PHY.K02UF Molecular and Solid State Physics

Drawing Wigner-Seitz Cells

In three dimensions, there are 14 Bravais lattices. To draw a Wigner-Seitz cells of these Bravais lattices, first the Bravais lattice vectors must be constructed from the primitive lattice vectors. The Bravais lattice vectors are, $\vec{T}_{hkl}=h\vec{a}_1+k\vec{a}_2+l\vec{a}_3$, where $\vec{a}_1,\,\vec{a}_2,\,\vec{a}_3$ are the primitive lattice vectors and $h,\,k$ and $l$ are integers. If $h$, $k$, and $l$ are restricted to the values 1, 0, and -1; there are 26 Bravais lattice points around the point at 0,0,0 to consider.

The Bravais lattice vectors are,

$$\vec{T}_{hkl}=h\vec{a}_1+k\vec{a}_2+l\vec{a}_3=\left(ha_{1x}+ka_{2x}+la_{3x}\right)\hat{x}+\left(ha_{1y}+ka_{2y}+la_{3y}\right)\hat{y}+\left(ha_{1z}+ka_{2z}+la_{3z}\right)\hat{z}.$$

A plane normal to each Bravais lattice vector is drawn that passes through $\vec{T}_{hkl}/2$. All of the points that can be reached from the origin without crossing any of these planes is in the Wigner-Seitz cell. Before we continue, there is a brief review of how you find the intersection point of three planes.

A brief review of vectors and planes
The set of planes perpendicular to a vector \(A_x\hat{x}+A_y\hat{y}+A_z\hat{z}\) is,

$$A_xx+A_yy+A_zz = C,$$

where \(C\) is any constant. If a point \( (x_0,y_0,z_0)\) on the plane is known, \(C\) can be calculated,

$$C=A_xx_0+A_yy_0+A_zz_0.$$

To find a point where three planes intersect, write down the equations that describe these three planes. These will be three linear equations with three unknowns. This is a standard linear algebra problem. The app below will solve for the three unknowns and determine the point of intersection of the planes.

 $x+$  $y+$  $z=$ 
 $x+$  $y+$  $z=$ 
 $x+$  $y+$  $z=$ 

The form below takes the primitive lattice vectors in real space as input and calculates the Bravais lattice vectors $\vec{T}_{hkl} $, the planes $(hkl)$ that form the Wigner-Seitz cell boundaries, and the corners of the Wigner-Seitz cell.

Primitive lattice vectors:

 $\vec{a}_1=$ $\hat{x}+$ $\hat{y}+$ $\hat{z}$ [Å] 
 $\vec{a}_2=$ $\hat{x}+$ $\hat{y}+$ $\hat{z}$ [Å]
 $\vec{a}_3=$ $\hat{x}+$ $\hat{y}+$ $\hat{z}$ [Å]

A boundary of the Wigner-Seitz cell is a plane normal to $\vec{T}_{hkl}$, that passes through the point $\frac{\vec{T}_{hkl}}{2}$. For the planes that make up the Wigner-Seitz cell boundary, the distance from $\frac{\vec{T}_{hkl}}{2}$ to the origin is smaller than the distance from $\frac{\vec{T}_{hkl}}{2}$ to any of the other Bravais lattice vectors. By computing these distances, the planes that make up the Wigner-Seitz cell can be determined.

Once the planes are known, the points at the corners of the Wigner-Seitz cell can be determined by considering the intersections of the planes. The formula for the $(hkl)$ plane is,

$$T_{hkl,x}x+T_{hkl,y}y+T_{hkl,z}z=\frac{T_{hkl,x}^2}{2}+\frac{T_{hkl,y}^2}{2}+\frac{T_{hkl,z}^2}{2}.$$

By solving the sets of linear equations, the corners can be determined.

The edges of the Wigner-Seitz cell can be determined by considering all pairs of corners. If a pair of corners share two planes, there is an edge between these two corners.

Rotatable model of the Wigner-Seitz cell


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