球對稱位勢 乃是一種只與徑向距離有關的位勢 。許多描述宇宙交互作用的基本位勢，像重力勢 、電勢 ，都是球對稱位勢。這條目只講述，在量子力學 裏，運動於球對稱位勢中的粒子 的量子行為。這量子行為，可以用薛丁格方程式 表達為
−
ℏ
2
2
μ
∇
2
ψ
+
V
(
r
)
ψ
=
E
ψ
{\displaystyle -{\frac {\hbar ^{2}}{2\mu }}\nabla ^{2}\psi +V(r)\psi =E\psi }
其中，
ℏ
{\displaystyle \hbar }
普朗克常數 ，
μ
{\displaystyle \mu }
質量 ，
ψ
{\displaystyle \psi }
波函數 ，
V
{\displaystyle V}
位勢 ，
r
{\displaystyle r}
E
{\displaystyle E}
能量 。
由於球對稱位勢
V
(
r
)
{\displaystyle V(r)}
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
球坐標
(
r
,
θ
,
ϕ
)
{\displaystyle (r,\ \theta ,\ \phi )}
分離變數法 ，可以將薛丁格方程式 分為兩部分，徑向部分與角部分。
採用球坐標
(
r
,
θ
,
ϕ
)
{\displaystyle (r,\ \theta ,\ \phi )}
拉普拉斯算子
∇
2
{\displaystyle \nabla ^{2}}
−
ℏ
2
2
μ
r
2
{
∂
∂
r
(
r
2
∂
∂
r
)
+
1
sin
2
θ
[
sin
θ
∂
∂
θ
(
sin
θ
∂
∂
θ
)
+
∂
2
∂
ϕ
2
]
}
ψ
+
V
(
r
)
ψ
=
E
ψ
{\displaystyle -{\frac {\hbar ^{2}}{2\mu r^{2}}}\left\{{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)+{\frac {1}{\sin ^{2}\theta }}\left[\sin \theta {\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right]\right\}\psi +V(r)\psi =E\psi }
滿足薛丁格方程式的本徵函數
ψ
{\displaystyle \psi }
ψ
(
r
,
θ
,
ϕ
)
=
R
(
r
)
Θ
(
θ
)
Φ
(
ϕ
)
{\displaystyle \psi (r,\ \theta ,\ \phi )=R(r)\Theta (\theta )\Phi (\phi )}
其中，
R
(
r
)
{\displaystyle R(r)}
Θ
(
θ
)
{\displaystyle \Theta (\theta )}
Φ
(
ϕ
)
{\displaystyle \Phi (\phi )}
Θ
(
θ
)
{\displaystyle \Theta (\theta )}
Φ
(
ϕ
)
{\displaystyle \Phi (\phi )}
球諧函數 ，
Y
l
m
(
θ
,
ϕ
)
=
Θ
(
θ
)
Φ
(
ϕ
)
{\displaystyle Y_{lm}(\theta ,\ \phi )=\Theta (\theta )\Phi (\phi )}
ψ
{\displaystyle \psi }
ψ
(
r
,
θ
,
ϕ
)
=
R
(
r
)
Y
l
m
(
θ
,
ϕ
)
{\displaystyle \psi (r,\ \theta ,\ \phi )=R(r)Y_{lm}(\theta ,\ \phi )}
參數為天頂角
θ
{\displaystyle \theta }
ϕ
{\displaystyle \phi }
Y
l
m
{\displaystyle Y_{lm}}
−
1
sin
2
θ
[
sin
θ
∂
∂
θ
(
sin
θ
∂
∂
θ
)
+
∂
2
∂
ϕ
2
]
Y
l
m
(
θ
,
ϕ
)
=
l
(
l
+
1
)
Y
l
m
(
θ
,
ϕ
)
{\displaystyle -{\frac {1}{\sin ^{2}\theta }}\left[\sin \theta {\frac {\partial }{\partial \theta }}{\Big (}\sin \theta {\frac {\partial }{\partial \theta }}{\Big )}+{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right]Y_{lm}(\theta ,\phi )=l(l+1)Y_{lm}(\theta ,\phi )}
其中，非負整數
l
{\displaystyle l}
角動量 的角量子數 。
m
{\displaystyle m}
−
l
≤
m
≤
l
{\displaystyle -l\leq m\leq l}
投影 。不同的
l
{\displaystyle l}
m
{\displaystyle m}
Y
l
m
{\displaystyle Y_{lm}}
Y
l
m
(
θ
,
ϕ
)
=
(
i
)
m
+
|
m
|
(
2
l
+
1
)
4
π
(
l
−
|
m
|
)
!
(
l
+
|
m
|
)
!
