Question

Derive the connection formulas at a downward-sloping turning point, and confirm equation8.50.

$\mathit{\psi }\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}\frac{D\text{'}}{\sqrt{\left|p\right(x\left)\right|}}exp\left[-\frac{1}{h}{\int }_x^{x_1}\left|p\right(x\text{'}\left)\right|dx\text{'}\right].ifx<x_1\\ \frac{2D\text{'}}{\sqrt{p\left(x\right)}}sin\left[-\frac{1}{h}{\int }_{x_1}^{x}p(x\text{'})dx\text{'}+\frac{\pi }{4}\right].ifx<x_1\end{array}$

Short answer

The downward-sloping turning point is,

$\mathrm{\psi }=\left\{\begin{array}{l}\frac{1}{\sqrt{\left|\mathrm{p}\right|}}{\mathrm{De}}^{\left(-\frac{1}{\mathrm{h}}{\int }_{\mathrm{x}}^{{\mathrm{x}}_1}\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}\right)}\mathrm{x}<{\mathrm{x}}_1\\ \frac{2\mathrm{D}}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\sin \left[{\int }_{{\mathrm{x}}_1}^{\mathrm{x}}\mathrm{p}\mathrm{dx}\text{'}/\hslash +\mathrm{\pi }/4\right]\mathrm{x}>{\mathrm{x}}_1\end{array}\right.$

Step-by-step solution

To derive the downward sloping point.

In this problem we need to derive the connection formulas at a downward-sloping turning point.

The wave function has the form:

$\psi \left(x\right)\approx \frac{1}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\left({\mathrm{Be}}^{\mathrm{i}{\int }_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/\hslash }+{\mathrm{Ce}}^{-\mathrm{i}{\int }_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/\hslash }\right).......\left(1\right)$

If the particle energy is greater than the potential (that is $\mathrm{E}>\mathrm{V}\left(\mathrm{x}\right)$ ), the momentum is:

$\mathrm{p}\left(\mathrm{x}\right)=\sqrt{2\mathrm{m}(\mathrm{E}-\mathrm{V}(\mathrm{x}\left)\right)}$

at$E<V\left(x\right)$(the tunneling case) the WKB approximation gives:

$\psi \left(x\right)\approx \frac{1}{\sqrt{\left|\mathrm{p}\right|}}\left({\mathrm{De}}^{-\frac{1}{\hslash }{\int }_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}+F{e}^{-\frac{1}{\hslash }{\int }_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}\right)$ ……………………(2)

Now in the decreasing potential case, we have $\mathrm{E}<\mathrm{V}\left(\mathrm{x}\right)\mathrm{for}\mathrm{x}<{\mathrm{x}}_1\mathrm{and}\mathrm{E}>\mathrm{V}\left(\mathrm{x}\right)$ for $x>x_1$, as before we move the turning point so that $x_1=0$, for $x>0$the integration limits of the integrals in equation (1) will be from 0 to x and the integration limits of the integrals in equation (2) will be from x to 0 , thus:

role="math" localid="1658400827653" $\mathrm{\psi }=\left\{\begin{array}{l}\frac{1}{\sqrt{\left|\mathrm{p}\right|}}\left({\mathrm{De}}^{-\frac{1}{\hslash }{\int }_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}+F{e}^{-\frac{1}{\hslash }{\int }_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}\right)\mathrm{x}<0\\ \frac{1}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\left({\mathrm{Be}}^{\mathrm{i}{\int }_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/\hslash }+{\mathrm{Ce}}^{-\mathrm{i}{\int }_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/\hslash }\right)\mathrm{x}>0\end{array}\right.$

for $x<0$we must set $F=0$to keep finite as $x\to -\infty$, since the integral for $x<0$is positive , and it is in the exponential function. Thus:

role="math" localid="1658400203138" $\mathrm{\psi }=\left\{\begin{array}{l}\frac{1}{\sqrt{\left|\mathrm{p}\right|}}\left({\mathrm{De}}^{-\frac{1}{\hslash }{\int }_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}\right)\mathrm{x}<0\\ \frac{1}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\left({\mathrm{Be}}^{\mathrm{i}{\int }_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/\hslash }+{\mathrm{Ce}}^{-\mathrm{i}{\int }_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/\hslash }\right)\mathrm{x}>0\end{array}\right..........\left(3\right)$

Assume that  x=0   and find potential.

