Electronic Properties of Solids
R.J. Nicholas
Electronic Properties:
Combination of :
•
•
•
•
•
•
•
Metals
Semiconductors
Insulators
Paramagnets
Diamagnets
Ferromagnets
Superconductors
Crystal Structure
Atomic Structure
Free electron theory of metals
• Metals are good conductors (both electrical and thermal)
• Electronic heat capacity has an additional (temperature
dependent) contribution from the electrons.
• Why are some materials metals and others not?
Simple approximation: treat electrons as free
to move within the crystal
Metals – HT10 – RJ Nicholas
1
Free electron theory of metals
• Alkali metals (K, Na, Rb) and Noble metals (Cu, Ag,
Au) have filled shell + 1 outer s-electron.
• Atomic s-electrons are delocalised due to overlap of
outer orbits.
• Crystal looks like positive ion cores of charge +e
embedded in a sea of conduction electrons
• Conduction electrons can interact with each other and
ion cores but these interactions are weak because:
(1) Periodic crystal potential (ion cores) is orthogonal to conduction
electrons - they are eigenstates of total Hamiltonian e.g. for Na conduct.
electrons are 3s states, but cores are n=1 and n=2 atomic orbitals.
(2) Electron-electron scattering is suppressed by Pauli exclusion
principle.
Assumptions:
(i) ions are static - adiabatic approx.
(ii) electrons are independent - do not interact.
(iii) model interactions with ion cores by using an “effective mass” m*
(iv) free electrons so we usually put m* = me
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2
Free Electron Model
L
Put free electrons into a very wide potential well the
same size as the crystal i.e. they are 'de-localised'
Free electron properties
Free electron Hamiltonian has
only kinetic energy operator:
Free electrons are plane waves
= 2 ∂ 2ψ
Eψ = −
2m ∂x 2
ψ = A e ± ikx
with:
Momentum:
i=
∂ψ
= ± =k ψ
∂x
Metals – HT10 – RJ Nicholas
Energy:
−
= 2 ∂ 2ψ
=2k 2
ψ
=
2m ∂x 2
2m
Group velocity:
∂ω
1 ∂E
=k
=
=
∂k
= ∂k
m
3
Free Electron Model – Periodic
boundary conditions
L
L
Add a second piece of crystal the same size:
The properties must be the same.
Density of states
Calculate allowed values of k.
Use periodic (Born-von Karman)
boundary conditions:
L = size of crystal
ψ ( x) = ψ ( x + L)
∴ e ikx = e ik ( x + L )
∴ e ikL = 1
2π
4π
, ±
L
L
2π
∴δk =
L
∴ k = 0, ±
Density of allowed states in reciprocal (k-) space is:
ΔK
in 1 − D
δ k
Metals – HT10 – RJ Nicholas
or
ΔVK
in 3 − D
δ k3
x 2 for spin states
4
Density of states (2)
States have energies
ε to ε + dε
g (ε )d ε = g (k )dk = 4π k 2 dk ×
dk
dε
dε
∴ g (ε )d ε = g (k )
=2 k 2
ε =
2m
⎛ 2mε ⎞
k =⎜ 2 ⎟
⎝ = ⎠
=
1
dk ⎛ 2m ⎞
=⎜
⎟
dε ⎝ =2 ⎠
2
1
=
2
1
2ε
1
2
8π
⎛ 2π ⎞
⎜
⎟
⎝ L ⎠
k
3
4π
⎛ 2π ⎞
⎜
⎟
⎝ L ⎠
2
δ k3
⎛ 2m ⎞
⎜ 2 ⎟
⎝= ⎠
2
1
2ε
2mε ⎛ 2m ⎞
⎜
⎟
=2 ⎝ =2 ⎠
3
⎛ 2m ⎞
= 4π ⎜ 2 ⎟
⎝h ⎠
3
2
ε
1
1
2
2
1
1
dε
2
1
2
ε
1
2
dε
dε × V
Fermi Energy
Electrons are Fermions
N =
∞
∫ g (ε ) f
F −D
(ε ) d ε
0
μ
at T = 0
N =
∫ g (ε ) d ε
0
μ at T = 0 is known as the
Fermi Energy, EF
Metals – HT10 – RJ Nicholas
N
8π ⎛ 2mEF ⎞
n=
=
⎜
⎟
V
3 ⎝ h2 ⎠
⎛ 3N ⎞
EF = ⎜
⎟
⎝ 8π V ⎠
2
3
3
2
h2
2m
5
Typical value for EF e.g. Sodium (monatomic)
crystal structure: b.c.c.
crystal basis: single Na atom
lattice points per conventional (cubic) unit cell: 2
conduction electrons per unit cell
2
∴ electrons per lattice point = 1
lattice constant (cube side) = a = 0.423 nm
∴ density of electrons n = N/V= 2/a3 = 2.6 x 1028 m-3
∴ EF = 3.2 eV
Fermi Temperature TF?
kBTF = EF
∴ TF = 24,000 K
Finite Temperatures and Heat Capacity
Fermi-Dirac distribution function fF-D = 1/(eE-μ/kBT + 1)
electrons are excited by an energy ~ kBT
Number of electrons is ≈ kBT g(EF)
∴ ΔE ≈ kB2T2 g(EF)
∴ CV = ΔE/ ΔT ≈ 2kB2T g(EF)
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∴ ln n = 3 ln E + const.
