1000 Solved Problems in Modern Physics

184 3 Quantum Mechanics – II

Fig. 3.13

For (a) the inside and outside wave functions are as in the deuteron

Problem 3.19. For (b) the inside wave function is similar but the outside

function becomes constant (W 1 =0) and is a horizontal line.

3.36 (a) Class I: Refer to Problem 3.25

ψ 1 =Aeβx(−∞<x<−a)

ψ 2 =Dcosax(−a<x<+a)

ψ 3 =Ae−βx(a<x<∞)

Normalization implies that

∫−a

−∞

|ψ 1 |^2 dx+

∫a

−a

|ψ 2 |^2 dx+

∫∞

a

|ψ 3 |^2 dx= 1

∫−a

−∞

A^2 e^2 βxdx+

∫a

−a

D^2 cos^2 αxdx+

∫∞

a

A^2 e−^2 βxdx= 1

A^2 e−^2 βa

2 β

+D^2

[

a+

sin(2αa)

2 α

]

+

A^2 e

− 2 βa

2 β

= 1

Or

A^2 e−^2 βa/β+D^2 (a+sin(2αa)/ 2 α)=1(1)

Boundary condition atx=agives

Dcosαa=ae−βa (2)

Combining (1) and (2) gives

D=

(

a+

1

β

)− 1

A=eβacosαa

(

a+

1

β

)− 1