1000 Solved Problems in Modern Physics

200 3 Quantum Mechanics – II

3.52u 0 =

[

√α

π

]

e−ξ

(^2) / 2
H 0 (ξ);ξ=αx

P= 1 −

∫a

−a

|u 0 |^2 dx= 1 − 2

∫a

0

(α/

√

π)e−ξ

(^2) / 2
dx

= 1 −

2

√

π

∫aα

0

e−ξ

2

dξ

E 0 =^1 / 2 ka^2 =

ω

2

(n=0)

Thereforea^2 =kω=

(

k

)(k

m

) 1 / 2

=√km =α^12

Thereforeα^2 a^2 =1orαa= 1

P= 1 −

2

√

π

∫ 1

0

e−ξ^2 dξ

= 1 −

2

√

π

[

1 −ξ^2 +

ξ^4

2!

−

ξ^6

3!

+

ξ^8

4!

]

dξ

= 1 −

2

√

π

[

1 −

1

3

+

1

10

−

1

42

+

1

216

...

]

≈ 0. 16

Therefore,p≈16%

Fig. 3.21Probability of the
particle found outside the
classical limits is shown
shaded

3.53The potential is of the formV(r)=−V 0 +γ

(^2) r (^2) (1)
Schrodinger’s radial equation is given by,
d^2 u
dr^2

=

[

l(l+1)

r^2

+

2 μ

^2 (V(r)−E)

]

u (2)

Upon substituting (1) in (2), we obtain

d^2 u

dr^2

+

[

2 μ

^2 (V 0 +E−γ^2 r^2 )

−

l(l+1)

r^2

]

u=0(3)

The quantityγ^2 can be expressed in terms of the classical oscillator fre-

quency

γ^2 =

μω^2

2

(4)

Forr→0, (3) may be approximated to

d^2 u

dr^2

−

l(l+1)u

r^2

= 0

The solution of which is,

u(r)=arl+^1 +

b

r

withaandbas constants.