Sconyers 8p3 NearlyFreeElectronModelalan.ece.gatech.edu/ECE6451/Lectures/Student... · The Kronig-Penney Model (review) According to this model, discontinuities in allowed energies

Georgia Tech ECE 6451

Nearly-Free-Electron Model

Reading:Brennan Chapter 8.3

Lecture prepared by Christopher Sconyers


The Kronig-Penney Model (review)

According to this model, discontinuities in allowed energies appear at k=nπ/(a+b), where a and b relate to the periodicity of the rectangular potential.What about any arbitrary periodic potential? Will discrete bands of energy show up in the solution as expected?

The solution to the Schroedinger equation for a periodic rectangular barrier predicts that only discrete bands of energy are allowed.

Brennan, Figure 8.2.1. One dimensionalperiodic potential with period a+b.


The Schroedinger Equation

)()(2

2

2

202

xfxVm

kmE

γ=−

=

h

h

Defining a periodic potential

[ ]

0)(22

0)(2

222

2

22

2

=Ψ

−+Ψ

=Ψ−+Ψ

Ψ=Ψ

xVmmEdxd

xVEmdxd

EH

hh

h

Let’s start with the Schroedinger equation, and go from there:

To simplify this, we will make two helpful substitutions

Finally, the Schroedinger equation becomes:

[ ] 0)(202

2

=Ψ++Ψ xfk

dxd γ



But how do we define this new f(x)? Well, we know two things about our arbitrary periodic potential V(x):

• V(x) has a set lattice constant, or period, a.• V(x) can be described as a Fourier series.

Defining a periodic potential

[ ] 0)(202

2

=Ψ++Ψ xfk

dxd γ

So, let’s expand f(x), a periodic function with the same periodicity as V(x), in terms of a complex Fourier series.

factorscalingaisγand

)(1

)(

2

2

0dxexf

aCwhere

eCxf

anxi

anxi

a

n

nn

π

π

∫

∑

=

=−

+∞

−∞=



As seen before, any periodic potential will result in a periodic wave function of the form

Bloch functions and periodicity

)()(:ldimensiona one

)()(:ldimensiona-n

xuexruer

kikx

k

nkikr

nk

=Ψ

=Ψ

where uk(x) is periodic, with the same periodicity of the potential. Therefore, we can expand it in terms of another complex Fourier series.

∑

∑−

−

=Ψ

=+∞

−∞=

nn

ikx

nnk

anxi

anxi

ebex

ebxu

π

π

2

2

)(

)(

The wave function will retain the same magnitude each period a. Only its phase will change from one lattice point to the next. The properties of the Bloch function are retained.But, this is not very useful unless we can relate this to f(x). So we will make a few more observations about the nature of the wave function relating to the periodic potential, and then solve the Schroedinger equation. Hopefully, we can shine some light on this new wave function.



From before…Relating f(x) to uk(x)

[ ] 0)(202

2

=Ψ++Ψ xfk

dxd γ

If γ = 0, then V(x) is zero and the electron is free, which is represented as a simple plane wave.

xikebx

kdxd

00

202

2

)(

0

=Ψ

=Ψ+Ψ

If γ ≠ 0, we assume the electron to be traveling in a weak potential, and can thus expand the wave function into two parts. One corresponding to a plane wave, or a free electron, and another corresponding to a periodic correction factor. Hence the name, “nearly-free-electron model.”

∑≠

−

+=Ψ0

0

2

)(n

nikxikx a

nxi

ebeebxπ

γ

“free electron”plane wave

periodic correctionfactor (small)


Solving the Schroedinger Equation

Now we have a general solution for our arbitrary, periodic potential. Remember, we are trying to determine if any periodic potential will result in discrete energy bandgaps. To do this, we will want to solve for the energy versus k relationship, which means we first must solve the Schroedingerequation to find the coefficients bn. Plugging Ψ(x) into our simplified Schroedinger equation:

First Order Solution

∑≠

−

+=Ψ0

0

2

)(n

nikxikx a

nxi

ebeebxπ

γ

[ ] 0)(202

2

=Ψ++Ψ xfk

dxd γ

The full derivation is not shown here (Brennan p.419). After substituting, expanding the second-derivative term, and combining like terms, the result is:

( ) ( ) ( )

( ) ( )

( ) ( ) 00

44

0 0

2

00

0

220

2200

2

2

22

222

2

=−+

++

−+−

∑

∑ ∑∑

∑

≠

−

≠ ≠′

−−′

≠

−

≠

−

′

n

xkia

na

nn

n n

xkinn

n

xkin

n

xkin

ikx

an

an

an

an

an

ekb

ebCeCb

ebkkekkb

π

πππ

π

ππγ

γγ

γ

TOOCOMPLEX!!



The problem already seems to have blown up. However, we made the assumption that V(x) is weak, therefore γ must be small, and the γ2 term (fourth term) can be neglected. Also, we can further simplify the equation. Consider only the terms in γ.


We want to combine these into one simple term, and we can by defining the following relation:

( ) ( ) ( ) ( )

( )∑

∑∑

≠

−

≠

−

≠

−

+

−+−

00

0

44

0

220

2

2

2

222

n

xkin

n

xkia

na

nn

n

xkin

an

an

an

eCb

ekbebkk

π

ππ

γ

γγ ππ

an

nkk π2+=

Combining the first two terms and substituting in the above relation:( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )∑∑

∑∑

∑∑

≠

−

≠

−

≠

−

≠

−

≠

−

≠

−

+−++−−

+−+−−−

+−+−

00

0

8484220

00

0

4444220

00

0

44220

22

2

22

2

22

22

2

22

2

22

22

2

22

n

xkin

n

xkina

na

na

na

nn

n

xkin

n

xkina

na

na

nna

nn

n

xkin

n

xkina

na

n

an

an

an

an

an

an

eCbebkkkk

eCbebkkkk

eCbebkkk

ππ

ππ

ππ

γγ

γγ

γγ

ππππ

ππππ

ππ


Continued…

( ) ( ) ( )

( ) ( ) ( )

( )[ ] ( )∑

∑∑

∑∑

≠

−

≠

−

≠

−

≠

−

≠

−

+−

+−

+−++−−

00

220

00

0

220

00

0

8484220

2

22

22

2

22

2

22

n

xkinnn

n

xkin

n

xkinn

n

xkin

n

xkina

na

na

na

nn

an

an

an

an

an

eCbbkk

eCbebkk

eCbebkkkk

π

ππ

ππ

γ

γγ

γγ ππππ

Putting this back into our previous equation, we get a much simpler form to deal with:

( ) ( )[ ] ( ) 00

022

022

00

2

=+−+− ∑≠

−

n

xkinnn

ikx an

eCbbkkekkbπ

γ

This will make our job much easier. Now for another trick…

Solving the Schroedinger EquationFirst Order Solution



Let’s multiply the above equation by:


( ) ( )[ ] ( ) 00

022

022

00

2

=+−+− ∑≠

−

n

xkinnn

ikx an

eCbbkkekkbπ

γ

am

m

xik

kkwheree m

π2−=

−

Then we’ll integrate over a full period (0 a):

( ) ( )[ ] ( )0

000

2200

2200

22

=+−+− ∑ ∫∫≠

−

n

a

nnn

adxeCbbkkdxekkb a

xnmia

mxi ππ

γ

Note that the two integrals are of similar form, which has the following useful property:

=≠

=∫ 0 if 0 if 0

0

2

λλπλ

adxe

aa

xi



Case 1: m = 0


( ) ( )[ ] ( )0

000

2200

2200

22

=+−+− ∑ ∫∫≠

−

n

a

nnn

adxeCbbkkdxekkb a

xnmia

mxi ππ

γ

=≠

=∫ 0 if 0 if 0

0

2

λλπλ

adxe

aa

xi

Within the summation, n ≠ 0. Therefore the integral is always zero when m=0.

