Electrons Periodic Potential

8/6/2019 Electrons Periodic Potential

1/47

VI. Electrons in a Periodic Potential

A. Energy Bands and Energy Gaps in a Periodic Potential

B. Metals, Insulators, and Semiconductors

C. Energy Bands and Fermi Surfaces in 2-D and 3-D Systems

D. Blochs Theorem

E. Using Blochs Theorem: The Kronig-Penney ModelF. Empty Lattice Bands and Simple Metals

G. Density of States for a Periodic Potential

H. E(k) and N(E) for d-electron Metals

I. Dynamics of Bloch Electrons in a Periodic Potential

J. Effective Mass of Electrons

K. Electrons and Holes


2/47

A. Energy Bands and Energy Gaps in a Periodic Potential

Recall the electrostatic potential energy in a crystalline solid along a line passing

through a line of atoms:

bare ions

solid

Along a line parallel to this but running between atoms, the divergences of the periodicpotential energy are softened:

A simple 1D mathematical model thatcaptures the periodicity of such a potential is:

!

a

xUUxU

T2cos)( 10

U

x

010 UU


3/47

Electron Wavefunctions in a Periodic Potential

Consider the following cases:

Electrons wavelengths much larger than a, so wavefunctions andenergy bands are nearly the same as above

01 !U)( tkxi

Ae[

]

! m

k

E 2

22J

!

ak

U

T

{ 01

Wavefunctions are plane waves

and energy bands are parabolic:

ak

U

T!

{ 01 Electrons waves are strongly back-scattered (Bragg scattering) sostanding waves are formed:

? A ? A tiikxikxtkxitkxi eeeeeC [[[] s s!s! 21)()(

akU

Te{ 01 Electrons wavelengths approach a, so waves begin to be stronglyback-scattered :

)()( tkxitkxi Bee [[] s s! B


4/47

Electron Energies in a Periodic Potential

There are two such standing waves possible:

? Atitiikxikx ekxAeeeA [[]

!! )cos(22

1

2

1

? A titiikxikx ekxiAeeeA [[] !! )sin(22121

)(cos2 22*a

xA T]] !

)(sin2 22*axA T]] !

ak T!These two approximate solutions to the S. E. at have very different

potential energies. has its peaks at x = a, 2a, 3a, at the positions of

the atoms, where U is at its minimum (low energy wavefunction). The othersolution, has its peaks at x = a/2, 3a/2, 5a/2, at positions in betweenatoms, where U is at its maximum (high energy wavefunction).

]]*

]]*

We can do an approximate calculation of the energy difference between these

two states as follows. Letting U0 = 0 for simplicity, and remembering U1 < 0:

!!

!$a

x

ax

ax

ax

a

x

dxUAdxxUEE0

2221

2

0

** )(sin)(cos)cos(2)( TTT]]]]

axT2cosaA

1!


5/47

Origin of the Energy Gap

In between the two energiesthere are no allowed energies;i.e., an energy gap exists. Wecan sketch these 1-D resultsschematically:

!

a

x

ax

adxUEE

0

221

2 )(cos T Now use theidentity:

xx 2cos1cos 212 !

ga

axa

a

Ua

x

ax

a

UEUxdxEE !!!

!

104

4

0

4 )sin()cos(1 11 TT

T

E

kx

aT

aT

E-

E+

Eg

energy gap

The periodic potential U(x)splits the free-electron E(k) intoenergy bands separated bygaps at each BZ boundary.


6/47

Different Representations of E(k)

If we apply periodic boundaryconditions to the 1-D crystal, the

energy bands are invariant under areciprocal lattice translation vector:

iGkEGkEa

n )()( 2T!!XXXX

The bands can be graphically

displayed in either the (i) extendedzone scheme; (ii) periodic zonescheme; or (iii) reduced zonescheme.

