Yu-Jun Zhao

Department of Physics, SCUT

Lecture XX I

Energy bands

- Central equation and

Kronig-Penney model

www.compphys.cn/~zhaoyj/lectures/

The central equation

Consider a linear lattice with the lattice constant a. The potential

energy U(x) is invariant under a crystal lattice translation.

)()( xUaxU

We can expand U(x) as a Fourier series in the reciprocal lattice

vectors G.

G

iGx

GeUxU )(

In independent-electron approximation, the Schrödinger equation

is

)()(2

)()(2

22

xxeUm

pxxU

m

p

G

iGx

G

The wavefunction (x) can be expresses as the sum over all values

of the plane wave permitted in crystal,

k

ikx

k

k

ikx eCekCx )()(

where k = 2n/L due to the periodic boundary condition.

Substitute the wavefunction into the Schrödinger equation:

ikx

k

keCkm

xdx

d

mx

m

p 2

2

2

222

2)(

2)(

2

G k

xGki

kG

k

ikx

k

G

iGx

G eCUeCeUxxU )()()(

k

ikx

k

G k

xGki

kG

k

ikx

k eCeCUeCkm

)(22

2

02

22

k

ikx

G

GkGk eCUCm

k

i.e.

Therefore 0

G

GkGkk CUC

with the notationm

kk

2

22

the central equation

G

xGki

Gkk eCx )()(

k(x) is the wave packet which is a linear combination of the plane

waves with the wavecectors k+G.

No

differential

In principle, there are an infinite number of Ck to be determined.

However in practice a small number of Ck will suffice.

The central equation is the Schrödinger equation expressed in the

reciprocal space. Here we have a set of algebraic equations instead

of the differential equation.

0

G

GkGkk CUC

the central equation

2 2

-2

G k G

Gk

U C

Ck

m

What about 2 2 2 2( ')

=2 2

k k G

m m

Bragg reflection condition

Restatement of the Bloch theorem

The wavefunction is given as

)(

)( )(

xueeCe

eCx

k

ikx

G

iGx

Gk

ikx

G

xGki

Gkk

where we define .)(

G

iGx

Gkk eCxu

)(

)( )2()(

xueC

eCeCTxu

k

G

iGx

Gk

G

sGxi

Gk

G

TxiG

Gkk

Suppose T is a crystal lattice vector, TG = 2s.

Discussions about the Bloch theorem

(1) The Bloch function does not have the same periodicity

as the lattice, i.e.

)(rk

)exp()( )( RkirRrkk

As proved before,

Generally k is not a reciprocal lattice vector G,

1)exp( Rki

Therefore )( )( rRrkk

However )( )( rRrkk

Hence )()()()(22

rrRrRrkk

)( )( rRrkk

𝜌 𝑟 have the same periodicity as the lattice!

(2) The Bloch function has the same periodicity as the

reciprocal lattice:

)(rk

)()( rrkGk

G

rGki

GkkeCr

)()(

'

)'(

')(

G

rGGki

GGkGkeCr

Set G” = G’ G

)(

)(

"

")(

"

"

")(

"

'

)'(

'

reC

eC

eCr

kG

rGki

Gk

GG

rGki

Gk

G

rGGki

GGkGk

kGkEE

)()( rErHkkk

From the Schrödinger equation, one has

and )()( rErHGkGkGk

)()( rrkGk

)()(

)()()(

rErE

rErHrH

kGkGkGk

kkkGk

Therefore

(3) If the lattice potential vanishes, U(x) = 0.

The central equation reduces to (k )Ck = 0, so that all CkG

are zero except Ck, and uk(r) is constant. Thus we have

rki

ker

)(

(4) The crystal momentum of an electron

The quantity k enters in the conservation laws that govern

collision processes in crystal.

The crystal momentum of an electron k

If an electron absorbs in a collision a phonon of wavevector ,

the selection rule is

k

q

Gkqk

'

Solution of the central equation

1). For a given potential ,)( G

rGi

GeUrU

the central equation represents a set

of simultaneous linear equations.

0

G

GkGkk CUC

These equations are consistent if the determinant of the

coefficients vanishes.

2). The determinant in principle is infinite in extent, but in

practice a small number of Ck will suffice.

The values of the coefficients UG for actual crystal potentials

tend to decrease rapidly with increasing magnitude of G.

2017/5/11

0

G

GkGkk CUC

Assume that Ck+mG =0 for m = ±4, ±5, ±6,….

Suppose that the potential energy U(x) contains only a single

Fourier component Ug = Ug = U, where g denotes the shortest G.

Take five successive equations of the central equation. Then the

determinant of the coefficients is given by

0

0 0 0

0 0

0 0

0 0

0 0 0

2

2

gk

gk

k

gk

gk

U

UU

UU

UU

U

For a given k, the solution of the determinant gives a set of

energy eigenvalues nk, which lie on different bands.

