91影视

Write down the WKB wave functions in regions (i) x >$x_2$, (ii) $x_1>x>x_2$and (iii) $<x<x_1$. Impose the appropriate connection formulas at $x_1$and $x_2$, to show that

$$\begin{gathered} {蠄}_{WKB}\left(x\right)鈮�\left\{\begin{array}{l}\frac{D}{\sqrt{p}\overline{)\left(x\right)}}exp\frac{1}{h}{鈭�}_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\sin \frac{1}{h}{鈭�}_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx+\frac{蟺}{4}} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\left[2\cos 胃e\overline{)P}\overline{)\overline{)\left(x\text{'}\right)}}dx+\sin 胃e^{-\frac{1}{h}{鈭�}_x^{x_1}\overline{)p\overline{)\left(x\right)}dx\text{'}}}\right]\\ \end{array}\right. \end{gathered}$$

The WKB wave function takes the form,

$$\begin{gathered} {蠄}_{WKB}\left(x\right)鈮�\left\{\begin{array}{l}\frac{D}{\sqrt{p}\overline{)\left(x\right)}}exp\frac{1}{h}{鈭�}_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\sin \frac{1}{h}{鈭�}_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx+\frac{蟺}{4}}\\ \end{array}\right. \end{gathered}$$

To do the same thing for the turning point , we will follow the same procedure. But first let's rewrite localid="1658393003007" ${蠄}_{WKB}$. (ii) in such a form that help us effectively join it with ${蠄}_{WKB}$. (iii) at$X_1$. Using the definition of Equating 9.59, we rewrite the WKB wave function in region (ii) as

${蠄}_{WKB}\left(x\right)鈮�\frac{2D}{\sqrt{p}\left(x\right)}\sin \left[-\frac{1}{h}{鈭�}_{x_1}^{x}p\left(x\text{'}\right)dx\text{'}胃+\frac{蟺}{4}\right]$

The next step then is to patch the WKB solutions (in regions (ii) and (iii)) at. Let us shift the axes over so that the turning pointoccurs at, then we have

$$\begin{gathered} {蠄}_{WKB}\left(x\right)鈮�\left\{\begin{array}{l}\frac{-2D}{\sqrt{p}\overline{)\left(x\right)}}\sin \left[\frac{1}{h}{鈭�}_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx-胃-\frac{蟺}{4}}\right],x>0 \\ \frac{1}{\sqrt{p}\overline{)\left(x\right)}}F{e}^{\frac{1}{n}{鈭�}_x^0\overline{)P\overline{)\left(x\text{'}\right)}dx}}+G{e}^{-\frac{1}{h}{鈭�}_x^{x_1}\overline{)p\overline{)\left(x\right)}dx\text{'}}}X<0\\ \end{array}\right. \end{gathered}$$ 鈥︹�︹�︹�︹�� ()

The turning points$x_1$has a downward slope, so the patching wave function, in this case, assumes the form

${蠄}_p\left(x\right)=aAi\left(-ax\right)+bBi\left(-ax\right)$鈥︹�︹�︹�︹�︹�︹�︹�� ( 2)

Where $a={\left[2mV\left(0\right)/h^2\right]}^{1/3}$, and also, the momentum, in this case, takes the following form:

$p\left(x\right)=h{a}^{3/2}\sqrt{x}$

.

For,

${鈭�}_x^0\overline{)p\left(x\text{'}\right)\overline{)d}x\text{'}鈮�褯a^{3/2}{鈭�}_x^0\sqrt{-x\text{'}dx\text{'}}}=\frac{2}{3}褯{\left(-ax\right)}^{3/2}$

And therefore the WKB wave function (Equation (1)) can be written as

${蠄}_{WKB}\left(x\right)鈮�\frac{1}{h{a}^{3/4}{\left(-x\right)}^{1/4}}F{e}^{\frac{2}{3}}{\left(-ax\right)}^{3/2}+Ge-\frac{2}{3}{(-ax)}^{3/2}$

Meanwhile, using the large- z asymptotic form of the Airy function for large positive z (remember that$\left(z=-ax\right)$), the patching wave function (Equation (2)), for x<0, becomes