(i) extended zone scheme: plot E(k) from k = 0 through all possible BZs (bold curve)

(ii) periodic zone scheme: redraw E(k) in each zone and superimpose

(iii) reduced zone scheme: all states with |k| > T/a are translated back into 1st BZ

1G

X


7/47

B. Metals, Insulators, and Semiconductors

It is easy to show that the number of kvalues in each BZ is just N, the

number of primitive unit cells in thesample. Thus, each band can beoccupied by 2N electrons due to theirspin degeneracy.

E

kx

aT

aT

E-

E+

Eg

A monovalent element with one atom

per primitive cell has only 1 valenceelectron per primitive cell and thus Nelectrons in the lowest energy band.This band will only be half-filled.

EF

The Fermi energy is the energydividing the occupied and unoccupiedstates, as shown for a monovalentelement.


8/47

Metals, Insulators, and Semiconductors

For reasons that will be explained more fully later:

Metals are solids with incompletely filled energy bands

Semiconductors and insulators have a completely filled or empty bands and an energygap separating the highest filled and lowest unfilled band. Semiconductors have a smallenergy gap (Eg < 2.0 eV).

Quick quiz: Does this mean a divalent

element will always be an insulator?Answer: In 1-D, yes, but notnecessarily in 2-D or 3-D! Bandsalong different directions in k-spacecan overlap, so that electrons can

partially occupy both of the

overlapping bands and thus form ametal. But it is true that only crystals with an

even number of valence electrons in aprimitive cell can be insulators.


9/47

C. Energy Bands and Fermi Surfaces in 2-D and 3-D Lattices

Lets analyze a simple square latticeof atoms with interatomic distance a.

Its reciprocal lattice will also besquare, with reciprocal lattice basevector of length 2T/a.

The Brillouin zones of the reciprocallattice can be identified with a simpleconstruction:

1

2

2

2

233

3 3

3

33

3 2 T/a

2 T/a

Now consider what happens if wehave a weak periodic potential andhave atoms with 1, 2, and 3 valenceelectrons per atom. A truly freeelectron system would have a Fermi

circle to define the locus of states atthe Fermi energy.

Lets see how the shape of the Fermicircle is distorted by the periodic

petential.


10/47

2-D Energy Bands and Fermi Circles

This figure shows theFermi circles

corresponding to 2-Dcrystals with one, twoand 3 valence electrons

per atom.

Note how the divalent

crystal has a Fermicircle that cuts acrossthe 1st BZ boundary(in red).

The different Fermi

circles correspond tosystems withsuccessively higher EFvalues, as shown in this2D band diagram. EF1

EF2

EF3


11/47

Shape of 2-D Energy Bands and Fermi Circles

This 2D band diagram shows how aweak periodic potential introduces

gaps in the free-electron bands atthe BZ boundary. Here, bands areshown along the [10] and [11]directions in k-space.

Thus, there is a similardiscontinuity introduced into afree-electron Fermi circle any time

it crosses a BZ boundary. This isillustrated for a 2-D divalentcrystal:

EF2

Note that the Fermi circle does not completely fill the

1st BZ, so some electrons are in the 2nd BZ.


12/47

Schematic Shape of a 3-D Fermi Surface

In 3D crystals the periodic potentialdistorts the shape of a Fermi sphere

in the vicinity of the BZ boundary.A schematic example for a simplecubic lattice and a crude model E(k)function is shown here:

Note that the Fermi circle does notcompletely fill the 1st BZ but makes

contact with the 1st BZ boundaryalong the [100] directions.


13/47

Shape of 3-D Energy Bands in a Real Metal

In 3D the energy bands are plottedalong the major symmetry

directions in the 1st

BZ. Many ofthe high symmetry points on the 1st

BZ boundary are labeled by letters.

The gamma point ( + ) is always thezone center, where k = 0.

Free-electron

bands in an fcccrystal

Electronbands in Al


14/47

Shape of 3-D Fermi Surface in a Real Metal

Even if the free-electron Fermi sphere does not intersect a BZ boundary, its shape canstill be affected at points close to the boundary where the energy bands begin to

deviate from the free-electron parabolic shape. This is the case with Cu.

