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Periodic Potentials Kronig Penney - Quantum Mechanics Engineers | EEE 434, Assignments of Quantum Mechanics

Material Type: Assignment; Class: Quantum Mechanics Engineers; Subject: Electrical Engineering; University: Arizona State University - Tempe; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

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Periodic potentials - Kronig-Penney model
Electrons in a lattice see a periodic potential due to the presence of the atoms, which is of
the form shown in Figure 1.
a
Figure 1. Periodic potential in a one-dimensional lattice.
As will be shown shortly, this periodic potential will open gaps in the dispersion relation,
i.e. it will impose limits on the allowed energies. To simplify the problem, we will
assume that the width of the potential energy term goes to zero, i.e. we represent them as
δ-functions (see Fig. 2):
x
V(x)
a
V0
(1) (2)
Figure 2. Periodic δ-function potentials (a simplified model to the one from Figure 1).
We can express mathematically this potential energy term V(x) as:
VxVxna
n
( ) ( )=
=
0δ .
In region (1), the wavefunction is given by:
ψ1( )xAe Be
ikx ikx
=+ , where kmE
=2
2
h .
To connect this wavefunction to the one in region (2), we will use the so-called Bloch
theorem, which states that for a periodic potentials, the solutions to the TISE are of the
following form:
ψ( ) ( )xux eiKx
= ,
where u(x) is the Bloch periodic part that has the periodicity of the lattice, i.e.
u(x+a)=u(x), and the exponential term is the plane-wave component. Using Bloch
theorem, we have:
pf3
pf4

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Periodic potentials - Kronig-Penney model

Electrons in a lattice see a periodic potential due to the presence of the atoms, which is of the form shown in Figure 1.

a

Figure 1. Periodic potential in a one-dimensional lattice.

As will be shown shortly, this periodic potential will open gaps in the dispersion relation, i.e. it will impose limits on the allowed energies. To simplify the problem, we will assume that the width of the potential energy term goes to zero, i.e. we represent them as δ-functions (see Fig. 2):

x

V ( x )

a

V 0

Figure 2. Periodic δ-function potentials (a simplified model to the one from Figure 1).

We can express mathematically this potential energy term V(x) as:

V x V x na n

=− ∞

∞ 0 ∑δ^.

In region (1), the wavefunction is given by:

ψ 1 ( ) x = Ae ikx^ + Be −^ ikx , where k = 2 mE^2 h

To connect this wavefunction to the one in region (2), we will use the so-called Bloch theorem , which states that for a periodic potentials, the solutions to the TISE are of the following form:

ψ ( ) x = u x e ( ) iKx ,

where u ( x ) is the Bloch periodic part that has the periodicity of the lattice, i.e. u ( x + a )=u( x ), and the exponential term is the plane-wave component. Using Bloch theorem, we have:

ψ ψ

ψ ψ

ψ ψ

U

V

W

u a u a e

a u a e u ika e^ a^ e

iKa iKa iKa

Therefore, we can write the wavefunction in region (2) in terms of the one in region (1) using Bloch theorem, to get:

ψ 2 ( ) x = ψ 1 ( x − a e ) iKa^ = Ae ik^ b^ x^ −^ a^ g^ + Be −^ ik^ b^ x^ − a^ g^ eiKa.

We also know that for a wavefunction to be a proper function, it must satisfy the continuity requirement, i.e.

ψ 1 ( ) a = ψ 2 ( ) a ,

which gives:

b^ A^^ +^ B e g^ iKa^ =^ Ae^ ika^ +^ Be^ −^ ika^ →^ A e c^ iKa^ −^ e^ ika^ h= B e c^ − ika^ − eiKa h.^ (1)

The continuity of the first derivative is not satisfied when V ( x ) is a δ-function. This can be shown directly from the 1D TISE,

− + = → = − h h

2 2 2

2 2 2 2

m

d dx

V x x E x d dx

m V x E x ψ ψ ψ ψ ( ) ( ) ( ) ( ) ψ( ).

If this equation is integrated in the neighborhood of x = a , we get:

d dx

d dx

dx d dx

d dx m (^) V x a E x dx mV a

x a

x a

x a x a

x a

x a

= −

= +

= + = −

= −

= + →

z

z

F

HG^

I

KJ^

δ

δ

δ δ

δ

δ δ

ψ ψ ψ

(^2 2 0) δ ψ 2 0 22 0 ψ 2 h h

Using the expression for ψ 2 ( ) a , we arrive at a second equation that is relating coefficients A and B :

ike ike mV e A ike ike mV

L iKa − ika − iKa iKa ika e iKa B

NM^

O

QP^

= L − +

NM^

O

QP

2

0 h h^2.^ (2)

Dividing equations (1) and (2) and rearranging the terms leads to the following final expression:

cos( ) sin( ) Ka cos( ) mV ka ka

= + ka

h 2.

Let's consider the limiting cases of the above equation first:

(a) Free particle In this case, V 0 = 0 , which gives

cos( Ka ) cos( ka ) K k mE^ E K m

2 2 h

h (^).

As it is clear from the figure, the periodic potential introduces gaps in the free particle dispersion relation. If one starts from the other extreme (with infinite well, that has discrete energy levels), then we can say that the interactions between the wells lift the degeneracy of the energy levels and broaden them into bands. This can rather easily be demonstrated on a two-well problem, which was given as a homework problem last year (see the web-site to get to the solution of this problem). The dispersion relation for a particle in a periodic potential is shown in Figure 4.

K

E

− 2 π/ a − π/ a π/ a 2 π/ a (a)

K

E

− π/ a π/ a

First Brillouin zone

(b) Figure 4. Dispersion relation for a particle in a periodic potential. (a) expanded zone scheme. (b) reduced zone scheme, also known as the First Brillouin zone.