guanjihuan/guanjihuan.com
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

update

Browse Source
This commit is contained in:
guanjihuan
2021-06-09 04:38:18 +08:00
commit 14a297b604
137 changed files with 8965 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

102
academic_codes/2020.12.31_DOS_calculation_using_Dyson_equations/cubic_lattice/DOS_calculation_using_Dyson_equations_in_cubic_lattice.py Executable file
View File

@@ -0,0 +1,102 @@
"""
This code is supported by the website: https://www.guanjihuan.com
The newest version of this code is on the web page: https://www.guanjihuan.com/archives/7650
"""
import numpy as np
import matplotlib.pyplot as plt
from math import *
def matrix_00(width, length):
h00 = np.zeros((width*length, width*length))
for x in range(length):
for y in range(width-1):
h00[x*width+y, x*width+y+1] = 1
h00[x*width+y+1, x*width+y] = 1
for x in range(length-1):
for y in range(width):
h00[x*width+y, (x+1)*width+y] = 1
h00[(x+1)*width+y, x*width+y] = 1
return h00
def matrix_01(width, length):
h01 = np.identity(width*length)
return h01
def main():
height = 2 # z
width = 3 # y
length = 4 # x
eta = 1e-2
E = 0
h00 = matrix_00(width, length)
h01 = matrix_01(width, length)
G_ii_n_array = G_ii_n_with_Dyson_equation(width, length, height, E, eta, h00, h01)
for i0 in range(height):
print('z=', i0+1, ':')
for j0 in range(width):
print(' y=', j0+1, ':')
for k0 in range(length):
print(' x=', k0+1, ' ', -np.imag(G_ii_n_array[i0, k0*width+j0, k0*width+j0])/pi) # 态密度
def G_ii_n_with_Dyson_equation(width, length, height, E, eta, h00, h01):
dim = length*width
G_ii_n_array = np.zeros((height, dim, dim), dtype=complex)
G_11_1 = np.linalg.inv((E+eta*1j)*np.identity(dim)-h00)
for i in range(height): # i为格林函数的右下指标
# 初始化开始
G_nn_n_minus = G_11_1
G_in_n_minus = G_11_1
G_ni_n_minus = G_11_1
G_ii_n_minus = G_11_1
for i0 in range(i):
G_nn_n = Green_nn_n(E, eta, h00, h01, G_nn_n_minus)
G_nn_n_minus = G_nn_n
if i!=0:
G_in_n_minus = G_nn_n
G_ni_n_minus = G_nn_n
G_ii_n_minus = G_nn_n
# 初始化结束
for j0 in range(height-1-i): # j0为格林函数的右上指标，表示当前体系大小，即G^{(j0)}
G_nn_n = Green_nn_n(E, eta, h00, h01, G_nn_n_minus)
G_nn_n_minus = G_nn_n
G_ii_n = Green_ii_n(G_ii_n_minus, G_in_n_minus, h01, G_nn_n, G_ni_n_minus) # 需要求的对角分块矩阵
G_ii_n_minus = G_ii_n
G_in_n = Green_in_n(G_in_n_minus, h01, G_nn_n)
G_in_n_minus = G_in_n
G_ni_n = Green_ni_n(G_nn_n, h01, G_ni_n_minus)
G_ni_n_minus = G_ni_n
G_ii_n_array[i, :, :] = G_ii_n_minus
return G_ii_n_array
def Green_nn_n(E, eta, H00, V, G_nn_n_minus): # n>=2
dim = H00.shape[0]
G_nn_n = np.linalg.inv((E+eta*1j)*np.identity(dim)-H00-np.dot(np.dot(V.transpose().conj(), G_nn_n_minus), V))
return G_nn_n
def Green_in_n(G_in_n_minus, V, G_nn_n): # n>=2
G_in_n = np.dot(np.dot(G_in_n_minus, V), G_nn_n)
return G_in_n
def Green_ni_n(G_nn_n, V, G_ni_n_minus): # n>=2
G_ni_n = np.dot(np.dot(G_nn_n, V.transpose().conj()), G_ni_n_minus)
return G_ni_n
def Green_ii_n(G_ii_n_minus, G_in_n_minus, V, G_nn_n, G_ni_n_minus): # n>=i
G_ii_n = G_ii_n_minus+np.dot(np.dot(np.dot(np.dot(G_in_n_minus, V), G_nn_n), V.transpose().conj()),G_ni_n_minus)
return G_ii_n
if __name__ == '__main__':
main()

