= Solution
For highly ionized hydrogen in <photoionization equilibrium>, balance the <photoionization rate> against <radiative recombination>:
$$
n_{\rm HI}\Gamma_{\rm HI}=\alpha(T)n_en_{\rm HII}\simeq\alpha(T)n_H^2.
$$
Thus the neutral fraction is $n_{\rm HI}/n_H\simeq\alpha(T)n_H/\Gamma_{\rm HI}$ and the neutral <column density> across a proper path $L$ is
$$
N_{\rm HI}\simeq\frac{\alpha(T)n_H^2L}{\Gamma_{\rm HI}}.
$$
For an illustrative diffuse gas with $T\sim10^4\,\mathrm K$, $\alpha\sim4\times10^{-13}\,\mathrm{cm^3\,s^{-1}}$, $\Gamma_{\rm HI}\sim10^{-12}\,\mathrm{s^{-1}}$, and $n_H\sim10^{-5}\,\mathrm{cm^{-3}}$, the neutral fraction is only $4\times10^{-6}$. A proper path of $100$ kiloparsecs then gives $N_{\rm HI}\sim10^{13}\,\mathrm{cm^{-2}}$, enough to produce a weak <Lyman-alpha absorption> feature even though the gas is almost entirely ionized.
This hydrogen <number density> is of the order of the cosmic mean at redshift a few. Since the weak forest is common along most high-redshift <lines of sight>, substantial path length passes through such diffuse gas. Much denser material would give much larger <column density> for comparable path lengths and ionizing flux. These arguments support near-mean or modestly overdense <intergalactic medium> as the origin of weak forest lines, not the claim that every absorber has exactly the cosmic mean density: path length, temperature, and <ionizing background> must also be constrained.
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