Since
J
x
{\displaystyle J_{x}}
,
J
y
{\displaystyle J_{y}}
, and
J
z
{\displaystyle J_{z}}
are Hermitian,
J
x
=
J
x
†
{\displaystyle J_{x}=J_{x}^{\dagger }}
,
J
y
=
J
y
†
{\displaystyle J_{y}=J_{y}^{\dagger }}
, and
J
z
=
J
z
†
{\displaystyle J_{z}=J_{z}^{\dagger }}
, all of their eigenvalues are real, and there is a set of eigenvectors that forms a complete basis of the Hilbert space.
For any Hermitian operator, say
A
^
{\displaystyle {\hat {A}}}
, let its eigenbasis,
{
|
a
1
⟩
,
|
a
2
⟩
,
⋯
}
{\displaystyle \left\{|a_{1}\rangle ,|a_{2}\rangle ,\cdots \right\}}
. Then for any
|
ψ
⟩
{\displaystyle |\psi \rangle }
,
⟨
ψ
|
A
^
|
ψ
⟩
=
∑
i
,
j
⟨
ψ
|
a
i
⟩
⟨
a
i
|
A
^
|
a
j
⟩
⟨
a
j
|
ψ
⟩
=
∑
i
,
j
a
j
⟨
a
i
|
ψ
⟩
⟨
a
i
|
a
j
⟩
⟨
a
j
|
ψ
⟩
=
∑
i
,
j
a
j
δ
i
j
⟨
a
i
|
ψ
⟩
⟨
a
j
|
ψ
⟩
=
∑
i
a
i
|
⟨
a
i
|
ψ
⟩
|
2
∈
R
{\displaystyle \langle \psi |{\hat {A}}|\psi \rangle =\sum _{i,j}\langle \psi |a_{i}\rangle \langle a_{i}|{\hat {A}}|a_{j}\rangle \langle a_{j}|\psi \rangle =\sum _{i,j}a_{j}\langle a_{i}|\psi \rangle \langle a_{i}|a_{j}\rangle \langle a_{j}|\psi \rangle =\sum _{i,j}a_{j}\delta _{ij}\langle a_{i}|\psi \rangle \langle a_{j}|\psi \rangle =\sum _{i}a_{i}\left|\langle a_{i}|\psi \rangle \right|^{2}\in \mathbb {R} }
⟨
J
x
⟩
=
⟨
j
m
|
J
x
|
j
m
⟩
=
⟨
j
m
|
−
i
ℏ
[
J
y
,
J
z
]
|
j
m
⟩
=
−
i
ℏ
(
⟨
j
m
|
J
y
J
z
|
j
m
⟩
−
⟨
j
m
|
J
z
J
y
|
j
m
⟩
)
=
−
i
ℏ
(
m
ℏ
⟨
j
m
|
J
y
|
j
m
⟩
−
⟨
j
m
|
J
y
†
J
z
†
|
j
m
⟩
†
)
=
−
i
ℏ
(
m
ℏ
⟨
j
m
|
J
y
|
j
m
⟩
−
⟨
j
m
|
J
y
J
z
|
j
m
⟩
†
)
=
−
i
ℏ
(
m
ℏ
⟨
j
m
|
J
y
|
j
m
⟩
−
m
ℏ
⟨
j
m
|
J
y
|
j
m
⟩
†
)
=
m
Im
{
⟨
j
m
|
J
y
|
j
m
⟩
}
=
0
(
∵
⟨
j
m
|
J
y
|
j
m
⟩
∈
R
)
{\displaystyle {\begin{aligned}\langle J_{x}\rangle &=\langle jm|J_{x}|jm\rangle =\langle jm|-{i \over \hbar }\left[J_{y},J_{z}\right]|jm\rangle =-{i \over \hbar }\left(\langle jm|J_{y}J_{z}|jm\rangle -\langle jm|J_{z}J_{y}|jm\rangle \right)\\&=-{i \over \hbar }\left(m\hbar \langle jm|J_{y}|jm\rangle -\langle jm|J_{y}^{\dagger }J_{z}^{\dagger }|jm\rangle ^{\dagger }\right)=-{i \over \hbar }\left(m\hbar \langle jm|J_{y}|jm\rangle -\langle jm|J_{y}J_{z}|jm\rangle ^{\dagger }\right)\\&=-{i \over \hbar }\left(m\hbar \langle jm|J_{y}|jm\rangle -m\hbar \langle jm|J_{y}|jm\rangle ^{\dagger }\right)=m\ {\text{Im}}\left\{\langle jm|J_{y}|jm\rangle \right\}=0\left(\because \langle jm|J_{y}|jm\rangle \in \mathbb {R} \right)\end{aligned}}}
⟨
J
y
⟩
=
⟨
j
m
|
J
y
|
j
m
⟩
=
⟨
j
m
|
−
i
ℏ
[
J
z
,
J
x
]
|
j
m
⟩
=
−
i
ℏ
(
⟨
j
m
|
J
z
J
x
|
j
m
⟩
−
⟨
j
m
|
J
x
J
z
|
j
m
⟩
)
=
i
ℏ
(
m
ℏ
⟨
j
m
|
J
x
|
j
m
⟩
−
⟨
j
m
|
J
x
†
J
z
†
|
j
m
⟩
†
)
=
i
ℏ
(
m
ℏ
⟨
j
m
|
J
x
|
j
m
⟩
−
⟨
j
m
|
J
x
J
z
|
j
m
⟩
†
)
=
i
ℏ
(
m
ℏ
⟨
j
m
|
J
x
|
j
m
⟩
−
m
ℏ
⟨
j
m
|
J
x
|
j
m
⟩
†
)
=
−
m
Im
{
⟨
j
m
|
J
x
|
j
m
⟩
}
=
0
(
∵
⟨
j
m
|
J
x
|
j
m
⟩
∈
R
)
{\displaystyle {\begin{aligned}\langle J_{y}\rangle &=\langle jm|J_{y}|jm\rangle =\langle jm|-{i \over \hbar }\left[J_{z},J_{x}\right]|jm\rangle =-{i \over \hbar }\left(\langle jm|J_{z}J_{x}|jm\rangle -\langle jm|J_{x}J_{z}|jm\rangle \right)\\&={i \over \hbar }\left(m\hbar \langle jm|J_{x}|jm\rangle -\langle jm|J_{x}^{\dagger }J_{z}^{\dagger }|jm\rangle ^{\dagger }\right)={i \over \hbar }\left(m\hbar \langle jm|J_{x}|jm\rangle -\langle jm|J_{x}J_{z}|jm\rangle ^{\dagger }\right)\\&={i \over \hbar }\left(m\hbar \langle jm|J_{x}|jm\rangle -m\hbar \langle jm|J_{x}|jm\rangle ^{\dagger }\right)=-m\ {\text{Im}}\left\{\langle jm|J_{x}|jm\rangle \right\}=0\left(\because \langle jm|J_{x}|jm\rangle \in \mathbb {R} \right)\end{aligned}}}
⟨
J
x
2
⟩
+
⟨
J
y
2
⟩
=
⟨
J
x
2
+
J
y
2
⟩
=
⟨
J
2
−
J
z
2
⟩
=
⟨
j
m
|
J
2
−
J
z
2
|
j
m
⟩
=
j
(
j
+
1
)
ℏ
2
−
m
2
ℏ
2
{\displaystyle \langle J_{x}^{2}\rangle +\langle J_{y}^{2}\rangle =\langle J_{x}^{2}+J_{y}^{2}\rangle =\langle J^{2}-J_{z}^{2}\rangle =\langle jm|J^{2}-J_{z}^{2}|jm\rangle =j\left(j+1\right)\hbar ^{2}-m^{2}\hbar ^{2}}
By symmetry,
⟨
J
x
2
⟩
=
⟨
J
y
2
⟩
{\displaystyle \langle J_{x}^{2}\rangle =\langle J_{y}^{2}\rangle }
.
∴
⟨
J
x
2
⟩
=
⟨
J
y
2
⟩
=
1
2
ℏ
2
[
j
(
j
+
1
)
−
m
2
]
{\displaystyle \therefore \langle J_{x}^{2}\rangle =\langle J_{y}^{2}\rangle ={1 \over 2}\hbar ^{2}\left[j\left(j+1\right)-m^{2}\right]}
(
Δ
J
x
)
2
=
⟨
J
x
2
⟩
−
⟨
J
x
⟩
2
=
1
2
ℏ
2
[
j
(
j
+
1
)
−
m
2
]
{\displaystyle \left(\Delta J_{x}\right)^{2}=\langle J_{x}^{2}\rangle -\langle J_{x}\rangle ^{2}={1 \over 2}\hbar ^{2}\left[j\left(j+1\right)-m^{2}\right]}
and
(
Δ
J
y
)
2
=
⟨
J
y
2
⟩
−
⟨
J
y
⟩
2
=
1
2
ℏ
2
[
j
(
j
+
1
)
−
m
2
]
{\displaystyle \left(\Delta J_{y}\right)^{2}=\langle J_{y}^{2}\rangle -\langle J_{y}\rangle ^{2}={1 \over 2}\hbar ^{2}\left[j\left(j+1\right)-m^{2}\right]}
.
