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

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

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:

${\displaystyle H={\frac {p_{r}^{2}}{2m}}+{\frac {p_{\varphi }^{2}}{2mr^{2}}}-{\frac {e^{2}}{4\pi \epsilon _{0}r}}}$

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

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

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

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

${\displaystyle I_{r}={\frac {1}{\pi }}\int _{r_{1}}^{r_{2}}p_{r}dr={\frac {me^{2}}{4\pi \epsilon _{0}{\sqrt {-2mE}}}}-|L|}$

where we are integrating between the two classical turning points ${\displaystyle r_{1},r_{2}}$ (${\displaystyle \mu _{r}=2}$)

${\displaystyle E=-{\frac {me^{4}}{32\pi ^{2}\epsilon _{0}^{2}(I_{r}+I_{\varphi })^{2}}}}$

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

${\displaystyle I_{\varphi }=\hbar m\quad$ ;\quad m=0,1,2,\cdots }

${\displaystyle I_{r}=\hbar (n_{r}+1/2)\quad$ ;\quad n_{r}=0,1,2,\cdots }

${\displaystyle E=-{\frac {me^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}(n_{r}+m+1/2)^{2}}}}$

and by making ${\displaystyle n=n_{r}+m+1}$ the spectrum of the 2D hydrogen atom ^[7] is recovered :

${\displaystyle E_{n}=-{\frac {me^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}(n-1/2)^{2}}}\quad$ ;\quad n=1,2,3,\cdots }

Note that for this case ${\displaystyle I_{\varphi }=|L|}$ 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.