1

Lecture 4 1

Lecture 4Nearly Free Electron Model

References:1. Marder, Chapters 7-8

2. Kittel, Chapter 73. Ashcroft and Mermin, Chapter 94. Kaxiras, Chapter 3

5. Ibach, Chapter 7

4.1 Nearly Free Electron Model

4.1.1 Brilloiun Zone

4.1.2 Energy Gaps

4.2 Translational Symmetry – Bloch’s Theorem

4.3 Kronig-Penney Model

4.4 Tight-Binding Approximation

4.5 Examples

Lecture 4 2

Sommerfeld’s theory does not explain all…

Metal’s conduction electrons form highly degenerate Fermi gas

Free electron model: works only for metals- heat capacity, thermal and electrical conductivity, magnetic susceptibility, etc

Drawbacks: predicted electron mean path is too longincreases with temperature

positive values for the Hall coefficient, magnetotransport

difference between a good conductor (10-10 Ohm-cm) and a good insulator (10-22 Ohm-cm) – 1032 !!!

2

Lecture 4 3

Electron Occupancy of Allowed Energy Bands

• No electrons can move in an electric field (energy band is completely filled or empty) – insulator ;

• One or more bands are partly filled – conductor

Basic Assumptions: - crystal structure is periodic- periodicity leads to formation of energy bands (allowed energy levels)- energy bands are separated by energy gaps or band gaps (region in energy

for which no wavelike electron orbital exist)

Adapted from Kittel

Lecture 4 4

4.1 Nearly Free Electron Model

In free electron model: all energy values from 0 to infinity are allowed

( )2222

22

22 zyxkkkk

mk

m++== hrh

rε

Wavefunctions are in the form: ),exp()( rkirk

rrrr ⋅=ψ

where the components of the wavevector are:kr

;...4

;2

;0LL

kx

ππ ±±=

Nearly free electron model: weak perturbation of electrons by periodic potential of ions

3

Lecture 4 5

Nearly Free Electrons

Consider the effects due to a periodic crystal structure

Under condition , the

electron wave will undergo Bragg reflection

Energy gaps develop at these k due to these reflections

At k = n π/a the wavefunctions are notthe traveling wave of free electrons

a

nGk

π±=±=2

1

The region between - π/a and π/a :

first Brillouin zone of this 1D lattice

Lecture 4 6

4.1.1 Brilloiun Zone in 1D: extended, reduced and repeated

Extended

Reduced

Repeated

BZ boundaries

4

Lecture 4 7

Reduction to the first Brillouin zone

This general demand of periodicity implies that the possible electron states are not restricted to a single parabola in k-space, but can be found on any parabola shifted by any G-vector:

For 1D case:

22

2)()( Gk

mGkk

rrhrrr+=+= εε

aGG

π2=→r

Lecture 4 8

Brilloiun Zone in 3D : Wigner-Seitz cell of the reciprocal lattice

; ; ;such that vectorsbasic are ,, where

, 2 2 2

321

213

321

132

321

321321

332211

aaa

aab

aaa

aab

aaa

aabbbb

bnbnbnG

rrr

rrr

rrr

rrr

rrr

rrrrrr

rrrr

×⋅×=

×⋅×=

×⋅×=

++= πππ

Brilloiun Zone in 3D

Recall: reciprocal lattice vector

Some properties of reciprocal lattice:

The direct lattice is the reciprocal of its own reciprocal lattice

The unit cell of the reciprocal lattice need not be a paralellopiped, e.g., Wigner-Seitz cell

first Brilloin Zone (BZ) of the fcc lattice

5

Lecture 4 9

4.1.2 Origin of the Energy Gap

a

x

a

x

πψρ

ψ

πψρ

ψ

22

22

sin|)(|)(

:)( wavefor the

cos|)(|)(

:)( wavefor the

∝−=−

−

∝+=+

+

For plane waves the charge density is not constant:

Crystal Potential - U The probability density of the particle is

ψ*ψ = | ψ|2

For pure traveling wave:1)exp()exp( =−= ikxikxρ

Lecture 4 10

Magnitude of the Energy Gap

The potential energy due to the crystal can be approximated as:

This potential has the periodicity of the lattice, U(x) = U (x + a)

The wavefunctions at the Brillouin zone boundary k = π/a (normalized over unit length of line, a) are

The difference between the two standing wave states is

The gap is equal to the Fourier component of the crystal potentia l

a

xUxU

π2cos)( =

a

x

a

x ππsin2 and cos2

Udxa

x

a

x

a

xUdxxUEg =−=−−+= ∫∫ )sin(cos

2cos2]|)(||)()[|( 222

1

0

2 πππψψ

6

Lecture 4 11

4.2 Translational Symmetry – Bloch’s Theorem

Bloch’s theorem: the wave functions of the electrons in a crystal must be of a special form (the Bloch form)

uk(r) – the periodicity of the lattice (depends on the wave vector!)Note: the Bloch function can be decomposed into a sum of traveling waves In 1D: Consider a crystal of length L = N a (N primitive u. c. of length a on a ring)

