Phy5645/Hydrogen Atom WKB

The WKB approximation is given by

$\int _{r_{1}}^{r_{2}}p(r)\,dr=(n+{\tfrac {1}{2}})\pi \hbar ,$

where

$p(r)={\sqrt {2m(E-V_{\text{eff}}(r))}}={\sqrt {2m\left(E+{\frac {e^{2}}{r}}-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})^{2}}{2mr^{2}}}\right)}}.$

We may rewrite the above as

${\sqrt {2mE}}\int _{r_{1}}^{r_{2}}{\sqrt {1-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})^{2}}{2mEr^{2}}}+{\frac {e^{2}}{Er}}}}\,dr=(n+{\tfrac {1}{2}})\pi \hbar ,$

or, making the substitution,

$T=-{\frac {\hbar ^{2}(l+{\tfrac {1}{2}})}{2mE}}$ and $V={\frac {e^{2}}{E}},$

${\sqrt {2mE}}\int \limits _{r1}^{r2}{(1+{\frac {T}{r^{2}}}}+{\frac {V}{r}})^{1/2}dr=(n+{\frac {1}{2}})\pi \hbar {\text{ }}$

${\text{the relation r}}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)$

${\sqrt {2mE}}\int \limits _{r1}^{r2}{\left({\frac {(r_{1}-r)(r_{2}-r)}{r^{2}}}\right)^{1/2}dr=(n+{\frac {1}{2}})\pi \hbar }$

${\text{ the definition }}\int \limits _{r1}^{r2}{\left({\frac {(x-a)(x-b)}{x^{2}}}\right)^{1/2}dx}={\frac {\pi }{2}}({\sqrt {b}}-{\sqrt {a}})^{2}$

${\sqrt {2mE}}*{\frac {\pi }{2}}*({\sqrt {r_{2}}}-{\sqrt {r_{1}}})^{2}=(n+{\frac {1}{2}})\pi \hbar$

${\sqrt {2mE}}*{\frac {\pi }{2}}*(r_{2}+r_{1}-2{\sqrt {r_{1}r_{2}}})=(n+{\frac {1}{2}})\pi \hbar$

${\text{let r}}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})+r_{1}r_{2}$

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 \text{so V=}(r_{1}+r_{2})\text{ and T=}r_{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 \sqrt {2mE} *\frac{\pi }{2}*(V-2\sqrt {T} )=(n+\frac{1}{2})\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 \sqrt {2mE} \left ({-\frac{e^{2}}{E}-2\sqrt {-\frac{\hbar ^{2}l(l+1)}{2mE}} } \right )=(n+\frac{1}{2})\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 -e^{2}\sqrt {\frac{2m}{E}} -2\sqrt {\hbar ^{2}l(l+1)} =2\hbar (n+\frac{1}{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 \text{ }2\hbar (n+\frac{1}{2})+2\hbar \sqrt {l(l+1)} =e^{2}\sqrt {\frac{2m}{-E}} }

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{4\hbar ^{2}\left ({n+\frac{1}{2}+\sqrt {l(l+1)} } \right )^{2}}{2me^{4}}=\frac{1}{-E}\text{ }}

Then if we finally pull out E,

${\text{E=}}{\frac {-me^{4}}{2\hbar ^{2}\left({n+{\frac {1}{2}}+{\sqrt {l(l+1)}}}\right)^{2}}}$

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