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

1 MTLE-6120: Advanced Electronic Properties of Materials Fermi theory of metals Reading: ◮ Kasap: 4.6 - 4.7, 4.10 - 4.11

2 Band theory (vs. free electrons) ◮ Band energies E = E n ( � k ) with complex dependence (vs. E = � 2 k 2 / (2 m ) ) v n ( � v = � � ◮ Group velocity � k ) = ∇ � k E/ � (vs. � k/m ) k ) = � 2 � � − 1 (vs. m ∗ = m ) n ( � k E n ( � ◮ Effective mass tensor ¯ m ∗ ∇ � k ∇ � k ) ◮ Density of states g ( E ) = � � 2d � (2 π ) 3 δ ( E − E n ( � k k )) n � √ √ � 3 E 2 m (vs. g ( E ) = ) 2 π 2 � ◮ 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)

3 Fermi statistics ◮ At temperature T and chemical potential µ , each electronic state of energy E has average occupation 1 f ( E ) = 1 + exp E − µ k B T ◮ In contrast, classical occupation exp µ − E k B T f(E) Fermi Classical 1 0 E µ -2 k B T µ +2 k B T µ

4 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 = � d Eg ( E ) f ( E ) ◮ Classically, for a free electron gas √ � √ � 3 � ∞ E 2 m exp µ − E n = d E 2 π 2 k B T � 0 � √ � 3 � ∞ 1 2 m µ d EE 1 / 2 exp − E = exp 2 π 2 k B T k B T � 0 � √ � 3 1 2 m µ ( k B T ) 3 / 2 = exp k B T Γ(3 / 2) 2 π 2 � � �� � √ π/ 2 �� � 3 = 1 2 mk B T/π µ exp 4 k B T �

5 Chemical potential: classical ◮ Classically, at finite temperature T , �� � 3 n = 1 2 mk B T/π µ exp 4 � k B T ◮ Electron number density given, chemical potential varies with temperature  � 3  ��  1 2 mk B T/π µ ( T ) = − k B T ln  4 n � ◮ 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!

6 Electron number: quantum at T = 0 � � 1 + exp E − µ ◮ Fermi function f ( E ) = 1 / → 1 for E < µ − few k B T k B T and f ( E ) → 0 for E > µ + few k B T . � 1 , E < µ ◮ Therefore, for T → 0 , f ( E ) → Θ( µ − E ) ≡ 0 , E > µ ◮ Number of electrons at T = 0 is � µ n = d Eg ( E ) (general) √ � √ � 3 � µ E 2 m = d E (free electrons) 2 π 2 � 0 � √ 2 mµ � 3 1 = 3 π 2 � µ = � 2 2 m (3 π 2 n ) 2 / 3 , ⇒ a non-zero constant

7 Electron number: finite T corrections For constant n , let’s find the change in µ for small changes in T from T = 0 , � n = d Eg ( E ) f ( E ) � ∂f � � 0 = ∂n ∂T + ∂f ∂µ · ∂µ ⇒ ∂T = d Eg ( E ) ∂T � � � � � � � ∂ 1 + ∂ 1 · ∂µ = d Eg ( E ) 1 + exp E − µ 1 + exp E − µ ∂T ∂µ ∂T k B T k B T exp E − µ � � E − µ � k B T · ∂µ 1 k B T = d Eg ( E ) k B T 2 + � � 2 ∂T 1 + exp E − µ k B T � E − µ � � 1 + ∂µ = d Eg ( E ) 4 k B T cosh 2 E − µ T ∂T 2 k B T � �� � sharp peak at E = µ � E − µ � � d E g ( µ ) + g ′ ( µ )( E − µ ) + · · · + ∂µ = 4 k B T cosh 2 E − µ T ∂T 2 k B T

