Transcript Solid State Physics II - Politechnika Wrocławska

Lecture VIII
Band theory
dr hab. Ewa Popko
1
Band Theory
The calculation of the allowed electron states in a solid is
referred to as band theory or band structure theory.
To obtain the full band structure, we need to solve
Schrödinger’s equation for the full lattice potential. This
cannot be done exactly and various approximation
schemes are used. We will introduce two very different
models, the nearly free electron and tight binding models.
We will continue to treat the electrons as independent, i.e.
neglect the electron-electron interaction.
2
Bound States in atoms
Electrons in isolated
atoms occupy discrete
allowed energy levels
E0, E1, E2 etc. .
00
-1
Increasing
Binding
E1
Energy
E0
-2
The potential energy of
an electron a distance r
from a positively charge
nucleus of charge q is
 qe
V( r ) =
4  o r
2
V(r)
E2
-3
-4
-5
-8
-6
-4
-2
0
2
4
6
8
rr
3
Bound and “free” states in solids
The 1D potential energy
of an electron due to an 0
0
array of nuclei of charge
q separated by a distance
-1
a is
2
 qe
V (r ) = 
n 4  o r  na
-2
Where n = 0, +/-1, +/-2 etc. -3
V(r)
E2
E1
E0
V(r)
Solid
-3
This is shown as the
black line in the figure.
-4
V(r) lower in solid (work
function).
-5
-8
-6
-4
-2
Naive picture: lowest
binding energy states can +
+
become free to move
Nuclear positions
throughout crystal
00
r2
+r
4
+
6
8
+
a
4
Energy Levels and Bands
Isolated atoms have precise allowed energy levels.
In the presence of the periodic lattice potential bands of allowed
states are separated by energy gaps for which there are no allowed
energy states.
The allowed states in conductors can be constructed from
combinations of free electron states (the nearly free electron model)
or from linear combinations of the states of the isolated atoms (the
tight binding model).
E
+
position
+
+
+
+
5
Influence of the lattice periodicity
In the free electron model, the allowed energy states are
where for periodic boundary conditions
2n y
2nx
2nz
kx 
; ky 
; kz 
L
L
L
nx , ny and ny positive or negative integers.
E
2 2
E
(k x  k y2  k z2 )
2m
0 k
0
Periodic potential
Exact form of potential is complicated
-1
Has property V(r+ R) = V(r) where
-2
R = m1a + m2b + m3c
where m1, m2, m3 are integers and a ,b ,c
are the primitive lattice vectors.
E
-3
-4
-5
r
6
Waves in a periodic lattice
Wave moving to right
Recall X-ray scattering in Solid State:
nl = 2asina
Consider a wave, wavelength l moving
through a 1D lattice of period a.
Scattered waves
moving to left
Strong backscattering for nl = 2a
Backscattered
interfere.
waves
constructively
a
Scattering potential period a
Wave has wavevector k = 2/l.
1D lattice: Bragg condition is k = n/a (n – integer)
3D lattice: Scattering for k to k' occurs if k' = k + G
where G = ha1 + ka2 + la3 h,k,l integer and a1 ,a2 ,a3
k'
k
G
are the primitive reciprocal lattice vectors
7
Real and Reciprocal Lattice Spaces
• R for a crystal can be expressed in general as:
R=n1a1+n2a2+n3a3 where a1, a2 and a3 are the
primitive lattice vectors and n1,n2 and n3 are
integers
• Corresponding to a1, a2 and a3 there are three
primitive reciprocal lattice vectors: b1, b2 and
b3 defined in terms of a1, a2 and a3 by:
8
Bragg scattering & energy gaps
1D potential period a. Reciprocal lattice vectors G = 2n /a
A free electron of in a state exp( ix/a), ( rightward moving wave) will
be Bragg reflected since k = /a and a left moving wave exp( -ix/a)
will also exist.
In the nearly free electron model allowed un-normalised states for k =
/a are
 (+) = exp(ix/a) + exp( - ix/a) = 2 cos(x/a)
ψ(-) = exp(ix/a) - exp( - ix/a) = 2i sin(x/a)
N.B. Have two allowed states for same k which have different energies
E
+
position
+
+
a
+
+
9
Reciprocal lattice
Use of reciprocal lattice space:
Wave vectors k for Bloch waves lie in the reciprocal lattice
space.
Translation symmetry=> a Bloch wave can be characterized
by two wavevectors (or wavelengths) provided they differ by a
reciprocal lattice vector!
Example in 1D:
Suppose k’=k+(2/a) then Fk(x)=exp(ikx)u(x)
and Fk’(x)=exp(ik’x)u(x)=exp(ikx)exp(i2x/a)u(x) =exp(ikx)u’(x)
essentially have the same “wavelength”
10
Cosine solution lower energy than sine solution
Cosine solution ψ(+)
has maximum electron
probability density at
minima in potential.
Sine solution ψ(-) has
maximum electron
probability density at
maxima in potential.
In a periodic lattice the
allowed wavefunctions
have the property
 ( r  R)   (r )
2
where R is any real
lattice vector.
Cos(x/a)
Sin(x/a)
Cos2(x/a)
2
Sin2(x/a)
11
Magnitude of the energy gap
Let the lattice potential be approximated by V ( x)  V0 cos(2x / a)
Let the length of the crystal in the x-direction to be L. Note that L/a is
the number of unit cells and is therefore an integer. Normalising the
wavefunction  (+) = Acos(x/a) gives
1
2
L 2
2


