1000 Solved Problems in Modern Physics

204 3 Quantum Mechanics – II

Classically,E=ka

2

2 =

(

n+^12

)

ω(quantum mechanically)≈nω(n→∞)

Thereforea^2 =^2 nkω=

( 2 n

k

)(k

m

) 1 / 2

=√^2 kmn =^2 αn 2

ω=

√

k

m

or

a=

√

2 n

α

(6)

Sinβ=

ξ

√

2 n

=

αx

√

2 n

=

x

a

Therefore

cosβ=

(a^2 −x^2 )

(^12)
a

(7)

Using (6) and (7) in (5)

P(x)=

exp(−ξ^2 exp(2nβ^2 ))

π(a^2 −x^2 )^1 /^2

(8)

Now whenn→∞,sinβ→βand

β→ξ/

√

2 n,andexp(−ξ^2 )exp(2nβ^2 )→1)

ThereforeP(x)=π√a^12 −x 2 (classical)

3.56 One can expect the probability of finding the particle of massmat distancex
from the equilibrium position to be inversely proportional to the velocity

P(x)=

A

v

(1)

whereA=normalization constant. The equation for S.H.O. is

d^2 x

dt^2

+ω^2 x= 0

which has the solution

x=asinωt;(att= 0 ,x=0)

whereais the amplitude.

v=

dx

dt

=ω

√

a^2 −x^2 (2)

Using (2) in (1)

P(x)=A/ω

√

a^2 −x^2 (3)

We can find the normalization constant∫ A.

P(x)dx=

∫a

−a

Adx

ω

√

a^2 −x^2

=

Aπ

ω

= 1

Therefore,

A=

ω

π

(4)

Using (4) in (3), the normalized distribution is