1 MTLE-6120: Advanced Electronic Properties of Materials Electron transport: phonons and electron-phonon scattering Contents: ◮ Phonon density of states and heat capacity ◮ Thermal conductivity: electronic and lattice ◮ Electron-phonon momentum relaxation time Reading: ◮ Kasap: 4.10
2 Phonons in 1D ◮ 1D chain of atoms of mass M connected by springs K � � � K � sin ka ◮ Frequencies ω = 2 � M 2 � ◮ Group velocity at k → 0 is v L = a K M , the sound velocity 2( K/M ) 1/2 Frequency ω 0 - π / a 0 + π / a Wavevector k
3 Phonons in 3D (eg. Au) 20 Energy [meV] 10 0 X W L K Γ Γ ◮ 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 ( v L ) and transverse ( v T ) ◮ Two tranverse modes degenerate near k = 0 only (three polarizations) ◮ Compare energy scale against electrons: factor of 100 smaller
4 Phonon density of states ◮ Number of states per unit volume in � 1 k per unit volume = (2 π ) d ◮ Phonon dispersion relations ω = ω n ( � k ) (many bands) � ◮ Therefore density of states g ( ω ) = � (2 π ) d δ ( ω − ω n ( � d � k k )) n ◮ Need a simple model to evaluate analytically (analogous to free electrons)
5 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 � d � k (2 π ) d δ ( ω − v | � g ( ω ) = k | ) � x d k d − 1 d k = δ ( ω − vk ) (2 π ) d � x d w d − 1 d w = δ ( ω − w ) ( w = vk ) (2 πv ) d = x d ω d − 1 (2 πv ) d = 4 πω 2 (in 3D) (2 πv ) 3
6 Debye frequency ◮ For linear phonon dispersion in 3D g ( ω ) = 4 πω 2 (2 πv ) 3 � ◮ If this is true for all ω , then total number of modes is d ωg ( ω ) = ∞ ◮ Debye model: cutoff model at ω D to keep correct number of states ◮ Number of states per unit volume (one band) = n ion (# ions / volume) ◮ Therefore impose condition: � ω D � ω D d ω 4 πω 2 n ion = d ωg ( ω ) = (2 πv ) 3 0 0 4 πω 3 ω 3 D D = 3(2 πv ) 3 = 6 π 2 v 3 ω D = v (6 π 2 n ion ) 1 / 3 ⇒ ◮ Corresponding cutoff in wave vector, k D = ω D /v = (6 π 2 n ion ) 1 / 3 ◮ Note similarity to k F = (3 π 2 n ) 1 / 3 for electrons
7 Phonon density of states: real metals 300 300 DOS [10 29 eV -1 m -3 ] a) Al b) Ag Eq. 4 (this work) 240 240 Debye 180 180 120 120 60 60 DOS [10 29 eV -1 m -3 ] c) Au d) Cu 240 240 180 180 120 120 60 60 0 0 0 0.02 0.04 0.06 0 0.02 0.04 0.06 ε [eV] ε [eV] ◮ Using two Debye models for velocities v L and v T ◮ Note energy scale E D = k B T D = � ω D Phys. Rev. B 91 , 075120 (2016)
8 Bose statistics ◮ At temperature T each phononic state of energy E has average occupation 1 n ph ( E ) = E exp k B T − 1 ◮ For massless bosons like photons and phonons, µ = 0 because number not constrained f(E) Bose Classical Fermi 1 0 E µ -2 k B T µ +2 k B T µ