A 
A cos2 (x / a)dx  1
so
0
L

Solving Schrödinger’s equation with
V ( x)  V0 cos(2x / a)
H ()  E ()
 2 2

 
 V0 cos(2x / a) ()  E()
2
 2m x

 2 k 2  V0
E
2m 2
12
Gaps at the Brillouin zone boundaries
At points A ψ(+) = 2 cos(x/a)
and
E=(k)2/2me - V0/2 .
At points B ψ(-) = 2isin(x/a)
and
E=(k)2/2me + V0/2 .
13
Bloch States
In a periodic lattice the allowed wavefunctions have the property
 ( r  R)   (r )
2
Therefore
2
where R is any real lattice vector.
 (r  R)  eia ( R) (r)
where the function a(R) is real, independent of r, and dimensionless.
Now consider ψ(r + R1 + R2). This can be written
 (r  R1  R2 )  eia ( R R ) (r )
1
Or
 (r  R1  R 2 )  e
2
ia ( R1 )
 (r  R 2 )  e ia ( R ) e ia ( R ) (r)
1
2
Therefore
a (R1 + R2) = a(R1) + a(R2)
a(R) is linear in R and can be written a(R) = kxRx + kyRy + kzRz = k.R. where
kx, ky and kz are the components of some wavevector k so
 (r  R)  e ik.R (r)
(Bloch’s Theorem)
14
Alternative form of Bloch’s Theorem
(Bloch’s Theorem)
ψ(r  R)  eik.Rψ(r)
(1)
For any k one can write the general form of any
wavefunction as
 (r)  eik.ru(r)
(2)
where u(r) has the periodicity ( translational symmetry)
of the lattice. This is an alternative statement of Bloch’s
theorem.
Re [y(x)]
x
Real part of a Bloch function. ψ ≈ eikx for a large fraction of the crystal volume.
15
Bloch Wavefunctions: allowed k-states
ψ(r) = exp[ik.r]u(r)
Periodic boundary conditions. For a cube of side L we require
ψ(x + L) = ψ(x) etc.. So
eikx(x  L)u(x  L)  eikx xu(x  L)
but u(x+L) = u(x) because it has the periodicity of the lattice therefore
Therefore
eik x(x  L)  eik x x
i.e. kx = 2 nx/L
nx integer.
Same allowed k-vectors for Bloch states as free electron states.
Bloch states are not momentum eigenstates i.e.
p  k
The allowed states can be labelled by a wavevectors k.
Band structure calculations give E(k) which determines the
dynamical behaviour.
16
Nearly Free Electrons
Construct Bloch wavefunctions of electrons out of plane wave states.
Need to solve the Schrödinger equation. Consider 1D
 2  2

+
V
(x)
 (x)= E  (x)
2
 2m  x

write the potential as a Fourier sum V ( x) 
iGx
V
e
G
G
where G = 2n/a and n are positive and negative integers. Write a
general Bloch function
 ( x)  eikxu(r )  eikx  Ag eigx
g
where g = 2m/a and m are positive and negative integers. Note the
periodic function is also written as a Fourier sum
Must restrict g to a small number of values to obtain a solution.
For n= + 1 and –1 and m=0 and 1, and k ~ /a
E=(k)2/2me + or - V0/2
17
Tight Binding Approximation
NFE Model: construct wavefunction as a sum over plane waves.
Tight Binding Model: construct wavefunction as a linear combination
of atomic orbitals of the atoms comprising the crystal.
 (r) =  c j f (r - r
j
j
)
Where f(r) is a wavefunction of the isolated atom
rj are the positions of the atom in the crystal.
18
Molecular orbitals and bonding
Consider a electron in the ground, 1s, state of a hydrogen atom
1 3/2 -r/ao
i.e. f(r) =
where
a o is the Bohr Radius
ao e

2
-  2 2 a
e
where
a=
The Hamiltonian is H =
4  o
2m r
Solving Schrödinger’s equation :
E = E1s = -13.6eV
F(r)
E1s
V(r)
+
19
Hydrogen Molecular Ion
Consider the H2+ molecular ion in which
one electron experiences the potential
of two protons. The Hamiltonian is
-  2 2
-  2 2 a
a
H=
+ U( r ) =
- 2m
2m
r |r - R|
e-
r
p+
R
p+
We approximate the electron wavefunctions as
  ( r )= A [ f( r ) +f(| r - R |)]  A[f 1 +f 2 ]
and
  (r) = B [f(r)  f(| r - R |)]  B[f1  f2]
20
Bonding and
anti-bonding states
Solution:
E = E1s – g(R) for
1.2
 ( r )
1.0
0.8
0.6
0.4
0.2
 (r )

