solid state physics lecture

1. UEEP2024 Solid State Physics
Topic 3 Free Electron in Metals

2. Drude’s classical theory of Electrical
Conduction
• Drude assumed that a metal is composed of ions, which are
stationary, and valence electrons, which are free to move.
• If no voltage applied to the metal then at each collision the
electron is deflected in a different direction so that the
overall motion is quite random.
• The valence electrons appears to be similar to the
molecules in an idea gas.
• The velocity of the electrons with mass m at temperature T
is given by the equation
where KB is the Boltzmann’s constant.
TKmv B
2
3
2
1 2


3. Drude’s classical theory of Electrical
Conduction
• The mean free path  is the average distance
that an electron travels between collisions.
• The relaxation time  is the average time
duration between collisions.
• The mean speed of the electron
• The speed of electron at room temperature is
about 105 ms-1.
• The mean free path is about 1 nm and the
relaxation time about 10-14 s.
.


v

4. Drude’s classical theory of Electrical
Conduction
• When we apply an electric field to a sample,
the electrons are attracted towards the
positive end of the sample, a net flow of
electrons in this direction.
• If the electric field is E, then the force on each
electron is eE.
• Acceleration on each electron is .
• Change in velocity
• This quantity is called the drift velocity.
em
eE
a 
.
em
eE
av

 

5. Drude’s classical theory of Electrical
Conduction
• Electron mobility µ =e/m.
• The current density
where n is the number of valence electrons per unit
volume,
• Conductivity of the metal is given by
• Drube’s model is
a) consistent with ohm’s law.
b) Explain the phenomenon of electrical resistance.
c) Gives good values of conductivity.
.eEn
A
I
J 
.en 

6. Failures of Classical model
• Conductivity should be directly proportional to the valence electron
concentration, not in good agreement with experimental data.
• Electrical properties of alloys should be intermediate between the
values of corresponding pure materials. However many alloys have
resistivity which are considerably larger than those of either of the
pure constituents.
• The dependence of resistivity on temperature:
Drube’s model predicted resistivity should be proportional to T1/2,
experimental measurement show that resistivity is actually
proportional to T over a wide range of temperature.
• Drude ‘s model predict the molar specific heat capacity for
monovalent metal of 9R/2, 6R for divalent metal and 15R/2 for
trivalent metal. However, experiment results show that the molar
specific heat capacity at room temperature is approximately 3R,
regardless of the valency of the metal.

7. Example
Estimate the typical conductivity of a metal at
295 K assuming that the mean free path is
about 1 nm and the number of valency
electrons is about 1029 m-3.

8. Solution
• Thermal velocity
• Relaxation time
• Conductivity
15
31
23
1016.1
1011.9
)295)(1038.1(33 





 ms
m
Tk
v B
s
v
15
5
9
1062.8
1016.1
101 







    
117
31
15219292
1042.2
1011.9
1062.8106.110







m
m
ne 


9. Free Electron Fermi Gas
• Atoms bounded by “free electrons”. Good
example is alkali metals (Li, K, Na, etc.)
Electron in a metal can move freely in a straight path over many
atomic distances, undeflected by collisions with other conduction
electron or by collisions with the atom cores

10. Free Electron Fermi Gas
1. A conduction electron is not deflected by ion cores
arranged on a periodic lattice
2. A conduction electron is scattered only infrequently by
other conduction electrons
Consequence of the Pauli exclusion principle
Free electron Fermi gas – A gas of free electrons
subject to the Pauli principle

11. Free Electron Fermi Gas
),(ˆ),( trHtr
t
i 



For a general quantum system
(1)
For a single particle in three dimensions
),()(),(
2
),( 2
2
trrVtr
m
tr
t
i 

 
 (2)
Wave functionHamiltonian operator

12. Free Electron Fermi Gas
(1-D)
nn
n
E
dx
d
m


 2
22
2

Consider a free electron gas in 1-D (electron of mass m is confined
to a length L by infinite barriers), (2) can be expressed as
With boundary conditions
0)(0)0(  Lnn
(3)
Energy of the electron

13. Free Electron Fermi Gas
(1-D)
LnwherexA n
n
n 





 


