1000 Solved Problems in Modern Physics

(Romina) #1

162 3 Quantum Mechanics – II


3.12 First the wave function is normalized


N^2

∫∞

0

ψ∗ψdx= 1

N^2

∫∞

0

(√

2 e−

x
L

) 2

dx= 1

N= 1 /


L

The probability of finding the particle in the regionx≥1nmis
(
1
L

)∫∞

1

ψ∗ψdx=

∫∞

1

((

1

L

)^12

e−

xL

) 2

dx=

(

2

L

)∫∞

1

e−^2 x/Ldx

=−e−^2 x/L


∣∞

1 =e

− (^2) = 0. 135


3.3.2 Schrodinger Equation. .........................


3.13

(

d^2
dr^2

+

2

r

+ 2 E

)

F(r)=0(1)

(a) By usingF(r)=exp(−r/v)y(r), andE=− 21 ν 2 , it is easily verified that
d^2 y
dr^2

=

2

v

(

d
dr


v
r

)

y (2)

(b)y(r)=

∑∞

p= 0
aprp+^1 (3)

dy
dr

=


ap(p+1)rp (4)

d^2 y
dr^2

=


app(p+1)rp−^1 (5)

Substitute (3), (4) and (5) in (2)

Σapp(p+ 1 )rp−^1 =

2

v

Σap(p+1)rp− 2 Σaprp

Replacepbyp−1 in the RHS and simplify
Σapp(p+1)rp−^1 =

2

v

Σap− 1 (p−ν)rp−^1

Comparing the coefficients ofrp−^1 on both sides

p(p+1)ap=

2

v

(p−v)ap− 1 (6)

(c) The series in (3) will terminate whenν=nwherenis a positive integer.
Heren= 2
Using (3)

y(r)=

∑^1

0

ap rp+^1 =a 0 r+a 1 r^2
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