Handout 12 Energy Bands in Group IV and III-V SemiconductorsMost group VI and group III-V...

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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


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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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


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


• 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


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


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


kgVsp 23


kgVsp 33


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


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


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


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


• 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


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


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


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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Tight Binding Vs Pseudopotential Technique

Simplest TB Approach

A Little More Sophisticated ApproachNonlocal Pseudopotential Method

GaAs Energy Bands(Chelikowski and Cohen, 1976)

GaAs


Energy Bands of Silicon and Germanium

Silicon Energy Bands(Chelikowski and Cohen, 1976)

Germanium Energy Bands(Chelikowski and Cohen, 1976)

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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

,,


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

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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)


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:

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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


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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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.


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:

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rr

kErr

rVcm

irVm kn

knn

kn

knrr

r

,

,,

,

,22

222ˆ.ˆ

42


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

,

,,,



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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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


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


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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?


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

,,

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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

,,

,

,,,

,

,

,,, ,

,,


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

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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:


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


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

,,

Handout 12 Energy Bands in Group IV and III-V SemiconductorsMost group VI and group III-V...

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Transcript of Handout 12 Energy Bands in Group IV and III-V SemiconductorsMost group VI and group III-V...