Question

It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant $\mathbf{C}$ . First, use the continuity/smoothness conditions to eliminate $\mathbf{A}$, $\mathbf{B}$ , and$\mathbf{G}$ in favor of $\mathbf{C}$in (21). Then make the change of variables $\mathbf{z}\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{L}\mathbf{/}\mathbf{2}$ and use the trigonometric relations

$\mathbf{sin}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{sinacosb}\mathbf{+}\mathbf{cosasinb}$and

$\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{cosacosb}\mathbf{-}\mathbf{sinasinb}$on the

functions in region I, $\mathbf{-}\mathbf{L}\mathbf{/}\mathbf{2}\mathbf{<}\mathbf{z}\mathbf{<}\mathbf{L}\mathbf{/}\mathbf{2}$. The change of variables shifts the problem so that it is symmetric about $\mathit{z}\mathbf{=}\mathbf{0}$, which requires that the probability density be symmetric and thus that $\mathbf{\psi }\mathbf{\left(}\mathbf{z}\mathbf{\right)}$be either an odd or even function of $\mathit{z}$. By comparing the region II and region III functions, argue that this in turn demands that $\mathbf{(}\mathbf{\alpha }\mathbf{/}\mathbf{k}\mathbf{)}\mathbf{sinkL}\mathbf{+}\mathbf{coskL}$ must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as $\frac{\mathbf{\alpha }}{\mathbf{k}}\mathbf{=}\mathbf{tan}\frac{\mathbf{kL}}{\mathbf{2}}$ and $\frac{\mathbf{\alpha }}{\mathbf{k}}\mathbf{=}\mathbf{-}\mathbf{cot}\frac{\mathbf{kL}}{\mathbf{2}}$. Finally, plug these separately back into the region I solutions and show that

$\mathbf{\psi }\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{C}\mathbf{\times }\{\begin{array}{l}\begin{array}{c}{e}^{\alpha (z+L/2)}z<L/2\\ \frac{\mathrm{coskz}}{\cos \frac{\mathrm{kL}}{2}}-L/2<z<L/2\\ e^{-\alpha (z-L/2)}z>L/2\end{array}\\ \end{array}$

or

$\mathbf{\psi }\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{C}\mathbf{\times }\mathbf{}\{\begin{array}{c}{e}^{\alpha (z+L/2)}z<L/2\\ -\frac{\mathrm{sinkz}}{\sin \frac{\mathrm{kL}}{2}}-L/2<z<L/2\\ e^{-\alpha (z-L/2)}z>L/2\end{array}$

Note that $\mathbf{C}$is now a standard multiplicative normalization constant. Setting the integral of ${\mathbf{\left|}\mathbf{\psi }\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{\right|}}^{\mathbf{2}}$ over all space to 1 would give it in terms of $\mathbf{k}$and $\mathbf{\alpha }$ , but because we can’t solve (22) exactly for k(or E), neither can we obtain an exact value for $\mathbf{C}$.

Short answer

It is shown that the solution to the finite well potential is

$\mathbf{\psi }\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{C}\mathbf{\times }\mathbf{}\{\begin{array}{c}{e}^{\alpha (z+L/2)}z<L/2\\ \frac{\mathrm{coskz}}{\cos \frac{\mathrm{kL}}{2}}-L/2<z<L/2\\ e^{-\alpha (z-L/2)}z>L/2\end{array}$

when the solution is odd and

$\mathbf{\psi }\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{=}\mathbf{C}\mathbf{\times }\mathbf{}\{\begin{array}{c}{e}^{\alpha (z+L/2)}z<L/2\\ -\frac{\mathrm{sinkz}}{\sin \frac{\mathrm{kL}}{2}}-L/2<z<L/2\\ e^{-\alpha (z-L/2)}z>L/2\end{array}$

when the solution is even.

Step-by-step solution

Finite well potential solution

The solution for the finite well potential of width $L$ is

$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{}\{\begin{array}{c}{\mathrm{Ce}}^{\mathrm{\alpha x}}x<0\\ \mathrm{Asinkx}+\mathrm{Bcoskx}0<x<L\\ {\mathrm{Ge}}^{-\mathrm{\alpha x}}x>L\end{array}$ .....(I)

Determining the solution in terms of one constant

The function in equation (I) should be continuous at$\mathrm{x}=0$, that is

$\begin{array}{c}{\mathrm{Ce}}^0=\mathrm{Asin}0+\mathrm{Bcos}0\\ \mathrm{C}=\mathrm{B}\end{array}$.

The derivative of the function should be continuous at $x=0$, that is

$\begin{array}{c}{\mathrm{\alpha Ce}}^0=\mathrm{kAcos}0-\mathrm{kBsin}0\\ \mathrm{A}=\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{C}\end{array}$

The function should be continuous at $\mathrm{x}=\mathrm{L}$, that is

$\begin{array}{c}\mathrm{AsinkL}+\mathrm{BcoskL}={\mathrm{Ge}}^{-\mathrm{\alpha L}}\\ \frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{CsinkL}+\mathrm{CcoskL}={\mathrm{Ge}}^{-\mathrm{\alpha L}}\\ \mathrm{G}=\left(\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{CsinkL}+\mathrm{CcoskL}\right){\mathrm{e}}^{\mathrm{\alpha L}}\end{array}$

