Transcript FREE ELECTRON THEORY - West Virginia University

Chapter 1: Applied Drude model to understand current
density, the Hall coefficient, AC conductivity, heat
capacity and thermal conductivity in simple metals.
Required knowledge/determination of carrier density.
Chapter 2: Derive and show how Fermi-Dirac statistics
can be added to improve the Drude model.
Today’s Objectives
Energy level example
Derive Fermi Dirac distribution
Define/apply density of states
Review heat capacity with improved model
Dropping HW 2.4 to keep on schedule, Others discussed today
34 Electrons in 3D Infinite Well


(n1, n1, n1)
The lowest energy for this system is 3E0, which corresponds
to n1 = n2 = n3 = 1


Next energy level (6E0), for which one of n’s is 2 [211]


What are the combinations of n’s for this energy level?
Next energy level (11E0) [311]


A total of 6 (six) electrons can have this energy
Next energy level (9E0) [221]


Thus only 2 (two) electrons can have this energy: one with spin
 and one with spin 
Accommodates 6 electrons
Next energy level (12E0) is two-fold degenerate [222]
34 Electrons in 3D Infinite Well
So far we have placed 22 electrons, so we need to
add another 12 electrons
(3,2,2)
 What is the next energy level?
(3,2,1)
 The next energy level is 14E0
(2,2,2)
 What are n1, n2, and n3? 1, 2, 3
(3,1,1)

18
17
16



Energy (in units of E0)
15
14
13
12
11
10
What is the degeneracy?
9
This energy can be had by 12 electrons87
We have placed all 34 electrons!!
6
(2,2,1)
(2,1,1)
5
4


En1n2n3  E0 n1  n2  n3 , with E0 
2
2
2
 2 2
2mL2
3
2
1
0
(1,1,1)
34 Electrons in 3D Infinite Well
In this configuration,




What is the probability at T =0
that a level with energy 14E0 or
less will be occupied?
It is 1!
What is the probability that a
level with energy above 14E0
will be occupied at T=0?
It is 0!
18
17
(3,2,2)
16
15
Energy (in units of E0)

14
(3,2,1)
13
12
(2,2,2)
11
(3,1,1)
10
9
(2,2,1)
8
7
6
(2,1,1)
5
4
3
2
1
0
(1,1,1)
Fermi Energy
18
17
(3,2,2)
16
15
Energy (in units of E0)
In a metal:
 The highest filled energy is
called the Fermi Energy
 It is often denoted as EF
 In our case: EF = 14E0
 An electron with E = 14E0 is
said to be at the Fermi level
14
(3,2,1)
13
12
(2,2,2)
11
(3,1,1)
10
9
(2,2,1)
8
7
6
(2,1,1)
5
4
3
2
1
0
(1,1,1)
The Fermi Temperature

Only at temperatures above EF=kBTF will the free
electron gas behave like a classical gas.

Fermi momentum
PF  kF

And Fermi velocity
PF  meVF

These are the momentum and the velocity values
of the electrons at the states on the Fermi
surface of the Fermi sphere.
The Fermi Sphere plays important role on the
behavior of metals.

Example: monovalent potassium metal
 3 N 
EF 


2m  V 
2
2
 2.12eV
1/ 3
 3 N 
kF  

V


2
2/3
 0.746 A1
PF
VF 
 0.86 106 ms 1
me
EF
TF 
 2.46 104 K
kB
Thermal Properties
At T=0, all levels
up to EF are filled.
17
(3,2,2)
16
15
Energy (in units of E0)
What happens
when T is greater
than zero?
18
14
(3,2,1)
13
12
(2,2,2)
11
(3,1,1)
10
9
(2,2,1)
8
7
6
(2,1,1)
5
4
3
(1,1,1)
2
1
0
T=0
Probability with Temperature
At finite temperature T the probability that
state N is occupied depends on the
Boltzmann distribution:
 E / k BT
PN  e
 E / kBT
e
PN ( E ) 
e

