Logarithmic Potential in WKB

The Bohr-Sommerfeld quantization condition for this problem is

$\int _{0}^{x_{0}}{\sqrt {2m\left[E-V_{0}\ln \left({\frac {x}{a}}\right)\right]}}\,dx=(n-{\tfrac {1}{4}})\pi \hbar .$

Note that $E=V_{0}\ln \left({\frac {x_{0}}{a}}\right)$ defines $x_{0}.\!$ We may then rewrite the integral as

${\sqrt {2mV_{0}}}\int _{0}^{xr_{0}}{\sqrt {ln\left({\frac {x_{0}}{x}}\right)}}\,dx=(n-{\tfrac {1}{4}})\pi \hbar .$

Let us now make the substitution, $\xi =\ln \left({\frac {x_{0}}{x}}\right).$ We then obtain

${\sqrt {2mV_{0}}}x_{0}\int _{0}^{\infty }{\sqrt {x}}e^{-x}\,dx=(n-{\tfrac {1}{4}})\pi \hbar ,$

or, evaluating the integral,

${\tfrac {1}{2}}{\sqrt {2\pi mV_{0}}}x_{0}=(n-{\tfrac {1}{4}})\pi \hbar .$

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

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

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