1000 Solved Problems in Modern Physics

208 3 Quantum Mechanics – II

It is seen from the last column of the table that the degeneracy D is given by

the sum of natural numbers, that is,=n(n+1)/2, if we replacenbyN+1,

D=(N+1)(N+2)/2.

3.61 As the time evolves, the eigen function would be

ψ(x,t)=

∑

n= 0. 1 Cnψn(x)exp(−iEnt/)

=C 0 ψ 0 (x)exp(−iE 0 t/)+C 1 ψ 1 (x)exp(−iE 1 t/)

The probability density

|ψ(x,t)|^2 =C^20 +C 12 +C 0 C 1 ψ 0 (x)ψ 1 (x)[exp(i(E 1 −E 0 )t/)

−exp(−i(E 1 −E 0 )t/)]

=C 02 +C 12 + 2 C 0 C 1 ψ 0 (x)ψ 1 (x) cosωt

where we have used the energy differenceE 1 −E 0 =ω. Thus the probability

density varies with the angular frequency.

3.62=

∑

n= 1 , 2 , 3

|Cn|^2 En=C 02 E 0 +C 12 E 1 +C 22 E 2

=

(

1

2

)

·

ω

2

+

(

1

3

)

·

3 ω

2

+

(

1

6

)

·

5 ω

2

=

7 ω

6

.

3.63 (a)ψ 0 (x)=Aexp(−x^2 / 2 a^2 )
Differentiate twice and multiply by−^2 / 2 m

−

(

^2

2 m

)

d^2 ψ 0

dx^2

=

(

A^2

2 ma^2

)(

1 −

x^2

a^2

)

exp

(

−

x^2

2 a^2

)

=

(

^2

2 ma^2

)

ψ 0 −

(

^2 x^2

2 ma^4

)

ψ 0

or−

(

^2

2 m

)

d^2 ψ 0

dx^2

+

(

^2 x^2

2 ma^4

)

ψ 0 =

(

^2

2 ma^2

)

ψ 0

Compare the equation with the Schrodinger equation

E=

^2

2 ma^2

=

ω

2

ω=



ma^2

(1)

ora=

(



mω

) 1 / 2

Same relation is obtained by setting

V=

^2 x^2

2 ma^4

=

mω^2 x^2

2

(b)ψ 1 =Bxexp

(

−x

2

2 a^2

)

Differentiate twice and multiply by−

2

2 m

−

^2

2 m

d^2 ψ 1

dx^2

=

B^2 x^3 exp

(

x^2

a

)

2 ma^4

+

3 B^2 exp

(

−x

2

2 a^2

)

2 ma^2