MTLE-6120: Advanced Electronic Properties of Materials Fermi theory - - PowerPoint PPT Presentation

MTLE-6120: Advanced Electronic Properties of Materials Fermi theory of metals

Reading:

◮ Kasap: 4.6 - 4.7, 4.10 - 4.11

1

Band theory (vs. free electrons)

◮ Band energies E = En(

k) with complex dependence (vs. E = 2k2/(2m))

◮ Group velocity

vn( k) = ∇

kE/ (vs.

v = k/m)

◮ Effective mass tensor ¯

m∗

n(

k) = 2 ∇

k∇ kEn(

k) −1 (vs. m∗ = m)

◮ Density of states g(E) = n

• 2d

k (2π)3 δ(E − En(

k)) (vs. g(E) =

√ E 2π2

√

2m

• 3

)

◮ Gaps in energy, usually at high-symmetry points in Brillouin zone such as

• k = 0 or zone boundaries (vs. all E > 0 allowed)

◮ Metals if HOMO = LUMO and semiconductor/insulator if not

(vs. no gap ⇒ always metallic)

2

Fermi statistics

◮ At temperature T and chemical potential µ, each electronic state of energy

E has average occupation f(E) = 1 1 + exp E−µ

kBT ◮ In contrast, classical occupation exp µ−E kBT

1 µ-2kBT µ µ+2kBT E f(E) Fermi Classical

3

Electron number

◮ Number of states per energy per volume = g(E) ◮ Average occupation per state of energy E at temperature T = f(E) ◮ Average number of electrons per volume at temperature T, n =

• dEg(E)f(E)

◮ Classically, for a free electron gas

n = ∞ dE √ E 2π2 √ 2m

• 3

exp µ − E kBT = 1 2π2 √ 2m

• 3

exp µ kBT ∞ dEE1/2 exp −E kBT = 1 2π2 √ 2m

• 3

exp µ kBT Γ(3/2)

√π/2

(kBT)3/2 = 1 4

• 2mkBT/π
• 3

exp µ kBT

4

Chemical potential: classical

◮ Classically, at finite temperature T,

n = 1 4

• 2mkBT/π
• 3

exp µ kBT

◮ Electron number density given, chemical potential varies with temperature

µ(T) = −kBT ln   1 4n

• 2mkBT/π
• 3



◮ Classically, µ decreases with T as ∼ −T ln T, with µ → 0 as T → 0 ◮ This is the correct behavior for gases, but not electrons!

5

Electron number: quantum at T = 0

◮ Fermi function f(E) = 1/

• 1 + exp E−µ

kBT

• → 1 for E < µ− few kBT

and f(E) → 0 for E > µ+ few kBT.

◮ Therefore, for T → 0, f(E) → Θ(µ − E) ≡

• 1,

E < µ 0, E > µ

◮ Number of electrons at T = 0 is

n = µ dEg(E) (general) = µ dE √ E 2π2 √ 2m

• 3

(free electrons) = 1 3π2 √2mµ

• 3

⇒ µ = 2 2m(3π2n)2/3, a non-zero constant

6

Electron number: finite T corrections

For constant n, let’s find the change in µ for small changes in T from T = 0, n =

• dEg(E)f(E)

⇒ 0 = ∂n ∂T =

• dEg(E)

∂f ∂T + ∂f ∂µ · ∂µ ∂T

• =
• dEg(E)
• ∂

∂T

• 1

1 + exp E−µ

kBT

• + ∂

∂µ

• 1

1 + exp E−µ

kBT

• · ∂µ

∂T

• =
• dEg(E)

exp E−µ

kBT

• 1 + exp E−µ

kBT

2 E − µ kBT 2 + 1 kBT · ∂µ ∂T

• =
• dEg(E)

1 4kBT cosh2 E−µ

2kBT

• sharp peak at E = µ

E − µ T + ∂µ ∂T

• =
• dE g(µ) + g′(µ)(E − µ) + · · ·

4kBT cosh2 E−µ

2kBT

E − µ T + ∂µ ∂T

• 7

Electron number: finite T corrections (contd.)