2
dn 3 dE
∴
=
n
2 E
dn
3 n
=
= g ( EF )
dE
2 EF
Previously
we have n = AEF3/2
∴ C v = 3nk B
k BT
EF
= 3nk B
T
TF
∴ Heat Capacity is:
(i) less than classical value by factor ~kBT/EF
(ii) proportional to g(EF)
Is this significant?
Lattice
Room
Temperature
Low
Temperature
Electrons
3nat.kB
π2/2 nkB (kBT/EF)
12π4/5 nat.kB (T/ΘD)3
π2/2 nkB (kBT/EF)
C/T = βT2 + γ
Debye term
Metals – HT10 – RJ Nicholas
free electron term
7
Rigorous derivation
∞
U = ∫ ε g (ε
)
∂f F − D
?
∂T
f F − D (ε ) d ε
0
∞
∂U
∂f
= ∫ ε g (ε )
dε
∂T
∂
T
0
∴
2
= g ( EF ) k B T
∞
∫
− EF
k BT
=
π2
3
f =
2 x
x e dx
(e
x
)
+ 1
2
+ δ
≈−∞
2
g ( EF ) k B T
1
ε − μ
, x=
e + 1
kT
x
∂f
− ex
=
∂T
ex + 1
(
)
δ ∝
∞
∫
− EF
k BT
=
×
2
∂x
∂T
x e x dx
(e
x
)
+ 1
2
0 (why?)
Magnetic susceptibility
• Susceptibility for a spin ½ particle is:
χ=
μ B2 μ0
kT
/ electron
• This is much bigger than is found experimentally
- Why?
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Pauli paramagnetism
Separate density of states for spin up and spin down,
shifted in energy by ± ½gμBB (g=2)
Imbalance of electron moments Δn
Δn = ½ g(εF) × 2μBB
giving a magnetization M
M = μB Δn = μB2 g(εF) B
and a susceptibility
χ = M/H = μ0 μB2 g(εF) = 3nμ0 μB2 /2εF
k-space picture and the Fermi Surface
T=0 states filled up to EF
= 2k 2
∴
= EF
2m
Map of filled states in k-space
= Fermi surface
∴ kF =
2mEF
=2
N = 2×
4π k F3
3 ⎛⎜ 2π ⎞⎟ 3
or we can write:
⎝ L ⎠
∴ k F3 =
Metals – HT10 – RJ Nicholas
3π 2 N
V
9
k-space picture and the Fermi Surface
∴
T=0 states filled up to EF
E
= 2k 2
= EF
2m
∴ kF =
2mEF
=2
EF
k
kF
How big is Fermi surface/sphere compared to
Brillouin Zone?
Simple cubic structure
volume of Brillouin Zone = (2π/a)3
electron density n = 1/a3
volume of Fermi sphere = 4πkF3/3 = 4π3/a3
= half of one B.Z.
Metals – HT10 – RJ Nicholas
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Electron Transport - Electrical Conductivity
Equation of motion: Force = rate of change of momentum
=
∂k
= − e (E + B × v )
∂t
Apply electric field - electrons are accelerated to a steady
state with a drift velocity vd - momentum is lost by
scattering with an average momentum relaxation time τ
∴ momentum loss =
mvd
τ
∴ current j = nevd =
= −eE
ne 2τ
E
m
ne 2τ
∴ conductivity σ =
= neμ
m
μ is mobility with:
vd = μE
What happens in k-space?
All electrons in k-space are
accelerated by electric field:
On average all electrons
shifted by: δ k = − eEτ
=
= δ k = Fδ t = − eE δ t
E
EF
k
δk
Metals – HT10 – RJ Nicholas
kF
11
What happens in k-space?
All electrons in k-space are
accelerated by electric field:
= δ k = Fδ t = − eE δ t
On average all electrons shifted by:
δk = −
eEτ
=
Fermi sphere is shifted in k-space by δk << kF
∴ To relax electron momentum k must be changed by ~ kF
Scattering occurs at EF
∴ we need phonons with large value of k. But phonon energy is small
so only a small fraction of electrons kBT/εF can be scattered
Scattering processes
Basic Principle: Scattering occurs because of deviations
from perfect crystal arrangement
Electron scattering mechanisms:
(i) thermal vibrations i.e. phonons (vibrations of the atoms
are a deviation from perfect crystal structure)
(ii) presence of impurities - charged impurities are very
important - scattering is by Coulomb force i.e.
Rutherford scattering.