( ) ( )[ ]( ) ( )[ ]( )

( )

0

2200

00

220

2200

000

2200

2200

0

00

02

kk

akkb

Cbbkkakkb

dxeCbbkkdxkkb

nnnn

n

a

nnn

aa

nxi

=

=−

=+−+−

=+−+−

∑

∑ ∫∫

≠

≠

−

γ

γπ



Case 2: m ≠ 0


( ) ( )[ ] ( )0

000

2200

2200

22

=+−+− ∑ ∫∫≠

−

n

a

nnn

adxeCbbkkdxekkb a

xnmia

mxi ππ

γ

=≠

=∫ 0 if 0 if 0

0

2

λλπλ

adxe

aa

xi

( ) ( )[ ] ( )

( )( ) ( )[ ] ( )00

0

000

220

2200

000

2200

2200

2

22

=+−+−

=+−+−

∑ ∫

∑ ∫∫

≠

≠

−

−

n

a

nnn

n

a

nnn

a

dxeCbbkkkkb

dxeCbbkkdxekkb

axnmi

axnmi

amxi

π

ππ

γ

γ

The first term goes to zero for all m≠0, and the second term goes to zero for all m≠n. That only leaves the case when m=n. Lastly, we substitute k=k0 in the final step.

( )[ ]( )[ ]

( )220

022

0

0022

0

0

0

kkCbb

aCbbkk

dxCbbkk

m

mm

mmm

mn

a

nnn

−=

=+−

=+−∑ ∫=

γ

γ


Substituting the value for the coefficients into the wave function, and we’ve at last found a first order solution for our electron in a periodic potential.

Solving the Schroedinger EquationFirst Order Solution

( )

( )

−

−=Ψ

−+=Ψ

+=Ψ

∑

∑

∑

≠

≠

≠

−

−

−

0220

022

00

00

2

2

2

1)(

)(

)(

n n

nikx

n n

nikxikx

nn

ikxikx

anxi

anxi

anxi

ekk

Cebx

ekk

Cbeebx

ebeebx

π

π

π

γ

γ

γ

The implications of this wave function are immediately apparent. For n=0, the term inside the brackets becomes 1 and the electron is represented as a plane wave. For all other n, a small correction factor given by the summation and scaled by γ will slightly alter the plane wave periodically, with period a, as formerly predicted. This is a free electron with a small, periodic correction.However, the solution diverges when k=kn (the denominator goes to zero). A singularity occurs in this case. Let’s move on and find this wave’s energy.


What is the Energy of the Electron?

As before, we’ll start with a first-order solution (neglecting γ2). We found previously that k=k0 in the first-order case. Plugging this in for the energy solution:

First Order Energy Correction

( )

−

−=Ψ ∑≠

−

0220

2

1)(n n

nikx anxi

ekk

Cebxπ

γ

mk

mkE

22

20

222 hh==

This tells us that for a first-order approximation, the energy of the electron in a periodic potential is the same as the energy of a free electron. This must mean that the energy correction factor has gone to zero for first-order. While the first-order solution helped approximate the wave function, we will have to go back to the second-order (γ2) to determine the energy of the electron.

( ) ( )[ ] ( )

( ) 00 0

2

00

220

2200

22

2

=+

+−+−

∑ ∑

∑

≠ ≠′

−−′

≠

−

′

n n

xkinn

n

xkinnn

ikx

an

an

an

ebC

eCbbkkekkb

ππ

π

γ

γ

an

n kk π2−=Reminder



As before, multiply by e-ikx and integrate over a full period (0 a):

Second Order Energy Correction

( ) ( )[ ]( )

( ) 0

0

0

22200

0 00

2

000

2200

2200

2

2

=+−

=+

+−+−

∑

∑ ∑ ∫

∑ ∫∫

≠′′′−

≠ ≠′′

≠′+−

−

nnn

n n

a

nn

n

a

nnn

a

abCakkb

dxebC

dxeCbbkkdxkkb

axnni

anxi

γ

γ

γ

π

π

The second integral goes to zero, since n≠0 for all n. The third integral goes to zero in all cases except where n´ = -n. Now, -n´ has been substituted in for all n, and the double summation becomes a single summation. Also…

