Electrons in metals Modified 2008 - web.ift.uib.noweb.ift.uib.no/.../metals_Ex_2008/metals.pdf · The Drude Theory of metals ... Heat capacity of the electron gas Both the phonons

Electrons in metalsModified 2008

PHYS208

January 15, 2010

PHYS208 () Electrons in metals Modified 2008 January 15, 2010 1 / 23

The Drude Theory of metals

The Drude model is the application of ideal gas theory to electrons in a solid.


Basic assumptions

Electrons move in static lattice of ions

Free electron approximation (neglect ion-electron interaction)

Independent electron approximation (neglect electron-electron interaction)

Some scattering mechanism exists, erasing all information of the electronprevious to the scattering event, giving each a random velocity.

Probability of scattering during time interval ∆t: ∆t/τ , where τ is known asthe relaxation time.

Alternatively: the electrons move a distance ` before colliding (mean freepath).


Conductivity - static electric field

If the electrons in a metal are subjected to a staticelectric field, they will tend to drift in the oppositedirection to the field, with a drift velocity determined bythe field. Since an electron has approximately a time τ tobe accelerated by the field before a collision occurs, it willgain a velocity vdrift given by

vdrift = −∫ τ

0

eE

mdt = −eEτ

m. (1)

This net displacement of electrons results in a nonzero current density j, which is

charge density × velocity =charge

unit volume× velocity (2)


With N electrons of charge −e in a volume V , and an average drift velocity vdrift ,we arrive at

j = −eN

V×−eEτ

m=

Ne2τ

mVE. (3)

The proportionality constant is called the conductivity σ, ie

j = σE. (4)

Written in terms of electrons per unit volume n = N/V , the conductivity becomes

σ =ne2τ

m(5)


Thermal conductivityA temperature gradient in a slab of metal will induce a thermal current density. Ifit is not too large, Fourier’s law applies,

~j = −κ∇T . (6)

We want to calculate the thermal conductivity κ of metals.

We consider an oversimplified model, putting the z-direction along thetemperature gradient. An electron arriving at a point z will carry an energy givenby the temperature at its last collision site `, ε(T [z − ` cos θ])


Thermal conductivityThe z-component of the velocity of carrier i is

(~vi )z = −

(~i

`

)z

v = −v cos θi z , (7)

where v is the average velocity of the carriers.The energiflux due to carrier i is

~ji = (~vi )z∆u, (8)

where ∆U is the energy difference between to sites separated by ~.

~ji = −v cos θi(u(~r + ~)− u(~r)

)z (9)

= −v cos θi(∇u · ~

)= −v cos θi

(∂u

∂z` cos θi

)z

= −∂u

∂z`v cos2 θi z .


Thermal conductivity

To obtain the total energyflux we sum over i ,

~j = −∑

i

∂U

∂z`v cos2 θi z . (10)

In stead of summing, we integrate over all solide angles dΩ = sin θdθdφ,

~j = − 1∫dΩ

∫∂U

∂z`v cos2 θdΩz = − 1

4π

∂U

∂z`v

∫cos2 θdΩz , (11)

where∫

cos2 θdΩ = 4π3 and ∂U

∂z z = ∂U∂T

∂T∂z z = ∂U

∂T∇T = cv∇T


Thermal conductivity

Thus

~j = − 1∫dΩ

∫∂U

∂z`v cos2 θdΩ = −

(1

3cvv`

)∇T (12)

which, upon comparison with Fouriers law yields an expression for the thermalconductivity,

κ =1

3`vcv =

1

3v2τcv (13)


Wiedemann-Franz law

The Wiedemann-Franz law states that in a metal the ratio of thermal to electricalconductivity is proportional to the temperature,

κ

σ∝ T , (14)

where κ = 13v2τcv and σ = ne2τ

m as shown.

Ideal gas theory yields 〈 12mv2〉 = 3

2kBT , which gives the expression for spesificheat capacity,

cv =3

2

N

VkB . (15)

Using the results from the Drude model, we can write the Lorenz number κ/σT as

κ

σT=

3

2

(kB

e

)2

. (16)


Wiedemann-Franz law

Calculated Lorenz number:

κ

σT= 1.11× 10−8 W · Ω

K2, (17)

which is about halv of the experimental values.

Experimental Lorenz numbers:

Figure: From Tony Harker, Chiranijib Mitra and Andrew Horsfield, Solid State PhysicsNotes


The ideal Fermi gas

To adequately describe electrons in metals, their quantum nature must be takeninto account.

Classical (Boltzmann) statisics insufficient

Need to take into account Pauli exclusion principle

We consider a box of volume V = L3 containing N electrons.Under the assumption of no electron-electron interaction, thewavefunction of the system factors into N components,Ψ(r1, · · · , rN) = ψ(r1) · · ·ψ(rN), each of which should satisfythe time-independent Schrodinger equation,

− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)ψ(ri ) = εψ(ri ) (18)


The ideal Fermi gas

The solutions are plane waves,

ψk(r) ∝ eik·r, (19)

with (one-electron) energies

ε(k) =~2k2

2m, ki =

2πni

Li, ni = 0, 1, 2, ... (20)

The discrete energy levels are obtained by choosing periodic (Born von-Karman)boundary conditions.