${蠄}_p\left(x\right)鈮�\frac{1}{蟺{\left(-ax\right)}^{1/4}}\frac{a}{2}{e}^{-\frac{2}{3}{\left(-ax\right)}^{3/2}}+b{e}^{\frac{2}{3}\left(-ax\right)3/2}$

Comparing the WKB and patching wave functions in the region where x<0 , we find

$a=2\frac{蟺}{一a}G$and $b=\sqrt{\frac{蟺}{褯a}F}$ 鈥︹�︹�︹�︹�︹�︹��(3 )

For x>0 , we have

$$\begin{gathered} {鈭�}_0^{x}P\left(\text{'}x\right)dx\text{'}魔a^{3/2}{鈭�}_0^x\sqrt{x} \\ {鈭�}_0^{x}P\left(\text{'}x\right)dx=\frac{2}{3}魔{\left(ax\right)}^{3/2} \end{gathered}$$

and therefore the WKB wave function (Equation ()) can be written as$$\begin{gathered} {蠄}_{WKB}\left(X\right)鈮�\frac{-2D}{魔a^{3/4}{x}^{1/4}}\sin \frac{2}{3}{\left(ax\right)}^{3/2}-胃-\frac{蟺}{4} \\ {蠄}_{WKB}\left(X\right)鈮�\frac{-2D}{魔a^{3/4}{x}^{1/4}}\cos \frac{2}{3}{\left(ax\right)}^{3/2}+\frac{蟺}{4}-胃 \end{gathered}$$

Let $尾鈮�\left(2/3\right){\left(ax\right)}^{3/2}+蟺/4$, then using elementary trigonometry,

${蠄}_{WKB}\left(X\right)鈮�\frac{2D}{魔a^{3/4}{x}^{1/4}}\left[\cos 尾\cos 胃+\sin 尾\sin 胃\right]$

Meanwhile, using the large-z asymptotic form of the Airy function for large negative z (remember that z=-ax ), the patching wave function (Equation (2)) x>0 , for becomes

localid="1658823967085" ${蠄}_p\left(x\right)鈮�\frac{1}{\sqrt{蟺}{\left(ax\right)}^{1/4}}\left[a\sin 尾+b\cos 尾\right]$

(In terms of$尾$). Comparing the WKB and patching wave functions in the region where x>0, we find

$\frac{a}{\sqrt{蟺}}=\frac{2D}{魔a}\sin 胃$And $\frac{b}{\sqrt{蟺}}=\frac{2D}{\sqrt{ha}}\cos 胃$

Or, putting in Equation (3) for and :

$G=D\sin 胃$And $F=2D\cos 胃$

These are the connection formulas, joining the WKB solutions at either side of the turning point . Expressing everything in terms of the one normalization constant D , and shifting the turning point back from the origin to an arbitrary point $x_1$, we show that the WKB wave function in the three regions (i), (ii) and (iii) is

localid="1658824119486" $$\begin{gathered} {蠄}_{WKB}\left(x\right)鈮�\left\{\begin{array}{l}\frac{D}{\sqrt{p}\left(x\right)}exp-\frac{1}{h}{鈭�}_{x_2}^x\overline{)p\overline{)\left(x\right)}dx\text{'}} \\ \frac{2D}{\sqrt{p}\left(x\right)}\sin \frac{1}{h}{鈭�}_x^{x_2}p\left(x\text{'}\right)dx+\frac{蟺}{4} \\ \frac{D}{p\left(x\right)I}\left[2\cos 胃e^{\frac{1}{r}{鈭�}_x^{x_1}p\left(x\text{'}\right)dx}+\sin 胃e^{\frac{1}{r}{鈭�}_x^{x_1}\overline{)p}\left(x\text{'}\right)dx\text{'}}\right]\\ \end{array}\right. \end{gathered}$$ 鈥︹赌︹赌︹赌︹赌�(4)

Where,

$胃=1/魔{鈭�}_{x_1}^{x_2}p\left(x\right)dx$

Let's start by the easy case; the oddwave functions. It requires$蠄\left(0\right)=0$.

