Lecture Notes # 3 Understanding Density of States –Solve 1-D Schrödinger equation for particle- in-a-box –Extend to 3-D –Invoke periodicity requirement.

Lecture Notes # 3Lecture Notes # 3• Understanding Density of States

– Solve 1-D Schrödinger equation for particle-in-a-box

– Extend to 3-D

– Invoke periodicity requirement

– Solve for density of states

Review of Quantum MechanicsReview of Quantum Mechanics

H

• Often times you do not know or , but you have boundary conditions and want to solve for possible values of and a functional form of

electron that of

energies quantized allowableor Energy

space/timein electron of nature

depictingion wavefunctMathematic

operatorn Hamiltonia

H

Review of Quantum MechanicsReview of Quantum Mechanics

zyxUm

H ,,2

22

• Most general case: Time independent

Kinetic E. Potential E.

• How do we know first part is K.E?

22

2

22

2

22

1 So,

operator momentum quantum is ˆ22

1

yclassicall 2

1

mmv

ipm

pmv

mvKE

Review of Quantum MechanicsReview of Quantum Mechanics

ap

rikr

ˆ

expFor

rkrikirr

irip

ˆ

particle ofvelocity

m

kv

kmvp

Momentum operator Eigenvalue

a = Momentum Eigenvalue

• Free electron floating around in vacuum• Let’s impose some boundary – confine it to a region of space, a box or a unit cell in 1-D

Particle in a 1-D BoxParticle in a 1-D Box

UUU

U = 00 L x

Confined e-

Confine it by settingoutside the box andinside the box

U = 0

• Since U(x) = 0 for 0<x<L, we can drop U(x) out of the Hamiltonian, which becomes

22

2

mH

• Because outside the box, we know the e- CANNOT be there, so we get the boundary condition:

Particle in a 1-D BoxParticle in a 1-D BoxU

00 and

where

pickmust we,eigenvaluean back

givemust n Hamiltonia theBecause

00

Lff

xAfxfxf

Lxx

OK 0)sin( ,for

OK 0)0sin( ,0for

...4,3,2,1 ,sin

nLLx

x

nxL

nA

• We do not care what happens between 0 and L, so the simplest solution is just:

• Plug into the Schrödinger equation to make sure H=

Particle in a 1-D BoxParticle in a 1-D Box

xL

nA

L

n

m

xL

nA

L

n

m

xL

nA

L

n

dx

d

m

xL

nA

dx

d

m

sin2

1

sin2

cos2

sin2

2

2

222

2

2

22

Energy, E

• Energy values are quantized since n is an integer

• n=1 is lowest energy state, n=2 has higher energy, etc.

Particle in a 1-D BoxParticle in a 1-D Box

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4

2

222

2 x

mL

nE

0 L

E

9x

4x

n=4

n=3

n=2

n=1

n

• We can map out (x,n) vs. E

• Allowed energy states

• Now let’s fill up the states with electrons. Suppose we have N e - we want to pour into our 1-D box.

Particle in a 1-D BoxParticle in a 1-D Box

• For N e- you can calculate the energies since we know we can have 2e-/n states (two spins).

• So N electrons fills nF= N/2 states.

• The highest energy state, nF, gives F, the Fermi energy.

L

n

mF

F

2

2

• Fermi energy is well defined at T = 0 K because there is no thermal promotion

• At high T, there is thermalization, so F is not as clear

Fermi-Dirac DistributionFermi-Dirac Distribution

0

2

4

6

8

10

12

14

16

18

0 1 2 3 4

2

222

2 x

mL

nE

n

1/exp

1

Tkf

B

• Officially defined as the energy where the probability of finding an electron is ½• This definition comes from the Fermi-Dirac Distribution:

• This is the probability that an orbital (at a given energy) will be filled with an e - at a given temperature

• At T=0, =F and = F, so f(F)=1/2

F

• Let’s confine e- now to a 3-D box• Similar to a unit cell, but e- is confined

inside by outside the box• Schrödinger’s equation is now

Particle in a 3-D BoxParticle in a 3-D Box

zyxzyxzyxm

,,,,2 2

2

2

2

2

22

U

• You can show that the answer is:

z

L

ny

L

nx

L

nAzyx zyx

n

sinsinsin,,

• We now have 3 quantum numbers nx, ny, and nz that are totally independent

• (1,2,1) is energetically degenerate with (2,1,1) and (1,1,2)

• What’s different about this situation?– U(x,y,z)=0– No region where U = infinity

• So, there’s really no reason that

• We don’t need those boundary conditions anymore

• Now let’s repeat this box infinitely in each direction to get a repeated “unit cell”

Particle in a 3-D BoxParticle in a 3-D Box

LxUxU

Lxx

0 since

00

• For now, we don’t need such a strict boundary condition• Make sure is periodic with L, which would make each 3-D box identical• Because of this, we’ll have a periodic boundary condition such that

Periodic Boundary ConditionPeriodic Boundary Condition

zyxzyLx ,,,, • Wave functions that satisfy this periodic B.C. and are solutions to the

Schrödinger equation are TRAVELING WAVES (not a standing wave anymore)

