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# Free electron theory of metals

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```									                                 Electronic Properties of Solids
R.J. Nicholas

Electronic Properties:      •   Metals
•   Semiconductors
•   Insulators
•   Paramagnets
•   Diamagnets
•   Ferromagnets
•   Superconductors

Combination of :         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

Metals – HT10 – RJ Nicholas                                                                     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
∂ 2ψ
2
only kinetic energy operator:           Eψ = −
2m ∂x 2

Free electrons are plane waves              ψ = A e ± ikx

with:
Momentum:                   Energy:             Group velocity:
∂ψ                      2
∂ 2ψ   2 2
k            ∂ω   1 ∂E   k
i       = ± kψ          −           =     ψ            =      =
∂x                     2m ∂x 2   2m            ∂k     ∂k   m

Metals – HT10 – RJ Nicholas                                                                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.            ψ ( x) = ψ ( x + L)
Use periodic (Born-von Karman)            ∴ e ikx = e ik ( x + L )
boundary conditions:
∴ e ikL = 1
2π     4π
L = size of crystal                 ∴ k = 0, ±  , ±
L      L
2π
∴δk =
L

Density of allowed states in reciprocal (k-) space is:
ΔK                         ΔVK
in 1 − D          or        in 3 − D        x 2 for spin states
δ k                        δ k3

Metals – HT10 – RJ Nicholas                                                               4
Density of states (2)
States have energies
ε to ε + dε                                                                                            2
g (ε )d ε = g (k )dk = 4π k 2 dk ×
δ k3
dk
∴ g (ε )d ε = g (k )                dε
dε
2
k2                                            8π
1
ε =                                                                        ⎛ 2m ⎞              2
1
2m                                        =            3       k   2
⎜ 2 ⎟                               1
dε
⎛ 2π ⎞
⎜    ⎟                   ⎝    ⎠                  2ε              2
1                                    ⎝ L ⎠
⎛ 2mε ⎞       2
k =⎜ 2 ⎟                                                                                           1
⎝     ⎠                                        4π               2mε ⎛ 2m ⎞                          2
1
=            3         2 ⎜ 2 ⎟                                       1
dε
1                                    ⎛ 2π ⎞
⎜    ⎟               ⎝    ⎠                                  ε       2
dk ⎛ 2m ⎞       2
1                           ⎝ L ⎠
=⎜    ⎟
dε ⎝ 2 ⎠            2ε
1
2
⎛ 2m ⎞
3
2       1
= 4π ⎜ 2 ⎟                           ε       2
dε × V
⎝h ⎠

Fermi Energy
∞

Electrons are Fermions                      N =             ∫ g (ε ) f
0
F −D        (ε ) d ε

μ
at T = 0     N =             ∫ g (ε ) d ε
0

3
N   8π ⎛ 2mEF ⎞                                                2
n=   =    ⎜      ⎟
V    3 ⎝ h2 ⎠
μ at T = 0 is known as the                                                2
⎛ 3N ⎞                       3
h2
Fermi Energy, EF                            EF = ⎜      ⎟
⎝ 8π V ⎠                         2m

Metals – HT10 – RJ Nicholas                                                                                                                         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)

Metals – HT10 – RJ Nicholas                                                       6
∴ ln n = 3 ln E + const.
Previously                            2
we have n = AEF3/2            dn 3 dE
∴    =
n    2 E
dn    3 n
=       = g ( EF )
dE    2 EF
k BT               T
∴ C v = 3nk B            = 3nk B
EF                TF

∴ Heat Capacity is:
(i) less than classical value by factor ~kBT/EF
(ii) proportional to g(EF)

Is this significant?
Lattice                Electrons

Room
Temperature                  3nat.kB           π2/2 nkB (kBT/EF)

Low                    12π4/5 nat.kB (T/ΘD)3    π2/2 nkB (kBT/EF)
Temperature

C/T = βT2 + γ
Debye term         free electron term

Metals – HT10 – RJ Nicholas                                                              7
Rigorous derivation

