Kai Sun University of Michigan, Ann Arbor
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