Question

For spherically symmetrical potentials we can apply the WKB approximation to the radial part (Equation 4.37). In the case $\mathrm{I}=0$it is reasonable 15to use Equation 8.47in the form

${\mathbf{\int }}_{\mathbf{0}}^{{\mathbf{r}}_{\mathbf{0}}}\mathbf{p}\left(r\right)\mathbf{dr}\mathbf{=}\left(n-1/4\right)\mathbf{\pi h}\mathbf{.}$

Where ${\mathbf{r}}_{\mathbf{0}}$is the turning point (in effect, we treat $\mathbf{r}\mathbf{=}\mathbf{0}$as an infinite wall). Exploit this formula to estimate the allowed energies of a particle in the logarithmic potential.

$\mathbf{V}\left(r\right)\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{In}\left(r/a\right)$

(for constant ${\mathbf{V}}_{\mathbf{0}}$and $\mathbf{a}$). Treat only the case $\mathbf{I}\mathbf{=}\mathbf{0}$. Show that the spacing between the levels is independent of mass

Short answer

The radial part of the energy level E is,

$$\begin{gathered} {\mathrm{E}}_{\mathrm{n}}={\mathrm{V}}_0\mathrm{In}\left(\sqrt{\frac{2\mathrm{\pi }}{{\mathrm{mV}}_0}}\left(\mathrm{n}-\frac{1}{4}\right)\frac{\mathrm{h}}{\mathrm{a}}\right) \\ {\mathrm{E}}_{\mathrm{n}+1}-{\mathrm{E}}_{\mathrm{n}}={\mathrm{V}}_0\mathrm{In}\left(\frac{\mathrm{n}+3/4}{\mathrm{n}-1/4}\right) \end{gathered}$$

Step-by-step solution

To determine the energy by using radial equation. 

The radial equation for hydrogen is given by:

$-\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\left(\mathrm{V}\left(\mathrm{r}\right)+\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{\mathrm{I}\left(\mathrm{I}+1\right)}{{\mathrm{r}}^2}\right)\mathrm{u}={\mathrm{E}}_{\mathrm{u}}$

Where u(r) .The simplest case is when $I=0$, that is:

$-\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\mathrm{V}\left(\mathrm{r}\right)\mathrm{u}=\mathrm{Eu}$ …………………..(1)

The WKB functions on either side of the turning point, for an increasing potential are given by:

$\mathrm{u}\left(\mathrm{r}\right)=\left\{\begin{array}{l}\begin{array}{c}2\mathrm{D}\\ \sqrt{\mathrm{p}\left(\mathrm{r}\right)}\end{array}\sin \left[{\int }_{\mathrm{r}}^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right){\mathrm{dr}}^{\text{'}}/\mathrm{h}+\mathrm{\pi }/4\right]\mathrm{r}<{\mathrm{r}}_2\\ \frac{\mathrm{D}}{\sqrt{\left|\mathrm{p}\left(\mathrm{r}\right)\right|}}\exp \left[-{\int }_{{\mathrm{r}}_2}^{\mathrm{r}}\left|\mathrm{p}\left(\mathrm{r}\right)\right|{\mathrm{dr}}^{\text{'}}/\mathrm{h}\right]\mathrm{r}>{\mathrm{r}}_2\end{array}\right.$

Where $r_2$is the turning point. But $\mathrm{u}\left(\mathrm{r}\right)$must be zero at r=0 , so the sine function must equal zero at this point, the sine function equals zero when the value inside it is equal to $n_{\pi }$that is:

$$\begin{gathered} {\int }_0^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}/\mathrm{h}+\frac{\mathrm{\pi }}{4}=\mathrm{n\pi } \\ {\int }_0^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{\pi h}.......................\left(2\right) \end{gathered}$$

The potential that we want to apply the approximations on it is:

$\mathrm{V}\left(\mathrm{r}\right)={\mathrm{V}}_0\mathrm{In}\left(\frac{\mathrm{r}}{\mathrm{a}}\right)$ ……………………….(3)

The turning point can be determined by setting the potential at this point equal to the energy that is:

$\mathrm{E}={\mathrm{V}}_0\mathrm{In}\left(\frac{{\mathrm{r}}_2}{\mathrm{a}}\right)$$\mathrm{E}={\mathrm{V}}_0\mathrm{In}\left(\frac{{\mathrm{r}}_2}{\mathrm{a}}\right)$ …………………………….(4)

To find the integral value.

