# Lattice:Diamond

The diamond lattice, which consists of tetrahedrally-arranged atoms/particles, has simple cubic symmetry.

Along an appropriate axis, the lattice has six-fold symmetry (hexagonal).

## Canonical Diamond

A canonical diamond lattice (single atom/particle type arranged as shown above) has symmetry Fd3m. The atoms occupy the positions (0,0,0;1/4,1/4,1/4)+face-centering.

### Symmetry

• Crystal Family: Cubic
• Crystal System: Cubic
• Bravais Lattice: F (fcc)
• Crystal class: Hexoctahedral
• Point Group: d3m
• Space Group: Fd3m
• Particles per unit cell: $n=8$
• Volume of unit cell: $V_d=a^3$
• Dimensionality: $d=3$
• Projected d-dimensional volume: $v_d=a^3$
• Solid angle: $\Omega_d=4\pi$
• Nearest-neighbor distance: $d_{nn}=\sqrt{3}a/4 \approx 0.433 a$
• Assuming spherical particles of radius R:
• Particle volume fraction: $\phi=32 \pi R^3/\left(3a^3\right)$
• Maximum volume fraction: $\phi_{max}=\pi\sqrt{3}/16\approx0.340$ when $R=a\sqrt{3}/8$

### Structure

The lattice may be thought of as two interpenetrating FCC lattices.

### Particle Positions

There are 18 positions. In total there are 8 particles in the unit cell:

• $8 \, \mathrm{corners} \, \times \, \frac{1}{8} = 1$
• $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$
• $6 \, \mathrm{faces} \, \times \, \frac{1}{2} = 3$
• $\left(0,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},0,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},0\right),\left(1,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},1,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},1\right)$
• $4 \, \mathrm{internal \,\, tetrahedral \,\, sites} \, \times \, 1 = 4$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+(0,0,0)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(0,\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{4},\frac{3}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},0,\frac{1}{2}\right)=\left(\frac{3}{4},\frac{1}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},\frac{1}{2},0\right)=\left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right)$

### Examples

#### Elemental

6. Carbon (C) (a = 3.567 Å)
14. Silicon (Si) (a = 5.431 Å)
32. Germanium (Ge) (a = 5.657 Å)
50. Gray Tin (Sn) (a = 6.491 Å)

• TBD

## Diamond-like Two-particle

Also known as zincblende. This is effectively

## Double-filled Diamond-like Two-particle

The diamond lattice includes 8 "tetrahedral holes", with only 4 occupied in a 'normal' diamond structure. A two-particle lattice can be formed by filling all 8 internal holes with the 2nd particle-type, in which case the particles exist in a 1:2 ratio. This is often called a CaF2 lattice.

### Particle Positions

There are 22 positions. In total there are 12 particles in the unit cell:

• $8 \, \mathrm{corners} \, \times \, \frac{1}{8} = 1$
• $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$
• $6 \, \mathrm{faces} \, \times \, \frac{1}{2} = 3$
• $\left(0,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},0,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},0\right),\left(1,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},1,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},1\right)$
• $4 \, \mathrm{principal \,\, tetrahedral \,\, sites} \, \times \, 1 = 4$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+(0,0,0)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(0,\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{4},\frac{3}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},0,\frac{1}{2}\right)=\left(\frac{3}{4},\frac{1}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},\frac{1}{2},0\right)=\left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right)$
• $4 \, \mathrm{tetrahedral \,\, holes} \, \times \, 1 = 4$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-(0,0,0)=\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-\left(0,\frac{1}{2},\frac{1}{2}\right)=\left(\frac{3}{4},\frac{1}{4},\frac{1}{4}\right)$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-\left(\frac{1}{2},0,\frac{1}{2}\right)=\left(\frac{1}{4},\frac{3}{4},\frac{1}{4}\right)$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-\left(\frac{1}{2},\frac{1}{2},0\right)=\left(\frac{1}{4},\frac{1}{4},\frac{3}{4}\right)$

### Neighbors

There are 12 particles in the unit cell: 4 of type A, and 8 of type B. The type A particles each connect to 8 particles of type B. The type B particles each connect (tetrahedrally) to 4 particles of type A.

The lattice may also be thought of as having cubes of type B with a connected type A particle at the center. However the extended lattice is not BCC-like, since only every second "BCC cube" has a central particle of type A.

## Double Inter-penetrating Diamond Two-Particle

This lattice is formed from two diamond lattices that inter-penetrate. It is often called the NaTl lattice.