Just a slightly perturbed free-electron sphere!


15/47

D. Blochs Theorem and Bloch Wavefunctions

This theorem is one of the most important formal results in all of solid state physicsbecause it tells us the mathematical form of an electron wavefunction in the presence of

a periodic potential energy. We will prove the 1-D version, which is known as Floquetstheorem.

What exactly did Felix Bloch prove in 1928? In theindependent-electron approximation, the time-independent Schrodinger equation for an electron ina periodic potential is:

]] ErUm

!

)(2

22

XJ

)()( rUTrUXXX

!where the potential energyis invariant under a latticetranslation vector :T

X

Bloch showed that thesolutions to the SE are theproduct of a plane waveand a function with the

periodicity of the lattice:

rki

kkerurXX

XXXX ! )()(]

Bloch functions

)()( ruTrukk

XXXXX !where

cwbvauTXXXX

!and


16/47

Heisenberg and Bloch

Who says physicists dont have any fun?


17/47

Proof of Blochs Theorem in 1-D

1. First notice that Blochs theorem implies:

It is easy to show that this equation formallyimplies Blochs theorem, so if we can prove itwe will have proven Blochs theorem.

Tkirki

kk eeTruTr

XXXX

XX

XXXX

! )()(]Tkirki

k eeru

XXXX

X

X

! )(Tki

k er

XX

X

X

! )(]

Or just:Tki

kkerTr

XX

XXXXX ! )()( ]]

2. Prove the statement shown above in 1-D:

Consider N identical lattice points around a circular ring,each separated by a distance a. Our task is to prove:

ikaexax )()( ]] !

12 N

3)()( xNax ]] !Built into the ring model is theperiodic boundary condition:

The symmetry of the ringimplies that we can find asolution to the wave equation:

)()( xax ]] !


18/47

Proof of Blochs Theorem in 1-D: Conclusion

If we apply this translation N times we willreturn to the initial atom position:

Now that we know C we can rewrite

)()()( xxCNax N ]]] !!

This requires 1!NC And has the mostgeneral solution: ,...2,1,02

ss!! neC niN T

ikaNni eeC !! /2TOr: Where we define the Bloch wavevector:Na

nk

T2!

...)()()( DEQxexCax ika]]] !!

It is not hard to generalize this to 3-D:Tki

kkerTr

XX

XXXXX ! )()( ]]

But what do these functions look like?

rki

kkerurXX

XX XX ! )()(]


19/47

Bloch Wavefunctions

This result gives evidence to support the nearly-free electron approximation, in which theperiodic potential is assumed to have a very small effect on the plane-wave character of afree electron wavefunction. It also explains why the free-electron gas model is sosuccessful for the simple metals!


20/47

A Remarkable Consequence of Blochs Theorem:

In the case of lattice vibrations in a crystal, we can calculate the propagation speed(group velocity) of a wave pulse from a knowledge of the dispersion relation:

Similarly, it can be shown using Blochs theorem that the propagation speed of anelectron wavepacket in a periodic crystal can be calculated from a knowledge of theenergy band along that direction in reciprocal space:

dkdvg

[!group velocity: (1-D)

dk

dE

dk

dvg

J

1!!

[electron velocity: (1-D)

)()( kkvkgXXXX X[!(3-D)

)(1

)( kEkvkg

XX

J

XXX!(3-D)

This means that an electron (with a specified wavevector) moves through a perfectperiodic lattice with a constant velocity; i.e., it moves without being scattered or in any

way having its velocity affected!

But waitdoes Blochs theorem prove too much? If this result is true, then what is theorigin of electrical resistivity in a crystal?


21/47

E. Using Blochs Theorem: The Krnig-Penney Model

Blochs theorem, along with the use of periodic boundary conditions, allows us tocalculate (in principle) the energy bands of electrons in a crystal if we know the

potential energy function experienced by the electron. This was first done for a simplefinite square well potential model by Krnig and Penney in 1931:

Each atom is represented by a finitesquare well of width a and depth

U0. The atomic spacing is a+b.

