Building up a diamond lattice

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  • 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 = 8 m^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 \le i_a < 4m
  •  0 \le j_a < 4m
  •  0 \le 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 =  4 n \right. ; OR;  \left. i_a + j_a + k_a = 4 n + 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[edit]

  • 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