)()exp()( rurkir kk

rrrrr ⋅=ψ

)()( Truru kk

rrr +=

)2

exp()()( Therefore

1-N1,2,..., 0,s ;2

exp and|)(||(

)()(

:givessymmetry onal translati theofAddition

const.-C here ;)()(

thatdemanscondition boundary periodic The

22

Na

sxixux

N

siCxna)x|ψ

xCxNa)(x

axxC

k

N

πψ

πψ

ψψψ

ψψ

=

===+

==+

+=

Kittel, pp.179-180

Lecture 4 12

Bloch’s Theorem

Bloch wave functions are periodic functions u (r) modulated by a plane wave of a longer period

Periodic function u (r)

For non-interacting electrons moving in a periodic potential, U (r)

)()( rURrUrrr =+

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Lecture 4 13

Bloch’s solution

For non-interacting electrons moving in a periodic potential, U (r)

)(2

ˆ 2

RUm

pH +=

)()( rRrrrr ψψ =+ - tempting, but WRONG!

Lecture 4 14

Translation Operators

Let translate wave function by :RT rˆ R

rh

r

r

RPi

ReT

ˆ

ˆ −=

Theorem : if one has a collection of Hermitian operators that commute with one another, they can be diagonalized simultaneously

Any eigenvector of the Hamiltonian can be taken as an eigenfunction of all the translational operators as well:

Use theorem:

)()(

|||ˆˆ

rCRr

CeT

R

R

RPi

Rrrr

r

rh

r

r

ψψ

ψψψ

=+

==−

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Lecture 4 15

Translation Operators

Operating with eigenfunction of momentum:

ψψ kCkeR

Rkirr

r

rr

=

⇒ either or RkiR

eCrr

r = 0=ψkr

index Band :

momentum Crystal :

vectorBloch wave :

n

k

kr

h

r

For a given value of Bloch wavevector, there is still the possibility of many energy eigenvalues (can be labeled by the band index n)

The eigenfunctions made possible by periodicity is:

kn

Rki

knR

knknkn

eT

EH

r

rr

rr

rrr

ψψ

ψψ

=

=

ˆ

ˆ

)()(

or )()(

reru

reRr

kn

rki

kn

kn

Rki

kn

rr

rrr

r

rr

r

r

rr

r

ψ

ψψ−=

=+

Lecture 4 16

Energy Bands

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Lecture 4 17

Allowed values of k

If crystal is periodic with (macroscopic) dimensions then requiring to be periodic constrains to

332211 ,, aMaMaMrrr

]exp[ rkirr

⋅ kr

''31

3

1

2such that are ... where,0, llllllll l

l abbbMmbM

mk πδ=⋅≤≤=∑

=

rrrrrr

Periodic boundary condition place a condition on how small k can be

Demanding that be unique places conditions on how big k can beRkiR

eCrr

r =

Number of points in crystal equals number of unique Bloch wave vectors

Lecture 4 18

Energy Bands and Group Velocities

Velocity of electrons in the nth band with wave number k is:

knkknEv rrr

h

r ∇= 1

Note: this is similar to the solution of wave equations for a group velocity:

Wave packet:

kv

∂∂= ω

hrr

hrr

rr

rh

rr

rr

r

rr

rrrrr

/'/'

'

'

/'

''

'

')''(

')'(),,(

tiErkitiErki

rki

k

tiErki

kk

k

ekdkwe

kdeekkwtkrW

−−

−−

∝≈

≈⋅−=

∫

∫ ψ

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Lecture 4 19

4.3 Kronig-Penney Model

The wave equation is

In the region 0<x<a (U = 0), the eigenfunction is a linear combination of plane waves traveling to the right and to the left with energy

In the region –b < x < 0 within the barrier the solution is

εψψψ =+− )(2 2

22

xUdx

d

m

h

iKxiKx BeAe −+=ψm

K

2

22h=ε

m

QU

DeCe QxQx

2 where

,22

0

h=−

+= −

ε

ψ

Lecture 4 20

Kronig-Penney Model

Solution must be in the Bloch form:

The constants A, B, C, D are chosen so that wavefunction and its derivative are continuous at x = 0 and x = a

At x = 0 A + B = C + D

i K (A-B) = Q (C - D)

At x = a

Solution:

)()0()( baikexbbaxa +<<−=+<< ψψ

)(

)(

)()(

)(baikQbQbiKaiKa

baikQbQbiKaiKa

eDeCeQBeAeiK

eDeCeBeAe+−−

+−−

−=−+=+

kaKaKaKa

P

bakKaQbKaQbQKKQ

coscossin

1Qb andK Qlimit In the

)(coscoscoshsinsinh]2)[( 22

=+

<<>>+=+−

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Lecture 4 21

Functions and Energy for the K-P potential

Lecture 4 22

First Brillouin Zone for fcc lattice

12

Lecture 4 23

First Brillouin Zone for bcc lattice

a

π4

Lecture 4 24

First Brillouin Zone for hcp lattice

13

Lecture 4 25

Example: Nearly Free Electron in 1D

Electrons of mass m are confined to one dimension. A weak periodic potential is applied:

(a) Under what conditions will the nearly free-electron approximation work?