P
l
m
(
cos
θ
)
e
i
m
ϕ
{\displaystyle Y_{lm}(\theta ,\ \phi )=(i)^{m+|m|}{\sqrt {{(2l+1) \over 4\pi }{(l-|m|)! \over (l+|m|)!}}}\,P_{lm}(\cos {\theta })\,e^{im\phi }}
其中，
i
{\displaystyle i}
虛數單位 ，
P
l
m
(
cos
θ
)
{\displaystyle P_{lm}(\cos {\theta })}
伴隨勒讓德多項式 ，用方程式定義為
P
l
m
(
x
)
=
(
1
−
x
2
)
|
m
|
/
2
d
|
m
|
d
x
|
m
|
P
l
(
x
)
{\displaystyle P_{lm}(x)=(1-x^{2})^{|m|/2}\ {\frac {d^{|m|}}{dx^{|m|}}}P_{l}(x)}
而
P
l
(
x
)
{\displaystyle P_{l}(x)}
l
{\displaystyle l}
勒讓德多項式 ，可用羅德里格公式 表示為
P
l
(
x
)
=
1
2
l
l
!
d
l
d
x
l
(
x
2
−
1
)
l
{\displaystyle P_{l}(x)={1 \over 2^{l}l!}{d^{l} \over dx^{l}}(x^{2}-1)^{l}}
將角部分解答代入薛丁格方程式，則可得到一個一維的二階微分方程式：
{
−
ℏ
2
2
μ
r
2
d
d
r
(
r
2
d
d
r
)
+
ℏ
2
l
(
l
+
1
)
2
μ
r
2
+
V
(
r
)
}
R
(
r
)
=
E
R
(
r
)
{\displaystyle \left\{-{\hbar ^{2} \over 2\mu r^{2}}{d \over dr}\left(r^{2}{d \over dr}\right)+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}+V(r)\right\}R(r)=ER(r)}
(1) 設定函數
u
(
r
)
=
r
R
(
r
)
{\displaystyle u(r)=rR(r)}
−
ℏ
2
2
μ
d
2
u
(
r
)
d
r
2
+
ℏ
2
l
(
l
+
1
)
2
μ
r
2
u
(
r
)
+
V
(
r
)
u
(
r
)
=
E
u
(
r
)
{\displaystyle -{\hbar ^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}u(r)+V(r)u(r)=Eu(r)}
(2) 徑向方程式變為
−
ℏ
2
2
μ
d
2
u
(
r
)
d
r
2
+
V
e
f
f
(
r
)
u
(
r
)
=
E
u
(
r
)
{\displaystyle -{\hbar ^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+V_{\mathrm {eff} }(r)u(r)=Eu(r)}
(3) 其中，有效位勢
V
e
f
f
(
r
)
=
V
(
r
)
+
ℏ
2
l
(
l
+
1
)
2
μ
r
2
{\displaystyle V_{\mathrm {eff} }(r)=V(r)+{\frac {\hbar ^{2}l(l+1)}{2\mu r^{2}}}}
這正是函數為
u
(
r
)
{\displaystyle u(r)}
V
e
f
f
{\displaystyle V_{\mathrm {eff} }}
r
{\displaystyle r}
0
{\displaystyle 0}
∞
{\displaystyle \infty }
離心位勢 。
為了要更進一步解析方程式(2)，必須知道位勢的形式。不同的位勢有不同的解答。
在這裏，有四個很特別、很重要的實例。這些實例都有一個共同點，那就是，它們的位勢都是球對稱的。因此，它們的角部分解答都是球諧函數。這四個實例是：
V
(
r
)
=
0
{\displaystyle V(r)=0}
亥姆霍兹方程
(
∇
2
+
2
μ
E
ℏ
2
)
A
=
0
{\displaystyle (\nabla ^{2}+{\frac {2\mu E}{\hbar ^{2}}})A=0}
正交歸一基 ，解析眞空狀況實例。這實例可以做為別的實例的基礎。當
r
<
r
0
{\displaystyle r<r_{0}}
V
(
r
)
=
0
{\displaystyle V(r)=0}
V
(
r
)
=
∞
{\displaystyle V(r)=\infty }
V
(
r
)
∝
r
2
{\displaystyle V(r)\propto r^{2}}
均向性 諧振子 的實例。在量子力學裏，是少數幾個存在簡單的解析解 的量子模型。
V
(
r
)
∝
1
/
r
{\displaystyle V(r)\propto 1/r}
類氫原子 的束縛態 的實例，也有簡單的解析解。
思考
V
(
r
)
=
0
{\displaystyle V(r)=0}
k
=
d
e
f
2
μ
E
ℏ
2
{\displaystyle k\ {\stackrel {\mathrm {def} }{=}}\ {\sqrt {2\mu E \over \hbar ^{2}}}}
無因次 的變數
ρ
=
d
e
f
k
r
{\displaystyle \rho \ {\stackrel {\mathrm {def} }{=}}\ kr}
代入方程式(2)，定義
J
(
ρ
)
=
d
e
f
ρ
R
(
r
)
{\displaystyle J(\rho )\ {\stackrel {\mathrm {def} }{=}}\ {\sqrt {\rho }}R(r)}
貝塞爾方程式 ，一個二階常微分方程式 ：
ρ
2
d
2
J
d
ρ
2
+
ρ
d
J
d
ρ
+
[
ρ
2
−
(
l
+
1
2
)
2
]
J
=
0
{\displaystyle \rho ^{2}{d^{2}J \over d\rho ^{2}}+\rho {dJ \over d\rho }+\left[\rho ^{2}-\left(l+{\frac {1}{2}}\right)^{2}\right]J=0}
