\mathcal{M}= &\frac{g_{\chi}g_{e}}{(q^{2}-m_{V}^{2})}\bar{u}(p_{\chi ^{\prime}}, {s} ^{\prime})\gamma^{\mu}\gamma^{5}u(p_{\chi}, s)\bar{u}(p ^{\prime}_{e}, s^{\prime}_e)\gamma_{\mu}\gamma^{5}u(p_{e}, s_e)
\label{eq:amplitude}
\end{align}

We have assigned 4-momenta $p_{\chi}$ ($p^\prime_{\chi}$) and $p_e$ ($p^\prime_{e}$) to the dark matter and electron, respectively for the initial state (final state). We have also defined the momentum transfer $q$ as,

\begin{equation}
q^2 \equiv (p ^{\prime}_{\chi}-p_{\chi})^2= (p_{e}-p ^{\prime}_{e})^2. % \simeq -m_H^2 (\vec{v}_{\mathrm{in}}-\vec{v}_{\mathrm{out}})^2,
\end{equation}

In the non-relativistic limit we can show that the amplitude eq. <ref> can be written as [89],

\begin{align}
\mathcal{M} \simeq - 16 \frac{g_{\chi}g_{e}}{(q^{2}-m_{V}^{2})} m_\chi m_e \langle {s} ^{\prime}, s^{\prime}_e\vert \vec{S}_\chi \cdot \vec{S}_e\vert   {s} , s_e\rangle,
\end{align}

where $\vec{S} = \frac{1}{2}\vec{\sigma}$ denotes the spin projection operator and $\vec{\sigma}$ are Pauli matrices.
We can further write

\begin{align}
\vec{S}_{\chi}.\vec{S}_{e}=\frac{1}{4}\left(2({\sigma^{+}_{\chi}\sigma^{-}_{e}}+\sigma^{-}_{\chi}\sigma^{+}_{e})+\sigma^z_{\chi}\sigma^z_{e} \right),
\end{align}

where $\sigma^z$ is the diagonal Pauli matrix, and we have used the spin raising and lowering operators written in terms of the Pauli matrices $\sigma^{\pm}=\frac{1}{2}(\sigma^{x}\pm i \sigma^{y})$.
Thus, the matrix element in the scattering amplitude can be evaluated as,

\begin{align}
\langle {s} ^{\prime}, s^{\prime}_e\vert \vec{S}_\chi \cdot \vec{S}_e\vert   {s} , s_e\rangle =
\begin{blockarray}{*{4}{c} l}
\begin{block}{*{4}{>{$\footnotesize}c<{$}} l}
$\uparrow \uparrow$ & $\uparrow \downarrow$ & $\downarrow \uparrow$ & $\downarrow \downarrow$ & \\
\end{block}
\begin{block}{(*{4}{c})>{$\footnotesize}l<{$}}
\bigstrut[tb] \, \,\,\frac{1}{4} & \, \, \, 0 & \, \, \, 0 & 0 \bigstrut[tb]& $\uparrow \uparrow$ \\
0 & -\frac{1}{4} & \, \, \,\frac{1}{2} & 0 & $\uparrow \downarrow$  \\
0 & \, \, \, \frac{1}{2} & -\frac{1}{4} & 0  & $\downarrow \uparrow$ \\
\bigstrut[tb] \, \,\, \,0 & \, \, \, 0 & \, \, \, 0 & \frac{1}{4} \bigstrut[tb] & $\downarrow \downarrow$ \\
\end{block}
\end{blockarray},
\end{align}
where the column (row) labels refer to the spin of the outgoing (incoming) dark matter and electron respectively.
Using the $e \chi \rightarrow e \chi$ scattering amplitude, we can now evaluate the hydrogen excitation amplitude by using the bound state wave functions. In the non-relativistic limit we can identify the 4-vector momentum transfer as $q \simeq \left (0 , m_H (\vec{v}_0- \vec{v}_1) \right ) \equiv (0,\vec{q} \,) $.