2
12
sin
L
n
kwhere
m
k
E n
n
n


2
22

The solution of (3)
Energy En is given by
wavevector
0 /L 2/L
K-space
k
/L

14. Free Electron Fermi Gas
3-D
)()(
2
2
2
rEr
m
kkk 

Consider a free electron gas in 3-D (electron of mass m is confined
to a cube of edge L), (2) can be expressed as
With boundary conditions
0)(0)0(
0)(0)0(
0)(0)0(



Lzz
Lyy
Lxx
kk
kk
kk
(4)

15. Free Electron Fermi Gas
(3-D)
LnLnLn zzyyxx  
2
1
,
2
1
,
2
1




















 zyxA
zyx
k





 2
sin
2
sin
2
sin
The solution of (4)
where
(5)

16. Free Electron Fermi Gas
(3-D, periodic boundary conditions)
• 3-D system
• periodic boundary
conditions
,.....
8
,
6
,
4
,
2
,0,,
LLLL
kkk zyx


),,(),,(
),,(),,(
),,(),,(
zyxLzyx
zyxzLyx
zyxzyLx
kk
kk
kk



The solution of (4)
 rikArk  exp)( the components of k satisfy
One allowed value of k per volume (2/L)3

17. Free Electron Fermi Gas
(3-D, periodic boundary conditions)
 222
2
2
2
22
zyxk kkk
m
k
m
E 

Energy Ek is given by
K space
(6)

18. Free Electron Fermi Gas
(density of state)
 
2/3
22
3
3
3
2
3/2
3/4
2 







mEL
L
k
N


total number of allowed energy state
Density of states – number of allowed energy state per unit energy
range
2/1
2/3
22
3
2
2
)( E
mL
dE
dN
ED 







(7)
(8)

19. Free Electron Fermi Gas
(Fermi energy)
3/2
3
22
3
2 






L
N
m
EF

The Fermi energy is defined as the energy of the topmost filled
level in the ground state of the N electron system.
The state of the N electron system at absolute zero
From (7), we get
For N electron system

20. Free Electron Fermi Gas
(Fermi-Dirac distribution)
• The Fermi-Dirac distribution gives the probability that an
energy state with energy E is occupied by an electron
1exp
1
)(




 

kT
E
Ef

The electron density is given by
 dEEfEDn )()(
Fermi level – the energy at
which the probability of
occupation is 1/2

21. Example
Show that the probability for an electron
state at the Fermi energy is equal to 0.5
for all finite temperature.

22. Solution
• When E = EF,
.
2
1
11
1
1
1
1
1
1
1
)(
0
/)(/)(








 
e
ee
Ef TKEETKEE BFFBF

23. Example
Using the Fermi-Dirac distribution, determine
the values of energy corresponding to f = 0.9
and f = 0.1 at a temperature of 300 K.

24. Solution
TKEE
TKEE
TKEE
e
e
e
Ef
BF
BF
BF
TKEE
TKEE
TKEE
BF
BF
BF
20.2
20.2
)111.0ln(/)(
111.01111.1
9.0
1
1
9.0
1
1
)(
/)(
/)(
/)(












25. Fermi-Dirac Distribution
1.0
0.0
EF
Probability of occupation
f(E)
Energy

26. Solution
TKEE
TKEE
TKEE
e
e
e
Ef
BF
BF
BF
TKEE
TKEE
TKEE
BF
BF
BF
20.2
20.2
)9ln(/)(
9110
1.0
1
1
1.0
1
1
)(
/)(
/)(
/)(












27. Ohm’s Law
 BvqF

 
 qFmvmomentum 
In an electric field  and magnetic field B, the force on an
electron of charge –q is
If B = 0,


m
q
v 
thus
collision time

28. Ohm’s Law


m
nq
qnvJ
2

m
nq 

2

The electric current density is
If we define the electrical conductivity as
J
then
Ohm’s Law

29. Hall Effect
JBRB
qn
J
qvBqF
H



0B
Hall coefficient
JBRB
qn
J
qvBqF
H




30. Quantum Statistics
A single quantum mechanical system consists of N particles
constrained to some volume V
assumption
1. weakly interacting particles
2. the particle density is low enough so that the
energy of the system can be considered as the sum of
the individual particle energies





1
ˆ
i
ii
s
ii
s
i
EnE
EH
wavefunction of particle s
number of particles with
energy Ei

31. Quantum Statistics
• How to determine the most probable distribution of
finding n1 particles out of a total of N with energy E1, n2
with energy E2, and so on?
The most probable distribution (n1, n2 …,ns,….) is the one
associated with the largest number of microscopically
distinguishable arrangements
Thus, the plan of attack would be to derive an expression for P,
the total number of microscopically distinct arrangement
corresponding to a given arbitraty sequence (n1, n2 …,ns,….) and
then find the particular sequence that maximizes P.

