Hello,
I am using XXZ AFM spin-half chain with N sites and I set a state which has all spins up or down i.e. conserved qunatum number total Sz=N/2 . After DMRG calculations I expect ground state and first excited state energy to be same. But it comes different. The energy difference between which I get is equal to “weight” set in DMRG function. If it is not set then default “weight” is 1. It means I get gap. Can you please clarify?
The weight parameter w you set in excited state DMRG means that you run DMRG on a modified Hamiltonian H + w|\psi\rangle\langle\psi| for some state |\psi\rangle (if you specify a single state to orthogonalize against). So the energy of |\psi\rangle gets shifted by w temporarily within DMRG. Perhaps that is what you are seeing. However, if you measure the energy of the state \langle\psi|H|\psi\rangle it should give you the original energy, not shifted by w.
Matt’s answer is totally right, and the other thing to check in additiong to checking \bra{\psi}H\ket{\psi} afterward, is to check the overlap \bra{\psi_0}\ket{\psi_1}. My strong feeling is that this overlap will be nearly 1.0, rather than zero, meaning you are getting the same state twice. To avoid this problem, I think you’ll need to choose a larger weight and maybe do more sweeps of DMRG.
You should try out different values of the weight, experiment on small systems, and so on to get a feel for how to use this feature of ITensor.
Thanks for clarification.
I also checked overlap <psi|psi0>. This overlap is equal to 1. My question is what will be the excited state for saturated ground state? Since only one state is possible for S^z_total=N/2.
My guess is it would probably be a state with a single spin flipped, so total Sz=N/2-1/2 but I am not sure and it might depend on what Hamiltonian you are referring to.
I am using spin-1/2 xxz AFM model. I also calculated energies for Sz=N/2-1. I get finite gap. Ground state energy, first excited state energy, second excited state energy for Sz=N/2-1 are lower than ground state energy for Sz=N/2. Then how will single spin flip be first excited state?
Hi @vrushali.k, at this point the conversation is getting off topic for me and Matt, unfortunately. Due to time constraints, we are only able to answer questions on topics about the usage of the ITensor software. Others may be able to answer your question, however.