Crystals in reciprocal space

Exercises

Preliminary provocations

1. Calculate $\mathbf{a}_1 \cdot \mathbf{b}_1$ and $\mathbf{a}_2 \cdot \mathbf{b}_1$ using the definitions of the reciprocal lattice vectors given in the lecture. Is the result what you expected?

   We expect $\mathbf{a_i}\cdot\mathbf{b_j} = 2\pi\delta_{ij}$.

Exercise 1: Directions and Spacings of Miller planes

1. Explain the terms Miller planes and Miller indices.

   A Miller plane is a plane that intersects an infinite number of lattice points. Miller indices are a set of 3 integers which specify a set of parallel planes.

2. Consider a cubic crystal with one atom in the basis and a set of orthogonal primitive lattice vectors $a\hat{x}$, $a\hat{y}$ and $a\hat{z}$. Show that the direction $[hkl]$ in this crystal is normal to the planes with Miller indices $(hkl)$.

   The $(hkl)$ plane intersects the lattice vectors at position vectors of $\frac{\mathbf{a_1}}{h}, \frac{\mathbf{a_2}}{k}, \frac{\mathbf{a_1}}{l}$. Can you define two vectors inside the $(hkl)$ plane that span the plane? How do you compute a vector that is orthogonal to two other vectors?

3. Show that this is not true in general. Consider for instance an orthorhombic crystal, for which the primitive lattice vectors are still orthogonal but have different lengths.

   The procedure described in the previous subquestion also applies here.

4. Any set of Miller indices corresponds to a family of planes separated by a distance $d$. Show that the spacing $d$ of the $(hkl)$ set of planes in a cubic crystal with lattice parameter $a$ is $d = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$.

   Hint

   Recall that a family of lattice planes is an infinite set of equally separated parallel planes which taken all together contain all points of the lattice.

   Try computing the distance between the plane that contains the site $(0,0,0)$ of the conventional unit cell and a plane defined by the $(hkl)$ indices.

   We can calculate the distance between two planes by calculating the projection of any vector connecting the two planes onto a unit vector that is normal to the plane.

Exercise 2: The reciprocal lattice of the BCC and FCC lattices

The content on this page investigates how to construct the reciprocal lattice from a real-space lattice. We will now zoom in the properties of the FCC and BCC lattices, which are two of the most common lattices encountered in crystal structures. We analyse the reciprocal lattices and the shape of the first Brillouin zone. This helps us understanding e.g. the periodicity of 3D band structures and the Fermi surface database shown in the Attic.

We consider a bcc lattice of which the conventional unit cell has a side length $a$ and volume $V=a^3$.

1. Write down the primitive lattice vectors of the BCC lattice. Calculate the volume of the primitive unit cell. Is the volume the expected fraction of $V$?

   A possible set of BCC primitive lattice vectors is:

   $$\begin{align} \mathbf{a_1} & = \frac{a}{2} \left(-\hat{\mathbf{x}}+\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\ \mathbf{a_2} & = \frac{a}{2} \left(\hat{\mathbf{x}}-\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\ \mathbf{a_3} & = \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}}-\hat{\mathbf{z}} \right). \end{align}$$

   We expect that the volume of the primitive unit cell equals $a^3/2$ because the conventional unit cell has volume $a^3$ and contains two lattice points. We confirm this by calculating the volume of the primitive unit cell using $|\mathbf{a_1}\times\mathbf{a_2}\cdot\mathbf{a_3}| = a^3/2$.

2. Calculate the reciprocal lattice vectors associated with the primitive lattice vectors you found in the previous subquestion.

   $$\begin{align} \mathbf{b_1} & = \frac{2 \pi}{a} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right) \\ \mathbf{b_2} & = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\ \mathbf{b_3} & = \frac{2 \pi}{a} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right), \end{align}$$

3. Sketch the reciprocal lattice. Which type of lattice is it? What is the volume of its conventional unit cell?

   The reciprocal lattice forms an FCC lattice. Given the lattice vectors, the volume of the conventional unit cell is $(4\pi/a)^3$.