貝塞爾方程式的解答是第一類貝塞爾函數
J
l
+
1
/
2
(
ρ
)
{\displaystyle J_{l+1/2}(\rho )}
R
(
r
)
{\displaystyle R(r)}
球貝塞爾函數 邊界條件 要求原點的函數值有限，因此在原點趨於無窮的第二類球貝塞爾函數項的係數必須為零）：
R
(
r
)
=
j
l
(
k
r
)
=
d
e
f
π
/
(
2
k
r
)
J
l
+
1
/
2
(
k
r
)
{\displaystyle R(r)=j_{l}(kr)\ {\stackrel {\mathrm {def} }{=}}\ {\sqrt {\pi /(2kr)}}J_{l+1/2}(kr)}
(4) 在眞空裏，一個粒子的薛丁格方程（即自由空間中的齊次 亥姆霍兹方程 ）的解，以球坐標來表達，是球貝塞爾函數與球諧函數的乘積：
ψ
(
r
,
θ
,
ϕ
)
=
A
k
l
j
l
(
k
r
)
Y
l
m
(
θ
,
ϕ
)
{\displaystyle \psi (r,\ \theta ,\ \phi )=A_{kl}j_{l}(kr)Y_{lm}(\theta ,\phi )}
其中，歸一常數
A
k
l
=
2
π
k
{\displaystyle A_{kl}={\sqrt {\frac {2}{\pi }}}\,k}
l
{\displaystyle l}
m
{\displaystyle m}
−
l
≤
m
≤
l
{\displaystyle -l\leq m\leq l}
k
{\displaystyle k}
k
≥
0
{\displaystyle k\geq 0}
這些解答都是角動量確定態的波函數。這些確定態都有明確的角動量。
波函數的角部分已經歸一化，剩下來必須將徑向部分歸一化。徑向函數的歸一化條件為
1
=
A
k
l
2
∫
0
∞
r
2
j
l
2
(
k
r
)
d
r
{\displaystyle 1=A_{kl}^{2}\int _{0}^{\infty }\ r^{2}j_{l}^{2}(kr)\ dr}
根據球貝塞爾函數的封閉方程式 ，
∫
0
∞
x
2
j
α
(
k
1
x
)
j
α
(
k
2
x
)
d
x
=
π
2
k
1
2
δ
(
k
1
−
k
2
)
{\displaystyle \int _{0}^{\infty }\ x^{2}j_{\alpha }(k_{1}x)j_{\alpha }(k_{2}x)\ dx={\frac {\pi }{2k_{1}^{2}}}\delta (k_{1}-k_{2})}
其中，
α
>
0
{\displaystyle \alpha >0}
δ
k
{\displaystyle \delta _{k}}
克罗内克δ 。
所以，
1
=
A
k
l
2
π
2
k
2
{\displaystyle 1=A_{kl}^{2}{\frac {\pi }{2k^{2}}}}
A
k
l
=
2
π
k
{\displaystyle A_{kl}={\sqrt {\frac {2}{\pi }}}\,k}
球貝塞爾函數
j
l
(
x
)
{\displaystyle j_{l}(x)}
思考一個球對稱的無限深方形阱，阱內位勢為0，阱外位勢為無限大。用方程式表達：
V
(
r
)
=
{
0
,
if
r
≤
r
0
∞
,
if
r
>
r
0
{\displaystyle V(r)={\begin{cases}0,&{\mbox{if }}r\leq r_{0}\\\infty ,&{\mbox{if }}r>r_{0}\end{cases}}}
其中，
r
0
{\displaystyle r_{0}}
立刻，可以察覺，阱外的波函數是0；而由於阱內的薛丁格方程式與真空狀況的薛丁格方程式相同，波函數是球貝塞爾函數
R
(
r
)
=
j
l
(
k
r
)
{\displaystyle R(r)=j_{l}(kr)}
r
=
r
0
{\displaystyle r=r_{0}}
j
l
(
k
r
0
)
=
0
{\displaystyle j_{l}(kr_{0})=0}
設定
ξ
n
l
{\displaystyle \xi _{nl}}
l
{\displaystyle l}
j
l
{\displaystyle j_{l}}
n
{\displaystyle n}
k
n
l
r
0
=
ξ
n
l
{\displaystyle k_{nl}r_{0}=\xi _{nl}}
那麼，離散的能級
E
n
l
{\displaystyle E_{nl}}
E
n
l
=
ℏ
2
k
n
l
2
2
μ
=
ℏ
2
ξ
n
l
2
2
μ
r
0
2
{\displaystyle E_{nl}={\frac {\hbar ^{2}k_{nl}^{2}}{2\mu }}={\frac {\hbar ^{2}\xi _{nl}^{2}}{2\mu r_{0}^{2}}}}
薛丁格方程式的整個解答是
ψ
n
l
m
(
r
,
θ
,
ϕ
)
=
A
n
l
j
l
(
ξ
n
l
r
/
r
0
)
Y
l
m
(
θ
,
ϕ
)
{\displaystyle \psi _{nlm}(r,\ \theta ,\ \phi )=A_{nl}j_{l}(\xi _{nl}\,r/r_{0})\,Y_{lm}(\theta ,\ \phi )}
其中，歸一常數
A
n
l
=
(
2
r
0
3
)
1
/
2
1
j
l
+
1
(
ξ
n
l
)
{\displaystyle A_{nl}=\left({\frac {2}{r_{0}^{3}}}\right)^{1/2}{\frac {1}{j_{l+1}(\xi _{nl})}}}
波函數的角部分已經歸一化，剩下來必須將徑向部分歸一化。徑向函數的歸一化條件為
1
=
A
n
l
2
∫
0
r
0
r
2
j
l
2
(
k
n
l
r
)
d
r
{\displaystyle 1=A_{nl}^{2}\int _{0}^{r_{0}}\ r^{2}j_{l}^{2}(k_{nl}r)\ dr}
將球貝塞爾函數與第一類貝塞爾函數的關係方程式(4)代入積分：
1
=
A
n
l
2
∫
0
r
0
r
2
π
2
k
n
l
r
J
l
+
1
/
2
2
(
k
n
l
r
)
d
r
=
A
n
l
2
π
2
k
n
l
∫
0
r
0
r
