Einstein–Brillouin–Keller method

Examples

1D Harmonic oscillator

The Hamiltonian of a simple harmonic oscillator is given by

${\displaystyle H=\frac{p^2}{2m}+\frac{m{\omega }^2{x}^2}{2}}$

where ${\displaystyle p}$ is the linear momentum and ${\displaystyle x}$ the position coordinate. The action variable is given by

${\displaystyle I=\frac{2}{\pi }{\int }_0^{x_0}\sqrt{2mE-m^2{\omega }^2{x}^2}\mathrm{d}x}$

where we have used that ${\displaystyle H=E}$ is the energy and that the closed trajectory is 4 times the trajectory from 0 to the turning point ${\displaystyle x_0=\sqrt{2E/m{\omega }^2}}$.

The integral turns out to be

${\displaystyle E=I\omega }$,

which under EBK quantization there are two soft turning points in each orbit ${\displaystyle {\mu }_x=2}$ and ${\displaystyle b_x=0}$. Finally, that yields

${\displaystyle E=\hslash \omega (n+1/2)}$,

which is the exact result for quantization of the quantum harmonic oscillator.

2D hydrogen atom

The Hamiltonian for a non-relativistic electron (electric charge ${\displaystyle e}$) in a hydrogen atom is:

${\displaystyle H=\frac{p_r^2}{2m}+\frac{p_{\phi }^2}{2m{r}^2}-\frac{e^2}{4\pi {ϵ}_{0}r}}$

where ${\displaystyle p_r}$ is the canonical momentum to the radial distance ${\displaystyle r}$, and ${\displaystyle p_{\phi }}$ is the canonical momentum of the azimuthal angle ${\displaystyle \phi }$. Take the action-angle coordinates:

${\displaystyle I_{\phi }=\text{constant}=|L|}$

For the radial coordinate ${\displaystyle r}$:

${\displaystyle p_r=\sqrt{2mE-\frac{L^2}{r^2}+\frac{e^2}{4\pi {ϵ}_{0}r}}}$

${\displaystyle I_r=\frac{1}{\pi }{\int }_{r_1}^{r_2}{p}_{r}dr=\frac{m{e}^2}{4\pi {ϵ}_0\sqrt{-2mE}}-|L|}$

where we are integrating between the two classical turning points ${\displaystyle r_1,r_2}$ (${\displaystyle {\mu }_r=2}$)

${\displaystyle E=-\frac{m{e}^4}{32{\pi }^2{ϵ}_0^2(I_r+I_{\phi }{)}^2}}$

Using EBK quantization ${\displaystyle b_r={\mu }_{\phi }=b_{\phi }=0,n_{\phi }=m}$ :

${\displaystyle I_{\phi }=\hslash m;m=0,1,2,\cdots }$

${\displaystyle I_r=\hslash (n_r+1/2);n_r=0,1,2,\cdots }$

${\displaystyle E=-\frac{m{e}^4}{32{\pi }^2{ϵ}_0^2{\hslash }^2(n_r+m+1/2{)}^2}}$

and by making ${\displaystyle n=n_r+m+1}$ the spectrum of the 2D hydrogen atom is recovered :

${\displaystyle E_n=-\frac{m{e}^4}{32{\pi }^2{ϵ}_0^2{\hslash }^2(n-1/2{)}^2};n=1,2,3,\cdots }$

Note that for this case ${\displaystyle I_{\phi }=|L|}$ almost coincides with the usual quantization of the angular momentum operator on the plane ${\displaystyle L_z}$. For the 3D case, the EBK method for the total angular momentum is equivalent to the Langer correction.