MTLE-6120: Advanced Electronic Properties of Materials

Electron transport: phonons andelectron-phonon scattering

Reading:

I Kasap: 4.10

1

Phonons in 1D

I 1D chain of atoms of mass M connected by springs K

I Frequencies ω = 2√

KM

∣∣sin ka2

∣∣I Group velocity at k → 0 is vL = a

√KM , the sound velocity

0

2(K/M)1/2

-π/a 0 +π/a

Fre

quen

cy ω

Wavevector k

2

Phonons in 3D (eg. Au)

0

10

20

Γ X W L Γ K

Ener

gy [m

eV]

I Phonons have a band structure, just like electrons

I Key difference: linear near k = 0(Γ) rather than quadratic

I In 3D isotropic mateirals: two sound speeds: longitudinal (vL) andtransverse (vT )

I Two tranverse modes degenerate near k = 0 only (three polarizations)

I Compare energy scale against electrons: factor of 100 smaller

3

Phonon density of states

I Number of states per unit volume in ~k per unit volume = 1(2π)d

I Phonon dispersion relations ω = ωn(~k) (many bands)

I Therefore density of states g(ω) =∑n

∫d~k

(2π)dδ(ω − ωn(~k))

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

4

Debye model

I How many total phonon states in one band per unit cell? One!

I Debye model: keep linear dispersion with correct total number of states

I If ω = v|~k| for one band, density of states

g(ω) =

∫d~k

(2π)dδ(ω − v|~k|)

=

∫xdk

d−1dk

(2π)dδ(ω − vk)

=

∫xdw

d−1dw

(2πv)dδ(ω − w) (w = vk)

=xdω

d−1

(2πv)d

=4πω2

(2πv)3(in 3D)

5

Debye frequencyI For linear phonon dispersion in 3D

g(ω) =4πω2

(2πv)3

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

dωg(ω) = ∞I Debye model: cutoff model at ωD to keep correct number of states

I Number of states per unit volume (one band) = nion (# ions / volume)

I Therefore impose condition:

nion =

∫ ωD

0

dωg(ω) =

∫ ωD

0

dω4πω2

(2πv)3

=4πω3

D

3(2πv)3=

ω3D

6π2v3

⇒ ωD = v(6π2nion)1/3

I Corresponding cutoff in wave vector, kD = ωD/v = (6π2nion)1/3

I Note similarity to kF = (3π2n)1/3 for electrons

6

Phonon density of states: real metals

60

120

180

240

300

DO

S [10

29 e

V-1

m-3

]a) AlEq. 4 (this work)

Debye

60

120

180

240

300b) Ag

0

60

120

180

240

0 0.02 0.04 0.06

DO

S [10

29 e

V-1

m-3

]

ε [eV]

c) Au

0

60

120

180

240

0 0.02 0.04 0.06

ε [eV]

d) Cu

I Using two Debye models for velocities vL and vTI Note energy scale ED = kBTD = ~ωD

Phys. Rev. B 91, 075120 (2016)

7

Bose statistics

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

nph(E) =1

exp EkBT− 1

I For massless bosons like photons and phonons, µ = 0 because number notconstrained

0

1

µ-2kBT µ µ+2kBTE

f(E)Bose

Classical

Fermi

8

Lattice heat capacity

I Density of phonon modes g(ω) = 4πω2

(2πv)3 Θ(ωD − ω) (Debye model)

I Phonon modes occupied by Bose function nph(~ω) at temperature T

I Internal energy per unit volume:

U =

∫ ∞0

dω(~ω)g(ω)nph(ω)

=

∫ ∞0

dω(~ω)4πω2

(2πv)3Θ(ωD − ω)

1

exp ~ωkBT− 1

=~

2π2v3

∫ ωD

0

ω3dω

exp ~ωkBT− 1

9

Lattice heat capacity: high T (T TD)

I At high temperatures ~ω kBT ⇒ exp ~ωkBT− 1 ≈ ~ω

kBT

I Correspondingly, internal energy is approximately

U ≈ ~2π2v3

∫ ωD

0

ω3dω~ωkBT

=1

2π2v3

∫ ωD

0

ω2dωkBT

=1

2π2v3ω3D

3kBT

=k3D6π2

kBT = nionkBT

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

10

Lattice heat capacity: low T (T TD)I At low temperatures,

U ≈ ~2π2v3

∫ ωD

0

ω3dω

exp ~ωkBT− 1

=~

2π2v3

(kBT

~

)4 ∫ ~ωDkBT

0

x3dx

ex − 1x ≡ ~ω

kBT

≈ k4BT4

2π2~3v3

∫ ∞0

x3dx

ex − 1

~ωDkBT

=TDT 1

=k4BT

4

2π2~3v3· π

4

15=π2k4BT

4

30~3v3

I Accounting for one longitudinal and two transverse velocities:

U = T 4π2k4B

30~3

(1

v3L+

1

v3T1

+1

v3T2

)∝ T 4

I And corresponding heat capacity

CV ≡dU

dT= T 3 2π2k4B

15~3

(1

v3L+

1

v3T1

+1

v3T2

)∝ T 3

11

Lattice heat capacity: real metals

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

10

20

30

40

Cl [

10

5 J

/m3K

] a) Al

Eq. 5 (this work)Debye

10

20

30

40b) Ag

0

10

20

30

0 0.5 1 1.5 2

Cl [

10

5 J

/m3K

]

Tl [103 K]

c) Au

0

10

20

30

0 0.5 1 1.5 2

Tl [103 K]

d) Cu

Phys. Rev. B 91, 075120 (2016)

12

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/m

3K

]

a) Al

TotalLattice

Electronic 102

103

104

105

106 b) Ag

102

103

104

105

106

100

101

102

103

C [

J/m

3K

]

T [K]

c) Au

102

103

104

105

106

100

101

102

103

T [K]

d) Cu

13

Electronic thermal conductivity setupI Energy current due to one electron E~v

(contrast with −e~v for charge current)

I At energy E, energy flux g(E)f(E)(Ev(E)) in random directions,averages to zero

I Non-uniform temperature ⇒ average energy flow in one direction

I Assume constant temperature gradient T (x) = T0 + xdTdx

I What is the net flux of energy across x = 0?

qx =

∫dx

∫dEg(E)f(E, T (x))(Ev(E))

·

∫ 1

02πd cos θ

4π · e−(−x/ cos θ)/λ

λ , x < 0

−∫ 0

−12πd cos θ

4π · e−(−x/ cos θ)/λ

λ , x > 0

14

Electronic thermal conductivity derivation

qx =

∫dx

∫dEg(E)f(E, T (x))(Ev(E))

∫ 1

0d cos θ

2 · ex/(λ cos θ)

λ , x < 0

−∫ 0

−1d cos θ

2 · ex/(λ cos θ)

λ , x > 0

=

∫dx

∫dEg(E)

(f(E, T0)︸ ︷︷ ︸∫→0 (equilibrium)

+∂f(E, T

∂Tx

dT

dx

)(Ev(E))

·

∫ 1

0d cos θ

2 · ex/(λ cos θ)

λ , x < 0∫ −10

d cos θ2 · e

x/(λ cos θ)

λ , x > 0

=dT

dx

∫dEg(E)v(E)E

∂f(E, T )

∂T

∫ 1

0d cos θ

2

∫ 0

−∞ dxx ex/(λ cos θ)

λ

−∫ −10

d cos θ2

∫∞0

dxx ex/(λ cos θ)

λ

=dT

dx

∫dEg(E)v(E)E

∂f(E, T )

∂T

−∫ 1

0d cos θ

2 λ cos2 θ

−∫ −10

d cos θ2 λ cos2 θ

= −dT

dx

∫dEg(E)v(E)E

∂f

∂Tλ/3

κ =τ

3

∫dEg(E)v2(E)E

∂f(E, T )

∂T(λ = vτ)

15

Electronic thermal conductivity: result

Substitute Fermi function in thermal conductivity expression:

κ =τ

3

∫dEg(E)v2(E)E

∂f(E, T )

∂T

≈ v2F τ

3

∫dEg(E)E

∂f(E, T )

∂T

(∂f

∂Tsharply peaked

)=v2F τ

3CV =

v2F τ

3· g(EF )

π2k2BT

3

=v2F g(EF )τ

3· π

2k2BT

3

Note similarity to electrical conductivity σ =v2F g(EF )τ

3 · e2.Therefore expect Lorenz number

L ≡ κ

σT=π2k2B3e2

≈ 2.44× 10−8(V/K)2

to be same temperature-independent constant across metals(not just free-electron-like ones)

16

Wiedemann-Franz law

L ≡ κ

σT=π2k2B3e2

≈ 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

17

Lattice thermal conductivityI Phonons cannot transport charge, but they can transport heat.

I Our derivation for electrons didn’t assume electrons till this point:

κe =τ

3

∫dEg(E)v2(E)E

∂f(E, T )

∂T

I Lattice contribution: phonon DOS and Bose occupations instead

κL =τph3

∫dωg(ω)v2(ω)~ω

∂nph(ω, T )

∂T

I Which is larger?

I For semiconductors / insulators: few electrons (and holes)⇒ κL dominates

I For metals:I n ≈ nion

I vF ∼ 106 m/s vL, vT ∼ 103 − 104 m/s (electrons win)I τph ∼ ps, while τe ∼ 10 fs (phonons win)

I Net result: κe dominates at room temperature, κL important at high T

I Small κL ⇒ Lorenz number (based only on κe) close to ideal

18

Electron-phonon scattering rateI All transport coefficients depend on scattering time τ

I For electrons: τ determined by electron-phonon scattering

I Crude argument during Drude discussion:I Electrons only scatter against displaced ions (now we know why: band

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

ne2τ∝ T

I Why is this not true at low T?

