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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
| | or, evaluating the integral, |
| \end{cases} | | |
| </math> | | <math>\tfrac{1}{2}\sqrt{2\pi mV_{0}}x_{0}=(n-\tfrac{1}{4})\pi \hbar.</math> |
| <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> | | <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> |
Revision as of 03:10, 13 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,
![{\displaystyle 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}}}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/811accf4e7e6538978f5319a6fd89f74638979ab)
,which is indeed independent of m (and a).
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