We can then write the resulting amplitudes for all spin combinations in terms of the following master amplitude which we denote as $\mathcal{M}_0$,
\begin{equation}
\mathcal{M}_0 = 4\sqrt{2} g_\chi g_e \frac{m_H m_\chi}{q^2-m_V^2 } F(q^2)
\end{equation}
where $F(q^2)$ is the hydrogen atom form factor,
\begin{align}
F(q^2) &= \int\frac{d^{3}k}{(2\pi)^{3}}\tilde{\psi}_{1s}(\vec{k})\tilde{\psi}_{1s}^{\ast}(\vec{k}+\vec{q}), \\
& = \left(\frac{1}{1+\frac{q^2 a_0^2}{4}}\right )^2,\\
&\rightarrow 1 \textrm{\, for \,} q^2 \ll \frac{1}{a_0^2}.
\end{align}
The full amplitudes are non-zero only for the following spin combinations,
\begin{align}
\mathcal{M}(\chi (s=\uparrow)+H_{0}\rightarrow \chi(s^\prime=\downarrow)+H_{1}(s ^{\prime}_H=1)) & =    \mathcal{M}_0, \\
\mathcal{M}(\chi (s=\downarrow)+H_{0}\rightarrow \chi(s^\prime=\uparrow)+H_{1}(s ^{\prime}_H=-1)) &=  - \mathcal{M}_0, \\
\mathcal{M}(\chi (s=\uparrow)+H_{0}\rightarrow \chi(s^\prime=\uparrow)+H_{1}(s ^{\prime}_H=0)) &=  -\frac{1}{\sqrt{2}}  \mathcal{M}_0, \\
\mathcal{M}(\chi (s=\downarrow)+H_{0}\rightarrow \chi(s^\prime=\downarrow)+H_{1}(s ^{\prime}_H=0)) &=  \frac{1}{\sqrt{2}}  \mathcal{M}_0,
\end{align}
where $s ^{\prime}_H$ denotes the $z$ component of the spin of the final state triplet.
Squaring the amplitude and taking a spin-sum over final states and averaging over initial states we get,
\begin{align}
\frac{1}{(2 S_\chi+1)(2 S_0+1)} \sum\limits_{\{\textrm{spins} \}} \vert \mathcal{M} \vert^2 = \frac{3}{2} \vert \mathcal{M}_0 \vert^2,
\label{eq:ex_amp_squared}
\end{align}
where $S_\chi=1/2$, $S_0=0$ are the spins of $\chi$ and $H_0$ respectively.
\subsubsection{Amplitude for de-excitation}
For the de-excitation process, we can similarly show that the spin-summed and averaged amplitude is given by
\begin{align}
\frac{1}{(2 S_\chi+1)(2 S_1+1)} \sum\limits_{\{\textrm{spins} \}} \vert \mathcal{M} \vert^2 = \frac{1}{2} \vert \mathcal{M}_0 \vert^2,
\end{align}
where  $S_1=1$ is the spin of the triplet $H_1$.

