How to use

How to use

What are "flavor", "site", "spin"?

Flavor is the composite index of site and spin. We assume that sites and spins are ordered as (site0, up), (site0, down), (site1, up), (site1, down), .... Let us think about a two-orbital Hubbard-type impurity model. You can use either "model.sites=2, model.spins=2" or "model.sites=4, model.spins=1". The difference is only one thing. If you use the former setting, the solver automatically attempts global spin flips to reduce autocorrelation time.

Eliminating high-energy states

As described in the paper, we offer two options of eliminating high-energy states which do not contribute to low-temperature physics: "inner-outer-state cutoff" and "outer-state cutoff". Here, we discuss how to use this funcitonality. The following is typical output message of the impurity solver (for a three-orbital impurity model with spin-orbit coupling). Since the total electron number computes with the local Hamiltonian, the local Hilbert space can be split into at least seven subspaces. In the first block of the output messages, we can see the lowest and highest eigenvalues in each subspace. The dimensions of the subspaces are 1, 6, 15, 20, 15, 6, 1. Our code takes the advantage of additional symmetries if available.

You can set a cutoff energy (measured from the lowest eigenvalue) through the two control parameters, "model.outer_cutoff_energy", "model.inner_outer_cutoff_energy", respectively. In this example, we used "model.inner_outer_cutoff_energy = 10" and "model.outer_cutoff_energy = 0.1". First, high energy states above "model.outer_cutoff_energy" are eliminated. The result is shown in the second block. Only four sectors have non-zero active eigenstates.

Then, we apply "model.outer_cutoff_energy". As shown in the third block, you can see that only one symmetry sector is taken into account in the calculations.

dim of Hilbert space 64
# of blocks 7
# of sectors 7
Sector 0 : dim = 1, min energy = -25.8801, max energy = -25.8801
Sector 1 : dim = 6, min energy = -23.9361, max energy = -23.2863
Sector 2 : dim = 15, min energy = -21.1593, max energy = -19.6872
Sector 3 : dim = 20, min energy = -17.1634, max energy = -15.8047
Sector 4 : dim = 15, min energy = -12.2906, max energy = -10.9523
Sector 5 : dim = 6, min energy = -6.59378, max energy = -5.94392
Sector 6 : dim = 1, min energy = 0, max energy = 0
 Max eigen energy = 0
 Min eigen energy = -25.8801

 Throwing away high energy states...
Sector 0 : dim = 1, min energy = -25.8801, max energy = -25.8801
Sector 1 : dim = 6, min energy = -23.9361, max energy = -23.2863
Sector 2 : dim = 15, min energy = -21.1593, max energy = -19.6872
Sector 3 : dim = 18, min energy = -17.1634, max energy = -16.0201
 Max eigen energy = -16.0201
 Min eigen energy  = -25.8801
Reference energy -25.8801

Dim of ket: sector 0 inner 1 outer 1