在這條目裏，我們會解析四個很特別，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\,\!}
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 )\,\!}
貝塞爾函數 ；而
R
(
r
)
{\displaystyle R(r)\,\!}
球貝塞爾函數 ：
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式的解，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，我們察覺，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})\,\!}
廣義拉格耳多項式 ，
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，我們會將徑向方程式轉換為廣義拉格耳微分方程式。這方程式的解是廣義拉格耳多項式。再將廣義拉格耳多項式歸一化以後，就是我們所要的答案。
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) ，就可得到廣義拉格耳方程式：
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\,\!}
應用廣義拉格耳多項式的正交歸一性 ，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)]!}}\,\!}
這裏，我們用到了雙階乘 (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)\,\!}
拉格耳多項式 ，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
)
!
2
n
[
(
n
+
l
)
!
]
3
e
−
Z
r
n
a
0
(
2
Z
r
n
a
0
)
l
L
n
−
l
−
1
2
l
+
1
(
2
Z
r
n
a
0
)
{\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)\,\!}
能量是
E
=
−
Z
2
2
n
2
E
h
=
−
Z
2
2
n
2
m
e
(
e
2
4
π
ε
0
ℏ
)
2
,
n
=
1
,
2
,
…
{\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 \,\!}
![{\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 \,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a53bc6ec3be92b1adf64f5a00ded060b8cf7f82)
![{\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 )\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e54685aa1da1217f6726c108fcbf67bbc921e623)
![{\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\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3c0d7006beae67364a022b559f7ee0926a6d8a1f)
.svg)
![{\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\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/87a39de17d0ca233769d631e80b90ab0fd4af534)
![{\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}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f67c10d90c1579d41a7b784a323c6824b9e092c8)
![{\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}-y^{2}+2n-3\right]v(y)=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6774f82512731bd15701246acc3bc008c60c0bc)
![{\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}\right]v(y)=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4b5fab77f404bdef90626d3c798810b6595cdc1)
![{\displaystyle \left[{d^{2} \over dy^{2}}-y^{2}\right]v(y)=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1787a7720c63d7bef6f883fccf50e811431d9ed4)
![{\displaystyle \left[{d^{2} \over dy^{2}}+2\left({\frac {l+1}{y}}-y\right){\frac {d}{dy}}+2n-2l\right]f(y)=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83cb68f2d3920c12c5e9a2155beeb62827b4c386)
![{\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/470b0d0bd604b102bd4f4ef9df0f0510e7cbded0)
![{\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\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69c95af76320e6f8d6c1e94cb84f7e11f82dba55)
![{\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/d4453239fc097f63016c77927fad4801db1946fd)
![{\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/717ec7a669a8e3f68b23cba2c975e7c8a28faf88)
![{\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/33d101ca8297d9993ad29918724efbba732a5da5)
![{\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}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f51518cfbd9c9401466d887a698022c3465e17df)
![{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}+{\frac {2}{\alpha x}}-{\frac {1}{4}}\right]u_{l}=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c0df8105c8286f35e0261fe9dc1c7f2138f17bb)
![{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}\right]u_{l}=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8fb90886cc376e262746560c8b867bcf224c8cd)
![{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {1}{4}}\right]u_{l}=0\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b73c49d9a959f9330a5d082f15ecf11e8d260328)
![{\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/d9af05b91cb0baa291dc8fec8c6b82fef04d12f4)
![{\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/cb731fc3df53b13a3c399c1cd21f12b835403aa8)
