Dear Itensorians,
I’m considering the t-j model (3 local states per site) with t=1 and J=0. In this limit, one recovers the U=\infty Hubbard model, where double occupation of a given site excluded from the Hilbert space. Suppose I consider open chain of L sites. Then, I have U(1) symmetry of particle number conservation and conservation of the spin pattern (ITensors allow only conservation of the total Sz). The latter symmetry implies that the Hamiltonian conserves not only the total Sz, but also a given spin configuration(pattern). To be more clearer, suppose I have 8 sites and 4 particles: 2 up and 2 down. Due to the infinite repulsion U=\infty one can consider different spin subsectors, which all have total Sz=0:
|\uparrow, \downarrow, \uparrow, \downarrow>
|\uparrow, \uparrow, \downarrow ,\downarrow>
and others.
This particularly implies that if I prepare an initial state
|\psi>= |\uparrow, o, \downarrow, o, \uparrow, o, \downarrow, o >
and find the ground state via the DMRG routine of ITensors, I expect the found state should also have the same Neel pattern. However, the found state not always has the same pattern. The reason is that Itensors conserves only Sz=0 and does not “know” about the new symmtery. My question is: Is it possible to embed this symmetry onto the tjSites? If yes, what is the best way to do it?
You are correct that the default set of QNs chosen by ITensor only involves conserving total Sz. It is possible to conserve any other quantum number associated with an Abelian, on-site symmetry (e.g. fermion parity, particle number, etc.).
Does the kind of “spin pattern” you are wanting to conserve correspond to a symmetry of your Hamiltonian? If so, what is this symmetry and how is it described mathematically?
In general I would expect the t-J model to allow particles to move around if there are some “hole” sites, so I would not expect the value of Sz on a given site to remain the same from the initial state to the optimized ground state found by DMRG. Is there a mathematical reason to expect this?
Yes, the “spin pattern” corresponds to a symmetry of the t-J Hamiltonian with J=0; The
Hamiltonian is invariant under the transformations generated by the
algebra [u(2)]^N. The exact form of this algebra is given here: The one-dimensional Hubbard model for large or infiniteU | SpringerLink
I think you did not get my point regarding this symmetry. Although the mathematical formulation of this symmetry looks redundant, its physical picture is very simple: Imagine you have L sites in an open chain with 1 particle with spin up and 1 particle with spin down; The basis is formed from the following set of basis vectors:
where |0> denotes a hole in the system. The kinetic part of the t-J hamiltonian implies that these holes move these two particles through the chain. However, these two particles never exchange their order (if you count from the left of the chain to the right you always have spin up first, then spin down), i.e you DO NOT HAVE the following set of vectors in your basis:
|00..00\downarrow 00..000 \uparrow00..000>
|0\downarrow0..00 00..000 \uparrow00..000>
|00..0\downarrow0 00..000 \uparrow00..000>
… |00..00\downarrow 00..000 00..00\uparrow0>
This is because infinite repulsion between the particles on a given site does not allow them to exchange their order.
These two subspaces do not mix.