We have numerically investigated parametric variations of transmission peaks of symmetric rectangular double barrier in nontunneling regime. We have compared the variations with those for tunneling regime. One of the three variations in nontunneling regime is completely different from that for tunneling regime warranting rapid dissemination. The book contains background on Quantum Mechanics, Microelectronics and Nanostructure Physics to enable readers assimilate the book completely.
Index
i
Oscillatory transmission through nontunneling regime of symmetric rectangular
double barrier: parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
Section
Page number
Index
iii
Chapter
I
Background on Quantum Mechanics
1
25
1.1 Wave equation of a free particle: Schrödinger equation
2
1.2 Schrödinger equation of a particle subject to a conservative
mechanical force
4
1.3 Conservation of probability and probability current density
5
1.4 Timeindependent Schrödinger equation and stationary state
7
1.5 Continuous and discontinuous function
9
1.6 Finite and infinite discontinuity
11
1.7 Admissibility conditions on wavefunction
12
1.8 Free particle: eigenfunctions and probability current density
14
1.9 Single rectangular tunnel barrier
15
1.9.1 Calculation of transfer matrix and investigation of its properties
16
1.9.2 Calculation of transmission coefficient
21
Index
ii
Chapter
II
Background on Microelectronics
26
42
2.1 Insulator and its band model
27
2.2 Intrinsic semiconductor and its band model
28
2.3 Elemental and compound semiconductors
31
2.4 Alloy semiconductors: ternary and quaternary semiconductors
35
2.5 Bandgap engineering
37
2.6 Substrate and epitaxial layer
40
2.7 Semiconductor heterostructure and heterojunction
40
Chapter
III
Background on Nanostructure Physics
43
66
3.1.1 Tunnel barrier: structure and band model
44
3.1.2 Transport of electron or hole through tunnel barrier
45
3.2 Quantum
Well
(QW)
50
3.3 Symmetric
double
barrier
56
Chapter
IV
Numerical investigation of parametric variations
of transmission peaks of
symmetric rectangular double barrier for nontunneling regime
67
130
4.1 Transmission
through
symmetric
rectangular double barrier
68
4.2 Description of the problem
71
4.3 Tables and Figures showing parametric variations of energy of
transmission peaks of symmetric rectangular double barrier for
nontunneling regime
74
4.4 Conclusions about the parametric variations
130
References 131
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
1
Chapter I
Background on Quantum Mechanics
Chapter I Background on Quantum Mechanics
2
1.1 Wave equation of a free particle: Schrödinger equation
If we associate the wave packet
)
t
,
x
(
=
2
1
dk
e
)
k
(
a
)
t
kx
(
i
³


where
a(k) =
2
1
dx
e
)
t
,
x
(
)
t
kx
(
i
³



with a free material particle, we can use de Broglie's equations p =
! k and E =
!
and rewrite the above two equations as
)
t
,
x
(
=
!
2
1
)
k
(
d
e
)
k
(
a
)
t
kx
(
i
!
!
³


or,
)
t
,
x
(
=
!
2
1
dp
e
)
p
(
a
)
Et
px
(
i
³


!
where a(p) =
!
)
k
(
a
or, a(p)
=
!
2
1
dx
e
)
t
,
x
(
)
Et
px
(
i
³



!
In three dimensions
)
t
,
r
(
&
=
3
)
2
(
1
!
³

p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
(1.1)
where
r
.
p
&
&
=
z
p
y
p
x
p
z
y
x
+
+
Equation (1.1) gives
x
=
3
x
)
2
(
p
i
!
!
p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

and
2
2
x
=
3
2
2
x
)
2
(
p
!
!

p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

Similarly,
2
2
y
=
3
2
2
y
)
2
(
p
!
!

p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

all momentum space
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
3
and
2
2
z
=
3
2
2
z
)
2
(
p
!
!

p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

2
=
3
2
2
z
2
y
2
x
)
2
(
/
)
p
p
p
(
!
!
+
+

p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

=
3
2
2
)
2
(
p
!
!

p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

or,
2
=
2
2
p
!

or,
m
2
2
!

