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 )}
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}}
克羅內克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})}}}
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})}
廣義拉蓋爾多項式 ，
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}
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 }
Abramowitz, Milton; Stegun, Irene A. (編), 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 . 
![{\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 }](http://wikimedia.org/api/rest_v1/media/math/render/svg/addb25894f6807c87280d4c9c5c643c0a45449c0)
![{\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 )}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7dd400bacbaaa42d613d3fd3156fda00fbc0ee0a)
![{\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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/5717b4612192caf988f58172c5146c8528ffcba3)
.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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/7229bd958ff4051b255c88d044243dd23c85b97d)
![{\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}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b6dfb09870848ad209de17555904f7700cddd503)
![{\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}-y^{2}+2n-3\right]v(y)=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b454519dcce7b44e0f32791ebd4333c2b6558ee9)
![{\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}\right]v(y)=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/02c4d9d5912b0a2a39b36de21cbe2464d65a52ed)
![{\displaystyle \left[{d^{2} \over dy^{2}}-y^{2}\right]v(y)=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/dfeac834e89a4ffcb40791654f8f509b8d088003)
![{\displaystyle \left[{d^{2} \over dy^{2}}+2\left({\frac {l+1}{y}}-y\right){\frac {d}{dy}}+2n-2l\right]f(y)=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1680cb62f9b06d7f8d6d2c043c4ceec7edc8cf0d)
![{\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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/9590c999a4e6ab01d6df133dedcfc7d200fda39c)
![{\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}](http://wikimedia.org/api/rest_v1/media/math/render/svg/65ddd9dc8f2455f15d2ed683a98bff6a82c99ea7)
![{\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)]!}}}](http://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}}](http://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)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/981d8108775d9c0ef8bddb13cd0707f818569d74)
![{\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}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/3ae2acd3604904a7511493735f7d1c22d4361160)
![{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}+{\frac {2}{\alpha x}}-{\frac {1}{4}}\right]u_{l}=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/56e2ccc36d6f7e7074d65815e0393de597e8918c)
![{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}\right]u_{l}=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/8113d3ef5a47acca2ee3e89837d7a6992308b9a0)
![{\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {1}{4}}\right]u_{l}=0}](http://wikimedia.org/api/rest_v1/media/math/render/svg/cb969f09a3da00398dcd128930d873e76a723e07)
![{\displaystyle \left[x{\frac {d^{2}}{dx^{2}}}+(2l+2-x){\frac {d}{dx}}+(\nu -l-1)\right]f_{l}(x)=0}](http://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)}](http://wikimedia.org/api/rest_v1/media/math/render/svg/2bcdef141d55e1a3d3b5288e87aa3048f8b67849)
