80
CHAPTER 10 PARTICLE IN POTENSIAL BARRIER
A. Introduction
In this chapter it will be studied important thing about particle in potential barrier i.e.:
1. Particle in step potential with energy E V
o
. 2.
Particle in step potential with energy E V
o
. 3.
Particle in barrier potential with energy E V
o
. 4.
Particle in barrier potential with energy E V
o
. To study this topics it is required comprehension about Schrodinger equation,
continuity of wave function, and its physical interpretation. Analysis is conducted by describe eigen function in each area and determinate the transmission coefficient
transmutation T. B.
Particle in Step Potential with Energy E V
o
In Figure 10.1 is described particle in step potential with energy E V
o
:
Figure 10.1. Particle in step potential with energy E V
o
According to classical mechanic, particle with energy E V
o
will be reflected in left direction. Based on this view, particle is not possible in area of x0.
According quantum mechanics, in area x0 with Vx=0 full filled equation:
2
2 2
2
x E
dx x
d m
10.1 It has general solution:
Vx V
o
Vx=0
x
81
x ik
x ik
Be Ae
x
1 1
10.2
where
mE k
2 1
1
In area x0 with Vx= V
o
full filled:
2
2 2
2
x E
x V
dx x
d m
o
2
2 2
2
x V
E dx
x d
m
o
2
2 2
2
x E
V dx
x d
m
o
where V
o
-E 0. 10.3 Equation 10.3 has general solution:
x k
x k
De Ce
x
2 2
10.4
where
2 1
2
E V
m k
o
Then at ψx is applied boundary condition in order to the wave function has well
behave in all interval of x. 1.
ψx finite, hence C=0. If C≠0 then ψx=∞ for x→∞. 2.
ψx continue at x=0, it full filled relation among the coeficients: A + B = D 10.5
3. ψ’x also continue at x=0: In area x0 it is obtained:
x ik
x ik
Be ik
Ae ik
x
1 1
1 1
In area x0 it is obtained:
x k
De k
x
2
2
Continuity condition at x=0 produce:
C k
B ik
A ik
2 1
1
10.6 Based on equation 10.5 and 10.6 it can be obtained coefficient A and B which
stated in D as follow: Multiply equation 10.5 with
ik
1
then add with equation 10.6:
ik
1
A +
ik
1
B =
ik
1
D
82
C k
B ik
A ik
2 1
1
It is obtained: 2
2 1
1
k ik
D A
ik
1 2
1
2 ik
k ik
D A
1 2
1 2
ik k
D A
i
i ik
k D
A
1 2
1 2
1 2
1 2
k ik
D A
10.7
Coefficient of B is determined by this way:
1 2
1 2
k ik
D D
A D
B
1 2
1
2 k
ik k
D D
B
1 2
1 1
2 2
k D
ik k
D k
B
1 2
1
2k D
ik D
k B
1 2
1
2 k
ik k
D B
1 2
1 2
k ik
D B
10.8
Boundary condition that ψx must be
well behaved function
give solution: 1.
In area x 0 it has wave function:
x ik
x ik
e k
ik D
e k
ik D
x
1 1
1 2
1 2
1 2
1 2
10.9
2. In area x
0 it has wave function:
83
x k
De x
2
10.10 Total wave function:
iEt
e x
t x
,
Total wave function in area x
0 is:
t x
k i
t x
k i
e k
ik D
e k
ik D
t x
1 1
1 2
1 2
1 2
1 2
,
10.11 where
E
The first part of equation 10.11 represents wave which propagates in right
direction incident wave. The second part of equation 10.11 represent wave which propagates in left direction reflected wave.
Solution for area x
0 give total wave function:
t i
x k
e De
t x
2
, 10.12
Reffection Coefficienti:
1 1
2 1
2 1
2 1
2
1 2
1 2
1 2
1 2
k ik
D k
ik D
k ik
D k
ik D
A A
B B
R
10.13
This is according to classical description, all incident particle will be reflected all. In left side of x=0 or area x
0, it is full filled:
x k
De D
t x
t x
2
2
, ,
10.14
This will give approximation value of zero for x→∞. In area near x=0, there is
value of
t x
t x
, ,
. This is very different from classical viev that reject the attendance of particle in area x
0.
, ,
t x
t x
10.15 Equation 10.15 has a meaning that it will be penetration of potential wall
penetration effect if x is small enough.
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C. Particle in Step Potential With Energy E V
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