Lecture Notes # 3 - sunlight.caltech.edusunlight.caltech.edu/chem140a/Chem140alecture2.pdf ·...

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 ψ or ε, butyou have boundary conditions and want tosolve for possible values of ε and afunctional form of ψ

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

2

2

+!"

=!

• 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 ˆ

22

1

yclassicall 2

1

!"

=

!"=

=

=

mmv

ip

m

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

!=U

!=U!=U

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) outof the Hamiltonian, which becomes

2

2

2!

"=

mH

!

• Because outside the box, we know the e-

CANNOT be there, so we get the boundarycondition:

Particle in a 1-D BoxParticle in a 1-D Box!=U

( ) ( )

( ) ( ) ( )

( ) ( ) 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 ψ into the Schrödinger equation to makesure Hψ = Εψ


!"#

$%&

!"#

$%&

='(

)*+

,!"#

$%&-

-

='(

)*+

,!"#

$%&-

=!!"

#$$%

&!"#

$%&-

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 ψ

• Energy values are quantized since nis an integer

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


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 ψ(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 our1-D box.


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

• So N electrons fills nF= N/2 states.• The highest energy state, nF, gives εF, the Fermi

energy.!"#

$%&

=L

n

m

F

F

'(

2

2!

• Fermi energy is well defined atT = 0 K because there is nothermal promotion

• At high T, there isthermalization, so εF is not asclear

Fermi-Fermi-DiracDirac Distribution 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

Bµ""

• Officially defined as the energy where the probability of finding anelectron is ½

• This definition comes from the Fermi-Dirac Distribution:

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

• At T=0, µ=εF and ε = εF, so f(εF)=1/2

εF

• 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


( ) ( )zyxzyxzyxm

,,,,2

2

2

2

2

2

22

!"" =##$

%&&'

())

+))

+))* !

!=U

• You can show that the answer is:

( ) !"#

$%&

!!"

#$$%

&!"#

$%&

= zL

ny

L

nx

L

nAzyx zyx

n

'''( sinsinsin,,

• We now have 3 quantum numbers nx, ny, and nz that aretotally 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 arepeated “unit cell”


( ) ( )

( ) ( ) !"===

====

LxUxU

Lxx

0 since

00 ##

• For now, we don’t need such a strict boundary condition• Make sure ψ 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 aresolutions to the Schrödinger equation are TRAVELINGWAVES (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 2nπ/Lwhere 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

kF kz

kx

Fermi level

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

• Quantization of k in each direction leadsto discrete states within the sphere

• Satisfy the periodic boundary conditionsat ± 2π/L along one direction

• There is 1 allowed wave vector k, withdistinct kx, ky, kz quantum #s for thevolume element (2π/L)3 in k-space

• So, sphere has a k-space volume of

2

2

2FFk

m

!=! Vector in 3-D space

NOTE: This is a sphere only ifkx=ky=kz. Otherwise, we havean ellipsoid and have torecalculate everything. That canbe a mess.

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

3

3

4

FkV !=

• Number of quantum states is

• Since there are 2 e- per quantum state

Number of Quantum StatesNumber of Quantum States

( )33

2

34

state quantized allowed 1 of volume

volumetotal

L

kF

!

!

=

( )

31

2

3

2

3

2

3

3

3

3

,for Solve

3

32

34

2

!!"

#$$%

&=

=

=!!!

"

#

$$$

%

&=

V

Nk

k

kV

N

kL

L

k

N

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

2

2

3

2

2

!!"

#$$%

&=

=

V

N

m

km

F

F

'(

(

!

"!

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

23

22

2

3!"#

$%&

=!

'(

mVN

( ) 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!

"! #

$%

&'(

=!

mD

( ) ( ) 21

23

2

*

2

2

2

1CB

eEE

mD !""

#

$%%&

'=

!()

( ) ( ) 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

nC

e /exp2

223

2

*

!""#

$%%&

'= µ

(!

( ) ( )!"

==

CE

ee dEEfEDn fun

23

2

*

22 !!

"

#$$%

&=

!'kTm

Ne

C

( )[ ]kTEENnFCC/exp !!=

“EF”

Effective density of states in CB

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

Concentration of HolesConcentration of Holes

( )[ ]kTEkTm

p Vh /exp2

223

2

*

µ!

"##$

%&&'

(=

!

( ) ( )!"#

=VE

hh dEEfEDp

23

2

*

22 !!

"

#$$%

&=

!'kTm

Nh

V

( )[ ]kTEENp VFV /exp !!=

Effective density of states in VB

eh ff !=1

Intrinsic Carrier ConcentrationIntrinsic Carrier Concentration( ) ( )

!"

#$%

& ''''=(

kT

EEEENNpn VFFCVC exp

[ ]kTENNpn GVC /exp !="

gap band the,GVCEEE =!

( )!"

#$%

& ''=(

kT

EENNpn VCVC exp

Entropy term Enthalpy term2

inpn =!

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 = ni

( ) ( )!"#

$%& ''

=!"#

$%& ''

=kT

EEN

kT

EENn

Vi

V

iC

Ciexpexp

Intrinsic Fermi LevelIntrinsic Fermi Level

( ) ( )!!"

#$$%

&+!

"#

$%& ''

=!"#

$%& ''

C

VViiC

N

N

kT

EE

kT

EEln

( ) ( )!"#

$%& ''

=!"#

$%& ''

=kT

EEN

kT

EENn

Vi

V

iC

Ciexpexp

!!"

#$$%

&+'='

C

V

iVCi

N

NkTEEEE ln

!!"

#$$%

&++=

C

V

CVi

N

NkTEEE ln2

( ) !!"

#$$%

&+'='

C

V

VCVi

N

NkTEEEE ln2

( ) !!"

#$$%

&+='

C

VG

Vi

N

NkTEEE ln

22

Intrinsic Fermi LevelIntrinsic Fermi Level( ) !!

"

#$$%

&+='

C

VG

Vi

N

NkTEEE ln

22

• This says that the intrinsic Fermi level (relative to thevalence 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

NkTEG (eV)

1.12

1.420.67

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

• All of this has been intrinsic with no dopants

DopantsDopants

DCBDEE !="

++=

DiNnn carriers

• Let’s consider adding dopants

ECB

Ei

EVB

ED

n-type

Dependson EG, T

Depends on εD, 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

++!>>

DiDNnnN then , if

iinnn !>> then ,ND

DopantsDopants+

+=

=!

Di

i

Nnn

npn substitutejust holds, still 2

316316316

2

101010Say

densitiesdopant lfor typica re, temperaturoomat

!!+!"#$#=#$#=

%+

=

cmncmNcmN

NnNn

np

DD

D

Di

i

( )[ ]kTEENnFCC/exp !!=

( )[ ]

down meV 180 mV 180 mV 60 x decades 3

mV/decade 60about /exp10

10319

316

==

!!==!

!

kTEEcm

cm

N

nFC

C

Lecture Notes # 3 - sunlight.caltech.edusunlight.caltech.edu/chem140a/Chem140alecture2.pdf ·...

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