Konstantin Unruh

Waves in Crystals

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Introduction

Basics

To begin let’s just define some basic principals that are important to understand Solid State Physics.

Structure and Unit Cells

Lattice

Points in space that are ordered in a specific way so you can describe the position of every point by a combination of lattice vectors

r⃗=ua⃗+vb⃗+wc⃗\vec{r} = u \vec{a} + v \vec{b} + w \vec{c}

is called a lattice. The lattice vectors are not distinct and can be chosen in different ways.

The lattice points are only points in space, to describe a structure we need to add a basis that contains the position of the atoms relative to the lattice. The basis can only contain the null-vector so that the atoms lie exactly on the lattice points. If we on the other hand had two basis vectors p1⃗=(0,0)\vec{p_1} = (0, 0) and p2⃗=(12,12)\vec{p_2} = (\frac{1}{2}, \frac{1}{2}) we would get the following structure

Structure

One can define a unit cell for a lattice which then describes the whole lattice by translation of the cell, e.g. the following for a simple lattice

Unit Cell

However there would have been many possibilities. There is a definition for a distinct cell that is the Wigner-Seitz Unit Cell. It consists of all points that are closer to the lattice point than to any other lattice point.

Wigner-Seitz Cell

Common lattices

Now let’s look at actual lattices in three dimensions. There are three basic ones everyone should know.

Simple Cubic (SC) lattice

As the name suggests this lattice contains just the corners of a cube. The lattice vectors are a1⃗=(a,0,0)\vec{a_1} = (a, 0, 0), a2⃗=(0,a,0)\vec{a_2} = (0, a, 0) and a3⃗=(0,0,a)\vec{a_3} = (0, 0, a).

Simple Cubic

Face Centered Cubic (FCC) lattice

Another important lattice similar to the SC-lattice but with additional lattice points at each face of the cube. The lattice vectors are a1⃗=(a2,a2,0)\vec{a_1} = (\frac{a}{2}, \frac{a}{2}, 0), a2⃗=(a2,0,a2)\vec{a_2} = (\frac{a}{2}, 0, \frac{a}{2}) and a3⃗=(0,a2,a2)\vec{a_3} = (0, \frac{a}{2}, \frac{a}{2}).

Face Centered Cubic

Body Centered Cubic (BCC) lattice

Lastly there is the SC-lattice but with one additional lattice point at the center of the cube. The lattice vectors are a1⃗=(a,0,0)\vec{a_1} = (a, 0, 0), a2⃗=(0,a,0)\vec{a_2} = (0, a, 0) and a3⃗=(a2,a2,a2)\vec{a_3} = (\frac{a}{2}, \frac{a}{2}, \frac{a}{2}).

Body Centered Cubic

Miller Indices

There are many different planes one can consider in a crystal. To be able to distinctly differentiate between different planes one uses the Miller Indices h,k,lh, k, l.

Miller Indices Intersections

The plane then intercepts the axes at a⃗h\frac{\vec{a}}{h}, b⃗k\frac{\vec{b}}{k} and c⃗l\frac{\vec{c}}{l} and does not intercept an axis at all if the corresponding index is 00. We can now relate the axes intersections to the distance between the planes.

a⃗h⋅d⃗=∣a⃗∣hdcos⁡θ=d2\frac{\vec{a}}{h} \cdot \vec{d} = \frac{|\vec{a}|}{h} d \cos{\theta} = d^2

as d⃗\vec{d}, a⃗\vec{a} and the plane create a right triangle. This is of course true for all lattice vectors. After multiplying by 2π2\pi we get the following relation

a⃗2πd⃗d2=2πh\vec{a} \frac{2 \pi \vec{d}}{d^2} = 2 \pi h b⃗2πd⃗d2=2πk\vec{b} \frac{2 \pi \vec{d}}{d^2} = 2 \pi k c⃗2πd⃗d2=2πl\vec{c} \frac{2 \pi \vec{d}}{d^2} = 2 \pi l

This for now seems kind of arbitrary but we will need it later.

Miller Planes Examples

The Reciprocal Lattice

We know that we can write the direct lattice as arbitrary combinations of the lattice vectors R⃗=ua⃗+vb⃗+wc⃗\vec{R} = u\vec{a} + v\vec{b} + w\vec{c}. We can calculate the reciprocal lattice vectors as follows

a⃗∗=2π(b⃗×c⃗)a⃗(b⃗×c⃗)\vec{a}^* = \frac{2 \pi (\vec{b} \times \vec{c})}{\vec{a}(\vec{b} \times \vec{c})}

b⃗∗=2π(c⃗×a⃗)a⃗(b⃗×c⃗)\vec{b}^* = \frac{2 \pi (\vec{c} \times \vec{a})}{\vec{a}(\vec{b} \times \vec{c})}

c⃗∗=2π(a⃗×b⃗)a⃗(b⃗×c⃗)\vec{c}^* = \frac{2 \pi (\vec{a} \times \vec{b})}{\vec{a}(\vec{b} \times \vec{c})}

So that we can define the reciprocal lattice

G⃗h,k,l=ha⃗∗+kb⃗∗+lc⃗∗\vec{G}_{h,k,l} = h\vec{a}^* + k\vec{b}^* + l\vec{c}^*

The reciprocal lattice for common lattices we defined earlier can be seen in the following table

Direct Lattice Reciprocal Lattice
SC (side aa) SC (side 2πa\frac{2\pi}{a})
FCC (side aa) BCC (side 4πa\frac{4\pi}{a})
BCC (side aa) FCC (side 4πa\frac{4\pi}{a})

Waves in Crystals

Now let’s consider a plane (e.g. electromagnetic) wave that interacts with a certain lattice. The wave will scatter at the atoms and we want to determine in which directions the outgoing waves constructively interfere with each other. Let’s start by looking at only two lattice points. We have an incoming wave with wave vector k⃗\vec{k} and the vector connecting the two lattice points is d⃗\vec{d}.

