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Yu-Jun Zhao
Department of Physics, SCUT
Lecture XX I
Energy bands
- Central equation and
Kronig-Penney model
www.compphys.cn/~zhaoyj/lectures/
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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
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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
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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
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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
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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.
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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!
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(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
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kGkEE
)()( rErHkkk
From the Schrödinger equation, one has
and )()( rErHGkGkGk
)()( rrkGk
)()(
)()()(
rErE
rErHrH
kGkGkGk
kkkGk
Therefore
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(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
'
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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.
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2017/5/11
0
G
GkGkk CUC
Assume that Ck+mG =0 for m = ±4, ±5, ±6,….
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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
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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
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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.
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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
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Ratio of the coefficients2/1
2
2
22
UkGkkGk
-
-=
( - )
k
k G k
C U
C
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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
~
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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
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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
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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
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)(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.
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kaKaKaKa
Pcoscossin
m
K
2
22
plane waves with the energy
P = Q2ba/2
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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
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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
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)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
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Homework
2017/5/11
1. Problem 7.3 of textbook.