Since
J
x
=
J
+
+
J
−
2
{\displaystyle J_{x}={{J_{+}+J_{-}} \over 2}}
and
J
y
=
J
+
−
J
−
2
i
{\displaystyle J_{y}={{J_{+}-J_{-}} \over {2i}}}
,
J
x
|
j
m
⟩
=
1
2
(
J
+
|
j
m
⟩
+
J
−
|
j
m
⟩
)
=
1
2
(
J
+
|
j
m
⟩
+
J
−
|
j
m
⟩
)
=
1
2
[
ℏ
{
(
j
−
m
)
(
j
+
m
+
1
)
}
1
/
2
|
j
,
m
+
1
⟩
+
ℏ
{
(
j
+
m
)
(
j
−
m
+
1
)
}
1
/
2
|
j
,
m
−
1
⟩
]
J
y
|
j
m
⟩
=
1
2
i
(
J
+
|
j
m
⟩
−
J
−
|
j
m
⟩
)
=
1
2
i
(
J
+
|
j
m
⟩
−
J
−
|
j
m
⟩
)
=
1
2
i
[
ℏ
{
(
j
−
m
)
(
j
+
m
+
1
)
}
1
/
2
|
j
,
m
+
1
⟩
−
ℏ
{
(
j
+
m
)
(
j
−
m
+
1
)
}
1
/
2
|
j
,
m
−
1
⟩
]
{\displaystyle {\begin{aligned}J_{x}|jm\rangle &={1 \over 2}\left(J_{+}|jm\rangle +J_{-}|jm\rangle \right)={1 \over 2}\left(J_{+}|jm\rangle +J_{-}|jm\rangle \right)\\&={1 \over 2}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}|j,m+1\rangle +\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}|j,m-1\rangle \right]\\J_{y}|jm\rangle &={1 \over 2i}\left(J_{+}|jm\rangle -J_{-}|jm\rangle \right)={1 \over 2i}\left(J_{+}|jm\rangle -J_{-}|jm\rangle \right)\\&={1 \over 2i}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}|j,m+1\rangle -\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}|j,m-1\rangle \right]\end{aligned}}}
⟨
j
,
m
+
1
|
j
,
m
+
1
⟩
=
⟨
j
,
m
−
1
|
j
,
m
−
1
⟩
=
1
,
⟨
j
,
m
−
1
|
j
,
m
+
1
⟩
=
⟨
j
,
m
+
1
|
j
,
m
−
1
⟩
=
0
⇒
{\displaystyle \langle j,m+1|j,m+1\rangle =\langle j,m-1|j,m-1\rangle =1,\langle j,m-1|j,m+1\rangle =\langle j,m+1|j,m-1\rangle =0\Rightarrow }
⟨
j
m
|
J
x
J
y
|
j
m
⟩
=
(
J
x
†
|
j
m
⟩
)
†
J
y
|
j
m
⟩
=
(
J
x
|
j
m
⟩
)
†
J
y
|
j
m
⟩
=
1
2
[
ℏ
{
(
j
−
m
)
(
j
+
m
+
1
)
}
1
/
2
⟨
j
,
m
+
1
|
+
ℏ
{
(
j
+
m
)
(
j
−
m
+
1
)
}
1
/
2
⟨
j
,
m
−
1
|
]
×
1
2
i
[
ℏ
{
(
j
−
m
)
(
j
+
m
+
1
)
}
1
/
2
|
j
,
m
+
1
⟩
−
ℏ
{
(
j
+
m
)
(
j
−
m
+
1
)
}
1
/
2
|
j
,
m
−
1
⟩
]
=
ℏ
2
4
i
{
(
j
−
m
)
(
j
+
m
+
1
)
−
(
j
+
m
)
(
j
−
m
+
1
)
}
=
−
m
ℏ
2
2
i
{\displaystyle {\begin{aligned}\langle jm|J_{x}J_{y}|jm\rangle &=\left(J_{x}^{\dagger }|jm\rangle \right)^{\dagger }J_{y}|jm\rangle =\left(J_{x}|jm\rangle \right)^{\dagger }J_{y}|jm\rangle \\&={1 \over 2}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}\langle j,m+1|+\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}\langle j,m-1|\right]\times \\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {1 \over 2i}\left[\hbar \left\{\left(j-m\right)\left(j+m+1\right)\right\}^{1/2}|j,m+1\rangle -\hbar \left\{\left(j+m\right)\left(j-m+1\right)\right\}^{1/2}|j,m-1\rangle \right]\\&={{\hbar ^{2}} \over {4i}}\left\{\left(j-m\right)\left(j+m+1\right)-\left(j+m\right)\left(j-m+1\right)\right\}\\&=-{{m\hbar ^{2}} \over {2i}}\end{aligned}}}
⟨
J
x
⟩
2
⟨
J
y
⟩
2
−
|
⟨
j
m
|
J
x
J
y
|
j
m
⟩
|
2
=
ℏ
4
4
[
j
(
j
+
1
)
−
m
2
]
2
−
m
2
ℏ
4
4
=
ℏ
4
4
{
j
(
j
+
1
)
−
m
2
+
m
}
{
j
(
j
+
1
)
−
m
2
−
m
}
=
ℏ
4
4
(
j
+
m
)
(
j
−
m
+
1
)
(
j
+
m
+
1
)
(
j
−
m
)
=
ℏ
4
4
(
j
2
−
m
2
)
{
(
j
+
1
)
2
−
m
2
}
{\displaystyle {\begin{aligned}\langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}-\left|\langle jm|J_{x}J_{y}|jm\rangle \right|^{2}&={\hbar ^{4} \over 4}\left[j\left(j+1\right)-m^{2}\right]^{2}-{{m^{2}\hbar ^{4}} \over {4}}={\hbar ^{4} \over 4}\left\{j\left(j+1\right)-m^{2}+m\right\}\left\{j\left(j+1\right)-m^{2}-m\right\}\\&={\hbar ^{4} \over 4}\left(j+m\right)\left(j-m+1\right)\left(j+m+1\right)\left(j-m\right)={\hbar ^{4} \over 4}\left(j^{2}-m^{2}\right)\left\{\left(j+1\right)^{2}-m^{2}\right\}\end{aligned}}}
j
≥
0
{\displaystyle j\geq 0}
and
j
≥
m
≥
−
j
{\displaystyle j\geq m\geq -j}
. So
(
j
+
1
)
2
≥
j
2
≥
m
2
{\displaystyle \left(j+1\right)^{2}\geq j^{2}\geq m^{2}}
∴
⟨
J
x
⟩
2
⟨
J
y
⟩
2
≥
|
⟨
j
m
|
J
x
J
y
|
j
m
⟩
|
2
{\displaystyle \therefore \langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}\geq \left|\langle jm|J_{x}J_{y}|jm\rangle \right|^{2}}
When
m
=
±
j
{\displaystyle m=\pm j}
,
j
2
−
m
2
=
0
{\displaystyle j^{2}-m^{2}=0}
. Then
⟨
J
x
⟩
2
⟨
J
y
⟩
2
−
|
⟨
j
m
|
J
x
J
y
|
j
m
⟩
|
2
=
0
{\displaystyle \langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}-\left|\langle jm|J_{x}J_{y}|jm\rangle \right|^{2}=0}
∴
⟨
J
x
⟩
2
⟨
J
y
⟩
2
=
|
⟨
j
,
±
j
|
J
x
J
y
|
j
,
±
j
⟩
|
2
{\displaystyle \therefore \langle J_{x}\rangle ^{2}\langle J_{y}\rangle ^{2}=\left|\langle j,\pm j|J_{x}J_{y}|j,\pm j\rangle \right|^{2}}
−
ℏ
2
(
1
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
+
1
sin
2
θ
∂
2
∂
ϕ
2
)
ψ
α
β
(
θ
,
ϕ
)
=
α
ψ
α
β
(
θ
,
ϕ
)
{\displaystyle -\hbar ^{2}\left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{1 \over {\sin ^{2}\theta }}{\partial ^{2} \over {\partial \phi ^{2}}}\right)\psi _{\alpha \beta }(\theta ,\phi )=\alpha \psi _{\alpha \beta }(\theta ,\phi )}
⇒
−
ℏ
2
(
1
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
+
1
sin
2
θ
∂
2
∂
ϕ
2
)
(
P
α
β
(
θ
)
e
i
β
ϕ
)
=
α
P
α
β
(
θ
)
e
i
β
ϕ
{\displaystyle \Rightarrow -\hbar ^{2}\left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{1 \over {\sin ^{2}\theta }}{\partial ^{2} \over {\partial \phi ^{2}}}\right)\left(P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }\right)=\alpha P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }}
⇒
−
ℏ
2
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
P
α
β
(
θ
)
e
i
β
ϕ
+
β
2
ℏ
2
sin
2
θ
P
α
β
(
θ
)
e
i
β
ϕ
=
α
P
α
β
(
θ
)
e
i
β
ϕ
{\displaystyle \Rightarrow -{{\hbar ^{2}} \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }+{{\beta ^{2}\hbar ^{2}} \over {\sin ^{2}\theta }}P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }=\alpha P_{\alpha }^{\beta }(\theta )e^{i\beta \phi }}
⇒
1
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
P
α
β
(
θ
)
−
β
2
sin
2
θ
P
α
β
(
θ
)
=
−
α
ℏ
2
P
α
β
(
θ
)
{\displaystyle \Rightarrow {1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}P_{\alpha }^{\beta }(\theta )-{{\beta ^{2}} \over {\sin ^{2}\theta }}P_{\alpha }^{\beta }(\theta )=-{\alpha \over {\hbar ^{2}}}P_{\alpha }^{\beta }(\theta )}
⇒
(
1
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
+
α
ℏ
2
−
β
2
sin
2
θ
)
P
α
β
(
θ
)
=
0
{\displaystyle \Rightarrow \left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{\alpha \over {\hbar ^{2}}}-{{\beta ^{2}} \over {\sin ^{2}\theta }}\right)P_{\alpha }^{\beta }(\theta )=0}
∴
(
1
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
+
α
ℏ
2
−
m
2
sin
2
θ
)
P
α
m
(
θ
)
=
0
{\displaystyle \therefore \left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{\alpha \over {\hbar ^{2}}}-{{m^{2}} \over {\sin ^{2}\theta }}\right)P_{\alpha }^{m}(\theta )=0}
d
d
θ
P
α
m
=
d
u
d
θ
d
d
u
P
α
m
=
−
sin
θ
d
d
u
P
α
m
{\displaystyle {d \over {d\theta }}P_{\alpha }^{m}={{du} \over {d\theta }}{d \over {du}}P_{\alpha }^{m}=-\sin \theta {d \over {du}}P_{\alpha }^{m}}
⇒
1