\subsection{Excitation and de-excitation rates}
\label{sec:appendixb}
We will now use the amplitudes that we have computed to work out the excitation and de-excitation rates.
The rate for excitations $\chi+H_{0}\rightarrow \chi +H_{1}$ is given by,
\begin{align}
D_{01}  \equiv & \,   n_\chi \langle  \sigma_{01} v_{\textrm{rel}}\rangle  \nonumber \\
= & \,   n_\chi  (2\pi)^6 \int \frac{d^3p_{i}}{(2\pi)^3 2 E_{i}} \frac{d^3p_{0}}{(2\pi)^3 2 E_{0}} \frac{d^3p_{f}}{(2\pi)^3 2 E_{f}}  \frac{d^3p_{1}}{(2\pi)^3 2 E_{1}} f(p_{i}) f(p_0)
\nonumber \\
& \,  \times \frac{1}{(2 S_\chi + 1)(2S_0 + 1)} \sum\limits_{\{\textrm{spins} \}} \vert \mathcal{M} \vert^2 (2\pi)^4 \delta^{(4)}(p_i + p_0 - p_f - p_1),
\label{eq:excitationrate}
\end{align}
where we have taken $p_i$ and $p_0$ to be the momenta of the initial dark matter particle and singlet $H_0$, respectively and $p_f$ and $p_1$ to be the momenta of the final state dark matter particle and triplet $H_1$. The distribution functions for the initial particles are taken to be of the form $f(p_{i}) = \frac{1}{(2\pi m_\chi T_\chi)^{3/2}} e^{-\frac{p_i^2}{2 m_\chi T_\chi}}$ and $f(p_{0}) = \frac{1}{(2\pi m_H T_K)^{3/2}} e^{-\frac{p_0^2}{2 m_H T_K}}$, where $T_\chi$ is the temperature of the dark matter and $T_K$ the gas temperature.
We can now change integration variables for the initial state particles by defining the relative momenta $\vec{p}$ and a conjugate momentum variable $\vec{p}_{m}$,
\begin{align}
\vec{p} \equiv \mu \vec{v}_{\textrm{rel}}  = \mu \left( \frac{\vec{p}_0}{m_H} - \frac{\vec{p}_i}{m_\chi} \right ), \\
\vec{p}_{m} = M \left (\frac{\frac{\vec{p}_0}{T_K} + \frac{\vec{p}_i}{T_\chi}}{\frac{m_H}{T_K} + \frac{m_\chi}{T_\chi} }   \right ),
\end{align}
where $\vec{v}_{\textrm{rel}}$ is the relative velocity between the initial state particles, $ \mu = m_H m_\chi /(m_H + m_\chi)$ is the reduced mass of dark matter and hydrogen and $M = m_H +m_\chi$ is their mass sum. Then we can replace,
\begin{align}
\int d^3 p_{i} d^3 p_{0}  f(p_{i}) f(p_0) \rightarrow  \int  d^3 p d^3 p_{m}  f(p) f({p}_{m})
\end{align}
where the effective distribution functions for $p$ and $p_m$ are given respectively by,
\begin{align}
f(p) =  \frac{1}{(2\pi \mu \Teff)^{3/2}} e^{-\frac{p^2}{2 \mu \Teff}}, \textrm{\, and} \\
f(p_m) =   \frac{1}{(2\pi M T_m)^{3/2}} e^{-\frac{p_{m}^2}{2 M T_m}},
\end{align}
where we have defined $\Teff = \mu  \left (\frac{T_\chi}{m_\chi} + \frac{T_K}{m_H} \right)$ and $T_m = \frac{T_K T_\chi}{\Teff}$.
Making this substitution in eq.~\ref{eq:excitationrate} and using the fact that the excitation cross-section only depends on the relative momentum $p$ and not on $p_m$, we can trivially perform the integration over $p_m$ and write
\begin{align}
D_{01}  =  n_\chi \langle \sigma_{01} v_{\textrm{rel}} \rangle = n_\chi \int  d^3p  f(p) \sigma_{01} v_{\textrm{rel}}.
\label{eq:exrate}
\end{align}
We can evaluate the cross-section in the center-of-momentum frame (COM) and then identify $p$ as the magnitude of incoming momentum of either particle in this frame. In terms of the COM frame scattering angle $\theta$, one can evaluate the cross-section as,
\begin{align}
\sigma_{01} v_{\textrm{rel}} = \frac{1}{2 E_i}\frac{1}{2 E_0}\frac{1}{16 \pi} \int d(\cos\theta) \frac{2 p^\prime}{E_i + E_0} \frac{1}{2} \sum\limits_{\{\textrm{spins} \}} \vert \mathcal{M} \vert^2,
\end{align}
where the final state momentum in the COM frame is $p^\prime \simeq \sqrt{p^2 - p_{\textrm{th}}^2}$  and  $p_{\textrm{th}} = \sqrt{2 \Delta \mu}$ is the excitation threshold momentum, with $\Delta$ being the energy splitting between the singlet and triplet states.
The momentum transfer $q^2$ can be written as $q^2 \simeq -p^2 -  p^{\prime 2} + 2 p p^\prime \cos \theta$. Substituting our expression for the amplitude-squared worked out in the previous sub-section eq.~\ref{eq:ex_amp_squared}, we obtain
\begin{align}
\sigma_{01} v_{\textrm{rel}} \simeq \frac{3}{2 \pi} g_\chi^2 g_e^2\mu p^\prime  \int_{-1}^{1}  d(\cos\theta) \left (\frac{1}{q^2-m_V^2} \right)^2 F^2(q^2).
\end{align}
The scattering process is dominated by forward scattering which has a low momentum transfer and thus we can approximate $q^2 \ll 1/a_0^2$, and we can then take the hydrogen form factor $F(q^2) \simeq 1$. The angular integral then simplifies to,

\begin{align}
\mathcal{I} & = \int_{-1}^{1}  d(\cos\theta) \left (\frac{1}{-p^2 -  p^{\prime 2}  - m_V^2 + 2 p p^\prime \cos \theta } \right)^2 ,\\
&= \frac{2}{( p_{\textrm{th}}^2 - m_V^2)^2 + 4 m_V^2 p^2} \, ,
\label{eq:spinflipcrossintegrand}
\end{align}
where we have also substituted for $p^\prime$ in terms of $p$ and $p_{\textrm{th}}$ in the last line.
Note that the angular integration would be singular in the limit that $m_V \rightarrow 0$ and  $p_{\textrm{th}}\rightarrow 0$. This is due to the usual forward elastic-scattering singularity with a massless mediator. The divergence in the angular integration is cut-off by both the finite mediator mass, as well as by the threshold momentum of the inelastic reaction.

