Question

Use the WKB approximation to find the bound state energy for the potential in problem .

Short answer

The bound state energy for the potential of E is,

$E=-0.418\frac{{ħ}^2{a}^2}{m}.$

Step-by-step solution

To find the bound state energy potential. 

In this problem we need to use the WKB approximation to find the allowed energies of the potential:

$V\left(x\right)=-\frac{{ħ}^2{a}^2}{m}sec{h}^2\left(ax\right)$ ………………….(1)

When we apply the WKB approximation to a potential well in problem.1, we get the condition

${\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}p\left(x\right)dx=\left(n-\frac{1}{2}\right)\mathrm{\pi h}$ …………………………(2)

Where $x_1$is the left turning point, $x_2$is the right turning point. At the turning point the values of the energy equal the potential, so to find these points we set E=V(x) , so we get:

$$\begin{gathered} x_1=-x_2 \\ \mathrm{Where}, \\ \mathrm{F}=-\frac{{\mathrm{h}}^2{\mathrm{a}}^2}{\mathrm{m}}{\mathrm{sech}}^2\left({\mathrm{ax}}_2\right) \\ \mathrm{Note}\mathrm{that},\mathrm{V}\left(\mathrm{x}\right)\mathrm{is}\mathrm{even}.\mathrm{Therefore},\mathrm{we}\mathrm{can}\mathrm{write}\left(2\right)\mathrm{as} \\ {\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=2{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\sqrt{2\mathrm{m}\left(\mathrm{E}+\frac{{\mathrm{h}}^2{\mathrm{a}}^2}{\mathrm{m}}{\mathrm{sech}}^2\left(\mathrm{ax}\right)\right)}\mathrm{dx} \\ \mathrm{Using}\mathrm{the}\mathrm{substitutions}, \\ \mathrm{z}={\mathrm{sech}}^2\left(\mathrm{ax}\right) \\ \mathrm{dz}=-2{\mathrm{asech}}^2\left(\mathrm{ax}\right)\tanh \left(\mathrm{ax}\right)\mathrm{dx} \\ \mathrm{dz}=-2\mathrm{az}\sqrt{1-\mathrm{zdx}}\mathrm{and}\mathrm{the}\mathrm{limits}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{z},\mathrm{will}\mathrm{be}: \\ \mathrm{x}=0\to \mathrm{z}=1 \end{gathered}$$

role="math" localid="1656057550047" $$\begin{gathered} x=x_2\to z_2=sec{h}^2\left(a{x}_2\right) \\ {z}_2=-\frac{mE}{{ħ}^2{a}^2} \\ {z}_2=b \end{gathered}$$

To find the integral value.

So the integral becomes:

$$\begin{gathered} {\int }_{x_1}^{x_2}p\left(x\right)dx=-2{\int }_1^b\sqrt{2m\left(E+\frac{{ħ}^2{a}^2}{m}z\right)}\frac{dz}{2az\sqrt{1-z}} \\ {\int }_{x_1}^{x_2}p\left(x\right)dx\sqrt{2ħ}{\int }_b^1\frac{\sqrt{z-b}}{z\sqrt{1-z}}dz \end{gathered}$$

Now we can do the integral using any program. (in my case I used Mathematical) providing that b>0 , since is proportional to the negative of the energy and the energy is less than zero, then b>0 . The value of the integral is:

${\int }_b^1\frac{\sqrt{z-b}}{z\sqrt{1-z}}dz=\mathrm{\pi }\left(1-\sqrt{\mathrm{b}}\right)$

Thus,

$$\begin{gathered} {\int }_{x_1}^{x_2}p\left(x\right)dx=\sqrt{2\mathrm{\pi }}ħ\left(1-\sqrt{b}\right) \\ \mathrm{Substitute}\mathrm{into}\left(2\right)\mathrm{we}\mathrm{get}: \\ \sqrt{2\mathrm{\pi h}}\left(1-\sqrt{\mathrm{b}}\right)=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{\pi h} \\ \mathrm{Solve}\mathrm{for}\mathrm{to}\mathrm{get}: \\ \mathrm{n}=\sqrt{2}\left(1-\sqrt{\mathrm{b}}\right)+\frac{1}{2}..............................\left(3\right) \\ \mathrm{The}\mathrm{largest}\mathrm{n}\mathrm{can}\mathrm{be}\mathrm{achieved}\mathrm{by}\mathrm{setting}\mathrm{b}=0,\mathrm{that}\mathrm{is}: \\ {\mathrm{n}}_{\max }=\sqrt{2}+\frac{1}{2}=1.914 \end{gathered}$$

To find state energy.

So the only possible value of n is 1 , substitute with this value into (3) and then solve for b, as:

$$\begin{gathered} \sqrt{2}\left(1-\sqrt{b}\right)=\frac{1}{2} \\ b={\left(1-\frac{1}{2\sqrt{2}}\right)}^2 \\ b=\left(\frac{9}{8}-\frac{1}{\sqrt{2}}\right) \\ b\approx 0.418 \\ \mathrm{But} \\ \mathrm{b}=-\frac{\mathrm{mE}}{{\mathrm{ħ}}^2{\mathrm{a}}^2} \\ \mathrm{thus}, \\ \mathrm{E}=-0.418\frac{{\mathrm{ħ}}^2{\mathrm{a}}^2}{\mathrm{m}}. \end{gathered}$$