)
t
,
r
(
2
&
=
m
2
p
2
)
t
,
r
(
&
(1.2)
Equation (1.1) gives
t
=
!
iE

3
)
2
(
1
!
p
d
e
)
p
(
a
)
Et
r
.
p
(
i
&
&
&
&
!
³

=
!
iE

or, i
!
t
= E
(1.3)
For a free particle, E =
m
2
p
2
. Hence equation (1.2) and (1.3) give
m
2
2
!

2
= i
!
t
(1.4)
Equation (1.4) is the differential equation for the matter wave of a free particle and
equation (1.4) is called wave equation or Schrödinger equation for a free particle.
Equation (1.4) is a linear equation and hence a monochromatic wave such as
)
Et
r
.
p
(
i
e
)
p
(
a

&
&
!
&
as well as a wave packet given by equation (1.1) satisfy it.
Chapter I Background on Quantum Mechanics
4
1.2 Schrödinger equation of a particle subject to a conservative mechanical
force
1) A comparison of
< x > =
³
x
*
dx and < p > =
³
x
i
*
!
dx
shows that expectation value of momentum, < p > of a particle having wavefunction
associated with it can be computed in the same way as that of its position < x > if
the operator
x
i
!
is substituted in place of x. This statement introduces operator
formalism in quantum mechanics. Thus the operator of x is x, operator of p is
x
i
!
.
By "operator of p is
x
i
!
", we in fact mean, the operator we need to calculate
expectation value of p is
x
i
!
.
2) Schrödinger equation of a free particle is

2
2
2
x
m
2
!
= i
t
!
. This equation can
be obtained from the classical equation
m
2
p
2
= E by the operator correspondence of p
and E as
x
i
!
and i
t
! respectively, and letting the operators operate on the
wavefunction
. Thus operator of E is i
t
! .
3) If a conservative force acts on a particle, total energy E =
m
2
p
2
+ V where V is
potential energy and hence V is a function of position only. Hence
>
<
)
x
(
V
=
³
+

dx
)
t
,
x
(
)
x
(
V
2
=
³
+

dx
)
x
(
V
*
Thus operator of V(x) is V(x).
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
5
4) Using the operator correspondences of E, p and V,
m
2
p
2
+ V = E gives
m
2
x
i
2
¸
¹
·
¨
©
§
!
+ V = i
t
! or,

2
2
2
x
m
2
!
+ V = i
t
!
or,

2
2
2
x
m
2
!
+ V
= i
t
!
or,
»
»
¼
º
«
«
¬
ª
+

V
x
m
2
2
2
2
!
= i
t
!
.
This is the Schrödinger equation of a particle moving in a potential V. In three
dimensions, [
m
2
2
!

2
+ V]
= i
t
!
is the Schrödinger equation of a particle
moving in a potential V. The operator [
m
2
2
!

2
+ V] is the operator of total energy
of a conservative system and is called Hamiltonian operator.
1.3 Conservation of probability and probability current density
Let us consider a particle under the action of a conservative mechanical force.
The wave equation of the associated matter wave is given by the Schroedinger
equation

m
2
2
!
2
+ V = i!
t
(1.5)
where V is the potential associated with the force. Complex conjugate of equation
(1.5) is

m
2
2
!
2
*
+ V
*
=  i!
t
*
(1.6)
V is real.
Multiplying equation (1.5) by
*
, we get

m
2
2
!
*
2
+ V
*
= i!
*
t
(1.7)
Multiplying equation (1.6) by
, we get

m
2
2
!
2
*
+ V
*
=
 i!
t
*
(1.8)
Chapter I Background on Quantum Mechanics
6
Equation (1.7) (1.8)

m
2
2
!
[
*
2

2
*
] = i! ¨
©
§
t
*
+
¸
¸
¹
·
t
*
or,

m
2
2
!
&
· (
*
&

&
*
) = i!
t
(
*
)
or,
&
· [
mi
2
!
(
*
&

&
*
)] = 
t
2
(1.9)
or,
&
· S
&
=

t
where
=
2
and S
&
=
mi
2
!
(
*
&

&
*
)
(1.10)
Equation (1.9)
³
·
v
d
S
&
&
=

t
³
v
d
over a volume v.
or,
³
· A
d
S
&
&
=

t
³
v
d
(1.11)
Using
Gauss's
divergence
theorem
The surface integral is over the closed surface that encloses the volume v. In
equation (1.11),
³
v
d is probability of presence of a particle in the volume v.