We choose to work in the limit $m_V \rightarrow 0$, where the divergence in the angular integral is dominantly cut-off by the threshold momentum for excitations rather than the mediator mass. To be precise, we need to work in the limit $m_V^2 \ll  p_{\textrm{th}}^4 /p_c^2$, where $p_c^2= 2\mu \Teff$ is the characteristic relative momentum in the thermal distribution\footnote{This condition is more stringent than the condition $m_V^2 \ll  p_{\textrm{th}}^2$ which allows us to neglect the mediator mass in the first term in brackets in the denominator of eq.~\ref{eq:spinflipcrossintegrand}, but not in the second.}. This condition can be expressed as,

\begin{align}
m_V \ll \sqrt{\frac{2\mu}{\Teff}} \Delta \simeq 2.3~\textrm{eV} \sqrt{\left(\frac{1000~\textrm{K}}{\Teff}\right)\left( \frac{\mu}{\textrm{0.1~GeV}}\right)}.
\label{eq:medmass_upper}
\end{align}
In this limit, we obtain the excitation cross-section as,
\begin{align}
\sigma_{01} v_{\textrm{rel}} \simeq \frac{3}{4\pi} \frac{g_\chi^2 g_e^2}{\Delta^2} \frac{p^\prime}{\mu}.
\end{align}
Upon performing thermal averaging, we find
\begin{align}
\langle \sigma_{01} v_{\textrm{rel}} \rangle  &\simeq \int  d^3p  f(p) \sigma_{01} v_{\textrm{rel}}, \\
&= \frac{3}{4\pi} \frac{g_\chi^2 g_e^2}{\Delta^2} \sqrt{\frac{8 \Teff}{\pi \mu}} e^{-\frac{\Delta}{\Teff}}.
\end{align}
Here,  $\langle v_{\textrm{rel}} \rangle = \sqrt{\frac{8 \Teff}{\pi \mu}}$ is the thermal average relative velocity and the exponential suppression factor is due to thermal suppression of the excitation reaction.
Similarly, for the de-excitation cross-section in the limit of negligible mediator mass we obtain,
\begin{align}
\langle  \sigma_{10} v_{\textrm{rel}} \rangle  \simeq  \frac{1}{4\pi} \frac{g_\chi^2 g_e^2}{\Delta^2} \sqrt{\frac{8 \Teff}{\pi \mu}}.
\end{align}
Note the key differences of absence of a factor of 3 (due to fewer final states) and absence of an exponential suppression factor (due to lack of a threshold energy for de-excitation).
The excitation and de-excitation rates $D_{01}$ and $D_{10}$ are found by simply multiplying these thermal cross-sections with the number density of dark matter particles at the relevant red-shift. The ratio of rates is given by,
\begin{align}
\frac{D_{01}}{D_{10}} = 3 e^{-\frac{\Delta}{\Teff}}.
\end{align}

The characteristic low momentum DM spin-flip interaction cross-section is given by,
\begin{align}
\sigma_{01}  \simeq 3 \sigma_{10}  &\simeq  12\pi    \frac{\alpha_\chi \alpha_e}{\Delta^2} \\
&\simeq  4.2\times 10^{-14}\left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right) ~\textrm{cm}^2 \simeq 4.2 \times10^{10}\left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right)~\textrm{barns},
\end{align}
where we have rewritten the coupling constants in terms of $\alpha_\chi = g_\chi^2/4\pi$ and $\alpha_e = g_e^2/4\pi$. The typical relative velocity between the DM and hydrogen is given by  $\langle v_{\textrm{rel}} \rangle =  \sqrt{8\Teff/\pi\mu } \simeq 1.4~\textrm{km/s}\sqrt{\left(\frac{\Teff}{10~\textrm{K}} \right) \left ( \frac{0.1 \textrm{~GeV}}{\mu}\right )}$.

These cross-sections are huge by particle physics standards, and this is all the more surprising given the benchmark couplings we have assume are so weak. The reason for the large cross-section is easy to understand however: \textbf{\textit{Since the mass splitting between the singlet and triplet states is tiny, the t-channel scattering of dark matter with these states is almost elastic and therefore has a nearly divergent scattering cross-section driven by the large probability for forward scattering. This divergence is cut-off by the tiny hyperfine mass-splitting between the singlet and triplet states and leads to a large cross-section for the spin-flip interaction.}}

