Georgia Tech ECE 6451
Nearly-Free-Electron Model
Reading:Brennan Chapter 8.3
Lecture prepared by Christopher Sconyers
Georgia Tech ECE 6451
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.
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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
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The Schroedinger Equation
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
∫
∑
=
=−
+∞
−∞=
Georgia Tech ECE 6451
The Schroedinger Equation
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.
Georgia Tech ECE 6451
The Schroedinger Equation
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)
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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!!
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Solving the Schroedinger Equation
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.
First Order Solution
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
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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
Georgia Tech ECE 6451
Solving the Schroedinger Equation
Let’s multiply the above equation by:
First Order Solution
( ) ( )[ ] ( ) 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
Georgia Tech ECE 6451
Solving the Schroedinger Equation
Case 1: m = 0
First Order Solution
( ) ( )[ ] ( )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
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Solving the Schroedinger Equation
Case 2: m ≠ 0
First Order Solution
( ) ( )[ ] ( )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
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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.
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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
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What is the Energy of the Electron?
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
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What is the Energy of the Electron?
Plugging in the coefficients bn found in the first-order solution:
Second Order Energy Correction
( ) 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)
Georgia Tech ECE 6451
What is the Energy of the Electron?
Continued…
Second Order Energy Correction
( )∑≠ −
+=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
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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
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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.
Georgia Tech ECE 6451
Does this predict the expected Bandgaps?
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−
−
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Does this predict the expected Bandgaps?
Plugging this wave function into the Schroedinger equation, expanding the second-derivative, and collecting like terms results in:
Energy Values at the Bandgaps
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
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Does this predict the expected Bandgaps?
Step 2: Multiply by e-iknx and integrate over a full period (0 a):
Energy Values at the Bandgaps
( ) ( )
( ) ( ) ( )( ) 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
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Does this predict the expected Bandgaps?
Continued…
Energy Values at the Bandgaps
( )( ) 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
±=
±=
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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