t
³
v
d is time rate of decrease of probability of presence of the particle in the
volume v. Since this rate is equal to
³
· A
d
S
&
&
, we can say that
S
&
is time rate of flow
out of the probability per unit area through the surface enclosing the volume v. The
probability changes or decreases because of change of
with time. Hence
S
&
is
called probability current density.
For a system of a large number of particles, < S > is average particle current
density, i.e. number of particles crossing per unit time through a unit area
perpendicular to the flow.
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
7
1.4 Timeindependent Schrödinger equation and stationary state
Schrödinger equation for a particle under conservative force is

m
2
2
!
2
)
t
,
r
(
&
+ V( r
&
)
)
t
,
r
(
&
= i
!
t
)
t
,
r
(
&
(1.12)
which is called timedependent Schroedinger equation. V( r
&
) is independent of time.
The Hamiltonian is also time independent. Equation (1.12) simplifies considerably.
Let us try
)
t
,
r
(
&
= u( r
&
) f(t) as a solution of equation (1.12). u is a function
of space coordinates only and f is a function of time only.
Equation (1.12)

m
2
2
!
2
[u( r& ) f(t)] + V( r& ) u( r& ) f(t) = i!
t
[u( r
&
) f(t)]
or,

m
2
2
!
f(t)
2
u( r& ) + V( r& ) u( r& ) f(t) = i! u( r& )
t
f(t)
or,

m
2
2
!
( )
( )
r
u
r
u
2
&
&
+ V( r
&
) =
t
)
t
(
f
)
t
(
f
i
!
(1.13)
Dividing by u( r
&
) f(t)
LHS of equation (1.13) is a function of position only and RHS of equation
(1.13) is a function of time only, because V( r
&
) is a function of position only. Since
space (coordinates) and time are independent (ignoring theory of relativity), equation
(1.13) makes sense only if both sides of equation (1.13) are equal to a constant, say
C. Thus

m
2
2
!
( )
( )
r
u
r
u
2
&
&
+ V( r
&
) = C
(1.14)
and
t
)
t
(
f
)
t
(
f
i
!
=
C
(1.15)
Equation (1.14)

m
2
2
!
2
u( r& ) + V( r& ) u( r& ) = C u( r& )
or, [

m
2
2
!
2
+ V( r& )] u( r& ) = C u( r& )
(1.16a)
or,
op
H
u( r
&
) = C u( r
&
)
(1.16b)
Chapter I Background on Quantum Mechanics
8
op
H
is Hamiltonian operator, i.e. operator of total energy. Equation (1.16) is
eigenvalue equation of total energy.
C is an eigenvalue of total energy (an
observable).
C is real. Let us denote C by E.
Equation (1.16)
[

m
2
2
!
2
+ V( r& )] u( r& ) = E u( r& )
(1.17a)
or,
op
H
u( r
&
) = E u( r
&
)
(1.17b)
Equation (1.17) is called timeindependent Schroedinger equation.
Equation (1.15)
i
!
t
f(t) = E f(t)
or,
dt
d
f(t) =
!
i
)
t
(
Ef
Let f(t) =
nt
e
n
nt
e
= E
!
i
1
nt
e
or, (n
 E
!
i
1
)
nt
e
= 0
n  E
!
i
1
= 0
nt
e
0
or, n = E
!
i
1
=

!
iE
=
 i
f(t) =
t
i
e

=
Et
i
e
!

( r& , t) = u( r& ) f(t) = u( r& )
Et
i
e
!

= u( r
&
)
t
i
e

(1.18)
2
)
t
,
r
(
&
=
*
( r& , t) ( r& , t) = u*( r& )
t
E
i
*
e
!
u( r
&
)
Et
i
e
!

