Kai Sun

University of Michigan, Ann Arbor

Cubic system

hexagonal

Primitive cell: a right prism based on a rhombus with an included angle of 120 degree.

Volume (primitive cell): =

= 21

23

2123 =

3

22

2D planes formed by equilateral trianglesStack these planes on top of each other

Sodium Chloride structure

Face-centered cubic latticeNa+ ions form a face-centered cubic latticeCl- ions are located between each two neighboring Na+ ions

Equivalently, we can say thatCl- ions form a face-centered cubic latticeNa+ ions are located between each two neighboring Na+ ions

Cesium chloride structure

Simple cubic latticeCs+ ions form a cubic latticeCl- ions are located at the center of each cube

Equivalently, we can say thatCl- ions form a cubic latticeCs+ ions are located at the center of each cube

Coordinates:Cs: 000

Cl: 1

2

1

2

1

2

Notice that this is a simple cubic latticeNOT a body centered cubic lattice For a bcc lattice, the center site is the

same as the corner sites Here, center sites and corner sites are

different

Hexagonal Close-Packed Structure (hcp)

hcp: ABABAB fcc: ABCABCABC

Carbon atoms can form 3 different crystals

Graphene (Nobel Prize carbon)

Diamond (money carbon/love carbon)

Graphite (Pencil carbon)

Cubic Zinc Sulfide Structure

Very similar to Diamond latticeNow, black and white sites are two different atomsfcc with two atoms in each primitive cell

Good choices for junctions

How to see atoms in a solid?

For conductors, we can utilize scanning tunneling microscope (STM) to see atoms(Nobel Prize in Physics in 1986)

Images from wikipedia

Limitations: (1) conductors only and (2) surface only

X-ray scattering

A single crystal Polycrystal: many small pieces of crystals

Scatterings and Diffraction

Crystal Planes

Crystal planes: we can consider a 3D crystal as layers of 2D planes. These 2D planes are crystal planes

It is a geometry concept, instead of mechanical. Sometimes, these 2D planes are weakly coupled (graphene) For other cases, the 2D planes are coupled very strongly.

Crystal Planes

Index System for Crystal Planes

Define three axes 1,2 and 3 using the three lattice vectors 1, 2 and 3The three lattice vectors could be primitive or nonprimitive lattice vectors

Find the intercepts on the axes in terms of lattice constants 1, 2 and 3:We find three numbersWe only consider the case that they are rational numberi.e. 1/1, 2/2 and 3/3 (for the figure: 3, 2, 2)

Take the reciprocals of these numbers 1/1, 2/2 and 3/3 (for the figure: 1/3,1/2,1/2)

Find the least common multiple of 1, 2 and 3: (for the figure: 6)

So, we can define three integers h = n1

1, k =

2

2and l = n

3

3

We use (hkl) to mark this crystal plane

Index System for Crystal Planes

If one of the integer is negative, we put a bar on top of it (h ) We use parentheses () to mark crystal planes We use squre braket to mark directions in a crystal: = 1 + 2 + 3 In cubic crystals, the direction is perpendicular to the plane (), but this is not

true for generic lattices

The Bragg law: mirror reflection by crystal planes

The path difference: 2 sin Constructive interference: 2 sin =

Intensity peaks: = arcsin

2

Constructiveinterference

Destructiveinterference

: Bragg angleThe direction of the beam is changed by 2

The Bragg law: mirror reflection by crystal planes

Braggs law: 2 sin = . So sin = /2

sin 1, so

2 1. In other words: 2 d

d~size of an atom (about 1 = 1010), so ~1

Visible light, ~ 4000-7000 , too large for Brag scatterings X-rays, electrons, or neutrons

Particles (v c): =2

2, so = 2

Particles are waves (QM) = /

=2

=2

=

2

Fourier Analysis: a very powerful tool for periodic functions

A 1D example:For a 1D function with periodicity , + = (), we can always write it as the sum of cos and sin functions:

= 0 +

>0

[ cos2

+ sin

2

]

with p being an positive integer.

It is easy to check that + = :

+ = 0 +

>0

{ cos2

+ + sin

2

+ }

= 0 +

>0

cos2

+ 2 + sin

2

+ 2

= 0 +

>0

cos2

+ sin

2

= ()

Fourier Analysis

Another way to write down the Fourier series:

=

exp 2

with p being an integer.

Equivalent to the cos and sin formula shown on the previous page ( = cos + sin ) If is a real function [ = ],

=

=

exp 2

= 0 +

>0

exp 2

+ exp

2

= 0 +

>0

cos2

+ sin

2

+ cos

2

sin

2

= 0 +

>0

( + ) cos 2

+ ( ) sin

2

= 0 +

>0

cos2

+ sin

2

Real function

=

exp 2

If is a real function [ = ], =

= 0 +

>0

( + ) cos 2

+ ( ) sin

2

= 0 +

>0

( + ) cos

2

+ (

) sin2

= 0 +

>0

2 cos 2

2() sin

2

= 0 +

>0

cos2

+ sin

2

So: = 2() and = 2()

=

exp

2

=

exp

2

Compare and , it is easy to see that if = , we must have =

Reciprocal lattice

Periodic function: lattice

2

: the reciprocal lattice

=

exp 2

Inversion of Fourier Series

=

exp 2

= 1 0

exp 2

Proof:

r.h.s. = 1 0

exp 2

= 1 0

exp 2

exp

2

= 1

0

exp 2

If , 0 exp

2

=

2 {exp

2

1}=0

If = , 0 exp

2

= 0

exp 0 = 0

1 =

r.h.s. = 1

, =

, = = . . .

