Three-step model of high harmonic generation

For an original classical calculation of the return step, neglect the residual ion's Coulomb force after release. Take a spatially uniform driving electric field $E(t)=E_0\cos \omega t$, an Electron of electric charge $-e$ and mass $m$, and zero initial position and velocity at $t_i$. With $\alpha =\omega t_i$ and $\beta =\omega t$, integrating Newton's second law givesA return requires $\beta >\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver kinetic energySearching the first-return solutions for release phases $0<\alpha <\pi /2$ gives a maximum near $\alpha =0.3134$ and $\beta =4.3990$, with $K_{\mathrm{return}}\simeq 3.1731U_p$. For a chosen ionization energy $I_p$, recombination into the initial state can therefore emit a photon with energy up to approximately $I_p+3.17U_p$ in this model. This is an upper return-energy scale, not a prediction of the emitted yield. It omits release probabilities, quantum interference, propagation, phase matching and depletion of the neutral gas.

The calculation explains why a long driving wavelength can increase the available return energy: at fixed $E_0$, $U_p$ grows as ${\omega }^{-2}$. It does not say that conversion efficiency improves. The Electron must return with a probability amplitude for emission that survives the microscopic dynamics and adds coherently across the gas.