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 $V={\frac {e^{2}}{E}},$

${\sqrt {2mE}}\int _{r_{1}}^{r_{2}}{\sqrt {1-{\frac {V}{r}}+{\frac {T}{r^{2}}}}}\,dr=(n+{\tfrac {1}{2}})\pi \hbar .$

Using the fact that $r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)\!$ and that

$\int _{r_{1}}^{r_{2}}{\sqrt {\frac {(x-r_{1})(x-r_{2})}{x^{2}}}}\,dx={\frac {\pi }{2}}({\sqrt {r_{2}}}-{\sqrt {r_{1}}})^{2},$

we obtain

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

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

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

${\text{so V=}}(r_{1}+r_{2}){\text{ and T=}}r_{1}r_{2}$

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

${\sqrt {2mE}}\left({-{\frac {e^{2}}{E}}-2{\sqrt {-{\frac {\hbar ^{2}l(l+1)}{2mE}}}}}\right)=(n+{\frac {1}{2}})\pi \hbar$

$-e^{2}{\sqrt {\frac {2m}{E}}}-2{\sqrt {\hbar ^{2}l(l+1)}}=2\hbar (n+{\frac {1}{2}})$

${\text{ }}2\hbar (n+{\frac {1}{2}})+2\hbar {\sqrt {l(l+1)}}=e^{2}{\sqrt {\frac {2m}{-E}}}$

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

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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