0

G

GkGkk CUC

Approximate solution near a zone boundary -- a simple case

Supposem

kUUG

2

22

For a wavevector exactly at the Brillouin zone boundary,

2/ i.e. ,)2/( 22 GkGk

so that at the zone boundary the kinetic energy of the two

component waves k = G/2 are equal, i.e. G/2 = G/2.

Suppose CG/2 are important coefficients at the zone boundary

and neglect all others.

Then we have only two equations with k = G/2 :

0)(

0)(

2/2/

2/2/

GG

GG

UCC

UCC

Here mG 2/)2/( 22

For a nontrivial solution

0

U

U

whence UG

mU

22

22

The ratio of the C:

12/

2/

UC

C

G

G

Then the wavefunction at the zone boundary is

)2/exp()2/exp()( iGxiGxx

Here one solution gives the wavefunction at the bottom of the

energy gap, and the other gives the wavefunction at the top of

the energy gap.

Solutions near the zone boundary G/2

With the same two component approximation, the wavefunction

is:xGki

Gk

ikx

k eCeCx )()(

From the central equation we have

0)(

0)(

kGkGk

Gkkk

UCC

UCC

The determinant equals zero:

0

Gk

k

U

U

The energy 2/1

2

2

22

UkGkkGk

Ratio of the coefficients2/1

2

2

22

UkGkkGk

-

-=

( - )

k

k G k

C U

C

m

K

UUKG

m

Um

KKG

mK

2

~2

1~

4/2

2

~4~

4/2

22

2

222

2/1

222

222

~

If we expand the energy in terms of a quantity in

the region :

2/~

GkK

UmKG 2/~2

UG

m

22

22)(

At the Brillouin zone boundary

Therefore

Um

KK

21

2

~

)()(22

~

Kronig-Penney model – another application

Assume the periodic potential is the square-well periodic potential.

As shown in the figure, in one period

the square-well potential is

0for ,)(

0for ,0)(

0 xbUxU

axxU

The periodicity of U(x) is a + b.

The wave equation is

)()()()(2 2

22

xxxUxdx

d

m

iKxiKx BeAex )(

In the region 0 < x < a, U(x) = 0, the eigenfunction is a linear

combination of plane waves with the energy m

K

2

22

In the region b < x < 0, U(x) = U0, the eigenfunction is

QxQx DeCex )(

withm

QU

2

22

0

From the Bloch theorem,

)()0()( baik

kk exbbaxa

)()()()(2 2

22

xxxUxdx

d

m

The constants A, B, C, and D are chosen so that and d/dx are

continuous at x = 0 and x = a.

At x = 0

iKxiKx

k BeAeax )0(

)()( DCQBAiK

DCBA

At x = a

)(

)(

)()(

)(

baikQbQbiKaiKa

baikQbQbiKaiKa

eDeCeQBeAeiK

eDeCeBeAe

)()()( baikQxQx

k eDeCebaxa

The determinant of the coefficients:

0

1 1 1 1

)()(

)()(

baikQbbaikQbiKaiKa

baikQbbaikQbiKaiKa

QeQeiKeiKe

eeee

QQiKiK

)(coscoscoshsinsinh2

22

bakKaQbKaQbQK

KQ

If we represent the potential by the periodic delta function

0 and 0 i.e. Ub

and set P = Q2ba/2,

In the limit Q >> K and Qb << 1, the equation reduces to

kaKaKaKa

Pcoscossin

Note: K is not the wavevector k of the Bloch function.

kaKaKaKa

Pcoscossin

m

K

2

22

plane waves with the energy

P = Q2ba/2

Kronig-Penney model in reciprocal space (by central equation)

We use the Kronig-Penney model of a periodic delta function

potential:

0

/1

0

cos2)()(G

G

a

s

GxUsaxAaxU

Where A is a constant and a the lattice spacing. The sum of s is

over all atoms in a unit length, which means over 1/a atoms.

Thus

A

GsaAa

GxsaxdxAa

GxxUdxU

a

s

a

s

s

sa

G

cos

cos)(

cos)(

/1

0

/1

0

)1(

0

We have the central equation:

0)( /2

n

ankkk CAC

here G = 2n/a.

Define Gk

n

ankk fCf /2

Then kk

k

k fmk

mAf

AC

22

2

/2

/2

kank fmank

mAC

22

2

/2/2)/2(

/2

Sum both side over n

n

k

n

ank mankmAfC 1222

/2 ]/2)/2[(/2

)cos(cos4

sin

]/2)/2[(2/

2

1222

KakaKa

Kaa

mankmAn

Then we have

where we write 22 /2 mK

Then the final result is kaKaKa

KamAacoscos

sin

2 2

2

Homework

2017/5/11

1. Problem 7.3 of textbook.