a wavefunction to be smooth is that the wave function and its gradient, both must be continuous at both the boundaries of the well. If the three regions are named as A, B and C, with B inside the well and A and C on the right and left side of the well. If the well has two walls , with lying between and between . Then-

At boundary 1,

$$\begin{gathered} {\mathbf{\Psi }}_{\mathbf{A}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{\Psi }}_{\mathbf{B}}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}\mathbf{\left(}{\mathbf{\Psi }}_{\mathbf{A}}\left(x\right)\mathbf{\right)}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{d}\mathbf{\left(}{\mathbf{\Psi }}_{\mathbf{B}}\left(x\right)\mathbf{\right)}}{\mathbf{dx}} \end{gathered}$$

At boundary 2,

$$\begin{gathered} {\mathbf{\Psi }}_{\mathbf{B}}\left(x\right)\mathbf{=}{\mathbf{\Psi }}_{\mathbf{C}}\left(x\right) \\ \frac{\mathbf{d}\left({\Psi }_B\left(x\right)\right)}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{d}\left({\Psi }_C\left(x\right)\right)}{\mathbf{dx}} \end{gathered}$$

In region II $x<0$, we have,$\Psi \left(x\right)=C{e}^{+\alpha x}+D{e}^{-\alpha x}$

In region III $x<L$, we have, $\Psi \left(x\right)=F{e}^{+\alpha x}+G{e}^{-\alpha x}$

The necessary condition for the wave function to be smooth is that the wave function $\left(\Psi \left(x\right)\right)$and its first derivative with respect to x, $\frac{d\left(\Psi \left(x\right)\right)}{dx}$ , both should be continuous at both the walls to the well. Thus a total of four equations that must be satisfied, for the wave function to be smooth. Let the wave function in region I $\left(0<x<L\right)$, be $\Psi \left(x\right)=A\sin kx+B\cos kx$.The boundary between region I and II is $x=0$ and the boundary between region I and III is$x=L$.

So, the smoothness conditions are given as-

For $x=0$, we have $C+D=B$ and $\alpha \left(C-D\right)=kA$

For $x=L$, we have $A\sin kL+B\cos kL=F{e}^{\alpha L}+G{e}^{-\alpha L}$and $k(AcoskL-BsinkL)=\alpha (F{e}^{\alpha L}-G{e}^{-\alpha L})$.

It is impossible to obtain six variables from four equations. Thus we cannot obtain an equation in terms of $\alpha$andk alone.

As $\left|x\right|\to \infty ,$ it is easily observable that the terms including D and F will diverge.

Using both the smoothness conditions on boundary $x=0$, we have-

$$\begin{gathered} C=B-D \\ \alpha \left(C-D\right)=kA \\ \alpha \left(B-D-D\right)=kA \\ B-2D=\frac{kA}{\alpha } \end{gathered}$$

Solving further,

$D=\frac{1}{2}\left(B-\frac{kA}{\alpha }\right)$

Adding both the equations representing the smoothness condition for boundary .

$$\begin{gathered} \left(\left(A\sin kL+B\cos kL\right)+k(AcoskL-BsinkL)=\left(F{e}^{\alpha L}+G{e}^{-\alpha L}\right) \\ +\alpha (F{e}^{\alpha L}-G{e}^{-\alpha L})\right) \\ F=\frac{1}{2}{e}^{-\alpha L}\left(\left(A-B\frac{k}{\alpha }\right)\sin kl+\left(A\frac{k}{\alpha }+B\right)\cos kl\right) \end{gathered}$$

If $B=\frac{kA}{\alpha }$, D becomes zero. Inserting in above equation, we get-

$$\begin{gathered} F=\frac{1}{2}{e}^{-\alpha L}A\left(\left(1-\frac{k^2}{{\alpha }^2}\right)\sin kl+\left(2\frac{k}{\alpha }\right)\cos kl\right) \\ 0=\frac{1}{2}{e}^{-\alpha L}A\left(\left(1-\frac{k^2}{{\alpha }^2}\right)\sin kl+\left(2\frac{k}{\alpha }\right)\cos kl\right) \end{gathered}$$

Divide the given equation by role="math" localid="1660140664085" $\sin kl$and multiply $\frac{\alpha }{k}$, this gives

.$\frac{\alpha }{k}-\frac{k}{\alpha }+2\cot kl=0$

Hence, the required equation is proved.