WKB in Spherical Coordinates: Difference between revisions
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==WKB method for the Coulomb Potential == | |||
For the coulomb potential, the potential is given by: | |||
:<math> V(r) = -\frac{-Ze^2}{r} </math> | |||
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. | |||
The energies of the s-state (<math> \ell = 0 \!</math>) can be obtained from: | |||
:<math> \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr = n\pi\hbar </math> | |||
Using the change of variable: <math> x = \frac{a}{r} </math> | |||
:<math> | |||
\begin{align} | |||
\int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr &= \sqrt{-2mE} \int_0^a dr \sqrt{\frac{a}{r} - 1} \\ | |||
&= a\sqrt{-2mE} \int_0^1 dx\sqrt{\frac{1}{x} - 1} \\ | |||
&= \frac{\pi}{2}a\sqrt{-2mE} \\ | |||
&= -Ze^2\pi\sqrt{-\frac{2m}{E}} | |||
\end{align} | |||
</math> | |||
Where I have used the integral | |||
:<math> \int_0^1\sqrt{\frac{1}{x} -1} = \frac{\pi}{2} </math> | |||
Thus we have the expression: | |||
:<math>-Ze^2\pi\sqrt{-\frac{2m}{E}} = n\pi\hbar </math> | |||
:<math>\Rightarrow E_n = -\frac{mZ^2e^4}{\hbar^2} = -\frac{Z^2e^2}{2a_0}</math> | |||
Where <math> a_0\!</math> is the Bohr radius. Notice that this is the correct expression for the energy levels of a Coulomb potential. | |||
[[Phy5645/Gamowfactor|Calculation of Gamow factor using WKB Aprroximation Method]] |
Revision as of 15:06, 14 August 2013
It is possible to apply the WKB approximation to the radial equation using a method by R. E. Langer (1937).
Recall: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ u(r)=rR(r)} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}
Now apply the transformations: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ r = e^{s};} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ u(r) = W(s)e^{\frac{1}{2}s}}
Note that for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ r } varying from 0 to infinity, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ s } will vary from minus infinity to plus infinity.
The radial equation then transforms into:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \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}
In this case the Bohr-Sommerfeld quantization rule becomes:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ \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 }
?
For a central potential:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_r^2 = E - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \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 \\ &= \left(n + \frac{1}{2}\right)\pi\hbar \end{align} }
WKB method for the Coulomb Potential
For the coulomb potential, the potential is given by:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r) = -\frac{-Ze^2}{r} }
Since the electron is bound to the nucleus, it can be veiwed as moving between two rigid walls at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = 0 \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = a \!} with energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = V(a), a = -\frac{-Ze^2}{E}\!} . Thus, the energy of the electron is negative.
The energies of the s-state (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell = 0 \!} ) can be obtained from:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr = n\pi\hbar }
Using the change of variable: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \frac{a}{r} }
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr &= \sqrt{-2mE} \int_0^a dr \sqrt{\frac{a}{r} - 1} \\ &= a\sqrt{-2mE} \int_0^1 dx\sqrt{\frac{1}{x} - 1} \\ &= \frac{\pi}{2}a\sqrt{-2mE} \\ &= -Ze^2\pi\sqrt{-\frac{2m}{E}} \end{align} }
Where I have used the integral
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^1\sqrt{\frac{1}{x} -1} = \frac{\pi}{2} }
Thus we have the expression:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -Ze^2\pi\sqrt{-\frac{2m}{E}} = n\pi\hbar }
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow E_n = -\frac{mZ^2e^4}{\hbar^2} = -\frac{Z^2e^2}{2a_0}}
Where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_0\!} is the Bohr radius. Notice that this is the correct expression for the energy levels of a Coulomb potential.