1000 Solved Problems in Modern Physics

202 3 Quantum Mechanics – II

Br^2 =x (8)

Substitute (8) in (7) and simplify, to obtain

xd^2 v

dx^2

+

dv

dx

(C−x)−Av=0(9)

which is the familiar confluent hyper geometric equation whose solution

which is regular atx=0is;

V(A,C,x)=α 0

[

1 +

Ax

C

+

A(A+1)x^2

C(C+1)2!

+

A(A+1)(A+2)x^3

C(C+1)(C+2)3!

+···.

]

V(A,C,Br^2 )=α 0

[

1 +

ABr^2

C

+

A(A+1)B^2 r^4

C(C+1)2!

+

A(A+1)(A+2)B^3 r^6

C(C+1)(C+2)3!

+···

]

The asymptotic solutionV(r)→0, whiler→∞implies that the series

must break off for finite powers ofBr^2 sinceα 0

 =0. This means thatAmust

equal a negative integer−p; wherep= 0 , 1 , 2 , 3 ...

Therefore− 4 p= 2 l+ 3 −^2 ω(V 0 +E)

Where we have used the definition ofA(Eq. 6) from this we find, the energy

eigen values,

Ep,l=−V 0 +ω

(

2 p+l+

3

2

)

(p= 0 , 1 , 2 ,....)

Settingn= 2 p+l

En=−V 0 +ω

(

n+^32

)

(which is different from one-dimensional harmonic

oscillator)

E 0 =−V 0 +^3  2 ωcorresponds to ground state.

It is a single state (not degenerate)

sincen=0 can be formed only by the combinationl= 0 ,p=0.

3.54 When the oscillator is in the lowest energy state

<H>=<V+T>=

(

mω^2

2

)

<x^2 >+

(

1

2 m

)

<P^2 >

Now, ifa, bandcare three real numbers such thata+b=c, then

ab=

c^2

4

−

(

a−b

2

) 2

or

ab≤

c^2

4

Apply this inequality to

(

mω^2

2

)

<x^2 >,

( 1

2 m

)

<p^2 >and 2 ω

<Δx>^2 =<x^2 >−<x>^2 =<x^2 >and <x>= 0