32. Three types of quantum particles
1. Identical but distinguishable particles
• harmonic oscillators
2. Identical indistinguishable particles of half-odd integral spin
(Fermions)
• electrons
• protons
3. Identical indistinguishable particles of integral spin (Bosons)
• photons
• phonons
The particles of concern to us fall into one of three categories
obey Pauli exclusion principle
Pauli exclusion principle not apply

33. N particle system
,.......,......,, 21 sEEE
Our model system is taken to contain N particles, each with a
spectrum of allowed energy levels
We divide the single-particle energy spectrum into energy “bins”.
Bin s, as an example, represents all the elementary quantum states
whose energies lie within some arbitrarily chosen interval Es
centered about Es. The number of quantum states in bin s is
denoted by gs.

34. Identical but distinguishable particles
  1
1
1
1
11
1
1
!!
!
nN
n
n
Cgor
nNn
Ng
P 


Determine P of N particles such that bin 1 contains n1 particles,
bin 2 n2 particles, and so on.
For bin 1, the total number of distinguishable choices is
For bin 2, the total number of distinguishable choices is
 
  21
2
2
2
212
12
2
!!
!
nnN
n
n
Cgor
nnNn
nNg
P 




35. Identical but distinguishable particles
the total number of distinguishable choices in which bin 1
contains n1 particles, bin 2 n2 particles, and so on, is
 





1
2121
!
!
........,....,....,,
s s
n
s
ss
n
g
N
PPPnnnP
s

36. Identical indistinguishable particles of half-
odd integral spin (Fermions)
  11
!!
!
111
1
1 ng Cor
ngn
g
P


Determine P of N particles such that bin 1 contains n1 particles,
bin 2 n2 particles, and so on.
For bin 1, the total number of distinguishable choices is
For bin 2, the total number of distinguishable choices is
  22
!!
!
222
2
2 ng Cor
ngn
g
P



37. Identical indistinguishable particles of half-
odd integral spin (Fermions)
the total number of distinguishable choices in which bin 1
contains n1 particles, bin 2 n2 particles, and so on, is
 
 

 


1
2121
!!
!
........,....,....,,
s sss
s
ss
nng
g
PPPnnnP

38. Identical indistinguishable particles of
integral spin (Bosons)
 
 !1!
!1
11
11
1



gn
gn
P
Determine P of N particles such that bin 1 contains n1 particles,
bin 2 n2 particles, and so on.
For bin 1, the total number of distinguishable choices is
For bin 2, the total number of distinguishable choices is
 
 !1!
!1
22
22
2



gn
gn
P

39. Identical indistinguishable particles of
integral spin (Bosons)
the total number of distinguishable choices in which bin 1
contains n1 particles, bin 2 n2 particles, and so on, is
 
 
 

 



1
2121
!!1
!1
........,....,....,,
s ss
ss
ss
ng
gn
PPPnnnP

40. Lagrange method
constEnE
constNn
s
ss
s
s








1
1
we wish to find the set of ns for which P is maximized
subjected to conditions :
We constrain our solution using Lagrange multipliers
forming the function:
  











 



 11
21 )ln(,.......,,
s
ss
s
ss nEEnNPnnnF 

41. Lagrange method
sallfor
n
F
s
0


0lnln  sss Eng 
For Identical but distinguishable particles
 s
s
s
E
g
n
 

exp
Maxwell-Boltzmann

42. Lagrange method
  0lnln  ssss Enng 
For Identical indistinguishable particles of half-odd integral
spin (Fermions)
  1exp 

s
s
s
E
g
n

Fermi-Dirac

43. Lagrange method
  0lnln  ssss Enng 
For Identical indistinguishable particles of half-odd integral
spin (Fermions)
  1exp 

s
s
s
E
g
n

Bose-Einstein

44. Energy Band
Electron in a metal can move freely in a straight path over many
atomic distances, undeflected by collisions with other conduction
electron or by collisions with the atom cores