The number density of dark matter at a red-shift $z$ can be written as\footnote{Here, for simplicity we are considering the homogenous dark matter distribution only. At low redshifts there will be a boost to the interaction rate (which can be parameterized as an effective scaling of the number density of dark matter) due to the formation of halos.}
\begin{align}
n_\chi  &=  \frac{f \Omega_{\textrm{DM}} \rho_c}{m_\chi} (1+z)^3, \\
&\simeq  1.2 \times 10^{-44} \left( \frac{f}{0.1} \right) \left( \frac{0.1~\textrm{GeV}}{m_\chi} \right) \left( \frac{1+z}{1+10} \right)^3~\textrm{GeV}^{3}, \\
&\simeq 1.5 \times 10^{-3} \left( \frac{f}{0.1} \right) \left( \frac{0.1~\textrm{GeV}}{m_\chi} \right) \left( \frac{1+z}{1+10} \right)^3~\textrm{cm}^{-3}.
\end{align}
where $\rho_c = 3.63 \times 10^{-47}$~GeV$^4$ is the critical density and $\Omega_{\textrm{DM}}  = 0.26$ is the present day DM relic density fraction and we have allowed for the possibility that the species we are considering makes up only a fraction $f$ of the dark matter.
Thus, we can write the excitation and de-excitation rates as,
\begin{align}
\label{eq:excitationanddeexrate}
D_{10}  &\simeq \frac{1}{3}D_{01}, \nonumber \\
& = n_\chi 4 \pi \frac{\alpha_\chi \alpha_e}{\Delta^2} \sqrt{\frac{8 \Teff}{\pi \mu}} \nonumber \\
& =1.99 \times  10^{-36} \left( \frac{f}{0.1} \right)  \left( \frac{0.1~\textrm{GeV}}{m_\chi} \right) \left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right)  \left(\frac{0.1~\textrm{GeV}}{\mu} \right)^{\frac{1}{2}} \left(\frac{\Teff}{10~\textrm{K}} \right)^{\frac{1}{2}}\left( \frac{1+z}{1+10} \right)^3~\textrm{GeV},  \nonumber \\
&= 3.01\times 10^{-12}\left( \frac{f}{0.1} \right)  \left( \frac{0.1~\textrm{GeV}}{m_\chi} \right)  \left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right)    \left(\frac{0.1~\textrm{GeV}}{\mu} \right)^{\frac{1}{2}} \left(\frac{\Teff}{10~\textrm{K}} \right)^{\frac{1}{2}} \left( \frac{1+z}{1+10} \right)^3~\textrm{s}^{-1}.
\end{align}

\subsection{Kinetic energy transfer rate}
\label{sec:appendixc}
The same spin-flip interactions that we had considered $\chi_{s}+H_{0}\leftrightarrows \chi_{{s}^{\prime}} +H_{1, {s^{\prime}_H}}$, can lead to kinetic energy transfer from the DM to neutral hydrogen and vice-versa. It is also possible to have kinetic energy transfer via the elastic reaction $\chi_{s} + H_{1} \rightarrow \chi_{{s}^{\prime} }+ H_{1}$\footnote{This reaction proceeds only through the $s^{\prime}_H=\pm1$ states of $H_1$ without spin-flip. Both the other possible elastic scattering processes, involving the $s^{\prime}_H=0$ state of $H_1$ and the process $\chi + H_0 \rightarrow \chi + H_0$, have vanishing amplitudes.}, however we will argue that this contribution is sub-dominant.

In this sub-section we will compute the kinetic energy transfer rate and the effect on temperature evolution of the dark matter and neutral hydrogen species.

The temperature evolution of a species can be computed from the second moment of the Boltzmann equation assuming that scattering does not distort the distribution far away from thermal. For the DM species for example, we can write down the temperature evolution equation as,
\begin{equation}
\frac{d T_\chi}{dt} = - 2 H T_\chi + \frac{2}{3} \dot{Q}_\chi,
\end{equation}
where $\dot{Q}_\chi$ is the dark matter heating rate due to kinetic energy transfer from the hotter hydrogen gas.
We can write an expression for $\dot{Q}_\chi$  as,
\begin{align}
\dot{Q}_\chi  &\equiv \dot{Q}^{01}_\chi + \dot{Q}^{10}_\chi,\\
&\equiv    n_0 R_{01}  +   n_1 R_{10} ,
\label{eq:energytransferrate}
\end{align}
where the energy transfer weighted rate coefficient for excitation is $R_{01}$ and for de-excitations is $R_{10}$.
Symbolically, we may express the rate coefficient for excitations as,
\begin{align}
R_{01} \sim \langle E_T \overline{\sigma}_{01} v_{\textrm{rel}}\rangle,
\end{align}
where $E_T$ is the energy transfer and $\overline{\sigma}_{01}$ is the ``energy-transfer'' cross-section.
A more precise definition is as follows,
\begin{align}
R_{01}    =& \,  (2\pi)^6\int \frac{d^3p_{i}}{(2\pi)^3 2 E_{i}} \frac{d^3p_{0}}{(2\pi)^3 2 E_{0}} \frac{d^3p_{f}}{(2\pi)^3 2 E_{f}}  \frac{d^3p_{1}}{(2\pi)^3 2 E_{1}} f(p_{i}) f(p_0) \times (E_f-E_i) \nonumber \\
&\times \left( \frac{1}{(2 S_\chi + 1)(2S_0 + 1)} \sum\limits_{\{\textrm{spins} \}} \vert \mathcal{M} \vert^2  (2\pi)^4 \delta^{(4)}(p_i + p_0 - p_f - p_1) \right ).
\end{align}