Fermi-Dirac distribution

The distribution of electrons in thequantum gas is given by the Fermi-Diracdistribution,

f(ε) =1

e(ε−µ)/kBT + 1(21)

Special case: T = 0. Then f(ε) = 1 forε ≤ εF , zero elsewhere.

Very different from classical Boltzmanndistribution e−ε/kBT for lowtemperatures!


Fermi stuff

Fermi wave vector, kF , the wave vector of the electron in the highest energystate (T = 0).

Fermi momentumpF = ~kF (22)

Fermi energy (typically 1.5 - 15 eV i metals)

εF =~2k2

F

2m(23)

It can be shown that the fermi energy is proportional to the density of freeelectrons in the metal.

Fermi temperature (of the order 105 for many metals)

TF =εFkB

(24)


Density of states

For electrons (fermions), total number N is found by summing the meanoccupation number over all levels (factor 2 for spin),

N = 2∑

j

1

e(εj−µ)/kBT + 1(25)

If the box is very large, the energy spacing is very small. Thus we write,∑j

1

e(εj−µ)/kBT + 1=

1

(∆k)3

∑j

1

e(εj−µ)/kBT + 1(∆k)3 → L3

(2π)3

∫(...)d3k

(26)were we inserted ∆k = 2π

L outside the sum and taken the limit ∆k → 0. Since fonly depends on the norm of k (through ε), we introduce spherical coordinates,and integrate out angular dependence, yielding a factor 4π. We also replacing dεfor dk .


Density of states

Resulting expression for N:

N = 2V

(2π)2

(2m

~2

) 32∫ε

12 f(ε)dε (27)

We identify the density of states g(ε),

g(ε) = 2V

(2π)2

(2m

~2

) 32

ε12 (28)

→ Number of one-electron states in an energy interval dε is g(ε)dε.


Sommerfeld integrals

One routinely encounters integrals of the type,∫ ∞0

γ(ε)f(ε)dε (29)

when working with the Fermi-Dirac distribution. Often the temperatures one isinterested in are low enough that f(ε) is only non-constant in a very small interval

around ε = µ. In that case, we introduce γ(ε) = dΓ(ε)dε . Integrating by parts, we

find

Γ(ε)f(ε)|∞0 −∫ ∞

0

df(ε)

dεΓ(ε)dε (30)

The first part is zero since f (ε→∞) = 0, and since our case Γ(ε→ 0) = 0.


Sommerfeld integrals

Expanding Γ(ε) around ε = µ, and retaining only one term results in

∫ ∞0

dΓ(ε)

dεf(ε)dε = Γ(µ) +

π2

3

(kBT )2

2

d2Γ(ε)

dε2

∣∣∣∣∣ε=µ

(31)

which is sufficient for many applications.


Heat capacity of the electron gas

Both the phonons and the free electrons will contribute to the heat capacity ofmetals.

c totalv = celectr

v + cphononsv (32)

We can calculate the specific heat capacity of an electron gas using Fermi-Diracstatistics.

cv =1

V

(dU

dT

)V

. (33)

The total energy is calculated by the integral

U =

∫ ∞0

dεg(ε)f(ε)ε. (34)

We can evaluate the integral using Sommerfeld’s expansion.


Heat capacity of the electron gas

Now, γ(ε) ∝ ε 32 , Γ(ε) ∝ ε 5

2 and ∂2Γ∂ε2 ∝ ε

12 . We use the Sommerfeld expansion

with one term, ∫ ∞0

dΓ(ε)

dεf(ε)dε = Γ(µ) +

π2

3

(kBT )2

2

d2Γ(ε)

dε2

∣∣∣∣∣ε=µ

(35)

and differentiating the result, arriving at

celectrv =

π2

2

kBN

V

T

TF(36)

Notice linear temperature dependence.

Differs from classical result (ideal gas), cclassicv = 3

2NV kB , by a factor π2

3TTF

,which is ∼ 0.01 at room temperature.

celectrv has significant contribution at low temperature, when cphonons

v ∼ T 3.


Why is the heat capacity of electrons in metals small?

Unless temperature is very high, only relatively few electrons near the Fermisurface can change their state.

The remaining (majority) does not contribute to the heat capacity celectrv .

In the classical gas, all electrons contribute, yielding a higher heat capacity.

The energy of the excited electrons is about εF .


Lorenz number revisited

Lorenz number obtained from the new celectrv :

κ

σT=π2

3

(kB

e

)2

= 2.45× 10−8 W · ΩK2

. (37)

which is about the same as the experimental values.

Experimental Lorenz numbers:

Figure: From Tony Harker, Chiranijib Mitra and Andrew Horsfield, Solid State PhysicsNotes


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Electrons in metals Modified 2008 - web.ift.uib.noweb.ift.uib.no/.../metals_Ex_2008/metals.pdf · The Drude Theory of metals ... Heat capacity of the electron gas Both the phonons