VUMPS, Hopping term sign is important?

ITensor Development
,
1

Sign of hopping term of Hamiltonian is important in VUMPS or DMRG?

When the Hamiltonian is simply given by the hopping term and the U term,
Is there a difference when the sign of the hopping term is positive versus negative?

-t \sum_{ \langle ij \rangle } c^{\dagger} c + U \sum_{i} n_{i,\uparrow} n_{i,\downarrow }

vs

t \sum_{ \langle ij \rangle } c^{\dagger} c + U \sum_{i} n_{i,\uparrow} n_{i,\downarrow }

(U > 0)

2

If the lattice is bipartite, then you can do a unitary transformation (-1)^n on every alternate site, which transforms t \rightarrow -t. In that sense, the sign of the hopping doesn’t matter.

If the lattice is not bipartite (eg if you have a next-nearest neighbor hopping) then this transformation doesn’t work anymore (it flips the sign of hopping between different bipartitions, but the sign of hopping between the same bipartition stays the same).

1 Like
3

Also @sunam I just want to emphasize that the question of whether the sign of the hopping term in a Hamiltonian gives a different result is not specific to VUMPS or DMRG. It is a purely mathematical question about the properties (eigenvalues and eigenvectors) of two Hamiltonians and the answer would be the same for exact diagonalization, quantum Monte Carlo, or DMRG.

4

Thanks you !

5

Thanks you!

1 Like