Building up a diamond lattice

[EN CONSTRUCCION]

• Consider:

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

with $m$ being a positive integer

• The $\left.M\right.$ positions are those given by:

$\left\{{\begin{array}{l}x_{a}=i_{a}\times (\delta l)\\y_{a}=j_{a}\times (\delta l)\\z_{a}=k_{a}\times (\delta l)\end{array}}\right\}$

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

• $0\leq i_{a}<4m$
• $0\leq j_{a}<4m$
• $0\leq k_{a}<4m$,
• the sum of $\left.i_{a}+j_{a}+k_{a}\right.$ can have only the values: 0, 3, 4, 7, 8, 10, ...

i.e, $\left.i_{a}+j_{a}+k_{a}=4n\right.$; OR; $\left.i_{a}+j_{a}+k_{a}=4n+3\right.$, with $n$ being any integer number

• the indices $\left\{i_{a},j_{a},k_{a}\right\}$must be either all even or all odd.

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

Atomic position(s) on a cubic cell

• Number of atoms per cell: 8
• Coordinates:

Atom 1: $\left(x_{1},y_{1},z_{1}\right)=\left(0,0,0\right)$

Atom 2: $\left(x_{2},y_{2},z_{2}\right)=\left(0,{\frac {l}{2}},{\frac {l}{2}}\right)$

Atom 3: $\left(x_{3},y_{3},z_{3}\right)=\left({\frac {l}{2}},0,{\frac {l}{2}}\right)$

Atom 4: $\left(x_{4},y_{4},z_{4}\right)=\left({\frac {l}{2}},{\frac {l}{2}},0\right)$

Atom 5: $\left(x_{5},y_{5},z_{5}\right)=\left({\frac {l}{4}},{\frac {l}{4}},{\frac {l}{4}}\right)$

Atom 6: $\left(x_{6},y_{6},z_{6}\right)=\left({\frac {l}{4}},{\frac {3l}{4}},{\frac {3l}{4}}\right)$

Atom 7: $\left(x_{7},y_{7},z_{7}\right)=\left({\frac {3l}{4}},{\frac {l}{4}},{\frac {3l}{4}}\right)$

Atom 8: $\left(x_{8},y_{8},z_{8}\right)=\left({\frac {3l}{4}},{\frac {3l}{4}},{\frac {l}{4}}\right)$

Cell dimensions:

• $a=b=c=l$

• $\alpha =\beta =\gamma =90^{0}$