PH575 Spring 2014
Lecture #18 Free electron theory: Sutton Ch. 7 pp 132 -> 144; Kittel Ch. 6.
V 2m 3/2 2 2 " % 1/2 k D(E) = 2 $ 2 ' E E(k) = 2π # & 2m Assumption: electrons metal do not interact with each other, not with the potential due to the ions EXCEPT that the ion potential confines them to the region of space occupied by the material.
We will see that this assumption leads to a spherical Fermi surface and explains the properties of simple metals. It does not predict gaps in the band structure, but it is remarkable that the theory works as well as it does, and it improves the classical Drude model that is still used with success.
Paul Drude (1863-1906) Free electron bands & LCAO (tight binding) bands are qualitatively similar, with important exception of gaps. (Sutton, Fig. 7.6, Ge e.g.) So free electron model works well only for metals.
FE bands (Na) LCAO band (Na)
http://physics.njit.edu/~savrasov/Projects/Teaching/Phys%20485%20SPRING%202004/Lecture21Topics.htm Energy "Jellium" - free electrons a uniform positive BG
0 PE
+PBC
-V0 0 vacuum Metal, L vacuum L
ˆ 1 ik⋅r H ψ k = Ek ψ k ψ k (r ) = e V 2 # d 2 d 2 d 2 & − Similar to % 2 + 2 + 2 ( ψ k = Ek ψ k 2m $dx dy dz ' Bloch form
2 2 2 2 − " d d d % 1 ik⋅r r e $ 2 + 2 + 2 'ψ k (r ) = Ekψ k (r ) ψ k ( ) = 2m #dx dy dz & V r = xxˆ + yyˆ + zzˆ; r = r = x2 + y2 + z2 k = k xˆ + k yˆ + k zˆ; k = k = k 2 + k 2 + k 2 x y z x y z E k = ?; k = ? k = ? k = ? ( ) x y z 2 2 2π 2π 2π k k n ; k n ; k n E(k) = x = x y = y z = z 2m L L L
E0 kx
ky 2 Free electron bands: square lattice 2 k + G E = 2m E
kx (ky= 0) 2π 2π Real space Reciprocal G = a (0,0) G = a (±1,±1) 2π 2π 2π a1 = a(1,0) g1 = a (0,1) G = a (±1,0) G = a (0,2) a = a 0,1 g = 2π 1,0 2π 2π 2 ( ) 2 a ( ) G = a (0,1) G = a (±2,0) 2 Free electron bands: fcc lattice 2 k + G E = 2m
From Hummel, Electronic Properties of Materials 2 Free electron bands: fcc lattice 2 k + G Example Copper E = 2m
<- Calculated with pseudopotential methods. Note d-bands!
<- Free electron bands
From Tsymbal (UNL) There is a gap in the energy spectrum! This same result can be obtained by treating the periodic ion potential as a perturbation of the smooth potential (we won't do this, but it's discussed in Kittel, for example)
http://www.chembio.uoguelph.ca/educmat/chm753/properties/electronic/nearly_free_model.html Free electrons (plane waves) don't interact with the lattice much until the wave vector becomes comparable with 1/a, then they are Bragg reflected and there is interference between oppositely directed plane waves (standing waves).
Lower energy state piles up electron density on the atoms (stronger Coulomb attraction to cores), while the higher energy state piles up electron density between the atoms. Thus there is a gap in the energy spectrum!
eikx − e−ikx sin(kx)
eikx + e−ikx cos(kx)
http://www.chembio.uoguelph.ca/educmat/chm753/properties/electronic/nearly_free_model.html
Constant energy surfaces are spheres in k space Each state occupies 8π3/L3 in k-space When all N valence electrons have filled up k-states, reach the Fermi
Energy EF and the sphere has radius Fermi wave vector kF
3 3 2 2 8π 1 4πkF kF N ≡ EF = V 2 3 2m
Factor of 2 is from spin (2 electrons/state) 2 k 2 2π 2π 2π E(k) = kx = nx ; ky = ny ; kz = nz 2m L L L
E0 kx
ky Fermi parameters (see Kittel Ch.6, Table I): 1/3 " 2 % 8 -1 k 3 n kF = 0.92 × 10 cm : F = $ π ' # # e/vol& 2 k 2 F E = 3.2 eV: kinetic energy of Fermi level e EF = F 2m
EF TF = 37,500 K: temp. above which e obey classical TF = statistics (can ignore quantum effects) kB
k F v = 1.1 × 108 cm s-1 : 1% c vF = F m
Sodium: n = 2.65 × 1022 e-/cm3 Reminder: 3-D Density of states for free electron: dS S counts the states. D(E) = ⇒ dS = D(E)dE dE 2 k 2 dE 2 k E = ; = dS = D (k) × annular vol of k-space 2m dk m 2V 2 = 3 × 4πk dk (2π ) E0 E0 + dE 2V 2 D (E)dE = 3 4πk dk (2π ) 8πk 2V dk k x D (E) = 3 (2π ) dE 3/2 V " 2m% 1/2 D (E) = 2 $ 2 ' E 2π # & ky Example: Find total energy at T = 0 (homework) Will use this later for electronic specific heat.
