3.3 Solutions 191
Dividing (10) by (9) gives
ik(A−B)
A+B=−α (11)Diagrams forψat aroundx= 03.42k 1 =
(
2 mE
^2) 1 / 2
;k 2 =(
2 m(E−V 0 )
^2) 1 / 2
(1)
Boundary condition atx=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=
2 k 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
=
4 k 1 k 2
(k 1 +k 2 )^2(8)
Substituting the expressions fork 1 andk 2 from (1) and puttingE= 4 V 0 / 3
we find thatR= 1 /9 andT= 8 /9.
From (7) and (8) it is easily verified that
R+T=1(9)Fig. 3.16Graphs for
probability density