4. Describe the shape of the 1st Brillouin zone. How many sides does it have? (Note that the Brillouin zones are sketched in the Fermi surface periodic table in the Attic). Calculate the volume of the 1st Brillouin zone and check if it is the expected fraction of the volume you found in the previous subquestion.

   Because the 1st Brillouin Zone is the Wigner-Seitz cell of the reciprocal lattice, we need to construct the Wigner-Seitz cell of the FCC lattice. For visualisation, it is convenient to look at FCC lattice introduced in the previous section and count the nearest neighbours of each lattice point. We see that each lattice point contains 12 nearest neighbours and thus the Wigner-Seitz cell contains 12 sides! The volume of the 1st Brillouin zone is the same as the volume of any other primitive unit cell. Therefore, it is given by $\mathbf{b_1}\cdot(\mathbf{b_2}\times\mathbf{b_3})| = \tfrac{1}{4} (4\pi/a)^3$. This is one quarter of the conventional unit cell calculated in subquestion 3, which is as expected because the conventional unit cell of the FCC lattice contains 4 lattice points.

5. Based on the insight gained in this question, argue what lattice is the reciprocal lattice of the FCC lattice.

   The BCC and FCC lattices are reciprocal to each other.

Exercise 3: Miller planes and reciprocal lattice vectors

Consider a family of Miller planes $(hkl)$ in a crystal.

1. Prove that the reciprocal lattice vector $\mathbf{G} = h \mathbf{b}_1 + k \mathbf{b}_2 + l \mathbf{b}_3$ is perpendicular to the Miller plane $(hkl)$.

   Hint

   Choose two vectors that lie within the Miller plane and are not parallel to each other.

   To prove this, we can show that $\mathbf{G}$ is orthogonal to two non-parallel vectors in the Miller plane. We therefore first find two vectors in the Miller plane, such as $\mathbf{v_1} = \mathbf{a_1}/h-\mathbf{a_2}/k$ and $\mathbf{v_2} = \mathbf{a_2}/h - \mathbf{a_3}/l$. We then show that $\mathbf{G}\cdot \mathbf{v_{1,2}}=0$

2. Show that the distance between two adjacent Miller planes $(hkl)$ of any lattice is $d = 2\pi/|\mathbf{G}_\textrm{min}|$, where $\mathbf{G}_\textrm{min}$ is the shortest reciprocal lattice vector perpendicular to these Miller planes.

   We compute the distance between two parallel planes by projecting any vector connecting the two planes onto a unit vector perpendicular to the planes. Using the result of subquestion 1, the unit vector is

   $$\hat{\mathbf{n}} = \frac{\mathbf{G}}{|\mathbf{G}|}$$

   For lattice planes, there is always a plane intersecting the zero lattice point (0,0,0). As such, we can project the vector $\mathbf{a_1}/h$ onto $\mathbf{\hat{n}}$, yielding a distance:

   $$d = \hat{\mathbf{n}} \cdot \frac{\mathbf{a_1}}{h} = \frac{2 \pi}{|\mathbf{G}|}$$

3. In exercise 2, you derived the reciprocal lattice vectors of the BCC lattice from a set of primitive lattice vectors. Use these vectors to find the family of Miller planes that has the highest density of lattice points $\rho$. Use that $\rho = d/V$, where $V$ is the volume of the primitive unit cell and $d$ is the distance between adjacent planes derived in the previous subquestion. Formulate the Miller plane indices with respect to the primitive lattice vectors.

   Since $\rho=d / V$, we must maximize $d$ and therefore find the smallest $|\mathbf{G}|$. Using the primitive reciprocal lattice vectors derived in the previous question, we observe that all $\{\mathbf{b_i}\}$ have the same length and we cannot combine them to obtain an even shorter vector. Therefore, we have that the {100} family of planes (in terms of the BCC primitive lattice vectors) has the highest density of lattice points. In terms of the conventional lattice vectors, this is the {110} family of planes.

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Last update: October 30, 2023

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