1000 Solved Problems in Modern Physics

192 3 Quantum Mechanics – II

This is a direct result of the fact that the current density is constant for a

steady state.

Thus|A 0 |^2 v 1 =|A|^2 v 1 +|B|^2 v 2

wherev 1 =km^1 andv 2 =k^2 m

A^2 +B^2 =1 because the sum of the intensities of the reflected intensity and

transmitted intensities does not add up to unity. What is true is relation (9)

which is relevant to current densities.

3.43 (a) The wave function must be finite, single-valued and continous. At the

boundary this is ensured by requiring the magnitude and the first derivative

be equal.

(b)

Fig. 3.17Sketch of
ψ∼cos

( 3 πx

L

)

(c)

∫ L 2

−L 2

|ψ|^2 dx=A^2

∫ L 2

−L 2

cos^2

3 πx

L

dx= 1

or

A^2

∫ L 2

−L 2

(1+cos 6πx/L)dx=A^2 L= 1

ThereforeA= 1 /

√

L

(d)P

(

−

L

4

<x<

L

4

)

=A^2

∫ L

4

L 4

cos^2

(

3 πx

L

)

dx

=

(

1

L

)∫ L/ 4

−L/ 4

(^1) /
2 (1+cos

(

6 πx

L

)

dx=

1

4

+

1

6 π

= 0. 303

(e)

d^2 ψ

dx^2

=

d^2

dx^2

Acos

(

3 πx

L

)

=− 9 π^2

(

A

L

)

cos

(

3 πx

L

)

=−

(

9 π^2

L

)

ψ

Therefore,

(

−

^2

2 m

)

d^2 ψ

dx^2

=

(

9 π^2 ^2

2 mL

)

ψ−Eψ

Or

E=

9 π^2 ^2

2 mL