1000 Solved Problems in Modern Physics

3.3 Solutions 185

3.37 (a)un=

( 2

L

) 1 / 2

sin

(nπx

L

)

<x>=

∫ L

0

u∗nxundx=

(

2

L

)∫L

0

xsin^2

(nπx

a

)

dx

=

L

2

+

(

L

4 n^2 π^2

)

(cos(2nπ)−1)

The second term on the RHS vanishes for any integral value ofn. Thus

<x>=

L

2

Varx=σ^2 =<(x−<x>)^2 >=<x^2 >−<x>^2 =<x^2 >−

L^2

4

Now <x^2 >=

∫L

0

u∗nx^2 undx=

(

2

L

)∫L

0

x^2 sin^2

(nπx

L

)

dx

=

L^2

3

−

L^2

2 n^2 π^2

σ^2 =<x^2 >−<x>^2 =

L^2

3

−

L^2

2 n^2 π^2

−

L^2

4

=

(

L^2

12

)(

1 −

6

n^2 π^2

)

Forn→∞,<x>=L 2 ;σ^2 →L^2 / 12

(b) Classically the expected distribution is rectangular, that is flat.

The normalized function

f(x)=

1

L

<x>=

∫

xf(x)dx=

∫ L

0

xdx

L

=

L

2

σ^2 =<x^2 >−<x>^2

<x^2 >=

∫ L

0

x^2 f(x)dx=L^2 / 3

∴σ^2 =

L^2

3

−

L^2

4

=

L^2

12

Fig. 3.14

3.38 H=

(

−

^2

2 m

)

∇^2 +ar^2

[

1 −

5

6

sin^2 θcos^2 φ

]

(1)

In spherical coordinatesx=rsinθ cosφ. Therefore