Question

Consider the case of a symmetrical double well, such as the one pictured in Figure. We are interested in Figure 8.13bound states with E<V(0) .
(a) Write down the WKB wave functions in regions (i) $\mathit{x}\mathbf{>}{\mathit{x}}_{\mathbf{2}}$ , (ii) ${\mathit{x}}_{\mathbf{1}}\mathbf{<}\mathit{x}\mathbf{<}{\mathit{x}}_{\mathbf{2}}$ , and (iii) $\mathbf{0}\mathbf{<}\mathit{x}\mathbf{<}{\mathit{x}}_{\mathbf{1}}$. Impose the appropriate connection formulas at and (this has already been done, in Equation $\mathbf{8}\mathbf{.}\mathbf{46}$, for ${\mathit{x}}_{\mathbf{2}}$ ; you will have to work out ${\mathit{x}}_{\mathbf{1}}$ for yourself), to show that

$$\begin{gathered} \mathit{\psi }\left(x\right)\mathbf{=}\{\begin{array}{l}\frac{D}{\sqrt{P\overline{)\left(x\right)}}}exp\left[-\frac{1}{h}\overline{){\int }_{x_2}^x}p\left(x\text{'}\right)dx\text{'}\right] \\ \frac{2D}{\sqrt{P\overline{)\left(x\right)}}}sin\left[\frac{1}{h}\overline{){\int }_{x_2}^x}p\left(x\text{'}\right)dx\text{'}+\frac{\pi }{4}\right] \\ \frac{D}{\sqrt{P\overline{)\left(x\right)}}}\left[2cos\theta \overline{){}^{\frac{1}{h}{\int }_v^{x_1}}}p\overline{)\left(x\right)}dx+sin\theta e\frac{1}{h}{\int }_v^{x_1}\overline{)p}\overline{)\left(x\right)}dx\right]\\ \end{array} \end{gathered}$$

FIGURE 8.13: Symmetrical double well; Problem 8.15.
Where

$\mathit{\theta }\mathbf{\equiv }\frac{\mathbf{1}}{\mathbf{h}}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}_{\mathbf{2}}}\mathit{p}\left(x\right)\mathit{d}\mathit{x}$ .

(b) Because v(x) is symmetric, we need only consider even $\mathbf{(}\mathbf{+}\mathbf{)}$and odd (-)wave functions. In the former case $\mathit{\psi }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ , and in the latter case $\mathit{\psi }\mathbf{=}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ . Show that this leads to the following quantization condition:

$\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{=}\mathbf{\pm }\mathbf{2}{\mathit{e}}^{\mathbf{f}}$.

Where

$\mathit{f}\mathbf{\equiv }\frac{\mathbf{1}}{\mathbf{h}}{\mathbf{\int }}_{\mathbf{-}{\mathbf{x}}_{\mathbf{1}}}^{{\mathbf{x}}^{\mathbf{1}}}\overline{)\mathbf{p}\overline{)\left(x\right)}\mathbf{d}\mathbf{x}}$

Equation 8.59determines the (approximate) allowed energies (note that Ecomes into 1 and ${\mathit{x}}_{\mathbf{2}}$, so $\mathit{\theta }$ and $\mathit{\varphi }$ are both functions of E ).

(c) We are particularly interested in a high and/or broad central barrier, in which case $\mathit{\varphi }$ is large, and ${\mathit{e}}^{\mathbf{\varphi }}$is huge. Equation 8.59 then tells us thatA $\mathit{\theta }$must be very close to a half-integer multiple of $\pi$ . With this in mind, write localid="1658823154085" $\left(\theta =n+1/2\right)\mathit{\pi }\mathbf{+}\mathit{o}$ . where localid="1658823172105" $|\in |\mathbf{<}\mathbf{<}\mathbf{1}$, and show that the quantization condition becomes

localid="1658823190448" $\mathit{\theta }\mathbf{=}\left(n+\frac{1}{2}\right)\mathit{\pi }\mathit{m}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{f}}$.

(d) Suppose each well is a parabola: 16

localid="1658823244259" $$\begin{gathered} \mathit{v}\left(x\right)\mathbf{=}\{\begin{array}{l}\frac{1}{2}m{\omega }^2{\left(x+a\right)}^2,ifx<0 \\ \frac{1}{2}m{\omega }^2{\left(x-a\right)}^2,ifx<0\\ \end{array} \end{gathered}$$ .

