# 球對稱位勢

${\displaystyle -{\frac {\hbar ^{2}}{2\mu }}\nabla ^{2}\psi +V(r)\psi =E\psi }$

## 薛丁格方程式

${\displaystyle -{\frac {\hbar ^{2}}{2\mu r^{2}}}\left\{{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)+{\frac {1}{\sin ^{2}\theta }}\left[\sin \theta {\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right]\right\}\psi +V(r)\psi =E\psi }$

${\displaystyle \psi (r,\ \theta ,\ \phi )=R(r)\Theta (\theta )\Phi (\phi )}$

${\displaystyle \psi (r,\ \theta ,\ \phi )=R(r)Y_{lm}(\theta ,\ \phi )}$

## 角部分解答

${\displaystyle -{\frac {1}{\sin ^{2}\theta }}\left[\sin \theta {\frac {\partial }{\partial \theta }}{\Big (}\sin \theta {\frac {\partial }{\partial \theta }}{\Big )}+{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right]Y_{lm}(\theta ,\phi )=l(l+1)Y_{lm}(\theta ,\phi )}$

${\displaystyle Y_{lm}(\theta ,\ \phi )=(i)^{m+|m|}{\sqrt {{(2l+1) \over 4\pi }{(l-|m|)! \over (l+|m|)!}}}\,P_{lm}(\cos {\theta })\,e^{im\phi }}$

${\displaystyle P_{lm}(x)=(1-x^{2})^{|m|/2}\ {\frac {d^{|m|}}{dx^{|m|}}}P_{l}(x)}$

${\displaystyle P_{l}(x)}$ ${\displaystyle l}$ 勒讓德多項式，可用羅德里格公式表示為

${\displaystyle P_{l}(x)={1 \over 2^{l}l!}{d^{l} \over dx^{l}}(x^{2}-1)^{l}}$

## 徑向部分解答

${\displaystyle \left\{-{\hbar ^{2} \over 2\mu r^{2}}{d \over dr}\left(r^{2}{d \over dr}\right)+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}+V(r)\right\}R(r)=ER(r)}$ (1)

${\displaystyle -{\hbar ^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}u(r)+V(r)u(r)=Eu(r)}$ (2)

${\displaystyle -{\hbar ^{2} \over 2\mu }{d^{2}u(r) \over dr^{2}}+V_{\mathrm {eff} }(r)u(r)=Eu(r)}$ (3)

## 實例

1. ${\displaystyle V(r)=0}$ ：原方程變為亥姆霍兹方程${\displaystyle (\nabla ^{2}+{\frac {2\mu E}{\hbar ^{2}}})A=0}$ ，使用球諧函數為正交歸一基，解析眞空狀況實例。這實例可以做為別的實例的基礎。
2. ${\displaystyle r 時，${\displaystyle V(r)=0}$ ；否則，${\displaystyle V(r)=\infty }$ ：這實例比第一個實例複雜一點，可以描述三維的圓球形盒子中的粒子的量子行為。
3. ${\displaystyle V(r)\propto r^{2}}$ ：研討三維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。
4. ${\displaystyle V(r)\propto 1/r}$ ：關於類氫原子束縛態的實例，也有簡單的解析解。

### 真空狀況實例

${\displaystyle \rho \ {\stackrel {\mathrm {def} }{=}}\ kr}$

${\displaystyle \rho ^{2}{d^{2}J \over d\rho ^{2}}+\rho {dJ \over d\rho }+\left[\rho ^{2}-\left(l+{\frac {1}{2}}\right)^{2}\right]J=0}$

（真空解的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）：

${\displaystyle R(r)=j_{l}(kr)\ {\stackrel {\mathrm {def} }{=}}\ {\sqrt {\pi /(2kr)}}J_{l+1/2}(kr)}$ (4)

${\displaystyle \psi (r,\ \theta ,\ \phi )=A_{kl}j_{l}(kr)Y_{lm}(\theta ,\phi )}$