sin
θ
d
d
θ
sin
θ
d
d
θ
P
α
m
=
−
1
sin
θ
d
d
θ
sin
2
θ
d
d
u
P
α
m
=
−
1
sin
θ
d
d
θ
{
(
1
−
u
2
)
d
d
u
P
α
m
}
=
−
1
sin
θ
d
u
d
θ
d
d
u
{
(
1
−
u
2
)
d
d
u
P
α
m
}
=
(
1
−
u
2
)
d
2
d
u
2
P
α
m
−
2
u
d
d
u
P
α
m
{\displaystyle {\begin{aligned}\Rightarrow {1 \over {\sin \theta }}{d \over {d\theta }}\sin \theta {d \over {d\theta }}P_{\alpha }^{m}&=-{1 \over {\sin \theta }}{d \over {d\theta }}\sin ^{2}\theta {d \over {du}}P_{\alpha }^{m}=-{1 \over {\sin \theta }}{d \over {d\theta }}\left\{\left(1-u^{2}\right){d \over {du}}P_{\alpha }^{m}\right\}\\&=-{1 \over {\sin \theta }}{{du} \over {d\theta }}{d \over {du}}\left\{\left(1-u^{2}\right){d \over {du}}P_{\alpha }^{m}\right\}=\left(1-u^{2}\right){d^{2} \over {du^{2}}}P_{\alpha }^{m}-2u{d \over {du}}P_{\alpha }^{m}\end{aligned}}}
For
m
=
0
{\displaystyle m=0}
,
(
1
sin
θ
∂
∂
θ
sin
θ
∂
∂
θ
+
α
ℏ
2
)
P
α
0
(
θ
)
=
(
1
−
u
2
)
d
2
d
u
2
P
α
0
−
2
u
d
d
u
P
α
0
+
α
ℏ
2
P
α
0
=
0
{\displaystyle \left({1 \over {\sin \theta }}{\partial \over {\partial \theta }}\sin \theta {\partial \over {\partial \theta }}+{\alpha \over {\hbar ^{2}}}\right)P_{\alpha }^{0}(\theta )=\left(1-u^{2}\right){d^{2} \over {du^{2}}}P_{\alpha }^{0}-2u{d \over {du}}P_{\alpha }^{0}+{\alpha \over {\hbar ^{2}}}P_{\alpha }^{0}=0}
⇒
(
1
−
u
2
)
∑
n
=
2
∞
C
n
n
(
n
−
1
)
u
n
−
2
−
2
u
∑
n
=
1
∞
C
n
n
u
n
−
1
+
α
ℏ
2
∑
n
=
0
∞
C
n
u
n
=
0
{\displaystyle \Rightarrow \left(1-u^{2}\right)\sum _{n=2}^{\infty }C_{n}n\left(n-1\right)u^{n-2}-2u\sum _{n=1}^{\infty }C_{n}nu^{n-1}+{\alpha \over {\hbar ^{2}}}\sum _{n=0}^{\infty }C_{n}u^{n}=0}
⇒
∑
n
=
0
∞
C
n
+
2
(
n
+
2
)
(
n
+
1
)
u
n
−
∑
n
=
2
∞
C
n
n
(
n
−
1
)
u
n
−
2
∑
n
=
1
∞
C
n
n
u
n
+
α
ℏ
2
∑
n
=
0
∞
C
n
u
n
=
0
{\displaystyle \Rightarrow \sum _{n=0}^{\infty }C_{n+2}\left(n+2\right)\left(n+1\right)u^{n}-\sum _{n=2}^{\infty }C_{n}n\left(n-1\right)u^{n}-2\sum _{n=1}^{\infty }C_{n}nu^{n}+{\alpha \over {\hbar ^{2}}}\sum _{n=0}^{\infty }C_{n}u^{n}=0}
⇒
∑
n
=
0
∞
{
C
n
+
2
(
n
+
2
)
(
n
+
1
)
−
C
n
n
(
n
−
1
)
−
2
C
n
n
+
α
ℏ
2
C
n
}
u
n
=
0
(
∵
n
=
1
⇒
n
(
n
−
1
)
=
0
,
n
=
0
⇒
n
(
n
−
1
)
=
n
=
0
)
{\displaystyle \Rightarrow \sum _{n=0}^{\infty }\left\{C_{n+2}\left(n+2\right)\left(n+1\right)-C_{n}n\left(n-1\right)-2C_{n}n+{\alpha \over {\hbar ^{2}}}C_{n}\right\}u^{n}=0\left(\because n=1\Rightarrow n\left(n-1\right)=0,n=0\Rightarrow n\left(n-1\right)=n=0\right)}
⇒
C
n
+
2
(
n
+
2
)
(
n
+
1
)
=
(
n
2
+
n
−
α
ℏ
2
)
C
n
{\displaystyle \Rightarrow C_{n+2}\left(n+2\right)\left(n+1\right)=\left(n^{2}+n-{\alpha \over {\hbar ^{2}}}\right)C_{n}}
∴
lim
n
→
∞
C
n
+
2
C
n
=
lim
n
→
∞
n
2
+
n
−
α
/
ℏ
2
(
n
+
2
)
(
n
+
1
)
=
1
{\displaystyle \therefore \lim _{n\to \infty }{C_{n+2} \over C_{n}}=\lim _{n\to \infty }{n^{2}+n-\alpha /\hbar ^{2} \over \left(n+2\right)\left(n+1\right)}=1}
Let
a
0
=
1
{\displaystyle a_{0}=1}
, and
a
n
+
1
=
4
n
2
+
2
n
−
α
/
ℏ
2
(
2
n
+
2
)
(
2
n
+
1
)
a
n
{\displaystyle a_{n+1}={4n^{2}+2n-\alpha /\hbar ^{2} \over {\left(2n+2\right)\left(2n+1\right)}}a_{n}}
.
∑
a
n
{\displaystyle \sum a_{n}}
diverges if for any
N
∈
N
{\displaystyle N\in \mathbb {N} }
, there is
n
>
N
{\displaystyle n>N}
such that
a
n
≠
0
{\displaystyle a_{n}\neq 0}
.
C
0
C
1
>
0
⇒
|
P
α
0
(
0
)
|
≥
∑
|
C
0
|
a
n
C
0
C
1
<
0
⇒
|
P
α
0
(
π
)
|
≥
∑
|
C
1
|
a
n
C
0
=
0
⇒
|
P
α
0
(
π
)
|
≥
∑
|
C
1
|
a
n
C
1
=
0
⇒
|
P
α
0
(
0
)
|
≥
∑
|
C
0
|
a
n
{\displaystyle {\begin{aligned}C_{0}C_{1}>0&\Rightarrow \left|P_{\alpha }^{0}(0)\right|\geq \sum \left|C_{0}\right|a_{n}\\C_{0}C_{1}<0&\Rightarrow \left|P_{\alpha }^{0}(\pi )\right|\geq \sum \left|C_{1}\right|a_{n}\\C_{0}=0&\Rightarrow \left|P_{\alpha }^{0}(\pi )\right|\geq \sum \left|C_{1}\right|a_{n}\\C_{1}=0&\Rightarrow \left|P_{\alpha }^{0}(0)\right|\geq \sum \left|C_{0}\right|a_{n}\\\end{aligned}}}
Since
C
0
{\displaystyle C_{0}}
and
C
1
{\displaystyle C_{1}}
cannot both be zero at the same time, one of
P
α
0
(
0
)
{\displaystyle P_{\alpha }^{0}(0)}
or
P
α
0
(
π
)
{\displaystyle P_{\alpha }^{0}(\pi )}
diverges unless the series terminates.
α
ℏ
2
=
l
(
l
+
1
)
⇒
C
n
+
2
(
n
+
2
)
(
n
+
1
)
=
(
n
2
+
n
−
l
2
−
l
)
C
n
⇒
C
n
+
2
(
n
+
2
)
(
n
+
1
)
=
(
n
+
l
+
1
)
(
n
−
l
)
C
n
⇒
for all k
∈
N
,
C
l
+
2
k
=
0
{\displaystyle {\begin{aligned}{\alpha \over {\hbar ^{2}}}=l\left(l+1\right)&\Rightarrow C_{n+2}\left(n+2\right)\left(n+1\right)=\left(n^{2}+n-l^{2}-l\right)C_{n}\\&\Rightarrow C_{n+2}\left(n+2\right)\left(n+1\right)=\left(n+l+1\right)\left(n-l\right)C_{n}\\&\Rightarrow {\text{for all k}}\in \mathbb {N} ,C_{l+2k}=0\end{aligned}}}
If
l
{\displaystyle l}
is even and
C
1
=
0
{\displaystyle C_{1}=0}
,
P
α
0
{\displaystyle P_{\alpha }^{0}}
is even. If
l
{\displaystyle l}
is odd and
C
0
=
0
{\displaystyle C_{0}=0}
,
P
α
0
{\displaystyle P_{\alpha }^{0}}
is odd. For other values of
C
0
{\displaystyle C_{0}}
or
C
1
{\displaystyle C_{1}}
the series diverges and lacks any physical meaning.
For
l
=
0
{\displaystyle l=0}
,
C
2
k
−
1
=
C
2
k
=
0
{\displaystyle C_{2k-1}=C_{2k}=0}
for all
k
∈
N
⇒
P
0
=
1
{\displaystyle k\in \mathbb {N} \Rightarrow P_{0}=1}
.
For
l
=
1
{\displaystyle l=1}
,
C
2
k
−
2
=
C
2
k
+
1
=
0
{\displaystyle C_{2k-2}=C_{2k+1}=0}
for all
k
∈
N
⇒
P
1
=
u
{\displaystyle k\in \mathbb {N} \Rightarrow P_{1}=u}
.
For
l
=
2
{\displaystyle l=2}
,
C
2
k
+
2
=
C
2
k
−
1
=
0
{\displaystyle C_{2k+2}=C_{2k-1}=0}
for all
k
∈
N
{\displaystyle k\in \mathbb {N} }
and
2
C
2
=
−
6
C
0
⇒
P
2
=
3
u
2
−
1
{\displaystyle 2C_{2}=-6C_{0}\Rightarrow P_{2}=3u^{2}-1}
.
They are equal to
Y
0
0
=
(
4
π
)
-
1
/
2
,
Y
1
0
=
3
4
π
u
,
Y
2
0
=
5
16
π
(
3
u
2
−
1
)
{\displaystyle Y_{0}^{0}=\left(4\pi \right)^{{\text{-}}1/2},Y_{1}^{0}={\sqrt {3 \over {4\pi }}}u,Y_{2}^{0}={\sqrt {5 \over {16\pi }}}\left(3u^{2}-1\right)}
if we ignore the scaling factors.
The eigenvalue equation on
U
{\displaystyle U}
is
[
−
ℏ
2
2
μ
d
2
d
r
2
+
V
(
r
)
+
l
(
l
+
1
)
ℏ
2
2
μ
r
2
]
U
=
E
U
{\displaystyle \left[-{\hbar ^{2} \over {2\mu }}{d^{2} \over {dr^{2}}}+V(r)+{l\left(l+1\right)\hbar ^{2} \over {2\mu r^{2}}}\right]U=EU}
Since
l
=
0
{\displaystyle l=0}
,
{
−
ℏ
2
2
μ
d
2
d
r
2
U
=
E
U
r
<
r
0
−
ℏ
2
2
μ
d
2
d
r
2
U
=
(
E
−
V
0
)
U
r
>
r
0
{\displaystyle {\begin{cases}-{\hbar ^{2} \over {2\mu }}{d^{2} \over {dr^{2}}}U=EU&r<r_{0}\\-{\hbar ^{2} \over {2\mu }}{d^{2} \over {dr^{2}}}U=\left(E-V_{0}\right)U&r>r_{0}\end{cases}}}
Let
k
′
=
2
μ
E
ℏ
{\displaystyle k'={{\sqrt {2\mu E}} \over \hbar }}
,
κ
=
2
μ
(
V
0
−
E
)
ℏ
{\displaystyle \kappa ={{\sqrt {2\mu \left(V_{0}-E\right)}} \over \hbar }}
, and
U
E
(
r
)
=
{
U
1
r
<
r
0
U
2
r
>
r
0
{\displaystyle U_{E}(r)={\begin{cases}U_{1}&r<r_{0}\\U_{2}&r>r_{0}\end{cases}}}
.
