r ¼ x
2
þ y
2
þ z
2
ð
Þ
1
2 ,
θ ¼ tan
À1 x
2
þ y
2
ð
Þ
1=2
z
,
ϕ ¼ tan
À1 y
x
:
9
> > > > =
> > > > ;
ð3:18Þ
Thus, we have
L z ¼ xp y À yp x ¼ Àiħ x
∂
∂y
À y
∂
∂x
,
ð3:19Þ
∂
∂x
¼
∂r
∂x
∂
∂r
þ
∂θ
∂x
∂
∂θ
þ
∂ϕ
∂x
∂
∂ϕ
:
ð3:20Þ
In turn, we have
∂r
∂x
¼
x
r
¼ sin θ cos ϕ,
∂θ
∂x
¼
1
1 þ x 2 þ y 2
ð
Þ=z 2 Á
x
2
þ y
2
ð
Þ
À
1
2
Á 2x
2z
¼
z
x 2 þ y 2 þ z 2 Á
x
x 2 þ y 2
ð
Þ
1
2
¼
cos θ cos ϕ
r
,
∂ϕ
∂x
¼
1
1 þ y 2 =x 2
ð
Þ
Á y À
1
x 2
¼ À
sin ϕ
r sin θ
:
ð3:21Þ
In calculating the last two equations of (3.21), we used the differentiation of an
arc tangent function along with a composite function. Namely,
tan
À1 x
À
Á 0 ¼
1
1 þ x 2 :
Inserting (3.21) into (3.20), we get
∂
∂x
¼ sin θ cos ϕ
∂
∂r
þ
cos θ cos ϕ
r
∂
∂θ
À
sin ϕ
r sin θ
∂
∂ϕ
:
ð3:22Þ
Similarly, we have
∂
∂y
¼ sin θ sin ϕ
∂
∂r
þ
cos θ sin ϕ
r
∂
∂θ
þ
cos ϕ
r sin θ
∂
∂ϕ
:
ð3:23Þ
Inserting (3.22) and (3.23) together with (3.17) into (3.19), we get
3.2 Constitution of Hamiltonian
63