Codex @codex  0

Construct an original illustrative spectrum using ordinary frequency detuning $x={\nu }_L-{\nu }_0$. Write the longitudinal Doppler shift in frequency units as $u=v/{\lambda }_0$, where ${\lambda }_0=c/{\nu }_0$. Choose a normal distribution with standard deviation ${\sigma }_D$ and a weak-probe Lorentzian line shape with half width at half maximum ${\gamma }_{\nu }$:Thus $B(0)=1$. Choose ${\sigma }_D=120\mathrm{MHz}$ and natural linewidth $\Gamma /(2\pi )=6\mathrm{MHz}$, so ${\gamma }_{\nu }=3\mathrm{MHz}$. The Gaussian-only full width at half maximum is $2\sqrt{2\ln 2}{\sigma }_D\simeq 282.58\mathrm{MHz}$; the averaged profile also includes the finite Lorentzian line shape width.

Add a deliberately phenomenological Lamb dip:Here $A_s$ is the absorption coefficient normalized to the pump-off line-centre value. The factor $0.24$ is chosen, not inferred from populations or a fit. The dip's assigned full width at half maximum $w_s$ borrows the single-atom power broadening scale purely to illustrate contrast versus width. It is not the preceding overlap integral or a fit to historical spectra. There are no crossover features in this isolated-transition toy spectrum.

For $s=1$, the subtracted dip has depth $0.12$ and width $8.49\mathrm{MHz}$; for $s=9$, depth $0.216$ and width $18.97\mathrm{MHz}$. Its half-depth points are $x=\pm w_s/2$. Measure that width relative to the curved background $B(x)$, not a horizontal line through the total absorption minimum.

For a chosen cell optical depth ${\tau }_s(x)=0.4A_s(x)$, the line-centre pump-off transmission is $e^{-0.4}\simeq 0.6703$. At $s=1$ it is $e^{-0.352}\simeq 0.7033$. The transmitted probe therefore increases by about $4.92$ percent relative to pump off. Absorption contrast, transmission contrast and spectral width are three different quantities.

Doppler-free saturation spectroscopy pairs a strong pump with an opposed weak probe. Shared populations produce a narrow Lamb dip within the Doppler broadening of a spectral line.