Lanczos

Here we show how to use PauliStrings.jl to run the recursion method and compute lanczos coefficients. We will focus on reproducing the X in XX results from figure 2 of Parker 2019.

Start by importing PauliStrings :

using PauliStrings
import PauliStrings as ps

Define the XX Hamiltonian $H = \sum_i X_iX_{i+1}+Y_iY_{i+1}$ on a 1D chain with periodic boundary conditions

function XX(N)
    H = ps.Operator(N)
    for j in 1:(N - 1)
        H += "X",j,"X",j+1
        H += "Z",j,"Z",j+1
    end
    H += "X",1,"X",N
    H += "Z",1,"Z",N
    return H
end

and the X operator $\sum_{i=1}^N X_i$

function X(N)
    H = ps.Operator(N)
    for j in 1:N
        H += "X",j
    end
    return H
end

Initialize a Hamiltonian and an operator:

N = 50 # system size
H = XX(N) #hamiltonian
O = X(N) #operator

Compute and plot the lanczos coefficients for different truncation. For each level of truncation, we keep only 2^p with the highest weight at each step of the lanczos algorithm.

ioff()#pyplot
# nterms is the max pauli string length
for p in (14,16,18,20)
    @time bs = ps.lanczos(H, O, 20, 2^p; keepnorm=true)
    plot(bs, label="trim: 2^$p")
end
legend()
ylabel(L"$b_n$")
xlabel(L"$n$")
title("X in XX, N=$N spins")
savefig("lanczos_example.png")
show()

Tacking advantage of translation symmetry

If the problem is 1D and has translation symmetry, we can take advantage of the symmetry to save time and memory. Just pass $O$ and $H$ as OperatorTS1D to lanczos. For example :

Hts = OperatorTS1D(H)
Ots = OperatorTS1D(O)
bs = ps.lanczos(Hts, Ots, 20, 2^p; keepnorm=true)

Check the translation symmetry tutorial.