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.