= u*( r
&
) u( r
&
)
E is real,
E* = E.
=
2
)
r
(
u
&
(1.19)
From equation (1.19), we find that
2
)
t
,
r
(
&
is independent of time. Thus the states
given by equation (1.18) are stationary states.
Expectation value of any observable A is
< A > =
³
d
)
t
,
r
(
A
)
t
,
r
(
op
*
&
&
=
³

d
]
e
)
r
(
u
[
A
e
)
r
(
u
Et
i
op
t
E
i
*
*
!
!
&
&
using
equation
(1.18)
=
³

d
)
r
(
u
A
e
e
)
r
(
u
op
Et
i
Et
i
*
&
&
!
!
E is real.
If
op
A
does not explicitly contain the variable t (time).
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
9
=
³
d
)
r
(
u
A
)
r
(
u
op
*
&
&
For stationary states, expectation value of any observable A is independent of time,
provided the operator
op
A
itself does not depend explicitly on t.
In hydrogen atom, e.g., the potential V(r) =
0
4
1

r
e
is a function of position
only. Thus solution of equation (1.12) will give stationary state. Thus
2
for the
electron and expectation values of all observables (e.g. energy) of the electron remain
independent of time. This explains why hydrogen atom is stable; thus we get an
explanation of one of the ad hoc assumptions of Bohr that the electron in hydrogen
atom stays in stationary state.
1.5 Continuous and discontinuous function
f(x) above is a continuous function. It is single valued at every value of x.
dx
df
is also continuous and single valued.
x
f(x)
O
Chapter I Background on Quantum Mechanics
10
f(x) above is a discontinuous function of x. The discontinuity is at x =
0
x
where the function is many valued. f(
0
x ) is unspecified.
1
f < f(
0
x ) <
2
f .
0
Lt
f(
0
x

) =
2
f
0
Lt
f(
0
x +
) =
1
f .
dx
df
at x =
0
x
 is tan
1
.
x
f(x)
O
x
0
1
2
x
f(x)
O
f
1
f
2
x
0
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
11
dx
df
at x =
0
x +
is tan
2
. Here 0.
1
and
2
may or may not be equal, depending on the nature of f(x). Thus the
derivative
dx
df
may or may not be continuous.
At x =
0
x ,
dx
df
= tan 90
°
=
.
1.6 Finite and infinite discontinuity
The discontinuity considered above is finite discontinuity, because
1
f
and
2
f
are finite. If a < f(
0
x ) < b where either a or b (or both) is +
or

, the
discontinuity is infinite discontinuity.
x
f(x)
O
90
°
x
0
Chapter I Background on Quantum Mechanics
12
0
Lt
f(
0
x +
) = a
0
Lt
f(
0
x
 ) = +
a < f(
0
x ) <
.
1.7 Admissibility conditions on wavefunction
1.
(x , t)
0 as x
± , because
³
d
2
= 1 is finite.
2.
(x , t) must be finite, single valued and continuous function of x for all time t;
this is because of probability interpretation of
.
3.
t
is a continuous function of x, because otherwise
t
= c where
1
c < c <
2
c at
say x =
0
x .
x
f(x)
O
x
0
a
all space
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
13
This means
= c t which is many valued at x =
0
x . But
must be single valued
according to condition (2).
4.
x
must be continuous function of x if V(x , t) is continuous. Because: if V(x, t)
is continuous, V(x, t)
(x, t) is continuous.
t
is also continuous (condition (3))
function of x. Hence Schroedinger equation
i
!
t
t)
(x,
=

m
2
2
!
2
2
x
(x , t) +V(x , t) (x , t)
gives
2
2
x
is continuous. Thus
x
is continuous function of x; otherwise
2
2
x
becomes infinite at points (x) where
x
is discontinuous.
5.
x
must be continuous function of x if V does not have infinite discontinuity. In
Schroedinger equation i
!
t
=

m
2
2
!
2
2
x
+ V(x, t) (x, t),
t
is always
x
t
O
c
1
c
2
x
0
Chapter I Background on Quantum Mechanics
14
continuous function of x (condition 3). If
x
has any discontinuity say at x = x
0
, the
x
versus x curve becomes vertical at x =
0
x and hence its slope
2
2
x
becomes
infinite at x =
0
x . Thus

