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PHY.K02UF Molecular and Solid State Physics

## Drawing Brillouin Zones

To draw the first Brillouin zone corresponding to a Bravais lattice, the first step is to find the primitive lattice vectors in reciprocal space. Using the primitive lattice vectors, the reciprocal lattice vectors can be constructed, $\vec{G}_{hkl}=h\vec{b}_1+k\vec{b}_2+l\vec{b}_3$. A plane normal to each reciprocal lattice vector is drawn that passes through $\vec{G}_{hkl}/2$. All of the points that can be reached from $\Gamma$ without crossing any of these planes is in the first Brillouin zone. The form below takes the primitive lattice vectors in real space as input and calculates the primitive lattice vectors in reciprocal space ($\vec{b}_1$, $\vec{b}_2$, $\vec{b}_3$), the reciprocal lattice vectors $\vec{G}_{hkl}$, the planes $(hkl)$ that form the first Brillouin zone boundaries, and the corners of the first Brillouin zone boundary.

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]

Primitive reciprocal lattice vectors

 $\vec{b}_1=2\pi\frac{\vec{a}_2\times\vec{a}_3}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}=$ $\hat{k}_x+$ $\hat{k}_y+$ $\hat{k}_z$ [Å-1] $\vec{b}_2=2\pi\frac{\vec{a}_3\times\vec{a}_1}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}=$ $\hat{k}_x+$ $\hat{k}_y+$ $\hat{k}_z$ [Å-1] $\vec{b}_3=2\pi\frac{\vec{a}_1\times\vec{a}_2}{\vec{a}_1\cdot\left(\vec{a}_2\times\vec{a}_3\right)}=$ $\hat{k}_x+$ $\hat{k}_y+$ $\hat{k}_z$ [Å-1]

Using the primitive lattice vectors in reciprocal space, the reciprocal lattice vectors can be constructed,

$$\vec{G}_{hkl}=h\vec{b}_1+k\vec{b}_2+l\vec{b}_3=\left(hb_{1x}+kb_{2x}+lb_{3x}\right)\hat{k}_x+\left(hb_{1y}+kb_{2y}+lb_{3y}\right)\hat{k}_y+\left(hb_{1z}+kb_{2z}+lb_{3z}\right)\hat{k}_z.$$

A Brillouin zone boundary is a plane normal to $\vec{G}_{hkl}$, that passes through the point $\frac{\vec{G}_{hkl}}{2}$. For the planes that make up the first Brillouin zone boundary, the distance from $\frac{\vec{G}_{hkl}}{2}$ to $\Gamma$ is smaller than the distance from $\frac{\vec{G}_{hkl}}{2}$ to any of the other reciprocal lattice vectors. By computing these distances, the planes that make up the first Brillouin zone boundary can be determined.

Once the planes are known, the points at the corners of the first Brillouin zone boundary can be determined by considering the intersections of the planes. The formula for the $(hkl)$ plane is,

$$G_{hkl,x}k_x+G_{hkl,y}k_y+G_{hkl,z}k_z=\frac{G_{hkl,x}^2}{2}+\frac{G_{hkl,y}^2}{2}+\frac{G_{hkl,z}^2}{2}.$$

This plane will form part of the first Brillouin zone if $\frac{\vec{G}_{hkl}}{2}$ is closer to $\Gamma$ than it is to all of the other reciprocal lattice points.

The edges of the Brillouin zone 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 Brillouin zone

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