ABSTRACT:
We perform the first measurement of the thermal and ionization state of the intergalactic medium (IGM) across $0.9 \lt z \lt 1.5$ using 301 Ly $\,\alpha$ absorption lines fitted from 12 archival Hubble Space Telescope Space Telescope Imaging Spectrograph quasar spectra. We employ the machine-learning-based inference method that uses joint Doppler parameter–column density ($b{-}N_{{\rm {H\,{\small I}}}{}}$) distributions obtained from Ly$\,\alpha$ forest decomposition. Our results show that the H i photoionization rates, $\Gamma _{{\rm {H\,{\small I}}}{}}$, agree with recent ultraviolet background synthesis models, with $\log (\Gamma _{{\rm{H\,{\small I}}}}/\text{s}^{-1})={-11.79}^{+0.18}_{-0.15}$, ${-11.98}^{+0.09}_{-0.09}$, and ${-12.32}^{+0.10}_{-0.12}$ at $z=1.4$, 1.2, and 1, respectively. We obtain the IGM temperature at the mean density, $T_0$, and the adiabatic index, $\gamma$, as $[\log (T_0/\text{K}), \gamma ]=$ [${4.13}^{+0.12}_{-0.10}$, ${1.34}^{+0.10}_{-0.15}$], $[{3.79}^{+0.11}_{-0.11}$, ${1.70}^{+0.09}_{-0.09}]$, and $[{4.12}^{+0.15}_{-0.25}$, ${1.34}^{+0.21}_{-0.26}]$ at $z=1.4$, 1.2, and 1. Our measurements of $T_0$ at $z=1.4$ and 1.2 are consistent with the trend predicted from previous $z\lt 3$ temperature measurements and theoretical expectations, where the IGM cools down after He ii reionization in the absence of any non-standard heating. However, our $T_0$ measurement at $z=1$ shows unexpectedly high IGM temperature. Given the relatively large uncertainty in these measurements, where $\sigma _{T_0} \sim 5000$ K, mostly emanating from the limited size of our data set, we cannot conclude whether the IGM cools down as expected. Lastly, we generate mock data sets to test the constraining power of future measurement with larger data sets. The results demonstrate that, with redshift path-length $\Delta z \sim 2$ for each redshift bin, three times the current data set, we can constrain the $T_0$ of IGM within 1500 K, which would be sufficient to constrain the IGM thermal history at $z \lt 1.5$ conclusively.