m
2
2
!
2
2
x
=

. This forces V to become + to keep
the Schroedinger equation valid. Since
is finite, continuous and single valued
everywhere, V is forced to be +
at x =
0
x . Thus unless V has an infinite
discontinuity,
x
must be a continuous function of x.
Any finite discontinuity of V makes V
finite discontinuous. This is adjusted
by a finite discontinuity of
2
2
x
because
t
is always continuous [see
Schroedinger equation]. Finite discontinuity of
2
2
x
means
x
is continuous,
otherwise
2
2
x
gets infinite discontinuity there.
1.8 Free particle: eigenfunctions and probability current density
If the potential is constant, V(x) =
0
V , force acting on the particle F(x) =

x
V
= 0; so the particle is free.
0
V can be taken to be zero, without any loss of generality.
Timeindependent Schroedinger equation is

m
2
2
!
2
2
dx
)
x
(
u
d
+ 0 = E u(x)
or,
2
2
dx
u
d
+
2
mE
2
!
u = 0
or,
2
2
dx
u
d
+
2
k
u = 0.
If
0
V
0, k =
)
V
E
(
m
2
0
2

!
.
u(x) = A
ikx
e
(x , t) = A
)
t
kx
(
i
e

= A
)
Et
px
(
i
e

!
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
15
where k =
2
mE
2
!
, E =
m
2
k
2
2
!
.
Probability current density corresponding to the free particle eigenfunction
=A
)
Et
px
(
i
e

!
is
S =
mi
2
!
[
x
*

x
*
]
=
mi
2
!
[
*
A
)
Et
px
(
i
e


!
x
(A
)
Et
px
(
i
e

!
)
 A
)
Et
px
(
i
e

!
x
(
*
A
)
Et
px
(
i
e


!
)]
=
mi
2
!
2
A [
!
i
p (

!
i
p)] =
mi
2
!
2
A 2
!
i
p =
m
p
2
A =
m
k
!
2
A = v
2
A
and
2
=
2
A .
1.9 Single rectangular tunnel barrier
 a
+a x
If we have variation of potential V(x) as shown in the above figure, we have a
onedimensional, single, rectangular potential barrier. If the width and height of the
barrier are finite and small, we have a tunnel barrier of width 2a and height V
0
. The
barrier is defined as
V(x) = 0 for
a
x
>
= V
0
for
a
x
a
+
<
<

V(x)
(0, 0)
Region I
Region II
Region III
V
0
Chapter I Background on Quantum Mechanics
16
There are two finite discontinuities of V(x), one at x =
a and another at x = +a.
There are three regions as shown. According to the choice of origin, V(x) is zero in
two of the three regions and is constant (V
0
) in region II. We now proceed to obtain
transfer matrix of the barrier and find its properties. We shall use waves and match
them at the potential discontinuities using the boundary conditions described in
section 1.7.
1.9.1 Calculation of transfer matrix and investigation of its properties (E < V
0
)
 a
+a x
Solutions of timeindependent Schrödinger equation
[

m
2
2
!
2
2
dx
d
+ V(x)] u(x) = E u(x)
or,
2
2
dx
u
d
+
0
u
))
x
(
V
E
(
m
2
2
=

!
in the three regions are u
1
, u
2
and u
3
given by
ikx
ikx
1
Be
Ae
)
x
(
u

+
=
where
2
2
mE
2
k
!
=
x
x
2
De
Ce
)
x
(
u

+
=
where
2
0
2
)
E
V
(
m
2
!

=
ikx
ikx
3
He
Ge
)
x
(
u

+
=
The expressions for u
2
and
2
imply that we are considering free electrons of kinetic
energy less than V
0
impinging on the barrier from the left.
V(x)
(0, 0)
Region I
Region II
Region III
V
0
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
17
Using the boundary condition
u
1
= u
2
at x =
a, we get
ika
ika
Be
Ae
+

=
a
a
De
Ce

+
(1.20)
Again, the boundary condition
dx
du
dx
du
2
1
=
at x =
a gives
ik
ikx
ikx
ikBe
Ae