J
l
+
1
/
2
2
(
k
n
l
r
)
d
r
{\displaystyle 1=A_{nl}^{2}\int _{0}^{r_{0}}\ r^{2}\ {\frac {\pi }{2k_{nl}r}}\ J_{l+1/2}^{2}(k_{nl}r)\ dr=A_{nl}^{2}{\frac {\pi }{2k_{nl}}}\int _{0}^{r_{0}}\ rJ_{l+1/2}^{2}(k_{nl}r)\ dr}
設定變數
x
=
r
/
r
0
{\displaystyle x=r/r_{0}}
1
=
A
n
l
2
π
r
0
2
2
k
n
l
∫
0
1
x
J
l
+
1
/
2
2
(
k
n
l
r
0
x
)
d
x
=
A
n
l
2
π
r
0
3
2
ξ
n
l
∫
0
1
x
J
l
+
1
/
2
2
(
ξ
n
l
x
)
d
x
{\displaystyle 1=A_{nl}^{2}{\frac {\pi r_{0}^{2}}{2k_{nl}}}\int _{0}^{1}\ xJ_{l+1/2}^{2}(k_{nl}r_{0}x)\ dx=A_{nl}^{2}{\frac {\pi r_{0}^{3}}{2\xi _{nl}}}\int _{0}^{1}\ xJ_{l+1/2}^{2}(\xi _{nl}x)\ dx}
根據貝塞爾函數的正交歸一性 方程式 ，
∫
0
1
x
J
α
(
x
ξ
m
α
)
J
α
(
x
ξ
n
α
)
d
x
=
δ
m
n
2
J
α
+
1
(
ξ
n
α
)
2
{\displaystyle \int _{0}^{1}xJ_{\alpha }(x\xi _{m\alpha })J_{\alpha }(x\xi _{n\alpha })dx={\frac {\delta _{mn}}{2}}J_{\alpha +1}(\xi _{n\alpha })^{2}}
其中，
α
>
−
1
{\displaystyle \alpha >-1}
δ
m
n
{\displaystyle \delta _{mn}}
克罗内克δ ，
ξ
n
α
{\displaystyle \xi _{n\alpha }}
J
α
(
x
)
{\displaystyle J_{\alpha }(x)}
n
{\displaystyle n}
注意到
j
l
(
x
)
{\displaystyle j_{l}(x)}
n
{\displaystyle n}
ξ
n
l
{\displaystyle \xi _{nl}}
J
l
+
1
/
2
(
x
)
{\displaystyle J_{l+1/2}(x)}
n
{\displaystyle n}
1
=
A
n
l
2
π
r
0
3
4
ξ
n
l
J
l
+
3
/
2
2
(
ξ
n
l
)
=
A
n
l
2
r
0
3
2
j
l
+
1
2
(
ξ
n
l
)
{\displaystyle 1=A_{nl}^{2}\ {\frac {\pi r_{0}^{3}}{4\xi _{nl}}}\ J_{l+3/2}^{2}(\xi _{nl})=A_{nl}^{2}\ {\frac {r_{0}^{3}}{2}}\ j_{l+1}^{2}(\xi _{nl})}
取平方根，歸一常數
A
n
l
=
(
2
r
0
3
)
1
/
2
1
j
l
+
1
(
ξ
n
l
)
{\displaystyle A_{nl}=\left({\frac {2}{r_{0}^{3}}}\right)^{1/2}{\frac {1}{j_{l+1}(\xi _{nl})}}}
三維均向諧振子 的位勢為
V
(
r
)
=
1
2
μ
ω
2
r
2
{\displaystyle V(r)={\tfrac {1}{2}}\mu \omega ^{2}r^{2}}
其中，
ω
{\displaystyle \omega }
角頻率 。
用階梯算符 的方法，可以證明N維諧振子的能量是
E
n
=
ℏ
ω
(
n
+
N
2
)
with
n
=
0
,
1
,
…
,
∞
,
{\displaystyle E_{n}=\hbar \omega (n+{\tfrac {N}{2}})\quad {\hbox{with}}\quad n=0,1,\ldots ,\infty ,}
所以，三維均向諧振子的徑向薛丁格方程式是
[
−
ℏ
2
2
μ
d
2
d
r
2
+
ℏ
2
l
(
l
+
1
)
2
μ
r
2
+
1
2
μ
ω
2
r
2
−
ℏ
ω
(
n
+
3
2
)
]
u
(
r
)
=
0
{\displaystyle \left[-{\hbar ^{2} \over 2\mu }{d^{2} \over dr^{2}}+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}+{\frac {1}{2}}\mu \omega ^{2}r^{2}-\hbar \omega (n+{\frac {3}{2}})\right]u(r)=0}
(5) 設定常數
γ
{\displaystyle \gamma }
γ
≡
μ
ω
ℏ
{\displaystyle \gamma \equiv {\frac {\mu \omega }{\hbar }}}
回想
u
(
r
)
=
r
R
(
r
)
{\displaystyle u(r)=rR(r)}
歸一化 的解答：
R
n
l
(
r
)
=
N
n
l
r
l
e
−
1
2
γ
r
2
L
1
2
(
n
−
l
)
(
l
+
1
2
)
(
γ
r
2
)
{\displaystyle R_{nl}(r)=N_{nl}\,r^{l}\,e^{-{\frac {1}{2}}\gamma r^{2}}\;L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(\gamma r^{2})}
其中，函數
L
k
(
α
)
(
γ
r
2
)
{\displaystyle L_{k}^{(\alpha )}(\gamma r^{2})}
广义拉盖尔多项式 ，
N
n
l
{\displaystyle N_{nl}}
N
n
l
=
[
2
n
+
l
+
2
γ
l
+
3
2
π
1
2
]
1
2
[
[
1
2
(
n
−
l
)
]
!