Laue Conditions

The wave’s path difference between the two lattice points is a+ba+b. We can calculate their length in relation to d⃗\vec{d} and the angles α\alpha and β\beta.

The incoming wave vector is k⃗=2πλn⃗\vec{k} = \frac{2\pi}{\lambda} \vec{n} and the outgoing wave vector k′⃗=2πλn′⃗\vec{k'} = \frac{2\pi}{\lambda} \vec{n'}.

The length aa can be expressed as

a=∣d⃗∣cos⁡α=d⃗⋅n⃗a = |\vec{d}| \cos \alpha = \vec{d} \cdot \vec{n}

and similarly bb

b=∣d⃗∣cos⁡β=−d⃗⋅n′⃗b = |\vec{d}| \cos \beta = -\vec{d} \cdot \vec{n'}

So to get constructive interference the sum of aa and bb has to be a multiple of the wavelength λ\lambda

d⃗(n⃗−n′⃗)=mλ\vec{d} (\vec{n} - \vec{n'}) = m \lambda

To get the relation between the wave vector itself we multiply with 2πλ\frac{2\pi}{\lambda} and thus get

d⃗(k⃗−k′⃗)=d⃗⋅K⃗=2πm\vec{d} (\vec{k} - \vec{k'}) = \vec{d} \cdot \vec{K} = 2 \pi m

We have to consider the whole lattice so d⃗=ua⃗+vb⃗+wc⃗\vec{d} = u\vec{a} + v\vec{b} + w\vec{c} so that the conditions for all different d⃗\vec{d} must be met. These are the

K⃗⋅a⃗=2πH\vec{K} \cdot \vec{a} = 2 \pi H K⃗⋅b⃗=2πK\vec{K} \cdot \vec{b} = 2 \pi K K⃗⋅c⃗=2πL\vec{K} \cdot \vec{c} = 2 \pi L

As we discussed earlier the reciprocal lattice G⃗\vec{G} is given by

G⃗h,k,l=ha⃗∗+kb⃗∗+lc⃗∗\vec{G}_{h,k,l} = h \vec{a}^* + k \vec{b}^* + l \vec{c}^*

One can prove that all scattering vectors that are equal to a reciprocal lattice vector

K⃗=G⃗h,k,l\vec{K} = \vec{G}_{h,k,l}

fulfill the Laue conditions, which means

G⃗h,k,l⋅a⃗=2πh\vec{G}_{h,k,l} \cdot \vec{a} = 2 \pi h G⃗h,k,l⋅b⃗=2πk\vec{G}_{h,k,l} \cdot \vec{b} = 2 \pi k
G⃗h,k,l⋅c⃗=2πl\vec{G}_{h,k,l} \cdot \vec{c} = 2 \pi l

We showed earlier that

a⃗∗=2π(b⃗×c⃗)a⃗(b⃗×c⃗)\vec{a}^* = \frac{2 \pi (\vec{b} \times \vec{c})}{\vec{a}(\vec{b} \times \vec{c})}

b⃗∗=2π(c⃗×a⃗)a⃗(b⃗×c⃗)\vec{b}^* = \frac{2 \pi (\vec{c} \times \vec{a})}{\vec{a}(\vec{b} \times \vec{c})}

c⃗∗=2π(a⃗×b⃗)a⃗(b⃗×c⃗)\vec{c}^* = \frac{2 \pi (\vec{a} \times \vec{b})}{\vec{a}(\vec{b} \times \vec{c})}

So we can conclude a relation between the physical distance between planes and the reciprocal lattice vector

G⃗h,k,l=2πd⃗d2\vec{G}_{h,k,l} = \frac{2 \pi \vec{d}}{d^2}

This means G⃗h,k,l\vec{G}_{h,k,l} is parallel to d⃗\vec{d} and perpendicular to the h,k,lh,k,l-plane.

We can describe the incoming wave with wavevector k⃗\vec{k} and angular frequency ω\omega

Aei(k⃗⋅r⃗−ωt)A e^{i(\vec{k} \cdot \vec{r} - \omega t)}

The outgoing wave with wavevector k′⃗\vec{k'} and same frequency (we assume elastic scattering) is then

Aei(k′⃗⋅r⃗−ωt)f0+Aei(k′⃗⋅r⃗−K⃗⋅r1⃗−ωt)f1+Aei(k′⃗⋅r⃗−K⃗⋅r2⃗−ωt)f2+…A e^{i(\vec{k'} \cdot \vec{r} - \omega t)} f_0 + A e^{i(\vec{k'} \cdot \vec{r} - \vec{K} \cdot \vec{r_1} - \omega t)} f_1 + A e^{i(\vec{k'} \cdot \vec{r} - \vec{K} \cdot \vec{r_2} - \omega t)} f_2 + \dots

where fnf_n are the scattering probabilities. It simplifies to

Atotei(k′⃗⋅r⃗−ωt)A_{tot} e^{i(\vec{k'} \cdot \vec{r} - \omega t)}

where

Atot=∑ne−i(K⃗⋅rn⃗)fn=∑le−i(K⃗⋅rl⃗)∑pe−i(K⃗⋅rp⃗)fnA_{tot} = \sum_n e^{-i(\vec{K} \cdot \vec{r_n})} f_n = \sum_l e^{-i(\vec{K} \cdot \vec{r_l})} \sum_p e^{-i(\vec{K} \cdot \vec{r_p})}f_n

We call

S=∑pe−i(K⃗⋅rp⃗)fpS = \sum_p e^{-i(\vec{K} \cdot \vec{r_p})}f_p

the structure factor.

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