We can solve the SE in each region of space: ]]]

ExUdx

d

m!

)(2 2

22J

0 < x < a xixiI eAex

OO] !)(m

E2

22OJ

!

-b < x < 0

U

x

0 a a+b2a+b 2(a+b)

U0

-b

QxQx

II DeCex!)(]

mEU

2

22

0

J!

Now do youremember howto proceed?


22/47

Boundary Conditions and Blochs Theorem

x = 0

The solutions of the SE require that the wavefunctionand its derivative be continuous across the potential

boundaries. Thus, at the two boundaries (which areinfinitely repeated):

xixi

I eAexOO] !)(

QxQxII DeCex !)(]

Now using Blochs theorem for aperiodic potential with period a+b:

x = a )(aBee IIaiai

]

OO !

DCB ! (1) )()( DCQBAi !O (2)

)()()( baikIIII eba!]] k = Bloch wavevector

Now we can write the boundary conditions at x = a:

)()( baikQbQbaiai eDeCeBeAe ! OO (3)

)()()( baikQbQbaiai eDeCeQBeAei ! OOO (4)

The four simultaneousequations (1-4) can be writtencompactly in matrix form


23/47

Results of the Krnig-Penney Model

Since the values of a and b are inputs to the model, and Q depends on U0 andthe energy E, we can solve this system of equations to find the energy E at anyspecified value of the Bloch wavevector k. What is the easiest way to do this?

0

1111

)()(

)()(!

D

C

B

A

eQeeQeeiei

eeeeee

QQii

baikQbbaikQbaiai

baikQbbaikQbaiai

OO

OO

OO

OO

Taking the determinant and setting it equal to zero gives:

? A)(coscoshcossinhsin2

22bakQbaQba

Q

Q !

OOO

O


24/47

Numerical Solution of the Krnig-Penney Model

Using the values of a, b, and U0 we can construct the energy bands as follows:

1. Choose a value of k

2. Solve for the allowed energies E1(k), E2(k), E3(k) numerically

3. Choose a new value of k

Using atomic units where , length is measured in Bohr radii (a0 = 0.529)and energy has units of Rydbergs (1 Ry = 13.6 eV), I chose arbitrary values of:

a = 5.0 a0 b = 0.5 a0 U0 = 5.0 Ry

and obtained the following set of energy bands:

12

2

|m

J

4. Repeat step 2


25/47

Numerical Results of the Krnig-Penney Model

The 1st BZ has amaximum k valuegiven by:

00

max 57.05.5 a

rad

abak !!

!

TT

You can see the same qualitative 1-D band structure wededuced from the approximate band gap calculation earlier!


26/47

F. Empty Lattice Bands and Simple Metals

We expect that for a very weak periodic potential the energy bands of a solid shouldlook like those of a free-electron gas system. The limit of a vanishing potential is called

the empty lattice, and the empty-lattice bands are often plotted for comparison withthe energy bands of real solids. But how are the empty lattice bands calculated? It isdesirable to plot them in the reduced zone scheme, so we need to represent all of thetranslations of the fully periodic E(k) back into the 1st BZ.

The symmetry of the reciprocal lattice

requires:

)()( 1 GkEkEXXX

s!Where k1 is a wavevector

lying in the 1st

BZ.

So to represent all of the periodicbands in the reduced zone scheme:

2

1

2

12

)()( Gkm

GkEkEXXJXXX

s!s!

The s sign is redundant, and Myers chooses to use only the minus sign. In principleall reciprocal lattice vectors G will be represented, but electrons can only propagatewhen the G vector satisfies the Bragg condition (i.e.,only those G that give nonzerostructure factors).