Sketch this potential, find (Equation 8.58 ), and show that

localid="1658823297094" ${\mathit{E}}_{\mathbf{n}}^{\mathbf{\pm }}\mathbf{=}\left(n+\frac{1}{2}\right)\mathit{h}\mathit{\omega }\mathit{m}\frac{\mathbf{h}\mathbf{\omega }}{\mathbf{2}\mathbf{\pi }}{\mathit{e}}^{\mathbf{-}\mathbf{f}}$

Comment: If the central barrier were impenetrable localid="1658823318311" $\left(f\to \infty \right)$ , we would simply have two detached harmonic oscillators, and the energies, localid="1658823339386" ${\mathit{E}}_{\mathbf{n}}\mathbf{=}\left(n+1/2\right)\mathit{h}\mathit{\omega }$ , would be doubly degenerate, since the particle could be in the left well or in the right one. When the barrier becomes finite (putting the two wells into "communication"), the degeneracy is lifted. The even (localid="1658823360600" ${\mathit{\psi }}_{\mathbf{n}}^{\mathbf{+}}$) states have slightly lower energy, and the odd ones ${\psi }_n^{.}$ have slightly higher energy.
(e) Suppose the particle starts out in the right well-or, more precisely, in a state of the form

localid="1658823391675" $\mathit{\psi }\left(x,0\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\left({\psi }_n^{+}+{\psi }_n\right)$

which, assuming the phases are picked in the "natural" way, will be concentrated in the right well. Show that it oscillates back and forth between the wells, with a period

localid="1658823416179" $\mathit{\tau }\mathbf{=}\frac{\mathbf{2}{\mathbf{\pi }}^{\mathbf{2}}}{\mathbf{\omega }}{\mathit{e}}^{\mathbf{.}}$

(f) Calculate , for the specific potential in part (d), and show that for , localid="1658823448873" $\mathit{V}\left(0\right)\mathbf{.}\mathbf{>}\mathbf{>}\mathit{E}\mathbf{,}\mathit{\varphi }\mathbf{~}\mathit{m}\mathit{\omega }{\mathit{a}}^{\mathbf{2}}\mathbf{/}\mathit{h}$.

Short answer

Therefore,

(a) $$\begin{gathered} {\psi }_{WKB}\left(x\right)\approx \left\{\begin{array}{l}\frac{D}{\sqrt{p}\overline{)\left(x\right)}}exp\frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\sin \frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx+\frac{\pi }{4}} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\left[2\cos \theta e\overline{)P}\overline{)\overline{)\left(x\text{'}\right)}}dx+\sin \theta e^{-\frac{1}{h}{\int }_x^{x_1}\overline{)p\overline{)\left(x\right)}dx\text{'}}}\right]\\ \end{array}\right. \end{gathered}$$
(b) $\varphi =1/ħ{\int }_{-x_1}^x\overline{)p(x\text{'}\overline{))}dx}$.

(c)$\theta \approx \left(n+\frac{1}{2}\right)\pi \overline{+}\frac{1}{2}{e}^{-\varphi }$ .
(d) $E_n^{\overline{+}}=\left(n+\frac{1}{2}\right)ħ\omega \overline{+}\frac{ħ\omega }{2\pi }{e}^{-\varphi }$.
(e) $\tau =\frac{2{\pi }^2}{\omega }{e}^{\varphi }$.

(f) $\varphi ~\frac{m\omega a^2}{ħ}$ is showed.

Step-by-step solution

(a) By using wave function. 

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

$$\begin{gathered} {\psi }_{WKB}\left(x\right)\approx \left\{\begin{array}{l}\frac{D}{\sqrt{p}\overline{)\left(x\right)}}exp\frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\sin \frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx+\frac{\pi }{4}} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\left[2\cos \theta e\overline{)P}\overline{)\overline{)\left(x\text{'}\right)}}dx+\sin \theta e^{-\frac{1}{h}{\int }_x^{x_1}\overline{)p\overline{)\left(x\right)}dx\text{'}}}\right]\\ \end{array}\right. \end{gathered}$$

The WKB wave function takes the form,

$$\begin{gathered} {\psi }_{WKB}\left(x\right)\approx \left\{\begin{array}{l}\frac{D}{\sqrt{p}\overline{)\left(x\right)}}exp\frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx} \\ \frac{D}{\sqrt{p}\overline{)\left(x\right)}}\sin \frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx+\frac{\pi }{4}}\\ \end{array}\right. \end{gathered}$$

