Chapter 9, Problem 9.13P

To determine

Obtain the allowed energies of general power -law potential using the WKB approximation.

Answer to Problem 9.13P

Allowed energies are $E_n=\left(n-\frac{1}{2}\right)\hslash \omega \left(n=1,2,\mathrm{3......}\right)$.

Explanation of Solution

Write the quantization condition for quantized energies for a potential well with two sides.

  $\begin{array}{c}{\displaystyle \underset{x_1}{\overset{x_2}{\int }}p\left(x\right)}dx=\left(n-\frac{1}{2}\right)\pi \hslash \\ 2{\displaystyle \underset{0}{\overset{x_2}{\int }}\sqrt{2m\left(E-\alpha x^v\right)}}dx=\left(n-\frac{1}{2}\right)\pi \hslash \\ 2\sqrt{2mE}{\displaystyle \underset{0}{\overset{x_2}{\int }}\sqrt{\left(1-\frac{\alpha }{E}{x}^v\right)}}dx=\left(n-\frac{1}{2}\right)\pi \hslash \end{array}$        (I)

Consider the integral ${\displaystyle \underset{0}{\overset{x_2}{\int }}\sqrt{\left(1-\frac{\alpha }{E}{x}^v\right)}}dx$.

  Let $\begin{array}{c}\frac{\alpha }{E}{x}^v=z\\ x={\left(\frac{zE}{\alpha }\right)}^{\frac{1}{v}}\\ dx={\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{1}{v}{z}^{\frac{1}{v}-1}dz\end{array}$

Rewrite the left-hand side of equation (1) by substituting the above value of $dx$.

    $\begin{array}{c}2\sqrt{2mE}2{\displaystyle \underset{0}{\overset{x_2}{\int }}\sqrt{\left(1-\frac{\alpha }{E}{x}^v\right)}}dx=2\sqrt{2mE}{\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{1}{v}{\displaystyle \underset{0}{\overset{x_2}{\int }}{z}^{\frac{1}{v}-1}\sqrt{1-z}dz}\\ =2\sqrt{2mE}{\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{1}{v}\frac{\Gamma \left(1/v\right)\Gamma \left(3/2\right)}{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}\\ =2\sqrt{2mE}{\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{1}{v}\frac{\Gamma \left(1/v\right)\Gamma \left(3/2\right)}{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}\\ =2\sqrt{2mE}{\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{1}{v}\frac{\Gamma \left(\frac{1}{v}+1\right)\frac{1}{2}\sqrt{\pi }}{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}\\ =\sqrt{2\pi mE}{\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{\Gamma \left(\frac{1}{v}+1\right)}{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}\end{array}$

Equate with the right-hand side of equation (I).

    $\begin{array}{c}\sqrt{2\pi mE}{\left(\frac{E}{\alpha }\right)}^{\frac{1}{v}}\frac{\Gamma \left(\frac{1}{v}+1\right)}{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}=\left(n-\frac{1}{2}\right)\pi \hslash \\ E^{\frac{1}{v}+\frac{1}{2}}=\frac{\left(n-\frac{1}{2}\right)\pi \hslash }{\sqrt{2\pi m}}{\alpha }^{\frac{1}{v}}\frac{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}{\Gamma \left(\frac{1}{v}+1\right)}\\ E_n=\left(n-\frac{1}{2}\right)\hslash \sqrt{\frac{\pi }{2m\alpha }}{\left[\frac{\Gamma \left(\frac{1}{v}+\frac{3}{2}\right)}{\Gamma \left(\frac{1}{v}+1\right)}\right]}^{\left(\frac{2v}{v+2}\right)}\alpha \end{array}$

Substitute 2 for $v$ in the above equation.

    $\begin{array}{c}{E}_n=\left(n-\frac{1}{2}\right)\hslash \sqrt{\frac{\pi }{2m\alpha }}\left[\frac{\Gamma \left(2\right)}{\Gamma \left(\frac{3}{2}\right)}\right]\alpha \\ =\left(n-\frac{1}{2}\right)\hslash \sqrt{\frac{2\alpha }{m}}\end{array}$

Conclusion:

Substitute $\frac{1}{2}m{\omega }^2$ for $\alpha$ in the above equation.

    $\begin{array}{c}{E}_n=\left(n-\frac{1}{2}\right)\hslash \sqrt{\frac{\pi }{2m\alpha }}\left[\frac{\Gamma \left(2\right)}{\Gamma \left(\frac{3}{2}\right)}\right]\alpha \\ =\left(n-\frac{1}{2}\right)\hslash \omega \left(n=1,2,\mathrm{3......}\right)\end{array}$

Thus, the allowed energies are $E_n=\left(n-\frac{1}{2}\right)\hslash \omega \left(n=1,2,\mathrm{3......}\right)$.