9 Lattice heat capacity 4 πω 2 ◮ Density of phonon modes g ( ω ) = (2 πv ) 3 Θ( ω D − ω ) (Debye model) ◮ Phonon modes occupied by Bose function n ph ( � ω ) at temperature T ◮ Internal energy per unit volume: � ∞ U = d ω ( � ω ) g ( ω ) n ph ( ω ) 0 � ∞ d ω ( � ω ) 4 πω 2 1 = (2 πv ) 3 Θ( ω D − ω ) � ω exp k B T − 1 0 � ω D ω 3 d ω � = 2 π 2 v 3 � ω exp k B T − 1 0
10 Lattice heat capacity: high T ( T ≫ T D ) ◮ At high temperatures � ω ≪ k B T ⇒ exp � ω � ω k B T − 1 ≈ k B T ◮ Correspondingly, internal energy is approximately � ω D ω 3 d ω � U ≈ 2 π 2 v 3 � ω 0 k B T � ω D 1 ω 2 d ωk B T = 2 π 2 v 3 0 ω 3 1 D = 3 k B T 2 π 2 v 3 = k 3 D 6 π 2 k B T = n ion k B T ◮ Accounting for the three polarizations, U = 3 n ion k B T and C V = 3 n ion k B ◮ This is (of course) the equipartition result: high T → classical limit
11 Lattice heat capacity: low T ( T ≪ T D ) ◮ At low temperatures, � ω D ω 3 d ω � U ≈ � ω 2 π 2 v 3 exp k B T − 1 0 � ωD � k B T � 4 � x 3 d x x ≡ � ω � kBT = e x − 1 2 π 2 v 3 � k B T 0 � ∞ k 4 B T 4 x 3 d x � ω D k B T = T D ≈ ≫ 1 e x − 1 2 π 2 � 3 v 3 T 0 2 π 2 � 3 v 3 · π 4 k 4 B T 4 15 = π 2 k 4 B T 4 = 30 � 3 v 3 ◮ Accounting for one longitudinal and two transverse velocities: � 1 � U = T 4 π 2 k 4 1 1 B ∝ T 4 + + 30 � 3 v 3 v 3 v 3 L T 1 T 2 ◮ And corresponding heat capacity � 1 � d T = T 3 2 π 2 k 4 C V ≡ d U 1 1 B ∝ T 3 + + v 3 v 3 v 3 15 � 3 L T 1 T 2
12 Lattice heat capacity: real metals Debye model does a good job! Proportional to T 3 for T ≪ T D , constant 3 n ion k B for T ≫ T D (classical equipartition result) 40 40 a) Al b) Ag C l [10 5 J/m 3 K] 30 30 20 20 10 Eq. 5 (this work) 10 Debye c) Au d) Cu C l [10 5 J/m 3 K] 30 30 20 20 10 10 0 0 0 0.5 1 1.5 2 0 0.5 1 1.5 2 T l [10 3 K] T l [10 3 K] Phys. Rev. B 91 , 075120 (2016)
13 Heat capacity: electrons vs lattice Mostly lattice! Except at very low T ( < 10 K) when T wins over T 3 a) Al b) Ag 10 6 10 6 10 5 10 5 C [J/m 3 K] 10 4 10 4 10 3 10 3 Total Lattice 10 2 10 2 Electronic c) Au d) Cu 10 6 10 6 10 5 10 5 C [ J/m 3 K] 10 4 10 4 10 3 10 3 10 2 10 2 10 0 10 1 10 2 10 3 10 0 10 1 10 2 10 3 T [K] T [K]
14 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 ) = T 0 + x d T d x ◮ What is the net flux of energy across x = 0 ? � � q x = d x d Eg ( E ) f ( E, T ( x ))( Ev ( E )) �� 1 · e − ( − x/ cos θ ) /λ 2 π d cos θ , x < 0 4 π λ 0 · � 0 · e − ( − x/ cos θ ) /λ 2 π d cos θ − , x > 0 − 1 4 π λ
