PH575 Spring 2014 Lecture #18 Free electron theory: Sutton Ch. 7 pp 132 -> 144; Kittel Ch. 6.

E k( ) =

2k2

2m

D E( ) = V

2π 22m2

"#$

%&'

3/2

E1/2

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.

http://physics.njit.edu/~savrasov/Projects/Teaching/Phys%20485%20SPRING%202004/Lecture21Topics.htm

FE bands (Na) LCAO band (Na)

Metal, L vacuum

Energy

PE

vacuum

0

-V0

"Jellium" - free electrons a uniform positive BG

L

H ψ k = Ek ψ k

−2

2md 2

dx2+d 2

dy2+d 2

dz2#

$%

&

'( ψ k = Ek ψ k

0

+PBC

ψ kr( ) = 1

Veik ⋅r

Similar to Bloch form

−2

2md 2

dx2+d 2

dy2+d 2

dz2"

#$

%

&'ψ k (

r ) = Ekψ k (r )

ψ kr( ) = 1

Veik ⋅r

r = xx + yy + zz; r = r = x2 + y2 + z2k = kx x + ky y + kz z; k =

k = kx

2 + ky2 + kz

2

Ek( ) = ?; kx = ? ky = ? kz = ?

E k( ) =

2k2

2m

kx = nx2πL; ky = ny

2πL; kz = nz

2πL

E0 kx

ky

Free electron bands: square lattice

kx (ky= 0)

E E =2k +G

2

2m

Real spacea1 = a 1,0( )a2 = a 0,1( )

Reciprocalg1 = 2π

a 0,1( )g2 = 2π

a 1,0( )

G = 2π

a 0,0( )G = 2π

a ±1,±1( )G = 2π

a ±1,0( )G = 2π

a 0,2( )G = 2π

a 0,1( )G = 2π

a ±2,0( )

From Hummel, Electronic Properties of Materials

Free electron bands: fcc lattice

E =2k +G

2

2m

From Tsymbal (UNL)

Free electron bands: fcc lattice Example Copper

E =2k +G

2

2m

<- Free electron bands

<- Calculated with pseudopotential methods. Note d-bands!

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!

http://www.chembio.uoguelph.ca/educmat/chm753/properties/electronic/nearly_free_model.html

eikx − e− ikx

sin kx( )

eikx + e− ikx

cos kx( )

E k( ) =

2k2

2m

kx = nx2πL; ky = ny

2πL; kz = nz

2πL

E0 kx

ky

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

EF =

2kF2

2m

Factor of 2 is from spin (2 electrons/state)

N 8π3

V12≡4πkF

3

3

Fermi parameters (see Kittel Ch.6, Table I):

EF =

2kF2

2m

kF = 3π 2 n

# e/vol

"#$

%&'

1/3

TF =EF

kB

vF =

kFm

Sodium: n = 2.65 × 1022 e-/cm3

kF = 0.92 × 108 cm-1:

TF = 37,500 K: temp. above which e obey classical statistics (can ignore quantum effects)

EF = 3.2 eV: kinetic energy of Fermi level e

vF = 1.1 × 108 cm s-1 : 1% c

Reminder: 3-D Density of states for free electron:

S counts the states. D E( ) = dSdE

⇒ dS = D(E)dE

dS = D(k)× annular vol of k-space

=2V2π( )3 × 4πk2dk

D E( )dE = 2V2π( )3

4πk2dk

D E( ) = 8πk2V

2π( )3dkdE

D E( ) = V2π 2

2m2

"#$

%&'

3/2

E1/2

E0 + dEE0

kx

ky

E =

2k2

2m; dEdk

=2km

Example: Find total energy at T = 0 (homework) Will use this later for electronic specific heat.

ETot = ED E( )dE0

EF

∫

E0 + dEE0

kx

ky

ETot = E f E( )D E( )dE0

∞

∫

= E 1e E−EF( )/kBT +1

D E( )dE0

∞

∫

Total energy at T ≠ 0

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.

0.5 1.0 1.5 2.0E�eV⇥0.2

0.4

0.6

0.8

1.0

f �E⇥f E( ) = 1

eE− µkBT +1

µ = 1eV T = 0.015 eV = ?K T = 0.3 eV = ?K

kBT

kBT

0.0 0.5 1.0 1.5 2.00.0

0.1

0.2

0.3

0.4

0.5

kBT

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

0.5 1.0 1.5 2.0E�eV⇥

�1.0�0.5

0.5

1.0

1.5

2.0

many functions

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.

D E( ) = V

2π 22m2

"#$

%&'

3/2

E1/2

ETot = E f E( )D E( )dE0

∞

∫

= E 1e E−µ( )/kBT +1

D E( )dE0

∞

∫

df E,T( )dT

kBT

Cv =dETot

dT= E

df E,T( )dT

D E( )dE0

∞

∫

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?

0.5 1.0 1.5 2.0E�eV⇥

�1.0�0.5

0.5

1.0

1.5

2.0

many functions

EF = 1 eV kBT = 0.1 eV

Cv =dETot

dT= E

df E,T( )dT

D E( )dE0

∞

∫

D(E) varies slowly - remove from integral (but what value?

df E,T( )dT

= ?

Change variables: x = E − µ2kBT

Why can lower limit of integration be extended to -∞?

Helpful: x2

cosh2 x-∞

∞

∫ dx = π2

6

Cv =dETot

dT= E

df E,T( )dT

D E( )dE0

∞

∫

Cv =π 2

2D EF( )kB2T

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

Cv =dETot

dT= E

df E,T( )dT

D E( )dE0

∞

∫

Cv =π 2

2D EF( )kB2T

Cv,FE =π 2

2NkB

TTF

Much smaller than expected from classical theory:

Cv,class =32NkB

Cv,TOT = γTelectronic

+ AT 3

phonon

Temperature

Cv,TOT

Cv,TOT

T

T2

Slope A

Intercept γ

γexpt ���(mJ/mol/K2)

γfe (mJ/mol/K2)

Na 1.38 1.09

K 2.08 1.67

Cu 0.70 0.51

Zn 0.64 0.75

Ag 0.65 0.65

EXTRA SLIDES ABOUT FREE ELECTRON BANDS

Free electron bands: square lattice, reduced zone scheme

kx (ky=0)

E

E =2k +G

2

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.

http://physics.njit.edu/~savrasov/Projects/Teaching/Phys%20485%20SPRING%202004/Lecture21Topics.htm

FE bands (Na - bcc) LCAO band (Na)

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://www.tf.uni-kiel.de/matwis/amat/semi_en/kap_2/illustr/band_bcc_free.gif

Free electron bands: bcc lattice

E =2k +G

2

2m