0.0
 ( r )
-0.2
-0.4
-0.6
E = E1s + g(R) for

 (r )
-0.8
V(r)
-1.0
-1.2
-1.4
g(R) - a positive function
-6
-2
0
2
4
2
4
6
r
1.2
( r )
1.0
Two atoms: original 1s state
leads to two allowed electron
states in molecule.
-4
0.8
2
0.6
0.4
0.2
0.0
-0.2
-0.4
Find for N atoms in a solid
have N allowed energy states
-0.6
-0.8
-1.0
-1.2
-1.4
-6
-4
-2
0
r
6
21
The tight binding approximation for s states
Solution leads to the E(k) dependence!!
1D:
E (k) = - a - g (ei kxa  e- i kxa)  -a - 2g cos(kxa)
+
+
Nuclear positions
+
+
+
a
22
E(k) for a 3D lattice
(a,0,0); (0,a,0); (0,0,a)
Simple cubic: nearest neighbour atoms at
So
E(k) =  a 2g(coskxa + coskya + coskza)
Minimum E(k) =  a 6g
for kx=ky=kz=0
Maximum E(k) =  a 6g
for kx=ky=kz=+/-/2
0
-2
a 10
-4
g1
-6
E(k)
-8
Bandwidth = Emav- Emin = 12g
-10
For k << /a
cos(kxx) ~ 1- (kxx)2/2
-12
etc.
E(k) ~ constant + (ak)2g/2
c.f. E = (k
)2/me
F1
-14
-16
-18
-4
-2
/a
/a
0
2
4
k [111] direction
Behave like free electrons with “effective mass” /a2g
23
Each atomic orbital leads to a
band of allowed states in the solid
Band of allowed states
Gap: no allowed states
Band of allowed states
Gap: no allowed states
Band of allowed states
24
Independent Bloch states
F1
0
Solution of the tight binding model
is periodic in k. Apparently have
an infinite number of k-states for
each allowed energy state.
In fact the different k-states all
equivalent.
Bloch states  (r  R)  e
 (r)
-2
a 10
-4
g1
-6
E(k)
-8
-10
-12
-14
-16
ik.R
-18
-4
-2
Let k = ḱ + G where k is in the first Brillouin zone
and G is a reciprocal lattice vector.
/a
0
/a
2
4
k [111] direction
(r  R)  eik.ReiG.R(r )
But G.R = 2n, n-integer. Definition of the reciprocal lattice. So
eiG.R  1
and (r  R)  eik.R(r )
e ik.R  e ik.R
k is exactly equivalent to k.
The only independent values of k are those in the first Brillouin zone.
25
Reduced Brillouin zone scheme
The only independent values of k are those in the first Brillouin zone.
Discard for
|k| > /a
Results of tight binding calculation
2/a
-2/a
Displace into
1st B. Z.
Results of nearly free electron calculation
Reduced Brillouin zone scheme
26
Extended, reduced and periodic
Brillouin zone schemes
Periodic Zone
Reduced Zone
Extended Zone
All allowed states correspond to k-vectors in the first Brillouin Zone.
Can draw E(k) in 3 different ways
27
The number of states in a band
Independent k-states in the first Brillouin zone, i.e. kx < /a etc.
2n x
, n x  0,1,2,.... etc.
L
Monatomic simple cubic crystal, lattice constant a, and volume V.
Finite crystal: only discrete k-states allowed k x  
One allowed k state per volume (2)3/V in k-space.
Volume of first BZ is (2/a)3
Total number of allowed k-states in a band is therefore
 2 


 a 
3
2 3  V
V
a
3
N
Precisely N allowed k-states i.e. 2N electron states (Pauli) per band
This result is true for any lattice:
each primitive unit cell contributes exactly one k-state to each band.
28
Metals and insulators
In full band containing 2N electrons all states within the first B. Z. are
occupied. The sum of all the k-vectors in the band = 0.
A partially filled band can carry current, a filled band cannot
Insulators have an even integer number
of electrons per primitive unit cell.
With an even number of electrons per
unit cell can still have metallic behaviour
due to band overlap.
Overlap in energy need not occur
in the same k direction
EF
Metal due to
overlapping bands
29
EF
Empty Band
EF
Energy Gap
Partially
Filled Band
Full Band
Part Filled Band
Part Filled Band
Energy Gap
Full Band
INSULATOR
or SEMICONDUCTOR
METAL
METAL
or SEMI-METAL
30
Germanium
Bands in 3D
In 3D the band structure is
much more complicated
than in 1D because
crystals do not have
spherical symmetry.
Figure removed to
reduce file size
The form of E(k) is
dependent upon the
direction as well as the
magnitude of k.
31