91Ó°ÊÓ

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

$V\left(x\right)=\mathrm{Î\pm }{\left|x\right|}^v$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(1)

Where $\mathrm{Î\pm }$and v are positive constants, when we apply the WKB approximation to a potential well in problem 8.1, we get the condition

${∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}p\left(x\right)dx=\left(n-\frac{1}{2}\right)\mathrm{Ï€Ä\S }$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(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:

$$\begin{gathered} x_{1,2}=Â\pm {\left(\frac{E}{\mathrm{Î\pm }}\right)}^{1/v} \\ \mathrm{So}\mathrm{the}\mathrm{integral}\mathrm{in}\left(2\right)\mathrm{will}\mathrm{be}: \\ {∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}={∫}_{-{\left(EI\mathrm{Î\pm }\right)}^{\mathit{1}\mathit{/}\mathit{V}}}^{{\left(\mathrm{E}I\mathrm{Î\pm }\right)}^{1/\mathrm{V}}}\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{Î\pm ³\ae }}^{\mathrm{v}}\right)}\mathrm{dx} \\ {∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}={∫}_0^{{\left(EI\mathrm{Î\pm }\right)}^{1/\mathrm{V}}}\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{Î\pm ³\ae }}^{\mathrm{v}}\right)}\mathrm{dx} \\ \mathrm{Using}\mathrm{the}\mathrm{substitution}, \\ \mathrm{u}={\mathrm{x}}^{\mathrm{v}} \\ du\mathit{=}v{x}^{\mathit{1}\mathit{-}\mathit{1}\mathit{/}\mathit{v}}dx \\ du\mathit{=}\mathrm{ν}{u}^{\mathit{1}\mathit{-}\mathit{1}\mathit{/}\mathit{v}}dx \end{gathered}$$

$$\begin{gathered} dx=\frac{1}{v}{u}^{1/v-1}du \\ \mathrm{Now}\mathrm{the}\mathrm{turning}\mathrm{point}\mathrm{in}\mathrm{is}\mathrm{E}/\mathrm{Î\pm },\mathrm{thus}: \\ {∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{p}\left(\mathrm{x}\right)dx\mathit{=}\mathit{2}{∫}_0^{{\left(\mathrm{E}/\mathrm{Î\pm }\right)}^{\mathit{1}\mathit{/}\mathit{2}}}\sqrt{\mathit{2}\mathit{m}\left(E-\mathrm{Î\pm }{x}^v\right)}dx \\ {\mathit{∫}}_{{\mathit{x}}_{\mathit{1}}}^{{\mathit{x}}_{\mathit{2}}}\mathrm{p}\left(\mathrm{x}\right)dx\mathit{=}\frac{\mathit{2}\sqrt{\mathit{2}\mathit{m}}}{\mathit{v}}{∫}_0^{\mathrm{E}/\mathrm{Î\pm }}\sqrt{\mathit{E}\mathit{-}\mathit{Î\pm }\mathit{u}{\mathit{u}}^{\mathit{1}\mathit{/}\mathit{v}\mathit{-}\mathit{1}}}du \end{gathered}$$

Using any program we can find the integral, I used Mathematical to do so, and I got:

$$\begin{gathered} {∫}_0^{E/\mathrm{Î\pm }}\sqrt{E-\mathrm{Î\pm }u{u}^{1/v-1}}du=\frac{\sqrt{\mathrm{Ï€}}}{2{\mathrm{Î\pm }}^{1/v}}\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{v}}\right)}{\mathrm{Γ}\left({\displaystyle \frac{1}{v}}+{\displaystyle \frac{3}{2}}\right)}{E}^{\frac{1}{v}+\frac{1}{v}} \\ \mathrm{Thus}, \\ {∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\sqrt{2\mathrm{Ï€³¾}}}{\mathrm{v}}\frac{1}{{\mathrm{Î\pm }}^{1/\mathrm{v}}}\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{\mathrm{v}}}\right)}{\mathrm{Γ}\left(\frac{1}{\mathrm{v}}+{\displaystyle \frac{3}{2}}\right)}{\mathrm{E}}^{\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{v}}} \\ \mathrm{Using}\left(2\right)\mathrm{we}\mathrm{get}, \\ \frac{\sqrt{2\mathrm{Ï€³¾}}}{\mathrm{v}}\frac{1}{{\mathrm{Î\pm }}^{1/\mathrm{v}}}\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{\mathrm{v}}}\right)}{\mathrm{Γ}\left(\frac{1}{\mathrm{v}}+\frac{3}{2}\right)}{\mathrm{E}}^{\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{v}}}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{Ï€Ä\S } \\ \mathrm{Solve}\mathrm{for}\mathrm{E}, \\ {{\mathrm{E}}_{\mathrm{n}}}^{\frac{\mathrm{v}+2}{2\mathrm{v}}}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{Ï€Ä\S }\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{\mathrm{v}}+\frac{3}{2}}\right)}{\mathrm{Γ}\left(\frac{1}{\mathrm{v}}\right)}\frac{{\mathrm{\pm ¹Î\pm }}^{1/\mathrm{v}}}{\sqrt{2\mathrm{Ï€³¾}}} \\ {\mathrm{E}}_{\mathrm{n}}={\left[\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{Ï€Ä\S }\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{\mathrm{v}}+\frac{3}{2}}\right)}{\mathrm{Γ}\left(\frac{1}{\mathrm{v}}\right)}\frac{{\mathrm{\pm ¹Î\pm }}^{1/\mathrm{v}}}{\sqrt{2\mathrm{Ï€³¾}}}\right]}^{\frac{2\mathrm{v}}{\mathrm{v}-2}} \end{gathered}$$

