Considerata la componente
L
z
{\displaystyle L_{z}}
L
z
=
x
p
y
−
y
p
x
=
−
i
ℏ
(
x
∂
∂
y
−
y
∂
∂
x
)
{\displaystyle L_{z}=xp_{y}-yp_{x}=-i\hbar \left(x{\frac {\partial }{\partial y}}-y{\frac {\partial }{\partial x}}\right)}
risolvendo l'equazione agli autovalori :
L
z
ψ
(
x
,
y
,
z
)
=
m
ℏ
ψ
(
x
,
y
,
z
)
{\displaystyle L_{z}\psi (x,y,z)=m\hbar \psi (x,y,z)}
Riscriviamo l'operatore
L
z
{\displaystyle L_{z}}
coordinate sferiche :
{
x
=
r
cos
φ
sin
θ
y
=
r
sin
φ
sin
θ
z
=
r
cos
θ
{\displaystyle {\begin{cases}x=r\cos \varphi \sin \theta \\y=r\sin \varphi \sin \theta \\z=r\cos \theta \end{cases}}}
Allora le derivate parziali diventano:
{
∂
∂
x
=
cos
φ
sin
θ
∂
∂
r
+
1
r
cos
φ
cos
θ
∂
∂
θ
−
1
r
sin
φ
sin
θ
∂
∂
φ
∂
∂
y
=
sin
φ
sin
θ
∂
∂
r
+
1
r
sin
φ
cos
θ
∂
∂
θ
+
1
r
cos
φ
sin
θ
∂
∂
φ
∂
∂
z
=
cos
θ
∂
∂
r
−
1
r
sin
θ
∂
∂
θ
{\displaystyle {\begin{cases}{\frac {\partial }{\partial x}}=\cos \varphi \sin \theta {\frac {\partial }{\partial r}}+{\frac {1}{r}}\cos \varphi \cos \theta {\frac {\partial }{\partial \theta }}-{\frac {1}{r}}{\frac {\sin \varphi }{\sin \theta }}{\frac {\partial }{\partial \varphi }}\\{\frac {\partial }{\partial y}}=\sin \varphi \sin \theta {\frac {\partial }{\partial r}}+{\frac {1}{r}}\sin \varphi \cos \theta {\frac {\partial }{\partial \theta }}+{\frac {1}{r}}{\frac {\cos \varphi }{\sin \theta }}{\frac {\partial }{\partial \varphi }}\\{\frac {\partial }{\partial z}}=\cos \theta {\frac {\partial }{\partial r}}-{\frac {1}{r}}\sin \theta {\frac {\partial }{\partial \theta }}\end{cases}}}
L'operatore
L
z
{\displaystyle L_{z}}
L
z
=
−
i
ℏ
∂
∂
φ
{\displaystyle L_{z}=-i\hbar {\frac {\partial }{\partial \varphi }}}
L'equazione agli autovalori per
L
z
{\displaystyle L_{z}}
ψ
{\displaystyle \psi }
φ
{\displaystyle \varphi }
−
i
ℏ
∂
ψ
(
φ
)
∂
φ
=
m
ℏ
ψ
(
φ
)
{\displaystyle -i\hbar {\frac {\partial \psi (\varphi )}{\partial \varphi }}=m\hbar \psi (\varphi )}
questa è un'equazione differenziale al primo ordine, con soluzione generale:
ψ
(
φ
)
=
C
e
i
m
φ
{\displaystyle \psi (\varphi )=Ce^{im\varphi }}
Non resta che trovare il valore della costante, che deve essere tale che:
∫
0
2
π
|
ψ
(
φ
)
|
2
d
φ
=
|
C
|
2
∫
0
2
π
d
φ
=
1
{\displaystyle \int _{0}^{2\pi }|\psi (\varphi )|^{2}\,d\varphi =|C|^{2}\int _{0}^{2\pi }\,d\varphi =1}
