1000 Solved Problems in Modern Physics

(Tina Meador) #1

3.3 Solutions 157


3.3 Normalization condition is
∫∞

−∞

|ψ|^2 dx= 1

N^2

∫∞

−∞

(x^2 +a^2 )−^2 dx= 1

Putx=atanθ;dx=sec^2 θdθ
(
2 N^2
α^3

)∫π/ 2

0

cos^2 θdθ=N^2 π/ 2 a^3 = 1

ThereforeN=

(

2 a^3
π

) 1 / 2

3.4ψ=Aeikx+Be−ikx
The fluxJx=

(
2 im

)[
ψ∗ddψx−

(
dψ∗
dx

)
ψ

]

=

(

2 im

)
[(
Ae−ikx+Beikx

)
ik

(
Aeikx−Be−ikx

)

+ik

(
Ae−ikx−Beikx

)(
Aeikx+Be−ikx

)]

=

(
k
2 m

)
[
A^2 −B^2 −ABe−^2 ikx+ABe^2 ikx+A^2 −B^2 +ABe−^2 ikx−ABe^2 ikx

]

=

(
k
m

)(
A^2 −B^2

)

3.5 In natural units (=c=1) Klein – Gordon equation is

∇^2 φ−

∂^2 φ
dt^2

−m^2 φ=0(1)

The complex conjugate equation is

∇^2 φ∗−

∂^2 φ∗
∂t^2

−m^2 φ∗=0(2)

Multiplying (1) from left byφ∗and (2) byφand subtracting (1) from (2)

φ∇^2 φ∗−φ∗∇^2 φ−φ

∂^2 φ∗
∂t^2

−φ

∂^2 φ∗
∂t^2

+φ∗

∂^2 φ
∂t^2

= 0

∇.

(

φ∇φ∗∇φ

)



∂t

(

φ

∂φ∗
∂t

−φ∗

∂φ
∂t

)

= 0

Changing the sign through out and multiplying by 1/ 2 im
1
2 im

∇·

(

φ∗∇φ−φ∇φ∗

)


1

2 im


∂t

(

φ∗

∂φ
∂t

−φ

∂φ∗
∂t

)

= 0

∇·

[

1

2 im

(

φ∗∇φ−φ∇φ∗

)

]

+


∂t

[

i
2 m

(

φ∗

∂φ
∂t

−φ

∂φ∗
∂t

)]

= 0

Or∇·J+

∂ρ
∂t

= 0
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