EF E = E D E dE Tot ∫ ( ) 0
E E + dE 0 0 Total energy at T ≠ 0
∞ E = E f E D E dE Tot ∫ ( ) ( ) kx 0 ∞ 1 = E D(E)dE ∫ (E−EF )/kBT 0 e +1
ky Fermi Function • Probability that a state of energy E is occupied at temperature T.
• f(µ) = 0.5 • Varies sharply over width of ≈kBT about EF. Step function T = 0.
1 µ = 1eV f (E) = E − µ T = 0.015 eV = ?K e kBT + 1 T = 0.3 eV = ?K
f E 0.5 1.0
0.4 0.8 kBT ⇥ k T 0.6 B 0.3
0.4 0.2 0.2 kBT 0.1 E eV 0.5 1.0 1.5 2.0 0.0 0.0 0.5 1.0 1.5 2.0 E ⇥ Electronic specific heat • Electrons are a repository of energy in metals and contribute to the specific heat. Dominant contribution at low T where phonons are frozen out. • If electrons are a "classical gas" (no QM), equipartition theorem =>
electronic energy is Eelec = (3/2)kBT. Electronic specific heat is Celec = dEelec/dT= (3/2)kB • Actual values in metals at low temperatures are less than 1% of this, and are NOT temperature independent.
http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi2.html Electronic specific heat Nearly free electrons are a repository of energy in metals and contribute to the specific heat. At low temperatures, when phonons (lattice vibrations) are frozen out, this contribution dominates.
∞ E = E f E D E dE Tot ∫ ( ) ( ) df (E,T ) 0 dT ∞ 1 = E D(E)dE ∫ (E−µ)/kBT 0 e +1 many functions 2.0 1.5 3/2 1.0 V " 2m% 1/2 kBT 0.5 D(E) = 2 $ 2 ' E 2π # & E eV 0.5 1.0 1.5 2.0 0.5 1.0 ⇥ ∞ dETot df (E,T ) C = = E D E dE v ∫ ( ) dT 0 dT
Area under black curve is electronic specific heat. Very slight asymmetry is crucial! Easy enough to calculate numerically, but let's look for analytical insight. What is dominant contribution? Weakest?
many functions 2.0 1.5 EF = 1 eV 1.0 kBT = 0.1 eV 0.5 E eV 0.5 1.0 1.5 2.0 0.5 1.0 ⇥ ∞ dETot df (E,T ) C = = E D E dE v ∫ ( ) dT 0 dT
D(E) varies slowly - remove from integral (but what value? Why can lower limit of integration be extended to -∞? df (E,T ) = ? dT E − µ Change variables: x = 2kBT ∞ x2 π 2 Helpful: dx = ∫ 2 -∞ cosh x 6 ∞ dETot df (E,T ) C = = E D E dE v ∫ ( ) dT 0 dT
2 π 2 Cv = D(EF )kBT 2
Interpretation: Depends linearly on temperature - NOT independent Size depends only on density of states at Fermi surface - small for semimetals; large for so-called “heavy fermions” Also interpreted as each electron near Fermi surface
contributing roughly kB to heat capacity:
Cv = D(EF )kBT × kB ∞ dETot df (E,T ) C = = E D E dE v ∫ ( ) dT 0 dT
2 π 2 Cv = D(EF )kBT 2 π 2 T Cv,FE = NkB 2 TF
Much smaller than expected from classical theory:
3 C = Nk v,class 2 B electronic phonon C T AT 3 v,TOT = γ + γexpt γfe (mJ/mol/K2) (mJ/mol/K2) Cv,TOT
Na 1.38 1.09
K 2.08 1.67 Temperature Cu 0.70 0.51 Cv,T O T Zn 0.64 0.75 T Slope A Ag 0.65 0.65
2 Intercept γ T EXTRA SLIDES ABOUT FREE ELECTRON BANDS Free electron bands: square lattice, reduced zone scheme
E
kx (ky=0)
2 2 k + G E = 2m Free electron bands & LCAO (tight binding) bands are qualitatively similar, with important exception of gaps. (Sutton, Fig. 7.6, Ge e.g.) So free electron model works well only for metals.
FE bands (Na - bcc) LCAO band (Na)
http://physics.njit.edu/~savrasov/Projects/Teaching/Phys%20485%20SPRING%202004/Lecture21Topics.htm Free electron bands & LCAO (tight binding) bands are qualitatively similar, with important exception of gaps. (Sutton, Fig. 7.6, Ge e.g.) So free electron model works well only for metals.
FE bands (Na - bcc) LCAO band (Na) 2 Free electron bands: bcc lattice 2 k + G E = 2m
http://www.tf.uni-kiel.de/matwis/amat/semi_en/kap_2/illustr/band_bcc_free.gif