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| '''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:'''
| | The Bohr-Sommerfeld quantization condition for this problem is |
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| <math>\int_{0}^{r_{0}}p(r)dr = (n-1/4)\pi \hbar</math> | | <math>\int_{0}^{x_{0}}\sqrt{2m\left [E-V_{0}\ln\left (\frac{x}{a}\right )\right ]}\,dx=(n - \tfrac{1}{4})\pi\hbar.</math> |
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| '''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.''' | | Note that <math>E=V_{0}\ln\left (\frac{x_{0}}{a}\right )</math> ''defines'' <math>x_{0}.\!</math> We may then rewrite the integral as |
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| <math> V(r) = V_{0} ln(r/a) </math> | | <math>\sqrt{ 2m V_{0}} \int_{0}^{xr_{0}}\sqrt{ln\left (\frac{x_{0}}{x}\right )}\,dx=(n - \tfrac{1}{4})\pi\hbar.</math> |
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| '''(for constants''' <math> V_{0} </math> '''and a).'''
| | Let us now make the substitution, <math>\xi=\ln\left (\frac{x_{0}}{x}\right ).</math> We then obtain |
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| '''Treat only the case l = 0.
| | <math>\sqrt{2mV_{0}}x_{0}\int_{0}^{\infty}\sqrt{x}e^{-x}\,dx=(n-\tfrac{1}{4})\pi \hbar,</math> |
| Show the spacing between the levels is independent of mass.'''
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| | or, evaluating the integral, |
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| '''Answer:'''
| | <math>\tfrac{1}{2}\sqrt{2\pi mV_{0}}x_{0}=(n-\tfrac{1}{4})\pi \hbar.</math> |
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| <math> (n - \frac{1}{4})\pi h = \int_{0}^{r_{0}}\sqrt{2m[E-V_{0} ln(r/a)]}dr </math> | | Solving for <math>x_0,\!</math> we obtain |
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| <math>( E = V_{0} ln(r_{0}/a) '''defines''' r_{0} )</math> | | <math>x_{0}=\sqrt{\frac{\pi}{2mV_{0}}}(2n-\tfrac{1}{2})\hbar.</math> |
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| = <math> \sqrt{2m} \int_{0}^{r_{0} } \sqrt{ V_{0} ln(r_{0} / a) - V_{0} ln(r/a) } dr = \sqrt{ 2m V_{0}} \int_{0}^{r_{0}}\sqrt{ln(r_{0} / a)}dr</math>
| | The energy spectrum is thus |
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| '''Let''' <math> x\equiv ln(r_{0}/a)</math>
| | <math>E_n=V_0\ln\left [\sqrt{\frac{\pi}{2mV_{0}a^2}}(2n-\tfrac{1}{2})\hbar\right ].</math> |
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| '''so''' <math> e^{x} = r_{0}/r </math>'''or''' <math> r = r_{0}e^{-x} \Rightarrow dr = -r_{0}e^{-x}dx
| | If we now calculate the spacing between two adjacent energy levels, we obtain |
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| (n-\frac{1}{4})\pi \hbar = \sqrt{2mV_{0}}(-r_{0})\int_{x_{1}}^{x_{2}}\sqrt{x}e^{-e}dx </math> '''. Limits :''' <math> \begin{cases}
| | <math>E_{n+1}-E_{n}=V_{0}\ln\left (\frac{n+\tfrac{3}{4}}{n-\tfrac{1}{4}}\right ).</math> |
| & \text{ } r=0 \Rightarrow x_{1}=\infty \\
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| & \text{ } r=r_{0} \Rightarrow x_{2}=0
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| \end{cases} | |
| </math>
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| <math> (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} </math>
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| <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> | | We see that this spacing is indeed independent of mass (and, in fact, of <math>a\!</math> as well). |
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| <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).'''
| | Back to [[WKB Approximation#Problems|WKB Approximation]] |
Latest revision as of 13:37, 18 January 2014
The Bohr-Sommerfeld quantization condition for this problem is
![{\displaystyle \int _{0}^{x_{0}}{\sqrt {2m\left[E-V_{0}\ln \left({\frac {x}{a}}\right)\right]}}\,dx=(n-{\tfrac {1}{4}})\pi \hbar .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cf224782adf5aaca89c704abb854fc33b64e0be)
Note that 
defines 
We may then rewrite the integral as
Let us now make the substitution, 
We then obtain
or, evaluating the integral,
Solving for 
we obtain
The energy spectrum is thus
![{\displaystyle E_{n}=V_{0}\ln \left[{\sqrt {\frac {\pi }{2mV_{0}a^{2}}}}(2n-{\tfrac {1}{2}})\hbar \right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1408d7722ef5ab3b09db64514ed719b3bcac0b6)
If we now calculate the spacing between two adjacent energy levels, we obtain
We see that this spacing is indeed independent of mass (and, in fact, of 
as well).
Back to WKB Approximation