da cui:
C
=
1
2
π
{\displaystyle C={\frac {1}{\sqrt {2\pi }}}}
Per cui l'autofunzione di
L
z
{\displaystyle L_{z}}
ψ
(
φ
)
=
1
2
π
e
i
m
φ
{\displaystyle \psi (\varphi )={\frac {1}{\sqrt {2\pi }}}e^{im\varphi }}
Le espressioni di
L
x
{\displaystyle L_{x}}
L
y
{\displaystyle L_{y}}
{
L
x
=
i
ℏ
(
sin
φ
∂
∂
θ
+
cos
φ
tan
θ
∂
∂
φ
)
L
y
=
i
ℏ
(
−
cos
φ
∂
∂
θ
+
sin
φ
tan
θ
∂
∂
φ
)
{\displaystyle {\begin{cases}L_{x}=i\hbar \left(\sin \varphi {\frac {\partial }{\partial \theta }}+{\frac {\cos \varphi }{\tan \theta }}{\frac {\partial }{\partial \varphi }}\right)\\L_{y}=i\hbar \left(-\cos \varphi {\frac {\partial }{\partial \theta }}+{\frac {\sin \varphi }{\tan \theta }}{\frac {\partial }{\partial \varphi }}\right)\end{cases}}}
ovviamente non vi è nessun motivo particolare per scegliere la componente
L
z
{\displaystyle L_{z}}
stato quantistico , possiamo sempre immaginare di porci con l'asse
z
{\displaystyle z}
z
{\displaystyle z}
Riscriviamo il momento angolare (quadrato)
L
→
2
{\displaystyle {\vec {L}}^{2}}
L
→
2
=
−
ℏ
2
[
1
sin
θ
∂
∂
θ
(
sin
θ
∂
∂
θ
)
+
1
sin
2
θ
∂
2
∂
φ
2
]
{\displaystyle {\vec {L}}^{2}=-\hbar ^{2}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right]}
e gli operatori di scala sempre in coordinate polari sferiche:
L
±
=
ℏ
e
±
i
φ
(
±
∂
∂
θ
+
i
cot
θ
∂
∂
φ
)
{\displaystyle L_{\pm }=\hbar e^{\pm i\varphi }\left(\pm {\frac {\partial }{\partial \theta }}+i\cot \theta {\frac {\partial }{\partial \varphi }}\right)}
Sappiamo che l'autofunzione è simultanea di
L
→
2
{\displaystyle {\vec {L}}^{2}}
L
z
{\displaystyle L_{z}}
L
z
{\displaystyle L_{z}}
Y
l
,
m
(
θ
,
φ
)
=
Θ
(
θ
)
⋅
Φ
(
φ
)
=
Θ
(
θ
)
⋅
e
i
m
φ
{\displaystyle Y_{l,m}(\theta ,\varphi )=\Theta (\theta )\cdot \Phi (\varphi )=\Theta (\theta )\cdot e^{im\varphi }}
dove si è sostituita la soluzione per
L
z
{\displaystyle L_{z}}
Θ
(
θ
)
{\displaystyle \Theta (\theta )}
θ
{\displaystyle \theta }
l
(
l
+
1
)
{\displaystyle l(l+1)}
−
ℏ
2
[
1
sin
θ
∂
∂
θ
(
sin
θ
∂
∂
θ
)
+
1
sin
2
θ
∂
2
∂
φ
2
]
Y
l
,
m
(
θ
,
φ
)
=
ℏ
2
l
(
l
+
1
)
Y
l
,
m
(
θ
,
φ
)
{\displaystyle -\hbar ^{2}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right]Y_{l,m}(\theta ,\varphi )=\hbar ^{2}l(l+1)Y_{l,m}(\theta ,\varphi )}
