Help regarding chern number calculation
MS. PRIYANKA SINHA
sinhapriyanka2016 at iitg.ac.in
Fri May 31 08:41:15 CEST 2019
Dear Sir,
I am a new user of z2 pack and trying to calculate chern number following the journal https://journals.aps.org/prb/pdf/10.1103/PhysRevB.85.115439 . I have defined the Hamiltonian as given in equation 14. I should get non-zero chern number. But instead of that I am getting zero. I am attaching my code below. Kindly help me in this regard.
*********************************************************************
#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# Author: Dominik Gresch <greschd at gmx.ch>
# Date: 16.08.2016 14:25:39 CEST
# File: haldane.py
import json
import logging
import z2pack
import numpy as np
import matplotlib.pyplot as plt
from numpy import linalg as LA
logging.getLogger('z2pack').setLevel(logging.WARNING)
# defining pauli matrices
identity = np.identity(2, dtype=complex)
pauli_x = np.array([[0, 1], [1, 0]], dtype=complex)
pauli_y = np.array([[0, -1j], [1j, 0]], dtype=complex)
pauli_z = np.array([[1, 0], [0, -1]], dtype=complex)
Sx= 1./2. * np.array([[0, 1], [1, 0]], dtype=complex)
Sy= 1./2. * np.array([[0, -1j], [1j, 0]], dtype=complex)
Sz= 1./2. * np.array([[1, 0], [0, -1]], dtype=complex)
def Hamilton(k, m, t1, VR):
kx, ky,_= k
H = (3. *t1 /2. * ( pauli_x * kx + pauli_y * ky)) * identity
H += (3. *t1 /2. * (- pauli_x * kx + pauli_y * ky)) * identity
H += 3. *VR /2. * ( pauli_x * Sy - pauli_y * Sx)
H += 3. *VR /2. * (- pauli_x * Sy - pauli_y * Sx)
#H3 = n * 3. *sqrt3 * t2 * pauli_z *Sz
H += (m * Sz) * identity
#H5 = u * pauli_z * identity
return H
def get_chern( m, t1, VR, **settings):
system = z2pack.hm.System(lambda k: Hamilton(k, m, t1, VR), bands=1)
result = z2pack.surface.run(system=system, surface=lambda t1, t2: [t1, t2, 1], **settings)
return z2pack.invariant.chern(result)
if __name__ == "__main__":
# Task a)
print(get_chern(0.3, 1., 0.5 ))
#print(get_chern(0.5, 1., 1. / 3., -0.5 * np.pi))
#print(Hamilton(0.1, 1, 0.3,0.5))
********************************************************************
Thanking You,
Priyanka Sinha
IIT Guwahati
Research Scholar
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