15 Electronic thermal conductivity derivation �� 1 � � · e x/ ( λ cos θ ) d cos θ , x < 0 0 2 λ q x = d x d Eg ( E ) f ( E, T ( x ))( Ev ( E )) � 0 · e x/ ( λ cos θ ) d cos θ − , x > 0 2 λ − 1 � � � � + ∂f ( E, T x d T = d x d Eg ( E ) f ( E, T 0 ) ( Ev ( E )) ∂T d x � �� � � → 0 (equilibrium) �� 1 · e x/ ( λ cos θ ) d cos θ , x < 0 0 2 λ · � − 1 · e x/ ( λ cos θ ) d cos θ , x > 0 0 2 λ � � � 1 � 0 −∞ d xx e x/ ( λ cos θ ) � d cos θ = d T d Eg ( E ) v ( E ) E ∂f ( E, T ) 0 2 λ � − 1 � ∞ d xx e x/ ( λ cos θ ) d x ∂T d cos θ − 0 2 0 λ � � � 1 � λ cos 2 θ d cos θ = d T d Eg ( E ) v ( E ) E ∂f ( E, T ) − 0 2 � − 1 λ cos 2 θ d cos θ d x ∂T − 0 2 � = − d T d Eg ( E ) v ( E ) E ∂f ∂T λ/ 3 d x � κ = τ d Eg ( E ) v 2 ( E ) E ∂f ( E, T ) ( λ = vτ ) 3 ∂T
16 Electronic thermal conductivity: result Substitute Fermi function in thermal conductivity expression: � κ = τ d Eg ( E ) v 2 ( E ) E ∂f ( E, T ) 3 ∂T � ∂f � � ≈ v 2 F τ d Eg ( E ) E ∂f ( E, T ) ∂T sharply peaked 3 ∂T = v 2 3 C V = v 2 · g ( E F ) π 2 k 2 F τ F τ B T 3 3 = v 2 · π 2 k 2 F g ( E F ) τ B T 3 3 Note similarity to electrical conductivity σ = v 2 F g ( E F ) τ · e 2 . 3 Therefore expect Lorenz number σT = π 2 k 2 L ≡ κ B ≈ 2 . 44 × 10 − 8 (V/K) 2 3 e 2 to be same temperature-independent constant across metals (not just free-electron-like ones)
17 Wiedemann-Franz law σT = π 2 k 2 L ≡ κ B ≈ 2 . 44 × 10 − 8 (V/K) 2 3 e 2 L at T = 100 K (V/K) 2 L at T = 273 K (V/K) 2 Metal 1 . 9 × 10 − 8 2 . 3 × 10 − 8 Copper 2 . 0 × 10 − 8 2 . 4 × 10 − 8 Gold 1 . 5 × 10 − 8 2 . 2 × 10 − 8 Aluminum 1 . 8 × 10 − 8 2 . 3 × 10 − 8 Zinc 2 . 0 × 10 − 8 2 . 5 × 10 − 8 Lead
18 Lattice thermal conductivity ◮ Phonons cannot transport charge, but they can transport heat. ◮ Our derivation for electrons didn’t assume electrons till this point: � κ e = τ d Eg ( E ) v 2 ( E ) E ∂f ( E, T ) 3 ∂T ◮ Lattice contribution: phonon DOS and Bose occupations instead � κ L = τ ph d ωg ( ω ) v 2 ( ω ) � ω ∂n ph ( ω, T ) 3 ∂T ◮ Which is larger? ◮ For semiconductors / insulators: few electrons (and holes) ⇒ κ L dominates ◮ For metals: ◮ n ≈ n ion ◮ v F ∼ 10 6 m/s ≫ v L , v T ∼ 10 3 − 10 4 m/s (electrons win) ◮ τ ph ∼ ps, while τ e ∼ 10 fs (phonons win) ◮ Net result: κ e dominates at room temperature, κ L important at high T ◮ Small κ L ⇒ Lorenz number (based only on κ e ) close to ideal
19 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 ∝ k B T (equipartition) ◮ Therefore τ − 1 ∝ T , and ρ = m ne 2 τ ∝ T ◮ Why is this not true at low T ? ◮ Equipartion no longer valid for T ≪ T D 4 i N % 2 + Resistivity [10 -8 m] 3 N i % 1 + 2 d e r k o w d o l C r e p p o c e u r P 1 0 0 100 200 300 T emperature [K]
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