1000 Solved Problems in Modern Physics

3.3 Solutions 211

u^2210 dτ=

(

1

8 πa^30

)[∫∞

0

y^4 e−y

(

dy

4 a^20

)

a^50

][

cos^3 θ

3

] 1

− 1

(2π)

=

(

1

24

)

×4!= 1

Similarlyu^221 ± 1 dτ=π(2^1 a 0 ) 3

∫e− 2 x

2 x

(^2) sin (^2) θe±iφe∓iφr (^2) sinθdθdφdr

=

(

1

8 πa^30

)∫∞

0

(

e

−ar 0

2

)(

r^4 dr

4 a^20

)∫+ 1

− 1

(1−cos^2 θ) d(cosθ)

∫ 2 π

0

dφ

=

(

a 05

64 πa^50

)[∫∞

0

y^4 e−ydy

](

4

3

)

(2π)

=

(

1

192

)

(4!)(8)= 1

3.67

∫

u 21 ± 1 u 210 dr=A

(^22)
√
2

∫

e−xxcosθe−xxsinθe+φr^2 sinθdθdφdr

The integral

∫ 2 π

0 e

±iφdφ= 0

Thereforeu 21 ± 1 andu 210 are orthogonal.

Further the integral

∫

u∗ 211 U 21 − 1 dτinvolves the integral

∫ 2 π

0

e−iφe−iφdφor

∫ 2 π

0

e−^2 iφdφ= 0

So, the functionsu 211 andu 21 − 1 are also orthogonal.

3.68 The degree of degenerating is given by 2n^2 .Soforn=1, degenerary is 2, for
n=2itis8,forn=3, it is 18 and forn=4, it is 32.

3.69 Parity of the state is determined by the factor (−1)l.For1s,l= 0 ,parity= +1,
for 2p,l= 1 ,parity=−1 and for 3d,l= 2 ,parity=+1.
3.70 To show that the probability density of the 3dstate is independent of the polar
angleθ.Weform
u∗u=u∗(3, 2 , 0)u(3, 2 ,0)+ 2 u∗(3, 2 ,1)u(3, 2 ,1)+ 2 u∗(3, 2 ,2)
u(3, 2 ,2)
The factor 2 takes care ofmvalues±1 and±2, as in Table 3.2. Inserting the
functions the azimuth part,eiφore−iφdrop off when we multiply with the
complex conjugate,

i.e.

(

eiφ

)∗

eiφ=1or(e−iφ)∗(e−iφ)= 1

u∗u=A^23 e−^2 xx^4

[(

1

18

)

(3 cos^2 θ−1)^2 +

(

2

3

)

sin^2 θcos^2 θ+

(

1

6

)

sin^4 θ

]