Esplicitiamo la soluzione per
Φ
(
φ
)
{\displaystyle \Phi (\varphi )}
−
ℏ
2
[
1
sin
θ
∂
∂
θ
(
sin
θ
∂
∂
θ
)
+
1
sin
2
θ
∂
2
∂
φ
2
]
e
i
m
φ
Θ
(
θ
)
=
ℏ
2
l
(
l
+
1
)
e
i
m
φ
Θ
(
θ
)
{\displaystyle -\hbar ^{2}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right]e^{im\varphi }\Theta (\theta )=\hbar ^{2}l(l+1)e^{im\varphi }\Theta (\theta )}
quindi eseguendo la derivata seconda sull'esponenziale al primo membro e semplificando
ℏ
2
{\displaystyle \hbar ^{2}}
[
−
1
sin
θ
∂
∂
θ
(
sin
θ
∂
∂
θ
)
+
m
2
sin
2
θ
]
Θ
(
θ
)
=
l
(
l
+
1
)
Θ
(
θ
)
{\displaystyle \left[-{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {m^{2}}{\sin ^{2}\theta }}\right]\Theta (\theta )=l(l+1)\Theta (\theta )}
Facciamo un cambio di variabile esprimendo tutto in termini di
cos
θ
{\displaystyle \cos \theta }
θ
→
cos
θ
⇒
∂
∂
θ
=
−
sin
θ
∂
∂
(
cos
θ
)
{\displaystyle \theta \to \cos \theta \,\,\,\,\,\Rightarrow \,\,\,{\frac {\partial }{\partial \theta }}=-\sin \theta {\frac {\partial }{\partial (\cos \theta )}}}
quindi:
[
m
2
1
−
cos
2
θ
−
∂
∂
(
cos
θ
)
(
(
1
−
cos
2
θ
)
∂
∂
(
cos
θ
)
)
]
Θ
(
cos
θ
)
=
l
(
l
+
1
)
Θ
(
cos
θ
)
{\displaystyle \left[{\frac {m^{2}}{1-\cos ^{2}\theta }}-{\frac {\partial }{\partial (\cos \theta )}}\left((1-\cos ^{2}\theta ){\frac {\partial }{\partial (\cos \theta )}}\right)\right]\Theta (\cos \theta )=l(l+1)\Theta (\cos \theta )}
Per
m
=
0
{\displaystyle m=0}
Liouville :
−
∂
∂
(
cos
θ
)
(
(
1
−
cos
2
θ
)
∂
∂
(
cos
θ
)
)
Θ
l
,
m
=
0
(
cos
θ
)
=
l
(
l
+
1
)
Θ
l
,
m
=
0
(
cos
θ
)
{\displaystyle -{\frac {\partial }{\partial (\cos \theta )}}\left((1-\cos ^{2}\theta ){\frac {\partial }{\partial (\cos \theta )}}\right)\Theta _{l,m=0}(\cos \theta )=l(l+1)\Theta _{l,m=0}(\cos \theta )}
con soluzione:
Θ
l
,
m
=
0
=
(
−
1
)
l
2
l
⋅
l
!
d
l
d
(
cos
θ
)
l
(
1
−
cos
2
θ
)
l
{\displaystyle \Theta _{l,m=0}={\frac {(-1)^{l}}{2^{l}\cdot l!}}{\frac {d^{l}}{d(\cos \theta )^{l}}}(1-\cos ^{2}\theta )^{l}}
La soluzione
Θ
l
,
m
{\displaystyle \Theta _{l,m}}
Θ
l
,
m
=
C
1
(
sin
θ
)
m
(
d
d
(
cos
θ
)
)
l
−
m
(
1
−
cos
2
θ
)
l
{\displaystyle \Theta _{l,m}=C{\frac {1}{(\sin \theta )^{m}}}\left({\frac {d}{d(\cos \theta )}}\right)^{l-m}(1-\cos ^{2}\theta )^{l}}
dove C è una costante di normalizzazione.
