We expect the transmission data to theoretically be a sum of multiple Lorentzian functions. With the transitions we measure, we expect to see three hyperfine peaks. However, due to crossovers, we actually see six. Thus we can find the transition frequencies associated with the hyperfine structure of \(^{85}Rb\) and \(^{87}Rb\) by fitting the raw transmission data to a function comprised of six Lorentzians and an offset. The equation for a Lorentzian is, \[\label{eq2} T(\omega,\omega_{r})=\left( \frac{\Gamma}{2}\right )^{2} \frac{A}{(\omega-\omega_{r})^{2}+\frac{\Gamma}{2}^{2}}\] where \(\omega_{r}\) is the resonant frequency of a given hyperfine transition. Since we have a scan with 6 Lorentzians, our fit model is \[\label{eq1} T(\omega)=T(\omega,\omega_{1})+T(\omega,\omega_{2})+\cdots +T(\omega,\omega_{6})+\text{offset}\] where the offset is a polynomial fit in time up to order 2 and \(\omega_{1},\cdots,\omega_{6}\) are the resonant frequency of the hyperfine peaks in the scan ordered by time. Thus the transition frequency of the different peaks should theoretically be given by, \(\omega_{1},\cdots,\omega_{6}\). The offset accounts for the background Gaussian distribution, which the Lorentzian functions are superimposed onto.

We used a \(\tilde{\chi}_\nu^2\) test to determine if the proposed fit model (Equation \ref{eq2}), is in good agreement with the data. From the four plots of transmission data and their corresponding fits shown in Figure \ref{fig:HyperfineFits}, the values of \(\tilde{\chi}_\nu^2\) ranged from \(0.08\) to \(0.17\). Since we expect that for a reasonable fit, \(\tilde{\chi}_\nu^2\approx 1\) we reject the null hypothesis. In this specific case we do not believe that this indicates a disagreement between our proposed model and the data, since it qualitatively appears to be a good fit. Instead we believe \(\tilde{\chi}_\nu^2 \ll 1\) due to an error in the uncertainty calculations. Thus we believe that our error bars on the plot of \(\frac{\delta t}{\delta \omega}\) versus the order of the ratios as they appear in time (Fig. \ref{fig:Ratios}) should be much larger. Therefore it is likely that given larger error bars, the plots of \(\frac{\delta t}{\delta \omega}\) should either fit a line with a slope equal to zero or some non zero slope, which would indicate that our frequency scan is non-linear.