1

ECE 407 Spring 2009 Farhan Rana Cornell University

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
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
Examples, discussed in detail later in the course, are shown below:
Semiconductor Quantum Wells:
GaAs
GaAs
InGaAs
quantum well
(1-10 nm)
Graphene:
Semiconductor quantum
wells can be composed of
pretty much any
semiconductor from the
groups II, III, IV, V, and VI of
the periodic table
Graphene is a single atomic layer
of carbon atoms arranged in a
honeycomb lattice TEM
micrograph
STM
micrograph
2
ECE 407 Spring 2009 Farhan Rana Cornell University
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):
GaAs
InGaAs
Nanowire
GaAs
Semiconductor Quantum
Point Contacts
(Electrostatic Gating):
GaAs
InGaAs
Quantum well
Carbon Nanotubes
(Rolled Graphene
Sheets):
metal
metal
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:
( ) ( ) ( ) ( ) r E r r V r
m

= + V
2
2
2
Consider a large metal sheet of area A= L
x
L
y
:
x
L
y
L
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
The electron states inside the sheet are given
by the Schrodinger equation
( )
( ) sheet the outside for
sheet the inside for 0
r r V
r r V


=
=
free electrons
(experience no
potential when inside
the sheet)
y x
L L A =
3
ECE 407 Spring 2009 Farhan Rana Cornell University
Born Von Karman Periodic Boundary Conditions in 2D
( ) ( ) r E r
m

= V
2
2
2
Solve:
Use periodic boundary conditions:
( ) ( )
( ) ( ) z y x z L y x
z y x z y L x
y
x
, , , ,
, , , ,


= +
= +
These imply that each
edge of the sheet is
folded and joined to
the opposite edge
Solution is: ( )
( ) y k x k i
r k i y x
e
A
e
A
r
+
= =
1 1
.

The boundary conditions dictate that the allowed values of k

x
, and k
y
are such
that:
( )
( )
y
y
L k i
x
x
L k i
L
m k e
L
n k e
y y
x x
t
t
2
1
2
1
= =
= =
n = 0, 1, 2, 3,.
m = 0, 1, 2, 3,.
x
L
y
L
y x
L L A =
x
y
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
( )
m
k
k E
2
2 2

=
( )
r k i
k
e
A
r

.
1
=
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:
V =
i
p

( ) ( ) ( ) r k r
i
r p
k k k

= V =

Normalization: The wavefunction is properly normalized: ( ) 1

2
2
=
}
r r d
k

Orthogonality: Wavefunctions of two different states are orthogonal:

( ) ( )
( )
k k
r k k i
k k
A
e
r d r r r d

, '
. '
2 *
'
2
o =
}
=
}

Velocity:
Velocity of eigenstates is: ( ) ( ) k E
m
k
k v
k

V = =
1
4
ECE 407 Spring 2009 Farhan Rana Cornell University
States in 2D k-Space
x
L
t 2
y
L
t 2
k-space Visualization:
The allowed quantum states states can be
visualized as a 2D grid of points in the entire
k-space
y
y
x
x
L
m k
L
n k
t t 2 2
= =
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:
( )
A L L
y x
2
2 2 2 t t t
=
|
|
.
|

\
|
|
|
.
|

\
|
( )
2
2t
A
There are grid points per unit area of k-space
Very important
result
n, m = 0, 1, 2, 3, .
x
k
y
k
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?
Suppose T~0K and we are interested in a filling scheme
that gives the lowest total energy.
x
k
y
k
N
The energy of a quantum state is:
( )
( )
m
k
m
k k
k E
y x
2 2
2 2
2 2 2

=
+
=
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
When we are done, all filled (i.e. occupied)
quantum states correspond to grid-points that are
inside a circular region of radius k
F
F
k
x
L
y
L
y x
L L A =
x
y
5
ECE 407 Spring 2009 Farhan Rana Cornell University
x
k
y
k
F
k
Fermi circle
Each grid-point can be occupied by two electrons (spin
up and spin down)
All filled quantum states correspond to grid-points that
are inside a circular region of radius k
F
Area of the circular region =
Number of grid-points in the circular region =
2
F
k t
( )
2
2
2
F
k
A
t
t

