lecture 4lgonchar/courses/p9812/lecture4... · 2010. 2. 3. · lecture 4 nearly free electron model...
TRANSCRIPT
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 =+
7
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
=+
==−
8
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
9
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
10
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
=+
<<>>+=+−
11
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
14
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
17
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
18
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
19
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