1000 Solved Problems in Modern Physics

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)