Handout 3
Free Electron Gas in 2D and 1D
In this lecture you will learn:
• Free electron gas in two dimensions and in one dimension
• Density of States in k-space and in energy in lower dimensions
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electron Gases in 2D
• In several physical systems electron are confined to move in just 2
dimensions
STM
micrograph
• Examples, discussed in detail later in the course, are shown below:
Semiconductor Quantum Wells:
GaAs
Graphene:
InGaAs
quantum well
(1-10 nm)
GaAs
Semiconductor quantum
wells can be composed of
pretty much any
semiconductor from the
groups II, III, IV, V, and VI of
the periodic table
TEM
micrograph
Graphene is a single atomic layer
of carbon atoms arranged in a
honeycomb lattice
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
1
Electron Gases in 1D
• In several physical systems electron are confined to move in just 1 dimension
• Examples, discussed in detail later in the course, are shown below:
Semiconductor Quantum
Wires (or Nanowires):
Semiconductor Quantum
Point Contacts
(Electrostatic Gating):
GaAs
metal
Carbon Nanotubes
(Rolled Graphene
Sheets):
metal
InGaAs
Quantum well
InGaAs
Nanowire
GaAs
GaAs
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Electrons in 2D Metals: The Free Electron Model
The quantum state of an electron is described by the time-independent
Schrodinger equation:




2 2 
  r   V r  r   E  r 
2m
Consider a large metal sheet of area A= Lx Ly :
Use the Sommerfeld model:
• The electrons inside the sheet are confined in a
two-dimensional infinite potential well with zero
potential inside the sheet and infinite potential
outside the sheet

V r   0

V r   
A  Lx Ly
Ly
Lx

for r inside the sheet

for r outside the sheet
• The electron states inside the sheet are given
by the Schrodinger equation
free electrons
(experience no
potential when inside
the sheet)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
2
Born Von Karman Periodic Boundary Conditions in 2D
Solve:


2 2 
  r   E  r 
2m
Use periodic boundary conditions:

 r  
Solution is:
y
These imply that each
edge of the sheet is
folded and joined to
the opposite edge
  x  Lx , y , z     x , y , z 
 x , y  Ly , z     x , y , z 
1 i k . r
e

A
A  Lx Ly
x
Ly
Lx
1 i k x x  k y y 
e
A
The boundary conditions dictate that the allowed values of kx , and ky are such
that:
e i k x Lx   1
e

i k y Ly
1

kx  n
2
Lx
n = 0, ±1, ±2, ±3,…….

ky  m
2
Ly
m = 0, ±1, ±2, ±3,…….
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Born Von Karman Periodic Boundary Conditions in 2D
Labeling Scheme:
All electron states and energies can be labeled by the corresponding k-vector

 k r  
1 i k . r
e
A

 2k 2
Ek 
2m

Normalization: The wavefunction is properly normalized:
Orthogonality:
 2
2
 d r  k r   1
Wavefunctions of two different states are orthogonal:
2
d r


 i k  k '  . r
r   k r    d 2r e
  k ' , k
A
 
 k* '

Momentum Eigenstates:
Another advantage of using the plane-wave energy eigenstates (as opposed to the
“sine” energy eigenstates) is that the plane-wave states are also momentum
eigenstates

̂ 





Momentum operator: p  
 pˆ   r     r   k   r 
i
Velocity:
Velocity of eigenstates is:
k
i
k
k


  k 1 
vk 
 k E k
m 


ECE 407 – Spring 2009 – Farhan Rana – Cornell University
3
States in 2D k-Space
2
Lx
ky
k-space Visualization:
The allowed quantum states states can be
visualized as a 2D grid of points in the entire
“k-space”
kx  n
2
Lx
ky  m
2
Ly
kx
2
Ly
n, m = 0, ±1, ±2, ±3, …….
Density of Grid Points in k-space:
Looking at the figure, in k-space there is only one grid point in every small
area of size:
 2

 Lx
 There are
A
2 2
 2

 Ly
 2 2


A

Very important
result
grid points per unit area of k-space
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 2D at Zero Temperature - I
• Suppose we have N electrons in the sheet.
• Then how do we start filling the allowed quantum states?
N
y
• Suppose T~0K and we are interested in a filling scheme
that gives the lowest total energy.
The energy of a quantum state is:



 2 k x2  k y2
2k 2

Ek 
2m
2m

A  Lx Ly
x
Ly
Lx
ky
Strategy:
• Each grid-point can be occupied by two electrons
(spin up and spin down)
• Start filling up the grid-points (with two electrons
each) in circular regions of increasing radii until
you have a total of N electrons
kx
kF
• When we are done, all filled (i.e. occupied)
quantum states correspond to grid-points that are
inside a circular region of radius kF
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
4
The Electron Gas in 2D at Zero Temperature - II
ky
• Each grid-point can be occupied by two electrons (spin
up and spin down)
kF
• All filled quantum states correspond to grid-points that
are inside a circular region of radius kF
Area of the circular region =
kx
 kF2
A
Number of grid-points in the circular region =
Number of quantum states (including
spin) in the circular region =
2
2 
2
A
2 
2
Fermi circle
  kF2
  kF2 
A 2
kF
2
But the above must equal the total number N of electrons inside the box:
N
A 2
kF
2

n  electron density 

kF  2 n  2
1
N kF2

A 2
Units of the electron
density n are #/cm2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 2D at Zero Temperature - III
ky
• All quantum states inside the Fermi circle are filled (i.e.
occupied by electrons)
• All quantum states outside the Fermi circle are empty
kF
Fermi Momentum:
The largest momentum of the electrons is: kF
This is called the Fermi momentum
Fermi momentum can be found if one knows the electron
1
density:
kx
Fermi circle
kF  2 n  2
Fermi Energy:
 2kF2
The largest energy of the electrons is:
This is called the Fermi energy EF :
Also:
EF 
2 n
m
2m 2 2
 kF
EF 
2m
or
n
m
 2
EF
Fermi Velocity:
kF
The largest velocity of the electrons is called the Fermi velocity vF : v F 
m
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
5
The Electron Gas in 2D at Non-Zero Temperature - I
ky
A
Recall that there are
space
2 2
 So in area dk x dk y
grid points is:
A
2 
2
dk x
dk y
grid points per unit area of k-
kx
of k-space the number of
dk x dk y 
A
2 
2

d 2k
 The summation over all grid points in k-space can be replaced by an area integral
 
A
all k
Therefore:

d 2k
2 2



d 2k
N  2   f k  2  A 
f k
2
2 
all k




f k is the occupation probability of a quantum state
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 2D at Non-Zero Temperature - II



The probability f k that the quantum state of wavevector k is occupied by an
electron is given by the Fermi-Dirac distribution function:

f k 

1

E k  Ef K T
1 e   
Therefore:
N  2 A 

d 2k
2 2

Where:



 2 k x2  k y2
 2k 2
Ek 

2m
2m



d 2k
1

f k  2 A 
2


E
k
2  1  e  Ef  KT

Density of States:
The k-space integral is cumbersome. We need to convert into a simpler form – an
energy space integral – using the following steps:

d 2k  2 k dk
and
E
 2k 2
 2k
dk
 dE 
m
2m
Therefore:
2 A 

d 2k
2 
2

k
dk
0

A


m
0
2
A
dE
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
6
The Electron Gas in 2D at Non-Zero Temperature - III
N  2 A 
Where:

d 2k

1
2  1  e E k  Ef  KT

2
g2D E  
 A  dE g2D E 
0
m
1

E  E f  KT
1 e
Density of states function is constant
(independent of energy) in 2D
 2
g2D(E) has units: # / Joule-cm2
ky
The product g(E) dE represents the number of
quantum states available in the energy interval
between E and (E+dE) per cm2 of the metal
Suppose E corresponds to the inner circle
from the relation:
kx
 2k 2
2m
E
And suppose (E+dE) corresponds to the outer
circle, then g2D(E) dE corresponds to twice the
number of the grid points between the two
circles
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 2D at Non-Zero Temperature - IV

N  A  dE g2D E 
0

1
 A  dE g2D E  f E  Ef 

E  E f  KT
1 e
0
Where: g2D E  
m
 2
f E  Ef 
The expression for N can be visualized as the
integration over the product of the two functions:
Check: Suppose T=0K:
Ef
Ef

f E  Ef 
0
A
T = 0K
Ef
E

m
 2
EF

0
m
 2
n
Compare with the previous result at T=0K:
n
E
N  A  dE g2D E  f E  Ef   A  dE g2D E 
1
0
g2D E 
Ef
m
 2
Ef
At T=0K (and only at T=0K) the Fermi level
Ef is the same as the Fermi energy EF
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
7
The Electron Gas in 2D at Non-Zero Temperature - V
For T ≠ 0K:
Since the carrier density is known, and does not change with temperature, the
Fermi level at temperature T is found from the expression

n   dE g2D E 
0
1
1  e E  E f  KT

Ef 

1  e K T 
K
T
log


 2


m
In general, the Fermi level Ef is a function of temperature and decreases from EF as
the temperature increases. The exact relationship can be found by inverting the
above equation and recalling that:
n
m
 2
EF
to get:

 EF
Ef T   KT loge KT  1




ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Total Energy of the 2D Electron Gas
The total energy U of the electron gas can be written as:





d 2k
U  2   f k E k  2  A 
f k Ek
2


2

all k
 
 

Convert the k-space integral to energy integral:U  A  dE g2D E  f E  Ef  E
0
The energy density u is:
u
U 
  dE g2D E  f E  Ef  E
A 0
Suppose T=0K:
EF
m
0
2  2
u   dE g2D E  E 
Since:
n
We have: u 
m
 2
EF2
EF
1
n EF
2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
8
2D Electron Gas in an Applied Electric Field - I
ky


f k


 e 
k t     k 
E

e 
E

ky

E  E x xˆ
kx
kx
Electron distribution in k-space
when E-field is zero
Electron distribution is shifted in
k-space when E-field is not zero



Distribution function: f  k 
Distribution function: f k

e 
E


Since the wavevector of each electron is shifted by the same amount in the
presence of the E-field, the net effect in k-space is that the entire electron
distribution is shifted as shown

E
Ly
Lx
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
2D Electron Gas in an Applied Electric Field - II

Current density (units: A/cm)


d 2k   e     
J  2 e  
f k 
Ev k
 
2 2 

e 
E

ky

E
Do a shift in the integration variable:

d k
2
   e 
f k v k 


2 
  e


  k  
d 2k
J  2 e  
f k
m
2 2

 e 2 
 
d 2k
J
f k E
2  
2
m 
2 

2
 ne  

J
E  E
m

J  2 e  
2


kx

E


E

Electron distribution is shifted in
k-space when E-field is not zero

Distribution function: f  k 

e 
E



Where:

n e 2
m
electron density = n (units: #/cm2)
Same as the Drude result - but
units are different. Units of  are
Siemens in 2D
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
9
Electrons in 1D Metals: The Free Electron Model
The quantum state of an electron is described by the time-independent
Schrodinger equation:

2  2
  x   V  x   x   E   x 
2 m x 2
L
Consider a large metal wire of length L :
Use the Sommerfeld model:
• The electrons inside the wire are confined in a
one-dimensional infinite potential well with zero
potential inside the wire and infinite potential
outside the wire
V x   0
V x   
for x inside the wire
for x outside the wire
free electrons
(experience no
potential when inside
the wire)
• The electron states inside the wire are given by
the Schrodinger equation
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Born Von Karman Periodic Boundary Conditions in 1D

Solve:
2  2
 x   E  x 
2 m x 2
Use periodic boundary conditions:
  x  L, y , z     x , y , z 
Solution is:
 x  
L
These imply that each
facet of the sheet is
folded and joined to
the opposite facet
1 i k x x 
e
L
The boundary conditions dictate that the allowed values of kx are such that:
e i k x L   1

kx  n
2
L
n = 0, ±1, ±2, ±3,…….
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
10
States in 1D k-Space
k-space Visualization:
The allowed quantum states states can be
visualized as a 1D grid of points in the entire
“k-space”
kx  n
2
L
2
L
kx
0
n = 0, ±1, ±2, ±3, …….
Density of Grid Points in k-space:
Looking at the figure, in k-space there is only one grid point in every small
length of size:
 2 


 L 
 There are
L
2
Very important
result
grid points per unit length of k-space
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 1D at Zero Temperature - I
• Each grid-point can be occupied by two electrons (spin
up and spin down)
• All filled quantum states correspond to grid-points that
are within a distance kF from the origin
Length of the region = 2kF
Number of grid-points in the region =
kF
kF
kx
0
Fermi points
L
 2k F
2
Number of quantum states (including
L
2
 2k F
spin) in the region =
2
But the above must equal the total number N of electrons in the wire:
NL

2k F

n  electron density 

kF 
 n
N 2k F

L

Units of the electron
density n are #/cm
2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
11
The Electron Gas in 1D at Zero Temperature - II
• All quantum states between the Fermi points are filled (i.e.
occupied by electrons)
• All quantum states outside the Fermi points are empty
Fermi Momentum:
The largest momentum of the electrons is: kF
This is called the Fermi momentum
Fermi momentum can be found if one knows the electron
density:
kF 
kx
0
Fermi points
 n
2
Fermi Energy:
 2kF2
The largest energy of the electrons is:
2m 2 2
 kF
EF 
2m
This is called the Fermi energy EF :
EF 
Also:
 2 2 n2
8m
or
n
8m
 
