M.C. Chang

Dept of Phys

Chap 7

Non-interacting electrons in a periodic potential• Bloch theorem

• The central equation

• Brillouin zone

• Rotational symmetry

Bloch recalled,

The main problem was to explain how the electrons could sneak by

all the ions in a metal so as to avoid a mean free path of the order of

atomic distances. Such a distance was much too short to explain the

observed resistances, which even demanded that the mean free path

become longer and longer with decreasing temperature.

By straight Fourier analysis I found to my delight that the wave differed

from the plane wave of free electrons only by a periodic modulation.

This was so simple that I didn't think it could be much of a discovery,

but when I showed it to Heisenberg he said right away: "That's it!"

Hoddeson L – Out of the Crystal Maze, p.107

• Translation operator( )1 2 3

exp , , , mutually commutea a aR

iT p R T T T⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

• Translation symmetry of the lattice → , 0RH T⎡ ⎤ =⎣ ⎦

• Simultaneous eigen-states

,R R

HT C

αβ α αβ

αβ αββ

ψ ε ψ

ψ ψ

=

=

' '

,

1;

( ) ( )

R R R R R

i RR

i R

C C C C

C e

r R e r

ββ

βαβ αβψ ψ

+

⋅

⋅

= =

⇒ =

∴ + = [ ]'' ' '

nk nk nk

ik R

nnnk

R

n

k n

k

n

k

k

k

H

T e

Vψ ψ δ

ψ ψ

ψ ψ

δ

ε⋅

=

×

=

=

22 ( )

2

( ) ( )

U rm

U r R U r

ψ εψ⎡ ⎤− ∇ + =⎢ ⎥⎣ ⎦

+ =

Lattice Hamiltonian

|Ψ(x)|2 is the same in each unit cell

Bloch energy, Bloch state

a

U

Re-label

orthogonality

The electron states in a periodic potential can be written as

where uk(r)= uk(r+R) is a cell-periodic function

Bloch theorem (1928)

The cell-periodic part unk(x) depends on the form of the potential.

( ) ( )ik rnk nkr e u rψ ⋅=

define ( ) ( )

then from ( ) ( )

( ) ( ).

ik rnk nk

ik Rnk nk

nk nk

u r e r

r R e r

u r R u r

ψ

ψ ψ

− ⋅

⋅

=

+ =

⇒ + =

• Effective Hamiltonian for u(r)

Pf:

22

within one unit cell

where

( )

( ) ( )2

nk nk nk

ik r ik r

H k u u

H k e He k U rm i

ε

− ⋅ ⋅

=

∇⎛ ⎞≡ = + +⎜ ⎟⎝ ⎠

(1023 times less effort than the original Schrodinger eq.)

1N

Marder’s has

a factor

Allowed values of k are determined by the B.C.

31 21 2 3

1 2 3

( ) ( ), =1,2,3

1,

2 , ,

i i

i ink nk

iN k a

i i i i

r N a r i

e i

mm mk b b bN N N

N k a m m Z iπ

⋅

=

Ψ + = Ψ

→ = ∀

+ +

→ ⋅ = ∈ ∀

→

3 31 21 2 3

1 2 3

3

1 2 3

3

1 ( )

1 (2 ) , ( )

(2 ) , as in the free-electron case.

bb bk b b bN N N N

Va a a vN v N

V

π

π

⎛ ⎞Δ = ⋅ × = ⋅ ×⎜ ⎟

⎝ ⎠

= ⋅ × = =

=

Periodic B.C.(3-dim case)

Therefore, there are N k-points in a unit cell (of reciprocal lattice),

where N = total number of primitive cells in the crystal.

1 2 33

( )b b b Nk

⋅ ×=

Δ

important

1 2 3=N N N N

If f(r) has lattice translation symmetry, f(r)=f(r+R), for any lattice vector R

[eg. f(r) can be the lattice potential], then it can be expanded as,

f r e fiG rG

G

( ) = ⋅∑all

where G is the reciprocal lattice vector.

k hb kb lb

G h k lhkl

= + +

∀1 2 3

= , ,

Pf: Fourier expanson,

Therefore, for

f r e f k

f r R e e f k f r

e R

ik r

k

ik R ik r

k

ik R

( ) ( )

( ) ( ) ( ).

=

+ = =

= ∀

⋅

⋅ ⋅

⋅

∑

∑1

The expansion above is very general, it applies to

• all types of Bravais lattice (e.g. bcc, fcc, tetragonal, orthorombic...)

