Logarithmic Potential in WKB: Difference between revisions

Revision as of 03:16, 13 January 2014

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,

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

Solving for $x_{0},\!$ we obtain

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

The energy spectrum is thus

$E_{n}=V_{0}\ln \left({\sqrt {\frac {\pi }{2mV_{0}a^{2}}}}(2n-{\tfrac {1}{2}})\hbar \right).$

If we now calculate the spacing between two adjacent energy levels, we obtain

$E_{n+1}-E_{n}=V_{0}\ln \left({\frac {n+{\tfrac {3}{4}}}{n-{\tfrac {1}{4}}}}\right).$

We see that this spacing is indeed independent of mass (and, in fact, of $a\!$ as well).

@@ Line 15: / Line 15: @@
 <math>\tfrac{1}{2}\sqrt{2\pi mV_{0}}x_{0}=(n-\tfrac{1}{4})\pi \hbar.</math>
-<math> 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}}}] </math>
+Solving for <math>x_0,\!</math> we obtain
-<math> 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}) </math> ''',which is indeed independent of m (and a).'''
+<math>x_{0}=\sqrt{\frac{\pi}{2mV_{0}}}(2n-\tfrac{1}{2})\hbar.</math>
+The energy spectrum is thus
+<math>E_n=V_0\ln\left (\sqrt{\frac{\pi}{2mV_{0}a^2}}(2n-\tfrac{1}{2})\hbar\right ).</math>
+If we now calculate the spacing between two adjacent energy levels, we obtain
+<math>E_{n+1}-E_{n}=V_{0}\ln\left (\frac{n+\tfrac{3}{4}}{n-\tfrac{1}{4}}\right ).</math>
+We see that this spacing is indeed independent of mass (and, in fact, of <math>a\!</math> as well).
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