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.