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

${\psi }_{WKB}\left(x\right)\approx \frac{2D}{\sqrt{p}\left(x\right)}\sin \left[-\frac{1}{h}{\int }_{x_1}^{x}p\left(x\text{'}\right)dx\text{'}\theta +\frac{\pi }{4}\right]$

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

$$\begin{gathered} {\psi }_{WKB}\left(x\right)\approx \left\{\begin{array}{l}\frac{-2D}{\sqrt{p}\overline{)\left(x\right)}}\sin \left[\frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\text{'}\right)}dx-\theta -\frac{\pi }{4}}\right],x>0 \\ \frac{1}{\sqrt{p}\overline{)\left(x\right)}}F{e}^{\frac{1}{n}{\int }_x^0\overline{)P\overline{)\left(x\text{'}\right)}dx}}+G{e}^{-\frac{1}{h}{\int }_x^{x_1}\overline{)p\overline{)\left(x\right)}dx\text{'}}}X<0\\ \end{array}\right. \end{gathered}$$ …………… ()

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

${\psi }_p\left(x\right)=aAi\left(-ax\right)+bBi\left(-ax\right)$…………………… ( 2)

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

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

.

In particular, for x<0.

For,

${\int }_x^0\overline{)p\left(x\text{'}\right)\overline{)d}x\text{'}\approx ћa^{3/2}{\int }_x^0\sqrt{-x\text{'}dx\text{'}}}=\frac{2}{3}ћ{\left(-ax\right)}^{3/2}$

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

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

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

${\psi }_p\left(x\right)\approx \frac{1}{\pi {\left(-ax\right)}^{1/4}}\frac{a}{2}{e}^{-\frac{2}{3}{\left(-ax\right)}^{3/2}}+b{e}^{\frac{2}{3}\left(-ax\right)3/2}$

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

$a=2\frac{\pi }{һa}G$and $b=\sqrt{\frac{\pi }{ћa}F}$ …………………(3 )

In particular, for x>0.

For x>0 , we have

$$\begin{gathered} {\int }_0^{x}P\left(\text{'}x\right)dx\text{'}ħa^{3/2}{\int }_0^x\sqrt{x} \\ {\int }_0^{x}P\left(\text{'}x\right)dx=\frac{2}{3}ħ{\left(ax\right)}^{3/2} \end{gathered}$$

and therefore the WKB wave function (Equation ()) can be written as$$\begin{gathered} {\psi }_{WKB}\left(X\right)\approx \frac{-2D}{ħa^{3/4}{x}^{1/4}}\sin \frac{2}{3}{\left(ax\right)}^{3/2}-\theta -\frac{\pi }{4} \\ {\psi }_{WKB}\left(X\right)\approx \frac{-2D}{ħa^{3/4}{x}^{1/4}}\cos \frac{2}{3}{\left(ax\right)}^{3/2}+\frac{\pi }{4}-\theta \end{gathered}$$

Let $\beta \equiv \left(2/3\right){\left(ax\right)}^{3/2}+\pi /4$, then using elementary trigonometry,

${\psi }_{WKB}\left(X\right)\approx \frac{2D}{ħa^{3/4}{x}^{1/4}}\left[\cos \beta \cos \theta +\sin \beta \sin \theta \right]$

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" ${\psi }_p\left(x\right)\approx \frac{1}{\sqrt{\pi }{\left(ax\right)}^{1/4}}\left[a\sin \beta +b\cos \beta \right]$

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

$\frac{a}{\sqrt{\pi }}=\frac{2D}{ħa}\sin \theta$And $\frac{b}{\sqrt{\pi }}=\frac{2D}{\sqrt{ha}}\cos \theta$

Or, putting in Equation (3) for and :

$G=D\sin \theta$And $F=2D\cos \theta$

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" $$\begin{gathered} {\psi }_{WKB}\left(x\right)\approx \left\{\begin{array}{l}\frac{D}{\sqrt{p}\left(x\right)}exp-\frac{1}{h}{\int }_{x_2}^x\overline{)p\overline{)\left(x\right)}dx\text{'}} \\ \frac{2D}{\sqrt{p}\left(x\right)}\sin \frac{1}{h}{\int }_x^{x_2}p\left(x\text{'}\right)dx+\frac{\pi }{4} \\ \frac{D}{p\left(x\right)I}\left[2\cos \theta e^{\frac{1}{r}{\int }_x^{x_1}p\left(x\text{'}\right)dx}+\sin \theta e^{\frac{1}{r}{\int }_x^{x_1}\overline{)p}\left(x\text{'}\right)dx\text{'}}\right]\\ \end{array}\right. \end{gathered}$$ ……………(4)

Where,

$\theta =1/ħ{\int }_{x_1}^{x_2}p\left(x\right)dx$

(b) Odd wave function with quantization condition.

