= 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> gives
$$
v(t)=-\frac{eE_0}{m\omega}(\sin\beta-\sin\alpha),\qquad
x(t)=\frac{eE_0}{m\omega^2}\left[\cos\beta-\cos\alpha+(\beta-\alpha)\sin\alpha\right].
$$
A return requires $\beta>\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver <kinetic energy>
$$
U_p=\frac{e^2E_0^2}{4m\omega^2},\qquad
K_{\rm return}=2U_p(\sin\beta-\sin\alpha)^2.
$$
Searching the first-return solutions for release phases $0<\alpha<\pi/2$ gives a maximum near $\alpha=0.3134$ and $\beta=4.3990$, with $K_{\rm return}\simeq3.1731U_p$. 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>.
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