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I am trying to find the classical turning points in semi-parabolic coordinates for the hydrogen atom when an electric field is being applied to it in the y-axis. I am reading an article for those who are interested called Classical, semiclassical, and quantum dynamics in the lithium Stark System published in Physical Review A Volume 51, Number 5. It gives me the following three separated equations based on the Hamiltonian:

.5 * (p

_{u})^2 + .5*F*u^4 - E(u^2) = e

_{u}

.5 * (p

_{v})^2 - .5*F*u^4 - E(u^2) = e

_{v}

and eu + e

_{v}= 2

u and v are semi-parabolic position coordinates. eu and ev are separation constants.

The article also says that for the purposes of finding closed orbits, that's what I am looking for, E = scaled energy and F = 1. Scaled energy can be picked to be an arbitrary number. Scaled energy = E / (F^.5)

I know that at the classical turning point the kinetic energy is equal to 0 so the equations above simplify to

.5*u^4 - (scaled_energy)*u^2 = e

_{u}

-.5*v^4 - (scaled_energy)*v^2 = e

_{v}

Finding the classical turning point will hopefully help me determine closed orbits for hydrogen as they give me two integrals for determining when an orbit is closed if the electron is launched at a certain angle in a Field=1 and and Energy=Scaled Energy.

I have figured out that in the primitive orbit of hydrogen where the electron is launched vertically in the y-axis, parallel to the field, the electron will have u = 0 position and a velocity in the u direction = 0, so the quartic equation

0.5v^4 - (scaled_energy)*v^2 = e

_{v}= 2

will determine the classical turning point for that orbit.

Thank you