7Condensed MatterI

8/13/2019 7Condensed MatterI

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PHYS 7500Condensed Matter

Physics1

Professor Sen Cadogan

oom

Allen Building

Notes #1

1

Course Information

A selection of good CMP books

Kittel, Hook & Hall, Ashcroft & Mermin,Myers, Dekker, Rosenberg

Assessment:Exam 60%

Assignments 40%

Web page available

Diagrams are taken from Hook & Hallunless otherwise noted

Most problems will be taken from theabove books

2

Possible Topics

Crystal structures & dynamics (Ch 1-2 :

Hook & Hall)

Electrons in metals (Ch 3)

Energy bands (electrons in a periodic

potential) (Ch 4)

Semiconductors (Ch 5)

Superconductivity (Ch 10)

Brief Introduction to Mssbauer

Spectroscopy

Brief Introduction to Neutron Scattering

3

Crystals

Periodicity (regular arrangement of atoms)

Symmetry

Infinite 3D array

Lattice (mathematical array of identicalpoints)

+ Basis (1 or more atoms)

= Crystal

The lattice shows the crystal symmetry(e.g. mirror planes, rotation axes 2/n-foldrotation)

Lattice vectors a, b and c

Any point can be reached by a translationvector (u,v & w are integers)

cbar wvu ++=4


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Unit Cell

A generating cell, building block

Repetition of the cell in 3D builds up the

whole lattice Infinite number of possible cells

Primitive cell has minimum volume

one lattice point

Bravais (1845) 14 lattice types in 7 crystal

classes

( )cba =V5

7 crystal classes

Cubic, hexagonal,tetragonal,orthorhombic,trigonal,monoclinic,

c

Describe with 3axes and 3 angles

e.g. cubic : a=b=cand === 90o

a

b

6

Simple Cubic

Lattice points on the corners

8 points each point is shared between 8neighbouring cubes

So, 1 point/cube (primitive)

e.g. CsCl7

Body-Centred Cubic

Lattice points on the corners + 1 in thecentre of the cube

So, 2 points/cube e.g. Fe, Na, Cr

(8 x 1/8) + 1 = 28


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Face-Centred Cubic

Lattice points on the corners + 1 in thecentre of each face

Face atoms are shared betrween 2 cubes

So, 4 points/cube e.g. Al, Ni, Cu

(8 x 1/8) + (6 x 1/2) = 49

Other important structures

Hexagonal Close Packed

Diamond (e.g. Si, Ge) 2 fcclattices with a relativedisplacement of (a/4, a/4, a/4)

Zincblende (ZnS) e.g. GaAs

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Atomic Positions

e.g the body-centred position inthe bcc structure is (a/2, b/2,c/2)

Notation is (0.5, 0.5, 0.5)c

a

b

11

We can also draw this centred position as aprojection

e.g. project into the ab plane

The point is c/2 above the ab plane

b

a

(1/2)

12


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Braggs Law Elastic scattering no energy loss so

kin = kout (in = out)

Path difference = 2 d sin

Constructive interference when pathdifference = n

n = 2 d sin

d

17

Notation

1. Directions e.g. [231]

2. Planes: axial intercepts, then reciprocals, thenintegers.

cbaT ++= 32

= Miller indices e.g. (463)

3. Index = 0 if the plane does not cut a particularaxis

Q. Why reciprocals ?

A. All planes are describable within the unit cell

18

Interplanar spacing

d(a, b, c, , , , h, k, l)

e.g. cubic

Miller Indices of the plane

222 lkh

ad

++=

X-rays scatter from the electrons Scattering power is element-specific (# of

electrons) e.g. Fe (Z=26) & Co (Z=27) hard to

distinguish e.g. Fe (Z=26) & Pt (Z=78) easy to

distinguish Hard to see light elements.

19

Scattering AmplitudeScattering Amplitude

Sum the scattering over all atoms in thestructure

is the atomic-specific form factorf

( )

++

i

lzkyhxi

iiiief

2

Indexing an XRD pattern

(hkl) Line positions () crystal class &lattice parameters

Relative peak intensities atomicpositions within cell

[ ]0= Zfi

20


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Systematic absencesSystematic absences

Simple Cubic: atom position (0,0,0)

So, all (hkl) peaks appear in the XRD pattern

Body-Centred Cubic: atoms at (0,0,0) and(, , )

( ) 100002 ==++ ee lkhi

++ lkhi2 111

So, peaks with h+k+l=odd e.g (120), (111) are

absent

( )

=++

=++=

+=

+++

oddlkh

evenlkh

ee

ee

lkhi

0

2

0

21

Neutron DiffractionNeutron Diffraction

Neutrons can also be used to solve crystalstructures

De Broglie

~ 1.5 at RT

Neutrons have no electric charge so minimal

p

h=

absorption

Neutrons scatter mainly from nuclei so cansee light elements well (unlike X-rays)

Can also distinguish neighbouring elements(e.g. Fe and Co)

Neutrons have a magnetic moment so you can

also determine magnetic structures.

22

WignerWigner--Seitz cellSeitz cell

Another way of constructing a unit cell

Draw a line from the central point to eachsurrounding point

Draw the perpendicular bisectors to each line

The enclosed volume is the Wigner-Seitz cell(a unit cell).

