Georgia Tech ECE 6451
Nearly-Free-Electron Model
Reading:
Brennan Chapter 8.3
Lecture prepared by Christopher Sconyers
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Solving the Schroedinger Equation for a Periodic Potential: Energy Solutions and Bandgaps, Summaries of Chemistry
A detailed solution to the Schroedinger equation for a periodic potential, deriving the energy solutions and discussing the implications for bandgaps. the Schroedinger equation, defining a periodic potential, expanding the wave function in terms of Bloch functions, and deriving the energy solutions using a first-order and second-order approximation.
What you will learn
- What are Bloch functions and how are they used to expand the wave function for a particle in a periodic potential?
- What are the energy solutions for a particle in a periodic potential using a first-order approximation?
- What is the Schroedinger equation and how is it used to find the energy solutions for a particle in a potential?
- What is a periodic potential and how does it affect the wave function and energy solutions?
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ECE 6451
Nearly-Free-Electron Model
Reading: Brennan Chapter 8.
Lecture prepared by Christopher Sconyers
ECE 6451
The Kronig-Penney Model (review)
According to this model, discontinuities in allowed energies appear at
k=n
π /(a+b)
,
where
a^ and
b^ relate to the periodicity of the rectangular potential.
The solution to the Schroedinger equation for a periodic rectangular barrier predictsthat only discrete bands of energy are allowed. What about any arbitrary periodic potential? Will discrete bands of energy show upin the solution as expected?
Brennan, Figure 8.2.1.
One dimensional
periodic potential with period
a+b.
ECE 6451
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 )(
(^20)
2 2
= Ψ
Ψ^
xf
k
d dx
γ
So, let’s expand
f(x)
, a periodic function with the same periodicity as V(x), in
terms of a complex Fourier series.
factor
scalinga isγ and
) ( 1 ) (
2 2
dx ex f a C
where
e C
x f
nxi a nxi a
a
n
n
n
π π
∑ ∫ =^ =
−
ECE 6451
The Schroedinger Equation
As seen before, any periodic potential will result in a periodic wave functionof the form
Bloch functions and periodicity
)(
)( :l
dimensiona one
)(
)( :l
dimensiona- n
x u e x
r u e r
k ikx k
nk ikr
nk
= Ψ
= Ψ
where
u^ (x)k
is periodic, with the same periodicity of the potential. Therefore,
we can expand it in terms of another complex Fourier series.
∑^ ∑
− −
= Ψ
=
+∞ −∞=^ n
n n ikx
n
k
nxi a nxi a eb
e x
eb
x u
π 2 π^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 theBloch 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 theperiodic potential, and then solve the Schroedinger equation. Hopefully, wecan shine some light on this new wave function.
ECE 6451
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 indiscrete energy bandgaps. To do this, we will want to solve for the energyversus
k^ relationship, which means we first must solve the Schroedinger equation to find the coefficients
b. Pluggingn
Ψ(x) into our simplified
Schroedinger equation:
First Order Solution
∑≠
−
= Ψ^
0
0
2
)(
n
n ikx ikx^
nxi a eb
e eb x
π
γ [^
]^
0 )(
(^20)
2 2
= Ψ
Ψ^
x f k
d dx
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:
(^
)^
(^
)^
(^
)
(^
)^
(^
)
(^
()
)^
0
2 (^222)
2 2
2
2
−
−
− ∑
∑ ∑
∑
∑
≠^ −
′
n
x ki n a
n a n
n^
n
x
ki nn
n
x ki n
n
x ki n
ikx
na
na na
na
na
e
k b
e b C e C b eb k k e k k b π
π π
π
π
π π
γ
γ
γ
γ
TOO COMPLEX!!
ECE 6451
Solving the Schroedinger Equation
The problem already seems to have blown up. However, we made theassumption that V(x) is weak, therefore
γ^ must be small, and the
(^2) γ term
(fourth term) can be neglected. Also, we can further simplify the equation.Consider only the terms in
First Order Solution γ.
We want to combine these into one simple term, and we can by defining thefollowing relation:
(^
)^
(^
)^
(^
()
)
(^
)
∑
∑
∑
− +
−
−
2
2 (^222)
2
n
x ki n n
x ki n a
n a n
n
x ki n
na
na
na
e C b
e
k b
eb k k
π
π
π γ
γ
γ
n a kn k
=
Combining the first two terms and substituting in the above relation:
(^
)^
(^
)^
(^
)
(^
)^
(^
)^
(^
)
(^
)^
(^
)^
(^
)
∑
∑
∑
∑
∑
∑
− + + − −
− + − − −
−
−
2
2
(^222)
(^222)
2
2
(^222)
(^222)
2
2
(^222)
n
x ki n
n
x ki n n a
n a n a
n a n
n
x ki n
n
x ki n n a
n a n a n n a n
n
x ki n
n
x ki n n a
n a
na
na
na
na
na
na
e Cb
eb
k
k
k k
e Cb
eb
k
k
k k
e C b
eb
k
k k
π
π
π
π
π
π
γ
γ
γ
γ
γ
γ^
ECE 6451
Solving the Schroedinger Equation
Let’s multiply the above equation by:
First Order Solution
(^
)^
(^
)
[^
(]
)^
0
2
=
−
−^
∑≠
n
x ki n n n
ikx^
na e Cb b k k e k k b
π
γ
m a
m
xik k k where
e^
m
π 2 − − =
Then we’ll integrate over a full period (
0 Æ
a):
(^
)^
(^
)
[^
]^
(^
)
0
2
2
=
−
−^
∑^
∫
∫^
−
n
a n n n
a
dx e Cb b k k
dx e k k b^
a
xn m i
mxi a
π
π^
γ
Note that the two integrals are of similar form, which has the following usefulproperty:
≠ =
=
∫^
0 if
0 if 0
2
λ^ λ
πλ
a dx a^ e
xi a
ECE 6451
Solving the Schroedinger Equation
Case 1: m
=^0
First Order Solution
(^
)^
(^
)
[^
]^
(^
)
0
2
2
=
−
−^
∑^
∫
∫^
−
n
a n n n
a
dx e Cb b k k
dx e k k b^
a
xn m i
mxi a
π
π^
γ
≠ =
=
∫^
0 if
0 if 0
2
λ^ λ
πλ
a dx a^ e
xi a
Within the summation,
n^ ≠
- Therefore the integral is always zero when
m=
.