0 = ∂n ∂T =

• dE g(µ) + g′(µ)(E − µ) + · · ·

4kBT cosh2 E−µ

2kBT

E − µ T + ∂µ ∂T

• ≡ 1

T (I1g(µ) + I2g′(µ)) + ∂µ ∂T (I0g(µ) + I1g′(µ)) where In ≡

• dE

(E − µ)n 4kBT cosh2 E−µ

2kBT

=      1, n = 0 0, n = 1 (πkBT)2/3, n = 2 Therefore ∂µ ∂T = − I2g′(µ) I0Tg(µ) = −(πkB)2g′(µ)T 3g(µ) ⇒ µ(T) = µ(0)−(πkBT)2g′(µ) 6g(µ) For g(E) ∝ √ E, g′(µ)/g(µ) = 1/(2µ) ⇒ µ(T) = µ(0)

• 1 − (πkBT)2

12µ(0)2

• 8

Electron chemical potential: typical numbers

◮ Sodium: BCC 4.23 ˚

A, 1 valence electron/atom ⇒ n ≈ 2.6 × 1028 m-3

◮ Aluminum: FCC 4.05 ˚

A, 3 valence electrons/atom ⇒ n ≈ 1.8 × 1029 m-3

◮ µ(0) = 2 2m(3π2n)2/3 ≈ 3.2 eV for Na, and ≈ 12 eV for Al ◮ In comparison, kBT ≈ 0.026 eV at 300 K and ≈ 0.26 eV at 3000 K

(> Tmelt)

◮ Therefore µ(T) = µ(0)

• 1 − (πkBT )2

12µ(0)2

• is

essentially constant over relevant T range!

◮ Zero temperature chemical potential µ(0) ≡ EF , Fermi energy

9

Properties at the Fermi energy

◮ Fermi energy EF separates occupied states and unoccupied states at T = 0 ◮ For free electrons, EF = µ0 = 2 2m(3π2n)2/3 ◮ With band structure E = En(

k), Fermi surface ≡ set of k with E = EF

◮ For free electrons E(

k) = 2k2/2m, the Fermi surface is a sphere of radius kF = √2mEF / = (3π2n)1/3

◮ Fermi velocity vF = average magnitude of group velocity on Fermi surface ◮ For free electrons, vF = kF /m ◮ Many electronic properties of metals determined by Fermi properties alone

(exclusively a function of electron density for free electrons)

◮ Fermi-energy density of states g(EF ) ◮ For free electrons, g(ǫF ) = √EF 2π2

√

2m

• 3

=

3n 2EF

10

Electronic heat capacity: classical

Average energy in free electron gas of density n: CV ≡ dU dT = d dT

• dEEg(E)f(E)

= d dT

• dEEg0

√ Ee(µ−E)/(kBT ) = d dT g0eµ/(kBT )

• dEE3/2e−E/(kBT )

= d dT g0eµ/(kBT )Γ(5/2)(kBT)5/2 n ≡

• dEg(E)f(E) = g0eµ/(kBT )Γ(3/2)(kBT)3/2

⇒ CV = d dT

• nΓ(5/2)(kBT)5/2

Γ(3/2)(kBT)3/2

• = 3

2nkB A constant which looks familiar because equipartition theorem!

11

Electronic heat capacity: quantum

CV =

• dEEg(E)

∂f(E) ∂T + ∂f(E) ∂µ · dµ dT

• (just extra E than in n)

=

• dE

Eg(E) 4kBT cosh2 E−µ

2kBT

E − µ T + dµ dT

• =
• dE µg(µ) + (µg(µ))′(E − µ) + · · ·

4kBT cosh2 E−µ

2kBT

E − µ T − I2g′(µ) I0Tg(µ)

• = µg(µ)

I1 T − I0 I2g′(µ) I0Tg(µ)

• + (µg(µ))′

I2 T − I1 I2g′(µ) I0Tg(µ)

• + · · ·

(In defined in finite T corrections to µ derivation) = 0 − µg′(µ)I2 T + (µg(µ))′ I2 T − 0 + · · · (since I1 = 0) = g(µ)I2 T + · · · = g(EF )π2k2

BT

3 + O(T 2) (general g(Ef)) = 3 2nkB π2kBT 3EF (free-electron g(EF ))

12

Electronic heat capacity: comparison

◮ Classical: CV = 3 2nkB (equipartition) ◮ Quantum: CV = 3 2nkB π2kBT 3EF

(equipartition)

◮ Quantum mechanical result reduced by factor ∼ kBT/EF because only

electrons near Fermi energy ‘participate’

◮ Same reason for relative constancy of µ in quantum case ◮ Electrons in metal behave classically only when kBT ∼ EF , which is

∼ 3 × 104 K for Na and ∼ 1.4 × 105 K for Al, i.e. never!