[
1
2
(
n
+
l
)
]
!
(
n
+
l
+
1
)
!
]
1
2
{\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{\frac {3}{2}}}}{\pi ^{\frac {1}{2}}}}\right]^{\frac {1}{2}}\left[{\frac {[{\frac {1}{2}}(n-l)]!\;[{\frac {1}{2}}(n+l)]!}{(n+l+1)!}}\right]^{\frac {1}{2}}}
本徵能級
E
n
{\displaystyle E_{n}}
R
n
l
{\displaystyle R_{nl}}
Y
l
m
(
θ
,
ϕ
)
{\displaystyle Y_{lm}(\theta ,\phi )}
ψ
n
l
m
=
R
n
l
(
r
)
Y
l
m
(
θ
,
ϕ
)
{\displaystyle \psi _{nlm}=R_{nl}(r)\,Y_{lm}(\theta ,\ \phi )}
其中
l
=
n
,
n
−
2
,
…
,
l
m
i
n
{\displaystyle l=n,\ n-2,\ \ldots ,\ l_{\mathrm {min} }}
n
{\displaystyle n}
l
m
i
n
=
0
{\displaystyle l_{\mathrm {min} }=0}
l
m
i
n
=
1
{\displaystyle l_{\mathrm {min} }=1}
在這導引裏，徑向方程式會被轉換為广义拉盖尔微分方程式。這方程式的解是广义拉盖尔多项式。再將广义拉盖尔多项式歸一化以後，就是所要的答案。
首先，將徑向坐標無因次化 ，設定變數
y
=
γ
r
{\displaystyle y={\sqrt {\gamma }}r}
γ
≡
μ
ω
ℏ
{\displaystyle \gamma \equiv {\frac {\mu \omega }{\hbar }}}
[
d
2
d
y
2
−
l
(
l
+
1
)
y
2
−
y
2
+
2
n
−
3
]
v
(
y
)
=
0
{\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}-y^{2}+2n-3\right]v(y)=0}
(6) 其中，
v
(
y
)
=
u
(
y
/
γ
)
{\displaystyle v(y)=u\left(y/{\sqrt {\gamma }}\right)}
當
y
{\displaystyle y}
[
d
2
d
y
2
−
l
(
l
+
1
)
y
2
]
v
(
y
)
=
0
{\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}\right]v(y)=0}
所以，
v
(
y
)
{\displaystyle v(y)}
y
l
+
1
{\displaystyle y^{l+1}}
又當
y
{\displaystyle y}
[
d
2
d
y
2
−
y
2
]
v
(
y
)
=
0
{\displaystyle \left[{d^{2} \over dy^{2}}-y^{2}\right]v(y)=0}
因此，
v
(
y
)
{\displaystyle v(y)}
e
−
y
2
/
2
{\displaystyle e^{-y^{2}/2}}
為了除去
v
(
y
)
{\displaystyle v(y)}
v
(
y
)
{\displaystyle v(y)}
v
(
y
)
=
y
l
+
1
e
−
y
2
/
2
f
(
y
)
{\displaystyle v(y)=y^{l+1}e^{-y^{2}/2}f(y)}
經過一番運算，這個替換將微分方程式(6)轉換為
[
d
2
d
y
2
+
2
(
l
+
1
y
−
y
)
d
d
y
+
2
n
−
2
l
]
f
(
y
)
=
0
{\displaystyle \left[{d^{2} \over dy^{2}}+2\left({\frac {l+1}{y}}-y\right){\frac {d}{dy}}+2n-2l\right]f(y)=0}
(7)
設定變數
x
=
y
2
{\displaystyle x=y^{2}}
d
d
y
=
d
x
d
y
d
d
x
=
2
y
d
d
x
=
2
x
d
d
x
{\displaystyle {\frac {d}{dy}}={\frac {dx}{dy}}{\frac {d}{dx}}=2y{\frac {d}{dx}}=2{\sqrt {x}}{\frac {d}{dx}}}
d
2
d
y
2
=
d
d
y
(
2
y
d
d
x
)
=
4
x
d
2
d
x
2
+
2
d
d
x
{\displaystyle {\frac {d^{2}}{dy^{2}}}={\frac {d}{dy}}\left(2y{\frac {d}{dx}}\right)=4x{\frac {d^{2}}{dx^{2}}}+2{\frac {d}{dx}}}
代入方程式(7)，就可得到广义拉盖尔方程式：
x
d
2
g
d
x
2
+
(
(
l
+
1
2
)
+
1
−
x
)
d
g
d
x
+
1
2
(
n
−
l
)
g
(
x
)
=
0
{\displaystyle x{\frac {d^{2}g}{dx^{2}}}+{\Big (}(l+{\tfrac {1}{2}})+1-x{\Big )}{\frac {dg}{dx}}+{\tfrac {1}{2}}(n-l)g(x)=0}
其中，函數
g
(
x
)
≡
f
(
x
)
{\displaystyle g(x)\equiv f({\sqrt {x}})}
假若，
k
≡
(
n
−
l
)
/
2
{\displaystyle k\equiv (n-l)/2}
g
(
x
)
=
L
k
(
l
+
1
2
)
(
x
)
{\displaystyle g(x)=L_{k}^{(l+{\frac {1}{2}})}(x)}
因為
k
{\displaystyle k}