(^
)^
(^
)
[^
]
(^
)^
(^
)
[^
]( )
(^
2 )^0 (^20) (^00)
0
0 0
0
2
k k
a k k b
Cb b k k a k k b
dx e Cb b k k
dx k k b
n
n n n
n
a n n n
a^
nxi a
=
= −
=
−
−
=
−
−
∑^ ∑
∫
∫
−
γ γ
π
ECE 6451
Substituting the value for the coefficients into the wave function, and we’ve atlast found a first order solution for our electron in a periodic potential.
Solving the Schroedinger Equation
First Order Solution
(^
)
(^
)^
−
−
= Ψ
−
= Ψ
= Ψ
∑ ∑
∑ ≠ ≠
≠
− − −
0
2 2
0
0
2 (^02)
0
0
0
(^22) 2
1
)( )(
)(
n^
n n
ikx
n^
n
n
ikx ikx
n
n ikx ikx
nxi a nxi a nxi a e k C k
eb x
e k k
Cb
e eb x
eb
e eb x
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 planewave. 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=k
(the denominator goes to zero). A n^
singularity occurs in this case. Let’s move on and find this wave’s energy.
ECE 6451
What is the Energy of the Electron?
As before, we’ll start with a first-order solution (neglecting
(^2) γ ). We found
previously that
k=k
in the first-order case. Plugging this in for the energy 0
solution:
First Order Energy Correction
(^
)^
−
−
= Ψ^
∑≠
−
0
2 2
0
2
1
)(
n^
n n
ikx^
nxi a e k C k
eb x
π
γ
k m
k m E^
2
2
(^220)
(^22)
h
h^
=
=
This tells us that for a first-order approximation, the energy of the electron in aperiodic potential is the same as the energy of a free electron. This must meanthat the energy correction factor has gone to zero for first-order. While thefirst-order solution helped approximate the wave function, we will have to goback to the second-order (
(^2) γ ) to determine the energy of the electron.
(^
)^
(^
)
[^
(]
)
(^
)^
0
2 2
2
=
−
−
∑ ∑
∑ ≠^
′
n^
n
x
ki nn n
x ki n n n
ikx
na na
na
e b C
e Cb b k k e k k b π π
π
γ
γ
n a
n^
k k
π 2 − Reminder =
ECE 6451
What is the Energy of the Electron?
Plugging in the coefficients
b^ n^
found in the first-order solution: Second Order Energy Correction^ (^
)^
0
=
−
= ∑
n
nn n n
ab C
a k k b
C C
γ
Continued…
(^
)
(^
)^
(^
)^
0
= −
−
−
∑≠n^
n
n n
n
n
n
k k
C C
ab
k ka b
k k
Cb
b
γ
Dividing out
ba^0
and solving for (
(^2) k) (^0) , we get:
(^
)^
(^
) (^
)
∑≠ ∑≠
−
=
= −
−
0
n^
n n n
n^
n
n n
k k
C C
k k
k k
C C
k k
γ^ γ
Note: the singularityhas shown up again.(at^
k=k
) n
ECE 6451
What is the Energy of the Electron?
Continued…
Second Order Energy Correction
(^
∑≠^
=^
n^
n n
n k
k
C
C
k
k
Multiply both sides by
2 ħ /2m
to get:
(^
(^
[^
]
(^
[^
]^
n m
n
n^
π na m
m
n
n^
n na n
n a n n
n n n
C
V
where
k
k
V
k m
k
E
k
k
C
C
m
k m
E
k
k
k m
E
k
k
C C m k m k m
(^22)
2
2
and
γ
π π
h
h
h
h
h
h
h
h
h
h
Recall
ECE 6451
So now we know that a discontinuity exists at all points where k is an integermultiple 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 rectangularbarrier). Recall, that model predicts bandgaps will occur at:
Does this predict the expected Bandgaps?
Bandgap Location
n
n k^
a^
integer all for π =
(^
)b na k^
=
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. Thegeneral solution for an arbitrary, periodic potential accurately predicts thesolution to the Kronig-Penney model. Why is this?For any crystalline structure, as well as any periodic potential, Braggreflection at the Brillouin zone edges allow electrons to have only a certainenergy values.
These zone edges occur at
k=n
π/a
, where
a^ is the lattice
constant (period) of the periodic potential.
ECE 6451
Does this predict the expected Bandgaps?
Now we need to show that a Bandgap occurs at
k=n
π /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
∑≠
−
= Ψ^
0
0
2
)(
n
n ikx ikx^
nxi a eb
e eb x
π
γ
We’ll focus only on the points of interest: where
k=k
. At these points,n
b^ n
blows up, so we will approximate the sum as only one term:
b^ .n
(^
)
(^
n)a
n
xik n ikx
x ki n ikx
n ikx ikx
k k where
eb
eb x
eb
eb x
eb e eb x
nxi a na n π γ^ γ γ
π π 2
0 0 0 ) )( )( (
(^22) −
- =
= Ψ
= Ψ
= Ψ
− −