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| Worked problem - MehmetYesiltas- on behalf of Team 5
| | The WKB approximation is given by |
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| | <math>\int_{r_1}^{r_2} p(r)\,dr=(n+\tfrac{1}{2})\pi \hbar,</math> |
| Use WKB approximation to estimate energy spectrum for Hydrogen atom.
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| Hints:
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| <math>\text{use the relation r}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)</math>
| | where |
| and <math>\text{ the definition }\int\limits_{r1}^{r2} {\left ({\frac{(x-a)(x-b)}{x^{2}}} \right )^{1/2}dx}=\frac{\pi }{2}(\sqrt {b} -\sqrt {a} )^{2}
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| </math> | | <math>p(r)=\sqrt {2m(E-V_{\text{eff}}(r))} =\sqrt {2m\left (E+\frac{e^{2}}{r}-{\frac{\hbar ^{2}(l+\tfrac{1}{2})^2}{2mr^{2}}}\right )}.</math> |
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| The approximation is:
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| <math>\int {P(r)} dr=(n+\frac{1}{2})\pi \hbar</math>
| | We may rewrite the above as |
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| <math>\text{P(r)=}\sqrt {2m(E-V(r))} =\sqrt {2m(E-\left ({\frac{\hbar ^{2}l(l+1)}{2mr^{2}}-\frac{e^{2}}{r}} \right )} )</math> | | <math>\sqrt{2mE}\int_{r_1}^{r_2}\sqrt{1-\frac{\hbar^{2}(l+\tfrac{1}{2})^2}{2mEr^{2}}+\frac{e^{2}}{Er}}\,dr=(n+\tfrac{1}{2})\pi \hbar,</math> |
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| <math>\int\limits_{r1}^{r2} {\sqrt {2m(E-\frac{\hbar ^{2}l(l+1)}{2mr^{2}}+\frac{e^{2}}{r})} }dr=(n+\frac{1}{2})\pi \hbar </math> where r1 and r2 are turning points in this case.
| | or, making the substitution, |
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| <math>\sqrt {2mE} \int\limits_{r1}^{r2} {(1-}\frac{\hbar ^{2}l(l+1)}{2mr^{2}E}+\frac{e^{2}}{Er})^{1/2}dr=(n+\frac{1}{2})\pi \hbar </math> | | <math>T=-\frac{\hbar^{2}(l+\tfrac{1}{2})}{2mE}</math> and <math>V=\frac{e^{2}}{E},</math> |
| if we do this substitution:
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| <math>\text{let T=}-\frac{\hbar ^{2}l(l+1)}{2mE}\text{ and V=}-\frac{e^{2}}{r}\text{ }</math> | |
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| <math>\sqrt {2mE} \int\limits_{r1}^{r2} {(1+\frac{T}{r^{2}}}+\frac{V}{r})^{1/2}dr=(n+\frac{1}{2})\pi \hbar \text{ }</math> | | <math>\sqrt{2mE}\int_{r_1}^{r_2}\sqrt{1-\frac{V}{r}+\frac{T}{r^{2}}}\,dr=(n+\tfrac{1}{2})\pi \hbar.</math> |
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| <math>\text{the relation r}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)</math> | | Using the fact that <math>r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)\!</math> and that |
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| <math>\sqrt {2mE} \int\limits_{r1}^{r2} {\left ({\frac{(r_{1}-r)(r_{2}-r)}{r^{2}}} \right )^{1/2}dr=(n+\frac{1}{2})\pi \hbar }</math> | | <math>\int_{r_1}^{r_2}\sqrt{{\frac{(x-r_1)(x-r_2)}{x^{2}}}}\,dx=\frac{\pi }{2}(\sqrt {r_2} -\sqrt {r_1} )^{2},</math> |
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| <math>\text{ the definition }\int\limits_{r1}^{r2} {\left ({\frac{(x-a)(x-b)}{x^{2}}} \right )^{1/2}dx}=\frac{\pi }{2}(\sqrt {b} -\sqrt {a} )^{2}</math>
| | we obtain |
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| <math>\sqrt {2mE} *\frac{\pi }{2}*(\sqrt {r_{2}} -\sqrt {r_{1}} )^{2}=(n+\frac{1}{2})\pi \hbar </math> | | <math>\frac{\pi}{2}\sqrt {2mE}(\sqrt {r_{2}} -\sqrt {r_{1}} )^{2}=(n+\tfrac{1}{2})\pi \hbar.</math> |
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| <math>\sqrt {2mE} *\frac{\pi }{2}*(r_{2}+r_{1}-2\sqrt {r_{1}r_{2}} )=(n+\frac{1}{2})\pi \hbar </math> | | We now observe that <math>r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})r+r_{1}r_{2},\!</math> so that |
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| <math>\text{let r}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})+r_{1}r_{2}</math> | | <math>V=r_{1}+r_{2}\!</math> and <math>T=r_{1}r_{2}.\!</math> |
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| <math>\text{so V=}(r_{1}+r_{2})\text{ and T=}r_{1}r_{2}</math>
| | We thus obtain |
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| <math>\sqrt {2mE} *\frac{\pi }{2}*(V-2\sqrt {T} )=(n+\frac{1}{2})\pi \hbar </math> | | <math>\frac{\pi }{2}\sqrt {2mE}(V-2\sqrt {T} )=(n+\tfrac{1}{2})\pi \hbar,</math> |
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| <math>\sqrt {2mE} \left ({-\frac{e^{2}}{E}-2\sqrt {-\frac{\hbar ^{2}l(l+1)}{2mE}} } \right )=(n+\frac{1}{2})\pi \hbar </math>
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| <math>-e^{2}\sqrt {\frac{2m}{E}} -2\sqrt {\hbar ^{2}l(l+1)} =2\hbar (n+\frac{1}{2})</math> | | <math>\sqrt {2mE} \left ({-\frac{e^{2}}{E}-2\sqrt {-\frac{\hbar ^{2}(l+\tfrac{1}{2})^2}{2mE}} } \right )=(n+\tfrac{1}{2})\pi \hbar.</math> |
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| <math>\text{ }2\hbar (n+\frac{1}{2})+2\hbar \sqrt {l(l+1)} =e^{2}\sqrt {\frac{2m}{-E}} </math>
| | This equation simplifies to |
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| <math>\frac{4\hbar ^{2}\left ({n+\frac{1}{2}+\sqrt {l(l+1)} } \right )^{2}}{2me^{4}}=\frac{1}{-E}\text{ }</math> | | <math>-e^{2}\sqrt {-\frac{2m}{E}} -(2\ell+1)\hbar =(2n+1)\hbar.</math> |
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| Then if we finally pull out E,
| | We may now easily solve for <math>E,\!</math> obtaining |
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| <math>\text{E=}\frac{-me^{4}}{2\hbar ^{2}\left ({n+\frac{1}{2}+\sqrt {l(l+1)} } \right )^{2}}</math> | | <math>E=-\frac{me^{4}}{2\hbar ^{2}(n+\ell+1)^{2}}.</math> |
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| | Note that this is exactly the spectrum that we would obtain from an exact solution of the Coulomb problem. |
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| | Back to [[WKB in Spherical Coordinates#Problem|WKB in Spherical Coordinates]] |
The WKB approximation is given by
where
We may rewrite the above as
or, making the substitution,
and 
Using the fact that 
and that
we obtain
We now observe that 
so that
and 
We thus obtain
or
This equation simplifies to
We may now easily solve for 
obtaining
Note that this is exactly the spectrum that we would obtain from an exact solution of the Coulomb problem.
Back to WKB in Spherical Coordinates