I Equipartion no longer valid for T TD

0 100 200 300

Temperature [K]

0

1

2

3

4

Resi

stiv

ity [

10

-8 m

]

Pure copperCold worked

+ 1% Ni

+2% Ni

19

Electron-phonon scattering rate estimationI Process: electron in one state absorbs (or emits) a phonon to go into

another state

I 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 )

I Ingredient 1: sum over states∑j

I Count up electronic states using∫

dEjg(Ej)

I State must be unoccupied (Fermi statistics): (1− f(Ej))(was not relevant before because only one electron)

I Also need to sum over phonons available to absorb:∫

dωg(ω)nph(ω)

I Net result:∑j

→∫

dEjg(Ej)(1− f(Ej))

∫dωg(ω)nph(ω)

20

Electron-phonon scattering: selection ruleI So far assumed that final states counted as:∑

j

→∫

dEjg(Ej)(1− f(Ej))

∫dωg(ω)nph(ω)

I But many pairs of final states will not satisfy ~~ki + ~~kph = ~~kjI Instead of all pairs, only count those with correct momentumI RHS on sphere of radius kj ≈ ki (phonon energies negligible)

I LHS on sphere of radius kph centered on ~kiI Total pairs of ~kj ,~kph: (4πk2j ) · (4πk2ph)I Momentum conserving set: 2πkphI Fraction conserving momentum:

2πkph(4πk2j ) · (4πk2ph)

=1

8πk2jkph≈ vph

8πk2i ω

I Net final state counting:∑j

→∫

dEjg(Ej)(1− f(Ej))

∫dωg(ω)nph(ω)

vph8πk2i ω

21

Electron-phonon scattering: momentum relaxation

I Does electron-phonon scattering randomize momentum?(We assumed this about τ in Drude theory and its quantum extension.)

I Change in angle θ between ~ki and ~kj given by

cos θ =k2i + k2j − k2ph

2kikj≈ 1−

k2ph2k2i

I Extent of randomization proportional to:

1− cos θ ≈ ω2

2k2i v2ph

I To calculate momentum-relaxtion time instead ofmean-free time, include this (1− cos θ) factor:

∑j

→∫

dEjg(Ej)(1− f(Ej))

∫dωg(ω)nph(ω)

vph8πk2i ω

· ω2

2k2i v2ph

22

Electron-phonon scattering: matrix elementI Matrix element:

〈j|H ′|i〉 =

∫d~rψ∗j (~r)∆Vph(~r)ψi(~r)

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

I For estimate, make similar to argument made in classical case

I Instead of equipartition, apply to single phonon mode

I Mean-squared atom displacement ∝ ~ωI ∆V (~r) ∝ displacement of atoms (not squared)

I Therefore assume

〈j|H ′|i〉 ∝√~ω ≈

√ce-ph~ω(say)

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

23

Electron-phonon scattering: Fermi golden ruleI Collecting ingredients together:

τ−1 ≈ 2π

~

∫dEjg(Ej)(1− f(Ej))

∫dωg(ω)nph(ω)

· vph8πk2i ω

· ω2

2k2i v2ph

·(√

ce-ph~ω)2δ(Ej − Ei − ~ω)

I Neglect phonon energies ~ω Ej , Ei (two orders smaller)

τ−1 ≈ 2π

~

∫dEjg(Ej)(1− f(Ej))

∫dωg(ω)nph(ω)

· ce-ph~ω2

16πk4i vphδ(Ej − Ei)

=2π

~g(Ei)(1− f(Ei))

∫dω g(ω)︸︷︷︸∝ω2

nph(ω)ce-ph~

16πk4i vphω2

∝ g(Ei)

∫ ωD

0

dωω4

exp ~ωkBT− 1

24

Electron-phonon scattering: temperature dependenceI For high temperatures T TD, exp ~ω

kBT− 1 ≈ ~ω

kBT:

ρ ∝ τ−1 ∝ g(EF )

∫ ωD

0

dωω4

~ωkBT

∝ g(EF )T

I For low temperature T TD,

ρ ∝ τ−1 ∝ g(EF )

∫ ωD

0

dωω4

exp ~ωkBT− 1

≈ g(EF )

(kBT

~

)5 ∫ ∞0

dxx4

ex − 1

∝ g(EF )T 5

25

Electron-phonon scattering: conclusions

I Temperature dependence:

ρ ∝ τ−1 ∝ g(EF )

T, T TD

T 5, T TD

explains low temperature deviations

I Reduction from classical result because few phonon quantaoccupied at low T (similar to blackbody spectrum reason)

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

0 100 200 300

Temperature [K]

0

1

2

3

4

Resi

stiv

ity [

10

-8 m

]

Pure copperCold worked

+ 1% Ni

+2% Ni

26

DFT + Fermi Golden ruleQuantitative 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)

27