n
≥
l
{\displaystyle n\geq l}
n
{\displaystyle n}
l
{\displaystyle l}
l
{\displaystyle l}
回憶到
u
(
r
)
=
r
R
(
r
)
{\displaystyle u(r)=rR(r)}
R
n
l
(
r
)
=
N
n
l
r
l
e
−
1
2
γ
r
2
L
1
2
(
n
−
l
)
(
l
+
1
2
)
(
γ
r
2
)
{\displaystyle R_{nl}(r)=N_{nl}\,r^{l}\,e^{-{\frac {1}{2}}\gamma r^{2}}\;L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(\gamma r^{2})}
其中，
N
n
l
{\displaystyle N_{nl}}
R
n
l
(
r
)
{\displaystyle R_{nl}(r)}
∫
0
∞
r
2
|
R
n
l
(
r
)
|
2
d
r
=
1
{\displaystyle \int _{0}^{\infty }r^{2}|R_{nl}(r)|^{2}\,dr=1}
設定
q
=
γ
r
2
{\displaystyle q=\gamma r^{2}}
R
n
l
{\displaystyle R_{nl}}
q
{\displaystyle q}
N
n
l
2
2
γ
l
+
3
2
∫
0
∞
q
l
+
1
2
e
−
q
[
L
1
2
(
n
−
l
)
(
l
+
1
2
)
(
q
)
]
2
d
q
=
1
{\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\int _{0}^{\infty }q^{l+{1 \over 2}}e^{-q}\left[L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(q)\right]^{2}\,dq=1}
應用广义拉盖尔多项式的正交歸一性 ，這方程式簡化為
N
n
l
2
2
γ
l
+
3
2
⋅
Γ
[
1
2
(
n
+
l
+
1
)
+
1
]
[
1
2
(
n
−
l
)
]
!
=
1
{\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\cdot {\frac {\Gamma [{\frac {1}{2}}(n+l+1)+1]}{[{\frac {1}{2}}(n-l)]!}}=1}
因此，歸一常數可以表達為
N
n
l
=
2
γ
l
+
3
2
(
n
−
l
2
)
!
Γ
(
n
+
l
2
+
3
2
)
{\displaystyle N_{nl}={\sqrt {\frac {2\,\gamma ^{l+{3 \over 2}}\,({\frac {n-l}{2}})!}{\Gamma ({\frac {n+l}{2}}+{\frac {3}{2}})}}}}
應用伽瑪函數 的數學特性，同時注意
n
{\displaystyle n}
l
{\displaystyle l}
Γ
[
1
2
+
(
n
+
l
2
+
1
)
]
=
π
(
n
+
l
+
1
)
!
!
2
n
+
l
2
+
1
=
π
(
n
+
l
+
1
)
!
2
n
+
l
+
1
[
1
2
(
n
+
l
)
]
!
{\displaystyle \Gamma \left[{1 \over 2}+\left({\frac {n+l}{2}}+1\right)\right]={\frac {{\sqrt {\pi }}(n+l+1)!!}{2^{{\frac {n+l}{2}}+1}}}={\frac {{\sqrt {\pi }}(n+l+1)!}{2^{n+l+1}[{\frac {1}{2}}(n+l)]!}}}
在這裏用到了雙階乘 (double factorial )的定義。
所以，歸一常數等於
N
n
l
=
[
2
n
+
l
+
2
γ
l
+
3
2
[
1
2
(
n
−
l
)
]
!
[
1
2
(
n
+
l
)
]
!
π
1
2
(
n
+
l
+
1
)
!
]
1
2
{\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{3 \over 2}}\,[{1 \over 2}(n-l)]!\;[{1 \over 2}(n+l)]!}{\;\pi ^{1 \over 2}(n+l+1)!}}\right]^{1 \over 2}}
類氫原子只含有一個原子核 與一個電子 ，是個簡單的二體系統 。兩個物體之間，互相作用的位勢遵守庫侖定律 ：
V
(
r
)
=
−
1
4
π
ϵ
0
Z
e
2
r
{\displaystyle V(r)=-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r}}}
其中，
ϵ
0
{\displaystyle \epsilon _{0}}
真空電容率 ，
Z
{\displaystyle Z}
原子序 ，
e
{\displaystyle e}
單位電荷量 ，
r
{\displaystyle r}
原子核 的徑向距離。
將位勢代入方程式(1)，
{
−
ℏ
2
2
μ
r
2
d
d
r
(
r
2
d
d
r
)
+
ℏ
2
l
(
l
+
1
)
2
μ
r
2
−
1
4
π
ϵ
0
Z
e
2
r
}
R
(
r
)
=
E
R
(
r
)
{\displaystyle \left\{-{\hbar ^{2} \over 2\mu r^{2}}{d \over dr}\left(r^{2}{d \over dr}\right)+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r}}\right\}R(r)=ER(r)}
這方程式的解答是
R
n
l
(
r
)
=
(
2
Z
n
a
μ
)
3
(
n
−
l
−
1
)
!
2
n
[
(
n
+
l
)
!