*

*** 2

2

2

)()()(

)()(

nn

nn

nn

nn

CC

eCxfxfxf

eCxfnneCxf

anxi

anxi

anxi

=∴

=⇒=

=⇒−=⇒=

−

∞+

−∞=

+∞

−∞=−∞+

−∞= ∑

∑∑

−

π

π

π



Plugging in the coefficients bn found in the first-order solution:


( ) 00

*22200

*

=+−

=

∑≠

−

nnn

nn

abCakkbCC

γ

Continued…

( )( ) ( ) 0

022

*2

022

00

220

=−

+−

−=

∑≠n n

nn

n

nn

kkCCabkkab

kkCbb

γ

Dividing out b0a and solving for (k0)2, we get:

( ) ( )

( )∑

∑

≠

≠

−+=

=−

+−

022

*222

0

022

*222

0 0

n n

nn

n n

nn

kkCCkk

kkCCkk

γ

γNote: the singularityhas shown up again.(at k=kn)



Continued…


( )∑≠ −

+=0

22

*222

0n n

nn

kkCCkk γ

Multiply both sides by ħ2/2m to get:

( )

( )[ ]

( )[ ] nmnn a

πnmm

n

n an

nn

an

n

n n

nn

CVwherekk

VmkkE

kkCC

mmkE

kkmkE

kkCC

mk

mk

m

20

222

22

222

0222

*2222

220

2

022

*2

22

220

2

22

222)(

22

and 2

222

γ

π

π

γ

γ

h

hh

h

hh

h

hhh

−=−−

+=

−−+=

−==

−+=

∑

∑

∑

≠

≠

≠

Recall


Does this predict the expected Bandgaps?

We’ve finally derived a solution for the energy of the electron in relation to k. What we really want to know is if this properly predicts Bandgaps. In other words, in any arbitrary, periodic potential, are there certain discrete bands of energy that electrons are not allowed to populate? Recall…

Bandgap Location

( )[ ] nmnn a

πnmm

n CVwherekk

VmkkE 2

022

22

2

22222

222)( γh

hh

h−=

−−+= ∑

≠

[ ] an

nn n

n kkwherekk

VmkE π2

022

222

2−=

−+= ∑

≠

h

We can see that a singularity appears in the energy at k2=(kn)2, or at k=±kn. At these points, the correction term blows up to infinity, which is clearly not possible. We can identify the location of these points in terms of k:

002

2

=

=

−=

+=

n

kkkk

an

an

n

π

π

a

an

an

n

nkk

kkkk

π

π

π

=

=

−=−

−=

2

2

2


So now we know that a discontinuity exists at all points where k is an integer multiple of π/a. We’re not through yet. Lastly, we must ensure that at these points there is a jump in allowed energy, or a gap of disallowed energy values.First, let’s compare this to the Kronig-Penney model (periodic rectangular barrier). Recall, that model predicts bandgaps will occur at:

Does this predict the expected Bandgaps?Bandgap Location

nnk a integer allfor π=

( )bank += π

Where a and b are the width of the barrier and the spacing between barriers. Adding the two together gives the period of the periodic potential. The general solution for an arbitrary, periodic potential accurately predicts the solution to the Kronig-Penney model. Why is this?For any crystalline structure, as well as any periodic potential, Bragg reflection at the Brillouin zone edges allow electrons to have only a certain energy values. These zone edges occur at k=nπ/a, where a is the lattice constant (period) of the periodic potential.



Now we need to show that a Bandgap occurs at k=nπ/a. To do this, we must show that at these values of k, there are multiple values for energy.Let’s go back to the wave function:

Energy Values at the Bandgaps

∑≠

−

+=Ψ0

0

2

)(n

nikxikx a

nxi

ebeebxπ

γ

We’ll focus only on the points of interest: where k=kn. At these points, bnblows up, so we will approximate the sum as only one term: bn.

