The document discusses constructing a face-centered cubic (FCC) lattice and its reciprocal lattice in Mathematica. It defines the FCC lattice using basis vectors and plots the points. It then finds the basis vectors of the reciprocal lattice by solving equations. The reciprocal lattice is constructed and plotted similarly. High symmetry points and directions in the reciprocal lattice are identified. Finally, the dispersion relation along the [100] direction of the FCC lattice is plotted.

1. Homework 1
Dai - Nam Le
In[212]:= ClearAll "Global` "
1) Define a FCC lattice a×a×a with a = 2
In mathematica, there are lattice data for all Bravais lattices.
In[213]:= LatticeData "FaceCenteredCubic", "Image"
Out[213]=
2) Define basis vectors
Lattice data in mathematica also provide information about basis vectors of Bravais lattice
In[214]:= a1, a2, a3 LatticeData "FaceCenteredCubic", "Basis"
Out[214]= 1, 1, 0 , 1, 1, 0 , 0, 1, 1
Constructing FCC lattice via basis vectors

2. In[215]:= FCCPoints n1_, n2_, n3_ : n1 a1 n2 a2 n3 a3;
FCCList ;
n 6;
amax 2;
For n1 n, n1 n, n1 ,
For n2 n, n2 n, n2 ,
For n3 n, n3 n, n3 ,
If Table FCCPoints n1, n2, n3 i amax &&
FCCPoints n1, n2, n3 i amax, i, 1, 3 True, True, True ,
FCCList Join FCCList, FCCPoints n1, n2, n3
basis Graphics3D Arrowheads .05, 1 , Arrow Tube 0, 0, 0 , a1 , .025 ,
Arrow Tube 0, 0, 0 , a2 , .025 , Arrow Tube 0, 0, 0 , a3 , .025 ;
gridlines ParametricPlot3D 0, 0, t , 2, 0, t , 0, 2, t , 0, 2, t ,
2, 0, t , 2, 2, t , 2, 2, t , 2, 2, t , 2, 2, t , 0, t, 0 ,
2, t, 0 , 0, t, 2 , 0, t, 2 , 2, t, 0 , 2, t, 2 , 2, t, 2 ,
2, t, 2 , 2, t, 2 , t, 0, 0 , t, 2, 0 , t, 0, 2 , t, 0, 2 ,
t, 2, 0 , t, 2, 2 , t, 2, 2 , t, 2, 2 , t, 2, 2 ,
t, amax, amax , PlotRange amax, amax , amax, amax , amax, amax ,
PlotStyle AbsoluteThickness 1 , Black , AspectRatio 1 ;
FCCLattice ListPointPlot3D FCCList, PlotStyle AbsolutePointSize 10. ,
AspectRatio 1, PlotRange amax, amax , amax, amax , amax, amax ;
Show FCCLattice, basis, gridlines
Out[223]=
2
1
0
1
2
2
1
0
1
2
2
1
0
1
2
3) Find reciprocal lattice ’s basis vectors
To find reciprocal lattice ‘s basis vectors, we must solve the following system of equations
2 HW1_Nam.nb

3. In[224]:= b1 b1x, b1y, b1z ;
b2 b2x, b2y, b2z ;
b3 b3x, b3y, b3z ;
sol
Solve Flatten Table ai.bj 2 Π KroneckerDelta i, j , i, 1, 3 , j, 1, 3 ,
b1x, b1y, b1z, b2x, b2y, b2z, b3x, b3y, b3z ;
b1x b1x . sol 1 ;
b1y b1y . sol 1 ;
b1z b1z . sol 1 ;
b2x b2x . sol 1 ;
b2y b2y . sol 1 ;
b2z b2z . sol 1 ;
b3x b3x . sol 1 ;
b3y b3y . sol 1 ;
b3z b3z . sol 1 ;
In[237]:= b1, b2, b3
Out[237]= Π, Π, Π , Π, Π, Π , 0, 0, 2 Π
Comparing with BCC basis
In[238]:= LatticeData "BodyCenteredCubic", "Basis"
Out[238]= 2, 0, 0 , 0, 2, 0 , 1, 1, 1
4) Buid reciprocal lattice
HW1_Nam.nb 3