#### 波函數歸一化導引

${\displaystyle 1=A_{kl}^{2}\int _{0}^{\infty }\ r^{2}j_{l}^{2}(kr)\ dr}$

${\displaystyle \int _{0}^{\infty }\ x^{2}j_{\alpha }(k_{1}x)j_{\alpha }(k_{2}x)\ dx={\frac {\pi }{2k_{1}^{2}}}\delta (k_{1}-k_{2})}$

### 球對稱的三維無限深方形位勢阱

${\displaystyle V(r)={\begin{cases}0,&{\mbox{if }}r\leq r_{0}\\\infty ,&{\mbox{if }}r>r_{0}\end{cases}}}$

${\displaystyle j_{l}(kr_{0})=0}$

${\displaystyle E_{nl}={\frac {\hbar ^{2}k_{nl}^{2}}{2\mu }}={\frac {\hbar ^{2}\xi _{nl}^{2}}{2\mu r_{0}^{2}}}}$

${\displaystyle \psi _{nlm}(r,\ \theta ,\ \phi )=A_{nl}j_{l}(\xi _{nl}\,r/r_{0})\,Y_{lm}(\theta ,\ \phi )}$

#### 波函數歸一化導引

${\displaystyle 1=A_{nl}^{2}\int _{0}^{r_{0}}\ r^{2}j_{l}^{2}(k_{nl}r)\ dr}$

${\displaystyle 1=A_{nl}^{2}\int _{0}^{r_{0}}\ r^{2}\ {\frac {\pi }{2k_{nl}r}}\ J_{l+1/2}^{2}(k_{nl}r)\ dr=A_{nl}^{2}{\frac {\pi }{2k_{nl}}}\int _{0}^{r_{0}}\ rJ_{l+1/2}^{2}(k_{nl}r)\ dr}$

${\displaystyle 1=A_{nl}^{2}{\frac {\pi r_{0}^{2}}{2k_{nl}}}\int _{0}^{1}\ xJ_{l+1/2}^{2}(k_{nl}r_{0}x)\ dx=A_{nl}^{2}{\frac {\pi r_{0}^{3}}{2\xi _{nl}}}\int _{0}^{1}\ xJ_{l+1/2}^{2}(\xi _{nl}x)\ dx}$

${\displaystyle \int _{0}^{1}xJ_{\alpha }(x\xi _{m\alpha })J_{\alpha }(x\xi _{n\alpha })dx={\frac {\delta _{mn}}{2}}J_{\alpha +1}(\xi _{n\alpha })^{2}}$

${\displaystyle 1=A_{nl}^{2}\ {\frac {\pi r_{0}^{3}}{4\xi _{nl}}}\ J_{l+3/2}^{2}(\xi _{nl})=A_{nl}^{2}\ {\frac {r_{0}^{3}}{2}}\ j_{l+1}^{2}(\xi _{nl})}$

### 三維均向諧振子

${\displaystyle V(r)={\tfrac {1}{2}}\mu \omega ^{2}r^{2}}$

${\displaystyle E_{n}=\hbar \omega (n+{\tfrac {N}{2}})\quad {\hbox{with}}\quad n=0,1,\ldots ,\infty ,}$

${\displaystyle \left[-{\hbar ^{2} \over 2\mu }{d^{2} \over dr^{2}}+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}+{\frac {1}{2}}\mu \omega ^{2}r^{2}-\hbar \omega (n+{\frac {3}{2}})\right]u(r)=0}$ (5)

${\displaystyle \gamma \equiv {\frac {\mu \omega }{\hbar }}}$

${\displaystyle R_{nl}(r)=N_{nl}\,r^{l}\,e^{-{\frac {1}{2}}\gamma r^{2}}\;L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(\gamma r^{2})}$

${\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{\frac {3}{2}}}}{\pi ^{\frac {1}{2}}}}\right]^{\frac {1}{2}}\left[{\frac {[{\frac {1}{2}}(n-l)]!\;[{\frac {1}{2}}(n+l)]!}{(n+l+1)!}}\right]^{\frac {1}{2}}}$