13

Electronic density of states: Al free electron

10 20 30

• 10
• 5

5 10 E-EF [eV] g(E) [eV-1nm-3] All states Occupied states (T = 300 K) Occupied states (T = 3000 K) Occupied states (T = 30000 K) ◮ Parabolic DOS, EF ≈ 11.8 eV: now plotted relative to E − EF ◮ Heat stored by moving electrons ∼ kBT below Fermi level by ∼ kBT ◮ Therefore, U ∝ T 2 and CV ∝ T ◮ Only narrow window around Fermi level participates even at Tmelt ◮ Resembles Maxwell-Boltzmann (classical) distribution only for kBT ∼ EF

14

Electronic density of states: real metals

DOS resembles free electrons for an energy window around Fermi level for best conducting metals (also the plasmonic metals)

0.1 0.2 0.3 0.4 DOS [1029 eV-1 m-3]

a) Al

PBEsol+U (this work) Lin et al. 2008 free electron 0.5 1 1.5 2 2.5

b) Ag

0.5 1 1.5

• 10
• 5

5 10 DOS [1029 eV-1 m-3]

ε-εF [eV]

c) Au

1 2 3

• 10
• 5

5 10

ε-εF [eV]

d) Cu

• Phys. Rev. B 91, 075120 (2016)

15

Electronic heat capacity: real metals

Linear heat capacity till kBT accesses difference from free electron model

2 4 6 8

Ce [105 J/m3K]

a) Al

• Eq. 3 (this work)

Lin et al. 2008 Sommerfeld 4 8 12

b) Ag

4 8 12 2 4 6 8

Ce [105 J/m3K] Te [103 K]

c) Au

5 10 15 20 2 4 6 8

Te [103 K]

d) Cu

• Phys. Rev. B 91, 075120 (2016)

16

Fermi surfaces: real metals

Fermi surface somewhat spherical for best conducting metals

ACS Nano 10, 957 (2016)

17

Fermi surface: density of states

◮ Consider arbitrary shaped equi-energy surfaces in k-space ◮ (For E = EF , it be the Fermi surface) ◮ Let A(E) be the area in k-space of this surface (with elements dA) ◮ Number of states between E and E + dE is

dN = 2 (2π)3

• dAdk

= 2 (2π)3

• dA dk

dE dE = 2 (2π)3

• dA

1 v(k)dE = 2 (2π)3 A(E)dE 1 ¯ v(E) ≡ g(E)dE ⇒ g(E) = 2A(E) (2π)3¯ v(E)

18

Fermi surface: electronic current

◮ Apply electric field

E

◮ Force on electrons −e

E

◮ Average change in momentum −e

Eτ

◮ Average change in

k is K = −e

Eτ

• ◮ Effectively field displaces Fermi surface by

K

◮ Current without field must be zero

• j =
• d

k 2 (2π)3 Θ(EF − E)(−e v( k)) = 0

◮ Current only carried by difference between old and new Fermi surfaces! ◮ Normal displacement of Fermi surface ˆ

v · K, so

• j =

2 (2π)3

• dA(ˆ

v · K)(−e v)

19

Fermi surface: electronic conductivity

• j =

2 (2π)3

• dA(ˆ

v · K)(−e v) = 2 (2π)3 A(EF )(−e¯ v(EF ))ˆ v(ˆ v · K) = 2A(EF ) (2π)3 (−e¯ v(EF ))