=
x
x
De
Ce


at x =
a
or, ik
ika
ika
ikBe
Ae


=
a
a
De
Ce


or,
ika
ika
Be
ik
Ae
ik


=
a
a
De
Ce


(1.21)
The boundary condition u
2
= u
3
at x = +a gives
a
a
De
Ce

+
=
ika
ika
He
Ge

+
(1.22)
And
dx
du
dx
du
3
2
=
at x = +a gives
x
x
De
Ce


=
ikx
ikx
ikHe
ikGe


at x = +a
=>
a
a
De
Ce


=
ika
ika
ikHe
ikGe


=>
a
a
De
ik
Ce
ik


=
ika
ika
He
Ge


(1.23)
From equation (1.20) + (1.21), we have
ika
ika
a
Be
)
ik
1
(
Ae
)
ik
1
(
Ce
2

+
+
=


=>
a
ika
a
ika
Be
)
ik
1
(
2
1
Ae
)
ik
1
(
2
1
C
+
+


+
+
=
From equation (1.20)
 (1.21), we have
ika
ika
a
Be
)
ik
1
(
Ae
)
ik
1
(
De
2
+
+

=

=>
a
ika
a
ika
Be
)
ik
1
(
2
1
Ae
)
ik
1
(
2
1
D



+
+

=
Now in the form of matrix we can write
Chapter I Background on Quantum Mechanics
18
¸¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
©
§
¸¸
¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
¨
©
§
+


+
=
¸¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
©
§

+


+

B
A
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
D
C
a
ika
a
ika
a
ika
a
ika
From equation (1.22) + (1.23), we have
From equation (1.22)
 (1.23), we have
ika
a
ika
a
a
a
ika
De
)
ik
1
(
2
1
Ce
)
ik
1
(
2
1
H
De
)
ik
1
(
Ce
)
ik
1
(
He
2
+

+


+
+

=
=>
+
+

=
Now in the form of matrix we can write
¸¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
©
§
¸
¸
¸
¸
¹
·
¨
¨
¨
¨
©
§
+


+
=
¸¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
©
§
+



+

D
C
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
H
G
ika
a
ika
a
ika
a
ika
a
¸¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
©
§
¸
¸
¸
¸
¹
·
¨
¨
¨
¨
©
§
+


+
¸
¸
¸
¸
¹
·
¨
¨
¨
¨
©
§
+


+
=
¸¸
¸
¸
¸
¹
·
¨¨
¨
¨
¨
©
§

+


+

+



+

B
A
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
H
G
a
ika
a
ika
a
ika
a
ika
ika
a
ika
a
ika
a
ika
a
or,
¸¸¹
·
¨¨©
§
=
¸¸¹
·
¨¨©
§
22
12
21
11
M
M
M
M
H
G
¸¸¹
·
¨¨©
§
B
A
(1.24)
ika
a
ika
a
a
a
ika
De
)
ik
1
(
2
1
Ce
)
ik
1
(
2
1
G
De
)
ik
1
(
Ce
)
ik
1
(
Ge
2





+
+
=
=>

+
+
=
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
19
The M matrix is called transfer matrix. The elements of the matrix are calculated in
the following.
a
ika
ika
a
11
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
M
+


+
+
=
a
ika
ika
a
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1






+
ika
2
a
2
a
2
ika
2
a
2
a
2
e
]
e
)
1
ik
ik
1
(
e
)
1
ik
ik
1
[(
4
1
e
]
e
)
ik
1
)(
ik
1
(
e
)
ik
1
)(
ik
1
[(
4
1




+


+
+
+
+
=


+
+
+
=
ka
2
i
a
2
a
2
a
2
a
2
e
]
2
/
)
e
e
)(
ik
ik
(
2
1
)
e
e
(
2
1
[




+
+
+
=
ka
2
i
e
]
a
2
sinh
)
k
k
(
2
i
a
2
[cosh


+
=
(1.25)
a
ika
ika
a
22
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
M
+
+


=
a
ika
ika
a
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1

+

+
+
+
ika
2
a
2
a
2
e
]
e
)
ik
1
)(
ik
1
(
e
)
ik
1
)(
ik
1
[(
4
1

+
+
+


=
ka
2
i
a
2
a
2
a
2
a
2
ka
2
i
a
2
a
2
e
]
2
/
)
e
e
)(
ik
ik
(
2
1
)
e
e
(
2
1
[
e
]
e
)
1
ik
ik
1
(
e
)
1
ik
ik
1
[(
4
1