EF
Fermi Velocity:
kF
The largest velocity of the electrons is called the Fermi velocity vF : v F 
m
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 1D at Non-Zero Temperature - I
Recall that there are L
2
space
grid points per unit length of k-
dk x
 So in length dk x of k-space the number of
grid points is:
0
kx
L
dk x
2
 The summation over all grid points in k-space can be replaced by an integral
 
all k

dk x
  2
L 
Therefore:

dk x
f k x 
  2
N  2    f k x   2  L 
all k
f k x  is the occupation probability of a quantum state
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
12
The Electron Gas in 1D at Non-Zero Temperature - II
The probability f k x  that the quantum state of wavevector k x is occupied by an
electron is given by the Fermi-Dirac distribution function:
f k x  
1
1  e E k x  Ef  K T
Therefore:

2k x2
Ek 
2m

Where:

 dk
dk x
1
x
f k x   2  L 

E k x  E f  KT


2
2
1 e


N  2L 
Density of States:
The k-space integral is cumbersome. We need to convert into a simpler form – an
energy space integral – using the following steps:

 dk
dk x
 2L2 
  2
0 2
and
2L 
E
 2k 2
 2k
 dE 
dk
2m
m
Therefore:

dk x
  2
2L 

 L  dE
0
2m 1
E
 
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 1D at Non-Zero Temperature - III

dk x
1
1
 L  dE g1D E 




E
k

E
KT
x
f
1  e E  E f  KT
  2 1  e
0

N  2L 
Where:
g1D E  
2m 1
E
 
Density of states function in 1D
g1D(E) has units: # / Joule-cm
The product g(E) dE represents the number of
quantum states available in the energy interval
between E and (E+dE) per cm of the metal
Suppose E corresponds to the inner points
from the relation:
2 2
E
 k
2m
0
kx
And suppose (E+dE) corresponds to the outer
points, then g1D(E) dE corresponds to twice the
number of the grid points between the points
(adding contributions from both sides)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
13
The Electron Gas in 1D at Non-Zero Temperature - IV

N  L  dE g1D E 
0

1
 L  dE g1D E  f E  Ef 

E  E f  KT
1 e
0
g1D E 
2m 1
Where: g1D E  
  E
f E  Ef 
The expression for N can be visualized as the
integration over the product of the two functions:
Check: Suppose T=0K:
E

Ef
0
0
N  L  dE g1D E  f E  Ef   L  dE g1D E 
f E  Ef 
1
8m
L
Ef
 
T = 0K
0
Ef
Ef
E

n
8m
 
Ef
Compare with the previous result at T=0K:
n
8m
 
EF

At T=0K (and only at T=0K) the Fermi level
Ef is the same as the Fermi energy EF
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Electron Gas in 1D at Non-Zero Temperature - V
For T ≠ 0K:
Since the carrier density is known, and does not change with temperature, the
Fermi level at temperature T is found from the expression

n   dE g1D E 
0
1

E  E f  KT
1 e
In general, the Fermi level Ef is a function of temperature and decreases from EF as
the temperature increases.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
14
Total Energy of the 1D Electron Gas
The total energy U of the electron gas can be written as:

dk x
f k x  E k x 
  2
U  2    f k x  E k x   2  L 
all k

Convert the k-space integral to energy integral: U  L  dE g1D E  f E  Ef  E
0
U 
  dE g1D E  f E  Ef  E
L 0
The energy density u is:u 
Suppose T=0K:
EF
u   dE g1D E  E 
0
8m
Since: n 
We have:
 
u
32
8m E F
  3
EF
1
n EF
3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
1D Electron Gas in an Applied Electric Field - I
f k x 

k x t     k x 
e
Ex

e 
E


E  E x xˆ
kx
kx
Electron distribution in k-space
when E-field is zero
Distribution function: f k x 
Electron distribution is shifted in
k-space when E-field is not zero
e

Distribution function: f  k x 
Ex 



Since the wavevector of each electron is shifted by the same amount in the
presence of the E-field, the net effect in k-space is that the entire electron
distribution is shifted as shown

E
L
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
15
1D Electron Gas in an Applied Electric Field - II
Current (units: A)

dk x 
e

I  2 e  
f  kx 
E x  v k x 



  2

e
Ex


E  E x xˆ
Do a shift in the integration variable:
kx

dk x
e


f k x  v  k x 
Ex 



  2
e


 k x 
Ex 
 dk


x f k  
I  2 e  
x
m
2


I  2 e  
 dk
e 2 
x f k  E
2 
x  x

m 
2



2 

ne 
I
E  E
m
Electron distribution is shifted in
k-space when E-field is not zero
e

Distribution function: f  k x 
Ex 



I
Where:

n e 2
m
electron density = n (units: #/cm)
Same as the Drude result - but
units are different. Units of  are
Siemens-cm in 1D
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
16