99
academic_codes/2020.12.31_DOS_calculation_using_Dyson_equations/cubic_lattice/DOS_calculation_using_Dyson_equations_in_cubic_lattice_version_II.py Executable file
View File

@@ -0,0 +1,99 @@
"""
This code is supported by the website: https://www.guanjihuan.com
The newest version of this code is on the web page: https://www.guanjihuan.com/archives/7650
"""
import numpy as np
import matplotlib.pyplot as plt
from math import *
def matrix_00(width, length):
h00 = np.zeros((width*length, width*length))
for x in range(length):
for y in range(width-1):
h00[x*width+y, x*width+y+1] = 1
h00[x*width+y+1, x*width+y] = 1
for x in range(length-1):
for y in range(width):
h00[x*width+y, (x+1)*width+y] = 1
h00[(x+1)*width+y, x*width+y] = 1
return h00
def matrix_01(width, length):
h01 = np.identity(width*length)
return h01
def main():
height = 2 # z
width = 3 # y
length = 4 # x
eta = 1e-2
E = 0
h00 = matrix_00(width, length)
h01 = matrix_01(width, length)
G_ii_n_with_Dyson_equation_version_II(width, length, height, E, eta, h00, h01)
def G_ii_n_with_Dyson_equation_version_II(width, length, height, E, eta, h00, h01):
dim = length*width
G_11_1 = np.linalg.inv((E+eta*1j)*np.identity(dim)-h00)
for i in range(height): # i为格林函数的右下指标
# 初始化开始
G_nn_n_minus = G_11_1
G_in_n_minus = G_11_1
G_ni_n_minus = G_11_1
G_ii_n_minus = G_11_1
for i0 in range(i):
G_nn_n = Green_nn_n(E, eta, h00, h01, G_nn_n_minus)
G_nn_n_minus = G_nn_n
if i!=0:
G_in_n_minus = G_nn_n
G_ni_n_minus = G_nn_n
G_ii_n_minus = G_nn_n
# 初始化结束
for j0 in range(height-1-i): # j0为格林函数的右上指标，表示当前体系大小，即G^{(j0)}
G_nn_n = Green_nn_n(E, eta, h00, h01, G_nn_n_minus)
G_nn_n_minus = G_nn_n
G_ii_n = Green_ii_n(G_ii_n_minus, G_in_n_minus, h01, G_nn_n, G_ni_n_minus) # 需要求的对角分块矩阵
G_ii_n_minus = G_ii_n
G_in_n = Green_in_n(G_in_n_minus, h01, G_nn_n)
G_in_n_minus = G_in_n
G_ni_n = Green_ni_n(G_nn_n, h01, G_ni_n_minus)
G_ni_n_minus = G_ni_n
# 输出
print('z=', i+1, ':')
for j0 in range(width):
print(' y=', j0+1, ':')
for k0 in range(length):
print(' x=', k0+1, ' ', -np.imag(G_ii_n_minus[k0*width+j0, k0*width+j0])/pi) # 态密度
def Green_nn_n(E, eta, H00, V, G_nn_n_minus): # n>=2
dim = H00.shape[0]
G_nn_n = np.linalg.inv((E+eta*1j)*np.identity(dim)-H00-np.dot(np.dot(V.transpose().conj(), G_nn_n_minus), V))
return G_nn_n
def Green_in_n(G_in_n_minus, V, G_nn_n): # n>=2
G_in_n = np.dot(np.dot(G_in_n_minus, V), G_nn_n)
return G_in_n
def Green_ni_n(G_nn_n, V, G_ni_n_minus): # n>=2
G_ni_n = np.dot(np.dot(G_nn_n, V.transpose().conj()), G_ni_n_minus)
return G_ni_n
def Green_ii_n(G_ii_n_minus, G_in_n_minus, V, G_nn_n, G_ni_n_minus): # n>=i
G_ii_n = G_ii_n_minus+np.dot(np.dot(np.dot(np.dot(G_in_n_minus, V), G_nn_n), V.transpose().conj()),G_ni_n_minus)
return G_ii_n
if __name__ == '__main__':
main()

