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.
2.
3.
4.
Particle in step potential with energy E < Vo.
Particle in step potential with energy E > Vo.
Particle in barrier potential with energy E < Vo.
Particle in barrier potential with energy E > Vo.
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 < Vo
In Figure 10.1 is described particle in step potential with energy E < Vo:
V(x)
Vo
V(x)=0
Figure 10.1. Particle in step potential with energy E < Vo
x
0
According to classical mechanic, particle with energy E < Vo will be reflected in left
direction. Based on this view, particle is not possible in area of x>0.
According quantum mechanics, in area x<0 with V(x)=0 full filled equation:

 2 d 2 ( x )
 E ( x )
2m dx 2
(10.1)
It has general solution:
80
 ( x)  Aeik x  Be ik x
1
where k1 
1
(10.2)
1
2mE

In area x>0 with V(x)= Vo full filled:

 2 d 2 ( x)
 Vo ( x)  E ( x)
2m dx 2
 2 d 2 ( x )

 ( E  Vo ) ( x )
2m dx 2
 2 d 2 ( x )
 (Vo  E ) ( x ) where Vo-E > 0.
2m dx 2
(10.3)
Equation (10.3) has general solution:
 ( x)  Ce k x  De  k x
2
where k 2 
2
(10.4)
1
2m(Vo  E )

Then at ψ(x) is applied boundary condition in order to the wave function has well
behave in all interval of x.
1. ψ(x) finite, hence C=0. If C≠0 then ψ(x)=∞ for x→∞.
2. ψ(x) continue at x=0, it full filled relation among the coeficients:
A+B=D
3. ψ’(x) also continue at x=0:
(10.5)
In area x<0 it is obtained:  ' ( x)  ik1 Aeik1x  ik1 Be ik1x
In area x>0 it is obtained:  ' ( x)  k 2 De  k2 x
Continuity condition at x=0 produce:
ik 1 A  ik 1 B   k 2 C
(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 ik1 then add with equation (10.6):
ik1A + ik1B = ik1D
81
ik 1 A  ik 1 B   k 2 C
It is obtained:
2 ik 1 A  Dik 1  k 2 
A
D  ik 1  k 2

2  ik1



A
k 
D
1  2 
2  ik 1 
A
k
D
1  2
2  ik 1
i

i 
A
D  ik 2
1 
2 
k1



(10.7)
Coefficient of B is determined by this way:
B DA D
D  ik 2
1 
2 
k1






2k D  (k1  ik 2 ) D
B 1
2 k1
k D  ik 2 D
B 1
2k1
D(k1  ik 2 )
B
2 k1
B  D
B
D  k1  ik 2

2  k1
D  ik 2
1 
2 
k1



(10.8)
Boundary condition that ψ(x) must be well behaved function give solution:
1. In area x0 it has wave function:
 ik1x D  ik 2
e  1 
2
k1

2. In area x0 it has wave function:
 x  
D  ik 2
1 
2 
k1
 ik1x
e

(10.9)
82
 ( x)  De  k x
(10.10)
2
Total wave function:
 ( x, t )   ( x )e

iEt

Total wave function in area x0 is:
 ( x, t ) 
where   E
D  ik 2
1 
2 
k1
 i k1x t  D  ik 2
e
 1 
2
k1

 i k1x t 
e

(10.11)
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:
( x, t )  De  k2 x e it
(10.12)
Reffection Coefficienti:
 ik 2  D  ik 2
1 
 1 
k
k1
B B
1  2 
R   
A A D  ik 2  D  ik 2
1 
 1 
2 
k1  2 
k1

D
2


 1



(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, t )x, t   D  De 2k2 x
(10.14)
This will give approximation value of zero for x→∞. In area near x=0, there is
value of   ( x, t )x, t  . This is very different from classical viev that reject the
attendance of particle in area x0.
  ( x, t )x, t  0
(10.15)
Equation (10.15) has a meaning that it will be penetration of potential wall
(penetration effect) if x is small enough.
83
C. Particle in Step Potential With Energy E > Vo
In Figure 10.2 is described particle in step potential with energy E > Vo.
V(x)
E
V(x)=0
Vo
Figure 10.2. Particle in Step Potential With Energy E > Vo
0
According to quantum mechanics, in area
independent of Schrodinger equation:
x
x<0 with V(x)=0 full filled time
 2 d 2 ( x )

