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The Hamiltonian of Bose-Hubbard model reads as $$H=-J\sum\limits_{<i,j>}b_i^{\dagger}b_j+h.c.+\frac{U}{2}\sum\limits_{i}n_i(n_i-1)-\mu\sum\limits_in_i~~~~~~~~~(1)$$

For this we plot phase diagram in ( $J/U$, $\mu/U$ ) space.

Same way if I want to plot phase diagram of Hamiltonian which looks like $$H=-J\sum\limits_{<i,j>}b_i^{\dagger}b_j+h.c.+\frac{U}{2}\sum\limits_{i}n_i(n_i-1)~~~~~~~~~(2)$$

How to get phase diagram of such hamiltonian? I am solving this model Numerically by Exact Diagonalisation.

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  • $\begingroup$ please also explain how to get Phase Diagram of Bose-Hubbard Model Using Exact Diagonalisation $\endgroup$ – mpk Jul 16 '14 at 9:06
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In principle, it is very simple and straightforward.

The problem is to map out the region where the integer filling state is the ground state. Suppose you have $L$ sites. Take $N=L$ particles, find its ground state energy, which is denoted as $E_g(L)$. Note that here the Hamiltonian does not contain the $\mu $ term. Do it again for $N=L+1$, the ground state energy is $E_g(L+1)$. Then, you know below the line

$\mu_+ = E_g(L+1)- E_g(L)$

the $N=L$ state is the lower state with respect to the full hamiltonian containing $\mu$.

Do it once again with $N=L-1$, then you know above the line

$\mu_- = E_g(L)- E_g(L-1)$

the $N=L$ state is the lower state.

Therefore, between the two lines, the $N=L$ state is the lowest state. In this region, the unity filling state is the ground state. This is the first Mott lobe.

The idea is simple, but i really doubt you can get accurate results with ED. You had better do it with DMRG.

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