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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>
| | The Bohr-Sommerfeld quantization condition for this problem is |
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| <math>( E = V_{0} ln(r_{0}/a) '''defines''' r_{0} )</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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| = <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>
| | 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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| '''Let''' <math> x\equiv ln(r_{0}/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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| '''so''' <math> e^{x} = r_{0}/r </math>'''or''' <math> r = r_{0}e^{-x} \Rightarrow dr = -r_{0}e^{-x}dx
| | Let us now make the substitution, <math>\xi=\ln\left (\frac{x_{0}}{x}\right ).</math> We then 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>\sqrt{2mV_{0}}x_{0}\int_{0}^{\infty}\sqrt{x}e^{-x}\,dx=(n-\tfrac{1}{4})\pi \hbar,</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}
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| </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>
| | or, evaluating the integral, |
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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).''' | | <math>\tfrac{1}{2}\sqrt{2\pi mV_{0}}x_{0}=(n-\tfrac{1}{4})\pi \hbar.</math> |
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| Back to [[WKB in Spherical Coordinates]] | | Solving for <math>x_0,\!</math> we obtain |
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| | <math>x_{0}=\sqrt{\frac{\pi}{2mV_{0}}}(2n-\tfrac{1}{2})\hbar.</math> |
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| | The energy spectrum is thus |
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| | <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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| | If we now calculate the spacing between two adjacent energy levels, we obtain |
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| | <math>E_{n+1}-E_{n}=V_{0}\ln\left (\frac{n+\tfrac{3}{4}}{n-\tfrac{1}{4}}\right ).</math> |
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| | We see that this spacing is indeed independent of mass (and, in fact, of <math>a\!</math> as well). |
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| | 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