${\displaystyle \psi _{nlm}=R_{nl}(r)\,Y_{lm}(\theta ,\ \phi )}$

#### 導引

${\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}-y^{2}+2n-3\right]v(y)=0}$ (6)

${\displaystyle y}$ 接近0時，方程式(6)最顯著的項目是

${\displaystyle \left[{d^{2} \over dy^{2}}-{l(l+1) \over y^{2}}\right]v(y)=0}$

${\displaystyle \left[{d^{2} \over dy^{2}}-y^{2}\right]v(y)=0}$

${\displaystyle v(y)=y^{l+1}e^{-y^{2}/2}f(y)}$

${\displaystyle \left[{d^{2} \over dy^{2}}+2\left({\frac {l+1}{y}}-y\right){\frac {d}{dy}}+2n-2l\right]f(y)=0}$ (7)
##### 轉換為广义拉盖尔方程式

${\displaystyle {\frac {d}{dy}}={\frac {dx}{dy}}{\frac {d}{dx}}=2y{\frac {d}{dx}}=2{\sqrt {x}}{\frac {d}{dx}}}$
${\displaystyle {\frac {d^{2}}{dy^{2}}}={\frac {d}{dy}}\left(2y{\frac {d}{dx}}\right)=4x{\frac {d^{2}}{dx^{2}}}+2{\frac {d}{dx}}}$

${\displaystyle x{\frac {d^{2}g}{dx^{2}}}+{\Big (}(l+{\tfrac {1}{2}})+1-x{\Big )}{\frac {dg}{dx}}+{\tfrac {1}{2}}(n-l)g(x)=0}$

${\displaystyle g(x)=L_{k}^{(l+{\frac {1}{2}})}(x)}$

1. ${\displaystyle n\geq l}$
2. ${\displaystyle n}$ ${\displaystyle l}$ 同時為奇數或同時為偶數。這證明了前面所述${\displaystyle l}$ 必須遵守的條件。
##### 波函數歸一化

${\displaystyle R_{nl}(r)=N_{nl}\,r^{l}\,e^{-{\frac {1}{2}}\gamma r^{2}}\;L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(\gamma r^{2})}$

${\displaystyle R_{nl}(r)}$ 的歸一條件是

${\displaystyle \int _{0}^{\infty }r^{2}|R_{nl}(r)|^{2}\,dr=1}$

${\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\int _{0}^{\infty }q^{l+{1 \over 2}}e^{-q}\left[L_{{\frac {1}{2}}(n-l)}^{(l+{\frac {1}{2}})}(q)\right]^{2}\,dq=1}$

${\displaystyle {\frac {N_{nl}^{2}}{2\gamma ^{l+{3 \over 2}}}}\cdot {\frac {\Gamma [{\frac {1}{2}}(n+l+1)+1]}{[{\frac {1}{2}}(n-l)]!}}=1}$

${\displaystyle N_{nl}={\sqrt {\frac {2\,\gamma ^{l+{3 \over 2}}\,({\frac {n-l}{2}})!}{\Gamma ({\frac {n+l}{2}}+{\frac {3}{2}})}}}}$

${\displaystyle \Gamma \left[{1 \over 2}+\left({\frac {n+l}{2}}+1\right)\right]={\frac {{\sqrt {\pi }}(n+l+1)!!}{2^{{\frac {n+l}{2}}+1}}}={\frac {{\sqrt {\pi }}(n+l+1)!}{2^{n+l+1}[{\frac {1}{2}}(n+l)]!}}}$

${\displaystyle N_{nl}=\left[{\frac {2^{n+l+2}\,\gamma ^{l+{3 \over 2}}\,[{1 \over 2}(n-l)]!\;[{1 \over 2}(n+l)]!}{\;\pi ^{1 \over 2}(n+l+1)!}}\right]^{1 \over 2}}$