23

Reciprocal LatticeReciprocal Lattice

A family of crystal planes can berepresented by (i) the normal vector + (ii)the interplanar spacing

nhkl

Define a new lattice in Reciprocal Space

dhkl

hkl

hklhkl

d

nG 2=

24


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Each family of planes in REAL space onelattice point in RECIPROCAL space

Define lattice vectors in RECIPROCALspace

Relate to the REAL space lattice vectors

a*, b* and c*

*** cbaG lkhhkl ++=

0

2*

*

=

=

ba

aa

( )cbac

a

= 2*

25

Crystal BindingCrystal Binding

Cohesive Energy: separate a crystal into itscomponents

NaCl + 8.18 eV Na+ + Cl

Long-range attractive Coulomb forces

Short-range repulsive forces prevent overlap ofelectron shells (Pauli)

Repulsion

empirical repulsive potential (Lennard-Jones)

Quantum Mechanics (Born-Mayer)

B, andare parameters to be determinedexperimentally

( )

r

erUR

=

( )12r

BrUR =

26

Attraction

Coulomb (valence electrons, ions )

5 distinct categories

1 van der Waals

the inert gases e.g. Ar in their solid state

Closed electron shells neutral, spherical onaverage

Fluctuations instantaneous electric dipolemoments

Weak attraction between dipoles (E fields)

Attractive potential energy:

Total pair-potential energy is

Total potential energy of solid is

( ) 6r

A

rUA =

( )612

ijij

ijr

A

r

BrU =

=

ji ijij r

A

r

BU 6122

1

27

Lennard-Jones (van der Waals)

28


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2 Ionic

e.g. NaCl Na 11 e 1s22s22p63s1 = [Ne] 3s1

Cl 17 e [Ne] 3s23p5

Ionization Energy:

Na + 5.14 eV Na+ + e

Electron Affinity:

Cl + e Cl +3.61 eV

Na + Cl + 1.53 eV Na+ + Cl

+ ions

Coulomb potential energy

( )ijo

ijr

QrU

4

2

=

+ 2.82 U= 5.11 eV

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Total electrostatic energy = ?

ATTRACTION

+ + +

+ + +

+ + +

+ + +

+ + +

+ + +

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NaCl

Na+

has 6 Cl

nn at distance Ro+ 12 Na+ nnn at distance Ro2

+ 8 Cl nnnn at distance Ro 3

etc

Electrostatic energy per ion pair is

ooR

eU

4

2

=

is the Madelung constant depends onstructure

Convergence is a problem

L++=3

8

2

126

31

Madelung ConstantMadelung Constant

Structure

CsCl (sc) 1.76268

NaCl (fcc) 1.74757

ZnO (hexag) 1.63870

ZnS (diamond) 1.63806

Simple cubic has the largest so why arentall compounds simple cubic ?

Electrostatic energy varies as 1/Ro so whydoesnt the crystal structure collapse ?

Quantum , Pauli, repulsion etc

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3 Covalent

Sharing of valence electrons

e.g. H2, C, Si, Ge

(Hook & Hall)

Symmetric (+)

( )

( )21

21

+

+

Anti-Symmetric (-)

33 34

Symmetric: electrons provide an electrostatic

attraction between the + ions

So, U is lower

22middlemiddle +

>>

2 Also, K is lower

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4 Metallic

Not as identifiable as the others Atom (e.g. Na) donates 1 or more electrons

to the entire crystal

Approx. uniform electron gas

High electrical & thermal conductivity

Delocalization of the electron reduces thekinetic energy

Particle in a box problem

( )2

2

8ma

nhKn=

36


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5 Hydrogen-bonding

H atom forms link between 2 other atoms (F,O, N )

Partially covalent and partially ionic

Electric dipole moments

e.g. ice, glue, DNA

O

H H H

H

O

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Lattice VibrationsLattice Vibrations

Atoms in a crystal are never stationary evenat T=0 (Heisenberg, zero-point motion)

Assume small amplitudeHarmonic Limit

Basically,Hookes Law

Atoms connected by springs (spring constant)

21 xUxF ==

Levy

38

1D chain lattice of identical masses (spacing a)

Only nearest-neighbour forces Displacements from equilibrium positions =

n-1 n n+1 n+2

nu

Travelling-wave solution

( )nnnn

nnnnn

uuuum

ma

uuuu

211

11

+=

=

=

+

+

&&

Equation of Motion

( ) ( )tknain eAtu =

39

Substitute back into E uation of Motion

kv

nax

tkx

k

=

=

+

=2 Wavevector

Speed

( ) ( )tknain eAtu =

Solution is a Standing wave !

=

2sin22max

2 ka

m

4max=

40


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w

Dispersion relationDispersion relation

Plot kvs

k

3/ 3/2/ / 2//

Note: ALL possible values of arecontained in the range

1st Brillouin Zone

ak

a +

41

Low frequencies:

phasegroup

o

vv

kv

xx

k

=

=

sin

0

Homogeneous line

Many atoms participate in all displacements

Long- waves insensitive to the structuraldetails

Levy

42

Higher frequencies

kv

dk

dv

vv

a

ak

phasegroup

phasegroup

==

2

At the edge of the 1st

Brillouin Zone

is the physical quantity

No energy transfer (propagation)

0=

=

groupv

ak

groupv

43

w

k

3/ 3/2/ / 2//

44

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