I recently tried to use itensor to solve the Rydberg atomic problem on a two-dimensional kagome lattice. In fact I already have a handwritten kagome lattice function.
But I found that the dmrg ground state energy is not the same when my kagome is tilted to the right and tilted to the left. This is driving me crazy, because I feel that the distances between these lattice points are all symmetrical, and after my testing this is true, the list of lattice distances is exactly the same, so the difference in energy is puzzling.
I don’t know if I need to provide some of my code to fix this. Looking forward to getting some pointers and help from you.
So the way that DMRG works for 2D systems is that it does not know about spatial distances between points of your lattice. All it knows about is the graph topology of the interactions meaning which sites are connected to which other sites and by what strengths of couplings (e.g. the value of J for Heisenberg interactions).
Therefore the key question to ask is: are your two lattices the same in a graph sense? Meaning that the pattern of interactions connecting the sites are the same? If so, then DMRG must give the same result as the code cannot tell the two cases apart. If not, then for a finite-size system there is no reason a priori that DMRG should be giving the same results, and in general actually should give different results.
hi miles
Please forgive me for drawing my question in the picture in such a crude way.
As you can see in the picture, if I want to calculate a unit cell with Nx=2, Ny=3, I can take cells 1-6 and cells A-F. And both of these methods guarantee a total of 3x3x2 points, and the order of the grid points in each small triangle is also first white, then blue, and finally red.
My question is when I test the form of this lattice for calculations, should I get the same energy?
Thanks again for answering this question that bothers me.
So when you make you make a finite piece of your lattice into a system that you are going to do DMRG on, as you know what happens next is that the sites are ordered in a 1D fashion. Then the 2D Hamiltonian becomes a pseudo-1D Hamiltonian in this sense, with nearest-neighbor interactions in the 2D system becoming further-neighbor interactions in 1D.
The key thing to figure out to answer your question is whether this 1D Hamiltonian is the same or not for your two cases. If it isn’t the same, then there is no reason that the energies you get should be the same. If it is the same, then the energies have to be the same.