Abbiamo quindi trovato che l'autofunzione del momento angolare
L
→
2
{\displaystyle {\vec {L}}^{2}}
L
z
{\displaystyle L_{z}}
[
L
→
2
,
L
z
]
=
0
{\displaystyle [{\vec {L}}^{2},L_{z}]=0}
Y
l
,
m
(
θ
,
φ
)
=
Θ
(
θ
)
⋅
Φ
(
φ
)
=
Θ
(
θ
)
⋅
e
i
m
φ
{\displaystyle Y_{l,m}(\theta ,\varphi )=\Theta (\theta )\cdot \Phi (\varphi )=\Theta (\theta )\cdot e^{im\varphi }}
dove:
Φ
(
φ
)
=
e
i
m
φ
{\displaystyle \Phi (\varphi )=e^{im\varphi }}
e
Θ
l
,
m
=
C
1
(
sin
θ
)
m
(
d
d
(
cos
θ
)
)
l
−
m
(
1
−
cos
2
θ
)
l
{\displaystyle \Theta _{l,m}=C{\frac {1}{(\sin \theta )^{m}}}\left({\frac {d}{d(\cos \theta )}}\right)^{l-m}(1-\cos ^{2}\theta )^{l}}
dove inglobiamo le costanti di normalizzazione nel fattore C . Quindi la soluzione completa è data:
Y
l
,
m
(
θ
,
φ
)
=
C
e
i
m
φ
(
sin
θ
)
m
(
d
d
(
cos
θ
)
)
l
−
m
(
1
−
cos
2
θ
)
l
{\displaystyle Y_{l,m}(\theta ,\varphi )=C{\frac {e^{im\varphi }}{(\sin \theta )^{m}}}\left({\frac {d}{d(\cos \theta )}}\right)^{l-m}(1-\cos ^{2}\theta )^{l}}
queste soluzioni sono ben note alla fisica matematica e si chiamano armoniche sferiche , che dipendono ovviamente dai valori di
l
=
0
,
1
,
…
{\displaystyle l=0,1,\dots }
m
=
−
l
,
−
l
+
1
,
…
,
l
{\displaystyle m=-l,-l+1,\dots ,l}
parità , tra le quali:
Y
l
,
m
(
π
−
θ
,
φ
+
π
)
=
(
−
1
)
l
Y
l
,
m
{\displaystyle Y_{l,m}(\pi -\theta ,\varphi +\pi )=(-1)^{l}Y_{l,m}}
che ha un diretto significato fisico, essa rappresenta l'inversione spaziale delle coordinate polari sferiche.
![{\displaystyle {\vec {L}}^{2}=-\hbar ^{2}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61be2f18c9f7215e2d2b5661e5f73d77def863e7)
![{\displaystyle -\hbar ^{2}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right]Y_{l,m}(\theta ,\varphi )=\hbar ^{2}l(l+1)Y_{l,m}(\theta ,\varphi )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b895211f6f24f980c13a007527c2a6ebfc8bf68e)
![{\displaystyle -\hbar ^{2}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \varphi ^{2}}}\right]e^{im\varphi }\Theta (\theta )=\hbar ^{2}l(l+1)e^{im\varphi }\Theta (\theta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53a731a0190c58122adfb589b232617c17af09f2)
![{\displaystyle \left[-{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {m^{2}}{\sin ^{2}\theta }}\right]\Theta (\theta )=l(l+1)\Theta (\theta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdde976d8f94317f7ebf8f293320dc01a105c5ef)
![{\displaystyle \left[{\frac {m^{2}}{1-\cos ^{2}\theta }}-{\frac {\partial }{\partial (\cos \theta )}}\left((1-\cos ^{2}\theta ){\frac {\partial }{\partial (\cos \theta )}}\right)\right]\Theta (\cos \theta )=l(l+1)\Theta (\cos \theta )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c98081a06b3d8a3452c123628458037b2168daef)
![{\displaystyle [{\vec {L}}^{2},L_{z}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbce43eec2726d75670a2838e30c39b6d6baf518)
