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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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please also explain how to get Phase Diagram of Bose-Hubbard Model Using Exact Diagonalisation – mpk Jul 16 '14 at 9:06
up vote 1 down vote accepted

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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Thanks for explanations – mpk Jul 21 '14 at 14:44

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