Reciprocal Space and Brillouin Zone¶

Authors: Taylor James Baird, Dou Du and Giovanni Pizzi

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Source code: https://github.com/osscar-org/quantum-mechanics/blob/master/notebook/band-theory/brillouin_zone.ipynb

In this notebook, we explore two central concepts in electronic structure and band theory: those of reciprocal space and the Brillouin zone.

Goals¶

• Appreciate the nature of the relationship between the real and reciprocal space descriptions of a crystalline system.
• Use this understanding to explain the differences in the Brillouin zones observed for different crystal structures.

Background theory¶

More on the background theory.

Tasks and exercises¶

1. What is the relationship between real space primitive vectors and reciprocal space primitive vectors?
   Solution

   The definition of the reciprocal lattice vector $\vec{b}_1$ is: $$\vec{b}_1 = 2\pi \frac{\vec{a}_2 \times \vec{a}_3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)}$$ Hence, the dot products of each real space primitive vectors with $\vec{b}_1$ are computed as: $$\vec{a}_1 \cdot \vec{b}_1 = 2 \pi \vec{a}_1 \cdot \frac{\vec{a}_2 \times \vec{a}_3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = 2 \pi$$ $$\vec{a}_2 \cdot \vec{b}_1 =2 \pi \vec{a}_2 \cdot \frac{\vec{a}_2 \times \vec{a}_3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = 0$$ $$\vec{a}_3 \cdot \vec{b}_1 =2 \pi \vec{a}_3 \cdot \frac{\vec{a}_2 \times \vec{a}_3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = 0$$ Similarly, one can compute for other reciprocal lattice vector $\vec{b}_2$ and $\vec{b}_3$. In summary, we have the relation as $\vec{a}_i \cdot \vec{b}_j = 2\pi \delta _{ij}$.
2. What is the volume of the primitive cell in reciprocal space? How it is related to the volume of the real space primitive cell? Please give a mathematical proof.
   Solution

   The volume of the primitive cell in reciprocal space, $\Omega_b$, can be calculated as: $$\Omega_b = \vec{b}_1 \cdot (\vec{b}_2 \times \vec{b}_3) = 2\pi \frac{\vec{a}_2 \times \vec{a}_3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} (\vec{b}_2 \times \vec{b}_3)$$ According to the Lagrange's identity, one can show: $$(\vec{a}_2 \times \vec{a}_3) \cdot (\vec{b}_2 \times \vec{b}_3) = (\vec{a}_2 \cdot \vec{b}_2)(\vec{a}_3 \cdot \vec{b}_3) - (\vec{a}_2 \cdot \vec{b}_3)(\vec{a}_3 \cdot \vec{b}_2) = 2\pi \cdot 2\pi - 0 = 4\pi^2$$ Hence, $$\Omega_b = \frac{8\pi^3}{\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = \frac{8\pi^3}{\Omega_a}$$ where, $\Omega_a$ is the volume of the real space primitive cell.
3. What are the structures of the Wigner-Seitz primitive cells for the BCC and FCC lattices? How about their 1st Brillouin zones?
   Solution

   

   For a BCC structure, as shown in the Figure 1, the primitive lattice vectors can be constructed as: $$\vec{a}_1 = \frac{a}{2} (\vec{x}+\vec{y}-\vec{z})$$ $$\vec{a}_2 = \frac{a}{2} (-\vec{x}+\vec{y}+\vec{z})$$ $$\vec{a}_3 = \frac{a}{2} (\vec{x}-\vec{y}+\vec{z})$$