+

+
=
+
+
+
+
+


=
ka
2
i
e
]
a
2
sinh
)
ik
ik
(
2
1
a
2
[cosh
+

=
ka
2
i
e
]
a
2
sinh
)
k
k
(
2
i
a
2
[cosh


=
(1.26)
Chapter I Background on Quantum Mechanics
20
a
ika
ika
a
12
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
M
+


+
=
]
e
)
ik
1
)(
ik
1
(
e
)
ik
1
)(
ik
1
[(
4
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
a
2
a
2
a
ika
ika
a




+

+

+
=
+

+
]
e
)
1
ik
ik
1
(
e
)
1
ik
ik
1
[(
4
1
a
2
a
2



+
+

+

=
2
/
)
e
e
)(
ik
ik
(
2
1
a
2
a
2



=
a
2
sinh
)
ik
ik
(
2
1

=
a
2
sinh
)
k
k
(
2
i
+

=
(1.27)
a
ika
ika
a
21
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
M
+

+
+

=
]
e
)
ik
1
)(
ik
1
(
e
)
ik
1
)(
ik
1
[(
4
1
e
)
ik
1
(
2
1
e
)
ik
1
(
2
1
a
2
a
2
a
ika
ika
a



+


+
+
+

=

+
+
a
2
sinh
)
ik
ik
(
2
1
2
/
)
e
e
)(
ik
ik
(
2
1
]
e
)
1
ik
ik
1
(
e
)
1
ik
ik
1
[(
4
1
a
2
a
2
a
2
a
2

=


=

+

+


+
=


a
2
sinh
)
k
k
(
2
i
+
=
(1.28)
We find that the elements of the transfer matrix M obey the following properties.
11
*
22
M
M
=
22
*
11
M
M
,
or
=
(1.29)
21
*
12
M
M
=
or,
12
*
21
M
M
=
(1.30)
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
21
We now evaluate the determinant of the M matrix.
22
21
12
11
M
M
M
M
12
*
12
11
*
11
21
12
22
11
M
M
M
M
M
M
M
M

=

=
ka
2
i
ka
2
i
e
]
a
2
sinh
)
k
k
(
2
i
a
2
[cosh
e
]
a
2
sinh
)
k
k
(
2
i
a
2
[cosh


+


=
]
a
2
sinh
)
k
k
(
2
i
[
+

]
a
2
sinh
)
k
k
(
2
i
[
+

1
a
2
sinh
a
2
cosh
a
2
sinh
]
k
k
4
[
4
1
a
2
cosh
a
2
sinh
]
)
k
k
(
)
k
k
[(
4
1
a
2
cosh
a
2
sinh
)
k
k
(
4
1
a
2
sinh
)
k
k
(
4
1
a
2
cosh
2
2
2
2
2
2
2
2
2
2
2
2
2
=

=

+
=
+


+
=
+


+
=
Thus
1
M
M
M
M
22
21
12
11
=
(1.31)
i.e. the determinant of the transfer matrix is unity.
1.9.2 Calculation of transmission coefficient
 a
+a x
V(x)
(0 , 0)
Region I
Region II
Region III
A
B
V
0
G
H
C
D
Chapter I Background on Quantum Mechanics
22
We have solutions of timeindependent Schrödinger equation
[

m
2
2
!
2
2
dx
d
+ V(x)] u(x) = E u(x) or,
2
2
dx
u
d
+
0
u
))
x
(
V
E
(
m
2
2
=

!
in the three regions given by u
1
, u
2
and u
3
ikx
ikx
1
Be
Ae
)
x
(
u

+
=
where
2
2
mE
2
k
!
=
x
x
2
De
Ce
)
x
(
u

+
=
where
2
0
2
)
E
V
(
m
2
!