• K

3 = (¯ v(EF )g(EF ))(−e¯ v(EF ))−e Eτ 3 = e2v2

F g(EF )τ

3

• E

⇒ σ = e2v2

F g(EF )τ

3

◮ Note that ˆ

v(ˆ v · K) = K/3 is for isotropic Fermi surface

◮ In general case

j is not parallel to K ⇒ σ is a tensor

20

Fermi surface conductivity vs. Drude model

◮ Fermi surface result:

σ = e2v2

F g(EF )τ

3

◮ For free electron model with Fermi wave-vector kF :

σ = 1 3e2 kF m 2 √EF 2π2 √ 2m

• 3

τ = 1 3e2 kF m 2 √2mEF / 2π2 √ 2m

• 2

τ = 1 3e2 2k2

F

m2 kF 2π2 2m 2 τ = e2τ m k3

F

3π2 = ne2τ m which is exactly the Drude model result!

◮ Conceptual difference: in Drude model all electrons contribute to σ ◮ In Fermi theory, only electrons near the Fermi surface do!

21

Phonons in 1D

◮ 1D chain of atoms of mass M connected by springs K ◮ Frequencies ω = 2

• K

M

• sin ka

2

• ◮ Group velocity at k → 0 is vL = a
• K

M , the sound velocity

2(K/M)1/2

• π/a

+π/a Frequency ω Wavevector k

22

Phonons in 3D (eg. Au)

10 20 Γ X W L Γ K Energy [meV]

◮ Phonons have a band structure, just like electrons ◮ Key difference: linear near k = 0(Γ) rather than quadratic ◮ In 3D isotropic mateirals: two sound speeds: longitudinal (vL) and

transverse (vT )

◮ Two tranverse modes degenerate near k = 0 only (three polarizations) ◮ Compare energy scale against electrons: factor of 100 smaller

23

Phonon density of states

◮ Number of states per unit volume in

k per unit volume =

1 (2π)d ◮ Phonon dispersion relations ω = ωn(

k) (many bands)

◮ Therefore density of states g(ω) = n

• d

k (2π)d δ(ω − ωn(

k))

◮ Need a simple model to evaluate analytically (analogous to free electrons)

24

Debye model

◮ How many total phonon states in one band per unit cell? One! ◮ Debye model: keep linear dispersion with correct total number of states ◮ If ω = v|

k| for one band, density of states g(ω) =

• d

k (2π)d δ(ω − v| k|) = xdkd−1dk (2π)d δ(ω − vk) = xdwd−1dw (2πv)d δ(ω − w) (w = vk) = xdωd−1 (2πv)d = 4πω2 (2πv)3 (in 3D)

25

Debye frequency

◮ For linear phonon dispersion in 3D

g(ω) = 4πω2 (2πv)3

◮ If this is true for all ω, then total number of modes is

• dωg(ω) = ∞

◮ Debye model: cutoff model at ωD to keep correct number of states ◮ Number of states per unit volume (one band) = nion (# ions / volume) ◮ Therefore impose condition:

nion = ωD dωg(ω) = ωD dω 4πω2 (2πv)3 = 4πω3

D

3(2πv)3 = ω3

D

6π2v3 ⇒ ωD = v(6π2nion)1/3

◮ Corresponding cutoff in wave vector, kD = ωD/v = (6π2nion)1/3 ◮ Note similarity to kF = (3π2n)1/3 for electrons

26

Phonon density of states: real metals

60 120 180 240 300 DOS [1029 eV-1 m-3]

a) Al

• Eq. 4 (this work)

Debye 60 120 180 240 300

b) Ag

60 120 180 240 0.02 0.04 0.06 DOS [1029 eV-1 m-3]

ε [eV]

c) Au

60 120 180 240 0.02 0.04 0.06

ε [eV]

d) Cu

◮ Using two Debye models for velocities vL and vT ◮ Note energy scale ED = kBTD = ωD

• Phys. Rev. B 91, 075120 (2016)

27

Bose statistics

◮ At temperature T each phononic state of energy E has average occupation

nph(E) = 1 exp

E kBT − 1 ◮ For massless bosons like photons and phonons, µ = 0 because number not

constrained

1 µ-2kBT µ µ+2kBT E f(E) Bose Classical Fermi

28

Lattice heat capacity

◮ Density of phonon modes g(ω) = 4πω2 (2πv)3 Θ(ωD − ω) (Debye model) ◮ Phonon modes occupied by Bose function nph(ω) at temperature T ◮ Internal energy per unit volume:

U = ∞ dω(ω)g(ω)nph(ω) = ∞ dω(ω) 4πω2 (2πv)3 Θ(ωD − ω) 1 exp

ω kBT − 1

=

• 2π2v3

ωD ω3dω exp

ω kBT − 1

29

Lattice heat capacity: high T (T ≫ TD)

◮ At high temperatures ω ≪ kBT ⇒ exp ω kBT − 1 ≈ ω kBT ◮ Correspondingly, internal energy is approximately

U ≈

• 2π2v3

ωD ω3dω

ω kBT

= 1 2π2v3 ωD ω2dωkBT = 1 2π2v3 ω3

D

3 kBT = k3

D

6π2 kBT = nionkBT

◮ Accounting for the three polarizations, U = 3nionkBT and CV = 3nionkB ◮ This is (of course) the equipartition result: high T → classical limit

30

Lattice heat capacity: low T (T ≫ TD)

◮ At low temperatures,

U ≈

• 2π2v3

ωD ω3dω exp

ω kBT − 1

=

• 2π2v3

kBT

• 4

ωD kBT

x3dx ex − 1 x ≡ ω kBT ≈ k4

BT 4

2π23v3 ∞ x3dx ex − 1 ωD kBT = TD T ≫ 1 = k4

BT 4

2π23v3 · π4 15 = π2k4

BT 4

303v3

◮ Accounting for one longitudinal and two transverse velocities:

U = T 4 π2k4

B

303 1 v3

L

+ 1 v3

T 1

+ 1 v3

T 2

• ∝ T 4

◮ And corresponding heat capacity

CV ≡ dU dT = T 3 2π2k4

B

153 1 v3

L

+ 1 v3

T 1

+ 1 v3

T 2

• ∝ T 3

31

Lattice heat capacity: real metals

Debye model does a good job! Proptional to T 3 for T ≪ TD, constant 3nionkB for T ≫ TD (classical equipartition result)

10 20 30 40

Cl [105 J/m3K]

a) Al

• Eq. 5 (this work)

Debye 10 20 30 40

b) Ag

10 20 30 0.5 1 1.5 2

Cl [105 J/m3K] Tl [103 K]

c) Au

10 20 30 0.5 1 1.5 2

Tl [103 K]

d) Cu

• Phys. Rev. B 91, 075120 (2016)

32

Heat capacity: electrons vs lattice

Mostly lattice! Except at very low T (< 10 K) when T wins over T 3

102 103 104 105 106 C [J/m3K] a) Al Total Lattice Electronic 102 103 104 105 106 b) Ag 102 103 104 105 106 100 101 102 103 C [ J/m3K] T [K] c) Au 102 103 104 105 106 100 101 102 103 T [K] d) Cu

33

Electronic thermal conductivity setup

◮ Energy current due to one electron E

v (contrast with −e v for charge current)

◮ At energy E, energy flux g(E)f(E)(Ev(E)) in random directions,

averages to zero

◮ Non-uniform temperature ⇒ average energy flow in one direction ◮ Assume constant temperature gradient T(x) = T0 + x dT dx ◮ What is the net flux of energy across x = 0?

qx =

• dx
• dEg(E)f(E, T(x))(Ev(E))

· 1

2πd cos θ 4π

· e−(−x/ cos θ)/λ

λ

, x < 0 −

−1 2πd cos θ 4π

· e−(−x/ cos θ)/λ

λ

, x > 0

34

Electronic thermal conductivity derivation

qx =

• dx
• dEg(E)f(E, T(x))(Ev(E))

1

d cos θ 2

· ex/(λ cos θ)

λ

, x < 0 −

−1 d cos θ 2

· ex/(λ cos θ)

λ

, x > 0 =

• dx
• dEg(E)
• f(E, T0)
• →0 (equilibrium)

+∂f(E, T ∂T xdT dx

• (Ev(E))

· 1

d cos θ 2

· ex/(λ cos θ)