Higher dimensions

=

exp

= 1

exp

is a 3D periodic function: = + ,

is the lattice vector: = 1 1 + 2 2 + 3 3 1, 2 and 3 are three arbitrary integers 1, 2 and 3 are the lattice vectors,

is the reciprocal lattice vector: = 11 + 22 + 33 1, 2 and 3 are three integers

1 = 223

1(23), 2 = 2

31

1(23)and 3 = 2

12

1(23)

Another way to define 1, 2 and 3: = 2,

If , then

If = , then = 2

For orthorhombic and cubic lattices, || = 2/| |, but in general this is not ture.

=

exp 2

= 1 0

exp 2

Bravais Lattice and Reciprocal lattice

Reciprocal vectors plays the role of wave-vector (momentum) in Fourier transformations.

Three primitive lattice vectors 1, 2 and 3 defines a Bravais lattice with lattice sites

located at = 1 1 + 2 2 + 3 3 (lattice vectors)

These primitive lattice vectors also give us three reciprocal lattice vectors 1, 2 and 3 Using 1, 2 and 3 (as primitive lattice vectors ), we can define a lattice, which is called

the reciprocal lattice: = 11 + 22 + 33 (reciprocal lattice vectors). The primitive (conventional) unit cell in the reciprocal lattice is known the Brillouin zone.

= 1 1 + 2 2 + 3 3 = 11 + 22 + 33

Real space k-space/ momentum space

1

2 1

2

Diffraction Conditions

Scattering amplitude: = exp[( ) ] = exp

Path differenceScattering strength at

Number of particles: |()|2

=

Diffraction Conditions

Scattering amplitude: = exp[( ) ] = exp

is a periodic function: = + with = 1 1 + 2 2 + 3 3 So we know: = exp

Using , we can rewrite as:

=

exp exp =

exp ( )

If = , = =

If differs from slightly, will be very tiny (homework 1.4)

So there is a peak, whenever being a lattice vector.

Path differenceScattering strength at

Diffraction Conditions

For elastic scatterings (|| = ||)

|| = + = ||

So

( + ) ( + ) =

2 + 2 = 0

If is a reciprocal lattice, so is

2 2 = 0So, the diffraction condition:

2 = 2

= =

Q: Is this the same as the Braggs law?A: Yes

Separation between two neighboring layers for the surface

= 2/| |

where = 1 + 2 + 3.

From figure above, = 2 sin , so =2

2 sin . Or say 2 sin =

Laue Equations

=

Using the following two conditions: = = 11 + 22 + 33 and = 2,it is easy to note that

= where = 1,2 and 3.

Structure factor

Scattering amplitude: = exp

When the diffraction condition is satisfied = , one can prove that every unit cell contributes the same amount to the integral:

#1

exp = #2

exp

This is because

#2

exp = #1

+ exp +

= #1

exp

Here we used the fact that exp = 1.

Therefore,

=

exp =

where is the number of unit cells and is called the structure factor.

Atomic structure factor

=

exp

Similar to the scattering amplitude , but the integral is limited to 1 unit cell (integrates over the whole space): = /

Wavevector must be a reciprocal lattice vector (must satisfy the diffraction condition)

If a single cell contains s atoms ( = 1,2, , ), = =1 ( ) where is

the location of each atom and ( ) is the contribution from one atom. Therefore:

=

=1

( ) exp

=

=1

( ) exp[ ( )] exp

=

=1

exp ( ) exp =

=1

exp

Atomic structure factor

= where the structure factor is:

=

=1

exp

where the atomic structure factor is:

= ( ) exp

For same type of atom, they share the same the atomic structure factor!

Using bcc and fcc as an example

bcc

In one conventional cell, there are two identical atoms located at

(0,0,0) and (1

2,1

2,1

2)

=

=1

2

exp = {1 + exp 1

2 1 +1

2 2 +1

2 3 }

If = 11 + 22 + 33, we have

=

=1

2

exp = {1 + exp[ 1 + 2 + 3 ]}

If 1 + 2 + 3 = , =2 f

If 1 + 2 + 3 = , =0

Nave diffraction condition: a peak if = = 11 + 22 + 33Not quite true for bcc (conventional cell)

Half of (1 + 2 + 3 = ) has no peak. Instead, the amplitude is 0!

fcc

In one conventional cell, there are four identical atoms located at

(0,0,0), (1

2,1

2, 0), (

1

2, 0,1

2) and (0,

1

2,1

2)

=

=1

4

exp

= {1 + exp 1

2 1 +1

2 2 + exp

1

2 1 +1

2 3 + exp

1

2 2 +1

2 3 }

If = 11 + 22 + 33, we have

=

=1

2

exp = {1 + exp 1 + 2 + exp 1 + 3 + exp[ 2 + 3 ]}

For 1, 2 and 3 All odd or all even: =4 f

1 odd and 2 even, =0

2 odd and 1 even, =0Nave diffraction condition: a peak if = = 11 + 22 + 33Not true for fcc (conventional cell)

Some of (1 odd 2 even or 2 odd 1 even) has no peak. Instead, the amplitude is 0!

Atomic structure factor of an isotropic atom

= ( ) exp

Assume the atom is isotropic (approximation) ( ) is the density of electrons

= 2 2 sin exp cos

= 2 2 exp exp

= 4

sin

2

For very small , sin

1

= 4 2 =

where is the number of atomic electrons

Single crystals vs polycrystals

A single crystal Polycrystal: many small pieces of crystals