Physics in a nutshell

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# The Reciprocal Lattice

## Motivation

The key feature of crystals is their periodicity. Thus, it is evident that this property will be utilised a lot when describing the underlying physics. Accordingly, the physics that occurs within a crystal will reflect this periodicity as well. From this general consideration one can already guess that an aspect closely related with the description of crystals will be the topic of mechanical/electromagnetic waves due to their periodic nature. A concrete example for this is the structure determination by means of diffraction.

As will become apparent later it is useful to introduce the concept of the reciprocal lattice.

## Introduction of the Reciprocal Lattice

### Starting Point

As a starting point we consider a simple plane wave \begin{align} \Psi_k (r) = \Psi_0 \cdot e^{i\vec{k}\cdot\vec{r}} \end{align} with $\vec{k}$ being any arbitrary wave vector and a Bravais lattice which is the set of vectors \begin{align} \vec{R} = m \, \vec{a}_1 + n \, \vec{a}_2 + o \, \vec{a}_3 \end{align} with $m$, $n$ and $o$ being arbitrary integer coefficients and the vectors {$\vec{a}_i$} being the primitive translation vector of the Bravais lattice.

### Definition

Now we define the reciprocal lattice as the set of wave vectors $\vec{k}$ for which the corresponding plane waves $\Psi_k(\vec{r})$ have the periodicity of the Bravais lattice $\vec{R}$. Thus we are looking for all waves $\Psi_k (r)$ that remain unchanged when being shifted by any reciprocal lattice vector $\vec{R}$. Or, more formally written: \begin{align} \Psi_k(\vec{r}) &\overset{!}{=} \Psi_k (\vec{r} + \vec{R}) \\ \Leftrightarrow \quad \Psi_0 \cdot e^{ i \vec{k} \cdot \vec{r} } &= \Psi_0 \cdot e^{ i \vec{k} \cdot ( \vec{r} + \vec{R} ) }. \end{align} This results in the condition \begin{align} e^{i \vec{k}\cdot\vec{R} } & = 1 \quad \\ \Leftrightarrow \quad \vec{k}\cdot\vec{R} &= 2 \pi l, \quad l \in \mathbb{Z} \label{eq:reciprocalLatticeCondition} \end{align} which defines a set of vectors $\vec{k}$ with respect to the set of Bravais lattice vectors $\vec{R} = m \, \vec{a}_1 + n \, \vec{a}_2 + o \, \vec{a}_3$. The initial Bravais lattice of a reciprocal lattice is usually referred to as the direct lattice.[1] [2] [3] [4]

### Basis Representation of the Reciprocal Lattice Vectors

The definition is fine so far but we are of course interested in a more concrete representation of the actual reciprocal lattice. Since $\vec{R}$ is only a discrete set of vectors, there must be some restrictions to the possible vectors $\vec{k}$ as well. In order to find them we represent the vector $\vec{k}$ with respect to some basis $\vec{b}_i$ \begin{align} \vec{k} = p \, \vec{b}_1 + q \, \vec{b}_2 + r \, \vec{b}_3 \end{align} with $p$, $q$ and $r$ (the coordinates with respect to the basis) and the basis vectors {$\vec{b}_i$} initially not further specified. Now we can write eq. \eqref{eq:reciprocalLatticeCondition} in vector-matrix-notation : \begin{align} \Leftrightarrow \;\; \begin{pmatrix} p & q & r \end{pmatrix} \begin{pmatrix} \vec{b}_1 \cdot \vec{a}_1 & \vec{b}_1 \cdot \vec{a}_2 & \vec{b}_1 \cdot \vec{a}_3 \\ \vec{b}_2 \cdot \vec{a}_1 & \vec{b}_2 \cdot \vec{a}_2 & \vec{b}_2 \cdot \vec{a}_3 \\ \vec{b}_3 \cdot \vec{a}_1 & \vec{b}_3 \cdot \vec{a}_2 & \vec{b}_3 \cdot \vec{a}_3 \end{pmatrix} \begin{pmatrix} m \\ n \\ o \end{pmatrix} = 2 \pi l \quad \label{eq:matrixEquation} \end{align} Since we are free to choose any basis {$\vec{b}_i$} in order to represent the vectors $\vec{k}$, why not just the simplest one? If we choose a basis {$\vec{b}_i$} that is orthogonal to the basis {$\vec{a}_i$}, i.e. \begin{align} \vec{b}_i \cdot \vec{a}_j = 2 \pi \delta_{ij} \label{eq:orthogonalityCondition} \end{align} and divide eq. \eqref{eq:matrixEquation} by $2 \pi$, then the matrix in eq. \eqref{eq:matrixEquation} becomes the unit matrix and we can rewrite eq. \eqref{eq:matrixEquation} as follows: \begin{align} \Leftrightarrow \quad pm + qn + ro = l \end{align} Here $m$, $n$ and $o$ are still arbitrary integers and the equation must be fulfilled for every possible combination of them. Since $l \in \mathbb{Z}$ (eq. \eqref{eq:reciprocalLatticeCondition}), the LHS must always sum up to an integer as well no matter what the values of $m$, $n$, and $o$ are. That implies, that $p$, $q$ and $r$ must also be integers. Thus, the set of vectors $\vec{k}_{pqr}$ (the reciprocal lattice) forms a Bravais lattice as well![5] [6]

### Primitive Translation Vectors

This is a nice result. But we still did not specify the primitive-translation-vectors {$\vec{b}_i$} of the reciprocal lattice more than in eq. \eqref{eq:orthogonalityCondition}. Is there such a basis at all? Yes, there is and we can construct it from the basis {$\vec{a}_i$} which is given.