       
   The corresponding reciprocal lattice vectors are:
       $$\vec{b}_1 = 2\pi \frac{\vec{a}_2 \times \vec{a}_3}
       {\vec{a}_1 (\vec{a}_2 \times \vec{a}_3)} = \frac{2\pi}{\Omega}
       (\vec{a}_2 \times \vec{a}_3) = \frac{\pi a^2}{\Omega} (x + y)$$
       
       $$\vec{b}_2 = 2\pi \frac{\vec{a}_3 \times \vec{a}_1}
       {\vec{a}_1 (\vec{a}_2 \times \vec{a}_3)} = \frac{2\pi}{\Omega}
       (\vec{a}_3 \times \vec{a}_1) = \frac{\pi a^2}{\Omega} (y + z)$$
       
       $$\vec{b}_1 = 2\pi \frac{\vec{a}_1 \times \vec{a}_2}
       {\vec{a}_1 (\vec{a}_2 \times \vec{a}_3)} = \frac{2\pi}{\Omega}
       (\vec{a}_2 \times \vec{a}_3) = \frac{\pi a^2}{\Omega} (x + y)$$
       
   <div style="text-align:center">
   <img src="./images/FCC.png" alt="drawing" style="width:300px;"/>
   <figcaption>Figure 2.Demonstration of the procedure for constructing the primitive lattice vectors for the face-centered cubic (FCC) structure.</figcaption>
   </div>
       
   For an FCC structure as shown in Figure 2, the primitive lattice vectors can
   be constructed as:
       $$\vec{a}_1 = \frac{a}{2} (\vec{x}+\vec{y})$$
       $$\vec{a}_2 = \frac{a}{2} (\vec{y}+\vec{z})$$
       $$\vec{a}_3 = \frac{a}{2} (\vec{z}+\vec{x})$$
   
   The corresponding reciprocal lattice vectors are:
   
   $$\vec{b}_1 = 2\pi \frac{\vec{a}_2 \times \vec{a}_3}
       {\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = \frac{2\pi}{\Omega}
       (\vec{a}_2 \times \vec{a}_3) = \frac{\pi a^2}{2 \Omega}
       (\vec{x}+\vec{y}-\vec{z})$$
       
   $$\vec{b}_2 = 2\pi \frac{\vec{a}_3 \times \vec{a}_1}
       {\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = \frac{2\pi}{\Omega}
       (\vec{a}_3 \times \vec{a}_1) = \frac{\pi a^2}{2 \Omega}
       (-\vec{x}+\vec{y}+\vec{z})$$
       
   $$\vec{b}_1 = 2\pi \frac{\vec{a}_1 \times \vec{a}_2}
       {\vec{a}_1 \cdot (\vec{a}_2 \times \vec{a}_3)} = \frac{2\pi}{\Omega}
       (\vec{a}_2 \times \vec{a}_3) = \frac{\pi a^2}{2 \Omega}
       (\vec{x}-\vec{y}+\vec{z})$$
   
   One can see that the reciprocal lattice vectors of the BCC structure are simply 
   the real space primitive lattice vectors of the FCC structure. Conversely, the reciprocal lattice of the FCC structure
   is just that of BCC structure in real space. In other words, the 1st Brillouin zone of the BCC structure
   has the same structure as the Wigner-Seitz primitive cell of FCC.
   And the 1st Brillouin zone of the FCC lattice has the same structure as the
   the Wigner-Seitz primitive cell of BCC.
   </details>   
   </li>

Legend¶

(How to use the interactive visualization)

Interactive figures¶

There are two visualizers for the real space lattice structure (top left) and its 1st Brillouin zone in reciprocal space (top right) respectively. One can left-click and rotate both structures. Middle wheel of the mouse can zoom in and zoom out the structures. In the 1st Brillouin zone visualizer, the high symmetry points are also shown in the figure.

Controls¶

There are a collection of controls under the two visualizers. Firstly, one can use the "Crystal family" dropdown menu to choose from the possible crystal families in 3D, such as cubic, orthorhombic, tetragonal, monoclinic, etc. After having selected the crystal family one is interested in studying, the user may then select the type of centering they wish to investigate (primitive, face-centered, body-centered, base-centered). Of course, certain crystal families only possess a subset of these possible types of centering and therefore certain options shall be greyed out depending on the choice for the crystal family.

Finally, users can change the lattice parameters (lattice vector magnitudes and axial angles) on the bottom right panel. Recall that certain selections of the crystal family shall impose restrictions on the allowable lattice parameters and therefore some of the sliders shall be locked accordingly.