]
3
e
−
Z
r
/
n
a
μ
(
2
Z
r
n
a
μ
)
l
L
n
−
l
−
1
2
l
+
1
(
2
Z
r
n
a
μ
)
{\displaystyle R_{nl}(r)={\sqrt {{\left({\frac {2Z}{na_{\mu }}}\right)}^{3}{\frac {(n-l-1)!}{2n[(n+l)!]^{3}}}}}e^{-Zr/{na_{\mu }}}\left({\frac {2Zr}{na_{\mu }}}\right)^{l}L_{n-l-1}^{2l+1}\left({\frac {2Zr}{na_{\mu }}}\right)}
其中，
a
μ
=
4
π
ε
0
ℏ
2
μ
e
2
{\displaystyle a_{\mu }={{4\pi \varepsilon _{0}\hbar ^{2}} \over {\mu e^{2}}}}
a
μ
{\displaystyle a_{\mu }}
波耳半徑
a
0
{\displaystyle a_{0}}
a
μ
=
a
0
{\displaystyle a_{\mu }=a_{0}}
μ
=
m
e
{\displaystyle \mu =m_{e}}
L
n
−
l
−
1
2
l
+
1
{\displaystyle L_{n-l-1}^{2l+1}}
[ 1]
L
i
j
(
x
)
=
(
−
1
)
j
d
j
d
x
j
L
i
+
j
(
x
)
{\displaystyle L_{i}^{j}(x)=(-1)^{j}\ {\frac {d^{j}}{dx^{j}}}L_{i+j}(x)}
其中，
L
i
+
j
(
x
)
{\displaystyle L_{i+j}(x)}
拉盖尔多项式 ，可用羅德里格公式表示為
L
i
(
x
)
=
e
x
i
!
d
i
d
x
i
(
x
i
e
−
x
)
{\displaystyle L_{i}(x)={\frac {e^{x}}{i!}}\ {\frac {d^{i}}{dx^{i}}}(x^{i}e^{-x})}
為了滿足
R
n
l
(
r
)
{\displaystyle R_{nl}(r)}
n
{\displaystyle n}
能級
E
n
=
−
(
Z
2
μ
e
4
32
π
2
ϵ
0
2
ℏ
2
)
1
n
2
=
−
13.6
Z
2
n
2
(
e
V
)
{\displaystyle E_{n}=-\left({\frac {Z^{2}\mu e^{4}}{32\pi ^{2}\epsilon _{0}^{2}\hbar ^{2}}}\right){\frac {1}{n^{2}}}={\frac {-13.6Z^{2}}{n^{2}}}\ (eV)}
R
n
l
(
r
)
{\displaystyle R_{nl}(r)}
Y
l
m
{\displaystyle Y_{lm}}
遞迴關係 ，必須要求
l
<
n
{\displaystyle l<n}
知道徑向函數
R
n
l
(
r
)
{\displaystyle R_{nl}(r)}
Y
l
m
{\displaystyle Y_{lm}}
ψ
n
l
m
=
R
n
l
(
r
)
Y
l
m
(
θ
,
ϕ
)
{\displaystyle \psi _{nlm}=R_{nl}(r)\,Y_{lm}(\theta ,\phi )}
為了要簡化薛丁格方程式，設定能量與長度的原子單位 (atomic unit )
E
h
=
m
e
(
e
2
4
π
ε
0
ℏ
)
2
{\displaystyle E_{\textrm {h}}=m_{\textrm {e}}\left({\frac {e^{2}}{4\pi \varepsilon _{0}\hbar }}\right)^{2}}
a
0
=
4
π
ε
0
ℏ
2
m
e
e
2
{\displaystyle a_{0}={{4\pi \varepsilon _{0}\hbar ^{2}} \over {m_{\textrm {e}}e^{2}}}}
將變數
y
=
Z
r
/
a
0
{\displaystyle y=Zr/a_{0}}
W
=
E
/
(
Z
2
E
h
)
{\displaystyle W=E/(Z^{2}E_{\textrm {h}})}
[
−
1
2
d
2
d
y
2
+
1
2
l
(
l
+
1
)
y
2
−
1
y
]
u
l
=
W
u
l
{\displaystyle \left[-{\frac {1}{2}}{\frac {d^{2}}{dy^{2}}}+{\frac {1}{2}}{\frac {l(l+1)}{y^{2}}}-{\frac {1}{y}}\right]u_{l}=Wu_{l}}
(8) 這方程式有兩類解答：
W
<
0
{\displaystyle W<0}
束縛態 ，其本徵函數是平方可積函數 。量子化的
W
{\displaystyle W}
W
≥
0
{\displaystyle W\geq 0}
散射態 ，其本徵函數不是平方可積函數。這條目只講述第(1)類解答。設定正實數
α
≡
2
−
2
W
{\displaystyle \alpha \equiv 2{\sqrt {-2W}}}
x
≡
α
y
{\displaystyle x\equiv \alpha y}
[
d
2
d
x
2
−
l
(
l
+
1
)
x
2
+
2
α
x
−
1
4
]
u
l
=
0
{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}+{\frac {2}{\alpha x}}-{\frac {1}{4}}\right]u_{l}=0}
(9) 當
x
{\displaystyle x}
[
d
2
d
x
2
−
l
(
l
+
1
)
x
2
]
u
l
=
0
{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}\right]u_{l}=0}
所以，
u
l
(
x
)
{\displaystyle u_{l}(x)}
x
l
+
1
{\displaystyle x^{l+1}}
又當
x
{\displaystyle x}
[
d
2
d
x
2
−
1
4
]
u
l
=
0
{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {1}{4}}\right]u_{l}=0}
因此，
u
l
(
x
)
{\displaystyle u_{l}(x)}
e
−
x
/
2
{\displaystyle e^{-x/2}}
為了除去
u
l
(
x
)
{\displaystyle u_{l}(x)}
u
l
(
x
)
{\displaystyle u_{l}(x)}
u
l
(
x
)
=
x
l
+
1
e
−
x
/
2
f
l
(
x
)
{\displaystyle u_{l}(x)=x^{l+1}e^{-x/2}f_{l}(x)}
經過一番運算，得到
f
l
(
x
)
{\displaystyle f_{l}(x)}
[
x
d
2
d
x
2
+
(
2
l
+
2
−
x
)
d
d
x
+
(
ν
−
l
−
1
)
]
f
l
(
x
)
=
0
