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