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Matthiessen’s rule:
∴ρ =
Scattering rates (1/τ) add
m
ne 2
∑ 1τ
= ρ
T
+ ρ
i
Mean free path (λ):
electrons are moving with Fermi velocity vF
∴
λ = v F τ ( NOT v d τ )
Low temperature mean free paths can be very long as
electrons are only scattered by impurities
Hall Effect
In a magnetic Field B the electron experiences
a force perpendicular to its velocity.
A current j causes a build up of charge at the
edges which generates an Electric field E
which balances the Lorentz force
( − e ) ( E + v d × B ) y = 0;
Metals – HT10 – RJ Nicholas
E y = (vd ) x B z
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Hall Effect
Balance of forces:
In a magnetic Field B the electron
experiences a force perpendicular to its
velocity.
A current j causes a build up of charge at
the edges which generates an Electric
field E which balances the Lorentz force
( − e ) ( E + v d × B ) y = 0;
The Hall coefficient RH is:
j x = n ( − e) vd
E y = (vd ) x B z
RH =
⇒
Ey
RH = − 1
j x Bz
ne
Negative sign is sign of the charge on the electron
Metal
Charge/Atom (units of electron charge e)
Group
Hall Expt.
FE Theory
Lithium
-0.79
-1
I
Sodium
-1.13
-1
I
Potassium
-1.05
-1
I
Copper
-1.36
-1
IB
Silver
-1.18
-1
IB
Gold
-1.47
-1
IB
Beryllium
+0.1
-2
II
Magnesium
-0.88
-2
II
Calcium
-0.76
-2
II
Zinc
+0.75
-2
IIB
Cadmium
+1.2
-2
IIB
Aluminium
+1.0
-3
III
Indium
+1.0
-3
III
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Thermal conductivity
In metals heat is mainly carried by the electrons
Simple kinetic theory formula for thermal conductivity K:
K = 1/3CλvF
[C = π2/3 kB2T g(EF) = π2/2 nkB kBT/EF]
= π2/6 λvF nkB kBT/EF
[λ = vFτ ;
EF = ½mvF2]
= π2/3m n kB2 τ T
Scattering processes
• Low temperatures: defects, τ independent of T
• Intermediate temp. : Low Temp phonons - Debye
model τ ∝ T-3
• High temperatures: ‘classical’ phonons τ ∝ T-1
Wiedeman-Franz ratio
Electrical and Thermal conductivities of electrons are both
proportional to the relaxation time τ
Taking the ratio of the two should make this cancel so if we
define the Lorenz number as L = K/(σT) we have the
Wiedeman-Franz Law:
L = K
2 2
π
kB
=
σT
3e 2
Predicted value is absolute and the same for all metals.
Works well at high and low temps, - breaks down in ‘Debye’
region where energy and charge scattering are different
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Successes and Failures of Free Electron
Model
Successes:
• Temperature dependence of Heat Capacity
• paramagnetic (Pauli) susceptibility
• Ratio of thermal and electrical conductivities (Lorentz number)
• Magnitudes of heat capacities and Hall effect in simple metals
Failures:
•
•
•
•
Heat capacities and Hall effect of many metals are wrong
Hall effect can be positive
Does not explain why mean free paths can be so long
Does not explain why some materials are metals, some insulators
and some are semiconductors
Nearly Free Electron Approximation
Use a travelling wavefunction for an electron, e ikx, with kinetic
energy =2k2/2m
Assume that this is Bragg scattered by the wavevector G=2π/a to
give a second wave e i(k-G)x with energy =2(k-G)2/2m
Crystal potential is periodic in real space. Therefore we can
Fourier Transform the potential so that:
V ( x) =
∑
G
VG exp (iGx )
For a schematic solution we calculate what happens for a
single Fourier component VG so V(x) = VG(eiGx + e-iGx)
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Nearly Free Electron Approximation
Use a travelling wavefunction for an electron, e ikx, with kinetic
energy =2k2/2m
Bragg scattering
θ
d
Δk = 2|k|sinθ
= 4π/λ sinθ = 2π/d
2dsinθ = λ
Δk = G = ha* + kb* +lc*
a
With d =
2θ
h2 + k 2 + l 2
Formally what we are doing is to solve the Hamiltonian form of
Schrödinger equation
Hψ = Eψ
where ψ are the two travelling wave solutions. Expanding gives:
⎛ H11 − λ
⎜⎜
H 21
⎝
H11
H12 ⎞ ⎛ eikx ⎞
⎟ = (E − λ
⎟⎜
H 22 − λ ⎟⎠ ⎜⎝ ei (k −G ) x ⎟⎠
=2 ∂2
= ψ1 * −
ψ1
2m ∂x 2
= 2k 2
=
,
2m
⎛ eikx ⎞
) ⎜⎜ i (k −G )x ⎟⎟
⎝e
⎠
= 2 (k − G )
=
2m
2
H 22
H12 = ψ 1 * V ( x) ψ 2 = VG
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