Phy5645/Hydrogen Atom WKB

Use WKB approximation to estimate energy spectrum for Hydrogen atom. Hint: ${\text{use the relation r}}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)$ </math>

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The approximation is:

$\int {P(r)}dr=(n+{\frac {1}{2}})\pi \hbar$

${\text{P(r)=}}{\sqrt {2m(E-V(r))}}={\sqrt {2m(E-\left({{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}-{\frac {e^{2}}{r}}}\right)}})$

$\int \limits _{r1}^{r2}{\sqrt {2m(E-{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}+{\frac {e^{2}}{r}})}}dr=(n+{\frac {1}{2}})\pi \hbar$ where r1 and r2 are turning points in this case.

${\sqrt {2mE}}\int \limits _{r1}^{r2}{(1-}{\frac {\hbar ^{2}l(l+1)}{2mr^{2}E}}+{\frac {e^{2}}{Er}})^{1/2}dr=(n+{\frac {1}{2}})\pi \hbar$ if we do this substitution: ${\text{let T=}}-{\frac {\hbar ^{2}l(l+1)}{2mE}}{\text{ and V=}}-{\frac {e^{2}}{r}}{\text{ }}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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