∞                                                           ∂f F − D
U = ∫ ε g (ε            )    f F − D (ε ) d ε                                  ?
∂T
0

∂U
∞
∂f                                                 1        ε − μ
∴        = ∫ ε g (ε )    dε                                      f =   x
, x=
e + 1        kT
∂T   0
∂T
∂f     − ex                           ∂x
∞
x e dx  2 x                    =                              ×
∫                                         (              )
2
= g ( EF ) k B T
2
+ δ   ∂T   ex + 1                           ∂T
(e             )
2
x
− EF             + 1
k BT                                                ∞
x e x dx
π2                                           ≈−∞
δ ∝       ∫
(e               )
2                                                                              2
=            g ( EF ) k B T                                                   − EF
x
+ 1
3                                                                        k BT

=          0 (why?)

Magnetic susceptibility

• Susceptibility for a spin ½ particle is:

μ B μ0
2
χ=                         / electron
kT
• This is much bigger than is found experimentally
- Why?

Metals – HT10 – RJ Nicholas                                                                                                        8
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
2
k2
T=0 states filled up to EF          ∴     = EF
2m

Map of filled states in k-space
2mEF
= Fermi surface                     ∴ kF =            2

4π k F 3
N = 2×
or we can write:                              3 ⎛ 2π ⎞ 3
⎜    ⎟
⎝ L ⎠

3     3π 2 N
∴ kF =
V

Metals – HT10 – RJ Nicholas                                                      9
k-space picture and the Fermi Surface
2
k2
T=0 states filled up to EF             ∴       = EF
2m

E                               2mEF
∴ kF =         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                                                     10
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
mvd
∴ momentum loss =         = −eE
τ
ne 2τ
∴ current j = nevd =            E
m              μ is mobility with:
ne 2τ                       vd = μE
∴ conductivity σ =       = neμ
m

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                            E
shifted by: δ k = − eEτ
EF

k
kF
δk

Metals – HT10 – RJ Nicholas                                                                       11
What happens in k-space?

All electrons in k-space are
δ k = Fδ t = − eE δ t
accelerated by electric field:

eEτ
On average all electrons shifted by:   δk = −

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.

Metals – HT10 – RJ Nicholas                                                                12
Matthiessen’s rule:       Scattering rates (1/τ) add

m
∴ρ =
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;           E y = (vd ) x B z

Metals – HT10 – RJ Nicholas                                                      13
In a magnetic Field B the electron
Hall Effect                       experiences a force perpendicular to its
velocity.
A current j causes a build up of charge at
the edges which generates an Electric
Balance of forces:                 field E which balances the Lorentz force

( − e ) ( E + v d × B ) y = 0;       E y = (vd ) x B z

The Hall coefficient RH is:                   RH =
Ey
j x Bz

j x = n ( − e) vd      ⇒   RH = − 1
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
Aluminium                       +1.0                   -3              III
Indium                          +1.0                   -3              III

Metals – HT10 – RJ Nicholas                                                                        14
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

L = K
π 2kB
2
Wiedeman-Franz Law:
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

Metals – HT10 – RJ Nicholas                                                         15
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)

Metals – HT10 – RJ Nicholas                                                                 16
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
2dsinθ = λ                             = 4π/λ sinθ = 2π/d

Δk = G = ha* + kb* +lc*
a
2θ                                 With d =
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           H12 ⎞ ⎛ eikx ⎞                        ⎛ eikx ⎞
⎜                         ⎟⎜               ⎟ = (E − λ   ) ⎜ i ( k −G ) x ⎟
⎜    H 21        H 22 − λ ⎟ ⎜ ei (k −G ) x ⎟              ⎜e             ⎟
⎝                         ⎠⎝               ⎠              ⎝              ⎠

H11     = ψ1 * −
∂2 2
ψ1          =
k22
,       H 22 =
2
(k − G )2
2m ∂x 2               2m                             2m

H12 = ψ 1 * V ( x) ψ 2 = VG

Metals – HT10 – RJ Nicholas                                                                              17

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