(b) Sketch the three lowest energy bands in the first Brillouin zone. Number the energy bands (starting from one at the lowest band)

(c) Calculate (to first order) the energy gap at k = π /a (between the first and second band) and k = 0 (between the second and third band)

a

xV

a

xVVxV o

ππ 4cos

2cos)( 21 ++=

Lecture 4 26

Lecture 4, continued

Nth Brillioun zone: geometrical view

Procedure:

• perpendicular bisectors are drawn between the origin and all nearby reciprocal lattice points ⇒ zone boundaries

• the 1st, 2nd, and 3rd BZ are shaded in different color (same volume)

• electron response to the external electric field same as for free electron till it approaches a zone boundary plane

• an electron once in the nth BZ remains in the nth BZ

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Lecture 4 27

Example in two dimensions

Suppose 2D lattice has two conduction el. per lattice sites

The # of k-states in BZ = the # of lattice points

For a weak potential, shape of the energy surface ~ sphere

For lattice spacing a

The reciprocal lattice 2π/a

The volume of 1st BZ 4π2/a2

The Fermi sphere must have same volume

⇒ the Fermi surface slightly out of the 1st BZ

Lecture 4 28

Example in two dimensions

• Fermi surface completely enclosing the 1st BZ

• shape of surface is modified near the zone boundary

• portion of the Fermi surface in 2nd BZ is mapped back into the 1st zone

• portion of the Fermi surface in 3rd BZ is made continuous by translation through reciprocal lattice vectors

Harrison construction:

nth BZ mapped into 1st BZ

15

Lecture 4 29

Nearly Free Electron Fermi Surface Gallery

http://www.phys.ufl.edu/fermisurface/periodic_table .html

Lecture 4 30

Symmetry Properties

Basic idea: free molecule with N-atoms ⇒ 3N degrees of freedom

⇒ 3n−6 normal modes of vibration (or a linear molecule has 3n−5 normal modes of vibration because rotation about its molecular axis cannot be observed)

What are symmetry operations of the molecule?

Operations (reflection, rotation, etc.) which leave molecule invariant

16

Lecture 4 31

Symmetry operations:

1. Symmetry plane: σ

2. n-fold axis of symmetry Cn⇒ rotation by 2π/n

Lecture 4 32

Symmetry operations:

3. Inversion symmetry, i

4. Identity, E

5. n-fold rotation + reflection Sn ⇒ rotation by 2π/n + reflection through plane ⊥to rotational axis

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Lecture 4 33

Point Groups and Their Representations

Demonstrate concept by example with H2O moleculeConsider H2O molecule lying in xz-plane:

Which symmetry operations leave molecule unchanged?

As we shall see, these four operations form C2ν point group

To establish a group must consider how operations (i.e., elements) multiplyIdea: form product tableIf R and R’ -symmetry operation, define product R R’ as consecutive application of R and R’

x

y

z

Lecture 4 34

Group of Elements

Point group ⇒ at least one point is left unchanged under symmetry operation

Classes of conjugated elements:R, R’ are conjugate if R = S-1 R’ S, where S is another element in the group

Typically associates operations such as rotations or reflections, where S takes it about another plane or axis

existsR 4)

set ofmember is R' RProduct 3)

'R' )R' (R)'R' R(R' :tionmultiplica of rule eassociativObey 2)

set ofmember is E 1)

:satisfy that elements ofset a is elements of Group

element

1-

=

≡R

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Lecture 4 35

Matrix Representations

Matrix representation of element R of group [ Dij (R) ]

Character of representative matrixX (R) = trace [ Dij (R) ]

For the above representation of C2ν the characters are:

X (E)= X (C2)= X (σxz)= X (σyz)=

Note: matrix el-ts for the representations of all 4 operators of C2ν are ± 1

−−

−↔

100

010

001

2C

Lecture 4 36

Groups and Vibrations

General problem: how to connect representations of higher dimensions to those of lower dimensions

Define formal sum:

The 3 dimensional representation of C2ν that we have used can be expressed as

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Lecture 4 37

Groups and Vibrations

More common problem is to take high dimensional representation and reduce to sum of low dimensional representations

Proceed as guided by theorem:

1) Given any n-dimensional representation Γn, with matrices Dn (R1), Dn (R2), … and corresponding characters X(Dn (R1)), X(Dn (R2)), … where Ri is an operation of the point group.

If Dj0 (R1), Dj

n (R2), … , are the matrices and Xj(Djn (R1)), Xj(Dj

n (R2)), … the characters of the irreducible representations Γoj of the point group, then

2) The number of irreducible representations = number of classes

Lecture 4 38

Example

Let’s return back to H2O

20

Lecture 4 39

Digression: finding the aj’s

Lecture 4 40

Extended character table

Some of the modes characterize translational and rotational degrees of freedom, remainder are vibrational