( )

( )an

n

xikn

ikx

xkin

ikx

nikxikx

kkwhereebebx

ebebx

ebeebx

n

an

anxi

π

γ

γ

γπ

π

20

0

0

)(

)(

)(2

2

−=

+=Ψ

+=Ψ

+=Ψ−

−



Plugging this wave function into the Schroedinger equation, expanding the second-derivative, and collecting like terms results in:


xikn

ikx nebebx γ+=Ψ 0)(

( ) ( ) ( )

( ) ( ) 00

2

00

220

2200

222

2

=++

−+−

∑∑≠′

−−′

≠

−

−

′

n

xkinn

n

xkin

xkinn

ikx

an

an

an

an

eCbeCb

ekkbekkbπππ

π

γγ

γ

Now, we’ll perform two tricks on this equation to get two different equations.Step 1: Multiply by e-ikx and integrate over a full period (0 a):

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( )nnaCbakkb

dxeCbdxkkb

eCbeCbekkbkkb

nn

n

a

nn

an

nnn

nnn

axnni

axnni

anxi

anxi

−=′=+−

=+++−

=++−+−

∑ ∫∫

∑∑

≠′′

≠′′

≠′+−

′+−−−

0

000

0

*22200

00

2

0

2200

0

2

00

220

2200

2

222

γ

γ

γγγ

π

πππ



Step 2: Multiply by e-iknx and integrate over a full period (0 a):


( ) ( )

( ) ( ) ( )( ) 0

000

0

022

0

000

220

0

2

00

220

2200

22

=+−

=++−+

=++−+−

∫∫

∑∑≠′

′≠

′−−

aCbakkb

dxCbdxkkb

eCbCbkkbekkb

nnn

a

n

a

nn

nnn

nnnn

axni

anxi

γγ

γγ

γγγππ

We end up with two equations and two unknowns. If we collect the coefficients of the b0 and bn terms into a matrix, then the determinant of these coefficients must go to zero.

( )( )

( )( )

( )( ) 0

0

0

0

*2220

220

220

*2220

022

0

*22200

=−−−

=−

−

=+−

=+−

nnn

nn

n

nnn

nn

CCkkkk

kkCCkk

CbkkbCbkkb

γ

γ

γ



Continued…


( )( ) 0*2220

220 =−−− nnn CCkkkk γ

( )22*222220

40

*222220

220

40

02

0

nnnn

nnnn

kkCCkkkkkCCkkkkkkk

==−+−

=−+−−

γ

γ

If we treat the above as a second-order polynomial in (k0)2, we can solve for (k0)2 using the quadratic formula:

nn

nnn

nnn

CCkk

kkCCkkkk

CCkkkkk

*2220

*244220

*2224220

)(442

4442

γ

γ

γ

±=

=+−±=

+−±=

Multiply by ħ2/2m:

nn

nn

CCm

km

E

CCm

km

km

*22

42

2

*22

22

20

2

42

222

γ

γ

hh

hhh

±=

±=


Since we are solving for energy at the zone boundaries, then k=kn=nπ/a. Also, we can use the term Vn inside the square root.

Does this predict the expected Bandgaps?Energy Values at the Bandgaps

nn CCm

km

E *22

42

2

42γhh

±=

mV

an

mE

Vma

nm

E

CCm

V

n

n

nnn

22

22

2

222

2222

*22

2

hh

hh

h

±

=

±

=

=

π

π

γ

And so we have finally shown that not only do discontinuities occur at discrete values of k, but these discontinuities correspond to jumps in allowed electron energy, and this holds true for any periodic potential. The gap widths can be calculated from the Fourier coefficients of the potential.

mVE n 2

22h

=∆

two distinctenergy values

at the zone edge

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Sconyers 8p3 NearlyFreeElectronModelalan.ece.gatech.edu/ECE6451/Lectures/Student... · The Kronig-Penney Model (review) According to this model, discontinuities in allowed energies