+1 vote
asked by (680 points)

Hi Miles,

I'm trying to do some infinite DMRG without symmetry. As a first attempt I'm calculating a transeverse field Ising model by defining

W = IQTensor(dag(sites(n)),prime(sites(n)),row,col);

W += sites.op("Id",n) * row(1) * col(1); //ending state
W += sites
.op("Id",n) * row(2) * col(2); //starting state

W += sites.op("Sz",n) * row(3) * col(1);
W += sites
.op("Sz",n) * row(2) * col(3);

W += sites.op("Sp",n) * row(2) * col(1) * hx;
W += sites.op("Sm",n) * row(2) * col(1) * hx;

But the result does not make sense with ground state energy / site = -196089.678
I think maybe the problem is with QN, which I define as 0 due to lack of symmetry. Is this the right way of doing this? Thanks very much!

Best,
Chengshu

1 Answer

0 votes
answered by (70.1k points)
selected by
 
Best answer

Hi Chengshu,
First of all, there is an issue with the iDMRG code at the moment on the master branch. I am hoping to fix it very soon. But if you use version 2.0.11 then iDMRG should work perfectly fine.

Now about your specific question, if you are studying a model without any conserved quantities (or not any that you wish to conserve, because technically the transverse-field Ising model has a conserved parity) then the best way to proceed is to not use IQTensors (and IQMPOS and QNs, etc) at all, but rather just used ITensors and the associated MPO and MPS classes.

Would just setting up your calculation using ITensors and the MPO class give you what you are wanting?

Miles

commented by (680 points)
Thanks for your fast reply! Yes using MPO gives me a reasonable result.

However, the ultimate goal is to study a model with conserved QN on a sublattice (in which case a QN = 0 will be defined), so I'm wondering what's going on.
commented by (70.1k points)
I would guess that the problems you are seeing arise from setting all of the QNs to zero. That's not a case the QN and IQTensor system is intended for. So I would recommend doing your test on a different system that does have a conserved quantum number. It could be the TFIM with conserved parity, but a simpler case would be a spin chain that conserves total Sz such as the Heisenberg model.
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