4. In[239]:= ReciprocalPoints n1_, n2_, n3_ : n1 b1 n2 b2 n3 b3;
ReciprocalList ;
n 3;
bmax 2 Π;
For n1 n, n1 n, n1 ,
For n2 n, n2 n, n2 ,
For n3 n, n3 n, n3 ,
If Table FCCPoints n1, n2, n3 i bmax &&
FCCPoints n1, n2, n3 i bmax, i, 1, 3 True, True, True ,
ReciprocalList Join ReciprocalList, ReciprocalPoints n1, n2, n3
Reciprocalbasis
Graphics3D Arrowheads .05, 1 , Arrow Tube 0, 0, 0 , b1 , .025 ,
Arrow Tube 0, 0, 0 , b2 , .025 , Arrow Tube 0, 0, 0 , b3 , .025 ;
Reciprocalgridlines ParametricPlot3D 0, 0, t , 2 Π, 0, t ,
0, 2 Π, t , 0, 2 Π, t , 2 Π, 0, t , 2 Π, 2 Π, t ,
2 Π, 2 Π, t , 2 Π, 2 Π, t , 2 Π, 2 Π, t , 0, t, 0 ,
2 Π, t, 0 , 0, t, 2 Π , 0, t, 2 Π , 2 Π, t, 0 , 2 Π, t, 2 Π ,
2 Π, t, 2 Π , 2 Π, t, 2 Π , 2 Π, t, 2 Π , t, 0, 0 ,
t, 2 Π, 0 , t, 0, 2 Π , t, 0, 2 Π , t, 2 Π, 0 , t, 2 Π, 2 Π ,
t, 2 Π, 2 Π , t, 2 Π, 2 Π , t, 2 Π, 2 Π , t, bmax, bmax ,
PlotRange bmax, bmax , bmax, bmax , bmax, bmax ,
PlotStyle AbsoluteThickness 1 , Black , AspectRatio 1 ;
ReciprocalLattice ListPointPlot3D ReciprocalList,
PlotStyle AbsolutePointSize 10. , AspectRatio 1,
PlotRange bmax, bmax , bmax, bmax , bmax, bmax ;
Show ReciprocalLattice, Reciprocalbasis, Reciprocalgridlines
Out[247]=
5
0
5
5
0
5
5
0
5
4 HW1_Nam.nb

5. In[248]:= Clear n1, n2, n3
sol2 Solve ReciprocalPoints n1, n2, n3 kx, ky, kz , kx, ky, kz ;
kx kx . sol2 1 ;
ky ky . sol2 1 ;
kz kz . sol2 1 ;
Represent reciprocal vector in Cartesian coordinates
In[253]:= Print "on kx ", kx
Print "on ky ", ky
Print "on kz ", kz
on kx n1 Π n2 Π
on ky n1 Π n2 Π
on kz n1 Π n2 Π 2 n3 Π
5) [1,0,0]; [1,1,0] and [1,1,1]
a) on [1,0,0] direction
In[256]:= Clear n1, n2, n3, temp, min
temp Solve ky 0, kz 0 , n2, n3 ;
n2 n2 . temp 1 ;
n3 n3 . temp 1 ;
min Minimize kx2
ky2
kz2
, kx2
ky2
kz2
0, kx 0 , n1, Integers ;
n1 n1 . min 2 ;
In[262]:= kmin100 kx, ky, kz
Out[262]= 2 Π, 0, 0
b) on [1,1,0] direction
In[263]:= Clear n1, n2, n3, temp, min
temp Solve ky kx, kz 0 , n2, n3 ;
n2 n2 . temp 1 ;
n3 n3 . temp 1 ;
min Minimize kx2
ky2
kz2
, kx2
ky2
kz2
0, kx 0 , n1, Integers ;
n1 n1 . min 2 ;
In[269]:= kmin110 kx, ky, kz
Out[269]= Π, Π, 0
c) on [1,1,1] direction
In[270]:= Clear n1, n2, n3, temp, min
temp Solve ky kx, kz kx , n2, n3 ;
n2 n2 . temp 1 ;
n3 n3 . temp 1 ;
min Minimize kx2
ky2
kz2
, kx2
ky2
kz2
0, kx 0 , n1, Integers ;
n1 n1 . min 2 ;
In[276]:= kmin111 kx, ky, kz
Out[276]= Π, Π, Π
6) Dispersion relation on kx = [100] axis
HW1_Nam.nb 5

6. Define high symmetry points along kx axis: point and X point
In[277]:= point 0, 0, 0 ;
Xpoint
point kmin100
2
;
point
In[279]:= point
Out[279]= 0, 0, 0
Xpoint
In[280]:= Xpoint
Out[280]= Π, 0, 0
Dispersion relation from to X (a = 2, hbar = 1, m = 1)
In[281]:= m 1;
Plot Evaluate Flatten
Table
1
2
t, 0, 0 m1 b1 m2 b2 m3 b3 . t, 0, 0 m1 b1 m2 b2 m3 b3 ,
m1, m, m , m2, m, m , m3, m, m , t, point 1 , Xpoint 1 ,
PlotRange Full, Ticks 0,
Π
4
,
Π
2
,
3 Π
4
, Π , Automatic ,
PlotStyle ColorData 35, "ColorList" , Thick ,
PlotLegends Flatten Table " " ToString m1 " " ToString m2
" " ToString m3 " ", m1, m, m , m2, m, m , m3, m, m
Out[282]=
Π
4
Π
2
3 Π
4
Π
20
40
60
80
100 1 1 1
1 1 0
1 1 1
1 0 1
1 0 0
1 0 1
1 1 1
1 1 0
1 1 1
0 1 1
0 1 0
0 1 1
0 0 1
0 0 0
0 0 1
0 1 1
0 1 0
0 1 1
1 1 1
1 1 0
1 1
1 0
1 0 0
1 0 1
1 1
1 1 0
1 1 1
6 HW1_Nam.nb