Last Time - Purdue Universityerica/545/Lecture07.pdf · Last Time Free electron model Density of...
Transcript of Last Time - Purdue Universityerica/545/Lecture07.pdf · Last Time Free electron model Density of...
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