27/47

Empty Lattice Bands for bcc Lattice

Myers illustrates this procedure for the bcc lattice by plotting the empty lattice bandsalong the [100] direction in reciprocal space. (For HW you will do this for the fcc lattice

and the [111] direction)ClBkhGhklXXXX

!The general reciprocal lattice translation vector:

In our analysis of the structure factor, weused a simple cubic lattice, for which thereciprocal lattice is also simple cubic:

za

Cya

Bxa

2

2

2 TTT

!!!XXX

And thus the general reciprocallattice translation vector is:

zla

yka

xha

Ghkl 2

2

2 TTT

!X

We write the reciprocal lattice

vectors that lie in the 1st

BZ as:

zz

a

yy

a

xx

a

k 2

2

2

1

TTT!

X

The maximum value(s) of x, y, and z depend on the reciprocal lattice type and thedirection within the 1st BZ, as we can see..


28/47

k Values in 1st BZ for bcc Lattice

We write the reciprocal latticevectors that lie in the 1st BZ as:

zza

yya

xxa

k 2

2

2

1

TTT!

X

[100] 0 < x < 1

[110]0 < x < 0 < y <

The maximum value(s) of x, y, and z depend onthe reciprocal lattice type and the directionwithin the 1st BZ. For example:

Remember that the reciprocallattice for a bcc direct latticeis fcc! Here is a top view,from the + kz direction:

ky

kx

+

H

H

N

a

T2

a

T2


29/47

Empty Lattice Bands for bcc Lattice

Thus the empty lattice energy bands are given by:

We can enumerate the lowest few bands for the y = z = 0 case, usingonly G vectors that have nonzero structure factors (h + k + l = even):

{G} = {000}

? A222

222

1

2 2

22)( lzkyhxamGkmkE hkl

!!

TJXXJX

20222 2

2

xEx

am

E !

!

TJ

{G} = {110} ? A ? A1111 20220 !! xExEE)110()101()011()110(

)110()110()101()011(

? A ? A2112

0

222

0 !! xExEE

)200( ? A20 2! xEE

? A ? A1111 20220 !! xExEE)011()101()101()011(

{G} = {200}

)002( ? A20 2! xEE ? A ? A42 20220 !! xExEE)200()002()020()020(


30/47

Empty Lattice Bands for bcc Lattice: Results

Thus the lowest energy empty latticeenergy bands for the bcc lattice are: ? A22

2

0)( lkhxEkE !X

0

1

2

3

4

5

6

0 0.2 0.4 0.6 0.8 1

x

E/E0

Series1

Series2

Series3

Series4

Series6

Series7

Series8

These are identical to the bands on p. 196 of Myers, except one higherlying band is missing from my plot.


31/47

G. Density of States for a Periodic Potential

A weak periodic potential only perturbs the free-electron energy bands in the vicinity of the Brillouin

zone boundaries, so the density of states N(E) closelyresembles the ideal free-electron form:

m

kkE

2)(

22J

!

However, it is clear that the perturbationof free-electron bands near the BZ

boundaries will introduce sharp featuresin the N(E) where the gradient of E(k)goes to zero. Thus we canschematically plot N(E) for the simple

metals:

!

E

dSVEN

k

X38

2)(

T

for a 3-D system2/1)( CEEN !


32/47

H. Energy Bands and N(E) for Transition Metals

One new feature arises in the transitionmetalsthe appearance of a series of

partially filled d-electron bands. The s-bands overlap the d-bands in the first rowtransition metals, leading to bandhybridization, as we can see in the energy

bands of the transition metals:

For elements with the same type ofcrystal lattice, an increasing Z isaccompanied by an increase in theFermi energy, populating more andmore of the d-electron states in the

lattice.(Note the series V p Crp Fe and alsothe series Co pNi p Cu)


33/47

N(E) for Transition Metals

The d-bands are much narrower than s- and p-bands due to the greater localization of thed-electrons. As a result, the transition metals with partially filled d-bands have a much

higher N(EF) than the simple metals. This means that these transition metals:

are often very chemically reactive (especially with O)

have very large heat capacities and magnetic susceptibilities

often display superconductivity

make good catalysts (supply electrons to molecules that stick to metal surface)


34/47

I. Dynamics of Bloch Electrons in an E Field

Earlier we noted that metals have partially-filled upper bands, while semiconductors andinsulators have completely filled upper bands. What is the qualitative difference betweenfilled and partially-filled bands?

