In [1]:
import numpy as np
import matplotlib.pyplot as plt
# from __future__ import print_function
from ipywidgets import interact, interactive, fixed, interact_manual
import ipywidgets as widgets
%matplotlib inline
In [2]:
def band(nkl,kpoints,ham):
'''
:param k: wave number
'''
nk=32
# Define the position of the original lattice
#klines=np.zeros((nkl-1,nk))
n=0
d=np.size(ham([0.0,0.0]),1)
evk=np.zeros((nk*nkl-nk-1,d))
lenk=np.zeros((nk*nkl-nk-1,d))
for i in range(0,nkl-1):
#print(i)
klines=np.linspace(kpoints[i],kpoints[i+1],nk)
for j in range(0,nk):
mat=0
k=klines[j]
mat=ham(k)
n=i*nk+j
evk[n-1]=np.real(np.linalg.eigvalsh(mat))
lenk[n-1]=n
plt.plot(evk)
Square lattice¶
By analogy with 1D chains, we write the Hamiltonian for the square lattice with equal hopping integrals $t$ to the four nearest neighbor atoms (see the image below): $$ H(k)=t \exp(ik_x a)+t \exp(-ik_x a)+t \exp(ik_y a)+t \exp(-ik_y a)$$
Analytic solution¶
It is straighforward to arrive to the analytic solution $$E(k)=2t \cos(k_x a)+2t \cos(k_y a)$$
The script below plot the band structure evaluate for the k-point path $\Gamma$-M-R-$\Gamma$ in the first Brillouin which also has square shape.
In [3]:
def band_square(t):
nkl=4
d=2
kpoints=np.zeros((4,2))
kpoints[0]=[0.00,0.00]
kpoints[1]=[0.00,0.50]
kpoints[2]=[0.50,0.50]
kpoints[3]=[0.00,0.00]
ham1 = lambda k: square_2d(k,t)
plt.title(u'square lattice') #
plt.ylabel('Energy/t')
#plt.xlabel('')
plt.xticks([0,31,62,94],
['$\\Gamma$','M','R','$\\Gamma$'])
plt.ylim(-4.2, 4.2)
plt.plot([0,0],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([31,31],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([62,62],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([94,94],[-5,5],color='red', linewidth=1.0, linestyle='--')
band(nkl,kpoints,ham1)
def square_2d(kd,t):
ham=np.zeros((1,1))+0.j
k=[0.0,0.0]
k[0]=kd[0]*2*3.1415926
k[1]=kd[1]*2*3.1415926
#print(k)
k1=k[0]
k2=k[1]
ham[0][0]=0.0
ham[0][0]=2*t*np.cos(k1)+2*t*np.cos(k2)
#ham[1][1]=0.0
return(ham)
band_square(1.0)
Triangular lattice¶
Similarly, we can calculate the band structure of triangular lattice in which each site has six nearest neighbors.
In [4]:
def band_triangular(t):
nkl=4
d=2
kpoints=np.zeros((4,2))
kpoints[0]=[0.00,0.00]
kpoints[1]=[0.00,0.50]
kpoints[2]=[0.33,0.66]
kpoints[3]=[0.00,0.00]
ham1 = lambda k: triangular_2d(k,t)
plt.title(u'triangular lattice') #设置标题
plt.ylabel('Energy/t')
#plt.xlabel('')
plt.xticks([0,31,62,94],
['$\\Gamma$','M','K','$\\Gamma$'])
plt.ylim(-4.2, 7.2)
plt.plot([0,0],[-5,7],color='red', linewidth=1.0, linestyle='--')
plt.plot([31,31],[-5,7],color='red', linewidth=1.0, linestyle='--')
plt.plot([62,62],[-5,7],color='red', linewidth=1.0, linestyle='--')
plt.plot([94,94],[-5,7],color='red', linewidth=1.0, linestyle='--')
band(nkl,kpoints,ham1)
def triangular_2d(kd,t):
ham=np.zeros((1,1))+1.j
k=[0.0,0.0]
k[0]=kd[0]*2*3.1415926
k[1]=kd[1]*2*3.1415926
#print(k)
k1=k[0]
k2=k[1]
ham[0][0]=0.0
ham[0][0]=np.exp(k1*1.j)*t+np.exp(k2*1.j)*t+t*np.exp((k1-k2)*1.j)+np.exp(-k1*1.j)*t+np.exp(-k2*1.j)*t+t*np.exp((k2-k1)*1.j)
#ham[1][1]=0.0
return(ham)
band_triangular(1.0)
Honeycomb lattice¶
The honeycomb lattice is generated by two sublattices A and B. The Hamiltonian matrix is then a $2 \times 2$ matrix: $$H(k)=\left(\begin{array}{cc} 0 & t( \exp(ik\cdot \delta_1)+\exp(ik\cdot \delta_2)+\exp(ik\cdot \delta_3))\\ t(\exp(-ik\cdot \delta_1)+\exp(-ik\cdot \delta_2)+\exp(-ik\cdot \delta_3)) & 0 \end{array}\right)$$ where $\delta_i$ are vectors connecting nearest-neighbour atoms.
