Last Time

Free electron model

ε~gDensity of states in 3D

m

kE FF 2

22

=Fermi Surface kF =3π 2NV

⎛⎝⎜

⎞⎠⎟

1/3

( ) ( )Tfe

,1

1 εµεβ =+

= −

Fermi-Dirac Distribution Function

Debye Approximation.

Today

Measuring the occupied density of states

Effective Mass

Electrical Conductivity

Thermal Conductivity

Wiedemann-Franz Ratio

Heat Capacity 3BTATC +=

Electrons Phonons

Fermi-Dirac Distribution Function

http://ece-www.colorado.edu/~bart/book/distrib.htm#fermi

Becomes a step function at T=0.Low E: f ~ 1.High E: f ~ 0.

Go play with the Excel file “fermi.xls” at:

( ) ( ) 1

1,

+= −µεβεe

Tf

µ = chemical potential = “Fermi Level”

µ(T=0)=εF Fermi energy

Right at the Fermi level: f = 1/2.

Fermi-Dirac Distribution Function

http://ece-www.colorado.edu/~bart/book/distrib.htm#fermi

Becomes a step function at T=0.Low E: f ~ 1.High E: f ~ 0.

Go play with the Excel file “fermi.xls” at:

( ) ( ) 1

1,

+= −µεβεe

Tf

µ = chemical potential = “Fermi Level”

µ(T=0)=εF Fermi energy

Right at the Fermi level: f = 1/2.

n ε,T( ) = g ε( ) f ε,T( )

N = n ε,T( )0

∞

∫ dε = g ε( ) f ε,T( )dε = N0

∞

∫

µ T > 0( ) < εF

µ T( ) = εF 1− o T2( )⎡⎣ ⎤⎦

Number of electrons per energy range

Fermi functionDensity of states

Implicit equation for µ

N is conserved

Shaded areas are equal

⇒

0.01% @ room temp

Density of Occupied States

n ε,T( ) = g ε( ) f ε,T( )

N = n ε,T( )0

∞

∫ dε = g ε( ) f ε,T( )dε = N0

∞

∫

µ T > 0( ) < εF

µ T( ) = εF 1− o T2( )⎡⎣ ⎤⎦

Number of electrons per energy range

Fermi functionDensity of states

Implicit equation for µ

N is conserved

Shaded areas are equal

⇒

0.01% @ room temp

Density of Occupied States

Heat CapacityWidth of shaded region ~ kT

Room temp T ~ 300K, TF ~ 104 K→ Small width→ Few electrons thermally excited

How many electrons are excited thermally?

Shaded area ≈ triangle. Area = (base)(height)/2

Number of excited electrons: ≈ (g(εF)/2)(kT)/2 ≈ g(εF)(kT)/4

Excitation energy ≈ kT (thermal)

Total thermal energy in electrons: E ≈14g εF( )kT⎛

⎝⎜⎞⎠⎟kT( ) = 1

4g εF( ) kT( )2

C =dEdT

≈12g εF( )k2T =

34NεFk2T

C ~ THeat Capacity in a Metal

Heat CapacityWidth of shaded region ~ kT

Room temp T ~ 300K, TF ~ 104 K→ Small width→ Few electrons thermally excited

How many electrons are excited thermally?

Shaded area ≈ triangle. Area = (base)(height)/2

Number of excited electrons: ≈ (g(εF)/2)(kT)/2 ≈ g(εF)(kT)/4

Excitation energy ≈ kT (thermal)

Total thermal energy in electrons: E ≈14g εF( )kT⎛

⎝⎜⎞⎠⎟kT( ) = 1

4g εF( ) kT( )2

C =dEdT

≈12g εF( )k2T =

34NεFk2T

C ~ THeat Capacity in a Metal

Heat CapacityWidth of shaded region ~ kT

Room temp T ~ 300K, TF ~ 104 K→ Small width→ Few electrons thermally excited

How many electrons are excited thermally?

Shaded area ≈ triangle. Area = (base)(height)/2

Number of excited electrons: ≈ (g(εF)/2)(kT)/2 ≈ g(εF)(kT)/4

Excitation energy ≈ kT (thermal)

Total thermal energy in electrons: E ≈14g εF( )kT⎛

⎝⎜⎞⎠⎟kT( ) = 1

4g εF( ) kT( )2

C =dEdT

≈12g εF( )k2T =

34NεFk2T

C ~ THeat Capacity in a Metal

How you would do the real calculation:

N = n ε,T( )0

∞

∫ dε

E = εn ε,T( )0

∞

∫ dε = ε f ε,T( )g ε( )0

∞

∫ dε

dEdT

= C =π 2

2g εF( )kB2T

C = AT + BT 3

Implicit equation for µ → fully determines n(ε, T)

Then

In a metallic solid,

C ~ T is one of the signatures of the metallic state

Electrons Phonons

Correct in simple metals.

Heat Capacity

How you would do the real calculation:

N = n ε,T( )0

∞

∫ dε

E = εn ε,T( )0

∞

∫ dε = ε f ε,T( )g ε( )0

∞

∫ dε

dEdT

= C =π 2

2g εF( )kB2T

C = AT + BT 3

Implicit equation for µ → fully determines n(ε, T)

Then

In a metallic solid,

C ~ T is one of the signatures of the metallic state

Electrons Phonons

Correct in simple metals.