45. Energy Band
),(ˆ),( trHtr
t
i 



For a general quantum system
(1)
For a single particle in three dimensions
),()(),(
2
),( 2
2
trrVtr
m
tr
t
i 

 
 (2)
Wave functionHamiltonian operator

46. Energy Band
nn
n
E
dx
d
m


 2
22
2

free electron gas
(3)
Energy of the electron
V(r)=0
Free Electron Model

47. Energy Band
0)( rVif
Considering a periodic potential, V(r+T)=V(r)
)()()(
2
)( 2
2
rrVr
m
rE 

(2) can be expressed as (4)

48. Energy Band
 

G
riG
GeUrV )(

G
iGx
GeUxV )(
The periodic potential V(r) may be expanded as a Fourier series
in the reciprocal lattice vectors G
or
(5)
For 1-D system
Thus, (4) can be rewritten as
)(
)(
2
)( 2
22
xeU
dx
xd
m
xE
G
iGx
G 

 

(6)

49. Energy Band







G k
xGki
kG
k
ikx
k
eCUxxV
eCk
mdx
xd
m
)(
2
2
2
22
)()(
2
)(
2


k
ikx
k eCx)(
The wavefunction Ψ(x) may be expressed as a Fourier series
summed over all values of the wavevector permitted by the
boundary conditions
then
(7)

50. Energy Band
  
k
ikx
k
G k
xGki
kG
ikx
k
k
eCEeCUeCk
m
)(2
2
2

From (6), we get
 
m
k
where
CUCE
k
G
GkGkk
2
0
22


  


Each Fourier component must have the same
coefficient on both sides of the equation
(8)
central equation

51. Energy Band
 

G
xGki
Gk
ikx
kk eCeCx )(
)(
Once we determine The Ck from central equation, the
Ψ(x) is given as
(9)
Rearrange (9), we get
)(
)()(
xueCwhere
exueeCx
k
G
iGx
Gk
ikx
k
ikx
G
iGx
Gkk














(10)

52. Energy Band
(Bloch Theorem)
)()(
)exp()()(
Truru
rikrur
kk
kk


The solutions of the Schrodinger equation for a periodic
potential must be of a special form
The eigenfunctions of the wave equation for a periodic potential
are the product of a plane wave exp(ik•r) times a function uk(r)
with the periodicity of the crystal lattice
(11)
Translation vector
Bloch theorem

53. Energy Band
(Kronig-Penney Model)





 

0
0
)(
xbDeCe
axBeAe
x QxQx
iKxiKx
a0-b
Uo
If V(x) is
The solutions for (2) are
where
m
K
EandmQEUo
2
2
22
22 
 

54. Energy Band
(Kronig-Penney Model)
)()(
00
00
DCQBAik
dx
d
dx
d
DCBA
xx
xx







The constants A,B,C,D are chosen so that Ψ and dΨ/dx are
continuous at x=0 and x=a
)(
)0()( baik
exbbaxa 

Bloch theorem
(12)
(13)

55. Energy Band
(Kronig-Penney Model)
 
    )(
)(
)(
)(
baikQbQbiKaiKa
baik
bxax
baikQbQbiKaiKa
baik
bxax
eDeCeQBeAeiK
e
dx
d
dx
d
eDeCeBeAe
e














(14)
(15)

56. Energy Band
(Kronig-Penney Model)
)(coscoscoshsinsinh
2
22
bakKaQbKaQb
QK
KQ


 oUb
and limlim
0
From (12)-(15), we can get
If the potential is a periodic delta function
kaKaKaKaP coscossin)/( 
Then, (16) can reduce to
(16)
(17)
P=Q2ba/2

57. Energy Band
(Kronig-Penney Model)
-6 -4 -2 0 2 4 6
-2
-1
0
1
2
3
4
5
6
P=3/2
(P/Ka)sin(Ka)+cos(Ka)
Ka ()
Not allowed

58. Energy Band
(Kronig-Penney Model)
Energy gap
(Ka that is not
allowed)

59. Band Structure
Actual band structures are usually exhibited as plots of energy
versus wavevector in the first Brillouin zone. This is helpful in
visualization and economical of graph paper.
2/a