Now we need to find the integral in (2), where

$\mathrm{p}\left(\mathrm{r}\right)=\sqrt{2\mathrm{m}\left(\mathrm{E}-\mathrm{V}\right)}$

So we have:

$$\begin{gathered} {\int }_0^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{2\mathrm{m}}{\int }_0^{{\mathrm{r}}_2}\sqrt{\mathrm{E}-{\mathrm{V}}_0\mathrm{In}\left(\frac{\mathrm{r}}{\mathrm{a}}\right)}\mathrm{dr} \\ {\int }_0^{{\mathrm{r}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{2\mathrm{m}}{\int }_0^{{\mathrm{r}}_2}\sqrt{\hspace{0.17em}{\mathrm{V}}_0\mathrm{In}\left(\frac{\mathrm{r}}{\mathrm{a}}\right)-{\mathrm{V}}_0\mathrm{In}\left(\frac{\mathrm{r}}{\mathrm{a}}\right)}\mathrm{dr} \end{gathered}$$

By using the following substitutions,

$$\begin{gathered} v=In\frac{r^2}{r} \\ \mathrm{dv}=\frac{\mathrm{r}}{{\mathrm{r}}_0}\left(-\frac{{\mathrm{r}}^0}{{\mathrm{r}}^2}\right)\mathrm{dr} \\ \mathrm{v}=-\frac{1}{\mathrm{r}}\mathrm{dr} \\ \mathrm{dv}=-\frac{{\mathrm{e}}^{\mathrm{v}}}{{\mathrm{r}}_0}\mathrm{dr} \end{gathered}$$

And also change the limits $\mathrm{r}=0\to \mathrm{u}=\infty \mathrm{and}\mathrm{r}={\mathrm{r}}_2\to \mathrm{u}=0$we get:

$$\begin{gathered} {\int }_0^{{\mathrm{f}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\sqrt{2{\mathrm{mV}}_0}{\int }_0^{{\mathrm{f}}_2}\sqrt{\mathrm{In}\frac{{\mathrm{r}}_2}{\mathrm{r}}}\mathrm{dr} \\ {\int }_0^{{\mathrm{f}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}={\mathrm{r}}_2\sqrt{2{\mathrm{mV}}_0}\underset{\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\underbrace{{\int }_0^{\infty }\sqrt{\mathrm{v}}{\mathrm{e}}^{-\mathrm{v}}\mathrm{dv}}} \\ {\int }_0^{{\mathrm{f}}_2}\mathrm{p}\left(\mathrm{r}\right)\mathrm{dr}=\frac{\sqrt{2{\mathrm{\pi mV}}_0}{\mathrm{r}}_2}{2} \end{gathered}$$

Substitute into (2) we get:

$\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{\pi h}=\frac{\sqrt{2{\mathrm{\pi mV}}_0}{\mathrm{r}}_2}{2}$

To find spacing energy levels.

Solve for ${\mathrm{r}}_2$to get:

localid="1658383846287" ${\mathrm{r}}_2=\sqrt{\frac{2\mathrm{\pi }}{{\mathrm{mV}}_0}}\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{h}$ …………………(5)

Solve equation (4) for $r_2$, to get:

${\mathrm{r}}^2={\mathrm{ae}}^{\mathrm{E}/{\mathrm{V}}_0}$

Substitute into (5) with $r_2$, so we get:

${\mathrm{ae}}^{\mathrm{E}/{\mathrm{V}}_0}=\sqrt{\frac{2\mathrm{\pi }}{{\mathrm{mV}}_0}}\left(\mathrm{n}-\frac{1}{4}\right)\mathrm{h}$

Solve for E to get:

${\mathrm{E}}_{\mathrm{n}}={\mathrm{V}}_0\mathrm{In}\left(\sqrt{\frac{2\mathrm{\pi }}{{\mathrm{mV}}_0}}\left(\mathrm{n}-\frac{1}{4}\right)\frac{\mathrm{h}}{\mathrm{a}}\right)$

The spacing between energy levels with n and n+1 is:

${\mathrm{E}}_{\mathrm{n}+1}-\mathrm{E}={\mathrm{V}}_0\left[\mathrm{In}\left(\sqrt{\frac{2\mathrm{\pi }}{{\mathrm{mV}}_0}}\left(\mathrm{n}+\frac{3}{4}\right)\frac{\mathrm{h}}{\mathrm{a}}\right)-\mathrm{In}\left(\sqrt{\frac{2\mathrm{\pi }}{{\mathrm{mV}}_0}}\left(\mathrm{n}-\frac{1}{4}\right)\frac{\mathrm{h}}{\mathrm{a}}\right)\right]$

Using, (A) -In (B) =In(A/B) ,we get:

${\mathrm{E}}_{\mathrm{n}+1}-{\mathrm{E}}_{\mathrm{n}}={\mathrm{V}}_0\mathrm{In}\left(\frac{\mathrm{n}+3/4}{\mathrm{n}-1/4}\right)$Which is independent of m and a.