Electrons in metals Modified 2010 - Universitetet i...

Electrons in metalsModified 2010

PHYS208

May 21, 2010

PHYS208 () Electrons in metals Modified 2010 May 21, 2010 1 / 22

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 determined bythe temperatur at the given position.

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)

Ohm’s law gives us:j = σE. (4)

Where the proportionality constant is called the conductivity σ.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.

1

v1

l1

l1

u( r+ )

u( ) r

v

3

1

v2

v3

v4

v4

v2

v3

v1

l1

l2

l3

l4

2

14

v

The current density due to carrier i is

~ji = ~Jin − ~Jout = ∆~J = (~vi )z(ur+li − ur ) (8)

Where in represents the incomming current and out is the outgoing current,

~ji = (~vi )z∆u, (9)

where ∆u is the difference in energy density between to sites separated by ~.PHYS208 () Electrons in metals Modified 2010 May 21, 2010 7 / 22

Thermal conductivity

~ji = −v cos θi(u(~r + ~)− u(~r)

)z (10)

= −v cos θi(∇u · ~

)= −v cos θi

(∂u

∂z` cos θi

)z

= −∂u∂z`v cos2 θi z .

To obtain the total current density we sum over i ,

~j = −∑i

∂u

∂z`v cos2 θi z . (11)

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 , (12)

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 (13)

which, upon comparison with Fouriers law:

~j = −κ∇T . (14)

yields an expression for the thermal conductivity,

κ =1

3`vcv =

1

3v2τcv (15)


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 , (16)

where κ = 13v

2τcv and σ = ne2τm as shown.

Ideal gas theory yields 〈 12mv2〉 = 3

2kBT , which gives the expression for spesificheat capacity,

cv =3

2

N

VkB . (17)

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

κ

σT=

3

2

(kBe

)2

. (18)


Wiedemann-Franz law

Calculated Lorenz number:

κ

σT= 1.11× 10−8 W · Ω

K2, (19)

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 ) (20)


The ideal Fermi gas

The solutions are plane waves,

ψk(r) ∝ eik·r, (21)

By choosing periodic (Born von-Karman) boundary conditions, ie,

ψ(x + L) = ψ(x) (22)

one gets discrete (one-electron) energy levels.

ε(k) =~2k2

2m, ki =

2πniLi

, ni = 0, 1, 2, ... (23)


Fermi-Dirac distribution

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

f(ε) =1

e(ε−µ)/kBT + 1(24)

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

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


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 =

∫f(ε)g(ε)dε (27)

N = 2V

(2π)2

(2m

~2

) 32∫ε

12 f(ε)dε (28)

We identify the density of states g(ε),

g(ε) = 2V

(2π)2

(2m

~2

) 32

ε12 (29)

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


Fermi stuff

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

Fermi momentumpF = ~kF (30)

Fermi energy (typically 1.5 - 15 eV i metals)

εF =~2k2

F

2m(31)

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

(32)


Sommerfeld integralsIntegrals of the type, ∫ ∞

0

γ(ε)f(ε)dε (33)

may be subject to the Sommerfield expansion, if the temperatures are low enoughthat f(ε) is only non-constant in a very small interval around ε = µ. In that case,we introduce:

γ(ε) =dΓ(ε)

dε. (34)

Integrating by parts, we find

Γ(ε)f(ε)|∞0 −∫ ∞

0

df(ε)

dεΓ(ε)dε (35)

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

(36)


Heat capacity of the electron gas

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

c totalv = celectrv + cphononsv (37)

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

cv =1

V

(dU

dT

)V

. (38)

The total energy is calculated by the integral

U =

∫ ∞0

dεg(ε)f(ε)ε. (39)

We can evaluate the integral using Sommerfeld’s expansion.


Heat capacity of the electron gasWe 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

(40)

U =

∫ µ

0

εg(ε) +π2

3

(kBT )2

2[µg ′(µ) + g(µ)] (41)

U = U0 +π2

3

(kBT )2

2g(εF ), g(εF ) =

3

2

n

εF, TF =

εFKB

(42)

and differentiating the result, arriving at

celectrv =π2

2

kBN

V

T

TF(43)

Notice linear temperature dependence.

Differs from classical result (ideal gas), cclassicv = 32NV kB , by a factor π2

3TTF

,which is ∼ 0.01 at room temperature.celectrv has significant contribution at low temperature, when cphononsv ∼ 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

(kBe

)2

= 2.45× 10−8 W · ΩK2

. (44)

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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