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 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)
Metals – HT10 – RJ Nicholas
6
∴ 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?
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 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
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 ∴ 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:
