Post on 23-Oct-2020
1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Handout 12
Energy Bands in Group IV and III-V Semiconductors
In this lecture you will learn:
• The tight binding method (contd…)• The energy bands in group IV and group III-V semiconductors with FCC lattice structure• Spin-orbit coupling effects in solids
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
FCC Lattice: A Review
a
a
a1a2a
3a
zyaa ˆˆ21
zxaa ˆˆ22
yxaa ˆˆ23
Face Centered Cubic (FCC) Lattice:
x
y
z
Unit Cell
Most group VI and group III-V semiconductor, such as Si, Ge, GaAs, InP, etc have FCC lattices with a two-atom basis
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Lattices of Group IV Semiconductors (Silicon, Germanium, and Diamond)
• The underlying lattice is an FCC lattice with a two-point (or two-atom) basis.
• Each atom is covalently bonded to four other atoms (and vice versa) via sp3 bonds in a tetrahedral configuration
Diamond lattice (Si, Ge, and Diamond)
x
y
z
1,1,141an 1,1,1
42
an
1,1,143
an 1,1,144
an
Nearest neighbor vectors
1n
2n
3n
4n
Basis vectors01 d
1,1,1
42ad
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Lattices of III-V Binaries (GaAs, InP, InAs, AlAs, InSb, etc)
• The underlying lattice is an FCC lattice with a two-point (or two-atom) basis. In contrast to the diamond lattice, the two atoms in the basis of zincblende lattice are different – one belongs to group III and one belongs to group V
• Each Group III atom is covalently bonded to four other group V atoms (and vice versa) via sp3 bonds in a tetrahedral configuration
Zincblende lattice (GaAs, InP, InAs)
1,1,141an 1,1,1
42
an
1,1,143
an 1,1,144
an
Nearest neighbor vectors
x
y
z
1n
2n
3n
4n
Basis vectors01 d
1,1,1
42ad
3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
• Each Ga atom contributes one 4s-orbital and three 4p-robitals
• Each As atom also contributes one 4s-orbital and three 4p-robitals
Each primitive cell contributes a total of eight orbitals that participate in bonding
PAPzAPGPzG
PAPyAPGPyG
PAPxAPGPxG
SASASGSG
ErErErErErErErEr
One can write the trial tight-binding solution for wavevector as:k
Example: Tight Binding Solution for GaAs
x
y
z
1n
2n
3n
4n
1234
5678
m j j
mjjdki
mjjRki
k dRrceRrcNer
m 4
1
8
52
..
2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Example: Tight Binding Solution for GaAs
x
y
z
1n
2n
3n
4n
m j j
mjjdki
mjjRki
k dRrceRrcNer
m 4
1
8
52
..
2
Plug the solution above into the Schrodinger equation to get:
kckckckckckckckc
kE
kckckckckckckckc
H
8
7
6
5
4
3
2
1
8
7
6
5
4
3
2
1
4
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
SGE
0 PGE
0 0 0
0
0
0 0
0
0
PGE
PGE0
0
SAE
0 PAE
0 0 0
0
0
0 0
0
0
PAE
PAE0
0
kgVss
0 kgVsp
13
4321 ....0 nkinkinkinki eeeekg
4321 ....1 nkinkinkinki eeeekg
kgVsp 23
4321 ....2 nkinkinkinki eeeekg
kgVsp 33
4321 ....3 nkinkinkinki eeeekg
kgVsp 13
kgVsp 23
kgVsp 33
pppp VVV 32
31
1
kgV 01
kgV 01
kgV 01
pppp VVV 31
31
2
kgV 22 kgV
32
kgV 32 kgV
12
kgV 22 kgV
12
Hermitian
Tight Binding Solution for GaAs: The Matrix
H
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Tight Binding Solution for GaAs
Parameter values for GaAs:
Tight Binding Solution
eV 890eV 15.2eV 443eV 70.1
eV 91.7eV 90.4eV33.17eV 37.11
.VV
.VV
EEEE
ppsp
ppσss
PAPG
SASG
5
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Tight Binding Solution for GaAs: States at the -Point
At the -point:
400 kg
0321 kgkgkg
Energy eigenvalues can be found analytically
22
15 422
0 ssSGASGSASG VEEEEkE
The Bloch function of the lowest energy band and of the conduction band at -point are made up of ONLY s-orbitals from the Ga and As atoms
Two of the eigenvalues at the -point are:
m
mmkc dRrcRrcNr 255110,
1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
212
234678 422
0 VEEEEkE PAPGPAPG
Tight Binding Solution for GaAs: States at the -Point
The Bloch function of the highest three energy bands and of the three valence bands at -point are made up of ONLY p-orbitals from the Ga and As atoms
Six remaining eigenvalues at the -point are:
Each eignevalue above is triply degenerate
m
jmjj
jmjj
kvdRrc
Rrc
Nr
8
62
4
20,
1
6
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
• Need to include the effect of spin-orbit-coupling on the valence bandsSpin orbit coupling lifts the degeneracy of the valence bands• Need to include more orbitals (20 per primitive cell as opposed to 8 per primitive cell)• Use better parameter values
Simplest TB Approach Improved TB Approach with SO-Coupling(Figure not on the same scale)
Improved Tight Binding Approaches
HHLH
SO
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Spin-Orbit Interaction in SolidsAn electron moving in an electric field sees an effective magnetic field given by:
22mcPEBeff
The additional factor of 2 is coming from Thomas precession
The electron has a magnetic moment related to its spin angular momentum by:
Sg B
̂2
ˆ S ˆˆ
B22
gme
B
The interaction between the electron spin and the effective magnetic field adds a new term to the Hamiltonian:
PrVcm
PerV
mcBBH BeffBeffso
ˆˆ.ˆ4
ˆˆ
21.ˆ.ˆ.ˆ 222
S
zyx zyx ˆˆˆˆˆˆˆ
1001
ˆ0
0ˆ
0110
ˆ zyx ii
7
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Spin-Orbit Interaction in Solids: Simplified TreatmentNear an atom, where electrons spend most of their time, the potential varies mostly only in the radial direction away from the atom. Therefore:
LS
rrV
rcmL
rrV
rcm
PrrrV
rcmPrV
cmHso
ˆ.ˆ12
1ˆ.ˆ14
ˆˆ.ˆ14
ˆˆ.ˆ4
ˆ
2222
2222
PrL ˆˆˆ
is the orbital angular momentum of an electron near an atom
222222
ˆˆˆ21ˆ.ˆ
ˆ.ˆ2ˆˆˆ
ˆˆˆ
SLJLS
LSSLJ
SLJ
Recall from quantum mechanics that the total angular momentum is:Ĵ
Therefore: 22222 ˆˆˆ14
1ˆ SLJrrV
rcmHso
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
If the electron is in s-orbital then: 0ˆ0ˆˆˆ 222 soHSLJ
If the electron is in p-orbital then: 0ˆ0ˆˆˆ 222 soHSLJ
The energies of the Bloch states made up of p-orbitals (like in the case of the three degenerate valence bands at the point in GaAs) will be most affected by spin-orbit coupling
Spin-Orbit Interaction in Solids: Simplified TreatmentFor an electron in a p-orbital:
21ˆ 22 rLr pp
For an electron in a s-orbital:
01ˆ 22 rLr ss And we always have for an electron:
222431ˆ ssS
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Tight Binding Vs Pseudopotential Technique
Simplest TB Approach
A Little More Sophisticated ApproachNonlocal Pseudopotential Method
GaAs Energy Bands(Chelikowski and Cohen, 1976)
GaAs
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Energy Bands of Silicon and Germanium
Silicon Energy Bands(Chelikowski and Cohen, 1976)
Germanium Energy Bands(Chelikowski and Cohen, 1976)
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
rkErrVm knnknr
,,
22
2
Appendix: Spin-Orbit Interaction and Bloch FunctionsIn the absence of spin-orbit interaction we had:
Electron states with spin-up and spin-down were degenerate
rkErH knnkno
,,ˆ
In the presence of spin-orbit coupling the Hamiltonian becomes:
rrrsosoo
rVcm
iPrVcm
H
HHH
ˆ.ˆ4
ˆˆ.ˆ4
ˆ
ˆˆˆ
22
2
22
Since the Hamiltonian is now spin-dependent, pure spin-up or pure spin-down states are no longer the eigenstates of the Hamiltonian
The eigenstates can be written most generally as a superposition of up and down spin states, or:
rrr
rr knkn
kn
knkn
,,,
,,,
=Quantum number for the two spin degrees of freedom, usually taken to be +1 or -1
kEkE nn
,,
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Bloch Functions
rr
kErr
rVcm
irVm
rr
kErr
H
kn
knn
kn
knrr
r
kn
knn
kn
kn
,
,,
,
,22
222,
,,
,
,
ˆ.ˆ42
ˆ
For each wavevector in the FBZ, and for each band index, one will obtain two solutions of the above equation
We label one as = +1 and the other with = -1 and in general
n̂
kEkE nn
,,
These two solutions will correspond to spins pointing in two different directions (usually collinear and opposite directions). Let these directions be specified by at the location :
rrn
rrn
knkn
knkn
,,,,
,,,,
1ˆ.ˆ
1ˆ.ˆr
10
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Lattice SymmetriesIn the presence of spin-orbit interaction we have the Schrodinger equation:
Lattice Translation Symmetry:
rere
reRrRr
Rr knRki
knRki
knRki
kn
knkn
,,.
,.
,.
,
,,,
rr
kErr
rVcm
irVm kn
knn
kn
knrr
r
,
,,
,
,22
222ˆ.ˆ
42
Rotation Symmetry:
Let be an operator belonging to the rotation subgroup of the crystal point-group, such that:
Ŝ
unitaryˆˆ 1 SSrVrSV T
(The case of inversion symmetry will be treated separately)
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Suppose we have found the solution to the Schrodinger equation:
Appendix: Spin-Orbit Interaction and Rotation Symmetry
rr
kErr
rVcm
irVm kn
knn
kn
knrr
r
,
,,
,
,22
222ˆ.ˆ
42
We replace by everywhere in the Schrodinger equation:r
rSˆ
rSrS
kErSrS
rVScm
irVm
rSrS
kErSrS
rSVcm
irSVm
kn
knn
kn
knrr
r
kn
knn
kn
knrSrS
rS
ˆˆ
ˆˆ
ˆˆ.ˆ42
ˆˆ
ˆˆ
ˆˆ.ˆ4
ˆ2
,
,,
,
,22
222
,
,,
,
,ˆˆ22
22ˆ2
kErr
r nkn
knkn
,,
,,,
And the solution is:
11
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
rSrS
kErSrS
rVScm
irVm kn
knn
kn
knrr
r
ˆˆ
ˆˆ
ˆˆ.ˆ42 ,
,,
,
,22
222
The above equation does not look like the Schrodinger equation!
We define a unitary spin rotation operator that operates in the Hilbert space of spins and rotates spin states in the sense of the operator
Consider a spin vector pointing in the direction:n̂
SR ˆˆ
ba
Rba
RnS
ba
Rba
RRnR
ba
ba
RRn
ba
ba
n
SS
SSSS
SS
ˆˆ
ˆˆ1
ˆˆ
ˆ1
ˆ
ˆ1ˆˆˆ.ˆ
ˆ1ˆˆˆ.ˆˆ
1ˆˆˆ.ˆ
1ˆ.ˆ
The spin rotation operators have the property: nSRnRSS
ˆˆ.ˆˆˆ.ˆˆ 1ˆˆ
Ŝ
Appendix: Spin-Orbit Interaction and Rotation Symmetry
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
rSrS
kErSrS
rVScm
irVm kn
knn
kn
knrr
r
ˆˆ
ˆˆ
ˆˆ.ˆ42 ,
,,
,
,22
222
Start from:
Introduce spin rotation operator corresponding to the rotation generated by the matrix :
SR ˆˆŜ
rSrS
RkErSrS
RRrVScm
irVm
Rkn
knSn
kn
knSSrr
rS
ˆˆ
ˆˆˆ
ˆˆˆˆ.ˆ42
ˆ,
,1ˆ,
,
,1ˆˆ22
2221
ˆ
rSrS
RkErSrS
RrVcm
irVm kn
knSn
kn
knSrr
r
ˆˆ
ˆˆˆ
ˆˆ.ˆ42 ,
,1ˆ,
,
,1ˆ22
222
The above equation shows that the new state:
rSrS
Rkn
knS
ˆˆ
ˆ,
,1ˆ
satisfies the Schrodinger equation and has the same energy as the state:
rr
kn
kn
,
,
Appendix: Spin-Orbit Interaction and Point-Group Symmetry
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
kEkSE nn ,1', ˆ
Since:
rSrS
RerSrS
ReRrSRrS
Rkn
knS
RkSi
kn
knS
RSki
kn
knS
ˆˆ
ˆˆˆ
ˆˆˆ
ˆ,
,1ˆ
.ˆ
,
,1ˆ
ˆ.
,
,1ˆ
1
The new state is a Bloch state with wavevector kS1ˆ
Summary:
If is an operator for a point-group symmetry operation then the two states given by:Ŝ
r
rr
kn
knkn
,
,,,
rS
rSRr
kn
knSkSn
ˆˆ
ˆ,
,1ˆ',ˆ, 1
have the same energy:
Appendix: Spin-Orbit Interaction and Point-Group Symmetry
This represents a rotated (in space) version of the original Bloch state. Even the spin is rotated appropriately by the spin rotation operator.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Inversion Symmetry
Suppose we have found the solution to the Schrodinger equation:
rr
kErr
rVcm
irVm kn
knn
kn
knrr
r
,
,,
,
,22
222ˆ.ˆ
42
We replace by everywhere in the Schrodinger equation:r
r
rr
kErr
rVcm
irVm
rr
kErr
rVcm
irVm
kn
knn
kn
knrr
r
kn
knn
kn
knrr
r
,
,,
,
,22
222
,
,,
,
,22
222
ˆ.ˆ42
ˆ.ˆ42
kErr
r nkn
knkn
,,
,,,
And the solution is:
rVrV Suppose the crystal potential has inversion symmetry:
13
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
rr
kErr
rVcm
irVm kn
knn
kn
knrr
r
,
,,
,
,22
222ˆ.ˆ
42
Appendix: Spin-Orbit Interaction and Inversion Symmetry
The above equation shows that the new state:
rr
kn
kn
,
,
satisfies the Schrodinger equation and has the same energy as the state:
rr
kn
kn
,
,
Since:
rr
eRrRr
kn
knRki
kn
kn
,
,.
,
,
the new state is a Bloch state with wavevector k
rr
kn
kn
,
,
In most cases, the new state:
has the same spin direction as the state:
rr
kn
kn
,
,
So we can write:
rr
rkn
knkn
,
,,,
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Inversion Symmetry
kEkE nn
,,
Summary:
If the crystal potential has inversion symmetry then the two states given by:
r
rr
kn
knkn
,
,,,
rr
rkn
knkn
,
,,,
have the same energy:
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ECE 407 – Spring 2009 – Farhan Rana – Cornell University
rrr
rr knkn
kn
knkn
,,,
,,,
Consider the Bloch function:
Suppose the Bloch function corresponds to the spin pointing in the direction of the unit vector at the location : n̂
rrr
rr
nrn knkn
kn
kn
knkn
,,,
,
,
,,, 11ˆ.ˆˆ.ˆ
What if we want the state with the opposite spin at the same location?
The answer is:
r
rri
kn
knkny
,
,,, *
**ˆ
r
Proof:
riri
rnirni
rnirin
knykny
knyknyyyyy
knykny
,,,,
,,,,
,,,,
*ˆ1*ˆ
*ˆ.ˆˆ*ˆˆˆˆ.*ˆˆˆ
*ˆˆ.*ˆ*ˆˆ.ˆ
Appendix: Spin-Orbit Interaction and Time Reversal Symmetry
zyx zyx ˆˆˆˆˆˆˆ ˆˆˆˆˆˆˆ*ˆ zyx zyx
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
rr
kErr
rVcm
irVm kn
knn
kn
knr
,
,,
,
,22
222ˆ.ˆ
42
In the presence of spin-orbit interaction we have the Schrodinger equation:
We complex conjugate it:
rr
kErr
rVcm
irVm kn
knn
kn
knr
,
,,
,
,22
222
**
**ˆ.*ˆ
42
Suppose we have solved it and found the solution: kErr
r nkn
knkn
,,
,,,
It does not look like the original Schrodinger equation!
Note that:
ˆˆˆˆˆˆˆ*ˆ zyx zyx
zyx zyx ˆˆˆˆˆˆˆ
Appendix: Spin-Orbit Interaction and Time Reversal Symmetry
15
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Time Reversal SymmetryGiven an eigenvalue matrix equation:
One can always perform a unitary transformation with matrix T and obtain:
vAv
uBuTvTvTAT
1
TvuTATB
1
So try a transformation with the unitary matrix with the equation: yî
rr
ikErr
iirVcm
irVm
ikn
knyn
kn
knyy
ry
,
,,
,
,22
222
**
ˆ**
ˆˆˆ.*ˆ42
ˆ
rr
kErr
rVcm
irVm kn
knn
kn
knr
,
,,
,
,22
222
**
**ˆ.ˆ
42
r
rr
kn
knkn
,
,,,
We have found a new solution:
with the same energy as the original solution:
rr
kn
kn
,
,**
kEn
,
Question: What is the physical significance of the new solution?
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Time Reversal SymmetryUnder lattice translation we get for the new solution:
rr
eRrRr
kn
knRki
kn
kn
,
,.
,
,**
**
So the new solution is a Bloch state with wavevector k
r
rr
kn
knkn
,
,?,, *
*
Note that the new solution found can also be written as:
r
rri
kn
knkny
,
,,, *
**ˆ
But as shown earlier, the above state has spin opposite to the state
r
rr
kn
knkn
,
,,,
Therefore, the new solution is a Bloch state , i.e.: rkn
,,
r
rrir
kn
knknykn
,
,,,,, *
**ˆ
And we have also found that its energy is the same as that of the state : rkn
,,
kEkE nn
,,
16
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Time Reversal SymmetryIn the presence of spin-orbit interaction we have the time-dependent Schrodinger equation:
trtr
titr
trrV
cmirV
m knkn
kn
knr,,
,,ˆ.ˆ
42 ,,
,
,22
222
Lets see if we can find a solution under time-reversal (i.e. when t is replaced by –t):
trtr
titr
trrV
cmirV
m knkn
kn
knr,,
,,ˆ.ˆ
42 ,,
,
,22
222
The above does not look like a Schrodinger equation so we complex conjugate it:
trtr
titr
trrV
cmirV
m knkn
kn
knr,*,*
,*,*ˆ.*ˆ
42 ,,
,
,22
222
And it still does not look like the original Schrodinger equation!
Solution is:
tkiEkntkiE
kn
tkiEkn
kn
knkn
nn
n
erer
ertrtr
tr
,,
,
,,,
,
,
,,, ,
,,
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Time Reversal SymmetryGiven an eigenvalue matrix equation:
One can always perform a unitary transformation with matrix T and obtain:
vAv
uBuTvTvTAT
1
TvuTATB
1
So try a transformation with the unitary matrix with the equation:
trtr
titr
trrV
cmirV
m
trtr
it
itrtr
iirVcm
irVm
i
trtr
titr
trrV
cmirV
m
kn
kn
kn
knr
kn
kny
kn
knyy
ry
kn
kn
kn
knr
,*,*
,*,*ˆ.ˆ
42
,*,*
ˆ,*,*
ˆˆˆ.*ˆ42
ˆ
,*,*
,*,*ˆ.*ˆ
42
,
,
,
,22
222
,
,
,
,22
222
,
,
,
,22
222
yî
The above equation now looks like the time-dependent Schrodinger equation
17
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Spin-Orbit Interaction and Time Reversal Symmetry
tkiEkntkiE
kn
tkiEkn
kn
knkn
nn
n
erer
ertrtr
tr
,,
,
,,,
,
,
,,, ,
,,
Corresponding to the Bloch state:
with energy:
kEn
,
the time-reversed Bloch state is:
tkiEkntkiE
kn
tkiEkn
kn
kn nn
n
erer
ertrtr
,
,
,
,,,
,
,
,
*
*,*,*
and the time-reversed state has the same energy as the original state:
kEkE nn
,,
Summary:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Time reversal symmetry implies: kEkE nn
,,
Inversion symmetry implies: kEkE nn
,,
Appendix: Crystal Inversion Symmetry and Time Reversal Symmetry
In crystals which have inversion and time reversal symmetries the above two imply:
kEkE nn
,, There is spin degeneracy!
In crystals which do not have inversion symmetry the above two do not guarantee spin degeneracy. In general:
kEkE nn
,, Bands with different spins can have different energy dispersion relations
18
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Crystal Inversion Symmetry and Time Reversal SymmetryCartoon (and much exaggerated) sketches of the conduction bands of Ge and GaAsare shown below:
GaAsGe
k
EE
k
kEn
,
kEn
, kEn
,
kEn
,
kEkE nn
,, kEkE nn
,,