8 Electron number: finite T corrections (contd.) � E − µ � � d E g ( µ ) + g ′ ( µ )( E − µ ) + · · · 0 = ∂n + ∂µ ∂T = 4 k B T cosh 2 E − µ T ∂T 2 k B T ≡ 1 T ( I 1 g ( µ ) + I 2 g ′ ( µ )) + ∂µ ∂T ( I 0 g ( µ ) + I 1 g ′ ( µ )) where  1 , n = 0  � ( E − µ ) n  I n ≡ d E = 0 , n = 1 4 k B T cosh 2 E − µ  2 k B T  ( πk B T ) 2 / 3 , n = 2 Therefore ∂T = − I 2 g ′ ( µ ) I 0 Tg ( µ ) = − ( πk B ) 2 g ′ ( µ ) T µ ( T ) = µ (0) − ( πk B T ) 2 g ′ ( µ ) ∂µ ⇒ 3 g ( µ ) 6 g ( µ ) √ E , g ′ ( µ ) /g ( µ ) = 1 / (2 µ ) ⇒ For g ( E ) ∝ � � 1 − ( πk B T ) 2 µ ( T ) = µ (0) 12 µ (0) 2

9 Electron chemical potential: typical numbers A, 1 valence electron/atom ⇒ n ≈ 2 . 6 × 10 28 m -3 ◮ Sodium: BCC 4.23 ˚ A, 3 valence electrons/atom ⇒ n ≈ 1 . 8 × 10 29 m -3 ◮ Aluminum: FCC 4.05 ˚ ◮ µ (0) = � 2 2 m (3 π 2 n ) 2 / 3 ≈ 3 . 2 eV for Na, and ≈ 12 eV for Al ◮ In comparison, k B T ≈ 0.026 eV at 300 K and ≈ 0.26 eV at 3000 K ( > T melt ) � � 1 − ( πk B T ) 2 ◮ Therefore µ ( T ) = µ (0) is 12 µ (0) 2 essentially constant over relevant T range! ◮ Zero temperature chemical potential µ (0) ≡ E F , Fermi energy

10 Properties at the Fermi energy ◮ Fermi energy E F separates occupied states and unoccupied states at T = 0 ◮ For free electrons, E F = µ 0 = � 2 2 m (3 π 2 n ) 2 / 3 ◮ With band structure E = E n ( � k ) , Fermi surface ≡ set of � k with E = E F ◮ For free electrons E ( � k ) = � 2 k 2 / 2 m , the Fermi surface is a sphere of radius k F = √ 2 mE F / � = (3 π 2 n ) 1 / 3 ◮ Fermi velocity v F = average magnitude of group velocity on Fermi surface ◮ For free electrons, v F = � k F /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 ( E F ) � √ √ E F � 3 ◮ For free electrons, g ( ǫ F ) = 2 m 3 n = 2 π 2 � 2 E F

11 Electronic heat capacity: classical Average energy in free electron gas of density n : � C V ≡ d U d T = d d EEg ( E ) f ( E ) d T � √ = d Ee ( µ − E ) / ( k B T ) d EEg 0 d T � = d d T g 0 e µ/ ( k B T ) d EE 3 / 2 e − E/ ( k B T ) = d d T g 0 e µ/ ( k B T ) Γ(5 / 2)( k B T ) 5 / 2 � d Eg ( E ) f ( E ) = g 0 e µ/ ( k B T ) Γ(3 / 2)( k B T ) 3 / 2 n ≡ � � n Γ(5 / 2)( k B T ) 5 / 2 C V = d = 3 ⇒ 2 nk B Γ(3 / 2)( k B T ) 3 / 2 d T A constant which looks familiar because equipartition theorem!