The general solutions of each differential equations are
U
1
=
A
e
i
k
′
r
+
B
e
−
i
k
′
r
{\displaystyle U_{1}=Ae^{ik'r}+Be^{-ik'r}}
and
U
2
=
C
e
κ
r
+
D
e
−
κ
r
{\displaystyle U_{2}=Ce^{\kappa r}+De^{-\kappa r}}
.
Boundary conditions:
(1)
U
→
0
as
r
→
0
{\displaystyle U\to 0{\text{ as }}r\to 0}
A
+
B
=
0
{\displaystyle A+B=0}
(2)
U
→
0
as
r
→
∞
{\displaystyle U\to 0{\text{ as }}r\to \infty }
C
=
0
{\displaystyle C=0}
(3) continuity
A
e
i
k
′
r
0
−
A
e
−
i
k
′
r
0
=
D
e
−
κ
r
0
{\displaystyle Ae^{ik'r_{0}}-Ae^{-ik'r_{0}}=De^{-\kappa r_{0}}}
(4) continuity of the derivative
i
k
′
A
e
i
k
′
r
0
+
i
k
′
A
e
−
i
k
′
r
0
=
−
κ
D
e
−
κ
r
0
{\displaystyle ik'Ae^{ik'r_{0}}+ik'Ae^{-ik'r_{0}}=-\kappa De^{-\kappa r_{0}}}
i
k
′
A
e
i
k
′
r
0
+
i
k
′
A
e
−
i
k
′
r
0
=
−
κ
(
A
e
i
k
′
r
0
−
A
e
−
i
k
′
r
0
)
⇒
i
k
′
e
i
k
′
r
0
+
i
k
′
e
−
i
k
′
r
0
=
−
κ
(
e
i
k
′
r
0
−
e
−
i
k
′
r
0
)
⇒
2
i
k
′
cos
(
k
′
r
0
)
=
−
2
i
κ
sin
(
k
′
r
0
)
{\displaystyle {\begin{aligned}ik'Ae^{ik'r_{0}}+ik'Ae^{-ik'r_{0}}=-\kappa \left(Ae^{ik'r_{0}}-Ae^{-ik'r_{0}}\right)&\Rightarrow ik'e^{ik'r_{0}}+ik'e^{-ik'r_{0}}=-\kappa \left(e^{ik'r_{0}}-e^{-ik'r_{0}}\right)\\&\Rightarrow 2ik'\cos \left(k'r_{0}\right)=-2i\kappa \sin \left(k'r_{0}\right)\end{aligned}}}
∴
k
′
/
κ
=
−
tan
(
k
′
r
0
)
{\displaystyle \therefore k'/\kappa =-\tan \left(k'r_{0}\right)}
Since
k
′
2
+
κ
2
=
2
μ
V
0
ℏ
2
{\displaystyle k'^{2}+\kappa ^{2}={{2\mu V_{0}} \over \hbar ^{2}}}
, if we take
α
=
k
′
r
0
{\displaystyle \alpha =k'r_{0}}
and
β
=
κ
r
0
{\displaystyle \beta =\kappa r_{0}}
, then
α
2
+
β
2
=
2
μ
V
0
r
0
2
ℏ
2
{\displaystyle \alpha ^{2}+\beta ^{2}={{2\mu V_{0}{r_{0}}^{2}} \over \hbar ^{2}}}
and
β
=
−
α
cot
α
{\displaystyle \beta =-\alpha \cot \alpha }
.
The minimum value of
α
2
+
β
2
{\displaystyle \alpha ^{2}+\beta ^{2}}
on the first quadrant of the graph
β
=
−
α
cot
α
{\displaystyle \beta =-\alpha \cot \alpha }
is
π
2
4
{\displaystyle \pi ^{2} \over 4}
.
α
2
+
β
2
=
2
μ
V
0
r
0
2
ℏ
2
=
π
2
4
⇒
V
0
=
π
2
ℏ
2
8
μ
r
0
2
{\displaystyle \alpha ^{2}+\beta ^{2}={{2\mu V_{0}{r_{0}}^{2}} \over \hbar ^{2}}={\pi ^{2} \over 4}\Rightarrow V_{0}={{\pi ^{2}\hbar ^{2}} \over {8\mu {r_{0}}^{2}}}}
∴
{\displaystyle \therefore }
If
V
0
<
π
2
ℏ
2
8
μ
r
0
2
{\displaystyle V_{0}<{{\pi ^{2}\hbar ^{2}} \over {8\mu {r_{0}}^{2}}}}
there are no bound states.
v
′
=
(
l
+
1
)
y
l
∑
n
=
0
∞
C
n
y
n
+
y
l
+
1
∑
n
=
1
∞
n
C
n
y
n
−
1
=
y
l
∑
n
=
0
∞
(
n
+
l
+
1
)
C
n
y
n
{\displaystyle v'=\left(l+1\right)y^{l}\sum _{n=0}^{\infty }C_{n}y^{n}+y^{l+1}\sum _{n=1}^{\infty }nC_{n}y^{n-1}=y^{l}\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n}}
v
″
=
l
y
l
−
1
∑
n
=
0
∞
(
n
+
l
+
1
)
C
n
y
n
+
y
l
∑
n
=
1
∞
n
(
n
+
l
+
1
)
C
n
y
n
−
1
=
y
l
−
1
∑
n
=
0
∞
(
n
+
l
)
(
n
+
l
+
1
)
C
n
y
n
{\displaystyle v''=ly^{l-1}\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n}+y^{l}\sum _{n=1}^{\infty }n\left(n+l+1\right)C_{n}y^{n-1}=y^{l-1}\sum _{n=0}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}y^{n}}
v
″
−
2
y
v
′
+
[
2
λ
−
1
−
l
(
l
+
1
)
y
2
]
v
=
0
⇒
y
l
−
1
∑
n
=
0
∞
(
n
+
l
)
(
n
+
l
+
1
)
C
n
y
n
−
2
y
l
+
1
∑
n
=
0
∞
(
n
+
l
+
1
)
C
n
y
n
+
(
2
λ
−
1
)
y
l
+
1
∑
n
=
0
∞
C
n
y
n
−
l
(
l
+
1
)
y
l
−
1
∑
n
=
0
∞
C
n
y
n
=
0
⇒
∑
n
=
0
∞
(
n
+
l
)
(
n
+
l
+
1
)
C
n
y
n
−
2
∑
n
=
0
∞
(
n
+
l
+
1
)
C
n
y
n
+
2
+
(
2
λ
−
1
)
∑
n
=
0
∞
C
n
y
n
+
2
−
l
(
l
+
1
)
∑
n
=
0
∞
C
n
y
n
=
0
⇒
(
l
+
1
)
(
l
+
2
)
C
1
y
−
l
(
l
+
1
)
C
1
y
+
∑
n
=
2
∞
(
n
+
l
)
(
n
+
l
+
1
)
C
n
−
2
(
n
+
l
+
1
)
C
n
−
2
y
n
+
(
2
λ
−
1
)
C
n
−
2
y
n
−
l
(
l
+
1
)
C
n
y
n
=
0
⇒
2
(
l
+
1
)
C
1
y
+
∑
n
=
2
∞
{
n
(
n
+
2
l
+
1
)
C
n
−
(
2
n
+
2
l
−
2
λ
+
3
)
C
n
−
2
}
y
n
=
0
⇒
C
1
=
0
,
n
(
n
+
2
l
+
1
)
C
n
=
(
2
n
+
2
l
−
2
λ
+
3
)
C
n
−
2
⇒
for any
k
∈
N
,
C
2
k
−
1
=
0
{\displaystyle {\begin{aligned}v''&-2yv'+\left[2\lambda -1-{{l\left(l+1\right)} \over y^{2}}\right]v=0\\&\Rightarrow y^{l-1}\sum _{n=0}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}y^{n}-2y^{l+1}\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n}+\left(2\lambda -1\right)y^{l+1}\sum _{n=0}^{\infty }C_{n}y^{n}-l\left(l+1\right)y^{l-1}\sum _{n=0}^{\infty }C_{n}y^{n}=0\\&\Rightarrow \sum _{n=0}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}y^{n}-2\sum _{n=0}^{\infty }\left(n+l+1\right)C_{n}y^{n+2}+\left(2\lambda -1\right)\sum _{n=0}^{\infty }C_{n}y^{n+2}-l\left(l+1\right)\sum _{n=0}^{\infty }C_{n}y^{n}=0\\&\Rightarrow \left(l+1\right)\left(l+2\right)C_{1}y-l\left(l+1\right)C_{1}y+\sum _{n=2}^{\infty }\left(n+l\right)\left(n+l+1\right)C_{n}-2\left(n+l+1\right)C_{n-2}y^{n}+\left(2\lambda -1\right)C_{n-2}y^{n}-l\left(l+1\right)C_{n}y^{n}=0\\&\Rightarrow 2\left(l+1\right)C_{1}y+\sum _{n=2}^{\infty }\left\{n\left(n+2l+1\right)C_{n}-\left(2n+2l-2\lambda +3\right)C_{n-2}\right\}y^{n}=0\\&\Rightarrow C_{1}=0,\ n\left(n+2l+1\right)C_{n}=\left(2n+2l-2\lambda +3\right)C_{n-2}\\&\Rightarrow {\text{for any }}k\in \mathbb {N} ,C_{2k-1}=0\end{aligned}}}
If
C
0
=
0
{\displaystyle C_{0}=0}
, then for every
n
{\displaystyle n}
,
C
n
=
0
{\displaystyle C_{n}=0}
, which leads to a nonphysical zero wave function.
∴
C
0
≠
0
{\displaystyle \therefore C_{0}\neq 0}
For
y
=
1
{\displaystyle y=1}
,
v
=
∑
n
=
0
∞
C
n
=
∑
k
=
0
∞
C
2
k
{\displaystyle v=\sum _{n=0}^{\infty }C_{n}=\sum _{k=0}^{\infty }C_{2k}}
.