Number of quantum states (including
spin) in the circular region =
( )
2 2
2
2
2
2
F F
k
A
k
A
t
t
t
=
But the above must equal the total number N of electrons inside the box:
2
2
F
k
A
N
t
=
t 2
density electron
2
F
k
A
N
n = = =
( )
2
1
2 n k
F
t =
The Electron Gas in 2D at Zero Temperature - II
Units of the electron
density n are #/cm
2
ECE 407 Spring 2009 Farhan Rana Cornell University
x
k
y
k
F
k
Fermi circle
All quantum states inside the Fermi circle are filled (i.e.
occupied by electrons)
All quantum states outside the Fermi circle are empty
Fermi Momentum:
The largest momentum of the electrons is:
This is called the Fermi momentum
Fermi momentum can be found if one knows the electron
density:
F
k
( )
2
1
2 n k
F
t =
Fermi Energy:
The largest energy of the electrons is:
This is called the Fermi energy E
F
:
m
k
F
2
2 2

m
k
E
F
F
2
2 2

=
Fermi Velocity:
The largest velocity of the electrons is called the Fermi velocity v
F
:
m
k
v
F
F

=
The Electron Gas in 2D at Zero Temperature - III
m
n
E
F
t
2

=
or
F
E
m
n
2
t
=
Also:
6
ECE 407 Spring 2009 Farhan Rana Cornell University
( )
2
2t
A
Recall that there are grid points per unit area of k-
space
So in area of k-space the number of
grid points is:
y x
dk dk
( ) ( )
k d
A
dk dk
A
y x

2
2 2
2 2 t t
=
x
k
y
k
x
dk
y
dk
The summation over all grid points in k-space can be replaced by an area integral
( )
}

2
2
all 2t
k d
A
k

Therefore:
( )
( )
( ) k f
k d
A k f N
k

}
= =
2
2
all 2
2 2
t
The Electron Gas in 2D at Non-Zero Temperature - I
( ) k f

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 that the quantum state of wavevector is occupied by an
electron is given by the Fermi-Dirac distribution function:
k

( ) k f

( )
( ) ( ) T K E k E
f
e
k f

+
=

1
1
Therefore:
( )
( )
( )
( ) ( ) KT E k E
f
e
k d
A k f
k d
A N

+
}
=
}
=

1
1
2
2
2
2
2
2
2
2
t t
( )
( )
m
k
m
k k
k E
y x
2 2
2 2
2 2 2

=
+
=
Where:
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 k k d t 2
2
=

and
dk
m
k
dE
m
k
E
2 2 2
2

= =
Therefore:
( )
}

0
2
0
2
2
2
2 dE
m
A
dk k
A
k d
A

t
t
t
7
ECE 407 Spring 2009 Farhan Rana Cornell University
The Electron Gas in 2D at Non-Zero Temperature - III
( )
( ) ( )
( )
( ) KT E E
D
KT E k E f
f
e
E g dE A
e
k d
A N

+
}
=
+
}
=
1
1
1
1
2
2
2
0
2
2

t
Where: ( )
2
2
t
m
E g
D
=
Density of states function is constant
(independent of energy) in 2D
g
2D
(E) has units: # / Joule-cm
2
The product g(E) dE represents the number of
quantum states available in the energy interval
between E and (E+dE) per cm
2
of the metal
x
k
y
k
Suppose E corresponds to the inner circle
from the relation:
m
k
E
2
2 2

=
And suppose (E+dE) corresponds to the outer
circle, then g
2D
(E) dE corresponds to twice the
number of the grid points between the two
circles
ECE 407 Spring 2009 Farhan Rana Cornell University
( )
f
E E f
E
f
E
( ) E g
D 2
The expression for N can be visualized as the
integration over the product of the two functions:
The Electron Gas in 2D at Non-Zero Temperature - IV
( )
( )
( ) ( )
f D
KT E E
D
E E f E g dE A
e
E g dE A N
f