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

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

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

(Assuming$D\ne 0$), which leads to

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

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

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

$\tan \theta =-2e^{\varphi }$

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

$$\begin{gathered} {\psi }_{WKB}\left(x\right)=-\frac{1}{2Ip\left(x\right)I}\frac{dIP\left(x\right)I}{dx}{\psi }_{WKB}\left(x\right) \\ +\frac{D}{\sqrt{Ip\left(x\right)I}}2\cos \theta e^{\frac{1}{r}{\int }_x^{x_1}\overline{)p\left(x;\right)dx}}\left(-\frac{1}{h}Ip\left(x\right)I\right)+\sin \theta e^{\frac{1}{r}{\int }_x^{x_1}p\left(x\right)dx}\left(\frac{1}{h}Ip\left(x\right)I\right) \end{gathered}$$................. ( 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 ${\psi }_{WKB}\left(0\right)=0$ (Equation ( 5)) leads to

$-2\cos \theta e^{\varphi /2}+\sin \theta e^{-\varphi /2}=0$

$\tan \theta =2e^{\varphi }$ (Or)

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

$\varphi =\left(1/h\right){\int }_{-x}^{x_1}\overline{)p\overline{)\left(x\text{'}\right)}dx}$ …………………… ()

Where,

$\varphi =1/ħ{\int }_{-x1}^{x_1}\overline{)p\left(x\text{'}\right)dx\text{'}}$.

(c) Using quantization condition.

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

$\frac{tan\left(n+1/2\pi \right)+{\tan }_o}{1-\tan \left[\left(n+1/2\right)\pi \right]\tan o}=\pm 2e^{\varphi }$

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

$-\frac{1}{\tan o}\approx \frac{1}{o}=\pm 2e^{\varphi }$ (or) $o=\pm \frac{1}{2}{e}^{-\varphi }$

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

$\theta \approx \left(n+\frac{1}{2}\right)\pi \overline{+}\frac{1}{2}{e}^{-\varphi }$ ………………… (7 )

(d) Sketch the graph by using V(x).

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

$\theta =\frac{2}{h}{\int }_a^{x_2}p\left(x\right)dx$

Where the turning point$X_2$satisfies,

$E=V\left(x_2\right)=\frac{1}{2}m{\omega }^2{\left(x_2-a\right)}^2$…………….. (8 )

In this case,

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

Thus,

$\theta =\frac{2m\omega }{h}{\int }_a^{x_2}\left(x_2-a\right)={\left(x-a\right)}^{2}dx$

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:

$$\begin{gathered} \theta =\frac{2m\omega }{h}{\int }_0^{x_2-a}{\left(x_2-a\right)}^2-u^{2}du \\ \theta =\frac{\pi m\omega }{2h}{\left(x_2-a\right)}^2 \\ \theta =\frac{\pi }{h\omega }E \end{gathered}$$

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

$\left(n+\frac{1}{2}\right)\pi \pm \frac{1}{2}{e}^{-\varphi }=\frac{\pi }{h\omega }{E}_N^{\pm }$

(Or)

$E_n^{\pm }=\left(n+\frac{1}{2}\right)h\omega \overline{+}\frac{h\omega }{2\pi }{e}_{-\varphi }$ …………….(9)

To sketch a symmetric double parabola V(x).

(e) Sketch the graph.

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

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

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

${\psi }_{\tau }=\frac{1}{\sqrt{2}}{\psi }_N^{+}-{\psi }_N^{-}$

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

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

Where we define localid="1658464326203" $\delta \equiv \left(E_n^{-}-E_N^{+}\right)Iħ$It follows that

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

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

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

Here we notice something remarkable. Att =0

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

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

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

The particle hops into the left well. Atlocalid="1658467510407" $t=2\pi /\delta$, 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" $t=\frac{2\pi }{\delta }=\frac{2\pi ħ}{E_n^{-}-E_N^{+}}=\frac{2{\pi }^2}{\omega }{e}^{\varphi }$

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

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

To sketch the probability density att =0

(f) To evaluate the integral of.

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),

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

Where the turning points$x_1$satisfies,

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

Thus,

$\varphi =\frac{2m\omega }{h}{\int }_0^{x_1}{\left(a-x\right)}^2-{\left(a-x_1\right)}^{2}dx$

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

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

Where $y\equiv aI\left(a-x_1\right)>1$.

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

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