91影视

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

The wave function has the form:

$蠄\left(x\right)鈮�\frac{1}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\left({\mathrm{Be}}^{\mathrm{i}{鈭�}_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/鈩�}+{\mathrm{Ce}}^{-\mathrm{i}{鈭�}_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/鈩�}\right).......\left(1\right)$

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:

$蠄\left(x\right)鈮�\frac{1}{\sqrt{\left|\mathrm{p}\right|}}\left({\mathrm{De}}^{-\frac{1}{鈩�}{鈭�}_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}+F{e}^{-\frac{1}{鈩�}{鈭�}_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}\right)$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌�(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" $\mathrm{蠄}=\left\{\begin{array}{l}\frac{1}{\sqrt{\left|\mathrm{p}\right|}}\left({\mathrm{De}}^{-\frac{1}{鈩�}{鈭�}_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}+F{e}^{-\frac{1}{鈩�}{鈭�}_{\mathrm{x}}^0\left|\mathrm{p}\right(\mathrm{x}\text{'}\left)\right|\mathrm{dx}\text{'}}\right)\mathrm{x}<0\\ \frac{1}{\sqrt{\mathrm{p}\left(\mathrm{x}\right)}}\left({\mathrm{Be}}^{\mathrm{i}{鈭�}_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/鈩�}+{\mathrm{Ce}}^{-\mathrm{i}{鈭�}_0^{\mathrm{x}}\mathrm{p}(\mathrm{x}\text{'})\mathrm{dx}\text{'}/鈩�}\right)\mathrm{x}>0\end{array}\right.$

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

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

Assume that around, the potential is:

$V\left(x\right)鈮�E+V^{\text{'}}\left(0\right)x鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�.\left(4\right)$

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:

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

We define the following variables,

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

Thus,

$\frac{d^2{蠄}_y}{d{z}^2}=z{蠄}_p鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�\left(5\right)$

Which is the Airy's equation of:

$$\begin{gathered} {蠄}_p\left(z\right)=aAi\left(z\right)+bBi\left(z\right) \\ {蠄}_p\left(z\right)=aAi(-伪x)+bBi(-伪x)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�.\left(6\right) \end{gathered}$$

From table 8.1 , we have:

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

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

$$\begin{gathered} {\mathrm{蠄}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{2\sqrt{\mathrm{蟺}}{\mathrm{z}}^{1/4}}{\mathrm{e}}^{-2{\mathrm{z}}^{3/2/3}}+\frac{\mathrm{b}}{\sqrt{\mathrm{蟺}}{\mathrm{z}}^{1/4}}{\mathrm{e}}^{2{\mathrm{z}}^{3/2/3}} \\ {\mathrm{蠄}}_{\mathrm{p}}\left(\mathrm{x}\right)=\frac{\mathrm{a}}{2\sqrt{\mathrm{蟺}}(-\mathrm{伪虫}{)}^{1/4}}{\mathrm{e}}^{-2{(-\mathrm{伪虫})}^{3/2/3}}+\frac{\mathrm{b}}{\sqrt{\mathrm{蟺}}{\left(\mathrm{伪虫}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{\mathrm{e}}^{2{(-\mathrm{伪虫})}^{3/2)/3}}鈥�鈥�鈥�鈥�鈥�鈥�..\left(7\right) \end{gathered}$$

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

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

But,

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

so the integral is:

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

The wave function is therefore:

$\mathrm{蠄}\left(\mathrm{x}\right)=\frac{\mathrm{D}}{\sqrt{鈩�}{\mathrm{伪}}^{3/4}(-\mathrm{x}{)}^{1/4}}{\mathrm{e}}^{-2(-\mathrm{伪虫}{)}^{3/2/3}}$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌︹赌︹赌�.(8)

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

$$\begin{gathered} \frac{a}{2\sqrt{蟺}{伪}^{1/4}}=\frac{D}{\sqrt{鈩�}{伪}^{3/4}} \\ a=2\sqrt{\frac{蟺}{鈩�}}伪D \end{gathered}$$

The patching function is therefore:

${蠄}_p\left(x\right)=2\sqrt{\frac{蟺}{鈩�伪}}D{e}^{-2{(-伪x)}^{3/2/3}}$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌�.(9)