A similar expression holds for the de-excitation energy transfer rate coefficient.
It is instructive to explicitly show how to evaluate the spin-excitation energy-transfer rate, which we will do for excitations and we will just state the result for de-excitations. We begin by making the change of variables from $p_0$, $p_i$ $\rightarrow$ $p$, $p_m$ as before.
Then, taking the relative momentum to be along the $z$-axis without loss-of-generality, and defining $\theta$ as the scattering angle in the COM frame, the energy transfer can be written as,
\begin{equation}
(E_f-E_i) \simeq   \left(\frac{T_K - T_\chi}{\Teff} \right) \left(\frac{p^2}{M}\right) \left(  1 - \frac{p^\prime}{p} \cos \theta    \right) + \textrm{linear corrections in $p_m$}.
\end{equation}
Once again $p$ is the relative momentum in the COM frame of the incoming particles and $p^{\prime 2} \simeq p^2 - 2\mu \Delta$ is the outgoing momentum in the COM frame. The linear correction in $p_m$ to the energy transfer is unimportant since it does not contribute once we integrate over all possible directions of $p_m$.
The integration over $p_m$ can once again be performed trivially and we are thus left with an integral over relative momentum $p$,
\begin{align}
R_{01}  =  \int d^3p f(p) E_T \overline{\sigma}_{01} v_{\textrm{rel}}
\end{align}
where $f(p)=   \frac{1}{(2\pi \mu \Teff)^{3/2}} e^{-\frac{p^2}{2 \mu \Teff}} $ and we have schematically written the form of the integrand.
The energy transfer weighted rate before thermal averaging is given by,
\begin{align}
E_T \overline{\sigma}_{01} v_{\textrm{rel}} & \simeq  \left(\frac{T_K - T_\chi}{\Teff} \right)  \left(\frac{p^2}{M}\right)3 \frac{g_\chi^2 g_e^2}{2 \pi} \mu p^\prime  \int_{-1}^{1} d\cos\theta \frac{\left( 1-\frac{p^\prime}{p} \cos \theta \right ) }{\left (-p^2 - p^{\prime 2} -m_V^2 + 2 p p^{\prime} \cos \theta \right )^2} \\
& \simeq  \left(\frac{T_K - T_\chi}{\Teff} \right)  \left(\frac{p^2}{M}\right)3 \frac{g_\chi^2 g_e^2}{2 \pi} \mu p^\prime \left\{ \frac{1}{2\mu\Delta p^2}  + \textrm{Log}\left( \frac{4 p^4}{\mu^2 \Delta^2} \right)\right \}  \\
&\rightarrow   \left(\frac{T_K - T_\chi}{2} \right)  \frac{3}{4 \pi }  \frac{g_\chi^2 g_e^2}{ \Delta^2}\left(\frac{2\Delta \mu}{M \Teff}\right)  \frac{p^\prime}{\mu}. \label{eq:inelasticrate}
\end{align}
In the second line above, we see that the forward scattering divergence of the integral is cut-off by the inelastic hyperfine splitting parameter $\Delta$. The divergence has two parts, the leading divergence scales as $1/\Delta$ and a sub-leading piece which scales as $\textrm{Log} \, \Delta$. In the last line above we have dropped the sub-leading contribution.

Note that in the second line in the equation above we have neglected the mediator mass which could also cut-off the forward divergence. This can be justified if we assume the same upper bound on the mediator mass that we had assumed as in eq.~\ref{eq:medmass_upper}, when calculating the spin-flip rate. However, for the elastic scattering process $\chi + H_1 \rightarrow \chi + H_1$, the forward scattering divergence is cut-off only by the mediator mass. In that case, the corresponding integrand only has a logarithmic divergence, scaling as $\textrm{Log} \, m_V$, i.e. there is no $1/m_V$ type divergence (see a derivation in  Appendix~\ref{sec:appendixd}). Thus, the elastic scattering energy transfer cross-section is suppressed relative to the inelastic cross-section that we have considered here, and hence we will ignore this contribution to the energy transfer.

Now performing the thermal averaging of the rate in eq.~\ref{eq:inelasticrate} we get,
\begin{align}
R_{01}  & \simeq \left(\frac{T_K - T_\chi}{2} \right) \left(\frac{2\Delta \mu}{M \Teff}\right) \frac{3}{4 \pi }  \frac{g_\chi^2 g_e^2}{ \Delta^2} \sqrt{\frac{8\Teff}{\pi \mu}} e^{-\frac{\Delta}{\Teff}}.
\label{eq:excitationenergytransfer}
\end{align}
We can similarly evaluate the rate for de-excitation reactions as,
\begin{align}
R_{10}  & \simeq \left(\frac{T_K - T_\chi}{2} \right) \left(\frac{2\Delta \mu}{M \Teff}\right) \frac{1}{4 \pi }  \frac{g_\chi^2 g_e^2}{ \Delta^2} \sqrt{\frac{8\Teff}{\pi \mu}}.
\end{align}
Combining the energy transfer rates from both excitations and de-excitations we get,
\begin{equation}
\dot{Q}_\chi =  \Gamma_\chi  \left(T_K - T_\chi \right),
\end{equation}
where we have defined $\Gamma_\chi$ as the characteristic energy transfer rate. This would correspond to the inverse time-scale to transfer an $\mathcal{O}(1)$ fraction of the baryon kinetic energy to the dark matter.

