3.3 Solutions
191
Dividing (10) by (9) gives
ik( A − B)
A + B
= −α
(11)
Diagrams for ψ at around x = 0
3.42 k 1 =
2m E
2
1/2
; k 2 =
2m(E − V 0 )
2
1/2
(1)
Boundary condition at x = 0:
ψ 1 (0) = ψ 2 (0)
dψ 1
dx
x=0
=
dψ 2 (x)
dx
x=0
(2)
These lead to
A 0 + A = B
(3)
ik 1 ( A 0 − A) = ik 2 B
Or
k 1 ( A 0 − A) = k 2 B
(4)
Solving (3) and (4)
A =
k 1 − k 2
k 1 + k 2
A 0
(5)
B =
2k 1 A 0
k 1 + k 2
(6)
Reflection coefficient,
R =
|A|
2
| A 0 | 2 =
(k 1 − k 2 )
2
(k 1 + k 2 ) 2
(7)
Transmission coefficient,
T =
k 2
k 1
|B|
2
| A| 2 =
4k 1 k 2
(k 1 + k 2 ) 2
(8)
Substituting the expressions for k 1 and k 2 from (1) and putting E = 4V 0 /3
we find that R = 1/9 and T = 8/9.
From (7) and (8) it is easily verified that
R + T = 1
( 9 )
Fig. 3.16 Graphs for
probability density
191
Dividing (10) by (9) gives
ik( A − B)
A + B
= −α
(11)
Diagrams for ψ at around x = 0
3.42 k 1 =
2m E
2
1/2
; k 2 =
2m(E − V 0 )
2
1/2
(1)
Boundary condition at x = 0:
ψ 1 (0) = ψ 2 (0)
dψ 1
dx
x=0
=
dψ 2 (x)
dx
x=0
(2)
These lead to
A 0 + A = B
(3)
ik 1 ( A 0 − A) = ik 2 B
Or
k 1 ( A 0 − A) = k 2 B
(4)
Solving (3) and (4)
A =
k 1 − k 2
k 1 + k 2
A 0
(5)
B =
2k 1 A 0
k 1 + k 2
(6)
Reflection coefficient,
R =
|A|
2
| A 0 | 2 =
(k 1 − k 2 )
2
(k 1 + k 2 ) 2
(7)
Transmission coefficient,
T =
k 2
k 1
|B|
2
| A| 2 =
4k 1 k 2
(k 1 + k 2 ) 2
(8)
Substituting the expressions for k 1 and k 2 from (1) and putting E = 4V 0 /3
we find that R = 1/9 and T = 8/9.
From (7) and (8) it is easily verified that
R + T = 1
( 9 )
Fig. 3.16 Graphs for
probability density