 E ( x )
2m dx 2
It has general solution:
 ( x)  Aeik1x  Be ik1x
where k1 
1
2mE

in area x>0 with V(x)= Vo full filled:
 2 d 2 ( x)

 Vo ( x)  E ( x)
2m dx 2

 2 d 2 ( x )
 ( E  Vo ) ( x )
2m dx 2
(10.16)
Equation (10.16) has general solution:
 ( x)  Ceik x  De ik x
2
where k 2 
2
(10.17)
1
2m( E  Vo )

84
Then in ψ(x) is applied boundary condition in order wave function has well behave
in all interval of x.
1. Particle come from left side to the right side of step potential at x=0 may be
undergoes ressistance, hence reflection is happen. But in x>0 there is no
wave which propagate in left direction, hence coefficient D in equation
(10.17) equals to zero (D=0).
2. Condition of well behave function is ψ(x) continue at x=0, it is obtained:
A+B=C
(10.18)
3. Condition of well behave function ψ’(x) also continue at x=0:
In area x<0 it is obtained:  ' ( x)  ik1 Aeik1x  ik1 Be ik1x
In area x>0 it is obtained:  ' ( x)  ik 2 Ceik2 x
Continuity condition at x=0 produce:
ik 1 A  ik 1 B  ik 2 C
k1  A  B   k 2 C
(10.19)
Based on equation (10.18) and (10.19), it can be determined coefficient B and C
which are stated in A as follow:
B – C = -A
k1 B  k 2 C  k1 A
The equation can be written:
k 2 B  k 2 C  k 2 A
k1 B  k 2 C  k1 A
Sum of both equation produce:
( k 1  k 2 ) B  ( k1  k 2 ) A
k  k 2 
B 1
A
k1  k 2 
(10.20)
Furthermore coefficient C can be determined:
C  A B
k1  k 2 
A
k1  k 2 
k  k 2 A  k1  k 2 A
C 1
k1  k 2 
C  A
85
C
2 k1
A
k1  k 2 
(10.21)
The eigen function which determined are:
1. In area x<0:
k k 
 ( x)  Ae ik x  A 1 2 e ik x
 k1  k 2 
2. In area x>0:
1
 2 k1
 k1  k 2
 ( x)  A
1
 ik2 x
e

Refflection Coefficient R:
v1 B  B k1  k 2  A k1  k 2 A
R

v1 A A k1  k 2  A k1  k 2 A

k1  k 2 2
R
k1  k 2 2
(10.22)
where v1 is velocity of particle which has the same value with group velocity
Transmission Coefficient T:
v C C k
T 2   2
v1 A A k1
T

 2k1  A  2k1  A

  

 k1  k 2  A  k1  k 2  A
4 k1 k 2
(10.23)
k1  k 2 2
Sum of R and T produce:
R T 
R T 
R T 
k1  k 2 2
4 k1 k 2

2
k1  k 2  k1  k 2 2
k12  2k1 k 2  k 22  4k1 k 2
k1  k 2 2
k12  2k1 k 2  k 22
k1  k 2 2
86
R T 
k1  k 2 2
k1  k 2 2
1
(10.24)
Equation (10.24) shows the conservation of probability.
D. Particle in Barrier Potential with E<Vo
In Figure 10.3 is presented illustration about particle in barrier potential with
E < Vo. Potential barrier has wide of a and height of Vo.
Vo
E
V(x) = 0
V(x) = Vo
V(x) = 0
x=0
x=a
Area in a axis is devided in 3 part with eigen function as follow:
Figure 10.3. Particle in Barrier Potential With E<Vo
1
2mE

1
2. Area II, 0 < x < a:  ( x)  Ce k 2 x  De k 2 x where k 2 
2m(Vo  E )

