<aside> π‘
βItβs a trap!β - Admiral Ackbar
<Star Wars: Episode VI - Return of the Jedi> (1983), R. Marquand.
$$ U = \frac{\hbar \Omega^2}{4\Delta} = \frac{\hbar \Gamma}{8} \frac{I/I_{\rm sat}}{\Delta/\Gamma} $$
</aside>
Under the dressed state picture, far-off detuned light ($|\Delta| \gg \Omega$) generates potential (AC-Stark shift),
$$ \left|g\right>: E_g=0 \rightarrow -\frac{\hbar \Omega^2}{4|\Delta|} $$
$$ \left|g\right>: E_g=0 \rightarrow +\frac{\hbar \Omega^2}{4|\Delta|} $$
Under the red detuned light, atoms are attracted to the potential center, however, under the blue detuned light, atoms are repulsed to the potential center.
S. Chu, Nobel Lecture: The Manipulation of Neutral Particles, Rev. Mod. Phys. 70, 685 (1998).
The depth of the dipole trap is
$$ U = \frac{\hbar \Omega^2}{4\Delta} = \frac{\hbar \Gamma}{8} \frac{I/I_{\rm sat}}{\Delta/\Gamma} $$
where $I_{\rm sat}$ is saturation intensity, $I/I_{\rm sat} = 2\Omega^2/\Gamma^2$.
XXX
$w_0$ is a beam size, $z_R$ is Reyleigh length, and $\lambda$ is wavelength of the light. By assuming the Gaussian beam under the assumption $r,z\ll 1$, the intensity of the light is
$$ I(r,z) \sim I_0 \left(1-\left(\frac{z}{z_R}\right)^2-2\left(\frac{r}{w_0}\right)^2\right) $$