Keeping number of particles fixed at edges
Posted: 15 Apr 2021, 17:14
I am currently trying to calculate the charge gap \( E_c = E^{0}(N+1) + E^{0}(N-1) - 2 E^{0}(N) \) and the neutral gap \( E_n = E^{1}(N) - E^{0}(N) \) of the extended Boson Hubbard model in one dimension. (\( E^{1}(N) \) is the energy of the first excited state ,\( E^{0}(N) \) is the energy of the ground state. N is the number of bosons). The Hamiltonian of the model is given as:
\( H=-t \sum_{i} \left (b_{i}^{\dagger} b_{i+1}+\text { H.c. }\right)+ \frac{U}{2} \sum_{i} n_{i} \left(n_{i}-1\right)+\sum_{ \langle i,j \rangle} V n_{i} n_{j} \)
In order to prevent the effect of edge states on the calculations of the gap, I need to fix the number of bosons on the leftmost site to 0 and the number of bosons on the rightmost site to be 2 throughout the dmrg calculation. I am confused as to how do I implement this. I am using open boundary conditions.
\( H=-t \sum_{i} \left (b_{i}^{\dagger} b_{i+1}+\text { H.c. }\right)+ \frac{U}{2} \sum_{i} n_{i} \left(n_{i}-1\right)+\sum_{ \langle i,j \rangle} V n_{i} n_{j} \)
In order to prevent the effect of edge states on the calculations of the gap, I need to fix the number of bosons on the leftmost site to 0 and the number of bosons on the rightmost site to be 2 throughout the dmrg calculation. I am confused as to how do I implement this. I am using open boundary conditions.