Electronic Properties of R.J. Nicholas

Electronic Properties: • • • 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 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 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) 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 =22∂ ψ only kinetic energy operator: Eψ =− 2mx∂ 2

± Free electrons are plane waves ψ = A e ikx

with: Momentum: Energy: Group velocity:

∂ψ ==22∂ ψ 22k ∂ω 1 ∂E =k i= = ± =k ψ −=ψ = = ∂x 22mx∂ 2 m ∂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) ∴ eikx = eik ( x + L) boundary conditions: ∴ eikL = 1

2π 4π L = size of crystal ∴ k = 0, ± , ± L L 2π ∴ δ k = L Density of allowed states in reciprocal (k-) space is: ΔK ΔV in 1− D or K in 3 − D x 2 for states δ k δ k 3

Metals – HT10 – RJ Nicholas 4 (2) States have energies 2 ε to ε + dε gd()εε==× gkdkkdk () 4 π2 δ k 3 dk ∴=gd()εε gkd () ε dε 22 1 ε = = k π 2 = 8212 ⎛⎞m ε 3 kd 2m ⎛⎞2π ⎜⎟2 1 ⎜⎟ ⎝⎠= ε 2 1 ⎝⎠L 2 ⎛⎞2mε 2 k = 1 ⎜⎟2 42πεmm⎛⎞ 22 1 ⎝⎠= = dε π 3 22⎜⎟ 1 1 ⎛⎞2 == 2 2 ⎜⎟ ⎝⎠ε dk= ⎛⎞21 m ⎝⎠L ⎜⎟2 1 dε = 2 3 ⎝⎠2ε 2m 2 1 =×πεε⎛⎞ 2 4 ⎜⎟2 dV ⎝⎠h

Fermi Energy

∞ Ngfd= ()εεε () Electrons are ∫ FD− 0 μ at T==0() N∫ gεε d 0

3 N 8π 2mE 2 == ⎛⎞F n ⎜⎟2 Vh3 ⎝⎠ μ at T = 0 is known as the 2 3 2 = ⎛⎞3Nh , EF EF ⎜⎟ ⎝⎠82πVm

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

E-μ/k T Fermi-Dirac distribution function fF-D = 1/(e B + 1)

electrons are excited by an energy ~ kBT

≈ Number of electrons is kBTg(EF)

∴Δ ≈ 2 2 E kB T g(EF)

∴ Δ Δ ≈ 2 CV = E/ T 2kB T g(EF)

Metals – HT10 – RJ Nicholas 6 ∴=lnn3 ln E + const . Previously 2 3/2 dn3 dE we have n = AEF ∴= nE2 dn==3 n gE()F dE2 EF

∴=kTB = T C3nkvB 3nk B ETFF

∴ Heat Capacity is:

(i) less than classical value by factor ~kBT/EF

(ii) proportional to g(EF)

Is this significant?

Lattice Electrons

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

Low 12π4/5 n k (T/Θ )3 π2/2 nk (k T/E ) Temperature at. B D B B F

C/T = βT2 + γ

Debye term free electron term

Metals – HT10 – RJ Nicholas 7 Rigorous derivation

∞ ∂ f F −D Ugfd= εε() () εε ? ∫ FD− ∂T 0 ∞ 1 ε − μ ∂∂Uf fx==, ∴=εεgd() ε x ∂∂∫ ekT+ 1 TT0 ∂−f exx ∂ ∞ 2 x =× 2 xedx ∂∂x 2 =+gE() k T δ TT()e + 1 FB ∫ 2 − x EF ()e + 1 ∞ kTB x edxx 2 δ ∝ π 2 ≈−∞ ∫ 2 = − x + gE()FB k T EF ()e 1 3 kTB = 0(why?)

Magnetic susceptibility

• Susceptibility for a spin ½ particle is: μμ2 χ = B 0 / electron kT

• This is much bigger than is found experimentally -Why?

Metals – HT10 – RJ Nicholas 8 Pauli

Separate density of states for spin up and spin down, ± μ shifted in energy by ½g BB (g=2) Imbalance of electron moments Δn Δ ε × μ n = ½ g( F) 2 BB giving a magnetization M μ Δ μ 2 ε M = B n = B g( F) B and a susceptibility χ μ μ 2 ε μ μ 2 ε = M/H = 0 B g( F) = 3n 0 B /2 F

k-space picture and the

=2k 2 T=0 states filled up to E ∴ = E F 2m F

Map of filled states in k-space 2mE ∴ k = F = Fermi surface F =2

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

Metals – HT10 – RJ Nicholas 9 k-space picture and the Fermi Surface

=2k 2 T=0 states filled up to E ∴ = E F 2m F

2mE E ∴ k = F F =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

π 3 π3 3 volume of Fermi sphere = 4 kF /3 = 4 /a = 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 mv ∴==−momentum loss d eE τ 2τ ∴==ne current j nevd E m μ is mobility with: ne2τ ∴==conductivity σ neμ vd = μE m

What happens in k-space?

All electrons in k-space are accelerated by electric field: = δδkFt==− eEt δ

On average all electrons E eEτ shifted by: δ k =− = EF

k k δk F

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

All electrons in k-space are = δδkFt==− eEt δ accelerated by electric field:

eEτ On average all electrons shifted by: δ k =− =

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 with large value of k. But energy is small ε so only a small fraction of electrons kBT/ F 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

∴=ρρρm 1 = + ne2 ∑ τ Ti

Mean free path (λ):

electrons are moving with Fermi velocity vF ∴ λ = τ τ vF (NOT vd )

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 Bz

Metals – HT10 – RJ Nicholas 13 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 Balance of forces: field E which balances the Lorentz force

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

The Hall coefficient RH is: = Ey RH jx Bz

j = n(−e) v ⇒ R = − 1 x d H 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 -0.88 -2 II Calcium -0.76 -2 II Zinc +0.75 -2 IIB Cadmium +1.2 -2 IIB +1.0 -3 III Indium +1.0 -3 III

Metals – HT10 – RJ Nicholas 14 In metals heat is mainly carried by the electrons Simple kinetic theory formula for thermal conductivity K: 1 λ π2 2 π2 K = /3C vF [C = /3 kB T g(EF) = /2 nkB kBT/EF] π2 λ λ τ 2 = /6 vF nkB kBT/EF [ = vF ; EF = ½mvF ] π2 2 τ = /3m n kB T Scattering processes • Low temperatures: defects, τ independent of T • Intermediate temp. : Low Temp phonons - 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

π 2k 2 Wiedeman-Franz Law: L = K = B σ T 3e2

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) = V exp (iGx) ∑G G

For a schematic solution we calculate what happens for a iGx -iGx single Fourier component VG so V(x) = VG(e + e )

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 π λ θ= 2π 2dsinθλ = = 4 / sin /d

Δk = G = ha* + kb* +lc* a 2θ With d = hkl222++

Formally what we are doing is to solve the Hamiltonian form of Schrödinger equation H ψ = E ψ where ψ are the two travelling wave solutions. Expanding gives:

⎛ H − λ H ⎞ ⎛ eikx ⎞ ⎛ eikx ⎞ ⎜ 11 12 ⎟ ⎜ ⎟ = ()E − λ ⎜ ⎟ ⎜ − λ ⎟ ⎜ i()k −G x ⎟ ⎜ i()k −G x ⎟ ⎝ H 21 H 22 ⎠ ⎝e ⎠ ⎝e ⎠

=2 ∂ 2 =2k 2 =2 ()k − G 2 H = ψ * − ψ = , H = 11 1 2m ∂x2 1 2m 22 2m

= ψ ψ = H12 1 * V (x) 2 VG

Metals – HT10 – RJ Nicholas 17