### 類氫原子

${\displaystyle V(r)=-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r}}}$

${\displaystyle \left\{-{\hbar ^{2} \over 2\mu r^{2}}{d \over dr}\left(r^{2}{d \over dr}\right)+{\hbar ^{2}l(l+1) \over 2\mu r^{2}}-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Ze^{2}}{r}}\right\}R(r)=ER(r)}$

${\displaystyle R_{nl}(r)={\sqrt {{\left({\frac {2Z}{na_{\mu }}}\right)}^{3}{\frac {(n-l-1)!}{2n[(n+l)!]^{3}}}}}e^{-Zr/{na_{\mu }}}\left({\frac {2Zr}{na_{\mu }}}\right)^{l}L_{n-l-1}^{2l+1}\left({\frac {2Zr}{na_{\mu }}}\right)}$

${\displaystyle L_{i}^{j}(x)=(-1)^{j}\ {\frac {d^{j}}{dx^{j}}}L_{i+j}(x)}$

${\displaystyle L_{i}(x)={\frac {e^{x}}{i!}}\ {\frac {d^{i}}{dx^{i}}}(x^{i}e^{-x})}$

${\displaystyle \psi _{nlm}=R_{nl}(r)\,Y_{lm}(\theta ,\phi )}$

#### 導引

${\displaystyle E_{\textrm {h}}=m_{\textrm {e}}\left({\frac {e^{2}}{4\pi \varepsilon _{0}\hbar }}\right)^{2}}$
${\displaystyle a_{0}={{4\pi \varepsilon _{0}\hbar ^{2}} \over {m_{\textrm {e}}e^{2}}}}$

${\displaystyle \left[-{\frac {1}{2}}{\frac {d^{2}}{dy^{2}}}+{\frac {1}{2}}{\frac {l(l+1)}{y^{2}}}-{\frac {1}{y}}\right]u_{l}=Wu_{l}}$ (8)

1. ${\displaystyle W<0}$ ：量子態是束縛態，其本徵函數是平方可積函數。量子化的${\displaystyle W}$ 造成了離散的能量譜。
2. ${\displaystyle W\geq 0}$ ：量子態是散射態，其本徵函數不是平方可積函數。

${\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}+{\frac {2}{\alpha x}}-{\frac {1}{4}}\right]u_{l}=0}$ (9)

${\displaystyle x}$ 接近0時，方程式(9)最顯著的項目是

${\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {l(l+1)}{x^{2}}}\right]u_{l}=0}$

${\displaystyle \left[{\frac {d^{2}}{dx^{2}}}-{\frac {1}{4}}\right]u_{l}=0}$

${\displaystyle u_{l}(x)=x^{l+1}e^{-x/2}f_{l}(x)}$

${\displaystyle \left[x{\frac {d^{2}}{dx^{2}}}+(2l+2-x){\frac {d}{dx}}+(\nu -l-1)\right]f_{l}(x)=0}$

${\displaystyle L_{k}^{(2l+1)}(x),\qquad k=0,1,\ldots }$

${\displaystyle W=-{\frac {1}{2n^{2}}}}$

${\displaystyle R_{nl}(r)={\sqrt {\left({\frac {2Z}{na_{0}}}\right)^{3}\cdot {\frac {(n-l-1)!}{2n[(n+l)!]^{3}}}}}\;e^{-{\textstyle {\frac {Zr}{na_{0}}}}}\left({\frac {2Zr}{na_{0}}}\right)^{l}\;L_{n-l-1}^{2l+1}\left({\frac {2Zr}{na_{0}}}\right)}$

${\displaystyle E=-{\frac {Z^{2}}{2n^{2}}}E_{\textrm {h}}=-{\frac {Z^{2}}{2n^{2}}}m_{\textrm {e}}\left({\frac {e^{2}}{4\pi \varepsilon _{0}\hbar }}\right)^{2},\qquad n=1,2,\ldots }$

## 參考文獻

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.