1000 Solved Problems in Modern Physics

(Tina Meador) #1

190 3 Quantum Mechanics – II


ψ 2 =Cexp(ik 2 x)(5)

where

k^22 =

2 m(E−U 0 )
^2

(6)

It represents the transmitted wave to the right with reduced amplitude.
Note that the second term is absent in (5) as there is no reflected wave in
the regionx>0.
Case (ii),U 0 >E
Regionx< 0
ψ 3 =Aexp(ik 1 x)+Bexp(−ik 1 x)(7)
Regionx> 0
d^2 ψ
dx^2


2 mψ(U 0 −E)
^2

= 0

d^2 ψ
dx^2

−α^2 ψ= 0

ψ 4 =Ce−αx+Deαx
whereα^2 =^2 m(U^02 −E)
ψmust be finite everywhere including atx =−∞. We therefore set
D=0. The physically accepted solution is then
ψ 4 =Ce−αx (8)
(b) The continuity condition on the function and its derivative atx=0 leads
to Eqs. (9) and (10).
ψ 3 (0)=ψ 4 (0)
A+B=C (9)
dψ 3
dx





x= 0

=

dψ 4
dx





x= 0
ik 1 (A−B)=−Cα (10)

Fig. 3.15Case(i)

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