### Particle Positions

There are 36 positions. In total there are 16 particles in the unit cell, 8 of each type.

#### Particle Type A

• $8 \, \mathrm{corners} \, \times \, \frac{1}{8} = 1$
• $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$
• $6 \, \mathrm{faces} \, \times \, \frac{1}{2} = 3$
• $\left(0,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},0,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},0\right),\left(1,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},1,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},1\right)$
• $4 \, \mathrm{internal \,\, tetrahedral \,\, sites} \, \times \, 1 = 4$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+(0,0,0)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(0,\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{4},\frac{3}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},0,\frac{1}{2}\right)=\left(\frac{3}{4},\frac{1}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},\frac{1}{2},0\right)=\left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right)$

#### Particle Type B

• $12 \, \mathrm{edges} \, \times \, \frac{1}{4} = 3$
• $\left(\frac{1}{2},0,0\right) , \left(0,\frac{1}{2},0\right) , \left(0,0,\frac{1}{2}\right)$
• $\left(1,0,\frac{1}{2}\right) , \left(1,1,\frac{1}{2}\right) , \left(0,1,\frac{1}{2}\right) , \left(\frac{1}{2},1,0\right) , \left(1,\frac{1}{2},0\right) , \left(\frac{1}{2},0,1\right) , \left(0,\frac{1}{2},1\right), \left(1,\frac{1}{2},1\right) , \left(\frac{1}{2},1,1\right)$
• $1 \, \mathrm{central} \, \times \, 1 = 1$
• $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$
• $4 \, \mathrm{internal \,\, tetrahedral \,\, sites} \, \times \, 1 = 4$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-(0,0,0)=\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-\left(0,\frac{1}{2},\frac{1}{2}\right)=\left(\frac{3}{4},\frac{1}{4},\frac{1}{4}\right)$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-\left(\frac{1}{2},0,\frac{1}{2}\right)=\left(\frac{1}{4},\frac{3}{4},\frac{1}{4}\right)$
• $\left(\frac{3}{4},\frac{3}{4},\frac{3}{4}\right)-\left(\frac{1}{2},\frac{1}{2},0\right)=\left(\frac{1}{4},\frac{1}{4},\frac{3}{4}\right)$

### Examples

#### Atomics

• Sodium Thallium crystal (NaTl)

## Cristobalite

Cristobalite is a diamond-like lattice formed in silicon dioxide (SiO2). The Si atoms bond tetradrally, each to four O atoms. Each O atom is bonded to 2 Si atoms, effectively acting as a bridge.

### Structure

Can be thought of as a diamond lattice, where the tetrahedral species sit on the positions of the canonical diamond sites, and the two-bonded species sites at the midway point of each bond.

The tetrahedral species sit on the 14 "FCC" sites, plus 4 "internal" diamond sites. The two-bonded species sit on the 16 internal bonds in the unit cell. The tetrahedral species sit at a distance from one another of:

• Next-nearest-neighbor distance: $d_{nnn}=\sqrt{3}a/4 \approx 0.433 a$

The different species thus have a distance of half that:

• Nearest-neighbor distance: $d_{nn}=\sqrt{3}a/8 \approx 0.217 a$

### Particle Positions

There are 34 positions (24 in unit cell)

#### Particle Type A (bond tetrahedrally)

There are 18 positions. In total there are 8 particles in the unit cell:

• $8 \, \mathrm{corners} \, \times \, \frac{1}{8} = 1$
• $\left(0,0,0\right),(0,0,1),(0,1,0),(1,0,0),(0,1,1),(1,0,1),(1,1,0),(1,1,1)$
• $6 \, \mathrm{faces} \, \times \, \frac{1}{2} = 3$
• $\left(0,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},0,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},0\right),\left(1,\frac{1}{2},\frac{1}{2}\right),\left(\frac{1}{2},1,\frac{1}{2}\right),\left(\frac{1}{2},\frac{1}{2},1\right)$
• $4 \, \mathrm{internal \,\, tetrahedral \,\, sites} \, \times \, 1 = 4$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+(0,0,0)=\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(0,\frac{1}{2},\frac{1}{2}\right)=\left(\frac{1}{4},\frac{3}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},0,\frac{1}{2}\right)=\left(\frac{3}{4},\frac{1}{4},\frac{3}{4}\right)$
• $\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},\frac{1}{2},0\right)=\left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right)$

