1000 Solved Problems in Modern Physics

232 3 Quantum Mechanics – II

2 S(m=0);ψ 2 s(0)=(4π)−

(^12)

(

1

2 a

)^32 (

2 −

r

a

)

exp

(

−

r

a

)

2 P(m=0);ψ 2 p(0)=(4π)−

(^12)

(

1

2 a

) (^32) (
r
a

)

exp

(

−

r

2 a

)

cosθ

We can calculate

< 2 , 0 , 0 |z| 2 , 1 , 0 >=< 2 , 0 , 0 |rcosθ| 2 , 1 , 0 >

=

(

1

4 π

)(

1

2 a

) 3 (

1

a

)∫∞

0

r^4

(

2 −

r

a

)

exp

(

−

r

a

)

dr

∫π

0

cos^2 θsinθdθ

∫ 2 π

0

dφ

=− 3 a

Thus, the linear Stark effect splits the degeneratem=0 level into two

components, with the shift

ΔE=± 3 ae|E|

The corresponding eigen functions are√^12 (ψs(0)∓ψp(0))

The two components being mixed in equal proportion (Fig. 3.27).

Fig. 3.27Stark effect in
Hydrogen

3.102 E=

∫+a

−a

(

1

√

a

)

cos

(πx

2 a

)[(

−

^2

2 m

)

d^2

dx^2

+ 1 / 2 mω^2 x^2

]

1

√

a

cos

(πx

2 a

)

dx

=

π^2 ^2

8 ma^2

+

mω^2 a^4

10

+

8 a^5

π^2

(

1 −

6

π^2

)

The best approximation to the ground-state wave function is obtained by

setting∂∂αE=0. This gives

a=

[

3 π^2 ^2

5 m^2 ω^2 (π^2 −3)

] 1 / 4

3.103 The unperturbed wave function is

ψ^0 =ksin

(n

1 πx

a

)

sin(n 2 πy/a);H′=W 0

E=

(

π^2

2 ma^2

)

(

n^21 +n^22

)