(iii) (Equation (4)), we have

localid="1658402848392" $\sin 胃e^{-\frac{1}{r}{鈭�}_0^{x_1}\overline{)p\overline{)\left(x\right)}dx\text{'}}}=-2\cos {胃}^{\frac{1}{r}{鈭�}_0^{x_1}\overline{)p\overline{)\left(x\text{'}\right)}dx}}$

(Assuming$D鈮�0$), which leads to

$\tan 胃=-2e^{\frac{1}{r}{鈭�}_0^{x_1}\overline{)p\overline{)\left(x\text{'}\right)}dx}}$

$\tan 胃=-2e^{\frac{1}{r}{鈭�}_0^{x_1}\overline{)p\overline{)\left(x\text{'}\right)}dx}}$

Where the second equality is valid by the virtue of symmetry of v(x) (remember that ) $Ip\left(x\right)i=\sqrt{2mIE-v\left(x\right)I}$. Introducing whose definition is given by Equation 9.61, we can write

$\tan 胃=-2e^{蠒}$

Consider the case of even (+) wave functions. This requires$蠄\left(0\right)=0$. The derivative of${蠄}_{WKB}$(region (iii) in Equation (4)) is

$$\begin{gathered} {蠄}_{WKB}\left(x\right)=-\frac{1}{2Ip\left(x\right)I}\frac{dIP\left(x\right)I}{dx}{蠄}_{WKB}\left(x\right) \\ +\frac{D}{\sqrt{Ip\left(x\right)I}}2\cos 胃e^{\frac{1}{r}{鈭�}_x^{x_1}\overline{)p\left(x;\right)dx}}\left(-\frac{1}{h}Ip\left(x\right)I\right)+\sin 胃e^{\frac{1}{r}{鈭�}_x^{x_1}p\left(x\right)dx}\left(\frac{1}{h}Ip\left(x\right)I\right) \end{gathered}$$................. ( X)

At X=0 , the first term vanishes by the fact that

$$\begin{gathered} {\overline{)\frac{dIP\left(x\right)I}{dx}}}_{x-0}=\frac{d}{dx}{\left[\sqrt{2m\left(V\left(x\right)-E\right)}\right]}_{X-O} \\ {\overline{)\frac{dIP\left(x\right)I}{dx}}}_{x-0}=\frac{mV\text{'}\left(0\right)}{\sqrt{2mV\left(0\right)-E}}=0 \end{gathered}$$

(Indeed V(0)=0since V(x) since is symmetric). Thus ${蠄}_{WKB}\left(0\right)=0$ (Equation ( 5)) leads to

$-2\cos 胃e^{蠒/2}+\sin 胃e^{-蠒/2}=0$

$\tan 胃=2e^{蠒}$ (Or)

Putting the results together, we see that the requirements$蠄\left(0\right)=0$and$蠄\left(0\right)=0$leads to the following quantization condition:

$蠒=\left(1/h\right){鈭�}_{-x}^{x_1}\overline{)p\overline{)\left(x\text{'}\right)}dx}$ 鈥︹�︹�︹�︹�︹�︹�︹�� ()

Where,

$蠒=1/魔{鈭�}_{-x1}^{x_1}\overline{)p\left(x\text{'}\right)dx\text{'}}$.

The quantization condition (equation (6)), with$胃=\left(n+1/2\right)蟺+o$where$\left|o\right|<<1$, becomes

$\frac{tan\left(n+1/2蟺\right)+{\tan }_o}{1-\tan \left[\left(n+1/2\right)蟺\right]\tan o}=卤2e^{蠒}$

Dividing the numerator and denominator of the fraction on the left$\tan \left[\left(n+1/2\right)蟺\right]$by and realizing that$an(n+\$\$1/2)蟺鈫�鈭�$, the equation reduces to

$-\frac{1}{\tan o}鈮�\frac{1}{o}=卤2e^{蠒}$ (or) $o=卤\frac{1}{2}{e}^{-蠒}$

Expressing the result in terms ofyields the following form of quantization condition:

$胃鈮�\left(n+\frac{1}{2}\right)蟺\overline{+}\frac{1}{2}{e}^{-蠒}$ 鈥︹�︹�︹�︹�︹�︹�� (7 )

Figure shows the sketch of the potential V (X) . Since the right-side parabola (for X>0 ) is symmetric about X=A and the integration limits $X_1$and $X_2$are equidistant from a , then we can write

$胃=\frac{2}{h}{鈭�}_a^{x_2}p\left(x\right)dx$

Where the turning point$X_2$satisfies,

$E=V\left(x_2\right)=\frac{1}{2}m{蝇}^2{\left(x_2-a\right)}^2$鈥︹�︹�︹�︹��.. (8 )

In this case,

$$\begin{gathered} p\left(x\right)=2m\sqrt{E-\left(1/2\right)M{蝇}^2{\left(x-a\right)}^2} \\ p\left(x\right)=m蝇\sqrt{{\left(x_2-a\right)}^2-{\left(x-a\right)}^2} \end{gathered}$$

Thus,

$胃=\frac{2m蝇}{h}{鈭�}_a^{x_2}\left(x_2-a\right)={\left(x-a\right)}^{2}dx$

By using the substitution, u=x-a , this integral simply appears to be the area of a quarter circle with radius $x_2-a$, which is trivial to calculate:

$$\begin{gathered} 胃=\frac{2m蝇}{h}{鈭�}_0^{x_2-a}{\left(x_2-a\right)}^2-u^{2}du \\ 胃=\frac{蟺m蝇}{2h}{\left(x_2-a\right)}^2 \\ 胃=\frac{蟺}{h蝇}E \end{gathered}$$

(Equation ( 8)), and the quantization condition (Equation (7 )) yields

$\left(n+\frac{1}{2}\right)蟺卤\frac{1}{2}{e}^{-蠒}=\frac{蟺}{h蝇}{E}_N^{卤}$

(Or)

$E_n^{卤}=\left(n+\frac{1}{2}\right)h蝇\overline{+}\frac{h蝇}{2蟺}{e}_{-蠒}$ 鈥︹赌︹赌︹赌︹赌�.(9)

To sketch a symmetric double parabola V(x).

Suppose the particle starts out in the right well that is a state of the form

${蠄}_{蟿}=蠄\left(x,0\right)=\frac{1}{\sqrt{2}}\left({蠄}_n^{+}+{蠄}_N^{-}\right)$

It is also useful to define the state in which the particle resides in the left well;

${蠄}_{蟿}=\frac{1}{\sqrt{2}}{蠄}_N^{+}-{蠄}_N^{-}$

At arbitrary time, the state of the particle (tacking on the phases in the natural way) evolves to

$$\begin{gathered} 蠄\left(x,t\right)=\frac{1}{\sqrt{2}}\left({蠄}_n^{+}{e}^{-i{E}_n^{+}tir}+{蠄}_n^{-}{e}^{-i{E}_n^{-}tir}\right) \\ 蠄\left(x,t\right)=\frac{1}{\sqrt{2}}{e}^{-E_n^{+}tir}{蠄}_n^{+}+{蠄}_n^{-}{e}^{-ist} \end{gathered}$$

Where we define localid="1658464326203" $未鈮�\left(E_n^{-}-E_N^{+}\right)I魔$It follows that

$\mathrm{I}蠄\left(x,t\right){\mathrm{I}}^2=\frac{1}{2}{\overline{)\overline{){\mathrm{蠄}}_{\mathrm{n}}^{+}}}}^2+{\left|{\mathrm{蠄}}_{\mathrm{n}}^{-}\right|}^2+{\mathrm{蠄}}_{\mathrm{n}}^{+}{\mathrm{蠄}}_{\mathrm{n}}^{-}{\mathrm{e}}^{\mathrm{ist}}+{\mathrm{e}}^{-\mathrm{ist}}$

We used the fact and are real functions (see Equation (4 )). Since $\left(e^{ist}+e^{-ist}\right)/2=\cos \left(未t\right)$(Euler's formula), we can write

