Download Adding Fermi-Dirac Statistics to the Drude Model = Sommmerfield

Thermal conductivity
A material's ability to conduct heat.
Temperature
Electric current density
(je = I/A)
Heat current density
Fourier's Law for heat conduction.
 
E
 Thermal current density
jt  
 sec area 
jt   vn
 = Energy per particle
v = velocity
n = N/V
2l
Thermal conductivity
Heat current density
jt   vn
Heat Current Density jtot through the plane: jtot = jright - jleft
Limit as l gets small:
About half the particles are moving right, half to left.
x
Thermal conductivity
x
v
v
v
Thermal conductivity (expanding to 3d)
Heat current density
x

T
 T
x
v
2
 vx

1 2
jt   v cv T
jt  T
3
2
 vy
2
 vz
2
 3 vx
2
1 2
  v cv
3
How does / depend on temperature?
Thermal conductivity
1/3 cvv2
=
ne2/m
1/3 cvmv2
=
ne2
Drude applied kinetic theory of gases ½ mv2 = 3/2 kBT
=
cvkBT
ne2
The book jumps
through claiming a
value for cv
Classical Theory 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 ½ kT.
Therefore, the average energy per atom, regarded as
a 3D oscillator, is 3/2 kBT, and the total energy
density is 3/2 nkBT where n is the conduction
electron density and kB is Boltzmann constant.
Differentiation w.r.t temperature gives heat
capacity 3/2 n kB
=
3/2 n kB2T
ne2
=
3kB2T / 2e2
Thermal conductivity optimization
To maximize thermal conductivity, there are several
options:

Provide as many conduction electrons as possible


Make crystalline instead of amorphous


irregular atomic positions in amorphous materials scatter
phonons and diminish thermal conductivity
Remove grain boundaries


free electrons conduct heat more efficiently than phonons
(=lattice vibrations).
gb’s scatter electrons and phonons that carry heat
Remove pores (air is a terrible conductor of heat)
What happens?
The Seebeck Effect
A temperature gradient generates an
electric field E = QT,
where Q is known as the thermopower = -cv / 3ne
Seebeck and the reverse (Peltier) Effects
~ millivolts/K for (Pb,Bi)Te
The Seebeck effect is the conversion of temperature
differences directly into electricity.
Applications: Temperature measurement via
thermocouples; thermoelectric power generators;
thermoelectric refrigerators; recovering waste heat
Demo: https://www.youtube.com/watch?v=bt5o_rn0FmU
Many open questions:
Why does the Drude model work relatively
well when many of its assumptions seem so
wrong? In particular, the electrons don’t
seem to be scattered by each other. Why?
 Why is the actual heat capacity of metals
much smaller than predicted?

From Wikipedia: "The simple classical Drude model provides a
very good explanation of DC and AC conductivity in metals, the
Hall effect, and thermal conductivity (due to electrons) in
metals. The model also explains the Wiedemann-Franz law of
1853.
"However, the Drude model greatly overestimates the electronic
heat capacities of metals. In reality, metals and insulators have
roughly the same heat capacity at room temperature.“ It also
does not explain the positive charge carriers from the Hall effect.
Objectives
By the end of this section you should be able to:
 Apply Sch. Equation to a metal
 Apply periodic boundary conditions
 Start to understand k space
 Determine the density of states and Fermi energy
 Find the Fermi temperature, velocity, etc.
Improvement to the Drude Model
 Sommerfield
recognized we needed to
utilize Pauli’s exclusion principle
 Typically, this is the only difference
Electrons cannot all
be in the lowest
energy state, since
this would violate
the Pauli Principle.
Number of electrons per unit volume f(v)
Fermi-Dirac
Maxwell-Boltzmann
= N/V
Normalization condition
solves for constants
Another common way to
write is f(E)
Sommerfield still assumes the
free electron approximation
U(r)
U(r)
Neglect periodic potential & scattering
Reasonable for “simple metals” (Alkali Li,Na,K,Cs,Rb)
What does this remind you of?
The Quantum Analogy

These conduction electrons can be considered as
moving independently in a square well and the
edges of well corresponds to the edges of the
sample. (ignores periodic potential from atoms)

A metal with a shape of cube with edge length of L
Cube
V=L3
U
 2 d 2

 U ( x)  

2
 2m dx

in 1D
Inside U=0, for 3 dimensions:
0
L
How do we go about solving this?
U
0
L
Possible Boundary conditions
1. Common: Ψ(0)=0 and Ψ(L)=0
Standing waves. Wells aren’t really infinite
2. Periodic: Ψ(x,y,z)= Ψ(x+L,y,z)
Eigenstates

