On Jul 24, 2011, at 4:48 AM, zhian asadzadeh wrote:
Dear Matthias,
by \Sigma_{i,j}, I mean \sum_{i,j},
for the exact result, we know the model is integrable, so by fermionization and finally diagonalizing the Hamiltonian we will end up to an Hamiltonian with sum over independent modes, H=\sum_{k} E(k) (2 a_{k}^{dagger} a_{k}-1)
for the T=0 ground state energy, I have compared with DMRG results which are exactly same, so I am sure up to this point the form of E(k) and my summation over k space is correct.
so finally for E(T) simply we have: E(T)=\sum_{k} E(k)(2*n(k,T)-1) in which n(k,T)=1/(exp(E(k)/T)+1) ::with Boltzmann constant=1
Are you sure that you use the same units for temperature, etc. As the looper code is exact I wonder where the problem is. Can you please repeat this for a small system, e.g. 8 spins, where we can use fulldiag to compare. Also, which boundary conditions did you choose? DId you do the DMRG with periodic boundary conditions?
Matthias