• Bloch function

Periodic Boundary ConditionPeriodic Boundary Condition

rkirk

exp

;...4

;2

;0LL

kx

xik

Lnxi

niLnxi

LLxniLxik

x

x

exp

/2exp

2exp/2exp

/2expexp

• Wave vector k satisfies

• Etc. for ky and kz

• Quantum numbers are components of k of the form 2n/L where n=+ or - integer

• Periodicity satisfied

• Substitute

Back to SchrBack to Schrödinger Equationödinger Equation

• Important that kx can equal ky can equal kz or NOT

• The linear momentum operator

2222

22

2

2

2

2

2

22

22

gives

2

into

exp

zyxk

kkk

k

kkkm

km

rrzyxm

rkir

ip̂

m

kv

k

r

rkrirp

rkir

k

kkk

k

isk orbitalin velocity particle theand ,

of eigenvaluean with momentumlinear of

ioneigenfunctan is waveplane theso

ˆ

expfor

• Similarly, can calculate a Fermi level

Fermi Level in 3-DFermi Level in 3-D

ky

kFkz

kx

Fermi level

• Inside sphere k<kF, so orbitals are filled. k>kF, orbitals are empty

• Quantization of k in each direction leads to discrete states within the sphere• Satisfy the periodic boundary conditions at ± 2/L along one direction

• There is 1 allowed wave vector k, with distinct kx, ky, kz quantum #s for the volume element (2/L)3 in k-space

• So, sphere has a k-space volume of

22

2 FF km

Vector in 3-D space

NOTE: This is a sphere only if kx=ky=kz. Otherwise, we have an ellipsoid and have to recalculate everything. That can be a mess.

Sphere: GaAs (CB&VB), Si (VB)

Ellipsoid: Si (CB)

3

3

4FkV

• Number of quantum states is

• Since there are 2 e- per quantum state

Number of Quantum StatesNumber of Quantum States

33

23

4

state quantized allowed 1 of volume

volumetotal

L

kF

31

2

32

32

3

3

3

3

,for Solve3

323

42

V

Nk

k

kV

N

kL

L

kN

F

F

F

F

F

• Depends on e- concentration

• Plug kF into

• Relates Fermi energy to electron concentration

• Total number of electrons, N:

Density of StatesDensity of States

32

22

22

3

2

2

V

N

m

km

F

F

• Density of states is the number of orbitals per unit energy

23

22

2

3

mV

N

212

3

22

2

2

mV

d

dND

Relate to the surface of the sphere. For the next incremental growth in the sphere, how many states are in that additional space?

• Divide by V to get N/V which is electron density (#/cm3)• Volume density of orbitals/unit energy for free electron gas in periodic potential

Density of StatesDensity of States

212

3

22

2

2

1

m

D

21

23

2

*

2

2

2

1CB

e EEm

D

21

23

2

*

2

2

2

1EE

mD VB

h

Effective mass of e-

Effective mass of h+

Starting point energy

CB

VB

Start from VB and go down

Concentration of ElectronsConcentration of Electrons

kTEkTm

n Ce /exp

22

23

2

*

CE

ee dEEfEDn fun

23

2

*

22

kTm

N eC

kTEENn FCC /exp

“EF”

Effective density of states in CB

Writing it with a minus sign indicates that as E difference between EC and EF gets bigger, probability gets lower

Concentration of HolesConcentration of Holes

kTEkTm

p Vh /exp

22

23

2

*

VE

hh dEEfEDp

23

2

*

22

kTm

N hV

kTEENp VFV /exp

Effective density of states in VB

eh ff 1

Intrinsic Carrier ConcentrationIntrinsic Carrier Concentration

kT

EEEENNpn VFFC

VC exp

kTENNpn GVC /exp

gap band the,GVC EEE

kT

EENNpn VC

VC exp

Entropy term Enthalpy term2inpn

ionconcentratcarrier intrinsic in

Constant for a given temperature.Intrinsic = undoped

Intrinsic Carrier ConcentrationIntrinsic Carrier Concentration

kT

ENNn G

VCi 2exp

ni

Ge: 2.4 x 1013 cm-3

Si: 1.05 x 1010 cm-3

GaAs: 2 X 106 cm-3

At 300 K

it fromaway and mequilibriu

under holds which too,constant, is

thatmeans Tgiven afor constant

pn

ni

• At temperature T, n = p by conservation• Add a field and np = constant, but n does not equal p• As n increases, p decreases, and vice versa

• Useful to define Ei, which is Ei = EF when it is an intrinsic semiconductor (undoped), so n = p = n i

kT

EEN

kT

EENn Vi

ViC

Ci expexp

Intrinsic Fermi LevelIntrinsic Fermi Level

C

VViiC

N

N

kT

EE

kT

EEln

kT

EEN

kT

EENn Vi

ViC

Ci expexp

C

ViVCi N

NkTEEEE ln

C

VCVi N

NkTEEE ln2

C

VVCVi N

NkTEEEE ln2

C

VGVi N

NkTEEE ln

22

Intrinsic Fermi LevelIntrinsic Fermi Level

C

VGVi N

NkTEEE ln

22

• This says that the intrinsic Fermi level (relative to the valence band) is about mid-gap ± the (kT/2)ln(NV/NC) scaling factor

Gefor meV 7-

GaASfor meV 35

Sifor meV -13ln2

C

V

N

NkT EG (eV)1.12

1.420.67

• So Ei for Si and Ge is slightly below mid-gap. Ei for GaAs is slightly above. It is minor compared to EG, but just so you know

• All of this has been intrinsic with no dopants

DopantsDopants

DCBD EE

Di Nnn carriers

• Let’s consider adding dopants

ECB

Ei

EVB

ED

n-type

Depends on EG, T

Depends on D, T, and dopant density

At T=0, all donor states are filled. Hence, n = 0.But at room temperature in P doped Si, 99.96% of donor states are ionized.

At mid temperature,

At high temperature, such that

DiD NnnN then , if

ii nnn then ,ND

DopantsDopants

Di

i

Nnn

npn substitutejust holds, still 2

316316316

2

101010Say

densitiesdopant lfor typica re, temperaturoomat

cmncmNcmN

NnNn

np

DD

DDi

i

kTEENn FCC /exp

down meV 180 mV 180 mV 60 x decades 3

mV/decade 60about /exp10

10319

316

kTEEcm

cm

N

nFC

C