To calculate properties, we have to
take the average over all states
Helmholtz free energy F
F  U  TS
e
N
N
 EN / k BT
U=internal energy, S=entropy
 EN / k BT
Partition
function Z
e
 FN / k BT
Combining
PN ( E ) 
e
 E / k BT
e
N
e
 EN / k BT
 EN / k BT
e
 FN / k BT
N
 E / k BT
e
 ( E  FN ) / k BT
PN ( E )   FN / k BT  e
e
fi is probably of an electron
N
N
being in a particular one f
i  PN ( E )
electron level i
fi  1 PN (E )
N
N
Or one minus all
unoccupied states
What if we add one more electron
fi  1 PN (E )
N
N
18
17
16
Or more generically…
N 1
fi  1  PN (E
N
 i )
(3,2,2)
15
Energy (in units of E0)
The energy difference
between this 34
electron and a 35
electron system is the
energy in state (3,2,2)
i
14
(3,2,1)
13
12
(2,2,2)
11
(3,1,1)
10
9
(2,2,1)
8
7
6
(2,1,1)
5
4
3
2
1
0
(1,1,1)
Some algebra manipulation
N 1
PN (E
  i )  1  fi
N 1
PN ( E
N 1
PN ( E
N 1
PN ( E
  i )  e
 i )  e
 i )  e
N 1
PN ( E
  FN 1  FN
PN (E)  e
( E  FN ) / kBT
 (( E N 1  i )  FN ) / k BT
 i / k BT
( i  FN 1 ) / k BT
 i )  e
Chemical potential
N
e
e
 ( E N 1  FN ) / k BT
 ( E N 1  FN  FN 1 ) / k BT
( i  FN 1  FN ) / k BT
e
 ( E N 1  FN 1 ) / k BT
PN (EN 1   i )  e( i FN1  FN ) / kBT PN 1 (EN 1 )
Solving for fi
N 1
PN (E
  i )  1  fi
( i  ) / kBT
fi  1  e
N
fi  1  e
N
fi  1  e
N
fi (1  e
N
( i  ) / k BT
N 1
PN 1 (E )
( i  ) / k BT
The probably should not change
that much from one electron
fi
N 1
fi  fi
N
( i   ) / k BT
) 1
N
fi
N 1
N
fi 
1
N
1 e
(  i   ) / k BT
Temperature affect N
1
fi 
(   ) / k T
on probability
1 e
i
B
At the chemical potential, there is a 50%
chance of finding an electron
f ( )  1 / 2
As T0, EF
Fermi-Dirac distribution function at
low temperatures

When a metal is heated, electrons are transferred
from below EF to above EF. The rest of the electrons
deep inside the Fermi level are not effected.
Reminder: Summing over k space
Since the volume of k space is V/83,
summing any smooth function F(k) over k
space can be approximated as:
An extra factor of 2 because of spin
dk=k2sin dk d d
Carrier
density
n=
f
Calculating the carrier density
n=
f
f
dk=k2sin dk d d
g(E) is the density of levels per unit volume
N/V~g(E)
Let’s find g(E)!
(Just given in book)
f
Reminder: The Density of Levels
(Closely related to the density of states)
As we’ll see next time, we will often need to
know the number of allowed levels in k
space in some k-space volume 
If >>2/L, then the
number of states is
~ / (2/L)3
Or V/ (2)3
Then the density of those
levels is N/ or V/83
Density of Levels
The number of running waves N:
3
g (k )d k
2L/
kz
per dimension
4k2 dk
V 1
 s  k  d k  3  4 k 2 dk
N=g(k)
 8
2
Vk
2
s  k  d 3k  2 dk Times
N=g(k)
for spin!
2
3
dk
ky
k
kx
V 3
L 3 3
( ) d k 
d k
3
2
8
k Space
Result same for
standing waves!
2
N
k
 g (k ) 
2
V
2
Group:
Find dnk for n=1 and 2 dimensions
dnk
Finding the 3 dimensional density of states D(E)
g (k ) 
k
2
2
g (kg(k)
)dk  gg(E)
( E )dE
dE
g (k )  g ( E )
dk
g(k)
g(E)
You will do something similar to the
following in your homework.
Free electrons in 3D
Group: Find g(E) = g(k) dk/dE
g (k ) 
More than one way to approach
2
dk k
g ( E )  g (k )
 2
dE 

1

2
g(E)
k
2

2
Density of levels
 3 N 
EF 


2m  V 
2
2
2/3
The free electron gas
at T < Fermi temp
k

Big difference at
high temps
f FD 
1
1  e ( E  EF ) / k BT
The shaded area shows the change in
distribution between absolute zero and a
finite temperature.
n(E,T) number of free electrons per unit energy
range is just the area under n(E,T) graph.
n(E,T)
g(E)
Fermi-Dirac distribution
function is a symmetric
function, meaning:
T=0
T>0
EF
n( E, T )  g ( E) f FD ( E, T )
At low temperatures, the
same number of levels
below EF is emptied and
same number of levels
E
above EF are filled by
electrons.
How would do this in 2
dimensions (HW)?
Let’s review for 3 dimensions:
The number of allowed values of k
4k F
V
kF
N (
)( 3 )  2 V
3
8
6
3
3
Will need g(E) for 2D!
Since there are two spin states for each k
3
N kF
n  2
V 3
4rs
V 1
 