λ

, x < 0 −1

d cos θ 2

· ex/(λ cos θ)

λ

, x > 0 = dT dx

• dEg(E)v(E)E ∂f(E, T)

∂T

• 1

d cos θ 2 −∞ dxx ex/(λ cos θ) λ

− −1

d cos θ 2

∞ dxx ex/(λ cos θ)

λ

• = dT

dx

• dEg(E)v(E)E ∂f(E, T)

∂T

• −

1

d cos θ 2

λ cos2 θ − −1

d cos θ 2

λ cos2 θ

• = −dT

dx

• dEg(E)v(E)E ∂f

∂T λ/3 κ = τ 3

• dEg(E)v2(E)E ∂f(E, T)

∂T (λ = vτ)

35

Electronic thermal conductivity: result

Substitute Fermi function in thermal conductivity expression: κ = τ 3

• dEg(E)v2(E)E ∂f(E, T)

∂T ≈ v2

F τ

3

• dEg(E)E ∂f(E, T)

∂T ∂f ∂T sharply peaked

• = v2

F τ

3 CV = v2

F τ

3 · g(EF )π2k2

BT

3 = v2

F g(EF )τ

3 · π2k2

BT

3 Note similarity to electrical conductivity σ = v2

F g(EF )τ

3

· e2. Therefore expect Lorenz number L ≡ κ σT = π2k2

B

3e2 ≈ 2.44 × 10−8(V/K)2 to be same temperature-independent constant across metals (not just free-electron-like ones)

36

Wiedemann-Franz law

L ≡ κ σT = π2k2

B

3e2 ≈ 2.44 × 10−8(V/K)2 Metal L at T = 100 K (V/K)2 L at T = 273 K (V/K)2 Copper 1.9 × 10−8 2.3 × 10−8 Gold 2.0 × 10−8 2.4 × 10−8 Aluminum 1.5 × 10−8 2.2 × 10−8 Zinc 1.8 × 10−8 2.3 × 10−8 Lead 2.0 × 10−8 2.5 × 10−8

37

Lattice thermal conductivity

◮ Phonons cannot transportc harge, but they can transport heat. ◮ Our derivation for electrons didn’t assume electrons till this point:

κe = τ 3

• dEg(E)v2(E)E ∂f(E, T)

∂T

◮ Lattice contribution: phonon DOS and Bose occupations instead

κL = τph 3

• dωg(ω)v2(ω)ω ∂nph(ω, T)

∂T

◮ Which is larger? ◮ For semiconductors / insulators: few electrons (and holes) ⇒ κL dominates ◮ For metals:

◮ n ≈ nion ◮ vF ∼ 106 m/s ≫ vL, vT ∼ 103 − 104 m/s (electrons win) ◮ τph ∼ ps, while τe ∼ 10 fs (phonons win)

◮ Net result: κe dominates at room temperature, κL important at high T ◮ Small κL ⇒ Lorenz number (based only on κe) close to ideal

38

Electron-phonon scattering rate

◮ All transport coefficients depend on scattering time τ ◮ For electrons: τ determined by electron-phonon scattering ◮ Crude argument during Drude discussion:

◮ Electrons only scatter against displaced ions (now we know why: band

theory)

◮ Cross-section ∝ mean-squared displacement ∝ kBT (equipartition) ◮ Therefore τ −1 ∝ T, and ρ =

m ne2τ ∝ T

◮ Why is this not true at low T? ◮ Equipartion no longer valid for T ≪ TD

100 200 300

T emperature [K]

1 2 3 4

Resistivity [10-8 m]

P u r e c

• p

p e r C

• l

d w

• r

k e d + 1 % N i + 2 % N i

39

Electron-phonon scattering rate estimation

◮ Process: electron in one state absorbs (or emits) a phonon to go into

another state

◮ Phonon absorption rate of electron in state i (Fermi’s Golden rule):

τ −1 = 2π

• j

|j|H′|i|2δ(Ej − Ei − ω) (using j for final state to prevent mix up with Fermi energy EF )

◮ Ingredient 1: sum over states j ◮ Count up electronic states using

• dEjg(Ej)

