Building up a diamond lattice

• Consider:

1. a cubic simulation box whose sides are of length $L$
2. a number of lattice positions, $M$ given by $M=8m^3$,

with $m$ being a positive integer

• The $M$ positions are those given by:

$\{\begin{array}{c}{x}_a=i_a\times (\delta l)\\ y_a=j_a\times (\delta l)\\ z_a=k_a\times (\delta l)\end{array}\}$

where the indices of a given valid site are integer numbers that must fulfill the following criteria

• $0\le i_a<4m$
• $0\le j_a<4m$
• $0\le k_a<4m$,
• the sum of $i_a+j_a+k_a$ can have only the values: 0, 3, 4, 7, 8, 10, ...

i.e, $i_a+j_a+k_a=4n$; OR; $i_a+j_a+k_a=4n+3$, with $n$ being any integer number

• the indices $\{i_a,j_a,k_a\}$must be either all even or all odd.

with $\delta l=L/(4m)$

Atomic position(s) on a cubic cell

• Number of atoms per cell: 8
• Coordinates:

Atom 1: $(x_1,y_1,z_1)=(0,0,0)$

Atom 2: $(x_2,y_2,z_2)=(0,\frac{l}{2},\frac{l}{2})$

Atom 3: $(x_3,y_3,z_3)=(\frac{l}{2},0,\frac{l}{2})$

Atom 4: $(x_4,y_4,z_4)=(\frac{l}{2},\frac{l}{2},0)$

Atom 5: $(x_5,y_5,z_5)=(\frac{l}{4},\frac{l}{4},\frac{l}{4})$

Atom 6: $(x_6,y_6,z_6)=(\frac{l}{4},\frac{3l}{4},\frac{3l}{4})$

Atom 7: $(x_7,y_7,z_7)=(\frac{3l}{4},\frac{l}{4},\frac{3l}{4})$

Atom 8: $(x_8,y_8,z_8)=(\frac{3l}{4},\frac{3l}{4},\frac{l}{4})$

Cell dimensions:

• $a=b=c=l$

• $\alpha =\beta =\gamma =90^0$