$\mathrm{I}\mathrm{蠄}\left(\mathrm{x},\mathrm{t}\right){\mathrm{I}}^2=\frac{1}{2}\left[{\left|{\mathrm{蠄}}_{\mathrm{n}}^{+}\right|}^2+{\left|{\mathrm{蠄}}_{\mathrm{n}}^{-}\right|}^2+2{\mathrm{蠄}}_{\mathrm{n}}^{+}{\mathrm{蠄}}_{\mathrm{n}}^{-}\cos \left(\mathrm{未迟}\right)\right]$

Here we notice something remarkable. Att =0

$$\begin{gathered} \mathrm{滨蠄}{\left(\mathrm{x},0\right)}^2=\frac{1}{2}{\left|{\mathrm{蠄}}_{\mathrm{n}}^{+}+{\mathrm{蠄}}_{\mathrm{n}}^{-}\right|}^2 \\ \mathrm{滨蠄}{\left(\mathrm{x},0\right)}^2={\left|{\mathrm{蠄}}_{\mathrm{r}}\right|}^2 \end{gathered}$$

The particle starts out at the right well (as expected).localid="1658467491416" $Att=蟺/未$

localid="1658467501051" $$\begin{gathered} {\left|\mathrm{蠄}\left(\mathrm{x},\mathrm{蟺}/\mathrm{未}\right)\right|}^2=\frac{1}{2}{\left|{蠄}_n^{+}-{蠄}_n^{-}\right|}^2 \\ {\left|\mathrm{蠄}\left(\mathrm{x},\mathrm{蟺}/\mathrm{未}\right)\right|}^2={\left|蠄\right|}^2 \end{gathered}$$

The particle hops into the left well. Atlocalid="1658467510407" $t=2蟺/未$, the particle jumps back again into the right well, and we see the particle keeps oscillating back and forth between the wells, with a period

localid="1658467528233" $t=\frac{2蟺}{未}=\frac{2蟺魔}{E_n^{-}-E_N^{+}}=\frac{2{蟺}^2}{蝇}{e}^{蠒}$

To sketch for localid="1658467548354" ${蠄}_{e^{+}}$and localid="1658467540229" ${蠄}_{e-}$ (for x<0 ).

To sketch for ${蠄}_{e^{+}}$ and ${蠄}_{e^{-}}$(for ).

To sketch the probability density att =0

The specific potential in part (d) is symmetric, and so we need to deal only with the right parabola well then we double the answer. In this case (for x>0),

$$\begin{gathered} \mathrm{I}\mathrm{p}\left(\mathrm{x}\right)=2\mathrm{m}\mathrm{IE}-\frac{1}{2}{\mathrm{m蝇}}^2{\left(\mathrm{x}-\mathrm{a}\right)}^2 \\ \mathrm{I}\mathrm{p}\left(\mathrm{x}\right)\mathrm{I}=\mathrm{m蝇}\sqrt{{\left({\mathrm{x}}_1-\mathrm{a}\right)}^2-{\left(\mathrm{x}-\mathrm{a}\right)}^2} \end{gathered}$$

Where the turning points$x_1$satisfies,

$E=\frac{1}{2}m{蝇}^2{\left(x_1-a\right)}^2$

Thus,

$蠒=\frac{2m蝇}{h}{鈭�}_0^{x_1}{\left(a-x\right)}^2-{\left(a-x_1\right)}^{2}dx$

This integral can be done using the change of variables,$a-x=\left(a-x_1\right)\cos h胃$. It follows (check the answer yourself!)

$蠒=\frac{m蝇a}{h{y}^2}y\sqrt{y^2-1}-In\left(y+\sqrt{y^2-1}\right)$

Where $y鈮�aI\left(a-x_1\right)>1$.

For V(0)>>E, that is $a^2>>{\left(x_1-a\right)}^2$, or $y^2>>1$,

localid="1658824827027" $蠒~\frac{m蝇a^2}{h{y}^2}\left[y^2-2In\left(2y\right)\right]~\frac{m蝇a^2}{魔}$