• in every dimension (1, 2, and 3)

Fourier decomposition and reciprocal lattice vectors

All you need to do is find out the reciprocal lattice vectors G

important

A simple example:

electron density (or potential, or cell-periodic function) of a 1-dim lattice

a ax

b a x1

1 2

=

→ =

,

( / )π

G nb

G r nb xx n a xn

n

=

⋅ = ⋅ =1

1 2

,

( ) ( ) ( / )π(2 / )( ) i n a x

nn

x e πρ ρ∴ =∑

a

ρ(x)

x

Some useful formulas (for electrons in a lattice box)

,0

,0

( )2 2 /

2

2 ( )

k

k

L kkL

L

L k

δ δπ π

δπ

δ πδ

⎛ ⎞= ⎜ ⎟⎝ ⎠

→

∴ ↔

,01st BZ

,0

ik nan

k

ik nak

n

e N

e N

δ

δ

⋅

∈

⋅

=

=

∑

∑

( )

( )( )

3

3 ,0

3

,0

( )2 / 2 / 2 /2

2

2 ( )

x y z x x x

x y z

k

k

L L L k k kkL L L

V

V k

δ δ δ δπ π ππ

δπ

δ π δ

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠

→

∴ ↔

,01st BZ

,0

ik RR

k

ik Rk

R

e N

e N

δ

δ

⋅

∈

⋅

=

=

∑

∑

1D 3D

important

,0iGx

Gcell

dx e aδ=∫ 3,0

iG rG

cell

d r e vδ⋅ =∫

( ) ( ),03

3

for discrete note:

2 for continuous

kik r

all

V kd r e

k k

δ

π δ⋅

⎧⎪= ⎨⎪⎩

∫

Restrict k to the 1st BZ, Cf. Marder, App A.4

A+M, App. D

How do we determine uk(x) from the potential U(x)?

Schrodinger equation

Schrod. eq. in k-space aka. the central eq.

Fourier transform

1. the lattice potential

2. the wave function

G=2πn/a

k=2πn/L

( ) iG rG

G

U r U e ⋅= ∑

( ) ( ) iG rk k

G

u r C G e− ⋅= ∑

( )2

( ) ( ) ( )2 k k k

p kU r u r u r

mε

⎡ ⎤+⎢ ⎥+ =⎢ ⎥⎢ ⎥⎣ ⎦

important

( )2 2

0 0'

'

( ) ( ') 0, 2k kG Gk G k k

G

kC G U C Gm

ε ε ε−−− + = ≡∑

Keypoint: go to k-space to simplify the calculation

Matrix form of the central eq. (in 1D)

• For a given k, there are many eigen-energies εnk, with eigen-vectors Cnk.

02 2 3 4

02 3

02 2

03 2

04 3 2 2

k g k g g g g

g k g k g g g

g g k k g g

g g g k g k g

g g g g k g k

U U U UU U U UU U U UU U U UU U U U

ε εε ε

ε εε ε

ε ε

+ − − − −

+ − − −

− −

− −

−

⎛ −⎜

−⎜⎜ −

−−⎝

( 2 )( )(0) 0( )

(2 )

k

k

k

k

k

C gC gCC gC g

⎞ −⎛ ⎞⎟⎜ ⎟−⎟⎜ ⎟⎟⎜ ⎟ =

⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟⎝ ⎠⎠

G=ng (g=2π/a)

for a particular k

0 g k

εk

C(-g)=1 C(0)=1 C(g)=1U(x)=0

2g-2g -g ε1k

ε2k

ε3k

• when U(x) ≠0, for a particular k, unk is a linear combination of

plane waves, with coefficients Cnk : Also valid in higher dim

( ) ( ) iGxnk nk

Gu x C G e−= ∑

1st BZ

U(x) = 2U cos2πx/a

= U exp(2πix/a)+U exp(-2πix/a) (Ug=U-g=U)

Matrix form of the central eq.

02

0

0

0

02

( 2 )0 0 0( )0 0(0)0 0( )0 0

(2 )0 0 0

kk g k

kk g k

kk k

kk g k

kk g k

C gUC gU UCU UC gU UC gU

ε εε ε

ε εε ε

ε ε

+

+

−

−

⎛ ⎞ −− ⎛⎜ ⎟⎜ −−⎜ ⎟⎜⎜ ⎟⎜−⎜ ⎟⎜

−⎜ ⎟⎜⎜⎜ ⎟− ⎝⎝ ⎠

0

⎞⎟⎟⎟ =⎟⎟⎟⎠

,

,

( ) ( ) ( )

( ) ( )

( )

iGxn k G nk

n k

i

G n

Gxnk nk

G

k

u x e u xx x

u x C G e

ψ ψ

−

−+

+

⇒

=

=

=

∑

' ( ') ( )k G kC G G C G+ + =• From the central eq., one can see

• Bloch energy εn,k+G = εnk

Example.

a

A solvable model in 1-dim: The Kronig-Penny model (1930)(not a bad model for superlattice)

( ) ( ) 00

2 cosniG xn n n

s n nU x A a x sa U e U U G xδ

>

= − = = +∑ ∑ ∑

( )/2

/2

1n

aiG x

na

U dxU x e Aa

−

−

= =∫

( )0

0

0

0

1

k ng k n m n mm

n mmk k ng

n k k ng

C U C

AC C

A

ε ε

ε ε

ε ε

− −

−

−

− + =

⇒ =−

⇒ =−

∑

∑

∑( ) ( )