For a single electron: vnejXX

! For a collection of electrons: !k

kvneJX

XXX )(

For each electron: )(1

kEvk

XX

J

XX!

But the symmetry of theenergy bands requires: )()( kEkE

XX!

Thus we conclude: )()( kvkv XXX !

So for a filled band, which hasan equal number of electronswith k positive and negative,

0)( !! stBZ1

kvneJXXX Filled energy bands carry no

current! We will see that this istrue even when an electric field

is applied.

Note: the electrons in filled bands are not stationarythere are just the same number

moving in each direction, so the net current is zero.


35/47

Dynamics of Bloch Electrons in an E Field

The same argument that we made for filled bands also applies to a partially filled band inthe absence of an electric field. However, when an external electric field is applied to a

periodic solid, the electrons all experience a force F that causes a change in their k values:

The rate at which each electron absorbsenergy from the field is the absorbed power: dt

kdEvF

dt

rdF

dt

dWg

)(X

XXXX

!!

!

Substituting for the group velocity and

using the chain rule, we have:

By comparing both sides, we have:

dt

kdkE

dt

kdEkE

F kk

XXX

XXX

J

XXX !! )(

)()(

1

dt

kdF

X

JX

! The acceleration theorem

We could go on to write: dt

pd

F

c

XX

! where kpc

X

J

X

! crystal momentum

But rememberpc is not the electrons momentum, because F is only the external force anddoes not include the force of the lattice on the moving electron.


36/47

The Bottom Line

Now we see that the external electric field causes a change in the k vectors of all electrons:

Eedt

kdF

XX

J

X

!!

If the electrons are in a partially filled band, thiswill break the symmetry of electron states in the1st BZ and produce a net current. But if they are in

a filled band, even though all electrons change kvectors, the symmetry remains, so J = 0.

J

XXEe

dt

kd !

E

kx

aT

aT

v

kx

When an electron reaches the 1st BZ edge (at k =T/a) it immediately reappears at the opposite edge(k = -T/a) and continues to increase its k value.

As an electrons k value increases, its velocityincreases, then decreases to zero and then becomesnegative when it re-emerges at k = -T/a!!

Thus, an AC current is predicted to result from aDC field! (Bloch oscillations)


37/47

Do We EverObserve This?

Not until fairlyrecently, due to theeffect of collisionson electrons in a

periodic butvibrating lattice.

However

In both experiments the periodic potential was fabricated in anartificial way to minimize the effect of collisions and make it

possible to observe the Bloch oscillations of electrons (oratoms!).


38/47

But in Reality

Bloch oscillations are not routinely observed because the electrons in a periodic systemundergo collisions with ions in the lattice much too frequently. In practice this occurs onthe time scale of the collision time X (}10-14 s). Lets analyze the effect of an externalelectric field in this more realistic case:

X

kEe

dt

kdX

J

XX(

$

!

The steady-state situation can berepresented (for a nearly free electronmetal) by a very small shift in the Fermisphere:

This leads to a net current, since thereis no longer perfect cancellation ofterms corresponding to k values, aswhen E = 0.

XJ

xx

eEk

$(

kx

k

y

(kx Ex

!stBZ1

xx kvneJ )(X

We can also write this as:

Jx =density ofuncompensated e-

evelocity ofuncompensated e-

FxFx veEENJ (! )( !(E

energy shift of Fermi sphere


39/47


40/47

J. Band Effective Mass of an Electron

We can use our previous results to write the equation of motion of a Bloch electron in 1-D:

And we can write:dt

dk

dk

dv

dt

dv

a

x

x

xx

x!!

Also, from theacceleration theorem:

This gives:

This allows us to define a band effective mass

2

211

xxxx

g

x

x

dk

Ed

dk

dE

dk

d

dk

dv

dk

dv

JJ !

!!

J

xx F

dt

dk!