{\displaystyle \left[x{\frac {d^{2}}{dx^{2}}}+(2l+2-x){\frac {d}{dx}}+(\nu -l-1)\right]f_{l}(x)=0}
其中，
ν
=
(
−
2
W
)
−
1
2
{\displaystyle \nu =(-2W)^{-{\frac {1}{2}}}}
假若，
ν
−
l
−
1
{\displaystyle \nu -l-1}
k
{\displaystyle k}
L
k
(
2
l
+
1
)
(
x
)
,
k
=
0
,
1
,
…
{\displaystyle L_{k}^{(2l+1)}(x),\qquad k=0,1,\ldots }
採用Abramowitz and Stegun的慣例[ 1]
W
=
−
1
2
n
2
{\displaystyle W=-{\frac {1}{2n^{2}}}}
其中，主量子數
n
≡
k
+
l
+
1
{\displaystyle n\equiv k+l+1}
n
≥
l
+
1
{\displaystyle n\geq l+1}
l
≤
n
−
1
{\displaystyle l\leq n-1}
由於
α
=
2
/
n
{\displaystyle \alpha =2/n}
R
n
l
(
r
)
=
(
2
Z
n
a
0
)
3
⋅
(
n
−
l
−
1
)
!
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{\displaystyle R_{nl}(r)={\sqrt {\left({\frac {2Z}{na_{0}}}\right)^{3}\cdot {\frac {(n-l-1)!}{2n[(n+l)!]^{3}}}}}\;e^{-{\textstyle {\frac {Zr}{na_{0}}}}}\left({\frac {2Zr}{na_{0}}}\right)^{l}\;L_{n-l-1}^{2l+1}\left({\frac {2Zr}{na_{0}}}\right)}
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{\displaystyle E=-{\frac {Z^{2}}{2n^{2}}}E_{\textrm {h}}=-{\frac {Z^{2}}{2n^{2}}}m_{\textrm {e}}\left({\frac {e^{2}}{4\pi \varepsilon _{0}\hbar }}\right)^{2},\qquad n=1,2,\ldots }
^ 1.0 1.1 Abramowitz, Milton; Stegun, Irene A. (编), Chapter 22, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, New York: Dover, 1965, ISBN 0-486-61272-4
Griffiths, David J. Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. 2004. ISBN 0-13-111892-7 . 
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![{\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\int _{0}^{\infty }q^{l+{1 \over 2}}e^{-q}\left[L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(q)\right]^{2}\,dq=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9590c999a4e6ab01d6df133dedcfc7d200fda39c)
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![{\displaystyle \Gamma \left[{1 \over 2}+\left({\frac {n+l}{2}}+1\right)\right]={\frac {{\sqrt {\pi }}(n+l+1)!!}{2^{{\frac {n+l}{2}}+1}}}={\frac {{\sqrt {\pi }}(n+l+1)!}{2^{n+l+1}[{\frac {1}{2}}(n+l)]!}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ee79f528d1188d4c0399013bd4cc68b36ebd1cc)
![{\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{3 \over 2}}\,[{1 \over 2}(n-l)]!\;[{1 \over 2}(n+l)]!}{\;\pi ^{1 \over 2}(n+l+1)!}}\right]^{1 \over 2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e7bd1badc5c764354f20b86156f3b28aac8fbd5)
![{\displaystyle R_{nl}(r)={\sqrt {{\left({\frac {2Z}{na_{\mu }}}\right)}^{3}{\frac {(n-l-1)!}{2n[(n+l)!]^{3}}}}}e^{-Zr/{na_{\mu }}}\left({\frac {2Zr}{na_{\mu }}}\right)^{l}L_{n-l-1}^{2l+1}\left({\frac {2Zr}{na_{\mu }}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/981d8108775d9c0ef8bddb13cd0707f818569d74)
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![{\displaystyle \left[x{\frac {d^{2}}{dx^{2}}}+(2l+2-x){\frac {d}{dx}}+(\nu -l-1)\right]f_{l}(x)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/500fd63f6f1dae7be1d2c3110203d840e6f4403f)
![{\displaystyle R_{nl}(r)={\sqrt {\left({\frac {2Z}{na_{0}}}\right)^{3}\cdot {\frac {(n-l-1)!}{2n[(n+l)!]^{3}}}}}\;e^{-{\textstyle {\frac {Zr}{na_{0}}}}}\left({\frac {2Zr}{na_{0}}}\right)^{l}\;L_{n-l-1}^{2l+1}\left({\frac {2Zr}{na_{0}}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bcdef141d55e1a3d3b5288e87aa3048f8b67849)