}
=
+
}
=

2
0
2
0
1
1
Where: ( )
2
2
t
m
E g
D
=
Check: Suppose T=0K:
E
0
1
f
E
T = 0K
( ) ( ) ( )
f
f
D
E
f D
E
m
n
E
m
A
E g dE A E E f E g dE A N
f
2
2
2
0
2
0

t
t
=
=
}
=
}
=

Compare with the previous result at T=0K:
F
E
m
n
2
t
=

At T=0K (and only at T=0K) the Fermi level
E
f
is the same as the Fermi energy E
F
( )
f
E E f
8
ECE 407 Spring 2009 Farhan Rana Cornell University
The Electron Gas in 2D at Non-Zero Temperature - V
( )
( )
(
(

+ =
+
}
=

T K
E
KT E E
D
f
f
e T K
m
e
E g dE n 1 log
1
1
2
2
0
t
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
In general, the Fermi level E
f
is a function of temperature and decreases from E
F
as
the temperature increases. The exact relationship can be found by inverting the
above equation and recalling that:
( )
(
(

= 1 log
KT
E
f
F
e KT T E
F
E
m
n
2
t
=
to get:
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:
( ) ( )
( )
( ) ( ) k E k f
k d
A k E k f U
k

}
= =
2
2
all 2
2 2
t
Convert the k-space integral to energy integral: ( ) ( ) E E E f E g dE A U
f D

}
=

2
0
The energy density u is: ( ) ( ) E E E f E g dE
A
U
u
f D

}
= =

2
0
Suppose T=0K:
( )
2
2
2
0
2
F D
E
E
m
E E g dE u
F
t
=
}
=
F
E
m
n
2
t
= Since:
We have:
F
E n u
2
1
=
9
ECE 407 Spring 2009 Farhan Rana Cornell University
2D Electron Gas in an Applied Electric Field - I
x
k
y
k
Electron distribution in k-space
when E-field is zero
x
k
y
k
Electron distribution is shifted in
k-space when E-field is not zero
x E E
x
=

( ) E
e
k t k

t
= =
Distribution function: ( ) k f

Distribution function:
|
.
|

\
|
+ E
e
k f

t
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
( ) k f

x
L
y
L
E

ECE 407 Spring 2009 Farhan Rana Cornell University

2D Electron Gas in an Applied Electric Field - II
x
k
y
k
Electron distribution is shifted in
k-space when E-field is not zero
E

Distribution function:
|
.
|

\
|
+ E
e
k f

t
E
e

Current density (units: A/cm)

( )
( ) k v E
e
k f
k d
e J

|
.
|

\
|
+
}
=
t
t
2
2
2
2
Do a shift in the integration variable:
( )
( )
( )
( )
( )
( )
E E
m
e n
J
E k f
k d
m
e
J
m
E
e
k
k f
k d
e J
E
e
k v k f
k d
e J

o
t
t
t
t
t
t
t
= =
(

}
=
|
.
|

\
|

}
=
|
.
|

\
|

}
=
2
2
2 2
2
2
2
2
2
2
2
2
2
2
Where:
m
e n t
o
2
=
Same as the Drude result - but
units are different. Units of o are
Siemens in 2D
electron density = n (units: #/cm
2
)
10
ECE 407 Spring 2009 Farhan Rana Cornell University
Electrons in 1D Metals: The Free Electron Model
The quantum state of an electron is described by the time-independent
Schrodinger equation:
( ) ( ) ( ) ( ) x E x x V x
x
m
= +
c
c

2
2 2
2

Consider a large metal wire of length L :