Let us consider the vector $\vec{b}_1$. Eq. \eqref{eq:orthogonalityCondition} provides three conditions for this vector. Two of them can be combined as follows: \begin{align} \vec{b}_1 \cdot \vec{a}_2 = \vec{b}_1 \cdot \vec{a}_3 = 0 \\ \Rightarrow \quad \vec{b}_1 = c \cdot \vec{a}_2 \times \vec{a}_3 \label{eq:b1pre} \end{align} Here $c$ is some constant that must be further specified. Therefore we multiply eq. \eqref{eq:b1pre} by the vector $\vec{a}_1$ and apply the remaining condition $\vec{b}_1 \cdot \vec{a}_1 = 2 \pi$: \begin{align} \vec{a}_1 \cdot \vec{b}_1 = c \cdot \vec{a}_1 \cdot \left( \vec{a}_2 \times \vec{a}_3 \right) = 2 \pi \end{align} \begin{align} \Leftrightarrow \quad c = \frac{2\pi}{\vec{a}_1 \cdot \left( \vec{a}_2 \times \vec{a}_3 \right)} \end{align} Here ${V:=\vec{a}_1 \cdot \left( \vec{a}_2 \times \vec{a}_3 \right)}$ is the volume of the parallelepiped spanned by the three primitive translation vectors {$\vec{a}_i$} of the original Bravais lattice. It remains invariant under cyclic permutations of the indices.

The same can be done for the vectors $\vec{b}_2$ and $\vec{b}_3$ and one obtains \begin{align} \vec{b}_1 = 2 \pi \cdot \frac{\vec{a}_2 \times \vec{a}_3}{V} \label{eq:b1} \\ \vec{b}_2 = 2 \pi \cdot \frac{\vec{a}_3 \times \vec{a}_1}{V} \label{eq:b2} \\ \vec{b}_3 = 2 \pi \cdot \frac{\vec{a}_1 \times \vec{a}_2}{V} \label{eq:b3} \end{align} with ${V = \vec{a}_1 \cdot \left( \vec{a}_2 \times \vec{a}_3 \right)}$ as introduced above.[7] [8]

## General Properties

The main features of the reciprocal lattice are:

[9] [10] [11]

## Example: Reciprocal Lattice of the fcc Structure

Now we will exemplarily construct the reciprocal-lattice of the fcc structure. As a starting point we need to find three primitive translation vectors $\vec{a}_i$ such that every lattice point of the fcc Bravais lattice can be represented as an integer linear combination of these. One may be tempted to use the vectors which point along the edges of the conventional (cubic) unit cell but they are not primitive translation vectors. Those reach only the lattice points at the vertices of the cubic structure but not the ones at the faces.

Instead we can choose the vectors which span a primitive unit cell such as \begin{align} \vec{a}_1 &= \frac{a}{2} \cdot \left( \hat{y} + \hat {z} \right) \\ \vec{a}_2 &= \frac{a}{2} \cdot \left( \hat{x} + \hat {z} \right) \\ \vec{a}_3 &= \frac{a}{2} \cdot \left( \hat{x} + \hat {y} \right) . \end{align} Here $\hat{x}$, $\hat{y}$ and $\hat{z}$ denote the unit vectors in $x$-, $y$-, and $z$ direction. If the origin of the coordinate system is chosen to be at one of the vertices, these vectors point to the lattice points at the neighboured faces.

Now we apply eqs. \eqref{eq:b1} - \eqref{eq:b3} and obtain: \begin{align} \vec{b}_1 &= \frac{8 \pi}{a^3} \cdot \vec{a}_2 \times \vec{a}_3 = \frac{4\pi}{a} \cdot \left( - \frac{\hat{x}}{2} + \frac{\hat{y}}{2} + \frac{\hat{z}}{2} \right) \\ \vec{b}_2 &= \frac{8 \pi}{a^3} \cdot \vec{a}_3 \times \vec{a}_1 = \frac{4\pi}{a} \cdot \left( \frac{\hat{x}}{2} - \frac{\hat{y}}{2} + \frac{\hat{z}}{2} \right) \\ \vec{b}_3 &= \frac{8 \pi}{a^3} \cdot \vec{a}_1 \times \vec{a}_2 = \frac{4\pi}{a} \cdot \left( \frac{\hat{x}}{2} + \frac{\hat{y}}{2} - \frac{\hat{z}}{2} \right) \end{align} which turn out to be primitive translation vectors of the fcc structure. Thus, the reciprocal lattice of a fcc lattice with edge length $a$ is a bcc lattice with edge length $\frac{4\pi}{a}$.[12] [13] Accordingly, the reciprocal-lattice of a bcc lattice is a fcc lattice.[14]

## References

 [1] N. W. Ashcroft, N. D. Mermin Festkörperphysik Oldenbourg 2001 (p. 108) [2] R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch. 2.1.1) [3] E. N. Economou The Physics of Solids Springer 2010 (p. 74) [4] S. Hunklinger Festkörperphysik De Gruyter 2014 (ch. 4.3.1) [5] N. W. Ashcroft, N. D. Mermin Festkörperphysik Oldenbourg 2001 (pp. 109-110) [6] R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (ch. 2.1.3) [7] E. N. Economou The Physics of Solids Springer 2010 (p. 74) [8] S. Hunklinger Festkörperphysik De Gruyter 2014 (ch. 4.3.1) [9] N. W. Ashcroft, N. D. Mermin Festkörperphysik Oldenbourg 2001 (ch. 5) [10] R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (p. 59) [11] E. N. Economou The Physics of Solids Springer 2010 (p. 74) [12] N. W. Ashcroft, N. D. Mermin Festkörperphysik Oldenbourg 2001 (p. 111) [13] R. Gross, A. Marx Festkörperphysik De Gruyter 2014 (p. 60) [14] E. N. Economou The Physics of Solids Springer 2010 (p. 75)

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