Unless the series terminates,
as
k
→
∞
,
C
2
k
+
2
C
2
k
→
1
k
+
1
⇒
v
∼
y
l
+
1
e
y
2
/
2
{\displaystyle {\text{as }}k\to \infty ,{C_{2k+2} \over C_{2k}}\to {1 \over {k+1}}\Rightarrow v\sim y^{l+1}e^{y^{2}/2}}
, which goes to infinity as r increases.
This cannot be a physical solution, so the series must terminate at certain
k
⇒
4
k
+
2
l
−
2
λ
+
3
=
0
{\displaystyle k\Rightarrow 4k+2l-2\lambda +3=0}
∴
E
=
(
2
k
+
l
+
3
/
2
)
ℏ
ω
{\displaystyle \therefore E=\left(2k+l+3/2\right)\hbar \omega }
For each
n
{\displaystyle n}
, the allowed
l
{\displaystyle l}
values are
n
,
n
−
2
,
⋯
,
1
or
0
{\displaystyle n,n-2,\cdots ,1{\text{ or }}0}
.
For each
l
{\displaystyle l}
, there are
2
l
+
1
{\displaystyle 2l+1}
-fold degeneracy, so the total degeneracy is
(
2
n
+
1
)
+
(
2
n
−
3
)
+
⋯
+
3
or
1
{\displaystyle \left(2n+1\right)+\left(2n-3\right)+\cdots +3{\text{ or }}1}
.
If
n
=
2
k
{\displaystyle n=2k}
,
(
4
k
+
1
)
+
(
4
k
−
3
)
+
⋯
+
1
=
(
4
k
+
2
)
(
k
+
1
)
2
=
(
n
+
1
)
(
n
+
2
)
2
{\displaystyle \left(4k+1\right)+\left(4k-3\right)+\cdots +1={{\left(4k+2\right)\left(k+1\right)} \over 2}={{\left(n+1\right)\left(n+2\right)} \over 2}}
.
If
n
=
2
k
−
1
{\displaystyle n=2k-1}
,
(
4
k
−
1
)
+
(
4
k
−
5
)
+
⋯
+
3
=
(
4
k
+
2
)
k
2
=
(
n
+
1
)
(
n
+
2
)
2
{\displaystyle \left(4k-1\right)+\left(4k-5\right)+\cdots +3={{\left(4k+2\right)k} \over 2}={{\left(n+1\right)\left(n+2\right)} \over 2}}
.
∴
ψ
n
{\displaystyle \therefore \psi _{n}}
has
(
n
+
1
)
(
n
+
2
)
2
{\displaystyle {{\left(n+1\right)\left(n+2\right)} \over 2}}
-fold degeneracy
Also
Y
l
m
{\displaystyle Y_{l}^{m}}
are odd if
l
{\displaystyle l}
is odd and even if
l
{\displaystyle l}
is even.
∴
{\displaystyle \therefore }
parity of
ψ
n
{\displaystyle \psi _{n}}
is
(
-
1
)
n
{\displaystyle \left({\text{-}}1\right)^{n}}
The results are equal to the results obtained from the Cartesian solutions.
n
=
2
k
+
l
=
0
⇒
l
=
k
=
m
=
0
{\displaystyle n=2k+l=0\Rightarrow l=k=m=0}
Then
v
=
C
0
y
{\displaystyle v=C_{0}y}
because
C
k
=
0
{\displaystyle C_{k}=0}
for all
k
∈
N
{\displaystyle k\in \mathbb {N} }
.
To normalize
U
00
{\displaystyle U_{00}}
,
∫
0
∞
|
U
00
|
2
d
r
=
C
0
2
∫
0
∞
y
2
e
−
y
2
(
ℏ
μ
ω
)
1
/
2
d
y
=
C
0
2
(
ℏ
μ
ω
)
1
/
2
π
4
=
1
⇒
U
00
=
2
(
μ
ω
π
ℏ
)
1
/
4
y
e
−
y
2
/
2
{\displaystyle \int _{0}^{\infty }\left|U_{00}\right|^{2}dr={C_{0}}^{2}\int _{0}^{\infty }y^{2}e^{-y^{2}}\left({\hbar \over {\mu \omega }}\right)^{1/2}dy={C_{0}}^{2}\left({\hbar \over {\mu \omega }}\right)^{1/2}{{\sqrt {\pi }} \over 4}=1\Rightarrow U_{00}=2\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}ye^{-y^{2}/2}}
.
∵
∫
0
∞
x
2
e
−
x
2
=
[
−
x
2
e
−
x
2
]
0
∞
+
∫
0
∞
1
2
e
−
x
2
=
π
4
{\displaystyle \because \int _{0}^{\infty }x^{2}e^{-x^{2}}=\left[-{x \over 2}e^{-x^{2}}\right]_{0}^{\infty }+\int _{0}^{\infty }{1 \over 2}e^{-x^{2}}={{\sqrt {\pi }} \over 4}}
ψ
000
=
U
00
/
r
Y
0
0
=
2
(
μ
ω
π
ℏ
)
1
/
4
(
μ
ω
4
π
ℏ
)
1
/
2
exp
(
−
μ
ω
2
ℏ
r
2
)
=
(
μ
ω
π
ℏ
)
3
/
4
exp
(
−
μ
ω
2
ℏ
r
2
)
{\displaystyle \psi _{000}=U_{00}/rY_{0}^{0}=2\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}\left({\mu \omega \over {4\pi \hbar }}\right)^{1/2}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)=\left({{\mu \omega } \over {\pi \hbar }}\right)^{3/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)}
n
=
2
k
+
l
=
1
⇒
l
=
1
,
k
=
0
,
m
=
-
1
,
0
,
1
{\displaystyle n=2k+l=1\Rightarrow l=1,k=0,m={\text{-}}1,0,1}
Then
v
=
C
0
y
2
{\displaystyle v=C_{0}y^{2}}
because
C
k
=
0
{\displaystyle C_{k}=0}
for all
k
∈
N
{\displaystyle k\in \mathbb {N} }
.
To normalize
U
01
{\displaystyle U_{01}}
,
∫
0
∞
|
U
01
|
2
d
r
=
C
0
2
∫
0
∞
y
4
e
−
y
2
(
ℏ
μ
ω
)
1
/
2
d
y
=
C
0
2
(
ℏ
μ
ω
)
1
/
2
3
π
8
=
1
⇒
U
01
=
8
3
(
μ
ω
π
ℏ
)
1
/
4
y
2
e
−
y
2
/
2
{\displaystyle \int _{0}^{\infty }\left|U_{01}\right|^{2}dr={C_{0}}^{2}\int _{0}^{\infty }y^{4}e^{-y^{2}}\left({\hbar \over {\mu \omega }}\right)^{1/2}dy={C_{0}}^{2}\left({\hbar \over {\mu \omega }}\right)^{1/2}{{3{\sqrt {\pi }}} \over 8}=1\Rightarrow U_{01}={\sqrt {8 \over 3}}\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}y^{2}e^{-y^{2}/2}}
.
∵
∫
0
∞
x
4
e
−
x
2
=
[
−
x
3
2
e
−
x
2
]
0
∞
+
∫
0
∞
3
x
2
2
e
−
x
2
=
[
−
3
x
4
e
−
x
2
]
0
∞
+
∫
0
∞
3
4
e
−
x
2
=
3
π
8
{\displaystyle \because \int _{0}^{\infty }x^{4}e^{-x^{2}}=\left[-{x^{3} \over 2}e^{-x^{2}}\right]_{0}^{\infty }+\int _{0}^{\infty }{3x^{2} \over 2}e^{-x^{2}}=\left[-{3x \over 4}e^{-x^{2}}\right]_{0}^{\infty }+\int _{0}^{\infty }{3 \over 4}e^{-x^{2}}={{3{\sqrt {\pi }}} \over 8}}
ψ
11
±
1
=
U
01
/
r
Y
1
±
1
=
∓
8
3
3
8
π
(
μ
ω
π
ℏ
)
1
/
4
(
μ
ω
ℏ
)
r
exp
(
−
μ
ω
2
ℏ
r
2
)
sin
θ
e
±
ϕ
,
ψ
110
=
U
01
/
r
Y
1
0
=
8
3
3
4
π
(
μ
ω
π
ℏ
)
1
/
4
(
μ
ω
ℏ
)
r
exp
(
−
μ
ω
2
ℏ
r
2
)
cos
θ
,
{\displaystyle \psi _{11\pm 1}=U_{01}/rY_{1}^{\pm 1}=\mp {\sqrt {8 \over 3}}{\sqrt {3 \over {8\pi }}}\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}\left({\mu \omega \over {\hbar }}\right)r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\sin \theta e^{\pm \phi },\psi _{110}=U_{01}/rY_{1}^{0}={\sqrt {8 \over 3}}{\sqrt {3 \over {4\pi }}}\left({{\mu \omega } \over {\pi \hbar }}\right)^{1/4}\left({\mu \omega \over {\hbar }}\right)r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\cos \theta ,}
⇒
ψ
11
±
1
=
∓
π
(
μ
ω
π
ℏ
)
5
/
4
r
exp
(
−
μ
ω
2
ℏ
r
2
)
sin
θ
e
±
i
ϕ
,
ψ
110
=
2
π
(
μ
ω
π
ℏ
)
5
/
4
r
exp
(
−
μ
ω
2
ℏ
r
2
)
cos
θ
{\displaystyle \Rightarrow \psi _{11\pm 1}=\mp {\sqrt {\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\sin \theta e^{\pm i\phi },\psi _{110}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}r\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\cos \theta }
Cartesian solutions for
n
=
0
,
1
{\displaystyle n=0,1}
are:
ψ
000
′
=
ψ
0
(
x
)
ψ
0
(
y
)
ψ
0
(
z
)
=
A
0
3
exp
(
−
μ
ω
2
ℏ
r
2
)
=
(
μ
ω
π
ℏ
)
3
/
4
exp
(
−
μ
ω
2
ℏ
r
2