Assume that around, the potential is:

$V\left(x\right)\approx E+V^{\text{'}}\left(0\right)x\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\left(4\right)$

The potential is decreasing, which means the derivative of the potential is negative, that is$V^{\text{'}}\left(0\right)<0$. Now we define the patching function which satisfies the Schrodinger near at the potential V(x), that is:

$$\begin{gathered} \frac{d^2{\psi }_p}{d{x}^2}=\frac{2m{V}^{\text{'}}\left(0\right)}{{\hslash }^2}x{\psi }_p \\ \frac{d^2{\psi }_p}{d{x}^2}=\frac{2m{V}^{\text{'}}\left(0\right)}{{\hslash }^2}x{\psi }_p \end{gathered}$$

We define the following variables,

$$\begin{gathered} \alpha ={\left(\frac{2m\left|V^{\text{'}}\right(0\left)\right|}{{\hslash }^2}\right)}^{1/3} \\ z=-\alpha x \end{gathered}$$

Thus,

$\frac{d^2{\psi }_y}{d{z}^2}=z{\psi }_p\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(5\right)$

Which is the Airy's equation of:

$$\begin{gathered} {\psi }_p\left(z\right)=aAi\left(z\right)+bBi\left(z\right) \\ {\psi }_p\left(z\right)=aAi(-\alpha x)+bBi(-\alpha x)\dots \dots \dots \dots \dots \dots \dots \dots .\left(6\right) \end{gathered}$$

From table 8.1 , we have:

role="math" localid="1658406200718" $$\begin{gathered} \mathrm{Ai}\left(\mathrm{z}\right)~\left\{\begin{array}{l}\frac{1}{\sqrt{\mathrm{\pi }}{(-\mathrm{z})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\sin \left[\frac{2}{3}/3{(-\mathrm{z})}^{3/2}+\frac{\mathrm{\pi }}{4}\right]\mathrm{z}\ll 0\\ \frac{1}{2\sqrt{\mathrm{\pi }}{\mathrm{z}}^{1/4}}{\mathrm{e}}^{-2{\mathrm{z}}^{3/2}/3}\mathrm{z}\gg 0\end{array}\right. \\ \mathrm{Bi}\left(\mathrm{z}\right)~\left\{\begin{array}{l}\frac{1}{\sqrt{\mathrm{\pi }}{(-\mathrm{z})}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\cos \left[\frac{2}{3}/3{(-\mathrm{z})}^{3/2}+\frac{\mathrm{\pi }}{4}\right]\mathrm{z}\ll 0\\ \frac{1}{\sqrt{\mathrm{\pi }}{\mathrm{z}}^{1/4}}{\mathrm{e}}^{-2{\mathrm{z}}^{3/2}/3}\mathrm{z}\gg 0\end{array}\right. \end{gathered}$$

To find the wave function.

For $z=-\alpha x\gg 0$, we substitute with localid="1658465194049" $Ai\left(z\right)$and $Bi(z)$for$z\gg 0$, into equation (6) we get:

$$\begin{gathered} {\mathrm{\psi }}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{2\sqrt{\mathrm{\pi }}{\mathrm{z}}^{1/4}}{\mathrm{e}}^{-2{\mathrm{z}}^{3/2/3}}+\frac{\mathrm{b}}{\sqrt{\mathrm{\pi }}{\mathrm{z}}^{1/4}}{\mathrm{e}}^{2{\mathrm{z}}^{3/2/3}} \\ {\mathrm{\psi }}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{2\sqrt{\mathrm{\pi }}(-\mathrm{\alpha x}{)}^{1/4}}{\mathrm{e}}^{-2{(-\mathrm{\alpha x})}^{3/2/3}}+\frac{\mathrm{b}}{\sqrt{\mathrm{\pi }}{\left(\mathrm{\alpha x}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{\mathrm{e}}^{2{(-\mathrm{\alpha x})}^{3/2)/3}}\dots \dots \dots \dots \dots \dots ..\left(7\right) \end{gathered}$$

Now we need to work out the wave function in (3), for $z=-\alpha x\gg 0$(or $x<0$), and then compare it with (7), so:

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\sqrt{2\mathrm{m}(\mathrm{E}-\mathrm{V}(\mathrm{x}\left)\right)} \\ \mathrm{p}\left(\mathrm{x}\right)=\sqrt{-2{\mathrm{mV}}^{\text{'}}\left(0\right)\mathrm{x}} \\ \mathrm{p}\left(\mathrm{x}\right)=\hslash \sqrt{\mathrm{x}}{\mathrm{a}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

