quantum liquid crystal phases in strongly …sunkai/teaching/winter_2015/chapter02.pdf · primitive...
TRANSCRIPT
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Kai Sun
University of Michigan, Ann Arbor
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Cubic system
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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
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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
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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
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Hexagonal Close-Packed Structure (hcp)
hcp: ABABAB fcc: ABCABCABC
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Carbon atoms can form 3 different crystals
Graphene (Nobel Prize carbon)
Diamond (money carbon/love carbon)
Graphite (Pencil carbon)
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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
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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
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X-ray scattering
A single crystal Polycrystal: many small pieces of crystals
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Scatterings and Diffraction
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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.
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Crystal Planes
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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
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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
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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
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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
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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
= ()
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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
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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 =
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Reciprocal lattice
Periodic function: lattice
2
: the reciprocal lattice
=
exp 2
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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
, =
, = = . . .
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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
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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
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Diffraction Conditions
Scattering amplitude: = exp[( ) ] = exp
Path differenceScattering strength at
Number of particles: |()|2
=
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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
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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 =
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Laue Equations
=
Using the following two conditions: = = 11 + 22 + 33 and = 2,it is easy to note that
= where = 1,2 and 3.
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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.
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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
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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
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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!
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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!
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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
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Single crystals vs polycrystals
A single crystal Polycrystal: many small pieces of crystals