)
{\displaystyle \psi '_{000}=\psi _{0}(x)\psi _{0}(y)\psi _{0}(z)={A_{0}}^{3}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)=\left({{\mu \omega } \over {\pi \hbar }}\right)^{3/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)}
ψ
100
′
=
ψ
1
(
x
)
ψ
0
(
y
)
ψ
0
(
z
)
=
A
0
2
A
1
(
μ
ω
ℏ
)
1
/
2
x
exp
(
−
μ
ω
2
ℏ
r
2
)
=
2
π
(
μ
ω
π
ℏ
)
5
/
4
x
exp
(
−
μ
ω
2
ℏ
r
2
)
{\displaystyle \psi '_{100}=\psi _{1}(x)\psi _{0}(y)\psi _{0}(z)={A_{0}}^{2}A_{1}\left({\mu \omega \over {\hbar }}\right)^{1/2}x\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}x\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)}
By the same logic,
ψ
010
′
=
2
π
(
μ
ω
π
ℏ
)
5
/
4
y
exp
(
−
μ
ω
2
ℏ
r
2
)
{\displaystyle \psi '_{010}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}y\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)}
ψ
001
′
=
2
π
(
μ
ω
π
ℏ
)
5
/
4
z
exp
(
−
μ
ω
2
ℏ
r
2
)
{\displaystyle \psi '_{001}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}z\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)}
So
ψ
000
=
ψ
000
′
{\displaystyle \psi _{000}=\psi '_{000}}
and
ψ
11
±
1
=
∓
π
(
μ
ω
π
ℏ
)
5
/
4
exp
(
−
μ
ω
2
ℏ
r
2
)
r
sin
θ
(
cos
ϕ
±
i
sin
ϕ
)
=
∓
π
(
μ
ω
π
ℏ
)
5
/
4
exp
(
−
μ
ω
2
ℏ
r
2
)
(
x
±
i
y
)
=
∓
1
2
ψ
100
′
−
i
2
ψ
010
′
{\displaystyle \psi _{11\pm 1}=\mp {\sqrt {\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)r\sin \theta \left(\cos \phi \pm i\sin \phi \right)=\mp {\sqrt {\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)\left(x\pm iy\right)=\mp {1 \over {\sqrt {2}}}\psi '_{100}-{i \over {\sqrt {2}}}\psi '_{010}}
ψ
110
=
2
π
(
μ
ω
π
ℏ
)
5
/
4
z
exp
(
−
μ
ω
2
ℏ
r
2
)
=
ψ
001
′
{\displaystyle \psi _{110}={\sqrt {2\pi }}\left({{\mu \omega } \over {\pi \hbar }}\right)^{5/4}z\exp \left(-{\mu \omega \over {2\hbar }}r^{2}\right)=\psi '_{001}}
∴
ψ
000
=
ψ
000
′
,
ψ
111
=
−
1
2
ψ
100
′
−
i
2
ψ
010
′
,
ψ
110
=
ψ
001
′
,
ψ
11
-
1
=
1
2
ψ
100
′
−
i
2
ψ
010
′
{\displaystyle \therefore \psi _{000}=\psi '_{000},\psi _{111}=-{1 \over {\sqrt {2}}}\psi '_{100}-{i \over {\sqrt {2}}}\psi '_{010},\psi _{110}=\psi '_{001},\psi _{11{\text{-}}1}={1 \over {\sqrt {2}}}\psi '_{100}-{i \over {\sqrt {2}}}\psi '_{010}}
⟨
Ω
˙
⟩
=
−
i
ℏ
⟨
[
Ω
,
H
]
⟩
{\displaystyle \langle {\dot {\Omega }}\rangle =-{i \over \hbar }\langle \left[\Omega ,H\right]\rangle }
[
Ω
,
H
]
=
[
X
P
X
+
Y
P
Y
+
Z
P
Z
,
H
]
{\displaystyle \left[\Omega ,H\right]=\left[XP_{X}+YP_{Y}+ZP_{Z},H\right]}
[
X
P
X
,
H
]
=
1
2
m
[
X
P
X
,
P
X
2
+
P
Y
2
+
P
Z
2
]
−
e
2
[
X
P
X
,
1
R
]
{\displaystyle \left[XP_{X},H\right]={1 \over {2m}}\left[XP_{X},P_{X}^{2}+P_{Y}^{2}+P_{Z}^{2}\right]-e^{2}\left[XP_{X},{1 \over R}\right]}
[
X
P
X
,
P
X
2
+
P
Y
2
+
P
Z
2
]
=
[
X
P
X
,
P
X
2
]
=
X
P
X
3
−
P
X
2
X
P
X
=
X
P
X
3
−
P
X
X
P
X
2
+
P
X
X
P
X
2
−
P
X
2
X
P
X
=
[
X
,
P
X
]
P
X
2
+
P
X
[
X
,
P
X
]
P
X
=
2
i
ℏ
P
X
2
{\displaystyle {\begin{aligned}\left[XP_{X},P_{X}^{2}+P_{Y}^{2}+P_{Z}^{2}\right]&=\left[XP_{X},P_{X}^{2}\right]=XP_{X}^{3}-P_{X}^{2}XP_{X}\\&=XP_{X}^{3}-P_{X}XP_{X}^{2}+P_{X}XP_{X}^{2}-P_{X}^{2}XP_{X}=\left[X,P_{X}\right]P_{X}^{2}+P_{X}\left[X,P_{X}\right]P_{X}=2i\hbar P_{X}^{2}\end{aligned}}}
i
ℏ
⟨
x
′
,
y
′
,
z
′
|
[
X
P
X
,
1
R
]
|
x
,
y
,
z
⟩
=
x
′
δ
′
(
x
−
x
′
)
δ
(
y
−
y
′
)
δ
(
z
−
z
′
)
1
x
2
+
y
2
+
z
2
−
1
x
′
2
+
y
′
2
+
z
′
2
x
′
δ
′
(
x
−
x
′
)
δ
(
y
−
y
′
)
δ
(
z
−
z
′
)
{\displaystyle {i \over \hbar }\langle x',y',z'|\left[XP_{X},{1 \over R}\right]|x,y,z\rangle =x'\delta '(x-x')\delta (y-y')\delta (z-z'){1 \over {\sqrt {x^{2}+y^{2}+z^{2}}}}-{1 \over {\sqrt {x'^{2}+y'^{2}+z'^{2}}}}x'\delta '(x-x')\delta (y-y')\delta (z-z')}
i
ℏ
∫
⟨
n
,
l
,
m
|
x
′
,
y
′
,
z
′
⟩
⟨
x
′
,
y
′
,
z
′
|
[
X
P
X
,
1
R
]
|
x
,
y
,
z
⟩
⟨
x
,
y
,
z
|
n
,
l
,
m
⟩
d
x
d
y
d
z
d
x
′
d
y
′
d
z
′
=
∫
ψ
n
l
m
∗
x
∂
∂
x
1
x
2
+
y
2
+
z
2
ψ
n
l
m
−
ψ
n
l
m
∗
x
x
2
+
y
2
+
z
2
∂
∂
x
ψ
n
l
m
d
x
d
y
d
z
=
∫
ψ
n
l
m
∗
x
∂
∂
x
(
1
r
)
ψ
n
l
m
d
x
d
y
d
z
=
∫
−
ψ
n
l
m
∗
x
2
r
3
ψ
n
l
m
d
x
d
y
d
z
=
−
⟨
x
2
r
3
⟩
{\displaystyle {\begin{aligned}{i \over \hbar }\int \langle n,l,m&|x',y',z'\rangle \langle x',y',z'|\left[XP_{X},{1 \over R}\right]|x,y,z\rangle \langle x,y,z|n,l,m\rangle dxdydzdx'dy'dz'\\&=\int \psi _{nlm}^{*}x{\partial \over {\partial x}}{1 \over {\sqrt {x^{2}+y^{2}+z^{2}}}}\psi _{nlm}-\psi _{nlm}^{*}{x \over {\sqrt {x^{2}+y^{2}+z^{2}}}}{\partial \over {\partial x}}\psi _{nlm}dxdydz\\&=\int \psi _{nlm}^{*}x{\partial \over {\partial x}}\left({1 \over r}\right)\psi _{nlm}dxdydz\\&=\int -\psi _{nlm}^{*}{x^{2} \over r^{3}}\psi _{nlm}dxdydz=-\langle {x^{2} \over r^{3}}\rangle \end{aligned}}}
⟨
Ω
˙
⟩
=
−
i
ℏ
⟨
[
Ω
,
H
]
⟩
=
1
m
⟨
P
2
⟩
−
e
2
⟨
1
r
⟩
=
2
⟨
T
⟩
+
⟨
V
⟩
=
0
(
∵
⟨
x
2
r
3
⟩
+
⟨
y
2
r
3
⟩
+
⟨
z
2
r
3
⟩
=
⟨
1
r
⟩
)
{\displaystyle \langle {\dot {\Omega }}\rangle =-{i \over \hbar }\langle \left[\Omega ,H\right]\rangle ={1 \over m}\langle P^{2}\rangle -e^{2}\langle {1 \over r}\rangle =2\langle T\rangle +\langle V\rangle =0\left(\because \langle {x^{2} \over r^{3}}\rangle +\langle {y^{2} \over r^{3}}\rangle +\langle {z^{2} \over r^{3}}\rangle =\langle {1 \over r}\rangle \right)}
∴
⟨
T
⟩
=
−
1
2
⟨
V
⟩
{\displaystyle \therefore \langle T\rangle =-{1 \over 2}\langle V\rangle }
n
=
p
×
l
m
−
e
2
r
r
=
p
×
(
r
×
p
)
m
−
e
2
r
r
=
r
p
2
−
p
(
p
⋅
r
)
m
−
e
2
r
r
=
(
p
2
m
−
e
2
r
)
r
−
p
⋅
r
m
p
{\displaystyle \mathbf {n} ={{\mathbf {p} \times \mathbf {l} } \over m}-{e^{2} \over r}\mathbf {r} ={{\mathbf {p} \times \left(\mathbf {r} \times \mathbf {p} \right)} \over m}-{e^{2} \over r}\mathbf {r} ={{\mathbf {r} p^{2}-\mathbf {p} \left(\mathbf {p} \cdot \mathbf {r} \right)} \over m}-{e^{2} \over r}\mathbf {r} =\left({p^{2} \over m}-{e^{2} \over r}\right)\mathbf {r} -{\mathbf {p} \cdot \mathbf {r} \over m}\mathbf {p} }
By energy conservation,
p
2
2
m
−
e
2
r
=
E
{\displaystyle {p^{2} \over {2m}}-{e^{2} \over r}=E}
.