=
ikx
ikx
3
He
Ge
)
x
(
u

+
=
.
To calculate transmission coefficient of the tunnel barrier, we need to
recognize that A is amplitude of the plane wave incident on the barrier from the left.
B is that of the wave reflected from the barrier. G is that of transmitted plane wave. H
is that of the reflected wave (if any) in region III. We have equation (1.24) relating
amplitudes of the four plane waves.
¸
¸
¸
¹
·
¨
¨
¨
©
§
¸
¸
¸
¹
·
¨
¨
¨
©
§
=
¸
¸
¸
¹
·
¨
¨
¨
©
§
B
A
M
M
M
M
H
G
22
21
12
11
The matrix equation is equivalent to the following two equations.
B
M
A
M
G
12
11
+
=
(1.32)
and B
M
A
M
H
22
21
+
=
.
(1.33)
To obtain an expression for transmission coefficient, we set H = 0 recognizing that
we expect no reflection in region III. As such, equation (1.33) gives
A
M
M
B
22
21

=
(1.34)
which we can put in equation (1.32) to get
A
M
M
M
A
M
G
22
21
12
11

=
A
M
M
M
M
M
22
21
12
22
11

=
A
M
1
22
=
(1.35)
using
equation
(1.31)
Oscillatory transmission through nontunneling regime of symmetric rectangular double barrier:
Parametric variations of transmission peaks
Md. Abidur Rahman and Sujaul Chowdhury
23
Now transmission coefficient of the single barrier is given by
2
2
1
A
G
T
=
. With the aid
of equation (1.35), T
1
reduces to
2
22
1
M
1
T
=
(1.36)
Reflection coefficient of the single barrier is given by
2
2
1
A
B
R
=
.
With the aid of equation (1.34), R
1
reduces to
2
22
2
21
1
M
M
R
=
(1.37)
1
2
21
T
M
=
(1.38)
using equation (1.36).
We now use equation (1.36) to obtain an analytic expression for T
1
for E < V
0
,
The transmission coefficient is given by
2
22
1
M
1
T
=
22
*
22
M
M
1
=
ka
2
i
ka
2
i
e
]
a
2
sinh
)
k
k
(
2
i
a
2
[cosh
e
]
a
2
sinh
)
k
k
(
2
i
a
2
[cosh
1



+
=

using
equation
(1.26)
a
2
sinh
)
k
k
(
4
1
a
2
cosh
1
2
2
2

+
=
a
2
sinh
)
k
k
(
4
1
a
2
sinh
1
1
2
2
2

+
+
=
a
2
sinh
]
)
k
k
(
4
1
1
[
1
1
2
2

+
+
=
a
2
sinh
]
2
k
k
4
[
4
1
1
1
2
2
2
2
2

+
+
+
=
a
2
sinh
)
k
k
(
4
1
1
1
2
2
+
+
=
(1.39a)
Chapter I Background on Quantum Mechanics
24
a
2
sinh
2
k
k
4
1
1
1
2
2
2
»
»
¼
º
«
«
¬
ª
+
¸
¹
·
¨
©
§
+
¸
¹
·
¨
©
§
+
=
¸¸¹
·
¨¨©
§

»
¼
º
«
¬
ª
+

+

+
=
2
0
2
0
0
)
E
V
(
m
2
a
2
sinh
2
E
E
V
E
V
E
4
1
1
1
!
¸¸¹
·
¨¨©
§



+
+
=
2
0
2
0
2
0
)
E
V
(
m
2
a
2
sinh
)
E
V
(
E
))
E
V
(
E
(
4
1
1
1
!
¸¸¹
·
¨¨©
§


+
=
2
0
2
0
2
0
)
E
V
(
m
2
a
2
sinh
)
E
V
(
E
V
4
1
1
1
!
(1.39b)
Equation (1.39) is for E < V
0
.
For E > V
0
,
2
0
2
2
)
V
E
(
m
2

=


=
!
±
=
i
Using x
tan
i
ix
tanh
,
x
cos
ix
cosh
,
x
sin
i
ix
sinh
,
i
=
=
=
=
in equation (1.39a),
we get
a
2
i
sinh
)
k
i
i
k
(
4
1
1
1
T
2
2
1
+
+
=
)
a
2
sin
(
)
k
k
(
i
4
1
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