If we further make the approximation that $n_1 \simeq 3 n_0$ (for spin temperature $T_s \gg \Delta$) and using $n_H = n_0 + n_1$, we can write the energy transfer rate $\Gamma_\chi$ as,
\begin{align}
\label{eq:gammchianalytic}
\Gamma_\chi \simeq n_H \left(\frac{\Delta \mu}{2 M \Teff}\right) 12 \pi   \frac{\alpha_\chi \alpha_e}{ \Delta^2} \sqrt{\frac{8\Teff}{\pi \mu}}.
\end{align}
Upon comparing this with the expression for the rate $D_{01}$ for excitations (eq.~\ref{eq:excitationanddeexrate}) we see that the energy transfer rate is suppressed by a factor of $S\equiv\left(\frac{\Delta \mu}{2M \Teff}\right)  = 3.42\times 10^{-4} \left(\frac{10 \, \textrm{K}}{\Teff} \right ) \left(\frac{\mu}{0.1\textrm{~GeV}}\right) \left( \frac{1\textrm{~GeV}}{M} \right )$. This is because the energy transfer per collision is not of $\mathcal{O}(T_K - T_\chi)$, but rather  since the scattering process is dominantly forwards, the energy transfer is of the order of the mass splitting between the singlet and triplet states, i.e. it is $\mathcal{O}(\Delta)$. Thus, the timescale to transfer an $\mathcal{O}(1)$ fraction of the kinetic energy of the gas to the DM is longer than the interaction time scale by a factor of $\mathcal{O}(\Teff/\Delta)$.

Now we can use,
\begin{align}
n_H  &=  \frac{ \Omega_b \rho_c}{m_H} (1+z)^3, \\
&= 2.3 \times 10^{-45}  \left( \frac{1+z}{1+10} \right)^3~\textrm{GeV}^{3}, \\
& =  3.0\times 10^{-4}  \left( \frac{1+z}{1+10} \right)^3~\textrm{cm}^{-3}.
\end{align}
Here, we have used the present-day baryon density fraction $\Omega_b = 0.05$ and we have assumed for simplicity that all the baryons are in the form of neutral hydrogen at the relevant red-shifts.
Thus, we can write an expression for the rate $\Gamma_\chi$ as,
\begin{align}
\Gamma_\chi & =  3.97 \times  10^{-40} \left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right) \left( \frac{1~\textrm{GeV}}{M} \right) \left(\frac{\mu}{0.1~\textrm{GeV}} \right)^{\frac{1}{2}}  \left(\frac{10~\textrm{K}}{\Teff} \right)^{\frac{1}{2}} \left( \frac{1+z}{1+10} \right)^3~\textrm{GeV}, \\
&=  6.02 \times 10^{-16} \left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right) \left( \frac{1~\textrm{GeV}}{M} \right) \left(\frac{\mu}{0.1~\textrm{GeV}} \right)^{\frac{1}{2}}  \left(\frac{10~\textrm{K}}{\Teff} \right)^{\frac{1}{2}}   \left( \frac{1+z}{1+10} \right)^3~\textrm{s}^{-1}.
\end{align}