3
N n
3
Application:
Determining Metallic Properties
A knowledge of the thermal energy is fundamental to
obtaining an understanding many of the basic
properties of solids.
For example, the heat energy necessary to raise the
temperature of the material, called
Specific Heat or Heat Capacity
Want to Contrast with
Drude Model of Heat Capacity
When the solid is heated, the atoms vibrate
around their sites like harmonic oscillators.
The average energy for a 1D oscillator is kT.
Therefore, the average energy per atom, regarded
as a 3D oscillator, is 3kBT, and consequently the
energy per mole is
 = 3NkBT  3RT
where
N is Avagadro’s number, kB is
Boltzmann constant and R is the gas constant.
The differentiation w.r.t temperature gives heat capacity:
Cv  3R  3  6.02 1023 (atoms / mole) 1.38 1023 ( J / K )
Heat capacity of the free electron gas

From the diagram of n(E,T) the change in the
distribution of electrons can be resembled into
triangles of height ½ g(EF) and a base of 2kBT so
the area gives that ½ g(EF)kBT
electrons
increased their energy by kBT.
The difference in thermal energy from the value at T=0°K
E (T )  E (0) ~
1
g ( EF )(k BT ) 2
2
For an exact calculation:
E (T )  E (0) ~
1
g ( EF )(k BT ) 2
2
Differentiating with respect to T
gives the heat capacity at constant volume:

E
Cv 
 g ( EF )k B 2T
T
EF
0
2
N  EF g ( EF )
3
3 N
3N
g ( EF ) 

2 EF 2k B TF
N
g (E) 
g ( E )dE 
EF
V
 2
2
0
3N
Cv  g ( EF )kB T 
k B 2T
2kBTF
2
While this works great for metals, we
will find that it’s very different for
other materials.
V
2 2
3/ 2 1/ 2
(2
m
)
E
3
3/ 2 1/ 2
(2
m
)
E dE 
3
V
3 2
3/ 2
(2
mE
)
F
3
T 
3
Cv  Nk B  
2
 TF 
Heat capacity of
Free electron gas
Comparison with Data
T 
3
Cv  Nk B  
2
 TF 
TF=65,000 K
To improve upon this
model, we’ll have to
consider phonons
(later)
Classical Model suggests one constant value and same value for all materials
Cv  3R  3  6.02 1023 (atoms / mole) 1.38 1023 ( J / K )
Homework 2.3: When the Fermi-Dirac distribution is
close to the MB, what does that mean for rs?
fi 
1
N
Maxwell-Boltzmann
1 e
(  i   ) / k BT
Fermi-Dirac
rs – another measure of electronic density = radius of a sphere whose
volume is equal to the volume per electron (mean inter-electron spacing)
4rs
V 1
 
3
N n
3
Measuring specific heat on a budget
Step 1: Set-up the calorimeter
Energy conserved: qmetal = qwater + qcoffeecup
qcup can be ignored
cmetalmmetal Tsystem = cwatermwater Twater
Step 2: Boil the water
containing metal, Pour
Step 3: Stir while measuring
temperature
Modern calorimetry works on the same
principles, just looks more fancy.
constant volume or 'bomb'
calorimeter
Why Specific Heat?
Profile of Frances Hellman
Physics Professor at University of California, Berkeley
Previous chair of the physics department
“My research group is concerned with the
properties of novel magnetic and
superconducting materials especially in thin
film form. We use specific heat, magnetic
susceptibility, electrical resistivity, and other
measurements as a function of temperature
in order to test and develop models for
materials which challenge our understanding
of metallic behavior.
Current research includes: effects of spin on
transport and tunneling, including studies of
amorphous magnetic semiconductors and
spin injection from ferromagnets into Si;
finite size effects on magnetic and
thermodynamic properties…”
Strength of Metals (p. 39)
The book nicely derives (using P=-dE/dV) that
the pressure need to change the metal volume
P =(V/V) 2/3 n EF
As n is quite dense, need pressures over 10,000
atmospheres for a volume change of 1%!
This Fermi pressure is what makes solids stable.
(Bonding-antibonding suggests energy would
continue to decrease as bring atoms closer.)
Free Electron Models
Classical Model:
 Metal is an array of positive
ions with electrons that are
free to roam through the
ionic array
Electrons are treated as an
ideal neutral gas, and their
total energy depends on the
temperature and applied
field
In the absence of an
electrical field, electrons
move with randomly
distributed thermal
velocities

Quantum Mechanical
Model:
Electrons are in a potential
well with infinite barriers:
They do not leave metal,
but free to roam inside
 Electron energy levels
are quantized and well
defined, so average
energy of electron is not
equal to (3/2)kBT
 Electrons occupy energy
levels according to
Pauli’s exclusion
principle
Limitations of the FEG Model—and Beyond
The FEG model of Drude, augmented by quantum mechanics, was
extremely successful in accounting for many of the properties of
metals.
Some flawed assumptions behind the FEG model:
1. The free-electron approximation
The positive ions act only as scattering centers and is
assumed to have no effect on the motion of electrons
between collisions.
2. The independent electron approximation
Interactions between electrons are ignored.
Considerable progress comes from abandoning only the freeelectron approximation in order to take into account the effect
of the lattice on the conduction electrons.