of an additional excitation on top of the $n$th Mott insulator in the JCH model is given by $$\begin{gathered}
  E(nL+1) =   (L-1)  E^-_{n}  +E^-_{n+1}  +    F_n(k^\prime), \label{eq:effectiveEnergyPlus}\end{gathered}$$ where $$F_n(k)=   2\  B^-_n D^-_{n+1} \sum_d t_d \cos(2\pi\frac{   k  d}L)$$ and the momentum mode $k^\prime$ is chosen such that $F_n(k)\bigr|_{k^\prime}$ is minimal. It should be mentioned that the product $B_n^- D_{n+1}^-$ is positive for any $(\Delta,\omega,n)$, so the momentum mode is purely determined by the minimum of $\sum_d t_d \cos(2\pi\frac{   k  d}L)$.\
To calculate the energy for a hole in the $n$th Mott insulator, we follow exactly the same route. Now, the state $\left| \Downarrow\right\rangle$ is associated with $\left|- , n-1\right \rangle$ and $\left| \Uparrow\right\rangle$ with $\left|- , n\right \rangle$. The hopping operators act as $$\hat a^\dagger_{j+1} \hat a_j\left|\Downarrow\right \rangle_{j+1}\left|\Uparrow\right \rangle_j = B^-_{n-1} 
D^-_{n}\left|\Uparrow\right \rangle_{j+1}\left|\Downarrow\right \rangle_j$$ and the effective Hamiltonian is given by $$\begin{gathered}
{ \widetilde  H} = E^-_{n-1} \sum_j {\widetilde\sigma}^-_{j} {\widetilde\sigma}^+_{j}  +E^-_{n} \sum_j {\widetilde\sigma}^+_{j} {\widetilde\sigma}^-_{j} \\
 + B^-_{n-1} D^-_{n}\sum_d t_d \sum_j \left({\widetilde\sigma}^+_{j+d} {\widetilde\sigma}^-_{j}+
{\widetilde\sigma}^+_{j} {\widetilde\sigma}^-_{j+d}\right).\end{gathered}$$ Here the magnetization consist of one spin pointing downward. Again, after making use of a Jordan-Wigner transformation and subsequently a Fourier transformation, the energy of a single hole is given by $$\begin{gathered}
  E(nL-1) =    (L-1) E^-_{n}  +E^-_{n-1}+F_{n-1}(k^{\prime\prime})  