1000 Solved Problems in Modern Physics

3.3 Solutions 205

P(x)=

1

π

√

a^2 −x^2

(5)

3.57 Schrodinger’s equation in one dimension is

(

−

^2

2 m

)

d^2 ψ

dx^2

+V(x)ψ=Eψ (1)

Given

ψ=exp(−

1

2

ax^2 )(2)

Differentiating twice,

we get

d^2 ψ

dx^2

exp(−

1

2

ax^2 )(a^2 x^2 −a)(3)

Inserting (2) and (3) in (1), we get

V(x)=E+

(

^2

2 m

)

(a^2 x^2 −a)(4)

Minimum value ofV(x) is determined from

dV

dx

=

^2 a^2 x

m

= 0

Minimum ofV(x) occurs atx= 0

From (4) we find 0=E−

(^2) a
2 m
(a) Or the eigen valueE=
(^2) a
2 m
(b)V(x)=
(^2) a
2 m+

(

^2

2 m

)

(a^2 x^2 −a)=

(^2) a (^2) x 2
2 m

3.58

n=

En

2

<

P^2

2 m

>n=<H>n−<V>n=^1 / 2 En

<P^2 >n=mEn

Also<x>n=0;<P>n= 0

<(Δx)^2 >=<x^2 >n−<x^2 >n

<

(

ΔP^2

)

>=<P^2 >n−<Pn>^2 =<P^2 >n=mEn

But<x^2 >n=

∫∞

−∞u

∗

n(x)x

(^2) un(x)dx;ξ=αx