Substitute these in equation (I) to get

$\psi \left(x\right)=C\left\{\begin{array}{c}{e}^{\alpha x}x<0\\ \frac{\alpha }{k}\sin kx+\cos kx0<x<L\\ \left(\frac{\alpha }{k}\sin kL+\cos kL\right)e^{\alpha L}{e}^{-\alpha x}x>L\end{array}\right.$

Substitute $\mathrm{z}=\mathrm{x}-\mathrm{L}/2$ to get

$\mathrm{\psi }\left(\mathrm{z}\right)=\mathrm{C}\left\{\begin{array}{c}{\mathrm{e}}^{\mathrm{\alpha }(\mathrm{z}+\mathrm{L}/2)}\mathrm{x}<-\mathrm{L}/2\\ \frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{sink}(\mathrm{z}+\mathrm{L}/2)+\mathrm{cosk}(\mathrm{z}+\mathrm{L}/2)-\mathrm{L}/2<\mathrm{x}<\mathrm{L}/2\\ (\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{sinkL}+\mathrm{coskL}){\mathrm{e}}^{-\mathrm{\alpha }(\mathrm{z}-\mathrm{L}/2)}\mathrm{x}>\mathrm{L}\end{array}\right.$

Apply trigonometric identities to get

$\mathrm{\psi }\left(\mathrm{z}\right)=\mathrm{C}\left\{\begin{array}{c}{\mathrm{e}}^{\mathrm{\alpha }(\mathrm{z}+\mathrm{L}/2)}\mathrm{x}<-\mathrm{L}/2\\ \frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{sinkzcos}\frac{\mathrm{L}}{2}+\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{coskzsin}\frac{\mathrm{L}}{2}+\mathrm{coskzcos}\frac{\mathrm{L}}{2}-\mathrm{sinkzsin}\frac{\mathrm{L}}{2}-\mathrm{L}/2<\mathrm{x}<\mathrm{L}/2\\ \left(\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{sinkL}+\mathrm{coskL}\right){\mathrm{e}}^{-\mathrm{\alpha }(\mathrm{z}-\mathrm{L}/2)}\mathrm{x}>\mathrm{L}\end{array}\right.$

For $\psi \left(z\right)$ to be even the coefficient of the third function has to be 1, that is

$\begin{array}{c}\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{sinkL}+\mathrm{coskL}=1\\ \frac{\mathrm{\alpha }}{\mathrm{k}}=\mathrm{csckL}-\mathrm{cotkL}\\ =\tan \frac{\mathrm{kL}}{2}\end{array}$

Substitute this in the second function to get

$\begin{array}{c}\tan \frac{\mathrm{kL}}{2}\mathrm{sinkzcos}\frac{\mathrm{L}}{2}+\tan \frac{\mathrm{kL}}{2}\mathrm{coskzsin}\frac{\mathrm{L}}{2}+\mathrm{coskzcos}\frac{\mathrm{L}}{2}-\mathrm{sinkzsin}\frac{\mathrm{L}}{2}\\ =\frac{\mathrm{coskz}}{\cos \frac{\mathrm{kL}}{2}}\end{array}$

For $\mathrm{\psi }\left(\mathrm{z}\right)$ to be odd the coefficient of the third function has to be -1, that is

$\begin{array}{c}\frac{\mathrm{\alpha }}{\mathrm{k}}\mathrm{sinkL}+\mathrm{coskL}=-1\\ \frac{\mathrm{\alpha }}{\mathrm{k}}=-\mathrm{csckL}-\mathrm{cotkL}\\ =-\cot \frac{\mathrm{kL}}{2}\end{array}$

Substitute this in the second function to get

$\begin{array}{l}-\cot \frac{\mathrm{kL}}{2}\mathrm{sinkzcos}\frac{\mathrm{L}}{2}-\cot \frac{\mathrm{kL}}{2}\mathrm{coskzsin}\frac{\mathrm{L}}{2}+\mathrm{coskzcos}\frac{\mathrm{L}}{2}-\mathrm{sinkzsin}\frac{\mathrm{L}}{2}\\ =-\frac{\mathrm{sinkz}}{\sin \frac{\mathrm{kL}}{2}}\end{array}$

Thus the function becomes

$\mathrm{\psi }\left(\mathrm{z}\right)=\mathrm{C}\times \left\{\begin{array}{c}{\mathrm{e}}^{\mathrm{\alpha }(\mathrm{z}+\mathrm{L}/2)}\mathrm{z}<\mathrm{L}/2\\ \frac{\mathrm{coskz}}{\cos \frac{\mathrm{kL}}{2}}-\mathrm{L}/2<\mathrm{z}<\mathrm{L}/2\\ {\mathrm{e}}^{-\mathrm{\alpha }(\mathrm{z}-\mathrm{L}/2)}\mathrm{z}>\mathrm{L}/2\end{array}\right.$

for the even case and

$\mathrm{\psi }\left(\mathrm{z}\right)=\mathrm{C}\times \left\{\begin{array}{c}{\mathrm{e}}^{\mathrm{\alpha }(\mathrm{z}+\mathrm{L}/2)}\mathrm{z}<\mathrm{L}/2\\ -\frac{\mathrm{sinkz}}{\sin \frac{\mathrm{kL}}{2}}-\mathrm{L}/2<\mathrm{z}<\mathrm{L}/2\\ {\mathrm{e}}^{-\mathrm{\alpha }(\mathrm{z}-\mathrm{L}/2)}\mathrm{z}>\mathrm{L}/2\end{array}\right.$

for the odd case.