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
The electron states inside the wire are given by
the Schrodinger equation
( )
( ) wire the outside for
wire the inside for 0
x x V
x x V
=
=
free electrons
(experience no
potential when inside
the wire)
ECE 407 Spring 2009 Farhan Rana Cornell University
Born Von Karman Periodic Boundary Conditions in 1D
Solve:
Use periodic boundary conditions:
( ) ( ) z y x z y L x , , , , = +
These imply that each
facet of the sheet is
folded and joined to
the opposite facet
Solution is: ( )
( ) x k i
x
e
L
x
1
=
The boundary conditions dictate that the allowed values of k
x
are such that:
( )
L
n k e
x
L k i
x
t 2
1 = = n = 0, 1, 2, 3,.
( ) ( ) x E x
x
m
=
c
c

2
2 2
2

L
11
ECE 407 Spring 2009 Farhan Rana Cornell University
States in 1D k-Space
L
t 2
k-space Visualization:
The allowed quantum states states can be
visualized as a 1D grid of points in the entire
k-space
L
n k
x
t 2
=
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:
|
.
|

\
|
L
t 2
t 2
L
There are grid points per unit length of k-space
Very important
result
n = 0, 1, 2, 3, .
x
k
0
ECE 407 Spring 2009 Farhan Rana Cornell University
x
k
F
k
Fermi points
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 k
F
from the origin
Length of the region =
Number of grid-points in the region =
F
k 2
F
k
L
2
2

t
Number of quantum states (including
spin) in the region =
But the above must equal the total number N of electrons in the wire:
t
F
k
L N
2
=
t
F
k
L
N
n
2
density electron = = =
2
n
k
F
t
=
The Electron Gas in 1D at Zero Temperature - I
Units of the electron
density n are #/cm
0
F
k
F
k
L
2
2
2
t
12
ECE 407 Spring 2009 Farhan Rana Cornell University
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:
This is called the Fermi momentum
Fermi momentum can be found if one knows the electron
density:
F
k
Fermi Energy:
The largest energy of the electrons is:
This is called the Fermi energy E
F
:
m
k
F
2
2 2

m
k
E
F
F
2
2 2

=
Fermi Velocity:
The largest velocity of the electrons is called the Fermi velocity v
F
:
m
k
v
F
F

=
The Electron Gas in 1D at Zero Temperature - II
m
n
E
F
8
2 2 2
t
=
or
F
E
m
n
t
8
=
Also:
x
k
Fermi points
0
2
n
k
F
t
=
ECE 407 Spring 2009 Farhan Rana Cornell University
t 2
L
Recall that there are grid points per unit length of k-
space
So in length of k-space the number of
grid points is:
x
dk
x
dk
L
t 2
x
k
x
dk
The summation over all grid points in k-space can be replaced by an integral
}



t 2
all
x
k
dk
L

Therefore:
( ) ( )
x
x
k
x
k f
dk
L k f N
}
= =


t 2
2 2
all

The Electron Gas in 1D at Non-Zero Temperature - I
( )
x
k f is the occupation probability of a quantum state
0
13
ECE 407 Spring 2009 Farhan Rana Cornell University
The Electron Gas in 1D at Non-Zero Temperature - II
The probability that the quantum state of wavevector is occupied by an
electron is given by the Fermi-Dirac distribution function:
x
k ( )
x
k f
( )
( ) ( ) T K E k E
x
f x
e
k f

+
=
1
1
Therefore:
( )
( ) ( ) KT E k E
x
x
x
f x
e
dk
L k f
dk
L N

+
}
=
}
=
1
1
2
2
2
2
t t
( )
m
k
k E
x
2
2 2

= Where:
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:
and
dk
m
k
dE
m
k
E
2 2 2
2

= =
Therefore:
}

}

0
2
2 2
2
2
t t
dk
L
dk
L
x
E
m
dE L
dk
L
x
1 2
2
2
0
}

t t
ECE 407 Spring 2009 Farhan Rana Cornell University
0
The Electron Gas in 1D at Non-Zero Temperature - III
( ) ( )
( )
( ) KT E E
D
KT E k E
x
f f x
e
E g dE L
e
dk
L N