1
3. Area III, x > a:  ( x)  Ee ik1 x  Fe ik1x where k1 
2mE

In order to  (x)   at x   then coefficient E = 0, hence it is obtained:
1. Area I, x < 0:  ( x)  Ae ik1 x  Be  ik1x where k1 
 ( x)  Fe ik x
1
Continuity condition  (x) and  ' ( x) at x = 0 and di x = a, will give relation among
coefficients of A, B, C, D, and F. Based on the relation, we can determine
transmission coefficient T as follow:




2
v1 F  F 
e k2a  e k2a 
T
 1
v1 A A 
E
E 
 16 1   
Vo  Vo  


1

87


sinh 2 k 2 a

T  1

E
E
 4 1 
Vo  Vo





 
 
 2mVo2 a 2
where k 2 a  
2
 
1
(10.25)

E
1 
 Vo



1/ 2
Based on equation (10.25) show that quantum analysis for particle with energy less
than barrier potential prove the existence of particle in area III. It means that
penetration effect is occurred in barrier potential. In quantum view, less particle is
reflected in area I. This phenomenon is very different from classical analysis that it
does not probable particle in area III for energy of particle less than barrier potential.
According to classical mechanics, all particle will be deflected in area I.
E. Particle in Potential Barrier with E>Vo
In Figure 10.4 is presented illustration about particle in barrier potential with
E > Vo. Potential barrier has wide of a and height of Vo.
1
2mE

1
2. Area II, 0 < x < a:  ( x)  Ce k2 x  De  k 2 x where k 2 
2m( E  Vo )

1
3. Area III, x > a:  ( x)  Ee ik1 x  Fe ik1x where k1 
2mE

In order to  (x)   at x   then E = 0, hence it is obtained:
1. Area I, x < 0:  ( x)  Ae ik1 x  Be  ik1x where k1 
 ( x)  Fe ik x
1
E
Vo
V(x) = 0
V(x) = Vo
x=0
V(x) = 0
x=a
Figure 10.4. Particle in Barrier Potential With E>Vo
88
By applied continuity condition of  (x) and  ' ( x) at x = 0 and at x = a, it will give
realtion among coefficients of A, B, C, D, and F. Based on the relation, we can
determine transmission coefficient T as follow:



ik2 a
ik2 a 2 

v F F
e e

T  1   1 

v1 A A
E
E 
 16 1   
Vo  Vo  




sin 2 k 2 a

T  1

E
E
 4 1 
Vo  Vo

1





 
 
 2mVo2 a 2
where k 2 a  
2
 
1
(10.26)

E
1 
 Vo



1/ 2
Analysis of classical mechanics and quantum for particle with energy E > Vo can be
explained as followt:
o According to classical mechanics, particle with energy E > Vo always pass
through potential barrier.
o According to quantum mechanics, particle with energy E > Vo , part of particles
will be reflected and other parts will be transmitted. When particle pass through
barrier potential, kinetic energy of particle will decrease E-Vo. As consequence
momentum of particle decrease and de Broglie wave length of particle increase.
G. Problems
1. Electron and proton have the same energy (E), move close to step potential
which has height of V larger than E. Are both particle have the same probability
for passing through the barrier? Which particle does have bigger probability?
2. Electron with energy of 4 eV come to barrier potential which has height of 5 eV
and wide of 0,1 nm. Determine the probability of electron to penetrate the
barrier!
3. Electron beam comes to barrier potential which has height of 5 eV and wide of
0,2 nm. Determine energy of electron in order to the electron can penetrate the
barrier!
89
H. Reference
Beiser, 1990. Konsep Fisika Modern. Edisi ke-4. (Trans. The Houw Liong). Jakarta:
Erlangga.
Hariyadi Supangkat. 1993. Pemecahan Persamaan Schrodinger Bebas-Waktu Untuk
Beberapa Potensial yang Sederhana. ITB Bandung.
Yusman Wiyatmo. 2008. Fisika Atom dalam Perspektif Klasik, Semiklasik, dan
Kuantum. Yogyakarta: Pustaka Pelajar.
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