49
academic_codes/2020.12.31_DOS_calculation_using_Dyson_equations/cubic_lattice/get_DOS_by_direct_inverseion_of_Hamiltonian_of_cubic_lattice.py Executable file
View File

@@ -0,0 +1,49 @@
"""
This code is supported by the website: https://www.guanjihuan.com
The newest version of this code is on the web page: https://www.guanjihuan.com/archives/7650
"""
import numpy as np
import matplotlib.pyplot as plt
from math import *
def hamiltonian(width, length, height):
h = np.zeros((width*length*height, width*length*height))
for i0 in range(length):
for j0 in range(width):
for k0 in range(height-1):
h[k0*width*length+i0*width+j0, (k0+1)*width*length+i0*width+j0] = 1
h[(k0+1)*width*length+i0*width+j0, k0*width*length+i0*width+j0] = 1
for i0 in range(length):
for j0 in range(width-1):
for k0 in range(height):
h[k0*width*length+i0*width+j0, k0*width*length+i0*width+j0+1] = 1
h[k0*width*length+i0*width+j0+1, k0*width*length+i0*width+j0] = 1
for i0 in range(length-1):
for j0 in range(width):
for k0 in range(height):
h[k0*width*length+i0*width+j0, k0*width*length+(i0+1)*width+j0] = 1
h[k0*width*length+(i0+1)*width+j0, k0*width*length+i0*width+j0] = 1
return h
def main():
height = 2 # z
width = 3 # y
length = 4 # x
h = hamiltonian(width, length, height)
E = 0
green = np.linalg.inv((E+1e-2j)*np.eye(width*length*height)-h)
for k0 in range(height):
print('z=', k0+1, ':')
for j0 in range(width):
print(' y=', j0+1, ':')
for i0 in range(length):
print(' x=', i0+1, ' ', -np.imag(green[k0*width*length+i0*width+j0, k0*width*length+i0*width+j0])/pi) # 态密度
if __name__ == "__main__":
main()

95
academic_codes/2020.12.31_DOS_calculation_using_Dyson_equations/square_lattice/DOS_calculation_using_Dyson_equations_in_square_lattice.py Executable file
View File

