freedom can be easily accomplished via $$\begin{aligned}
  \hat a_{   j} &= \frac{1}{\sqrt{L}}\sum_{   k} e^{-2\pi i\frac{   k    j}{L}} \hat a_{   
k}\label{eq:FourierBoson}\\
 \hat c_{   j} &= \frac{1}{\sqrt{L}}\sum_{   k} e^{-2\pi i\frac{   k   j}{L}} \hat c_{   
k}\label{eq:FourierFermion}.\end{aligned}$$ Here $\hat a_{   k}$ and $\hat c _{   k}$ are operators in momentum space. Doing so, the JCH Hamiltonian transforms to that of uncoupled Jaynes-Cummings systems $$\begin{gathered}
  \hat H= \sum_{   k}  \omega_{   k}  \hat a_{   k} ^\dagger \hat a_{   k}  + \Delta \sum_{   k} \hat 
 c^\dagger_{   k}  \hat c_{   k}  + g\sum_{   k}  \left( \hat c ^\dagger_{   k}  \hat a_{   k}  + \hat a_{   k}^\dagger 
\hat c_{   k}\right),\label{eq:JCHMomentum}\end{gathered}$$ with $$\begin{gathered}
\omega_k = \omega + 2 \sum_{   d} t_{   d} \cos(2\pi\frac{   k    d}L). \label{eq:PhononEnergies}\end{gathered}$$ The ground state in any mode is given by the Jaynes-Cummings ground state (\[eq:JCGroundState\]) with frequency $\omega_k$. The energy of mode $   k$ with $n$ excitations is $$E^{n}_{   k} = (1-\delta_{n0})\left[ n\, \omega_{   k} + \frac{\Delta-\omega_{   k}}{2}-\frac 12\sqrt{ (\Delta-\omega_{   
k})^2+4 n g^2}\right]. \label{eq:JCEnergyMomentum}$$

Since the total number of excitations in the system $$\hat N=\sum_{   j} \left( \hat a_{   j}^\dagger\hat a_{   j}+ 
\hat \sigma^\dagger_{   j} \hat \sigma^-_{   j}\right)\mapsto\sum_{   
k} \left( \hat a_{   k}^\dagger\hat a_{   k}+  \hat c^\dagger_{   k} \hat c_k\right)$$ commutes with the Hamiltonian (\[eq:JCHMomentum\]), 