Use WKB approximation to estimate energy spectrum for Hydrogen atom.
The approximation is:
∫ P ( r ) d r = ( n + 1 2 ) π ℏ {\displaystyle \int {P(r)}dr=(n+{\frac {1}{2}})\pi \hbar }
P(r)= 2 m ( E − V ( r ) ) = 2 m ( E − ( ℏ 2 l ( l + 1 ) 2 m r 2 − e 2 r ) ) {\displaystyle {\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)}})}
∫ r 1 r 2 2 m ( E − ℏ 2 l ( l + 1 ) 2 m r 2 + e 2 r ) d r = ( n + 1 2 ) π ℏ {\displaystyle \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.
2 m E ∫ r 1 r 2 ( 1 − ℏ 2 l ( l + 1 ) 2 m r 2 E + e 2 E r ) 1 / 2 d r = ( n + 1 2 ) π ℏ {\displaystyle {\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 }
let T= − ℏ 2 l ( l + 1 ) 2 m E and V= − e 2 r {\displaystyle {\text{let T=}}-{\frac {\hbar ^{2}l(l+1)}{2mE}}{\text{ and V=}}-{\frac {e^{2}}{r}}{\text{ }}}
2 m E ∫ r 1 r 2 ( 1 + T r 2 + V r ) 1 / 2 d r = ( n + 1 2 ) π ℏ {\displaystyle {\sqrt {2mE}}\int \limits _{r1}^{r2}{(1+{\frac {T}{r^{2}}}}+{\frac {V}{r}})^{1/2}dr=(n+{\frac {1}{2}})\pi \hbar {\text{ }}}
use the relation let r 2 − V r + T = ( r 1 − r ) ( r 2 − r ) {\displaystyle {\text{use the relation let r}}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)}
2 m E ∫ r 1 r 2 ( ( r 1 − r ) ( r 2 − r ) r 2 ) 1 / 2 d r = ( n + 1 2 ) π ℏ {\displaystyle {\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 }}
use definition let ∫ r 1 r 2 ( ( x − a ) ( x − b ) x 2 ) 1 / 2 d x = π 2 ( b − a ) 2 {\displaystyle {\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}}
2 m E ∗ π 2 ∗ ( r 2 − r 1 ) 2 = ( n + 1 2 ) π ℏ {\displaystyle {\sqrt {2mE}}*{\frac {\pi }{2}}*({\sqrt {r_{2}}}-{\sqrt {r_{1}}})^{2}=(n+{\frac {1}{2}})\pi \hbar }
2 m E ∗ π 2 ∗ ( r 2 + r 1 − 2 r 1 r 2 ) = ( n + 1 2 ) π ℏ {\displaystyle {\sqrt {2mE}}*{\frac {\pi }{2}}*(r_{2}+r_{1}-2{\sqrt {r_{1}r_{2}}})=(n+{\frac {1}{2}})\pi \hbar }
let r 2 − V r + T = ( r 1 − r ) ( r 2 − r ) = r 2 − ( r 1 + r 2 ) + r 1 r 2 {\displaystyle {\text{let r}}^{2}-Vr+T=(r_{1}-r)(r_{2}-r)=r^{2}-(r_{1}+r_{2})+r_{1}r_{2}}
so V= ( r 1 + r 2 ) and T= r 1 r 2 {\displaystyle {\text{so V=}}(r_{1}+r_{2}){\text{ and T=}}r_{1}r_{2}}
2 m E ∗ π 2 ∗ ( V − 2 T ) = ( n + 1 2 ) π ℏ {\displaystyle {\sqrt {2mE}}*{\frac {\pi }{2}}*(V-2{\sqrt {T}})=(n+{\frac {1}{2}})\pi \hbar }
2 m E ( − e 2 E − 2 − ℏ 2 l ( l + 1 ) 2 m E ) = ( n + 1 2 ) π ℏ {\displaystyle {\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 2 m E − 2 ℏ 2 l ( l + 1 ) = 2 ℏ ( n + 1 2 ) {\displaystyle -e^{2}{\sqrt {\frac {2m}{E}}}-2{\sqrt {\hbar ^{2}l(l+1)}}=2\hbar (n+{\frac {1}{2}})}
2 ℏ ( n + 1 2 ) + 2 ℏ l ( l + 1 ) = e 2 2 m − E {\displaystyle {\text{ }}2\hbar (n+{\frac {1}{2}})+2\hbar {\sqrt {l(l+1)}}=e^{2}{\sqrt {\frac {2m}{-E}}}}
4 ℏ 2 ( n + 1 2 + l ( l + 1 ) ) 2 2 m e 4 = 1 − E {\displaystyle {\frac {4\hbar ^{2}\left({n+{\frac {1}{2}}+{\sqrt {l(l+1)}}}\right)^{2}}{2me^{4}}}={\frac {1}{-E}}{\text{ }}}
E= − m e 4 2 ℏ 2 ( n + 1 2 + l ( l + 1 ) ) 2 {\displaystyle {\text{E=}}{\frac {-me^{4}}{2\hbar ^{2}\left({n+{\frac {1}{2}}+{\sqrt {l(l+1)}}}\right)^{2}}}}
where r1 and r2 are turning points in this case.