The eigenvalues of this matrix have the following analytic form: $$E(k)=\pm \sqrt{3 + 2 \cos(k\cdot\delta_1) + 2 \cos(k\cdot\delta_2) + 2 \cos(k\cdot\delta_3)}$$
In [5]:
def band_honeycomb(t):
nkl=4
d=2
kpoints=np.zeros((4,2))
kpoints[0]=[0.00,0.00]
kpoints[1]=[0.00,0.50]
kpoints[2]=[0.33,0.66]
kpoints[3]=[0.00,0.00]
ham1 = lambda k: honeycomb_2d(k,t)
plt.title(u'honeycomb lattice') #设置标题
plt.ylabel('Energy/t')
#plt.xlabel('')
plt.xticks([0,31,62,94],
['$\\Gamma$','M','K','$\\Gamma$'])
plt.ylim(-4.2, 4.2)
plt.plot([0,0],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([31,31],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([62,62],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([94,94],[-5,5],color='red', linewidth=1.0, linestyle='--')
band(nkl,kpoints,ham1)
def honeycomb_2d(kd,t):
ham=np.zeros((2,2))+1.j
k=[0.0,0.0]
k[0]=kd[0]*2*3.1415926
k[1]=kd[1]*2*3.1415926
#print(k)
k1=k[0]
k2=k[1]
ham[0][0]=0.0
ham[0][1]=np.exp(k1*1.j)*t+np.exp(k2*1.j)*t+t
ham[1][0]=np.exp(-k1*1.j)*t+np.exp(-k2*1.j)*t+t
ham[1][1]=0.0
return(ham)
band_honeycomb(1.0)
In [6]:
def band_lieb(t):
nkl=4
d=2
kpoints=np.zeros((4,2))
kpoints[0]=[0.00,0.00]
kpoints[1]=[0.00,0.50]
kpoints[2]=[0.50,0.50]
kpoints[3]=[0.00,0.00]
ham1 = lambda k: lieb_2d(k,t)
plt.title(u'Lieb lattice') #
plt.ylabel('Energy/t')
#plt.xlabel('')
plt.xticks([0,31,62,94],
['$\\Gamma$','M','R','$\\Gamma$'])
plt.ylim(-4.2, 4.2)
plt.plot([0,0],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([31,31],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([62,62],[-5,5],color='red', linewidth=1.0, linestyle='--')
plt.plot([94,94],[-5,5],color='red', linewidth=1.0, linestyle='--')
band(nkl,kpoints,ham1)
def lieb_2d(kd,t):
ham=np.zeros((3,3))+0.j
k=[0.0,0.0]
k[0]=kd[0]*2*3.1415926
k[1]=kd[1]*2*3.1415926
#print(k)
k1=k[0]
k2=k[1]
ham[0][0]=0.0
ham[0][1]=t+t*np.exp(-k1*1.j)
ham[0][2]=t+t*np.exp(-k2*1.j)
ham[1][0]=t+t*np.exp(k1*1.j)
ham[1][1]=0.0
ham[1][2]=0.0
ham[2][0]=t+t*np.exp(k2*1.j)
ham[2][1]=0.0
ham[2][2]=0.0
return(ham)
band_lieb(1.0)
Questions¶
- Modify the code such that you can introduce unequal hopping integrals (e.g. $t_1 \ne t_2$ for the vertical and horizontal bonds in square lattice) and investigate their effect on the band structures.
In [ ]: