Chapter 9: Electronic Transport

Onsager

April 2, 2017

Contents

1 Quasiparticle Propagation 3

1.1 Quasiparticle Equation of Motion and Effective Mass . . . . . . . . . 6

2 Currents in Bands 9

2.1 Current in an Insulator . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.2 Currents in a Metal . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Scattering of Electrons in Bands 13

4 The Boltzmann Equation 18

4.1 Relaxation Time Approximation . . . . . . . . . . . . . . . . . . . . . 21

4.2 Linear Boltzmann Equation . . . . . . . . . . . . . . . . . . . . . . . 23

5 Conductivity of Metals 24

5.1 Drude Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.2 Conductivity Using the Linear Boltzmann Equation . . . . . . . . . . 25

6 Thermoelectric Effects 30

6.1 Linearized Boltzmann Equation . . . . . . . . . . . . . . . . . . . . . 31

6.2 Electric Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

1

6.3 Thermal and Energy Currents . . . . . . . . . . . . . . . . . . . . . . 34

6.4 Seebeck Effect, Thermocouples . . . . . . . . . . . . . . . . . . . . . 38

6.5 Peltier Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7 The Wiedemann-Franz Law (for good metals) 41

2

As we have seen, transport in insulators (of heat mostly)

is dominated by phonons. The thermal conductivity of some

insulators can be quite large (cf. diamond). However most

insulators have small and uninteresting transport properties.

Metals, on the other hand, with transport dominated by elec-

trons generally conduct both heat and charge quite well. In

addition the ability to conduct thermal, charge, and entropy

currents leads to interesting phenomena such as thermoelectric

effects.

1 Quasiparticle Propagation

In order to understand the transport of metals, we must un-

derstand how the metallic state propagates electrons: ie., we

must know the electronic dispersion ω(k). The dispersion is ob-

tained from band structure E(k) = h̄ω(k) in which the metal

is approximated as an almost free gas of electrons interacting

weakly with a lattice potential V (r), but not with each other.

The Bloch states of this system

φk(r) = Uk(r)eik·r, Uk(r) = Uk(r + rn) (1)

3

eV

V(r)

Figure 1: The dispersion is obtained from band structure E(k) = h̄ω(k) in which the

metal is approximated as an almost free gas of electrons interacting weakly with a

lattice potential V (r), but not with each other.

may be approximated as plane waves Uk(r) = Uk. Then, the

state describing a single quasiparticle may be expanded.

ψ(x, t) =1√2π

∫ ∞−∞

dkU(k)ei(k ·x −ω(k)t) (2)

If U(k) = cδ(k− k0) then ψ(x, t) ∝ ei(k0·x−ωt) and the quasi-

particle is delocalized. On the other hand, if U(k) = constant

Re ψ(x)

Figure 2: If U(k) = cδ(k − k0) then ψ(x, t) ∝ ei(k0·x−ωt) and the quasiparticle is

delocalized.

then ψ(x, t) ∝ δ(x) and the quasiparticle is perfectly localized.

This is an expression of the uncertainty principle

∆k ∆x ∼ 1 or ∆p ∆x ∼ h̄ , (3)

4

so that we cannot know both the momentum and location of

the quasiparticle to arbitrary precision.

ω(k) 6= constant, so the different components propagate with

different phase velocities, so the quasiparticle spreads as it prop-

agates. This is also the reason why the group velocity of the

quasiparticle is not the phase velocity. Consider the propaga-

tion of ψ(x, t) which when t = 0.

ψ(x, 0) =1√2π

∫ ∞−∞

dkU(k)eik·x . (4)

Suppose that U(k) has a well-defined dominant peak (See Fig.

U(k)

kk0

Figure 3: The distribution of plane wave state that make up a quasiparticle.

3) so that

ω(k) ' ω(k0) + ∇kω(k)|k0 · (k− k0)t (5)

5

then

ψ(x, t) ' 1√2π

∫ ∞−∞

dkU(k)ei(k·x−ω0t−∇kω(k)|k0(k−k0)t) (6)

ψ(x, t) ' ei(k0·∇kω(k)|k0−ω0)t√

2π

∫ ∞−∞

dkeik·(x−∇kω(k)|k0t)U(k)

' ψ(x− ∇kω(k)|k0 t, 0

)ei(k0∇kω(k)|k0−ω0)t (7)

Ie., aside from a phase factor, the quasiparticle travels along

with velocity ∇kω(k)|k0 = vg. (If we had considered higher

order terms, we would have seen the quasiparticle distorts as it

propagates. (c.f. Jackson p.305). In general,

vg = ∇kω(k) =1

h̄∇kE(k) (8)

1.1 Quasiparticle Equation of Motion and Effective Mass

We are ultimately interested in the transport; i.e. the response

of this quasiparticle to an external electric field E , from which

it gains energy.

δE = −eE · vδt (i.e. force × distance) (9)

6

This energy is reflected by the quasiparticle ascending to higher

energy k states.

δE = ∇kE(k) · δk = h̄v · δk (10)

So

h̄δk = −eEδt (11)

h̄k̇ = −eE E.O.M (12)

This equation of motion is identical to that for free electrons

(c.f. Jackson); however, it may be shown to be applicable to

general Bloch states provided that E is smaller than the atomic

fields, and it must not vary in space or time too fast.

We may put this EOM in a more familiar form

v̇i =1

h̄

d

dt(∇kE(k))i =

1

h̄

∑j

∂2E

∂ki∂kj

dkjdt

=1

h̄

∑j

∂2E

∂ki∂kj(−eEj) (13)

This will have the form F = ma, if we define the mass tensor(1

m∗

)ij

=

(1

m∗

)ji

=1

h̄2

∂E

∂ki∂kjsymmetric & real (14)

which may be diagonalized to define three principle axes. In

the simple cubic case, the matrix will have the same element

7

along each principle direction and

m∗ =h̄2

d2Edk2

. (15)

In this way the effective mass of electrons on a lattice can vary

strongly, the larger d2Edk2

is, the smaller m∗

m is. Consider the simple

1-d case (See Fig. 4)

E(k)

k

k

m*

dk2d E2

= 0

2dk

2d E > 0

2dk

2d E < 0

π/a-π/a

Figure 4:

8

2 Currents in Bands

Our previous discussion of the motion of an electron (or a quasi-

particle) in a metal under the influence of an applied field E ,

ignored the presence of other electrons and the Pauli principle.

2.1 Current in an Insulator

The Pauli principle insures that a full band of states is

insulating. Consider the electric current due to d3k states

dJ = −ev(k)

(L

2π

)3

d3k (16)

The current density is then

dj =−eh̄∇kE(k)

1

(2π)3d3k (17)

ie., different occupied states in the Brillouin zone contribute

differently to the current. The net current density j is then the

First B.Z.

Fermi surface

Figure 5: Different occupied states make different contributions to the current density.

9

integral over all occupied states, which for our full band is the

integral over the first Brillouin zone

j = − e

8π3h̄

∫1st B.Z.

∇kE(k)d3k . (18)

Thus for each k vector in the integral, there is also −k.

Now consider a lattice with inversion symmetry k→ −k so

that E(k) = E(−k). Alternatively, recall that time-reversal

invariance requires that

E↑(k) = E↓(−k) , (19)

but since E↑(k) = E↓(k) due to spin degeneracy, we must have

that E(k) = E(−k). Thus

v−k =1

h̄∇−kE(−k) = −∇kE(k) = −vk! (20)

i.e., for the insulator

j =−e

8π3h̄

∫1st B.Z.

d3k∇kE(k) ≡ 0 (21)

Now imagine that the band is not full (See Fig. 6, left). Then,

if we apply an external field E , so that

k̇ = −eEh̄

e > 0! (22)

10

occupied statesempty

kx

ky

empty occupied states

kx

ky

E x�

Figure 6: The Fermi sea of a partially filled band will shift under the influence of an

applied field E. This destroys the inversion symmetry of the Fermi sea, causing a net

current.

the electrons will redistribute as the Fermi surface shifts (See

Fig. 6 right).

j =−e8π3

∫koccupied

v(k)d3k

=−e8π3

∫1st B.Z.

d3kv(k)− −e8π3

∫empty

d3kv(k)

=e

8π3

∫empty

d3kv(k) (23)

Thus the current may be formally described as a current of

positive charge particles (holes) assigned to the unoccupied

states in the band.

2.2 Currents in a Metal

Now imagine that the band is almost full. Near the top of the

11

k

Ek

full states

E

holes

E

E

D(E)

Figure 7: Left: A nearly full simple band. States near the Fermi surface that can be

thermally excited have negative mass. Right: Density of states with holes at the top

which have positive charge and mass.

band d2Edk2

< 0, so the mass is negative and the dispersion at the

top of the band is always also parabolic, so

E(k) = E0 −h̄2k2

2∣∣∣m∗ˆ∣∣∣ k = deviation from top! (24)

or

v̇ =1

h̄

d

dt∇kEk = − h̄k̇∣∣∣m∗ˆ∣∣∣ =

eE∣∣∣m∗ˆ∣∣∣ (25)

This is the EOM of a positively charged particle with positive

mass in an electric field E . I.e., holes at the top of the band

have positive mass.

We have just shown that a material with full bands is an

insulator (See Fig. 8 left). Ie., it carries no current, as least at

T = 0 and for a small E . However we ignored the presence

of other bands. If there is a conductiong band, for T 6= 0,

12

D(E)

E

Valence band

Conduction band

EgEf

empty

full

D(E)

E

electrons

holes

T=0 T≠0

Figure 8: An insulator form when the fermi energy falls in a gap of D(E). As the

temperature is raised, electrons are promoted over the gap, and both the electrons and

holes contribute to the conductivity which increases with temperature.

and a reasonably small Eg, there will be conductivity due to

a small number of thermally excited holes and electrons n ∼

exp(−Eg/KBT ) (See Fig. 8 right). Thus perhaps a better

definition of an insulator is a material for which the conductivity

increases with T .

3 Scattering of Electrons in Bands

According to the EOM for hole at the top of a band

v̇ =e∣∣∣m∗ˆ∣∣∣E (26)

as long as E is finite, these holes will continue to accelerate and

j will increase accordingly. Of course, this does not happen.

13

Rather the material simply heats up (ie., has a finite R). In

addition, if E is returned to zero, then j likewise returns to

zero. Why?

In 1900 Drude assumed that the electrons scatter from the

lattice yielding resistivity. Of course, as we have seen the quasi-

e

Figure 9: Drude thought that electrons scatter off the lattice yielding resistivity. Bloch

showed this to be wrong.

particle state may be defined from a sum over Bloch waves (de-

scribed by k) each of which is a stationary state and describe

the unperturbed propagation of electrons. Thus a perfect lat-

tice yields no resistivity. We can get resistivity in two ways.

1. Deviations from a perfect lattice

(a) Defects (See Fig. 10a)

(b) Lattice vibrations = phonons (See Fig. 10b)

2. Electron - electron interactions (See Fig. 11)

14

e

e(a)

k

k + q

q

k

k + q

- q

(b)

or††c c a

k+q k -q†c c a k+q k q

c c (a + a )††k+q k -qq

⇓

E(k) - E(k+q) = hω(q)

Figure 10: Electrons do scatter from defects in the lattice or lattice vibrations. They

contribute to the resistivity, with the phonon contribution increasing with temperature,

and the defect contribution more-or-less constant.

Due to the strength of the electron-electron interaction and the

density of electrons, (2) should dominate. However, it is easy

to show, using the Pauli principle, that effect of (2) is quite

often negligible, so that we may return to regarding the pure

electronic system as a (perhaps renormalized) non-interacting

Fermi gas.

According to momentum and energy conservation Fig 11

E1 + E2 = E3 + E4 k1 + k2 = k3 + k4 . (27)

(Of course, momentum conservation is only up to a recipro-

cal lattice vector G,k1 + k2 = k3 + k4 = G; however, as with

15

1 2

3 4

E1

k1

E2

k2

E3

k3

E4

k4

Figure 11: E1 +E2 = E3 +E4 and k1 + k2 = k3 + k4. Electron-electron interactions

also contribute to the resistivity (from simple order of magnitude arguments based

on relative strengths of the interactions, their contribution should dominate–but due

to the Pauli principle, it does not).

phonon conductivity, these processes with finite G involve much

higher energies, and may be neglected near T = 0.) Further-

more, since all states up to EF are occupied, E3;E4 > EF !

Suppose E1 is (thermally) excited, so E1 > EF and it collides

with an occupied state E2 < EF . Then

(E1 −EF ) + (E2 −EF ) = (E3 −EF ) + (E4 −EF ) > 0 (28)

ε1 + ε2 = ε3 + ε4 > 0, ε3; ε4 > 0 (29)

or ε1 + ε2 > 0, However, since ε2 < 0, if ε1 is small, then

|ε2| ≤ ε1, is also small, so only states with ε2EF≤ ε1

EFstates

may scatter with the state k1 conserve energy and obey the

Pauli principle, thus restricting ε2 to a narrow shell of width ε1

16

around the Fermi surface.

Now consider the restrictions placed on the states 3 and 4 by

momentum conservation.

k1 − k3 = k4 − k2 (30)

I.e. k1 − k3 and k4 − k2 must remain parallel, and since k1

is fixed, this restriction on the final states further reduces the

scattering probability by a factor of ε1EF

.

3k - k1 3

k1 k

4

EF

2 1

4k - k4 2

E

D(E)

ky

kx

k3

k2

Figure 12: Momentum and energy conservation severley restrict the states that can

an electron can scatter with and into.

Thus the total scattering cross section σ is reduced from the

classical result σ0 by(ε1EF

)2

. If the initial excitation ε1 is due

to thermal effects, then ε1 ∼ kBT and

σ

σ0∼(kBT

EF

)2

� 1! (31)

17

The total scattering due to electron - electron repulsion is very

small. Therefore, unless EF can be made small, the dominant

contribution to a material’s resistivity is due to defects and

phonons.

4 The Boltzmann Equation

The nonequilibrium (but steady-state) situation of an electronic

current in a metal driven by an external field is described by

the Boltzmann equation.

L

V

e j

E

defect

Figure 13: Electronic transport due to an applied field E, is limited by inelastic colli-

sions with lattice defects and phonons.

This differs from the situation of a system in equilibrium in

that a constant deterministic current differs from random par-

ticle number fluctuations due to coupling to a heat and particle

18

bath. Away from equilibrium (E 6= 0) the distribution function

may depend upon r and t as well as k (or E(k)). Nevertheless,

when E = 0 we expect the distribution function of the particles

in V to return to

f0(k) = f (r,k, t)|E=0 =1

eβ(E(k)−EF ) + 1(32)

As indicated,

To derive a form for f (r,k, t), we will consider length scales

larger than atomic distances◦A, but smaller than distances in

which the field changes significantly. In this way the system

is considered essentially homogeneous with any inhomogeneity

driven by the external field. Now imagine that there is no scat-

tering (no defects, phonons), then since electrons are conserved

t - dt

r - dtdrdt

k - dtdkdt

rk

t

hk = -eE·

Figure 14: In lieu of scattering, particles flow without decay.

f (r,k, t) = f

(r− vdt,k +

eEh̄dt, t− dt

)(33)

Now consider defects and phonons (See Fig. 15) which can

19

scatter a quasiparticle in one state at r −v dt and time t−dt, to

another at r and time t, so that f (r,k, t) 6= f(r− vdt,k + eE

h̄ dt, tdt).

We will express this scattering by adding a term.

Figure 15: Scattering leads to quasiparticle decay.

f (r,k, t) = f

(r − v dt,k + eE dt

h̄, t− dt

)+

(∂f

∂t

)S

dt

(34)

For small dt we may expand

f (r,k, t) = f (r,k, t)−v · ∇rf + eE · ∇kf

h̄− ∂f∂t

+

(∂f

∂t

)S

(35)

or

∂f

∂t+v · ∇rf−

e

h̄E · ∇kf =

(∂f

∂t

)S

Boltzmann Equation

(36)

If the phonon and defect perturbations are small, time-independent,

and described by H , then the scattering rate from a Bloch state

k to k′ (occupied to unoccupied) is wk′k = 2πh̄ |〈k

′ |H|k〉|2.

20

Then(∂f (k)

∂t

)S

=

(L

2π

)3 ∫d3k′ {(1− f (k))wkk′f (k′)

− (1− f (k′))wk′kf (k)}(37)

Needless to say it is extremely difficult to solve these last two

coupled equations.

4.1 Relaxation Time Approximation

As a result we make a series of approximations and ansatze.

The first of these is the relaxation time approximation that the

rate at which a system returns to equilibrium f0 is proportional

to its deviation from equilibrium(∂f

∂t

)S

= −f (k)− f0(k)

τ (k). (38)

Here τ (k) is called the relaxation time (for a spatially inhomo-

geneous system τ will also depend upon r). Ie., we make the

assumption that scattering merely acts to drive a nonequilib-

rium system back to equilibrium.

If E 6= 0 for t < 0 and then at t = 0 it is switched off so

21

that for t > 0 E = 0, then for a homogeneous system

∂f

∂t=

(∂f

∂t

)S

= −f − f0

τ(39)

so that

f − f0 = (f (t = 0)− f0) e−tτ (40)

ie., τ is the time constant at which the system returns to equi-

librium.

Now consider the steady-state situation of a metallic system

in a time-independent external field E = E x̂. Then

∂f

∂t= 0 (41)

Furthermore since the system is homogeneous

∇rf = 0 (42)

then

−eh̄E · ∇kf (k) =

(∂f

∂t

)S

= −f (k)− f0(k)

τ (k)(43)

ie

f (k) = f0(k) +e

h̄τ (k)E · ∇kf (k) (44)

which may be solved iteratively, generating a power series in E

(or Ex).

22

4.2 Linear Boltzmann Equation

For small E (Ohmic conditions)

f (k) ' f0(k) +e

h̄τ (k)E · ∇kf0(k) linear Boltzmann Eqn.

(45)

I.e. the lowest order Taylor series of f (k). Or equivalently, if

E = Exx̂

f (k) ' f0

(k +

e

h̄τ (k)E

)(46)

Ie., the effect is to shift the Fermi surface from its equilibrium

position by an amount

δkx

1ST BZ

kx

ky

Figure 16: According to the linear Boltzmann equation, the effect of a field Ex is to

shift the Fermi surface by δkx = −eτ Exh̄

δkx = −eτ Exh̄

(47)

23

From the discussion in Sec.??, it is clear that a finite current

results.

Interesting! Note that elastic scattering |k| = |k′| cannot

restore equilibrium. Rather they would only cause the Fermi

surface to expand. Inelastic scattering (i.e. from phonons) is

needed to explain relaxation.

A

B

Figure 17: Note that elastic scattering |k| = |k′| cannot restore equilibrium. Rather

they would only cause the Fermi surface to expand.

5 Conductivity of Metals

5.1 Drude Approximation

As mentioned above, Drude calculated the conductivity of met-

als assuming that

• all free electrons participate, and

24

• electron-lattice scattering yields a scattering rate 1/τ .

Under these assumptions, the EOM is

mv̇ +m

τ

(v − vtherm

)= −eE (48)

where v− vtherm = vD, the drift velocity, and mτ vD is friction.

Again when E = 0, we again have an exponential decay of v

so τ is again the relaxation time.

In steady-state v̇ = 0

vD =−eτmE (49)

so that

j = −envD =ne2τ

mE (50)

or defining j = σE , and σ = µne,

σ =ne2τ

mµ =

eτ

m(51)

5.2 Conductivity Using the Linear Boltzmann Equation

Of course, this is wrong since all free electrons do not par-

ticipate in σ due to the Pauli principle. And a more careful

derivation, using the Boltzmann Equation, is required. Again,

the relationship between j and f (k) is

25

V

E = E x̂

Figure 18: To calculate the conductivity, we apply a field in the x-direction only and

use the linearized Boltzmann Eqn.

j =−e8π3

∫d3k v(k)f (k) (52)

' −e8π3

∫d3k v(k)

{f0(k) +

eτ (k)

h̄Ex∂f0

∂kx

}(53)

For an isotopic material jz = jy = 0, and the equation becomes

scalar. Furthermore, again∫v(k)f0(k)d3k = 0 (54)

since v−k = −vk. Then as

∂f0

∂kx=∂f0

∂E

∂E

∂kx=∂f0

∂Eh̄vx (55)

jx ' −e2

8π3Ex∫d3kv2

xτ (k)∂f0

∂E(56)

26

Then as ∂f0∂E ' −δ(E − EF ) for T � EF the integral in k is

confined to the surface of constant E, and

d3k = dSEdk⊥ = dSEdE

h̄v(k)(57)

then

constant energy surface

kx

ky

kz

k ∇ Ek⊥

dk =⊥ ∇ E k

dE

Figure 19:

σ =jxEx

=e2

8π3h̄

∫dSEdE

v2x(k)

v(k)τ (k)δ(E − EF ) (58)

=e2

8π3h̄

∫E=EF

dSEv2x(k)

v(k)τ (k) . (59)

As expected only the properties of the electrons on the Fermi

surface are relevant.

We can now calculate the conductivity of a metal by aver-

aging v2xv τ (k) over the Fermi surface. Consider a simple system

27

E

E = 0E ≠ 0

Figure 20: Only the electrons near the fermi surface participate in the transport. Far

below the Fermi surface, pairs of states k and −k are occupied. Their contribution

to the conductivity cancels, leaving contributions to only the occupied states near the

fermi surface.

with a spherical Fermi surface, then∫dSE

τ (k)v2x(k)

v(k)=

4π

3k2Fτ (EF )v(EF )

=4π

3k2Fτ (EF )

h̄kFm∗

(60)

or

σ =e2

8π3h̄

4π

3k2Fτ (EF )

h̄kFm∗

(61)

then as kBT � EF , N = 243π

k3F

(2πL )3 ⇒ k3

F = 3π2n we find

that

σ =e2τ (EF )

m∗n µ =

eτ (EF )

m∗(62)

For semiconductors where n is T dependent, and for more re-

alistic material where the Fermi surface 6= sphere, the formula

is more complicated.

28

However, for metals the temperature dependence of σ is dom-

inated by that of τ ; ie., by the temperature dependence of

phonons. However, before we can calculate σ(T ), we must first

disentangle the phonon from the defect scattering. Assuming

that the two mechanisms are independent, they must add

1

τ=

1

τph+

1

τdefect(63)

i.e.

ρ = ρph + ρdefect Matthiesen’s Rule (64)

The defect contribution is proportional to the defect cross sec-

tion Σdefect and the current, or v(EF ), 1τdefect

∝ Σdefectv(EF ).

Is is roughly temperature independent, since the cross section

Σdefect and v(EF ) are.

The phonon contribution, on the other hand, is highly tem-

perature dependent since at zero temperature, there are no

phonons. The scattering cross section is roughly proportional to

the rms phonon excursion⟨S2(q)

⟩. However, from the equipar-

tition theorem

1

2Mω2

q

⟨S2(q)

⟩=kBT

2T � θD . (65)

29

Thus1

τph∝⟨S2(q)

⟩∝ kBT

mω2q

(66)

Ie., at high temperatures, all modes contribute a linear in T

scattering to 1τph

. Therefore, at T � θD (θD = debye temper-

ature)

ρ = aT + ρdefect (67)

Cu + 3.32% Ni

Cu + 2.16% Ni

Cu + 1.12% Niρ

T

Cu

R

T ΘD

αT

αT 5

0.1

Figure 21: The phonon and defect contributions to the resistivity add (left), and the

phonon contribution is linear at high temperatures T � θD.

6 Thermoelectric Effects

Until now, we have assumed that the transport system is ther-

mally homogeneous. Of course this need not be the case since we

can obviously maintain both an electrical and a thermal current.

30

T1T ≠ T1 2

T2

x̂

Figure 22: Thermoelctric effects are important in systems with both electric potential

and thermal gradients. We will assume both are in the x-direction

Here, each electron can carry a charge current ∼ ev ∼ e2E

and a thermal current kTk∇T . In fact, a heat current can be

used to induce an electrical potential (Seebeck or thermoelectric

effect) and, conversely, an electric current can be used to move

heat (Peltier effect) which makes the solid state refrigeration

possible.

6.1 Linearized Boltzmann Equation

To allow for a thermal gradient ∇T , our formalism must be

modified. Imagine that ∇T and E are fixed in time, then the

Boltzmann equation becomes

∂f

∂t+v · ∇rf −

e

h̄E · ∇kf =

(∂f

∂t

)S

= −f (k)− f0(k)

τ (k)(68)

31

where in steady state ∂f∂t → 0. After linearizing (replacing f

by f0 in the left-hand side), we get

f (k) ' f0(k)− τ (k){

v · ∇rf0 −e

h̄E · ∇kf0

}(69)

Then, as before

e

h̄E · ∇kf0 =

e

h̄E · ∂f0

∂Eh̄v =

e

h̄Ex∂f0

∂Eh̄vx (70)

The spatial inhomogeneity is through ∇T , and in a semicon-

ductor for which EF depends strongly upon T , through ∇EF

v· ∇rf0 = v·{∂f0

∂T∇T +

∂f0

∂EF∇EF

}= v·

{∂f0

∂T∇T − ∂f0

∂E∇EF

}(71)

Apparently ∇EF only contributes a term which modifies the

electric field dependence

vx∂f0

∂E{eEx + (∇EF )x} ≡ vx

∂f0

∂EeE ′x (72)

Of course, in a metal E ′ = E .

6.2 Electric Current

Thus, we now have

f (k) = f0(k)− τ ∂f0

∂Tvx∂T

∂x+e

h̄τE ′xh̄vx

∂f0

∂E(73)

32

Then for

jx = − e

8π3

∫d3kvx(k)f (k) (74)

jx = − e

8π3

∫d3kvx(k)

{f0(k)− τ ∂f0

∂Tvx∂T

∂x+e

h̄τE ′x

∂f0

∂Eh̄vx

}recall that the last term yielded σ last time (and still will)

jx = σE ′x +e

8π3

∫d3kv2

xτ∂f0

∂T

∂T

∂x(75)

Again we will calculate the second term assuming a spherical

Fermi surface. The term ∂f0∂T confines the integral to the Fermi

E

T ≠ 0

T = 0

Ef

f0

Figure 23: The derivative ∂f0∂E

is only significant near the fermi surface.

sphere and so again effectively it amounts to a Fermi-surface

average, so v̄2x → 1

3v2 ≈ 2

3m∗12m∗v2 = 2

3m∗E or changing to an

integral over the DOS

jx = σE ′x +2

3

e

m∗

∫dEτ (E)ED(E)

∂f0

∂T

∂T

∂x(76)

33

Assuming that τ (E) ∼ τ (EF ), we get

jx = σE ′x +2

3

e

m∗τ (EF )

∂T

∂x

∫dEE

∂f0

∂TD(E) (77)

jx = σE ′x +2

3

e

m∗τ (EF )

∂T

∂xcv(T ) cv ∝ m∗ (78)

In general, this intuitive form is rewritten as

jx = σE ′x + L12xx

(−∂T∂x

)(79)

and from it we see that both an electric field (or the generalized

field strength E ′), and the thermal gradient contribute to the

electron current, jx.

6.3 Thermal and Energy Currents

Of course one can have a thermal current without having an

electric current (same number of electrons moving right and

left, but more of the hot ones moving right). Thermodynamics

is needed to quantify this though since these electrons will also

carry entropy as well as energy and heat.

Imagine that a small subsection of our material is in thermal

equilibrium and then some electrons are introduced/taken away

34

so that

dQ = TdS = dU − µdN First Law of Thermodynamics

(80)

in terms of particle flow

jQ = jE − EF jn − ejn = j (81)

where this equation defines jQ, the thermal current, and

jE =

∫d3k

8π3E(k)v(k)f (k, r) . (82)

Again one could work out the form of jQ for the spherical Fermi

surface using the linearized Boltzmann equation. However one

must obtain a form like

j = L11E ′ + L12(−∇T ) (83)

jQ = L21E ′ + L22(−∇T ) (84)

(The fact that L12 = L21 is referred as the Onsaser relation.)

These relationships between the L’s and the transport co-

efficients depends upon what experiment is being done. For

example in Fig. 24 there is a potential gradient (V 6= 0) but no

thermal gradient since T1 = T2. The electric field drives both

electric and thermal currents.

35

T1T = T1 2

T1

I ≠ 0

I

V ≠ 0

V

j = L E′11

j = L E′Q

12 11L = σ

j = σ E

ρ = σ-1

Figure 24: Here, there is a potential gradient (V 6= 0) but no thermal gradient since

T1 = T2. The electric field drives both electric and thermal currents. Thus, a heat

bath is required to keep both sides of the sample at the same temperature.

j = L11E

jQ = L12E (85)

Thus, we may identify

σ =neτ

m∗= L11 (86)

Note that since there is a thermal current induced by the po-

tential gradient, a heat bath is required to keep both sides of

the sample at the same temperature.

In Fig. 25 we maintain a thermal gradient, but turn off the

electric current. Here,

j = 0 = L11E ′ + L12(−∇T ) (87)

36

T1 T ≠ T1 2

T2

I = 0

IV

Figure 25: Here j = 0 = L11E ′ + L12(−∇T ) and jQ =(−L12

(L12L11

)+ L22

)(−∇T )

where −L12(L12L11

)+ L22 = κT

and

jQ = L21E ′ + L22(−∇T ) =

(−L21

(L12

L11

)+ L22

)(−∇T )

(88)

and since (you will show)

L12 = − 2eτ

3m∗cv = L21 (89)

κ = L22 −(L12)2

L11(90)

We could also measure the thermal conductivity by driving

a heat current through the sample, maintaining the ends at the

same potential (see Fig. 26 right). Here, we would find

κ = L22 . (91)

Thus, we can identify

κ = L22 or L22 −(L12)2

L11(92)

37

T ≠ T1 2

κ = L 22

j = L (-∇T)12

V ≠ V1 2

V1V

2

jQ

V1V

2

I

jQ

V = V1 2

j = L (-∇T)Q

22

11L

κ = L - 22

12(L )

Figure 26: Two methods for measuring κ.

depending upon the experiment. These are the same if the

sample is a good metal where L11 = σ is large. Many experi-

mentalists measure κ by the method of the left of Fig. 26. This

yields the more conventional definition of κ.

6.4 Seebeck Effect, Thermocouples

These relations result in some interesting physical effects. Con-

sider a bimetallic conducting loop with two junctions main-

tained at temperatures T1 T2. Let metal A be different than B,

so that LijA 6= LijB. If no current flows around the loop, then

j = 0 = L11Ex + L22(−∇T )⇒ Ex =

(L12

L11

)dT

dx(93)

38

V

metal A

metal Bmetal B T0

T1

T2

21

0

Figure 27: A bimetallic conducting loop with junctions maintained at T1 and T2. If

T1 6= T2, and L12A /L11

A 6= L12B /L11

B , then the heat current induces a potential V ∝ T2−T1

where S = L12L11 is called the thermopower and is a property of

a material.

The potential measured around the loop is given by

V =

∫ 1

0

EBdx +

∫ 2

1

EAdx +

∫ 0

2

EBdx (94)

or

V = SB

{∫ 1

0

∂T

∂xdx +

∫ 0

2

∂T

∂xdx

}+ SA

∫ 2

1

∂T

∂xdx (95)

= SB

∫ 1

2

∂T

∂xdx + SA

∫ 2

1

∂T

∂xdx (96)

= (SA − SB)

∫ T2

T1

dT = (SA − SB) (T2 − T1) (97)

Or if SA and SB are not T -independent V =∫ T2T1dT (SA − SB).

39

So if T1 6= T2 and SA 6= SB, then the heat current in-

duces an emf! This is called the Seebeck effect ⇒ (solid state

thermometer with ice H2O as a reference).

6.5 Peltier Effect

Now consider the inverse situation where an electrical current j

metal A

metal Bmetal B T0

21

0

j

Figure 28: An electrical current j is driven through the loop which is held at a fixed

temperature

is driven through the loop which is held at a fixed temperature(∂T∂x = 0

). Then

jQ = L21E j = L11E (98)

jQ =

(L21

L11

)j = πj (99)

40

This is known as the Peltier effect whereby heat is carried

from one junction to the other or an electric current is accom-

T0

0

Aπ j

Bπ j

Bπ j

jQ

jQ

=A

B B

=(π - π )j = j

BA Qj = (π - π )jA BQ

Figure 29: If dT/dx = 0 and πA = L21A /L11

A 6= L21B /L11

B = πB then the electric current

also induces a heat from one junction to another.

panied by a heat current. One may use this effect to create an

extremely simple (and similarly inefficient) refrigerator.

7 The Wiedemann-Franz Law (for good metals)

One may independently measure the thermal κ and electrical σ

conductivities. However, in general one expects that κ ∝ σT

j = σE

V ≠ V1 2

V1 V2

j

T1

T2

T = T1 2T ≠ T

1 2V1

V2

jQ V = V1 2

j = κ(-∇T)Q

T1T2

thermal field -∂T∂x

E

electric field -e/E

Figure 30: The thermal and electrical conductivities may be measured independently.

41

since in electrical conduction each election carries a charge e

and is acted on by a force −eE . The current per unit electric

field proportional to e2. In thermal conduction each electron

carries a thermal energy kBT and is acted on by a thermal

force −kB∇T . The heat current per unit thermal gradient is

proportional to k2BT , thus one expects

κ

σ∝ k2

B

e2T (100)

Due to the simplicity of these arguments, our formalism

should reproduce this relationship. As we discussed before

jQ = jE − EF jn (101)

=1

8π3

∫d3k (E − EF )v(k)f (k) (102)

In the linear approximation to the Boltzmann equation for E ′x =

0, we get

jQ =1

8π3

∫d3k (E − EF )

∂f0

∂Tv2xτ

(−∂T∂x

)(103)

where E−EF and ∂f0∂T are odd in (E−EF ), for the Fermi liquid

jQ ' −(∂T

∂x

)1

3v2FτF

∫d3k

3π3(E − EF )

∂f0

∂T(104)

42

EEF

F(E - E )

∂T

∂f0

Figure 31: The function (E−EF )∂f0∂T

is sharply peaked at the fermi surface and even.

in E − EF .

jQ =

(−∂T∂x

)1

3v3FτFcv (105)

κ =1

3v3FτFcv (106)

Now recall that for the Fermi liquid cv = kBπ2

2 nkBT

kBTFso that

κ =1

3

m∗v2F

m∗τFkB

π2

2nkB

T

EF=π2

3τFn

k2BT

m∗(107)

However, also for the Fermi liquid, we found that σ = e2τFnm∗ ,

soκ

σ=π2

3

(kBe

)2

T (108)

Of course this relationship only holds in a good metal. There

are two reasons for this. First we are neglecting terms like(L12)

2

σ in κ which are unimportant for a good metal (or if we

electrically short the sample.) Second, we are assuming that κ

43

is dominated by electronic transport.

44