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

${蠄}_p\left(x\right)=\frac{a}{\sqrt{蟺}{\left(伪x\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}sin\left[\frac{2}{3}(伪x{)}^{3/2}+\frac{蟺}{4}\right]$ 鈥︹赌︹赌︹赌︹赌︹赌︹赌�.(10)

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

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)=\sqrt{-2{\mathrm{mV}}^{\text{'}}\left(0\right)\mathrm{x}} \\ \mathrm{p}\left(\mathrm{x}\right)=鈩�{\mathrm{伪}}^{3/2}\sqrt{\mathrm{x}}{鈭�}_0^{\mathrm{x}}\mathrm{p}\left({\mathrm{x}}^{\text{'}}\right){\mathrm{dx}}^{\text{'}} \\ \mathrm{p}\left(\mathrm{x}\right)=\frac{2}{3}鈩�(\mathrm{伪虫}{)}^{3/2} \end{gathered}$$

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

localid="1658463172998" $\mathrm{蠄}\left(\mathrm{x}\right)=\frac{1}{\sqrt{鈩�}{\mathrm{伪}}^{3/4}(\mathrm{x}{)}^{1/4}}\left({\mathrm{Be}}^{2\mathrm{i}(\mathrm{伪虫}{)}^{3/2/3}}+{\mathrm{Ce}}^{-2\mathrm{i}(\mathrm{伪虫}{)}^{3/2/3}}\right)$ 鈥︹赌︹赌︹赌︹赌︹赌�..(11)

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

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

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

$$\begin{gathered} \frac{\mathrm{B}}{\sqrt{鈩�}{\mathrm{伪}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=\frac{{\mathrm{ae}}^{\mathrm{颈蟺}/4}}{2\mathrm{i}\sqrt{\mathrm{蟺}}{\mathrm{伪}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \mathrm{B}=\frac{{\mathrm{ae}}^{\mathrm{颈蟺}/4}\sqrt{鈩�\mathrm{伪}}}{2\mathrm{i}\sqrt{\mathrm{蟺}}} \\ \mathrm{B}=-{\mathrm{ie}}^{\mathrm{颈蟺}/4}\mathrm{D} \end{gathered}$$

And,

$$\begin{gathered} \frac{\mathrm{C}}{\sqrt{鈩�}{\mathrm{伪}}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}=-\frac{{\mathrm{ae}}^{-\mathrm{颈蟺}/4}}{2\mathrm{i}\sqrt{\mathrm{蟺}}{\mathrm{伪}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ \mathrm{C}=-\frac{{\mathrm{ae}}^{-\mathrm{颈蟺}/4}\sqrt{鈩�\mathrm{伪}}}{2\mathrm{i}\sqrt{\mathrm{蟺}}} \\ \mathrm{C}={\mathrm{ie}}^{-\mathrm{颈蟺}/4}\mathrm{D} \end{gathered}$$

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" $$\begin{gathered} {蠄}_{WKB}=\frac{1}{\sqrt{p\left(x\right)}}\left(-i{e}^{i蟺/4}D{e}^{i{鈭�}_0^{x}p\left(x^{\text{'}}\right)d{x}^{\text{'}}/鈩�}+i{e}^{-i蟺/4}D{e}^{-i{鈭�}_0^{x}p\left(x^{\text{'}}\right)d{x}^{\text{'}}/鈩�}\right) \\ {蠄}_{WKB}=\frac{-iD}{\sqrt{p\left(x\right)}}{e}^{i\left(鈭�pdx/鈩�+蟺/4\right)}-e^{-i\left(鈭�pdx/鈩�+蟺/4\right)} \\ {蠄}_{WKB}=\frac{2D}{\sqrt{p\left(x\right)}}\frac{1}{2i}(e^{i\left(鈭�pdx/鈩�+蟺/4\right)-e^{-i\left(鈭�pdx/鈩�+蟺/4\right)}} \\ {蠄}_{WKB}=\frac{2D}{\sqrt{p\left(x\right)}}sin\left[{鈭�}_0^{x}pd{x}^{\text{'}}/鈩�+蟺/4\right] \end{gathered}$$

So, the equation (3) becomes:

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

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:

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