.$$ Together with the energy of the system, one can thus write an effective Hamiltonian describing the upper boundary of the $n$th Mott lobe $$\begin{gathered}
{ \widetilde  H} = E^-_{n} \sum_j {\widetilde\sigma}^-_{j} {\widetilde\sigma}^+_{j}  +E^-_{n+1} \sum_j {\widetilde\sigma}^+_{j} {\widetilde\sigma}^-_{j} \\
 +B^-_n D^-_{n+1}  \sum_d t_d \sum_j \left( {\widetilde\sigma}^+_{j+d} {\widetilde\sigma}^-_{j}+ 
{\widetilde\sigma}^+_{j} {\widetilde\sigma}^-_{j+d}\right).\end{gathered}$$ This Hamiltonian is equivalent to $$\begin{gathered}
{\widetilde  H} = (L-1) E^-_{n}  +E^-_{n+1}  \\
 +B^-_n D^-_{n+1}  \sum_d t_d \sum_j \left( {\widetilde\sigma}^+_{j+d} {\widetilde\sigma}^-_{j}+ 
{\widetilde\sigma}^+_{j} {\widetilde\sigma}^-_{j+d}\right)\end{gathered}$$ since we are at fixed magnetization with only one spin pointing upwards. This Hamiltonian can be further simplified, by using a Jordan-Wigner transformation mapping the spin operators ${\widetilde\sigma}^-_{j}$ onto fermionic operators $\hat c_j$ and subsequently performing a Fourier transformation $$\hat c_{   j} = \frac{1}{\sqrt{L}}\sum_{   k} e^{-2\pi i\frac{   k   j}{L}} \hat c_{   k}.$$ Then, the ground state wave function factorizes, since the Hamiltonian decouples in momentum space $$\begin{gathered}
   {\widetilde  H} = (L-1) E^-_{n}  +E^-_{n+1} \\
   + 2 B^-_n D^-_{n+1}  \sum_d t_d \sum_k \cos(2\pi\frac{   k    d}L) \hat c_k^\dagger \hat c_k.
 \end{gathered}$$ This model is equivalent to free fermionic particles with hopping amplitudes given by $t_d$. In momentum space, a single fermion will occupy the mode with lowest energy. Thus the total energy of the single particle, and therefore the total energy 