Let the angle between
p
{\displaystyle \mathbf {p} }
and
r
{\displaystyle \mathbf {r} }
be
θ
{\displaystyle \theta }
. i.e.
p
⋅
r
=
p
r
cos
θ
{\displaystyle \mathbf {p} \cdot \mathbf {r} =pr\cos \theta }
n
=
(
2
E
+
e
2
r
)
r
−
p
⋅
r
m
p
=
(
2
E
+
e
2
r
)
r
−
(
r
˙
⋅
r
)
p
=
(
2
E
+
e
2
r
)
r
−
1
2
d
d
t
(
r
⋅
r
)
p
=
(
2
E
+
e
2
r
)
r
−
1
2
d
d
t
(
r
2
)
p
{\displaystyle \mathbf {n} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -{\mathbf {p} \cdot \mathbf {r} \over m}\mathbf {p} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -\left({\dot {\mathbf {r} }}\cdot \mathbf {r} \right)\mathbf {p} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -{1 \over 2}{d \over {dt}}\left(\mathbf {r} \cdot \mathbf {r} \right)\mathbf {p} =\left(2E+{e^{2} \over r}\right)\mathbf {r} -{1 \over 2}{d \over {dt}}\left(r^{2}\right)\mathbf {p} }
When the particle arrives at the maximum or minimum distance from the origin,
d
d
t
(
r
2
)
=
0
{\displaystyle {d \over {dt}}\left(r^{2}\right)=0}
.
So
n
=
r
max
(
2
E
+
e
2
r
max
)
=
r
min
(
2
E
+
e
2
r
min
)
{\displaystyle \mathbf {n} =\mathbf {r} _{\text{max}}\left(2E+{e^{2} \over r_{\text{max}}}\right)=\mathbf {r} _{\text{min}}\left(2E+{e^{2} \over r_{\text{min}}}\right)}
and
n
∥
r
max
∥
r
min
{\displaystyle \mathbf {n} \parallel \mathbf {r} _{\text{max}}\parallel \mathbf {r} _{\text{min}}}
.
If
n
{\displaystyle \mathbf {n} }
,
r
max
{\displaystyle \mathbf {r} _{\text{max}}}
, and
r
min
{\displaystyle \mathbf {r} _{\text{min}}}
are all pointing to the same direction,
r
max
=
r
max
r
^
,
r
min
=
r
min
r
^
{\displaystyle \mathbf {r} _{\text{max}}=r_{\text{max}}{\hat {r}},\mathbf {r} _{\text{min}}=r_{\text{min}}{\hat {r}}}
.
n
=
r
max
(
2
E
+
e
2
r
max
)
r
^
=
(
2
E
r
max
+
e
2
)
r
^
=
r
min
(
2
E
+
e
2
r
min
)
=
(
2
E
r
min
+
e
2
)
r
^
⇒
2
E
r
max
+
e
2
=
2
E
r
min
+
e
2
⇒
r
max
=
r
min
{\displaystyle \mathbf {n} =r_{\text{max}}\left(2E+{e^{2} \over r_{\text{max}}}\right){\hat {r}}=\left(2Er_{\text{max}}+e^{2}\right){\hat {r}}=r_{\text{min}}\left(2E+{e^{2} \over r_{\text{min}}}\right)=\left(2Er_{\text{min}}+e^{2}\right){\hat {r}}\Rightarrow 2Er_{\text{max}}+e^{2}=2Er_{\text{min}}+e^{2}\Rightarrow r_{\text{max}}=r_{\text{min}}}
This does not hold for general cases, so
r
max
=
−
r
max
r
^
{\displaystyle \mathbf {r} _{\text{max}}=-r_{\text{max}}{\hat {r}}}
.
n
=
−
r
max
(
2
E
+
e
2
r
max
)
r
^
=
−
(
2
E
r
max
+
e
2
)
r
^
=
r
min
(
2
E
+
e
2
r
min
)
=
(
2
E
r
min
+
e
2
)
r
^
⇒
−
2
E
r
max
−
e
2
=
2
E
r
min
+
e
2
⇒
r
max
+
r
min
=
−
e
2
E
{\displaystyle \mathbf {n} =-r_{\text{max}}\left(2E+{e^{2} \over r_{\text{max}}}\right){\hat {r}}=-\left(2Er_{\text{max}}+e^{2}\right){\hat {r}}=r_{\text{min}}\left(2E+{e^{2} \over r_{\text{min}}}\right)=\left(2Er_{\text{min}}+e^{2}\right){\hat {r}}\Rightarrow -2Er_{\text{max}}-e^{2}=2Er_{\text{min}}+e^{2}\Rightarrow r_{\text{max}}+r_{\text{min}}=-{e^{2} \over E}}
2
E
+
e
2
r
min
≥
0
{\displaystyle 2E+{e^{2} \over r_{\text{min}}}\geq 0}
, so
r
max
{\displaystyle \mathbf {r} _{\text{max}}}
and
r
min
{\displaystyle \mathbf {r} _{\text{min}}}
are antiparallel to each other.
For circular orbits,
r
max
+
r
min
=
2
r
=
−
e
2
E
⇒
2
E
+
e
2
r
=
0
{\displaystyle r_{\text{max}}+r_{\text{min}}=2r=-{e^{2} \over E}\Rightarrow 2E+{e^{2} \over r}=0}
.
∴
n
=
0
{\displaystyle \therefore \mathbf {n} =0}
[
X
^
,
P
^
x
]
=
[
Y
^
,
P
^
y
]
=
i
ℏ
,
[
X
^
,
Y
^
]
=
[
P
^
x
,
P
^
y
]
=
0
⇒
{\displaystyle \left[{\hat {X}},{\hat {P}}_{x}\right]=\left[{\hat {Y}},{\hat {P}}_{y}\right]=i\hbar ,\ \left[{\hat {X}},{\hat {Y}}\right]=\left[{\hat {P}}_{x},{\hat {P}}_{y}\right]=0\Rightarrow }
,
[
Q
^
1
,
P
^
1
]
=
1
2
(
X
^
+
a
2
ℏ
P
^
y
)
1
2
(
P
^
x
−
ℏ
a
2
Y
^
)
−
1
2
(
P
^
x
−
ℏ
a
2
Y
^
)
1
2
(
X
^
+
a
2
ℏ
P
^
y
)
=
1
2
(
[
X
^
,
P
^
x
]
−
[
P
^
y
,
Y
^
]
+
a
2
ℏ
[
P
^
y
,
P
^
x
]
−
ℏ
a
2
[
X
^
,
Y
^
]
)
=
i
ℏ
[
Q
^
2
,
P
^
2
]
=
1
2
(
X
^
−
a
2
ℏ
P
^
y
)
1
2
(
P
^
x
+
ℏ
a
2
Y
^
)
−
1
2
(
P
^
x
+
ℏ
a
2
Y
^
)
1
2
(
X
^
−
a
2
ℏ
P
^
y
)
=
1
2
(
[
X
^
,
P
^
x
]
−
[
P
^
y
,
Y
^
]
−
a
2
ℏ
[
P
^
y
,
P
^
x
]
+
ℏ
a
2
[
X
^
,
Y
^
]
)
=
i
ℏ
[
Q
^
1
,
Q
^
2
]
=
1
2
(
X
^
+
a
2
ℏ
P
^
y
)
1
2
(
X
^
−
a
2
ℏ
P
^
y
)
−
1
2
(
X
^
−
a
2
ℏ
P
^
y
)
1
2
(
X
^
+
a
2
ℏ
P
^
y
)
=
1
2
(
[
X
^
,
X
^
]
−
a
4
ℏ
2
[
P
^
y
,
P
^
y
]
−
a
2
ℏ
[
P
^
y
,
X
^
]
+
a
2
ℏ
[
X
^
,
P
^
y
]
)
=
0
[
P
^
1
,
P
^
2
]
=
1
2
(
P
^
x
−
ℏ
a
2
Y
^
)
1
2
(
P
^
x
+
ℏ
a
2
Y
^
)
−
1
2
(
P
^
x
+
ℏ
a
2
Y
^
)
1
2
(
P
^
x
−
ℏ
a
2
Y
^
)
=
1
2
(
[
P
^
x
,
P
^
x
]
−
ℏ
2
a
4
[
Y
^
,
Y
^
]
−
ℏ
a
2
[
Y
^
,
P
^
x
]
+
ℏ
a
2
[
P
^
x
,
Y
^
]
)
=
0
[
Q
^
1
,
P
^
2
]
=
1
2
(
X
^
+
a
2
ℏ
P
^
y
)
1
2
(
P
^
x
+
ℏ
a
2
Y
^
)
−
1
2
(
P
^
x
+
ℏ
a
2
Y
^
)
1
2
(
X
^
+
a
2
ℏ
P
^
y
)
=
1
2
(
[
X
^
,
P
^
x
]
+
[
P
^
y
,
Y
^
]
+
a
2
ℏ
[
P
^
y
,
P
^
x
]
+
ℏ
a
2
[
X
^
,
Y
^
]
)
=
0
[
Q
^
2
,
P
^
1
]
=
1
2
(
X
^
−
a
2
ℏ
P
^
y
)
1
2
(
P
^
x
−
ℏ
a
2
Y
^
)
−
1
2
(
P
^
x
−
ℏ
a
2
Y
^
)
1
2
(
X
^
−
a
2
ℏ
P
^
y
)
=
1
2
(
[
X
^
,
P
^
x
]
+
[
P
^
y
,
Y
^
]
−
a
2
ℏ
[
P
^
y
,
P
^
x
]
−
ℏ
a
2
[
X
^
,
Y
^
]
)
=
0
{\displaystyle {\begin{aligned}\left[{\hat {Q}}_{1},{\hat {P}}_{1}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]-\left[{\hat {P}}_{y},{\hat {Y}}\right]+{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]-{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=i\hbar \\\left[{\hat {Q}}_{2},{\hat {P}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]-\left[{\hat {P}}_{y},{\hat {Y}}\right]-{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]+{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=i\hbar \\\left[{\hat {Q}}_{1},{\hat {Q}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right)-{1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {X}}\right]-{a^{4} \over \hbar ^{2}}\left[{\hat {P}}_{y},{\hat {P}}_{y}\right]-{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {X}}\right]+{a^{2} \over \hbar }\left[{\hat {X}},{\hat {P}}_{y}\right]\right)\\&=0\\\left[{\hat {P}}_{1},{\hat {P}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right)\\&={1 \over 2}\left(\left[{\hat {P}}_{x},{\hat {P}}_{x}\right]-{\hbar ^{2} \over a^{4}}\left[{\hat {Y}},{\hat {Y}}\right]-{\hbar \over a^{2}}\left[{\hat {Y}},{\hat {P}}_{x}\right]+{\hbar \over a^{2}}\left[{\hat {P}}_{x},{\hat {Y}}\right]\right)\\&=0\\\left[{\hat {Q}}_{1},{\hat {P}}_{2}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}+{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}+{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]+\left[{\hat {P}}_{y},{\hat {Y}}\right]+{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]+{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=0\\\left[{\hat {Q}}_{2},{\hat {P}}_{1}\right]&={1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right){1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right)-{1 \over {\sqrt {2}}}\left({\hat {P}}_{x}-{\hbar \over a^{2}}{\hat {Y}}\right){1 \over {\sqrt {2}}}\left({\hat {X}}-{a^{2} \over \hbar }{\hat {P}}_{y}\right)\\&={1 \over 2}\left(\left[{\hat {X}},{\hat {P}}_{x}\right]+\left[{\hat {P}}_{y},{\hat {Y}}\right]-{a^{2} \over \hbar }\left[{\hat {P}}_{y},{\hat {P}}_{x}\right]-{\hbar \over a^{2}}\left[{\hat {X}},{\hat {Y}}\right]\right)\\&=0\\\end{aligned}}}