Heat Capacity

Measuring n(ε, T)

X-ray Emission

(1) Bombard sample with high energy electrons to remove some core electrons

(2) Electron from condition band falls to fill “hole”, emitting a photon of the energydifference

(3) Measure the photons -- i.e. the X-ray emission spectrum

N = n ε,T( )0

∞

∫ dε = g ε( ) f ε,T( )dε = N0

∞

∫ n(ε, T) is the actual numberof electrons at ε and T

Measuring n(ε, T)

X-ray Emission

N = n ε,T( )0

∞

∫ dε = g ε( ) f ε,T( )dε = N0

∞

∫ n(ε, T) is the actual numberof electrons at ε and T

Emission spectrum (how many X-rays come out as a function of energy) will look likethis.

Fine print: The actual spectrum is rounded by temperature, and subject to transition probabilities. Void in New Hampshire.

EFFECTIVE MASS

Real metals: electrons still behave likefree particles, but with “renormalized” effective mass m*

E =2k2

2m*

In potassium (a metal), assuming m* =1.25m gets the correct

(measured) electronic heat capacity

Physical intuition: m* > m, due to “cloud” of phonons and other excited electrons.

Fermi Surface

At T>0, the periodic crystal and electron-electroninteractions and electron-phonon interactionsrenormalize the elementary excitation to an “electron-likequasiparticle” of mass m*

EFFECTIVE MASS

Real metals: electrons still behave likefree particles, but with “renormalized” effective mass m*

E =2k2

2m*

In potassium (a metal), assuming m* =1.25m gets the correct

(measured) electronic heat capacity

Physical intuition: m* > m, due to “cloud” of phonons and other excited electrons.

Fermi Surface

At T>0, the periodic crystal and electron-electroninteractions and electron-phonon interactionsrenormalize the elementary excitation to an “electron-likequasiparticle” of mass m*

Electrical Conductivity

τvm

Eedt

dvmF

*

* −−==

v = −eτm*

E

*m

ee

τµ =

Em

nevnej

*

2τ=−=

Collisions cause drag

Electric Field Accelerates charge

τ ≈ mean time between collisions

0=vSteady state solution:

⇒

=mobility

Electric current density (charge per second per area)

Units: n=N/V ~ L-3 v ~ L/S

current per area

average velocity

Electrical Conductivity

τvm

Eedt

dvmF

*

* −−==

v = −eτm*

E

*m

ee

τµ =

Em

nevnej

*

2τ=−=

Collisions cause drag

Electric Field Accelerates charge

τ ≈ mean time between collisions

0=vSteady state solution:

⇒

=mobility

Electric current density (charge per second per area)

Units: n=N/V ~ L-3 v ~ L/S

current per area

average velocity

Electrical Conductivity

Em

nevnej

*

2τ=−=

Electric current density (charge per second per area)

current per area

Ej

σ≡

*

2

m

ne τσ = Electrical Conductivity

OHM’s LAW (V = I R )

n = N/V

me = mass of electron

e = charge on electron

τ = mean time between collisions

Electrical Conductivity

Em

nevnej

*

2τ=−=

Electric current density (charge per second per area)

current per area

Ej

σ≡

*

2

m

ne τσ = Electrical Conductivity

OHM’s LAW (V = I R )

n = N/V

me = mass of electron

e = charge on electron

τ = mean time between collisions

What Causes the Drag?

Bam!

Random

Collisions

On average,

I go about τ seconds between

collisions

with phonons

and impurities

electron

phonon

Bam!

Random

Collisions

On average,

I go about τ seconds between

collisions

with phonons

and impurities

electron

phonon

Scattering

It turns out that static ions do not cause collisions!

What causes the drag? (Otherwise metals would have infinite conductivity)

Electrons colliding with phonons (T > 0)

Electrons colliding with impurities

( ) ∞→= 0Tphτ

τimp is independent of T

Mathiesen’s Rule

( ) impphtot T τττ111 +=

how often electronsscatter total how often electrons scatter

from phonons

how often electrons scatter fromimpurities

Independent scattering processes means the RATES can be added.

5 phonons per sec. + 7 impurities per sec.

= 12 scattering events per second

Mathiesen’s Rule

( ) impphtot T τττ111 +=

how often electronsscatter total how often electrons scatter

from phonons

how often electrons scatter fromimpurities

Independent scattering processes means the RATES can be added.

5 phonons per sec. + 7 impurities per sec.

