Logarithmic Potential in WKB

For spherically symmetrical potentials, we can apply the WKB approximation to the radial equation. In the case l=0, it is reasonable to use the following equation:

$\int _{0}^{r_{0}}p(r)dr=(n-1/4)\pi \hbar$

where r_{0} is the turning point (in effect, we treat r = 0 as an infinite wall). Apply this formula to estimate the alowed energies of a particle in the logarithmic potential.

$V(r)=V_{0}ln(r/a)$

(for constants $V_{0}$ and a).

Treat only the case l = 0. Show the spacing between the levels is independent of mass.

Answer:

$(n-{\frac {1}{4}})\pi h=\int _{0}^{r_{0}}{\sqrt {2m[E-V_{0}ln(r/a)]}}dr$

$(E=V_{0}ln(r_{0}/a)'''defines'''r_{0})$

= ${\sqrt {2m}}\int _{0}^{r_{0}}{\sqrt {V_{0}ln(r_{0}/a)-V_{0}ln(r/a)}}dr={\sqrt {2mV_{0}}}\int _{0}^{r_{0}}{\sqrt {ln(r_{0}/a)}}dr$

Let $x\equiv ln(r_{0}/a)$

so $e^{x}=r_{0}/r$or $r=r_{0}e^{-x}\Rightarrow dr=-r_{0}e^{-x}dx(n-{\frac {1}{4}})\pi \hbar ={\sqrt {2mV_{0}}}(-r_{0})\int _{x_{1}}^{x_{2}}{\sqrt {x}}e^{-e}dx$ . Limits : ${\begin{cases}&{\text{ }}r=0\Rightarrow x_{1}=\infty \\&{\text{ }}r=r_{0}\Rightarrow x_{2}=0\end{cases}}$ $(n-{\frac {1}{4}})\pi \hbar ={\sqrt {2mV_{0}}}(r_{0})\int _{0}^{\infty }{\sqrt {x}}e^{-x}dx={\sqrt {2mV_{0}}}r_{0}\Gamma (3/2)={\sqrt {2mV_{0}}}r_{0}{\frac {\sqrt {\pi }}{2}}$

$r_{0}={\sqrt {\frac {2\pi }{mV_{0}}}}(n-{\frac {1}{4}})\Rightarrow E_{n}=V_{0}ln[{\frac {\hbar }{a}}{\sqrt {\frac {2\pi }{mV_{0}}}}(n-{\frac {1}{4}})]=V_{0}ln(n-{\frac {1}{4}})+V_{0}ln[{\frac {\hbar }{a}}{\sqrt {\frac {2\pi }{mV_{0}}}}]$

$E_{n+1}-E_{n}=V_{0}ln(n+{\frac {3}{4}})-V_{0}ln(n-{\frac {1}{4}})=V_{0}ln({\frac {n+3/4}{n-1/4}})$ ,which is indeed independent of m (and a).

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