[
X
^
,
P
^
y
]
=
[
Y
^
,
P
^
x
]
=
0
⇒
Q
1
^
2
−
Q
2
^
2
=
a
2
ℏ
X
^
P
^
y
,
P
1
^
2
−
P
2
^
2
=
−
ℏ
a
2
Y
^
P
^
x
{\displaystyle \left[{\hat {X}},{\hat {P}}_{y}\right]=\left[{\hat {Y}},{\hat {P}}_{x}\right]=0\Rightarrow {\hat {Q_{1}}}^{2}-{\hat {Q_{2}}}^{2}={a^{2} \over \hbar }{\hat {X}}{\hat {P}}_{y},\ {\hat {P_{1}}}^{2}-{\hat {P_{2}}}^{2}=-{\hbar \over a^{2}}{\hat {Y}}{\hat {P}}_{x}}
(
a
2
2
ℏ
P
1
^
2
+
ℏ
2
a
2
Q
1
^
2
)
−
(
a
2
2
ℏ
P
2
^
2
+
ℏ
2
a
2
Q
2
^
2
)
=
a
2
2
ℏ
(
P
1
^
2
−
P
2
^
2
)
+
ℏ
2
a
2
(
Q
1
^
2
−
Q
2
^
2
)
=
1
2
(
X
^
P
^
y
−
Y
^
P
^
x
)
=
L
^
z
{\displaystyle \left({a^{2} \over 2\hbar }{\hat {P_{1}}}^{2}+{\hbar \over 2a^{2}}{\hat {Q_{1}}}^{2}\right)-\left({a^{2} \over 2\hbar }{\hat {P_{2}}}^{2}+{\hbar \over 2a^{2}}{\hat {Q_{2}}}^{2}\right)={a^{2} \over 2\hbar }\left({\hat {P_{1}}}^{2}-{\hat {P_{2}}}^{2}\right)+{\hbar \over 2a^{2}}\left({\hat {Q_{1}}}^{2}-{\hat {Q_{2}}}^{2}\right)={1 \over 2}\left({\hat {X}}{\hat {P}}_{y}-{\hat {Y}}{\hat {P}}_{x}\right)={\hat {L}}_{z}}
Take
α
^
=
1
2
a
2
Q
^
1
+
i
a
2
2
ℏ
2
P
^
1
{\displaystyle {\hat {\alpha }}={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{1}+i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{1}}
,
β
^
=
1
2
a
2
Q
^
2
+
i
a
2
2
ℏ
2
P
^
2
{\displaystyle {\hat {\beta }}={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{2}+i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{2}}
, and
L
^
z
ℏ
=
l
^
z
{\displaystyle {{\hat {L}}_{z} \over \hbar }={\hat {l}}_{z}}
.
Then
α
^
†
=
1
2
a
2
Q
^
1
−
i
a
2
2
ℏ
2
P
^
1
{\displaystyle {\hat {\alpha }}^{\dagger }={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{1}-i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{1}}
and
β
^
†
=
1
2
a
2
Q
^
2
−
i
a
2
2
ℏ
2
P
^
2
{\displaystyle {\hat {\beta }}^{\dagger }={\sqrt {1 \over 2a^{2}}}{\hat {Q}}_{2}-i{\sqrt {a^{2} \over 2\hbar ^{2}}}{\hat {P}}_{2}}
, because
P
^
1
{\displaystyle {\hat {P}}_{1}}
,
P
^
2
{\displaystyle {\hat {P}}_{2}}
,
Q
^
1
{\displaystyle {\hat {Q}}_{1}}
, and
Q
^
2
{\displaystyle {\hat {Q}}_{2}}
are all hermitian operators.
α
^
†
α
^
=
1
2
a
2
Q
1
^
2
+
a
2
2
ℏ
2
P
1
^
2
+
i
2
ℏ
[
Q
^
1
,
P
^
1
]
=
(
a
2
2
ℏ
2
P
1
^
2
+
1
2
a
2
Q
1
^
2
)
−
1
2
β
^
†
β
^
=
1
2
a
2
Q
2
^
2
+
a
2
2
ℏ
2
P
2
^
2
+
i
2
ℏ
[
Q
^
2
,
P
^
2
]
=
(
a
2
2
ℏ
2
P
2
^
2
+
1
2
a
2
Q
2
^
2
)
−
1
2
⇒
l
^
z
=
α
^
†
α
^
−
β
^
†
β
^
{\displaystyle {\begin{aligned}{\hat {\alpha }}^{\dagger }{\hat {\alpha }}&={1 \over 2a^{2}}{\hat {Q_{1}}}^{2}+{a^{2} \over 2\hbar ^{2}}{\hat {P_{1}}}^{2}+{i \over {2\hbar }}\left[{\hat {Q}}_{1},{\hat {P}}_{1}\right]=\left({a^{2} \over 2\hbar ^{2}}{\hat {P_{1}}}^{2}+{1 \over 2a^{2}}{\hat {Q_{1}}}^{2}\right)-{1 \over 2}\\{\hat {\beta }}^{\dagger }{\hat {\beta }}&={1 \over 2a^{2}}{\hat {Q_{2}}}^{2}+{a^{2} \over 2\hbar ^{2}}{\hat {P_{2}}}^{2}+{i \over {2\hbar }}\left[{\hat {Q}}_{2},{\hat {P}}_{2}\right]=\left({a^{2} \over 2\hbar ^{2}}{\hat {P_{2}}}^{2}+{1 \over 2a^{2}}{\hat {Q_{2}}}^{2}\right)-{1 \over 2}\end{aligned}}\Rightarrow {\hat {l}}_{z}={\hat {\alpha }}^{\dagger }{\hat {\alpha }}-{\hat {\beta }}^{\dagger }{\hat {\beta }}}
Take
h
α
=
α
^
†
α
^
+
1
2
{\displaystyle h_{\alpha }={\hat {\alpha }}^{\dagger }{\hat {\alpha }}+{1 \over 2}}
and
h
β
=
β
^
†
β
^
+
1
2
{\displaystyle h_{\beta }={\hat {\beta }}^{\dagger }{\hat {\beta }}+{1 \over 2}}
. Then
l
^
z
=
h
α
−
h
β
{\displaystyle {\hat {l}}_{z}=h_{\alpha }-h_{\beta }}
[
Q
^
1
,
P
^
2
]
=
[
Q
^
2
,
P
^
1
]
=
0
{\displaystyle \left[{\hat {Q}}_{1},{\hat {P}}_{2}\right]=\left[{\hat {Q}}_{2},{\hat {P}}_{1}\right]=0}
, so
[
α
^
,
β
^
]
=
0
{\displaystyle \left[{\hat {\alpha }},{\hat {\beta }}\right]=0}
. Also,
[
α
^
,
α
^
†
]
=
α
^
α
^
†
−
α
^
†
α
^
=
1
{\displaystyle \left[{\hat {\alpha }},{\hat {\alpha }}^{\dagger }\right]={\hat {\alpha }}{\hat {\alpha }}^{\dagger }-{\hat {\alpha }}^{\dagger }{\hat {\alpha }}=1}
and
[
β
^
,
β
^
†
]
=
β
^
β
^
†
−
β
^
†
β
^
=
1
{\displaystyle \left[{\hat {\beta }},{\hat {\beta }}^{\dagger }\right]={\hat {\beta }}{\hat {\beta }}^{\dagger }-{\hat {\beta }}^{\dagger }{\hat {\beta }}=1}
.
Other commutation relations are:
[
α
^
,
h
α
]
=
α
^
α
^
†
α
^
−
α
^
†
α
^
2
=
[
α
^
,
α
^
†
]
α
^
=
α
^
{\displaystyle \left[{\hat {\alpha }},h_{\alpha }\right]={\hat {\alpha }}{\hat {\alpha }}^{\dagger }{\hat {\alpha }}-{\hat {\alpha }}^{\dagger }{\hat {\alpha }}^{2}=\left[{\hat {\alpha }},{\hat {\alpha }}^{\dagger }\right]{\hat {\alpha }}={\hat {\alpha }}}
,
[
α
^
†
,
h
α
]
=
α
†
^
2
α
^
−
α
^
†
α
^
α
^
†
=
−
α
^
†
[
α
^
,
α
^
†
]
=
−
α
^
†
{\displaystyle \left[{\hat {\alpha }}^{\dagger },h_{\alpha }\right]={\hat {\alpha ^{\dagger }}}^{2}{\hat {\alpha }}-{\hat {\alpha }}^{\dagger }{\hat {\alpha }}{\hat {\alpha }}^{\dagger }=-{\hat {\alpha }}^{\dagger }\left[{\hat {\alpha }},{\hat {\alpha }}^{\dagger }\right]=-{\hat {\alpha }}^{\dagger }}
,
[
β
^
,
h
β
]
=
−
β
^
{\displaystyle \left[{\hat {\beta }},h_{\beta }\right]=-{\hat {\beta }}}
, and
[
β
^
†
,
h
β
]
=
β
^
†
{\displaystyle \left[{\hat {\beta }}^{\dagger },h_{\beta }\right]={\hat {\beta }}^{\dagger }}
.
By the same logic with the harmonic oscillators,
E
α
=
n
α
+
1
2
{\displaystyle E_{\alpha }=n_{\alpha }+{1 \over 2}}
and
E
β
=
n
β
+
1
2
{\displaystyle E_{\beta }=n_{\beta }+{1 \over 2}}
. Since
α
^
{\displaystyle {\hat {\alpha }}}
and
β
^
{\displaystyle {\hat {\beta }}}
commute, an eigenstate of one operator is an eigenstate of another operator.
l
=
E
α
−
E
β
=
n
α
−
n
β
∈
N
{\displaystyle l=E_{\alpha }-E_{\beta }=n_{\alpha }-n_{\beta }\in \mathbb {N} }
∴
L
z
=
m
ℏ
{\displaystyle \therefore L_{z}=m\hbar }
where
m
∈
N
{\displaystyle m\in \mathbb {N} }