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

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

We now observe that $r^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})r+r_{1}r_{2},\!$ so that

$V=r_{1}+r_{2}\!$ and $T=r_{1}r_{2}.\!$

We thus obtain

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

or

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

This equation simplifies to

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

We may now easily solve for $E,\!$ obtaining

$E=-{\frac {me^{4}}{2\hbar ^{2}(n+\ell +1)^{2}}}.$

Note that this is exactly the spectrum that we would obtain from an exact solution of the Coulomb problem.

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