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| It is possible to apply the [[WKB Approximation|WKB approximation]] to the radial equation using a method by R. E. Langer (1937). | | It is possible to apply the [[WKB Approximation|WKB approximation]] to the radial equation using a method by R. E. Langer (1937). |
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| Recall: <math>\ u(r)=rR(r)</math>, | | Recall that |
| <math>\left[ -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V(r)-E\right] u(r)=0</math>
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| Now apply the transformations: <math>\ r = e^{s};</math> <math>\ u(r) = W(s)e^{\frac{1}{2}s}</math>
| | <math>u(r)=rR(r),\!</math> |
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| Note that for <math>\ r </math> varying from 0 to infinity, <math>\ s </math> will vary from minus infinity to plus infinity.
| | and that <math>u(r)\!</math> satisfies the effective one-dimensional [[Schrödinger Equation|Schrödinger equation]], |
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| | <math>\left[ -\frac{\hbar^2}{2m}\frac{d^2}{dr^2}+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}+V(r)-E\right] u(r)=0.</math> |
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| The radial equation then transforms into:
| | We now perform the following transformations: |
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| <math>\ \frac{d^{2}W}{ds^{2}}+\frac{2m}{\hbar^{2}}\left[E-V(e^{s})-\frac{\hbar^2}{2m}\left(l+\frac{1}{2}\right)^{2}e^{-2s}\right]e^{2s}=0</math> | | <math>\ r = e^{s};</math> <math>\ u(r) = W(s)e^{\frac{1}{2}s}</math> |
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| | Note that, for <math>0<r<\infty,\!</math> <math>-\infty<s<\infty.\!</math> The radial equation becomes |
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| In this case the [[WKB Approximation#Bohr-Sommerfeld Quantization Rule|Bohr-Sommerfeld quantization rule]] becomes:
| | <math>\ \frac{d^{2}W}{ds^{2}}+\frac{2m}{\hbar^{2}}\left[E-V(e^{s})-\frac{\hbar^2}{2m}\left(l+\tfrac{1}{2}\right)^{2}e^{-2s}\right]e^{2s}=0.</math> |
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| <math>\ \int_{r_1}^{r_2}\sqrt{2m\left(E_n - V(r) - \frac{\hbar^2}{2m}\frac{(\ell+\frac{1}{2})^2}{r^2}\right)}dr = \left(n + \frac{1}{2}\right)\pi\hbar </math>
| | In this case, the [[WKB Approximation#Bound States Within the WKB Approximation|Bohr-Sommerfeld quantization rule]] is as in the purely one-dimensional case, but with an effective potential, |
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| == ? == | | <math>V_{\text{eff}}(r)=V(r)+\frac{\hbar^2(\ell+\frac{1}{2})^2}{2mr^2}.</math> |
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| For a central potential:
| | ==Problem== |
| :<math> p_r^2 = E - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}</math>
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| :<math>
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| \begin{align}
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| \int_{r_1}^{r_2}p_r(r)dr &= \int_{0}^{\infty}\sqrt{2m\left(E_n - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)}dr \\
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| &= \left(n + \frac{1}{2}\right)\pi\hbar
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| \end{align}
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| </math>
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| | Use the WKB approximation to estimate the energy spectrum for a Hydrogen atom. |
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| [[Phy5645/WKBenergyspectrum|Worked Problem]]
| | ''Hint'': Use the relation, |
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| [[Worked by team]]
| | <math>r^{2}-Vr+T=(r_{1}-r)(r_{2}-r),\!</math> |
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| ==WKB method for the Coulomb Potential ==
| | where |
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| For the coulomb potential, the potential is given by:
| | <math>T=-\frac{\hbar^2(l+\tfrac{1}{2})^2}{2mE},\,V=\frac{e^2}{E},</math> |
| :<math> V(r) = -\frac{-Ze^2}{r} </math>
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| Since the electron is bound to the nucleus, it can be veiwed as moving between two rigid walls at <math> r = 0 \!</math> and <math> r = a \!</math> with energy <math> E = V(a), a = -\frac{-Ze^2}{E}\!</math>. Thus, the energy of the electron is negative.
| | and <math>r_1\!</math> and <math>r_2\!</math> are the classical turning points of the (effective) potential appearing in the WKB approximation for this problem, and the integral, |
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| The energies of the s-state (<math> \ell = 0 \!</math>) can be obtained from:
| | <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> \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr = n\pi\hbar </math>
| | [[Phy5645/Hydrogen Atom WKB|Solution]] |
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| Using the change of variable: <math> x = \frac{a}{r} </math>
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| :<math>
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| \begin{align}
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| \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr &= \sqrt{-2mE} \int_0^a dr \sqrt{\frac{a}{r} - 1} \\
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| &= a\sqrt{-2mE} \int_0^1 dx\sqrt{\frac{1}{x} - 1} \\
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| &= \frac{\pi}{2}a\sqrt{-2mE} \\
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| &= -Ze^2\pi\sqrt{-\frac{2m}{E}}
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| \end{align}
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| </math>
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| Where I have used the integral
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| :<math> \int_0^1\sqrt{\frac{1}{x} -1} = \frac{\pi}{2} </math>
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| Thus we have the expression:
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| :<math>-Ze^2\pi\sqrt{-\frac{2m}{E}} = n\pi\hbar </math>
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| :<math>\Rightarrow E_n = -\frac{mZ^2e^4}{\hbar^2} = -\frac{Z^2e^2}{2a_0}</math>
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| Where <math> a_0\!</math> is the Bohr radius. Notice that this is the correct expression for the energy levels of a Coulomb potential.
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| [[Phy5645/Gamowfactor|Calculation of Gamow factor using WKB Aprroximation Method]] | |
It is possible to apply the WKB approximation to the radial equation using a method by R. E. Langer (1937).
Recall that
and that 
satisfies the effective one-dimensional Schrödinger equation,
![{\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dr^{2}}}+{\frac {\hbar ^{2}}{2m}}{\frac {l(l+1)}{r^{2}}}+V(r)-E\right]u(r)=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f629e168f20969584778fde0f5b38fdf0f87767f)
We now perform the following transformations:
Note that, for 
The radial equation becomes
![{\displaystyle \ {\frac {d^{2}W}{ds^{2}}}+{\frac {2m}{\hbar ^{2}}}\left[E-V(e^{s})-{\frac {\hbar ^{2}}{2m}}\left(l+{\tfrac {1}{2}}\right)^{2}e^{-2s}\right]e^{2s}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba62255b0f90657ea8b34a7085c2dfe72cafb10d)
In this case, the Bohr-Sommerfeld quantization rule is as in the purely one-dimensional case, but with an effective potential,
Problem
Use the WKB approximation to estimate the energy spectrum for a Hydrogen atom.
Hint: Use the relation,
where
and 
and 
are the classical turning points of the (effective) potential appearing in the WKB approximation for this problem, and the integral,
Solution