= 12 scattering events per second

Mathiesen’s Rule ( ) impphtot T τττ111 +=

em

ne τσ2

=τ

ρ 12ne

me=

ph

eph ne

m

τρ 1

2=

imp

eimp ne

m

τρ 1

2=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=+=

impph

eimpphtot ne

m

ττρρρ 11

2

Resistivity

If the rates add, then resistivities also add:

Resistivities Add

(Mathiesen’s Rule)

Mathiesen’s Rule ( ) impphtot T τττ111 +=

em

ne τσ2

=τ

ρ 12ne

me=

ph

eph ne

m

τρ 1

2=

imp

eimp ne

m

τρ 1

2=

⎟⎟⎠

⎞⎜⎜⎝

⎛+=+=

impph

eimpphtot ne

m

ττρρρ 11

2

Resistivity

If the rates add, then resistivities also add:

Resistivities Add

(Mathiesen’s Rule)

Thermal conductivity

⎥⎦⎤

⎢⎣⎡

⋅=

area

Ejt sec

jt = εvn

Heat current density

ε = Energy per particle

v = velocity

n = N/V

Electric current density

Heat current density

Thermal conductivity

⎥⎦⎤

⎢⎣⎡

⋅=

area

Ejt sec

jt = εvn

Heat current density

ε = Energy per particle

v = velocity

n = N/V

Electric current density

Heat current density

Thermal conductivityHeat current density

x

Heat Current Density jtot through the plane: jtot = jright - jleft

jrightjleft

Heat energy per particle passing through the plane

started an average of “l” away.

About half the particles are moving right, and about half to the left.

x

Thermal conductivityHeat current density

x

Heat Current Density jtot through the plane: jtot = jright - jleft

jrightjleft

Heat energy per particle passing through the plane

started an average of “l” away.

About half the particles are moving right, and about half to the left.

x

Thermal conductivityHeat current density

x

Limit as l goes small:

Thermal conductivityHeat current density

x

Limit as l goes small:

Thermal conductivityHeat current density

x

Thermal conductivityHeat current density

x

Thermal conductivityHeat current density

x

Tx

T ∇→∂∂ 22222 3 xzyx vvvvv =++=

Tcvj vt ∇−=

τ2

3

1Tjt ∇−= κ vcv τκ 2

3

1=

How does it depend on temperature?

Thermal conductivityHeat current density

x

Tx

T ∇→∂∂ 22222 3 xzyx vvvvv =++=

Tcvj vt ∇−=

τ2

3

1Tjt ∇−= κ vcv τκ 2

3

1=

How does it depend on temperature?

Thermal conductivityvcv τκ 2

3

1=

2

2

1FeF vmE = cv =

π 2

2nkB

TTF

⎛⎝⎜

⎞⎠⎟

κ =132EF

me

⎛⎝⎜

⎞⎠⎟τ π

2

2nkB

TTF

⎛⎝⎜

⎞⎠⎟

=π 2nkB

2Tτ3me

=κ

=π 2kB

2T3

nτme

⎛⎝⎜

⎞⎠⎟

⇒

Thermal conductivityvcv τκ 2

3

1=

2

2

1FeF vmE = cv =

π 2

2nkB

TTF

⎛⎝⎜

⎞⎠⎟

κ =132EF

me

⎛⎝⎜

⎞⎠⎟τ π

2

2nkB

TTF

⎛⎝⎜

⎞⎠⎟

=π 2nkB

2Tτ3me

=κ

=π 2kB

2T3

nτme

⎛⎝⎜

⎞⎠⎟

⇒

Wiedemann-Franz Ratio

⎟⎟⎠

⎞⎜⎜⎝

⎛=

e

B

m

nTk τπκ3

22

⎟⎟⎠

⎞⎜⎜⎝

⎛=

em

ne

τσ 2

28

22

1045.23 κπ

σκ Ω×=⎟

⎠⎞⎜

⎝⎛= − W

e

k

TB

Fundamental Constants !

Cu: = 2.23 × 10-8 WΩ/κ2 (Good at low Temp)

Major Assumption: τthermal = τelectronic

Good @ very hi T & very low T

(not at intermediate T)

Wiedemann-Franz Ratio

⎟⎟⎠

⎞⎜⎜⎝

⎛=

e

B

m

nTk τπκ3

22

⎟⎟⎠

⎞⎜⎜⎝

⎛=

em

ne

τσ 2

28

22

1045.23 κπ

σκ Ω×=⎟

⎠⎞⎜

⎝⎛= − W

e

k

TB

Fundamental Constants !

Cu: = 2.23 × 10-8 WΩ/κ2 (Good at low Temp)

Major Assumption: τthermal = τelectronic

Good @ very hi T & very low T

(not at intermediate T)

Homework Problem 3 “rs”

Radius of sphere denoting volume perconduction electron

n=N/V=density of conduction electronsVN

≡1n=43πrs

3

rs =34πn

⎛⎝⎜

⎞⎠⎟1/3

In 3D

Defines rs

Homework Problem 3 “rs”

Radius of sphere denoting volume perconduction electron

n=N/V=density of conduction electronsVN

≡1n=43πrs

3

rs =34πn

⎛⎝⎜

⎞⎠⎟1/3

In 3D

Defines rs

Solid State Simulations

http://www.physics.cornell.edu/sss/

Go download these and play with them!

For this week, try the simulation “Drude”

Today

Measuring the occupied density of states

Effective Mass

Electrical Conductivity

Thermal Conductivity

Wiedemann-Franz Ratio

Heat Capacity 3BTATC +=

Electrons Phonons