◮ State must be unoccupied (Fermi statistics): (1 − f(Ej))

(was not relevant before because only one electron)

◮ Also need to sum over phonons available to absorb:

• dωg(ω)nph(ω)

◮ Net result:

• j

→

• dEjg(Ej)(1 − f(Ej))
• dωg(ω)nph(ω)

40

Electron-phonon scattering: matrix element

◮ Matrix element:

j|H′|i =

• d

rψ∗

j (

r)∆Vph( r)ψi( r) where ∆V ( r) is change in electron potential due to phonon

◮ For estimate, make similar to argument made in classical case ◮ Instead of equipartition, apply to single phonon mode ◮ Mean-squared atom displacement ∝ ω ◮ ∆V (

r) ∝ displacement of atoms (not squared)

◮ Therefore assume

j|H′|i ∝ √ ω ≈

• ce-phω(say)

(constant of proportionality ce-ph has dimensions of energy)

◮ Kasap 4.10.4 argues incorrectly that many e-ph scatters needed for

momentum relaxation; real ω dependence is from matrix element i.e. probability of scattering

41

Electron-phonon scattering: selection rule

◮ So far assumed that for any i, j

j|H′|i ≈

• ce-phω

◮ But many pairs of i, j are not allowed by momentum conservation:

• ki +

kph = kj

◮ So instead of all states at Ej, only count those with correct momentum ◮ Within free electron model, states of

constant Ej on sphere of radius kj

◮ LHS on sphere of radius kph centered on

ki

◮ Also note ki ≈ kj by energy conservation

(phonon energies negligible)

◮ ⇒ momentum conserving states ∝ 2πkph ∝ ω ◮ Assume probability of momentum conservation = c1ω

42

Electron-phonon scattering: Fermi golden rule

◮ Collecting ingredients together:

τ −1 ≈ 2π

• dEjg(Ej)(1 − f(Ej))
• dωg(ω)nph(ω)

· c1ω

• ce-phω

2 δ(Ej − Ei − ω)

◮ Neglect phonon energies ω ≪ Ej, Ei (two orders smaller)

τ −1 ≈ 2π

• dEjg(Ej)(1 − f(Ej))
• dωg(ω)nph(ω)

· c1ce-ph(ω)2δ(Ej − Ei) = 2π g(Ei)(1 − f(Ei))

• dω g(ω)
• ∝ω2

nph(ω)c1ce-ph(ω)2 ∝ g(Ei) ωD dω ω4 exp

ω kBT − 1

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Electron-phonon scattering: temperature dependence

◮ For high temperatures T ≫ TD, exp ω kBT − 1 ≈ ω kBT :

ρ ∝ τ −1 ∝ g(EF ) ωD dω ω4

ω kBT

∝ g(EF )T

◮ For low temperature T ≪ TD,

ρ ∝ τ −1 ∝ g(EF ) ωD dω ω4 exp

ω kBT − 1

≈ g(EF ) kBT

• 5 ∞

dx x4 ex − 1 ∝ g(EF )T 5

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Electron-phonon scattering: conclusions

◮ Temperature dependence:

ρ ∝ τ −1 ∝ g(EF )

• T,

T ≫ TD T 5, T ≪ TD explains low temperature deviations

◮ Reduction from classical result because few phonon quanta

• ccupied at low T (similar to blackbody spectrum reason)

◮ Transition metals ⇒ high d-band DOS at EF ⇒ high resistivity

100 200 300

T emperature [K]

1 2 3 4

Resistivity [10-8 m]

P u r e c

• p

p e r C

• l

d w

• r

k e d + 1 % N i + 2 % N i

45

DFT + Fermi Golden rule

Quantitative prediction of resistivities

Metal DFT ¯ τ [fs] DFT ρ [Ωm] Expt ρ [Ωm] Al 11.5 2.79 × 10−8 2.71 × 10−8 Cu 35.6 1.58 × 10−8 1.71 × 10−8 Ag 36.4 1.58 × 10−8 1.62 × 10−8 Au 26.3 2.23 × 10−8 2.26 × 10−8

Far from isotropic even for nominally free electron metals

ACS Nano 10, 957 (2016)

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