But,

$\mathrm{\psi }\left(\mathrm{x}\right)=\frac{1}{\sqrt{\left|\mathrm{p}\right|}}{\mathrm{De}}^{-\frac{1}{\mathrm{b}}{\int }_{\mathrm{x}}^0\left|\mathrm{p}\right({\mathrm{x}}^{\text{'}}\left)\right|{\mathrm{dx}}^{\text{'}}}$

so the integral is:

$$\begin{gathered} {\int }_0^{\mathrm{x}}\left|\mathrm{p}\right({\mathrm{x}}^{\text{'}}\left)\right|{\mathrm{dx}}^{\text{'}}=\sqrt{-2{\mathrm{mV}}^{\text{'}}\left(0\right)}{\int }_{\mathrm{x}}^0\sqrt{-{\mathrm{x}}^{\text{'}}}{\mathrm{dx}}^{\text{'}} \\ {\int }_0^{\mathrm{x}}\left|\mathrm{p}\right({\mathrm{x}}^{\text{'}}\left)\right|{\mathrm{dx}}^{\text{'}}=\sqrt{2\mathrm{m}\left|{\mathrm{V}}^{\text{'}}\right(0)}\left|\frac{2}{3}\right(-\mathrm{x}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ {\int }_0^{\mathrm{x}}\left|\mathrm{p}\right({\mathrm{x}}^{\text{'}}\left)\right|{\mathrm{dx}}^{\text{'}}=\frac{2}{3}\hslash {(-\mathrm{\alpha x})}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

The wave function is therefore:

$\mathrm{\psi }\left(\mathrm{x}\right)=\frac{\mathrm{D}}{\sqrt{\hslash }{\mathrm{\alpha }}^{3/4}(-\mathrm{x}{)}^{1/4}}{\mathrm{e}}^{-2(-\mathrm{\alpha x}{)}^{3/2/3}}$ ……………………….(8)

To find the Patching function.

Tocompare (8) with (7), we see that ,

$$\begin{gathered} \frac{a}{2\sqrt{\pi }{\alpha }^{1/4}}=\frac{D}{\sqrt{\hslash }{\alpha }^{3/4}} \\ a=2\sqrt{\frac{\pi }{\hslash }}\alpha D \end{gathered}$$

The patching function is therefore:

${\psi }_p\left(x\right)=2\sqrt{\frac{\pi }{\hslash \alpha }}D{e}^{-2{(-\alpha x)}^{3/2/3}}$ ………………….(9)

we follow the same method to find the patching function for the overlap region for, so we get:

${\psi }_p\left(x\right)=\frac{a}{\sqrt{\pi }{\left(\alpha x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}sin\left[\frac{2}{3}(\alpha x{)}^{3/2}+\frac{\pi }{4}\right]$ ………………….(10)

Now for the WKB function (at$x>0$) we start form (3), as:

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\sqrt{-2{\mathrm{mV}}^{\text{'}}\left(0\right)\mathrm{x}} \\ \mathrm{p}\left(\mathrm{x}\right)=\hslash {\mathrm{\alpha }}^{3/2}\sqrt{\mathrm{x}}{\int }_0^{\mathrm{x}}\mathrm{p}\left({\mathrm{x}}^{\text{'}}\right){\mathrm{dx}}^{\text{'}} \\ \mathrm{p}\left(\mathrm{x}\right)=\frac{2}{3}\hslash (\mathrm{\alpha x}{)}^{3/2} \end{gathered}$$

Note that we did these calculations before. Equation (3) forgives:

localid="1658463172998" $\mathrm{\psi }\left(\mathrm{x}\right)=\frac{1}{\sqrt{\hslash }{\mathrm{\alpha }}^{3/4}(\mathrm{x}{)}^{1/4}}\left({\mathrm{Be}}^{2\mathrm{i}(\mathrm{\alpha x}{)}^{3/2/3}}+{\mathrm{Ce}}^{-2\mathrm{i}(\mathrm{\alpha x}{)}^{3/2/3}}\right)$ ………………..(11)

Now we need to compare this equation with (10) but before we need to transform the sine function to an exponential, as:

${\mathrm{\psi }}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{2\mathrm{i}\sqrt{\mathrm{\pi }}{\left(\mathrm{\alpha x}\right)}^{1/4}}\left[{\mathrm{e}}^{\mathrm{i}\left[\frac{2}{3}{\left(\mathrm{\alpha x}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{\mathrm{\pi }}{4}\right]}-{\mathrm{e}}^{-\mathrm{i}\left[\frac{2}{3}{\left(\mathrm{\alpha x}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{\mathrm{\pi }}{4}\right]}\right]$

To find the constants B and C.