22

2 22

21 ,2 2

1 1 1 2 22

cot cot4 2 2

k

n

n

mKmA nK k

a

n nK K k K ka a

a a aK k K kK

επ

π π

= ≡⎛ ⎞− −⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟

= +⎜ ⎟⎜ ⎟+ − − +⎝ ⎠

⎡ ⎤= + + −⎢ ⎥⎣ ⎦

∑

∑

1 cotn

xx nπ

=+∑

2

...sin

2 2 cos cos

sin cos cos

a KamA K ka KaP Ka Ka kaKa

⇒ =−

⇒ + =2

2

mAaP⎛ ⎞

=⎜ ⎟⎝ ⎠

K has a real solution when

sin cosP Ka KaKa

+

Sometimes it is convenient to extend the domain of k with the following requirement

, '

, '

( ' 1st BZ) =k n k G

n k

kε ε

ε+= ∈

Fig from Dr. Suzukis’ note (SUNY@Albany)

1st

Brillouinzone

BZ

First Brillouin zone for 2D reciprocal lattice

reciprocal lattice

Ex: Hexagonal lattice

direct lattice

Wigner-Seitz cell of the reciprocal lattice

, ,

, ,( ) ( )n k G n k

n k G n kr r

ε ε

ψ ψ+

+

=

=

The first BZ of FCC lattice (its reciprocal lattice is BCC lattice)

4π/a

The first BZ of BCC lattice (its reciprocal lattice is FCC lattice)

xy

z

4π/a

Momentum and velocity of an electron in a Bloch state

1( )

1 =

( ) =

( ) =

n nk nk

nk nk

nk nk

n kk

v km i

u k um i

H ku uk

ε

ψ ψ= ∇

∇+

∂∂

∂∂

εn(k)

k

vn(k)

k

22

( )

( )2

( ) ( )

ik r ik r

nk n nk

H k e He

k U rm i

H k u k uε

− ⋅ ⋅=

∇⎛ ⎞= + +⎜ ⎟⎝ ⎠=

~ group velocity

• momentum

• velocity

ik rnk nk nkk e u

i iψ ψ ⋅∇ = + ∇

Not a momentum eigen-state

Crystal (quasi)-momentum

• In a perfect lattice, the motion is unhindered• velocity is zero at Brillouin zone boundary

( ),( ) i k G x

nk n k GG

x C eψ −−=∑

∴ for a Bloch state with crystal momentum ħk, the actual momentum can have, with various probabilities, an infinity of values.

A. Wilson (1932)

… Bloch, in showing that tightly bound electrons could in fact move

through the lattice, had "proved too much"—that all solids should be

metals. Were insulators simply very poor conductors? However,

implicit in Peierls's papers on the Hall effect lay the clue, not carried

further by Peierls, that a filled band would carry no current.

Hoddeson L – Out of the Crystal Maze, p.120

A filled band does not carry current (Peierls, 1929)

Electric current density

Inversion symmetry,εn(k)= εn(-k)

electrons with momenta ħk and -ħk have opposite velocities

no net current in equilibrium

a filled band carries no current even in an external field

3

3filled

( )1 1( )(2 )

n

k

kd kj ev eV k

επ

∂= − = −

∂∑ ∫

• If even number of electrons per primitive cell, then there are 2

possibilities: (a) no energy overlap or (b) energy overlap.

E.g., alkali earth elements can be conductor or insulator.

• If a solid has odd number of valence electron per primitive

cell, then the energy band is half-filled (conductor).

For example, all alkali metals are conductors

Difference between conductor and insulator (Wilson, 1931)

• There are N k-points in an energy band, each k-point can be

occupied by two electrons (spin up and down).

∴ each energy band has 2N “seats” for electrons.

• If a solid has even number of valence electron per primitive

cell, then the energy band is filled (insulator).

Wilson recalls Bloch's reply after hearing his arguments:

"No, it's quite wrong, quite wrong, quite wrong, not possible at all."

k

conductorE

insulator

k

E

k

conductorE

A nearly-filled band

filled

1st BZ unfilled

unfilled

k

k k

k

ej vV

e v vVe vV

∈

= −

⎛ ⎞= − −⎜ ⎟

⎝ ⎠

= +

∑

∑ ∑

∑∴ unoccupied states behave as +e charge carriers

If an electron of wavevector k0 is missing, then the sum over filled k, Σk= -k0.

Alternatively speaking, a hole with wavevector kh (= -k0). is produced.

• “momentum” of a hole:

• “charge” of a hole:

DOS for a Bloch band εnk

33( )

2 Shell

LD d d kε επ

⎛ ⎞= ⎜ ⎟⎝ ⎠ ∫

3

3

k

k

d k dS dk

d dk

dd k dS

ε

ε

ε ε

εε

⊥

⊥

=

= ∇

∴ =∇

3

( )2

k

dSLD εεπ ε

⎛ ⎞= ⎜ ⎟ ∇⇒

⎝ ⎠ ∫If 0, then there is

"van Hove singularity" (1953)k kv ε= ∇ =

Surface ε=const

Surface ε+dε =const

dk⊥dSε

important

For example, DOS of 1D energy bands

Van Hove singularity