role="math" localid="1656055283739" $$\begin{gathered} \mathrm{Using}\mathrm{the}\mathrm{identity}\mathrm{Γ}\left(\mathrm{z}+1\right)=\mathrm{³úΓ}\left(\mathrm{z}\right),\mathrm{so}\mathrm{we}\mathrm{get}: \\ {\mathrm{E}}_{\mathrm{n}}={\left[\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{Ï€Ä\S }\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{\mathrm{v}}+\frac{3}{2}}\right)}{\mathrm{Γ}\left(\frac{1}{\mathrm{v}}+1\right)}\frac{{\mathrm{Î\pm }}^{1/\mathrm{v}}}{\sqrt{2\mathrm{Ï€³¾}}}\right]}^{\frac{2\mathrm{v}}{\mathrm{v}-2}} \\ \mathrm{We}\mathrm{also}\mathrm{have}: \\ {\left({\mathrm{Î\pm }}^{1/\mathrm{v}}\right)}^{\frac{2\mathrm{v}}{\mathrm{v}+2}}={\mathrm{Î\pm }}^{{}^{\frac{2\mathrm{v}}{\mathrm{v}+2}}} \\ {\left({\mathrm{Î\pm }}^{1/\mathrm{v}}\right)}^{\frac{2\mathrm{v}}{\mathrm{v}+2}}={\mathrm{Î\pm }}^{{}^{\frac{\mathrm{v}+2-\mathrm{v}}{\mathrm{v}+2}}} \\ {\left({\mathrm{Î\pm }}^{1/\mathrm{v}}\right)}^{\frac{2\mathrm{v}}{\mathrm{v}+2}}=\mathrm{Î\pm }.{\mathrm{Î\pm }}^{-\frac{\mathrm{v}}{\mathrm{v}+2}} \\ {\left({\mathrm{Î\pm }}^{1/\mathrm{v}}\right)}^{\frac{2\mathrm{v}}{\mathrm{v}+2}}=\mathrm{Î\pm }.{\left[{\left(\mathrm{Î\pm }\right)}^{-1/2}\right]}^{\frac{2\mathrm{v}}{\mathrm{v}+2}} \end{gathered}$$

Thus,

$E_n=\mathrm{Î\pm }{\left[\left(n-\frac{1}{2}\right)\mathrm{Ä\S }\frac{\mathrm{Γ}\left({\displaystyle \frac{1}{v}}+{\displaystyle \frac{3}{2}}\right)}{\mathrm{Γ}\left({\displaystyle \frac{1}{v}}+1\right)}\frac{\sqrt{\mathrm{Ï€}}}{\sqrt{2m\mathrm{Î\pm }}}\right]}^{\frac{2v}{v+2}}$ …â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦(3)

Now we need to check this formula for the harmonic oscillator case, with a potential of:

$V\left(x\right)=\frac{1}{2}m{\mathrm{Ó\neg }}^2{x}^2$

Compare it with equation (1), we get v=2 and $\mathrm{Î\pm }=\frac{1}{2}m{\mathrm{Ó\neg }}^2$, so we need

$$\begin{gathered} \mathrm{Γ}\left(2\right)=1 \\ \mathrm{Γ}\left(\frac{3}{2}\right)=\frac{1}{2}\sqrt{\mathrm{Ï€}} \\ \mathrm{to}\mathrm{get}, \\ {\mathrm{E}}_{\mathrm{n}}=\left(\mathrm{n}-\frac{1}{2}\right)\mathrm{Ä\S Ó\neg }\mathrm{which}\mathrm{is}\mathrm{the}\mathrm{energy}\mathrm{of}\mathrm{the}\mathrm{harmonic}\mathrm{oscillator}. \end{gathered}$$