#### Particle Type B (two-fold bonded)

There are 16 positions (all 16 in the unit cell):

• $4 \, \mathrm{level \,\, one} \, \times \, 1 = 4$
• $\frac{ (0,0,0)+\left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right) }{2} =\left(\frac{1}{8},\frac{1}{8},\frac{1}{8}\right)$
• $\frac{ \left(\frac{1}{4},\frac{1}{4},\frac{1}{4}\right)+\left(\frac{1}{2},\frac{1}{2},0 \right) }{2} =\left(\frac{3}{8},\frac{3}{8},\frac{1}{8}\right)$
• $\frac{ \left(\frac{1}{2},\frac{1}{2},0 \right) + \left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right) }{2} =\left(\frac{5}{8},\frac{5}{8},\frac{1}{8}\right)$
• $\frac{ \left(\frac{3}{4},\frac{3}{4},\frac{1}{4}\right) + (1,1,0) }{2} =\left(\frac{7}{8},\frac{7}{8},\frac{1}{8}\right)$
• $4 \, \mathrm{level \,\, two} \, \times \, 1 = 4$
• $\frac{ \left(\frac{1}{4},\frac{1}{4},\frac{1}{4} \right) + \left(\frac{1}{2},0,\frac{1}{2} \right) }{2} =\left(\frac{3}{8},\frac{1}{8},\frac{3}{8}\right)$
• $\frac{ \left(\frac{1}{4},\frac{1}{4},\frac{1}{4} \right) + \left(0,\frac{1}{2},\frac{1}{2} \right) }{2} =\left(\frac{1}{8},\frac{3}{8},\frac{3}{8}\right)$
• $\frac{ \left(\frac{3}{4},\frac{3}{4},\frac{1}{4} \right) + \left(1,\frac{1}{2},\frac{1}{2} \right) }{2} =\left(\frac{7}{8},\frac{5}{8},\frac{3}{8}\right)$
• $\frac{ \left(\frac{3}{4},\frac{3}{4},\frac{1}{4} \right) + \left(\frac{1}{2},1,\frac{1}{2} \right) }{2} =\left(\frac{5}{8},\frac{7}{8},\frac{3}{8}\right)$
• $4 \, \mathrm{level \,\, three} \, \times \, 1 = 4$
• $\frac{ \left(\frac{3}{4},\frac{1}{4},\frac{3}{4} \right) + \left(\frac{1}{2},0,\frac{1}{2} \right) }{2} =\left(\frac{5}{8},\frac{1}{8},\frac{5}{8}\right)$
• $\frac{ \left(\frac{3}{4},\frac{1}{4},\frac{3}{4} \right) + \left(1,\frac{1}{2},\frac{1}{2} \right) }{2} =\left(\frac{7}{8},\frac{3}{8},\frac{5}{8}\right)$
• $\frac{ \left(\frac{1}{4},\frac{3}{4},\frac{3}{4} \right) + \left(0,\frac{1}{2},\frac{1}{2} \right) }{2} =\left(\frac{1}{8},\frac{5}{8},\frac{5}{8}\right)$
• $\frac{ \left(\frac{1}{4},\frac{3}{4},\frac{3}{4} \right) + \left(\frac{1}{2},1,\frac{1}{2} \right) }{2} =\left(\frac{3}{8},\frac{7}{8},\frac{5}{8}\right)$
• $4 \, \mathrm{level \,\, four} \, \times \, 1 = 4$
• $\frac{ (1,0,1) + \left(\frac{3}{4},\frac{1}{4},\frac{3}{4}\right) }{2} =\left(\frac{7}{8},\frac{1}{8},\frac{7}{8}\right)$
• $\frac{ \left(\frac{3}{4},\frac{1}{4},\frac{3}{4}\right) + \left(\frac{1}{2},\frac{1}{2},1 \right) }{2} =\left(\frac{5}{8},\frac{3}{8},\frac{7}{8}\right)$
• $\frac{ \left(\frac{1}{2},\frac{1}{2},1 \right) + \left(\frac{1}{4},\frac{3}{4},\frac{3}{4}\right) }{2} =\left(\frac{3}{8},\frac{5}{8},\frac{7}{8}\right)$
• $\frac{ \left(\frac{1}{4},\frac{3}{4},\frac{3}{4} \right) + (0,1,1) }{2} =\left(\frac{1}{8},\frac{7}{8},\frac{7}{8}\right)$