1 ikr
 (r ) 
e
V
Known as a running wave
with eigenvalues
 2k 2

2m
To Compare, Let’s Remind Ourselves of
the Standing Wave Solution
U
Boundary conditions
Ψ(0)=0 and Ψ(L)=0
0
L
Eigenstates
where
 ( x)  A sin( kx) with eigenvalues
n
k
L
 2k 2
E
2m
n y
nx
n z
, ky 
, kz 
Or in 3D: k x 
L
L
L
Similar idea for running waves:
e
Where nx, ny and
nz are integers



ik y L
ik x L
ik z L
1 e
How to find A?
e
Wavefunctions: Ideal Quantum Well
1D
 ( x)  A sin( kx)
 (x)
n
k
L
standing
waves
19
In your group, write the wavefunction for the lowest three energies.
Semiconductor Quantum Well
More about this diagram later today
In
CoFe
Optical Detection of Spin
hot Polarization in Quantum Wells
e
GaAs/InGaAs/GaAs
external
magnetic
field
CoFe
e
Alumina
tunnel
barrier
This is a very simple spin-selective device.
Electrons of one angular momentum are
favored as they travel past the Schottky
barrier due to the external magnetic field
and spin filtering in the CoFe. They then
fall into the quantum well and recombine
with holes. Emission from the quantum
well gives a good probe of spin.
hn
h
GaAs: n
InGaAs
i
i
p
The wave vector k is very important!
Eigenstates


1 ikr
 (r ) 
e
V
with eigenvalues
 2k 2

2m
To see why, note that  is an eigenstate
of the momentum operator p
 
k is the wave vector
ik  r
e
(Will explore more in Ch.5)
Momentum space or k-space is the set of all
wavevectors k, associated with particles - free and bound
All points in a crystal
that have an identical
environment are
described by one point
in k space.
This allows us to
dramatically reduce the
size of many atom
systems.
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 (in 3d)
Or V/ (2)3
Then the density of those
levels is N/ or V/83
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:
Will show an example later
Consider a spherical reason

Let’s find the number of allowed k values inside a
spherical shell of k-space of radius kF
3
4k F

3
The number of allowed values of k
4k F
V
kF
N (
)( 3 ) 
V
2
3
8
6
3
3
Since there are two spin states for each k
3
N kF
n  2
V 3
Warning: The Fermi level will be defined slightly differently for
nonmetals.
The Fermi Sphere
The k-space sphere with radius kF is called the Fermi sphere.
n
N kF
 2
V 3
2
 kF
F 
2m
1/ 3
3
kz
 3 2 N 
kF  

V


Fermi surface
kF
ky
kx
2
 3 N 
EF 


2m  V 
2
2
2/3
If we convert k-space to
energy space, the resulting
radius of the energy sphere
surface would give us the
cutoff between occupied and
unoccupied energy levels.
The surface of the Fermi sphere represent the
boundary between occupied and unoccupied k
states at absolute zero for the free electron gas.
Definition of the Work Function
Additional energy above the Fermi
level required to remove electrons
from the solid

EF
fermi level

=work function (3-4eV)
28
Fermi Energy in terms of the
Bohr radius
2
a0 
 0.529 Å – Bohr radius
2
me
2
Bohr radius = mean radius of
the orbit of an electron around
the nucleus of a hydrogen atom
at its ground state
2
2
 kF
e
2
F 
(
)( k F ao )
2m
2 ao
Recognize?
Ground state
energy of
hydrogen atom
2
e
 13.6eV  rydberg ( Ry )
2ao
Electrons in 3D Infinite Potential Well

Group: What is the ground state configuration of
many electrons in the 3D infinite potential well?
 Consider the case of solid with 34 electrons.
Determine the energy of each electron relative to
 2h2
Eo 
2mL2 .
 How many electrons are of each energy?
 Take the ground state to be when n1 = n2 = n3 = 1

Extra slides we may not have time to
cover (just extra examples)
Summing the energy density 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
Energy per Electron E/N
in the ground state
Combining:
3
N kF
n  2
V 3
E
V
2
2
E 3 hk
3

 F
N 10 m
5
Fermi Temperature
 F k BTF
The density of copper is 8.96 gm/cm3, and its atomic
weight is 63.5 gm/mole. (assume valence of 1)
 3 2 N 
EF 


2m  V 
2
(a)
(b)
2/3
Calculate the Fermi energy for copper.
Find the classical electron velocity from EF= ½ mvF2
and Fermi temperature. E  k T
F
B F
Pressure and Compressibility of an Electron Gas
(Skip if time low, result most important, in book)
 3 N 
EF 


2m  V 
2
2
2/3
Pressure is defined as E/ V for constant N,
so the pressure on an electron gas is
Reminder: Effective Radius
rs – another measure of electronic density = radius of a sphere whose
volume is equal to the volume per electron (mean inter-electron
3
spacing)
2
4rs
V 1
a0 
 0.529 Å – Bohr radius
 
2
me
3
N n
in metals rs ~ 1 – 3 Å (1 Å= 10-8 cm)
Combine with above:
1/ 3
 3 N 
kF  

 V 
2
3 1/ 3
k F  (3
)
3
4rs
1.92 3.63
kF 

rs
rs / ao
2
Å-1
rs/a0 ~ 2 – 6
9 1/ 3
( )
1.92
4


rs
rs