! JJx

xx

F

dk

Ed

a 2

21 Or, in the form ofNewtons 2nd law: x

x

x

a

dk

EdF

2

2

2J!

2

2

2

*

xdk

Edm

J|

By contrast to the free-electron mass, the band effective mass varies depending onthe electrons energy and thus its location in the band!

2

2

2

1

*

1

xdk

Ed

m J|or


41/47

Physical Meaning of the Band Effective Mass

Of course, for a free electron,

In a 3-D solid we would find that m* is asecond-order tensor with 9 components:

and

The effective mass concept if useful because it allows us to retain the notion of a free-electron even when we have a periodic potential, as long as we use m* to account forthe effect of the lattice on the acceleration of the electron.

m

kE x

2

22J

!

zyxjikk

E

m ji,,,

1

*

1 2

2!

xx

x|J

m

m

m !

|

2

2

*J

J

But what does it mean to have a varying effective mass for different materials?


42/47

Physical Meaning of the Band Effective Mass

Near the bottom of a nearly-free electron band m* is approximatelyconstant, but it increases dramatically near the inflection point andeven becomes negative (!) near the zone edge.

The effective mass isinversely proportional to thecurvature of the energy band.


43/47

K. Electrons and Holes

Remember that the Hall Effect gives us a way to measure the sign and concentration of thecharge carriers in a metal. If electrons are the charge carriers, we expect the Hallcoefficient R

H= 1/ne to be negative. However, for some elements (Zn, Cd, Bi) R

His

positive!

To understand this behavior, first consider theexcitation of an electron from a filled band intoan unfilled band:

This is what happens in a semiconductor. Theelectron in the upper band can now conductelectric current under a field. But how does thelower band now behave?

02/

1

!s

s!

N

i

ikXWhen the lower band was filled, we had a net

electron wavevector of zero:

By thebandsymmetry,

k-j = -kj


44/47

Electrons and Holes

So when the state +j is empty in a band, the (incomplete) band has effective wavevector k-j.

Now we analyze the current flow in theincomplete band under the influence of a field E:

02/

!s

{j

N

ji

i kkXX

This can also be written: jj

N

ji

i kkk

s

{

!!XXX2/

s

{

s

s!

!!2/2/

1

0N

ji

ji

N

i

i veveveXXX

jve

X

This shows that an incomplete band (with state +j empty) behaves just like a positivecharge moving with the same velocity an electron would have in that state.

Thus the properties of all of the remaining electrons in the incomplete band are equivalent

to those of the vacant state j if the vacant state has:

a. A k-vector k -j

b. A velocity v+j

c. A positive charge +e

We call this vacant state a positive hole (h+)


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Dynamics of Electrons and Holes

But earlier we deduced that the hole velocity is thesame as that of the corresponding missing electron:

If this hole is accelerated in an applied electric field:

So by equating the derivatives we find:

However, note that near the top of a band the band curvature is negative, so the effective

electron mass is also negative. The corresponding hole mass is then positive!

So the equation of motion of a holein an electromagnetic field is:

Eedt

vdm hh

XX

!

eh vvXX

!

The corresponding equation for the electron is: Eedt

vdm ee

XX!

eh m

Ee

m

EeXX

!eh mm !

BvEedt

kdF h

hXXX

X

JX

v!!This explains whysome metals have

positive RH!


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Dynamics of Electrons and Holes in K-Space

Remember that in a metal with the Fermi energy near the top of the band, we canEITHER describe the properties of the electrons OR the properties of the hole(s). The

hole has no tangible existence apart from the electrons in the band.

So we see that electrons and holes in the same band move rigidly in the same direction ink-space without altering their relative position--like beads on a string.

The motion of electrons and holes in k-space is tricky tounderstand, because a hole has the opposite wavevector asthe missing electron. So we need to use the followingnotation to describe hole motion in the presence of an

electric field:position of electron in k-space

position of hole at ke (site of hole)

position of hole at kh


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Documents

Electrons Periodic Potential