12 Electronic heat capacity: quantum � ∂f ( E ) � � + ∂f ( E ) · d µ C V = d EEg ( E ) (just extra E than in n ) ∂T ∂µ d T � E − µ � � Eg ( E ) + d µ = d E 4 k B T cosh 2 E − µ T d T 2 k B T � E − µ � � d E µg ( µ ) + ( µg ( µ )) ′ ( E − µ ) + · · · − I 2 g ′ ( µ ) = 4 k B T cosh 2 E − µ T I 0 Tg ( µ ) 2 k B T � I 1 � � I 2 � I 2 g ′ ( µ ) I 2 g ′ ( µ ) + ( µg ( µ )) ′ = µg ( µ ) T − I 0 T − I 1 + · · · I 0 Tg ( µ ) I 0 Tg ( µ ) ( I n defined in finite T corrections to µ derivation) = 0 − µg ′ ( µ ) I 2 T + ( µg ( µ )) ′ I 2 T − 0 + · · · (since I 1 = 0 ) T + · · · = g ( E F ) π 2 k 2 = g ( µ ) I 2 B T + O ( T 2 ) (general g ( E f ) ) 3 π 2 k B T = 3 2 nk B (free-electron g ( E F ) ) 3 E F

13 Electronic heat capacity: comparison ◮ Classical: C V = 3 2 nk B (equipartition) π 2 k B T ◮ Quantum: C V = 3 2 nk B (equipartition) 3 E F ◮ Quantum mechanical result reduced by factor ∼ k B T/E F because only electrons near Fermi energy ‘participate’ ◮ Same reason for relative constancy of µ in quantum case ◮ Electrons in metal behave classically only when k B T ∼ E F , which is ∼ 3 × 10 4 K for Na and ∼ 1 . 4 × 10 5 K for Al, i.e. never!

14 Electronic density of states: Al free electron g ( E ) [eV -1 nm -3 ] 30 20 All states Occupied states ( T = 300 K) Occupied states ( T = 3000 K) Occupied states ( T = 30000 K) 10 0 -10 -5 0 5 10 E-E F [eV] ◮ Parabolic DOS, E F ≈ 11 . 8 eV: now plotted relative to E − E F ◮ Heat stored by moving electrons ∼ k B T below Fermi level by ∼ k B T ◮ Therefore, U ∝ T 2 and C V ∝ T ◮ Only narrow window around Fermi level participates even at T melt ◮ Resembles Maxwell-Boltzmann (classical) distribution only for k B T ∼ E F

15 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) 2.5 DOS [10 29 eV -1 m -3 ] a) Al b) Ag 0.4 2 0.3 1.5 0.2 1 PBEsol+U (this work) 0.1 0.5 Lin et al. 2008 free electron DOS [10 29 eV -1 m -3 ] c) Au d) Cu 1.5 3 1 2 0.5 1 0 0 -10 -5 0 5 10 -10 -5 0 5 10 ε - ε F [eV] ε - ε F [eV] Phys. Rev. B 91 , 075120 (2016)

16 Electronic heat capacity: real metals Linear heat capacity till k B T accesses difference from free electron model 8 12 a) Al b) Ag C e [10 5 J/m 3 K] 6 8 4 4 Eq. 3 (this work) 2 Lin et al. 2008 Sommerfeld c) Au d) Cu C e [10 5 J/m 3 K] 12 20 15 8 10 4 5 0 0 0 2 4 6 8 0 2 4 6 8 T e [10 3 K] T e [10 3 K] Phys. Rev. B 91 , 075120 (2016)

17 Fermi surfaces: real metals Fermi surface somewhat spherical for best conducting metals ACS Nano 10 , 957 (2016)

18 Fermi surface: density of states ◮ Consider arbitrary shaped equi-energy surfaces in k -space ◮ (For E = E F , it be the Fermi surface) ◮ Let A ( E ) be the area in k -space of this surface (with elements d A ) ◮ Number of states between E and E + dE is � 2 d N = d A d k (2 π ) 3 � 2 d A d k = d E d E (2 π ) 3 � 2 1 = d A � v ( k )d E (2 π ) 3 2 1 = (2 π ) 3 A ( E )d E � ¯ v ( E ) ≡ g ( E )d E 2 A ( E ) ⇒ g ( E ) = (2 π ) 3 � ¯ v ( E )