@@ -0,0 +1,95 @@
"""
This code is supported by the website: https://www.guanjihuan.com
The newest version of this code is on the web page: https://www.guanjihuan.com/archives/7650
"""
import numpy as np
import matplotlib.pyplot as plt
from math import *
def matrix_00(width):
h00 = np.zeros((width, width))
for width0 in range(width-1):
h00[width0, width0+1] = 1
h00[width0+1, width0] = 1
return h00
def matrix_01(width):
h01 = np.identity(width)
return h01
def main():
width = 2
length = 3
eta = 1e-2
E = 0
h00 = matrix_00(width)
h01 = matrix_01(width)
G_ii_n_array = G_ii_n_with_Dyson_equation(width, length, E, eta, h00, h01)
for i0 in range(length):
# print('G_{'+str(i0+1)+','+str(i0+1)+'}^{('+str(length)+')}:')
# print(G_ii_n_array[i0, :, :],'\n')
print('x=', i0+1, ':')
for j0 in range(width):
print(' y=', j0+1, ' ', -np.imag(G_ii_n_array[i0, j0, j0])/pi) # 态密度
def G_ii_n_with_Dyson_equation(width, length, E, eta, h00, h01):
G_ii_n_array = np.zeros((length, width, width), complex)
G_11_1 = np.linalg.inv((E+eta*1j)*np.identity(width)-h00)
for i in range(length): # i为格林函数的右下指标
# 初始化开始
G_nn_n_minus = G_11_1
G_in_n_minus = G_11_1
G_ni_n_minus = G_11_1
G_ii_n_minus = G_11_1
for i0 in range(i):
G_nn_n = Green_nn_n(E, eta, h00, h01, G_nn_n_minus)
G_nn_n_minus = G_nn_n
if i!=0:
G_in_n_minus = G_nn_n
G_ni_n_minus = G_nn_n
G_ii_n_minus = G_nn_n
# 初始化结束
for j0 in range(length-1-i): # j0为格林函数的右上指标，表示当前体系大小，即G^{(j0)}
G_nn_n = Green_nn_n(E, eta, h00, h01, G_nn_n_minus)
G_nn_n_minus = G_nn_n
G_ii_n = Green_ii_n(G_ii_n_minus, G_in_n_minus, h01, G_nn_n, G_ni_n_minus) # 需要求的对角分块矩阵
G_ii_n_minus = G_ii_n
G_in_n = Green_in_n(G_in_n_minus, h01, G_nn_n)
G_in_n_minus = G_in_n
G_ni_n = Green_ni_n(G_nn_n, h01, G_ni_n_minus)
G_ni_n_minus = G_ni_n
G_ii_n_array[i, :, :] = G_ii_n_minus
return G_ii_n_array
def Green_nn_n(E, eta, H00, V, G_nn_n_minus): # n>=2
dim = H00.shape[0]
G_nn_n = np.linalg.inv((E+eta*1j)*np.identity(dim)-H00-np.dot(np.dot(V.transpose().conj(), G_nn_n_minus), V))
return G_nn_n
def Green_in_n(G_in_n_minus, V, G_nn_n): # n>=2
G_in_n = np.dot(np.dot(G_in_n_minus, V), G_nn_n)
return G_in_n
def Green_ni_n(G_nn_n, V, G_ni_n_minus): # n>=2
G_ni_n = np.dot(np.dot(G_nn_n, V.transpose().conj()), G_ni_n_minus)
return G_ni_n
def Green_ii_n(G_ii_n_minus, G_in_n_minus, V, G_nn_n, G_ni_n_minus): # n>=i
G_ii_n = G_ii_n_minus+np.dot(np.dot(np.dot(np.dot(G_in_n_minus, V), G_nn_n), V.transpose().conj()),G_ni_n_minus)
return G_ii_n
if __name__ == '__main__':
main()

41
academic_codes/2020.12.31_DOS_calculation_using_Dyson_equations/square_lattice/get_DOS_by_direct_inverseion_of_Hamiltonian_of_square_lattice.py Executable file
View File

@@ -0,0 +1,41 @@
"""
This code is supported by the website: https://www.guanjihuan.com
The newest version of this code is on the web page: https://www.guanjihuan.com/archives/7650
"""
import numpy as np
import matplotlib.pyplot as plt
from math import *
def hamiltonian(width, length):
h = np.zeros((width*length, width*length))
for i0 in range(length):
for j0 in range(width-1):
h[i0*width+j0, i0*width+j0+1] = 1
h[i0*width+j0+1, i0*width+j0] = 1
for i0 in range(length-1):
for j0 in range(width):
h[i0*width+j0, (i0+1)*width+j0] = 1
h[(i0+1)*width+j0, i0*width+j0] = 1
return h
def main():
width = 2
length = 3
h = hamiltonian(width, length)
E = 0
green = np.linalg.inv((E+1e-2j)*np.eye(width*length)-h)
for i0 in range(length):
# print('G_{'+str(i0+1)+','+str(i0+1)+'}^{('+str(length)+')}:')
# print(green[i0*width+0: i0*width+width, i0*width+0: i0*width+width], '\n')
print('x=', i0+1, ':')
for j0 in range(width):
print(' y=', j0+1, ' ', -np.imag(green[i0*width+j0, i0*width+j0])/pi)
if __name__ == "__main__":
main()