60. Band Structure
/a 2/a 3/a
k
E
If the band structure of A is :
/a
k
E

61. Band Structure
k
E
Eg
k
E
Eg
k
E
Eg
Direct Indirect Overlap
(negative bandgap)
Bandgap – the difference in energy between the lowest point of the
conduction band the highest point of the valence band.
valence band
conduction band

62. Band Structure
band structure of Si

63. Energy Band
(conductor and insulator)
A band is filled from low energy to high energy

64. Energy Band
(conductor and insulator)
 A full band cannot conduct electricity. A full band is always a full
band no matter what the external field are
 An empty band cannot conduct electricity because it does not
have charge carrier.
Only a partially filled band can
conduct electricity. The
occupied states are not
“balance” under an external
field. This unbalance causes
the current flow.

65. Energy Band
(Equation of motion and effective mass)
2
2
2
11
dk
Ed
me 

The equation of motion of an electron in an energy
band is :
dtdk
Edm
dk
d
dt
d
m
dt
dv
m
dt
dv
m
dt
dk
F
e
ee
e
2











From (1), we can get
(1)
effective mass
)( BvEq
dt
dk
F  

66. Energy Band
(Equation of motion and effective mass)
)(
2
22
0 A
m
k
EE


Approximate solution near a zone boundary
(2)
From (1), we get
A
m
mthus
m
A
dk
Ed
e 
2
22
2
2
11 

(3)
An electron of mass m when put into a crystal respond to
applied fields as if the mass were me (effective mass)
Electron rest mass

67. Example
By determine the number of grid points
contained beneath the surface of radius nmax,
show that the Fermi energy (i.e. the energy of
the highest occupied state at T = 0K ) is given
by
.
3
2
3/222







V
N
m
EF


68. Each allowed quantum state can be represented by a grid point
(nx, ny, nz).Grid points with the same energy are connected by a
surface of constant radius. In order to determine the Fermi
energy, find the values of the radius nmax, just contains sufficient
states to accommodate all the valence electrons in the crystal
ny
nx
nz
nmax

69. Solution
• The volume contained beneath the surface
corresponds to one-eight of a sphere of radius
nmax, the volume is
• Each grid point corresponding to a cube of unit
volume, this expression also gives the number of
grid points beneath the surface.
• As each grid point corresponding to a state which
can accommodate two electrons for a crystal
containing N electrons we require N/2 states.
•
63
4
8
1
3
max3
max
n
n

 
3/1
max
3
max 3
62








 N
n
nN

70. Solution
• By writing
• Energy
  2/1222
max zyx nnnn 
 
3/2223/2
3
32
3/222
2
max
22
222
2222
3
2
3
2
3
22
22







































V
N
m
h
L
N
m
h
N
Lm
h
n
Lm
h
nnn
Lm
h
m
kh
E zyx







71. Thermal Conductivity in Metals
• Thermal conductivity of a Fermi gas is
where n is the electron concentration
k is the Boltzmann constant
T is the temperature
m is them mass of electron
and  is the collision time.
m
Tnk
lv
mv
Tnk
K F
F
el
33
22
2
22



72. Thermal Conductivity in Metals
• Do the electrons or phonons carry the greater
part of the heat current in a metal?
• In pure metals the electronic contribution is
dominant at all temperatures.
• In impure metals or in disordered alloys, the
electron mean free path is reduced by
collisions with impurities, and phonon
contribution may be comparable with the
electronic contribution.

73. Wiedemann-Franz law
• The Wisdemann-Franz law states that for
metals at not too low temperatures the ratio
of the thermal conductivity to the electrical
conductivity is directly proportional to
temperature, with the value of the constant of
proportionality independent of the particular
metal.
T
e
k
m
ne
m
Tnk
K
22
2
22
3
3 











74. Lorenz number L
• The Lorenz number L is defined as
• The value of
• This remarkable result involves neither n nor m.
• Experimental values of L at 0oC and 100oC are in
good agreement.
T
KL

.ohm/deg-watt1045.2
3
28
22








e
k
L


75. Experimental Lorenz numbers
L ×108 watt-ohm/deg2
Metal 0oC 100oC
Ag 2.31 2.37
Au 2.35 2.40
Cd 2.42 2.43
Cu 2.23 2.33
Mo 2.61 2.79
Pb 2.47 2.56
Pt 2.51 2.60
Su 2.52 2.49
W 3.04 3.20
Zn 2.31 2.33