Now we can compare this equation with equation (11) in order to find the constants B and C as:

$$\begin{gathered} \frac{\mathrm{B}}{\sqrt{\hslash }{\mathrm{\alpha }}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=\frac{{\mathrm{ae}}^{\mathrm{i\pi }/4}}{2\mathrm{i}\sqrt{\mathrm{\pi }}{\mathrm{\alpha }}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \mathrm{B}=\frac{{\mathrm{ae}}^{\mathrm{i\pi }/4}\sqrt{\hslash \mathrm{\alpha }}}{2\mathrm{i}\sqrt{\mathrm{\pi }}} \\ \mathrm{B}=-{\mathrm{ie}}^{\mathrm{i\pi }/4}\mathrm{D} \end{gathered}$$

And,

$$\begin{gathered} \frac{\mathrm{C}}{\sqrt{\hslash }{\mathrm{\alpha }}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=-\frac{{\mathrm{ae}}^{-\mathrm{i\pi }/4}}{2\mathrm{i}\sqrt{\mathrm{\pi }}{\mathrm{\alpha }}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \mathrm{C}=-\frac{{\mathrm{ae}}^{-\mathrm{i\pi }/4}\sqrt{\hslash \mathrm{\alpha }}}{2\mathrm{i}\sqrt{\mathrm{\pi }}} \\ \mathrm{C}={\mathrm{ie}}^{-\mathrm{i\pi }/4}\mathrm{D} \end{gathered}$$

For $x>0$, the final form of the WKB function can be determined by substituting with these constants in the wave function of equation (3), for $x>0$, that is:

localid="1658465216214" $$\begin{gathered} {\psi }_{WKB}=\frac{1}{\sqrt{p\left(x\right)}}\left(-i{e}^{i\pi /4}D{e}^{i{\int }_0^{x}p\left(x^{\text{'}}\right)d{x}^{\text{'}}/\hslash }+i{e}^{-i\pi /4}D{e}^{-i{\int }_0^{x}p\left(x^{\text{'}}\right)d{x}^{\text{'}}/\hslash }\right) \\ {\psi }_{WKB}=\frac{-iD}{\sqrt{p\left(x\right)}}{e}^{i\left(\int pdx/\hslash +\pi /4\right)}-e^{-i\left(\int pdx/\hslash +\pi /4\right)} \\ {\psi }_{WKB}=\frac{2D}{\sqrt{p\left(x\right)}}\frac{1}{2i}(e^{i\left(\int pdx/\hslash +\pi /4\right)-e^{-i\left(\int pdx/\hslash +\pi /4\right)}} \\ {\psi }_{WKB}=\frac{2D}{\sqrt{p\left(x\right)}}sin\left[{\int }_0^{x}pd{x}^{\text{'}}/\hslash +\pi /4\right] \end{gathered}$$

To find the sloping point.

So, the equation (3) becomes:

$\psi =\left\{\begin{array}{l}\frac{1}{\sqrt{p\left(x\right)}}\left(D{e}^{-\frac{1}{\sout{h}}{\int }_x^0\left|p\left(x^{\text{'}}\right)\right|d{x}^{\text{'}}}\right)x<0\\ \frac{2D}{\sqrt{p\left(x\right)}}sin\left[{\int }_0^{x}pd{x}^{\text{'}}/\hslash +\pi /4\right]x>0\end{array}\right.$

By switching the origin back to${\mathrm{x}}_1$, just by replacing the 0 in the limits of the integrals by ${\mathrm{x}}_1$, so we get:

.$\mathrm{\psi }=\left\{\begin{array}{l}\frac{1}{\sqrt{\left|\mathrm{p}\left(\mathrm{x}\right)\right|}}\left({\mathrm{De}}^{-\frac{1}{\sout{\mathrm{h}}}{\int }_{\mathrm{x}}^{{\mathrm{x}}_1}\left|\mathrm{p}\left({\mathrm{x}}^{\text{'}}\right)\right|{\mathrm{dx}}^{\text{'}}}\right)\mathrm{x}<{\mathrm{x}}_1\\ \frac{2\mathrm{D}}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\sin \left[{\int }_{{\mathrm{x}}_1}^{\mathrm{x}}{\mathrm{pdx}}^{\text{'}}/\hslash +\mathrm{\pi }/4\right]\mathrm{x}>{\mathrm{x}}_1\end{array}\right.$