# Building up a diamond lattice

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

with ${\displaystyle m}$ being a positive integer

• The ${\displaystyle \left.M\right.}$ positions are those given by:
${\displaystyle \left\{{\begin{array}{l}x_{a}=i_{a}\times (\delta l)\\y_{a}=j_{a}\times (\delta l)\\z_{a}=k_{a}\times (\delta l)\end{array}}\right\}}$

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

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

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

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

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

## Atomic position(s) on a cubic cell

• Number of atoms per cell: 8
• Coordinates:

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

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

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

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

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

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

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

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

Cell dimensions:

• ${\displaystyle a=b=c=l}$
• ${\displaystyle \alpha =\beta =\gamma =90^{0}}$