We can also similarly work out the temperature evolution of the hydrogen kinetic temperature,
\begin{equation}
\frac{d T_K}{dt} = - 2 H T_\chi +\Gamma_c (\Tcmb-T_K) + \frac{2}{3} \dot{Q}_H,
\end{equation}
where the second term is the heating due to CMB and $\Gamma_c$ is the compton rate, which depends on the free electron fraction.
In the last term $\dot{Q}_H$ is the energy transfer rate from the dark matter fluid to the gas, which can be related to the heating rate of the DM $\dot{Q}_\chi$ as,
\begin{align}
\dot{Q}_H =& -\frac{n_\chi}{n_H} \dot{Q}_\chi, \\
=& \, \Gamma_H (T_\chi-T_K),
\end{align}
where we have defined the rate constant $\Gamma_H$ as,
\begin{align}
\Gamma_H =&\,  n_\chi \left(\frac{\Delta \mu}{2M \Teff}\right) \frac{3}{4 \pi }  \frac{g_\chi^2 g_e^2}{ \Delta^2} \sqrt{\frac{8\Teff}{\pi \mu}},  \nonumber \\
=& \, 2.05 \times  10^{-39} \left( \frac{f}{0.1}\right )  \left( \frac{0.1~\textrm{GeV}}{m_\chi}\right ) \left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right) \left( \frac{1~\textrm{GeV}}{M} \right) \left(\frac{\mu}{0.1~\textrm{GeV}} \right)^{\frac{1}{2}}  \left(\frac{10~\textrm{K}}{\Teff} \right)^{\frac{1}{2}}   \left( \frac{1+z}{1+10} \right)^3~\textrm{GeV},  \nonumber\\
=& \, 3.11\times 10^{-15} \left( \frac{f}{0.1} \right) \left( \frac{0.1~\textrm{GeV}}{m_\chi}\right ) \left( \frac{\alpha_\chi}{10^{-2}} \right) \left(\frac{\alpha_e}{10^{-14}}\right)  \left( \frac{1~\textrm{GeV}}{M} \right) \left(\frac{\mu}{0.1~\textrm{GeV}} \right)^{\frac{1}{2}}  \left(\frac{10~\textrm{K}}{\Teff} \right)^{\frac{1}{2}}   \left( \frac{1+z}{1+10} \right)^3~\textrm{s}^{-1},   \nonumber \\
\end{align}

\section{Elastic scattering energy-transfer cross-section}
\label{sec:appendixd}
We discuss here the forward scattering divergence of elastic scattering for processes such as $\chi + H_1 \rightarrow \chi + H_1$, $\chi + e^- \rightarrow \chi + e^-$ and $\chi + \chi \rightarrow \chi+ \chi$. The energy transfer cross-section for each of these processes can be defined as follows,
\begin{equation}
\overline{\sigma} = \int d\Omega \frac{d\sigma}{d\Omega} (1-\cos \theta),
\end{equation}
the integral over the polar scattering angle takes the form,
\begin{align}
\mathcal{I} & =  \int_{-1}^{1} d\cos\theta \frac{\left( 1- \cos \theta \right ) }{\left (1 -\cos \theta + \frac{m_V^2}{2 p^2} \right )^2} \\
& =\textrm{Log} \left( 1 +\frac{4p^2}{m_V^2}\right) -\frac{1}{1 +\frac{m_V^2}{4p^2}} \\
&\underset{m_V \to 0}{\longrightarrow} \textrm{Log} \left(\frac{4p^2}{m_V^2}\right),
\end{align}
where $p = \mu v$ is the relative momentum of the incoming particles, $\mu$ being the reduced mass and $v$ the relative velocity of the incoming particles. In the last expression we have taken the limit of small mediator mass, $m_V \rightarrow 0$. For elastic scattering, the forward divergence in the energy transfer cross-section is cut-off by the mediator mass and this gives rise to a logarithmic divergence. Similar expressions can be found in the literature, for example see ref.~[61] for DM elastic scattering through a light mediator.

For completeness, we present here the various energy transfer cross-sections for elastic scattering processes.

The process $\chi + e^- \rightarrow \chi + e^-$ is important for determining the pre-recombination initial conditions on the DM temperature. The energy transfer cross-section for this process is given by,
\begin{align}
\overline{\sigma}(\chi + e^- \rightarrow \chi + e^-) &= \frac{3}{8 \pi} \frac{g_\chi^2 g_e^2}{\mu^2_{\chi e}v^4} \textrm{Log} \left( \frac{4 \mu^2_{\chi e}v^2 }{m_V^2}\right),
\end{align}
where $\mu_{\chi e}$ is the reduced mass of the DM and electron.

DM self-scattering $\chi+ \chi \rightarrow \chi +\chi$ is important when considering astrophysical self-interaction constraints. For this process the energy transfer cross-section is given by,
% \blue{this formula only takes into account t-channel but not u-channel}
\begin{align}
\overline{\sigma}(\chi + \chi \rightarrow \chi +\chi) &\sim \frac{3}{2\pi} \frac{g_\chi^4}{m^2_{\chi}v^4} \textrm{Log} \left( \frac{m^2_{\chi}v^2 }{m_V^2}\right).
\end{align}

Finally, for elastic $\chi+ H_1 \rightarrow  \chi + H_1$ scattering, which only proceeds through the $s^{\prime}_H=\pm1$ states of $H_1$ without flipping the spin, we have the energy transfer cross-section of the form,
\begin{align}
\overline{\sigma}(\chi + H_1 \rightarrow \chi + H_1) &= \frac{1}{24 \pi} \frac{g_\chi^2 g_e^2}{\mu^2v^4} \textrm{Log} \left( \frac{4 \mu^2v^2 }{m_V^2}\right),
\end{align}
where $\mu$ is the reduced mass of the DM and hydrogen.

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\end{document}