+
}
=
+
}
=
1
1
1
1
2
2
1
0
t
Where:
Density of states function in 1D
g
1D
(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
x
k
Suppose E corresponds to the inner points
from the relation:
m
k
E
2
2 2

=
And suppose (E+dE) corresponds to the outer
points, then g
1D
(E) dE corresponds to twice the
number of the grid points between the points
(adding contributions from both sides)
( )
E
m
E g
D
1 2
1
t
=
14
ECE 407 Spring 2009 Farhan Rana Cornell University
( )
f
E E f
E
f
E
( ) E g
D 1
The expression for N can be visualized as the
integration over the product of the two functions:
The Electron Gas in 1D at Non-Zero Temperature - IV
Where:
Check: Suppose T=0K:
E
0
1
f
E
T = 0K
( ) ( ) ( )
f
f
D
E
f D
E
m
n
E
m
L
E g dE L E E f E g dE L N
f

t
t
8
8

1
0
1
0
=
=
}
=
}
=

Compare with the previous result at T=0K:

At T=0K (and only at T=0K) the Fermi level

E
f
is the same as the Fermi energy E
F
( )
( )
( ) ( )
f D
KT E E
D
E E f E g dE L
e
E g dE L N
f

}
=
+
}
=

1
0
1
0
1
1
( )
E
m
E g
D
1 2
1
t
=
F
E
m
n
t
8
=
( )
f
E E f
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
In general, the Fermi level E
f
is a function of temperature and decreases from E
F
as
the temperature increases.
( )
( ) KT E E
D
f
e
E g dE n

+
}
=
1
1
1
0
15
ECE 407 Spring 2009 Farhan Rana Cornell University
Total Energy of the 1D Electron Gas
The total energy U of the electron gas can be written as:
( ) ( ) ( ) ( )
x x
x
x
k
x
k E k f
dk
L k E k f U
}
= =


t 2
2 2
all

Convert the k-space integral to energy integral: ( ) ( ) E E E f E g dE L U
f D

}
=

1
0
The energy density u is: ( ) ( ) E E E f E g dE
L
U
u
f D

}
= =

1
0
Suppose T=0K:
( )
3
8
2 3
1
0
F
D
E
E m
E E g dE u
F
t
=
}
=
Since:
We have:
F
E n u
3
1
=
F
E
m
n
t
8
=
ECE 407 Spring 2009 Farhan Rana Cornell University
1D Electron Gas in an Applied Electric Field - I
x
k
Electron distribution in k-space
when E-field is zero
x
k
Electron distribution is shifted in
k-space when E-field is not zero
x E E
x
=

( )
x x x
E
e
k t k

t
= =
Distribution function:
( )
x
k f
Distribution function:
|
.
|

\
|
+
x x
E
e
k f

t
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
( )
x
k f
E

L
16
ECE 407 Spring 2009 Farhan Rana Cornell University
1D Electron Gas in an Applied Electric Field - II
Electron distribution is shifted in
k-space when E-field is not zero
Distribution function:
|
.
|

\
|
+
x x
E
e
k f

t
Current (units: A)
( )
x x x
x
k v E
e
k f
dk
e I |
.
|

\
|
+
}
=



t
t 2
2
Do a shift in the integration variable:
( )
( )
( )
E E
m
e n
I
E k f
dk
m
e
I
m
E
e
k
k f
dk
e I
E
e
k v k f
dk
e I
x x
x
x x
x
x
x x x
x

o
t
t
t
t
t
t
t
= =
(

}
=
|
.
|

\
|

}
=
|
.
|

\
|

}
=

2
2
2
2
2
2
2
2
Where:
m
e n t
o
2
=
Same as the Drude result - but
units are different. Units of o are
Siemens-cm in 1D
electron density = n (units: #/cm)
x
k
x E E
x
=

x
E
e

ECE 407 Spring 2009 Farhan Rana Cornell University