= Counterpropagating saturation resonance
{title2=$\delta-kv=\delta+kv=0$}
Take a single isolated optical transition of <angular frequency> $\omega_0$. Both beams have laboratory <angular frequency> $\omega_L$, with <laser detuning> $\delta=\omega_L-\omega_0$. Let the pump travel along $+z$, the probe along $-z$, $v$ denote an atom's $z$-<velocity>, and $k=\omega_0/c$ be the approximate vacuum <wavenumber>; $c$ is the <speed of light>. To first order in $v/c$, the <laser detunings> in the atom's frame are
$$
\Delta_{\rm pump}=\delta-kv,\qquad \Delta_{\rm probe}=\delta+kv.
$$
The pump addresses atoms near $v=\delta/k$, while the probe addresses atoms near $v=-\delta/k$. For the same transition, simultaneous exact resonance requires both expressions to vanish: adding and subtracting gives $\delta=0$ and $v=0$. The selected atoms can still have substantial transverse <velocity>. Finite <natural linewidth> admits a narrow interval of longitudinal <velocities>; a warm gas has not been cooled.
This is velocity selection, not cancellation of each atom's <Doppler shift> within a single absorption event. Changing <laser detuning> selects different atoms for ordinary <photon absorption>, but only near line centre do these two selected classes overlap. <Laser cooling> instead changes the distribution of the atoms' <velocities> through repeated momentum transfer.
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