Solid State TheorySolid State TheoryPhysics 545Physics 545

Electrons in metalsElectrons in metals

Electrons in metals

– Electronic: response of classical ‘free’ electrons in material:

Classical derivation of Ohm’s law and Drude conductivity.

Classical Hall effect, Hall constants.

– Failures of classical theory.

Energy levels– Energy levels.

– Fermi-Dirac distribution function.

– Sommerfeld theory.

Metals and insulatorsMeasured resistivities range over more than 30 orders of magnitude

Material Resistivity (Ωm) (295K)

Resistivity (Ωm) (4K)

10-12

“Pure”Metals

10-5Potassium

MetalsCopper

Semi- Ge (pure) 5 × 102 1012

2 × 10-6 10-10

Semi-Conductors

Ge (pure) 5 × 10 10

Insulators Diamond 1014

Polytetrafluoroethylene 1020

1014

1020y y(P.T.F.E)

Metals, insulators & semiconductors?

At low temperatures allmaterials are insulators

1020-Diamondmaterials are insulators

or metals.

P t l i ti it

1010-

ity (Ω

m)

GermaniumPure metals: resistivityincreases rapidly withincreasing temperature. 100 -

Res

istiv

i Germanium

10-10-R

CopperAs impurities are introduced the resistivity of metals increases and becomes less

Semiconductors: resistivity decreases rapidly with increasing

100 200 3000Temperature (K)

increases and becomes less temperature dependent.

Semiconductors: resistivity decreases rapidly with increasingtemperature.

Semiconductors have resistivities intermediate between metalsSemiconductors have resistivities intermediate between metalsand insulators at room temperature.

Core and Valence ElectronsThe most highly conducting metals are formed from atoms with partially filled atomic orbitals.e.g. Li, Na, and K etc, which have the electronic structure

y

Li 1s2 2s1 Na 1s2 2s2 2p6 3s1 K 1s2 2s2 2p6 3s2 3p6 4s1

The best insulators are formed from atoms with closed shells

e.g. Solid inert gases which have the electronic structure

He 1s2 Ne 1s2 2s2 2p6 Ar 1s2 2s2 2p6 3s2 3p6

I i l i di i i h b CORE l hi hIn a simple picture we distinguish between CORE electrons, which are tightly bound to the nuclei, and VALENCE electrons that move freely through the metals.

Energy Levels and BandsIsolated atoms have precise allowed energy levels.

I lid h l f i h l b d (hi h bi diIn solids the electron states of tightly bound (high binding energy) electrons are very similar to those of the isolated atoms.

Lo er binding electron states become bands of allo ed states

Band of allowed energy states

Lower binding electron states become bands of allowed states.

Band of allowed energy states.

E

+ + + + +position Electron level similar toposition Electron level similar to

that of an isolated atom

Why are metals good conductors?y gConsider a metallic Sodium crystal to comprise of a lattice of Na+

ions, containing the 10 electrons which occupy the 1s, 2s and 2pshells while the 3s valence electrons move throughout the crystalshells, while the 3s valence electrons move throughout the crystal.

The valence electrons form a very dense ‘electron gas’.

_ __ _ _+ +

+ + +

+ + +

+ + +

+

Na+ ions:Nucleus plus 10__ _ __

+ +

+ + +

+ + +

+ + +

+ 10 core electrons

__

__

__+ + + + + +

We might expect the negatively charged electrons to interact verystrongly with the lattice of negative ions and with each other.

In fact the valence electrons interact weakly with each other &electrons in a perfect lattice are not scattered by the positive ions.

Remove geometry of material V-voltageI-current

T

σ-conductivityL,T,W- length, thickness,width of the sample

V= IR= IR/ L R= L/(σTW)=(1/σ) L/A

T E=V/L electric fieldA=TW cross section areaJ=I/A current density

V= (1/σ) I L/A I/A =σ V/L J= σE

EJ

y

JEIsotropic materialEJ σ= JEIsotropic material

Anisotropic material JE

Materials infop J

zxzyxyxxxx EEEJ σ+σ+σ=

Free classical electronsLConsider a gas of free classical

electrons moving under theinfluence of electric and

L

Area Ainfluence of electric andmagnetic fields.

Ohms law and electron drift Electric field EOhms law and electron drift Electric field E

Drift velocity vd

Current density j = I/ACurrent density j = I/A

Volume element: area A, length dx and free electrondensity n per m3. Enclosed charge is

Area A

vd

dQ = -neAdx

I = dQ/dt (Coulomb s-1)

dx

dQ = -neAdx.

I = -neAdx/dt = -neAvd j = -nevd

Concept of mobility

J= σ E=- nev, by definition of flux through a cross- section

p y

J σ E nev, by definition of flux through a cross section

n= number of electrons per volume

v= velocity of the carriers due to electric field-->drift velocity

Define v = −E μ;

μ is mobility, since the electric field creates a force on the electron F=- eE

μ=nev/ E

σ =μneσ μne

Free classical electrons:AssumptionsWe will first consider a gas of free classical electrons subject toexternal electric and magnetic fields. Expressions obtained will beuseful when considering real conductorsuseful when considering real conductors

(i) FREE ELECTRONS: The valance electrons are not affectedb th l t i i t ti Th t i th i d i l b h i iby the electron-ion interaction. That is their dynamical behaviour isas if they are not acted on by any forces internal to the conductor.

(ii) NON-INTERACTING ELECTRONS: The valence electronsfrom a `gas' of non-interacting electrons. They behave asINDEPENDENT ELECTRONS; they do not show any `collective'; y ybehaviour.

(iii) ELECTRONS ARE CLASSICAL PARTICLES:(iii) ELECTRONS ARE CLASSICAL PARTICLES:

(iv) ELECTRONS ARE SCATTERED BY DEFECTS IN THELATTICE: ‘Collisions’ with defects limit the electrical conductivityLATTICE: Collisions with defects limit the electrical conductivity.This is considered in the relaxation time approximation.

Drude model: Sea of electronsMicroscopic Origin: Can we predict Conductivity of Metals?

– all electrons are bound to ion atom cores except valence electrons

Drude model: Sea of electrons

electrons

–ignore cores

l t–electron gas

– electrons obey Boltzmann statistics

+ + + +

+ + + +

+ + + +

Metal that contains free electrons wondering around positive cores.

FREE-ELECTRON APPROXIMATION

Does this microscopic picture of metals give us Ohm’s Law?

F= m aF=- eE

F= meame ( dv/ dt)=- eEv=-( eE/ me) t

J= σE=- nev= ne2Et/ mee

σ= ne2t/ me

Constant E gives ever- increasing J

N Oh ’ l b l f l i f l !No, Ohm’s law can not be only from electric force on electron!

Relaxation time approximationAt equilibrium, in the presence of an electric field, electrons in a conductor move with a constant drift velocity since scattering produces an effective frictional forcean effective frictional force.

Assumptions of the relaxation time approximation :p pp

1/ Electrons undergo collisions. Each collision randomises the electront i Th l t t ft tt i i i d d tmomentum i.e. The electron momentum after scattering is independent

of the momentum before scattering.

2/ Probability of a collision occurring in a time interval dt is dt/τ.τ is called the ‘scattering time’, or ‘momentum relaxation time’.

3/ τ is independent of the initial electron momentum & energy.

Momentum relaxation Consider electrons, of mass me, moving with a drift velocity vd due toan electric field E which is switch off at t=0. At t=0 the averageelectron momentum is F1electron momentum is

In a time interval dt the fractionalp(t = 0) = mevd(t = 0) 1.0

F1

change in the average electronmomentum due to collisions is 0.6

0.8

/p(t=

0)

integrating from t=0 to t then gives

dp/p(t) = - dt/τp ; dp/dt = -p(t)/τp

0.2

0.4p(t)/

integrating from t 0 to t then gives

p(t) = p(0)exp(-t/τp ) -1 0 1 2 3 40.0

t/τpτp is the characteristic momentum or drift velocity relaxation time.

If, in a particular conductor, the average time between scatteringt i d it t k 3 tt i t t d ievents is τs and it average takes 3 scattering event to randomise

the momentum. Then the momentum relaxation time is τp = 3τs.

Hydrodynamic representation of e-motion

p= momentum= mv

p(t)dp(t) ....)t(F)t(F 21 +++τ

−=p(t)

dtdp(t)

Response (ma) = Drag Driving force Restoring force

Add a drag term,i. e. the electrons have many collisions during drift

eE−τ

−=p(t)

dtdp(t)

1/τ represents a ‘viscosity’ in mechanical terms

τdt

1/τ represents a viscosity in mechanical terms

Steady state:y

0=dp(t) )e1(

tτ

−−= pp(t) τ−=∞ eEp0

dt)e1(∞pp(t) τ∞ eEp

mevavg=-eEτ

vavg =- eEτ/me

2τneOhm’s law

emτne

=σmeτ

=μe

em

S tt i b bilit i ti l t ti l t

Classical derivation of relaxation time

Scattering probability is proportional to cross sectional area atomtakes up when vibrating πa2.

i th l l it f l t d fi d f

2 3kT3kTvm

vth is thermal velocity of electron defined from Boltzmann statistics

eth

e

m3kTv

23kT

2vm

th =⇒=

Mean free path l=vth τt

Volume, in which scattering occurs πa2l=πa2 vth τ.

This volume should contain at least one scatter.

nπa2 vth τ =13kT

m2

e

anπ=τ

For Li N=0.47·1023 cm-3 and a=0.068 nm. Thus the classical theory gives τ= 13.2 fs and σ=17.5·106 (Ω·m)-1. The experimental value is 8.11 fs and 10.5 ·106 (Ω·m)-1.

Drude model successfully explains Ohm’s law.

Electrical ConductivityIn the absence of collisions, the average momentum of free electronssubject to an electric field E would be given by

pd ⎞⎛

The rate of change of the momentum due to collisions is

Ep edtd

Field−=⎟

⎠⎞

⎜⎝⎛

g

p/dtd

Collisionsτ−=⎟

⎠⎞

⎜⎝⎛ pp

At equilibrium + τ==⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ Εppp e- So0

dtd

dtd

CollisionsField

Now j = -nevd = -nep/me = (ne2τp /me) E

So the conductivity is σ = j/E = ne2τp /mey j p e

The electron mobility, μ, is defined as the drift velocity perunit applied electric field

μ = vd / E = eτp /me (units m2V-1s-1)

Classical Hall effectH

Vh

h

Lorentz force: FL=e v×H;

Electrostatic force: FE=Ee=Vhe/l

Vhd

At equilibrium FL= FE and

v×H= Vh/l I

Current density is related to the velocity as ven=I/(dh) =>dIHR

endHIV hh ==

en1Rh = Rh is Hall constant

Classical theory predicts that -Rhen=1

Is it true?

The Hall EffectEx, jx EyBz

vd = vx

An electric field Ex causes a current jx to flow.

A magnetic field Bz produces a force in the y-direction on theelectrons. Electrons accumulate on one face and positive chargeon the other producing a field E

• F = -e (E + v × B). In equilibrium jy = 0 so Fy = -e (Ey - vxBz) = 0

on the other producing a field Ey .

j = nev so E = j B /ne

• Therefore Ey = +vxBz

• The Hall resistivity is ρΗ = Ey/jx = -B/ne

jx = -nevx so Ey = -jxBz/ne

• The Hall coefficient is RH = Ey/jxBz = -1/ne

The Hall Effect

The Hall coefficient RH = E /j B = -1/neThe Hall coefficient RH Ey/jxBz 1/ne

The Hall angle is given by tan φ = Ey/Ex = ρH/ρg g y φ y x ρH ρ

For many metals RH is quiet well described by this expression which isuseful for obtaining the electron density, in some cases.

However the value of n obtained differs from the number of valenceHowever, the value of n obtained differs from the number of valenceelectrons in most cases and in some cases the Hall coefficient of ordinarymetals, like Pb and Zn, is positive seeming to indicate conduction by

iti ti l !positive particles!

This is totally inexplicable within the free electron model.This is totally inexplicable within the free electron model.

Drude model fails to predict Hall and conductivity for all metals except for alkaline metals

InAlAuAgKNaLiMetal

1/31/31 481 191 000 990 78R en -1/3-1/31.481.191.000.990.78-Rhen

Sign of Hall EffectHall Effect for free particles with charge +e ( “holes” )

Hall Effect for free particles with charge +e ( electrons )

Ex, jxEyBz

Ex, jx EB Bzvd

EyBzvd

E B BE B B Ey = +vxBz = - vd Bz

j = nev = ne v

Ey = +vxBz = vd Bz

j = nev = ne v jx = -nevx = ne vd

Ey = -jxBz/ne

jx = nevx = ne vd

Ey = jxBz/ne y jx z

RH = Ey/jxBz = -1/ne

y jx z

RH = Ey/jxBz = 1/ne RH Ey/jxBz 1/neRH Ey/jxBz 1/ne

Classical approach to heat ppcapacity of metals

Classical theory postulates kT/2 of energy per degree of freedom.degree of freedom.Heat capacity of metal=heat capacity of ion lattice (3R) + heat capacitance of electron gas.(3R) heat capacitance of electron gas. In practice all crystalline solids, metals and dielectrics alike, follow 3R rule (CP≈3R J/mole) at sufficientlyalike, follow 3R rule (CP 3R J/mole) at sufficiently high temperature.

Success and Failure of Free e-picture

Success Examples of Failure

–conductivity of monovalent metals

–Insulators, Semiconductors

–Conductivity of multivalent –Skin Depth Effect (conductivity at very high frequencies)

metals

–Thermoelectric effect–Wiedmann- Franz law ( in metals, thermal conductivity conductivity

–Colors of metals

is proportional to electrical conductivity)

Boltzmann statistics does not reflect energy distribution forBoltzmann statistics does not reflect energy distribution for electrons

Energy levelsEnergy levelsand

density of statesdensity of states.

Classical statistics treat energy levels as continuum.

Quantum mechanics postulates uncertainty principlep y p p

ΔxΔpx > h, which implies that energy levels are discreet.

Density of States:

Often, there many states with energies very close to each other, and we simply just want to know how many states, ΔN(E), for a given energy interval, ΔE, which of course depends on how densely the energy levels of the states are distributed. The density of depends on how densely the energy levels of the states are distributed. The density of states is defined as

.1)()(VE

ENEΔΔ

Δ=

ρ

i.e, the number of states per unit energy and per unit volume.

In one dimensional case density of states is defined as number of states per unit In one-dimensional case, density of states is defined as number of states per unit energy interval per unit length. In a similar way, the two-dimensional density of states is number of states per unit energy interval per unit area.

One-Dimensional Case: Let us consider a quantum well of width L, the energy is quantized and given byis quantized and given by

2

222

i 2E

Li π

= 22mL

The gap between ith and (i+1)th levels is

increases. Las 02

)12(E 2

22

i →π+

=ΔmL

i

L

ΔE ΔE ΔE

One-dimensional quantum well. As the width of the quantum well increases, the energy levels become more and more densely packed and it becomes necessary to energy levels become more and more densely packed, and it becomes necessary to use density of states to describe the energy levels.

How to determine density of states? An easier task is to calculate the number f f i l d d l h i l d If l i l of states for a momentum interval, dpx, and length interval, dx. If we plot a particle

momentum and position (px vs. x), each point represents a state of the particle. In classical physics, since a particle is allowed to take any values of momentum ( l it ) d iti t i t i th l t b bit il l d d it f (velocity) and position, two points in the plot can be arbitrarily close, and density of states is infinite! According to quantum physics, however, two points CANNOT be too close because of the uncertainty principle, ΔxΔpx > h. In other words, if we divide the entire p x plane into squares of area h each square can take point only divide the entire px-x plane into squares of area h, each square can take point only. So the number of states for an “area” of dpxdx in the p-x plane is

dxdpEN x=)( (1)

px px

hEN D =)(1

(1)

Δpx xxΔx

Δpx xx

Δx Δpx ~ h

dEmd 2/12

x )2(1pESi ⇒

dEdxEm

hEN

dEE

dpm

D

x

2/11

2/1x

)2(21)( 1, eq. toAccording

)(22

pESince

=

=⇒=

Eh2

1-D density of states is:

21 m 2/11 )2(

21)(

Em

hED =ρ (2)

22)( pdpdxdydxdydpdp

EN yxD

π⇒=

Two-dimensional case: The total number of states for a “volume” of dpxdxdpydy is

222 )(hh

EN D ⇒

mdEpdpm

pdpdEm

=⇒=⇒= pE Since2

2

mm2

222)(

hmdEdxdyEN D

π= (3)

222)(h

mEDπρ =2-D density of states is: (4)

Three-dimensional case: The total number of states for a “volume” of

3

2

334)( dpdxdydzpdxdydzdpdpdp

EN zyxD

π⇒=

dpxdxdpydydpzdz is

333 )(hh

EN D ⇒

2/12/328 dEdVE2

2/12/3

328)(

hdEdVEmEN D

π= (5)

3-D density of states is:

(6)3

2/12/3

328)(

hEmED

π=ρ

Energy dependence of d it f t t f 1 D density of states for 1-D (a), 2-D (b) and 3-D (c). The dependence is highly sensitive to the sensitive to the dimensionality of the system.

Schrödinger's equation for free electrong q

Consider electrons within a cube of metal of side of L.The shape does not matter, but the mathematics is simpler.

Must solve Schrödinger's equation with appropriateMust solve Schrödinger s equation with appropriateboundary conditions to obtain the allowed wavefunctions ψ(r).

2222 ⎞⎛ ∂∂∂ )r())rV( - (E = )r(z

+ y

+ x2m

-2

2

2

2

2

22

ψψ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

∂∂

∂∂

First set V(r) ≡ 0 since the electrons are considered to be free.

We could now set ψ(r) = 0 at the cube surface This gives the correctWe could now set ψ(r) = 0 at the cube surface. This gives the correctenergy states. Instead we choose Periodic Boundary Conditions i.e.we set the value of the wavefunction at opposite faces to be equalii.e.

Allowed k-statesPeriodic Boundary Conditions i.e. ψ(x, y, z+L) = ψ(x, y, z)

ψ(x, y+L, z) = ψ(x, y, z)ψ(x+L, y, z) = ψ(x, y, z)ψ y ψ y

Useful because the solutions are now travelling waves.

The unnormalised allowed states are ψ(r) = exp[ik.r] where

nL

2 = k ; nL

2 = k ; n L

2 = k zzyyxxπππ

and nx, ny, and nz = +/- 1,2,3,4……

The allowed k-states are uniformly spaced in k-space with

One state per ol me (2π/L)3 in k spaceOne state per volume (2π/L)3 in k-space.

k- space y pypFree Classical Electrons states x px

Defined by position (x,y,z) and momentum (px, py, pz) z pz

Free Quantum Electrons states kUniquely determined by the wavevector, k. Or equivalently by (px, py, pz) = ( kx, ky, kz).

ky

(px, py, pz) ( kx, ky, kz).

Equal probability of electron being anywhere in conductor. kx

Electron state defined by a point in k space

g y

a point in k-space kz

Lk-spaceky

2π/L

Each dot

p

ac dotrepresents and allowed k-state

kxAllowed k statesAllowed k-states are uniformly spaced in k-space.p p

With one state

Th k t t di t b t f l i d d t

per volume (2π/L)3

The k-states are discrete but for a normal sized conductorthere are ~1026 states so we can treat them as a continuum.

LDensity of State D(k)

2 /L

The density of states in k-space,i.e. the number of available statesbetween k and k+dk is the number

⎟⎠

⎞⎜⎝

⎛2

3/2

2

2mE

6V

2 = Nπ

kx2π/Lbetween k and k dk is the number

of states contained between asphere of radius k and a sphere ofradius k+dk This is

dkVk = 8V . dkk = dkkD

3π

24)(

22

radius k+dk. This is

kx

8 3 ππ 2

Number of k-statesper unit volume

Volume in k-spaceper unit volume

This expression for the density ofstates in k-space is true for all bulkstates in k space is true for all bulkelectron states in solids and notjust free electrons as we will showlater dD 2later.

dk2Vk = dk

dkdD = D(k)dk

2

2

π

Density of State D(E)For every k-state there is acorresponding energy state. (E

) dE

corresponding energy state.From the exclusion principleeach state can accommodatet o electrons of opposite spin

DE)

dE

two electrons of opposite spin.

D(E

The density of states, i.e. the number ofavailable states between E and E+dE is

Energy

available states between E and E+dE is

dEE2m

2V =

dEdN = D(E)dE

2

3/2

221

⎟⎠⎞

⎜⎝⎛

Thi i l t f f l t i h d

2dE 22 ⎠⎝π

This is only true for free electrons since we have assumedE = 2k2/2m.

Number of states per unit volume between two energies:

∫2

)6()(E

dEEρ

Number of states per unit volume between two energies:

1E

Example: The number of states available in the energy range from 2 to 3 eV in a Example: The number of states available in the energy range from 2 to 3 eV in a metal whose volume is 1 mm3.

∫eV32/13eV32/13eV3 2)2()2(2 mmE ⎤⎡

∫ ∫ ∫eV2

2/332

eV232

eV2 32)2()2()()(

2

1

EmdEEmdEEdEEE

=⎥⎦⎤

⎢⎣⎡

π=

π=ρ=ρ

2/3192/31956eV3

eV2

2/332

2/13

)106.12()106.13(321006.1

32)2( Em

=⎥⎦⎤

⎢⎣⎡ ××−×××=⎥⎦

⎤⎢⎣⎡

π= −−

319328 mm1007.11007.1 perstatesmperstates ×=×=

Fermi-Dirac distribution functionThe Density of States tells us what states are available. We now wishto know the occupancy of these statesto know the occupancy of these states.

Electrons obey the Pauli exclusion principle. So we may only have two electrons (one spin-up and one spin-down) in any quantum state.

E)f(

E

1The function giving the probability of occupation of a particular state is the Fermi-Dirac

E)For T=0 all states are occupied

pdistribution function, f(E).

E0

EF

f(E

up to an energy, called the Fermi energy, EF , and all statesabove E are empty EEF

Fermi-Dirac function for T=0.above EF are empty.

Classical and Quantum Statistics.

Let us consider a large number of particles in equilibrium at a temperature T.

The Nature of Particles:I l i l h i t id ti l ti l b di ti i h dIn classical physics, two identical particles can be distinguished

by, say, following their paths. In quantum mechanics, this seeming simple task CANNOT be carried be accomplished seeming simple task CANNOT be carried be accomplished because the uncertainty principle that does not allow us to follow the paths the particles – so identical particles are indistinguishable!

This indistinguishability has profound consequences on the behaviors of particles such as electrons in semiconductorsbehaviors of particles such as electrons in semiconductors.

Fermions and Bosons: In quantum physics, particles are divided into two q p y , pcategories, fermions and bosons, according to their spins that can be either integral values and half-integral values (in units of ). Particles with integral spin values are bosons while those with half-integral spin areintegral spin values are bosons, while those with half-integral spin are fermions. Particles such as electrons, protons and neutrons are fermions. Particles like photons, mesons are bosons. When two electrons bound together to form a composite particle it becomes a boson and it istogether to form a composite particle, it becomes a boson, and it is responsible for superconductivity. An atom can be regarded as a composite particle. Whether it is fermion or boson depends on the total spin.

Homework: Is your soccer ball a fermion or boson?

Fermions obey the Pauli Exclusion Principle, i.e., no two identical fermions can occupy the same state. However, bosons DO NOT obey Pauli Exclusion Principle. Any number of bosons can occupy a given state!p y py g

Basic task of Quantum and Classical Statistics is to determine the distribution function of a large number of particles in various states at thermal equilibrium To find function of a large number of particles in various states at thermal equilibrium. To find the distribution function, we need to consider the following two requirements:

• The distribution should be such that the free energy of the system is minimum;• The distribution should be such that the free energy of the system is minimum;• The distribution should respect special properties of the particles (fermion or boson).

The first requirement says that the free energy F whereThe first requirement says that the free energy, F, where

F=U-TS should be minimized. Here U is the total energy of the system of particles and the S is gy y pthe entropy of the system. Entropy measures the disorder of the system which has a tendency to increase (2nd law of thermodynamics). So minimizing F is the requirement of minimizing the energy, U, and maximum the entropy , S (note the negative sign). g gy py g gWe can see that minimization of U would like to produce an ordered system, where all particles occupy the lowest possible energy, while entropy term would like the particle to distribute as randomly as possible. The entropy is given by

S= kBlnW

where W is the total number of ways the particles can be distributed in all the states.

The second requirement is that we have to consider whether the particles are indistinguishable fermions and bosons or distinguishable classical particles. In the case indistinguishable particles, the Pauli exclusion principle prohibits two or more fermi particles from occupying the same states, while no such constraint for bose particles. Correspondingly, we have three type of distribution functions, Fermi-Dirac distribution for fermions, Bose-Einstein distribution for bosons, and Maxwell-Boltzman distribution for classical distinguishable particle.

Fermi-Dirac Distribution FunctionIf we have a system with allowed energy levels, ε1, ε2, ε3, …, and corresponding d 1 2 3 Th l f N i l h Th degeneracy g1, g2, g3, …. There are a total of N particles to occupy these states. The key in determining the distribution function of such a system is to calculate W, the total number of ways the particles can be distributed in all the states. Let us consider di t ib ti th t h N1 ti l i t t 1 N2 ti l i t t 2 d T a distribution that has N1 particles in state ε1, N2 particles in state ε2, and so on. To

determine W of such a distribution, we can first find the ways N1 particle can be distributed in g1 states of the energy level ε1:Particle 1 can be placed in g1 waysParticle 1 can be placed in g1 ways.Particle 2 can be placed in (g1-1)/2 because of the argument below:Since particle 1 has already occupied one of g1 states, particle 2 can occupy the remaining g1-1 states only due to Pauli exclusion principle In addition particles 1 remaining g1 1 states only due to Pauli exclusion principle. In addition, particles 1 and 2 are indistinguishable, so the total ways are reduced by a factor of 2.

Particle 3 can be placed in (g1-2)/3, based on the similar consideration given above.…….Particle N1 can be placed in (g1-N1+1)/N1.The total number for N1 particles in ε1 states is thus

)!11(!1!1

!1)111)...(21)(11(11

NgNg

NNggggP

−=

+−−−=

Similarly the total number for N2 particles in ε2 states is

)!22(!2!2

!2)112)...(22)(12(22

NgNg

NNggggP

−=

+−−−=

The total number of ways for the given distribution is

!∏ −=

i iii

i

NgNgW

)!(!!

(7)

The first requirement says the free energy must minimized The first requirement says the free energy must minimized,

0)!(!

!ln)()(0 =−=−⇒= ∏∑ iBii NN

gTkNTSUF εδδδ (7a))!(!

)()(−∏∑

i iiiBi

ii NgN

!We note that,

)!ln(!ln!ln)!(!

!ln iiiiii iii

i NgNgNgN

g−−−=

− ∑∏

xxxx −≈ ln!lnEquation (7a) becomes,Equation (7a) becomes,

0)]ln(ln[ =−+−−∑ iiiiBi

ii NNgNTkN δδε (8)

Since the total number of particles is fixed, or

0=⇒= ∑∑ ii NNN δ (8a)∑∑ii

We multiply Eq. (8a) by a constant –A and then add it to Eq. (8) to obtainWe multiply Eq. (8a) by a constant A and then add it to Eq. (8) to obtain

0)]ln(ln[)( =−+−−−∑ NNgNTkA iiiiBi δε

)ln( −=

−⇒

TkA

NNg

B

i

i

ii

i

ε

1)exp( +−

=⇒ AgN

TkN

i

ii

Bi

ε (9)1)exp( +

TkB

The occupation probability f (or distribution function) for a state with energy ε is

1)exp(

1)()(),(

+−

==

TkAg

NTf εεεε

TkB

When ε=A, f=1/2, and A is often denoted as EF, called Fermi Energy, so

1)(f1)exp(

1),(+

−=

Tk

Tf

B

Fεεε (10)Fermi-Dirac Distribution

Bose-Einstein Distribution: For bosons, the particles are indistinguishable, but p goccupy the same states, we have derive the distribution function in a similar way, which is called Bose-Einstein distribution,

1)( Tf ε1)exp(

),(−

−=

TkATf

B

εε(11)

Classical Maxwell-Boltzman Distribution: For classical particles, they are distinguishable, the corresponding distribution function is:g p g

)exp(),(Tk

CTfB

εε −= (12)B

It is easy to tell whether we should use Fermi Dirac distribution or Bose EinsteinIt is easy to tell whether we should use Fermi-Dirac distribution or Bose-Einstein distribution, but when should we use the classical Maxwell-Boltzman distribution? The answer is when the de Broglie wavelength is much smaller than the average distance between particles orbetween particles, or

.d<<λ

Electrons in Metals.F Q El d lFree Quantum Electron gas model.

Sommerfeld theory

The (Quantum)Free Electron model: ( )Assumptions

(i) FREE ELECTRONS Th l l t t ff t d b(i) FREE ELECTRONS: The valance electrons are not affected bythe electron-ion interaction. That is their dynamical behaviour is asif they are not acted on by any forces internal to the conductor.

(ii) NON-INTERACTING ELECTRONS: The valence electron from a`gas' of non-interacting electrons. That is they behave asg g yINDEPENDENT ELECTRONS that do not show any `collective'behaviour.

(iii) ELECTRONS ARE FERMIONS: The electrons obey Fermi-Diracstatistics.

(iv) ‘Collisions’ with imperfections in the lattice limit the electricalconductivity. This is considered in the relaxation time

i tiapproximation.

The quantum free-electron gas that follows Fermi-Di di t ib tiDirac distribution.

1]/)exp[(1),(

+−=

TkEETEf

+ + + +

1]/)exp[( +− TkEE BF

+ + + +

+ + + +

Metal that contains free electrons Free electron approximationwondering around positive cores.

Free-electron approximation

• This distribution function, as we have discussed previously, it gives the This distribution function, as we have discussed previously, it gives the probability that a state of given energy is OCCUPIED by an electron at a given temperature.

• By definition the probability of the same state NOT being occupied by an y p y g p yelectron is therefore just 1 – f(E).

• If T=0, f(E) is a simple step function with an edge at EF, i.e., all the states with energies below the Fermi energy EF, is completely occupied, and all the states with energies above EF is completely vacant.

111)0( ===< EEf 111]exp[

)0,( ==+−∞

=< FEEf

11

f(E)

1

011]exp[

1)0,( =∞

=++∞

=> FEEf

0EE

½

21

1]0exp[1)0,( =

+== FEEf

AT ABSOLUTE ZERO THE FERMI-DIRAC DISTRIBUTION FUNCTION YIELDS EXACTLY THE FORM WE SPECULATED

EARLIER

EEF

The effects of temperatureAt a temperature T theprobability that a state is

i d i i b th F i

Fermi-Dirac distribution function

p

occupied is give by the Fermi-Dirac function

⎞⎛ ⎞⎛-1

με

for TF =50,000K,approximate Fermi temperature of Copper

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛ 1 + Tk

- = )f(B

μεε expCopper.

h i th h i lwhere μ is the chemical potential. For kBT << EF μ is almost exactly equal to EF.

• At higher temperatures we have already mentioned that electrons should move to higher energy states in order to accommodate their INCREASED thermal energy

* This process is also described by the Fermi-Dirac distribution whose step edge develops a ROUNDED form with increasing temperature

⇒ The rounding indicates that certain states that were filled at absolute zero EMPTY as these electrons move to occupy HIGHER energy states

(E) d

E kBT

The thermal energy only

N( T=0

n(E)

dE

The thermal energy onlychanges the occupation ofavailable electron states ina range ~k T about E

EEF

T>0

na range ~kBT about EF.

• The temperature dependence of the Fermi-Dirac distribution function reveals that the main effect of temperature is to change the occupation of states NEAR the Fermi level

* When the temperature is increased states well below the Fermi level therefore remain filled while states well above the Fermi level remain empty

* Quite generally it can be shown that the main effect of temperature is to change stateoccupation in a NARROW window of width 3.5kBT that is centered on the Fermi energy

11( ) 27 %1 1f E E k T

F B e= + = ≈

+1( 3 ) 5%3 1

f E E k TF B e

= + = ≈+

1( 5 ) 1%5 1f E E k T

F B e= + = ≈

+

THE MAIN EFFECT OF FINITE TEMPERATURE IS TO MODIFY THE FILLING OFTHE MAIN EFFECT OF FINITE TEMPERATURE IS TO MODIFY THE FILLING OFSTATES WITHIN A FEW kBT OF THE FERMI SURFACE

The density of occupied states is the density of states that are actually occupied by y p y y p yelectrons, which is simply given by

)()()( EfEEocc ρ=ρ (13))()()( focc ρρ

Fermi-Energy: Fermi energy is an important parameter, which can be determined by the following method. If the total number of free electrons in a metal is N, then we have

⎮⌠

⎮⌠

∞∞

1)()()( dEEVdEEfEVN

g ,

(14)⎮⎮

⌡⎮⌡⌠

+−ρ=ρ=

00

1]/)exp[()()()( dE

TkEEEVdEEfEVN

BF

(14)

When kBT<<EF, or T<< EF/kB, f(E) is a step function. For metals, EF is a few eV and, EF/kB ~ 10000 K! So kBT<<EF holds even when the metal melts. Replacing f(E) with a step function, we have

⎮⌡

⌠ π=

π=ρ= ∫

FF

E

F

E

Emh

VdEEmh

VdEEVN 2/333

33 28

3228)(

(15)⎮⌡

∫ hh0

30

3 3

From Eq 15 we haveFrom Eq. 15, we have

[ ] 3/222

32

nEF π= (17)[ ]2mF

where n=N/V is the number density of electrons (number of electrons per unit volume).

[ ] 3/123/222

21 [ ] 3/123/222 )3(322

1 nm

vnm

Emv FFF ππ =⇒== (18)

It is important to note that even at 0K, the velocity of the highest occupied state is vF, not zero.

The Fermi Metals have a Fermi energy E

SurfaceMetals have a Fermi energy, EF.

The Fermi Temperature,TF, is thetemperature at which kBTF = EFtemperature at which kBTF = EF.

All the free electron states withina Fermi sphere in k-space area Fermi sphere in k space arefilled up to a Fermi wavevector,kF.

The surface of this sphere isThe surface of this sphere iscalled the Fermi surface.

On the Fermi surface the freeelectrons have a Fermi velocity vF= hkF/me.

A Fermi surface still exists when the states are not freeA Fermi surface still exists when the states are not free electron states but it need not be a sphere.

Example: A metal with an electron density of 5×1028 m-3 has the density of occupied states at an energy of 5 eV and at a temperature of 0 and 300 K given by:

[ ] eV96.432

3/222

== nm

EF π

00eV)5(K)0eV5(eV)5(K)0eV5( =×=×= ρρρ f

K)300eV,5(eV)5(K)300eV,5( focc ×= ρρ

00eV)5(K)0eV,5(eV)5(K)0eV,5( =×=×= ρρρ fOcc

314646 mJ1066.118.01048.9 −−×=××=[ ] 1/exp

122

1 2/12/3

22 +−×⎥⎦

⎤⎢⎣⎡=

TkEEEm

BFπ

ELECTRONS MOVE FROM LOWER ENERGY STATES WHEN THE TEMPERATURE IS INCREASED!WHEN THE TEMPERATURE IS INCREASED!

Internal energy and heat capacitance of free electron gas.

∫∫∞∞ π 2/12/3281)()( EdEEmEdEEEfU

Internal energy, U, (Helmholz energy) of electrons from (6) and (10)

∫∫π

+−=ρ=

03

0

81]/)exp[(

)()( EdEhm

TkEEEdEEEfU

BF

∫∞π 2/32/328 Em ⎟

⎞⎜⎛ ⎞⎛π∞ 2252/3 2 TkE

(19)

∫ +−π

=0

3 1]/)exp[(28 dE

TkEEE

hmU

BF⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛π+≈

+−∫ 2

0 452

1]/)exp[( F

BF

BF ETkEdE

TkEEE

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛π+

π=

2225

3

2/3

45228

F

BF E

TkEh

mU⎟⎠

⎜⎝ ⎠⎝45 FEh

TTkEmTUC BFV γ==

∂∂

= 23

2/3

2Heat capacitance is => Using (17)

T BFV γ∂ 32

( ) TTknmTnmC BV γ=π= 231

22 3

2),,(

g ( )

( ) BV γ22)(

For n=1023 cm-3 and T=300 K CV=27 kJ/(m3K) ≈ 0.16 J/(mole K)

Thermodynamic potential of free electron gas.

Assuming that Eq.(19) can be transformed in⎢⎣

⎡><

=f

f

EEifEEif

Ef01

)(

35

FFFF

E

nEEh

Emh

EmEdEh

EmUF

53

5216

521628

3

23

2/3

3

25

2/3

03

2/12/3

=π

=π

=π

= ∫ hhh 5550

potentialchemicalEU==

∂ 3 potentialchemicalEn F ==

∂ 5

Fermi energy is fully equivalent to chemical potential of electronselectrons

Electronic specific heat capacityConsider a monovalent metal i.e. one in which the number of freeelectrons is equal to the number of atoms.

If the conducting electrons behaved as a gas of classical particlesthe electron internal energy at a temperature T would be

U (k T/2) ( b f d f f d 3)U = (kBT/2) x n x (number of degrees of freedom = 3)

So the specific heat at constant volume CV = dU/dT= 3/2nkB.

At room temperature the lattice specific heat, 3nkB ( n harmonicoscillators with 6 degrees of freedom).

In most metals at room temperature C is very close to 3nkIn most metals, at room temperature, CV is very close to 3nkB.

The absence of a measurable contribution to CV was historicallythe major objection to the free classical electron model.the major objection to the free classical electron model.

If electrons are free to carry current why are they not free toabsorb heat energy? The answer is that they are Fermions.

Electronic Specific heatThe total energy of the electrons perm3 in a metal can be written as

dE

kBT

E = Eo(T=0) + ΔE(T).

Where Eo(T=0) is the value at T=0.

N(E

) d

T=0

Eo( )

T>0

n(E)

dE

At a temperature T only thoseelectrons within ~ kBT of EF have a

EEF

T>0

The number of electrons that increase their energy is

greater energy that they had at T=0.

The number of electrons that increase their energy is

~n(kBT/EF) where n is the number of electrons per m3

nkTk 2

Each of these increases its energy by ~kBT

T Enk = Tk.

ETk n E 2

F

BB

F

BΔ∴

)( 31 −−Δ mJKT . Enk2

dTE)d( =

dTdE = C

F

2B

elThe electronic specific heat is therefore

)mJK(.Tnk=C 312B

2

el−−πA full calculation gives )mJK(.T

E2CF

el

It is useful to express this in terms of TF

)mJK(TT

nk2 = C 31

FB

2

el−−π

For a typical metals this is ˜ 1% of the value for a classical gas ofelectrons E g Copper (T/T ) ~ 300/50 000 = 0 6%electrons. E.g. Copper (T/TF) ~ 300/50,000 = 0.6%

At t t th th h t ib ti d i tAt room temperature the the phonon contribution dominates.

Low Temperature Specific Heatp p)( 31 −−= mJKT.T

E2nk = C

2B

2

el γπPredicted electronic specific heat

At low temperature one finds that

E2 F

T + T = C 3V αγ where γ and α are constants

The first term is due to the electrons and the second to phonons. The linear T dependent term is observed for virtually all metals. H h i d f b diff f h fHowever the magnitude of γ can be very different from the free electron value.

1 1 1 1Metal γcalc. (JK-1 mol-1) γexpt(JK-1mol-1)Cu 5 × 10-7 7 × 10-7

Pb 1 5 × 10-6 3 × 10-6Pb 1.5 × 10 6 3 × 10 6

Dynamics of free quantum electronsClassical free electrons F = -e (E + v × B) = dp/dt and p =mev .

Quantum free electrons the eigenfunctions are ψ(r) = V-1/2 exp[i(k.r-ωt) ]The wavefunction extends throughout the conductor.

Can construct localise wavefunction i.e. a wave packets

The velocity of the wave packet is

)]t -exp[i(k.rA (r)k k ω∑=Ψ

The velocity of the wave packet is the group velocity of the waves

dEd 1ω pk

The expectation value of the momentum of the wave packet responds to a

for E = 2k2/2mekk

vddE

dd 1

==ω

ee mmpkv ==

The expectation value of the momentum of the wave packet responds to a force according to F = d<p>/dt (Ehrenfest’s Theorem)

Free quantum electrons have free electron dynamics

Conductivity & Hall effectConductivity & Hall effectFree quantum electrons have free electron dynamics

Free electron expressions for the Conductivity, Drift velocity, M bilit & H ll ff t t f t l tMobility, & Hall effect are correct for quantum electrons.

Current Density j = -nevd

Conductivity σ = j/E = ne2τp /me

Mobility μ = vd / E = eτp /me

Hall coefficient RH = E /j B = -1/neHall coefficient RH = Ey/jxBz = -1/ne

Will now consider some consequences of the electrons being Fermions on scattering of electrons.

(a) F=0 (b) F

Drift velocity: ( ) ( )

k-state picture

(i) An electric field E produces a drift velocity vd = -eEτ/me i.e. the electrons acquire a net momentum p = -eEτ

(ii) For free electrons p = k so this net momentum corresponds to shifting all the states in k-space by δk = -eEτ/ p y

ScatteringMobile electrons move ~107 lattice constants between scattering events in “pure” metals at low temperature.

Assume that electrons are not scattered by the periodic potential of a perfect lattice.

The scattering of electrons is due to defects, impurities and phonons (lattice vibrations).(lattice vibrations).

Collisions between mobile electrons do not contribute to the resistivity. Such collisions do not attenuate an electric current since the totalSuch collisions do not attenuate an electric current since the total momentum of the electrons is unchanged.

j d f i l ∑j = -nevd and for i electrons vd = ∑i

ivv3v1

El t l t lli iv4v2

Collision

Electron-electron collisions only redistribute momentum

Elastic & Inelastic CollisionsScattering of electrons is due to defects, and impurities.

Consider an electron colliding with a stationary atom, mass 30 amu.m m MMvi

Before

vf

u

After

me me MM

Before After

4ee2f

2ifi 10m4Mm4vvEE −−−

Check this!

If Ei is the initial electron energy and Ef is the final then

4e

e

e2i

fi

i

fi 10M)Mm(vE

≈≈+

== Check this!

Scattering of electrons is due to phonons (lattice vibrations).

Electron-phonon scattering involves an electron absorbing or emitting a phonon which has energy ~ kBT ~25meV @ 300K.

Typically EF~ 10eV. So for an electron at the Fermi surface the fractional change in energy is only ~ 2.5x10-3

Scattering of electrons is quasi-elastic

fractional change in energy is only 2.5x10

Scattering only at the Fermi surface(a) (b)

Electrons can only be scattered into available empty states.

Scattered electrons change energy by ΔE << EF during collisions.

Therefore only electrons with energy close to EF can be scattered

The mean free path for these electrons is Λ = vF τp

Use the momentum relaxation time, τp, at the Fermi surface in the free electron i f th l t i l d ti itexpression for the electrical conductivity.

Scattering ProbabilitiesThe dominant energy dependence of the probability of an electron being scattered comes from that of the Fermi function

Scattering almost elastic

Probability of an electron 1f(E) (1- f(E))

Probability of an electron being scattered from an occupied state of energy E to an unoccupied state ofto an unoccupied state of energy E is proportional to

f(E)*(1- f(E)) f(E)*(1- f(E))

( ) ( ( ))

Probability of a state being

Probability of a state beinga state being

occupied0 1 2

0

E/EF

a state being empty

Only have scattering FFermi functions corresponding to 2,400K for Copper (TF= 80,000K)

Only have scattering for states close to EF

Which Electrons Carry Current ?C d i j l l d iCurrent density j = nvde, n = total electron density.

Drift velocity corresponds to a shift of all states by δk in k-space.F i f ti T 2 400K T 80 000Kj = n δk e/m ( where mevd = δk )

111n(kx , j = 0) n(kx , j > 0)

Fermi functions T = 2,400K; TF= 80,000K

Consider states with ky = kz = 0D it f t t D(k ) i

n(k x

)Density of states D(kx) is aconstant for these states.

n(k ) = f(k ) D(k ) 000Fermi Dirac function in terms of kx

δk

n(kx) = f(kx) D(kx)

-1-1-1menkndkk

menkdk

mej FF ∫∫

+∞+∞

∞−

=Δ≈Δ=0

2. δ Δn = n(kx , j > 0) - n(kx , j = 0)

-2 -1 0 1 2kx/kF

-2 -1 0 1 2kx/kF

-2 -1 0 1 2kx/kF

j > 0; excess of electrons with k k and deficit of electrons with k k

c.f. j = n δk e/m

j > 0; excess of electrons with k~kF and deficit of electrons with k~ -kF.

Can consider the current to be carried by a fraction of electrons with E ~ EF.

Scattering of electrons by phononsExperimentally find ρ = ρ0 + ρL (Τ) ( Matthiessen’s rule). ρ0 is due todefects and impurities & ρL (T) is due to phonons.

Experimentally find ρL (T) proportional to T for T > TD

ρL (T) proportional to T5 for T << TDρL (T) proportional to T for T << TD

where the typical Debye temperature, TD, ~ 300K for metals.

Phonons have energy ω, and wavevector q where q ~ ω/vs where the typical sound velocity, vs is ~3000 ms-1

Scattering by phonons must conserve momentum (wavevector), Free electrons momentum p = k. Phonons momentum p = q

Phonon absorption : ki & kf initial & final electron wavevectors.q phonon wavevector kkkhkkq phonon wavevector.

Fifif kkkwhereqkk ≈=+=

Phonon scattering for T TD≥For T TD the number of phonons increases as T.≥

For T TD average phonon energy is ω ~ kBTD ~constant. ≥

T ~ TD ~ 300Kq ~ ω/vs ~ kBTD/ vs = 1010 m-1 ~ kF. kF

q

φ

Average scattering event randomises momentum

q s B D s F

kF

F φ

Average scattering event randomises momentum.

Average scattering angle φ ~ π/2

qkk +Scattering time, τs = Momentum relaxation time, τp.

Scattering rate, τs-1 = Momentum relaxation rate, τp

-1 .

qkk if +=

Resistivity proportional to number of phonons i.e. ρ proportional to T

Phonon scattering for T << TD

For T ~ 1 K; q ~ ω/vs ~ kBT/ vs ~ 108 m-1 << kF ~ 1010 m-1

T << TD number of phonons increases as T3. Average phonon energy 3kBT.

Average scattering angle φ << π/2 therefore it takes many scattering event to randomises the initial momentum.event to randomises the initial momentum.

Scattering rate, τs-1 >> Momentum relaxation rate, τp

-1.

An average scattering event changes thekF

q

An average scattering event changes thecomponent of the momentum of the electronin the original direction by a factor

kF

qφ

ΔkCheck this!Δk/kF = q2 /2kF

2 ~ T2

qkk if +=τp-1 = Δk/kF τs

-1

Resistivity proportional to (number of phonons x τp−1 ) i.e. to (T3 x T2) = T5

Cyclotron resonanceFree electrons in a magnetic field Bmove in circular trajectories with amove in circular trajectories with acyclotron radius rc & angular frequency ωc

ym ωc2 rc = Bev = Be ωc rc So ωc = eB/m

xrc

Mobile electrons can resonantly absorbenergy from an electromagnetic wave withangular frequency ω = ω Such Cyclotron vangular frequency ω = ωc. Such CyclotronResonance measurements give the massof the electrons.

Cyclotron resonance absorption is normally only observed at highmagnetic fields i.e. at high ωc. To get significant absorption thecprobability of a carrier executing one cyclotron orbit must besignificant. Find that one requires ωcτs> 1.

k-space motion in a magnetic fieldFree electrons BkBvF ×−=×−= )/( emee

The electrons move in circles in real space and in k-space.

The Lorentz force does not change the energy of the electrons.

Th l t i i l d th F i fThe electrons move in circles around the Fermi surface.kyy

x k

krc x kx

v

BzBz

Electronic Thermal ConductivityTreat conduction electrons as a gas. From kinetic theory the thermalconductivity, K, of a gas is given byy, , g g y

K = ⊄vCv/3 where v is the root mean square electron speed, ⊄ is theelectron mean free path & C is the electron heat capacity per m3electron mean free path & Cv is the electron heat capacity per m

For T<< TF we can set v = vF and Λ = vFτp and Cv = (π2/2) nkB (T/TF)F F F p v ( ) B ( F)

So K = π2nk2BτpT/3m now σ = ne2τp/m

Therefore K /σT = (π2/3)(kB/e)2 = 2.45 x 10-8 WΩK-2 (Lorentz number)

The above result is called the Wiedermann-Franz law

Measured values of the Lorentz number at 300K are Cu 2 23 In 2 49 PbMeasured values of the Lorentz number at 300K are Cu 2.23, In 2.49, Pb2.47, Au 2.35 x 10-8 WΩK-2 (very good agreement)

Heat currents and energy relaxation EFenergy relaxation

T + dT T

In a temperature gradient there is a net flow of hotter electrons towardscooler regions and cooler electrons to hotter regions.

T + dT T

cooler regions and cooler electrons to hotter regions.

In a metal an electronic heat current corresponds to a net drift ofelectrons with E > E in one direction and a net drift of electrons with E <electrons with E > EF in one direction and a net drift of electrons with E <EF in the opposite direction.

The energy flow is disrupted by any scattering event that converts a hotelectron into a cool electron i.e. changes the energy of an electron by ~kBT.

Energy relaxation and deviationsEnergy relaxation and deviations from the Wiedermann-Franz law

At high temperatures phonon scattering changes an electron’smomentum by ~kF and energy by ~kBT.momentum by kF and energy by kBT.

Phonon scattering is therefore equally effective at relaxingmomentum and energy and the Wiedermann Franz law is correctmomentum and energy and the Wiedermann-Franz law is correct.

At low temperatures phonons are ineffective at relaxing theelectrons momentum but still effectively relax the electrons energyand attenuate any heat current.

The Wiedermann-Franz law is therefore incorrect at lowtemperatures in pure metals at low temperatures.

Free electron model: SuccessesFree electron model: SuccessesIntroduces useful idea of a momentum relaxation time.

Give the correct temperature dependent of the electronic ifi h tspecific heat

Reasonable agreement with the observed Wiedermann FranzReasonable agreement with the observed Wiedermann-Franz Law

Observed magnitudes of the electronic specific heat and Hall coefficients are similar to the predicted values in many metal

Indicates that electrons are much more like free electrons than one might imaginethan one might imagine.

Free electron model: FailuresElectronic specific heats are very different from the free electron predictions in some metalselectron predictions in some metals

Hall coefficients can have the wrong sign indicating that theHall coefficients can have the wrong sign indicating that the electron dynamics can be far from free.

Masses obtained from cyclotron resonance are often very different from free electron mass and often observe multiple absorptions (masses) More than one type of electron?!absorptions (masses). More than one type of electron?!

Does not address the central problem of why some materialsDoes not address the central problem of why some materials are insulators and other metals.