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角動量算符

本页使用了标题或全文手工转换
维基百科,自由的百科全书
系列条目
量子力学
it|ψ(t)=H^|ψ(t){\displaystyle i\hbar {\frac {\partial }{\partial t}}|\psi (t)\rangle ={\hat {H}}|\psi (t)\rangle }

量子力學裏,角動量算符(英語:angular momentum operator)是一種算符,類比於經典的角動量。在原子物理學涉及旋轉對稱性rotational symmetry)的理論裏,角動量算符佔有中心的角色。角動量,動量,與能量是物體運動的三個基本特性[1]

簡介

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角動量促使在旋轉方面的運動得以數量化。在孤立系統裏,如同能量和動量,角動量是守恆的。在量子力學裏,角動量算符的概念是必要的,因為角動量的計算實現於描述量子系統的波函數,而不是經典地實現於一點或一剛體。在量子尺寸世界,分析的對象都是以波函數或量子幅來描述其機率性行為,而不是命定性(deterministic)行為。

數學定義

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經典力學裏,角動量L=(Lx, Ly, Lz){\displaystyle \mathbf {L} =(L_{x},\ L_{y},\ L_{z})\,\!} 定義為位置r=(x, y, z){\displaystyle \mathbf {r} =(x,\ y,\ z)\,\!} 與動量p=(px, py, pz){\displaystyle \mathbf {p} =(p_{x},\ p_{y},\ p_{z})\,\!}叉積

L =def r×p{\displaystyle \mathbf {L} \ {\stackrel {def}{=}}\ \mathbf {r} \times \mathbf {p} \,\!}

在量子力學裏,對應的角動量算符L^{\displaystyle {\hat {\mathbf {L} }}\,\!} 定義為位置算符r^{\displaystyle {\hat {\mathbf {r} }}\,\!}動量算符p^{\displaystyle {\hat {\mathbf {p} }}\,\!} 的叉積:

L^ =def r^×p^{\displaystyle {\hat {\mathbf {L} }}\ {\stackrel {def}{=}}\ {\hat {\mathbf {r} }}\times {\hat {\mathbf {p} }}\,\!}

由於動量算符的形式為

p^=i{\displaystyle {\hat {\mathbf {p} }}=-i\hbar \nabla \,\!}

角動量算符的形式為

L^=i(r^×){\displaystyle {\hat {\mathbf {L} }}=-i\hbar ({\hat {\mathbf {r} }}\times \nabla )\,\!}

其中,{\displaystyle \nabla \,\!}梯度算符。

角動量是厄米算符

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在量子力學裏,每一個可觀察量所對應的算符都是厄米算符。角動量是一個可觀察量,所以,角動量算符應該也是厄米算符。讓我們現在證明這一點,思考角動量算符的 x-分量L^x{\displaystyle {\hat {L}}_{x}\,\!}

L^x=y^p^zz^p^y{\displaystyle {\hat {L}}_{x}={\hat {y}}{\hat {p}}_{z}-{\hat {z}}{\hat {p}}_{y}\,\!}

伴隨算符Lx{\displaystyle L_{x}^{\dagger }\,\!}

L^x=(y^p^zz^p^y)=p^zy^p^y+z^{\displaystyle {\hat {L}}_{x}^{\dagger }=({\hat {y}}{\hat {p}}_{z}-{\hat {z}}{\hat {p}}_{y})^{\dagger }={\hat {p}}_{z}^{\dagger }{\hat {y}}^{\dagger }-{\hat {p}}_{y}^{+}{\hat {z}}^{\dagger }\,\!}

由於y^{\displaystyle {\hat {y}}\,\!}z^{\displaystyle {\hat {z}}\,\!}p^y{\displaystyle {\hat {p}}_{y}\,\!}p^z{\displaystyle {\hat {p}}_{z}\,\!} ,都是厄米算符,

L^x=p^zy^p^yz^{\displaystyle {\hat {L}}_{x}^{\dagger }={\hat {p}}_{z}{\hat {y}}-{\hat {p}}_{y}{\hat {z}}\,\!}

由於p^z{\displaystyle {\hat {p}}_{z}\,\!}y^{\displaystyle {\hat {y}}\,\!} 之間、p^y{\displaystyle {\hat {p}}_{y}\,\!}z^{\displaystyle {\hat {z}}\,\!} 之間分別相互對易,所以,

L^x=y^p^zz^p^y=L^x{\displaystyle {\hat {L}}_{x}^{\dagger }={\hat {y}}{\hat {p}}_{z}-{\hat {z}}{\hat {p}}_{y}={\hat {L}}_{x}\,\!}

因此,L^x{\displaystyle {\hat {L}}_{x}\,\!} 是一個厄米算符。類似地,L^y{\displaystyle {\hat {L}}_{y}\,\!}L^z{\displaystyle {\hat {L}}_{z}\,\!} 都是厄米算符。總結,角動量算符是厄米算符。

再思考L^2{\displaystyle {\hat {L}}^{2}\,\!} 算符,

L^2=L^x2+L^y2+L^z2{\displaystyle {\hat {L}}^{2}={\hat {L}}_{x}^{2}+{\hat {L}}_{y}^{2}+{\hat {L}}_{z}^{2}\,\!}

伴隨算符(L^2){\displaystyle ({\hat {L}}^{2})^{\dagger }\,\!}

(L^2)=(L^x2+L^y2+L^z2)=(L^x2)+(L^y2)+(L^z2){\displaystyle ({\hat {L}}^{2})^{\dagger }=({\hat {L}}_{x}^{2}+{\hat {L}}_{y}^{2}+{\hat {L}}_{z}^{2})^{\dagger }=({\hat {L}}_{x}^{2})^{\dagger }+({\hat {L}}_{y}^{2})^{\dagger }+({\hat {L}}_{z}^{2})^{\dagger }\,\!}

由於L^x2{\displaystyle {\hat {L}}_{x}^{2}\,\!} 算符、L^y2{\displaystyle {\hat {L}}_{y}^{2}\,\!} 算符、L^z2{\displaystyle {\hat {L}}_{z}^{2}\,\!} 算符,都是厄米算符,

(L^2)=L^x2+L^y2+L^z2=L^2{\displaystyle ({\hat {L}}^{2})^{\dagger }={\hat {L}}_{x}^{2}+{\hat {L}}_{y}^{2}+{\hat {L}}_{z}^{2}={\hat {L}}^{2}\,\!}

所以,L^2{\displaystyle {\hat {L}}^{2}\,\!} 算符是厄米算符。

對易關係

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主条目:角动量算符对易关系

兩個算符A^{\displaystyle {\hat {A}}\,\!}B^{\displaystyle {\hat {B}}\,\!}交換算符[A^, B^]{\displaystyle [{\hat {A}},\ {\hat {B}}]\,\!} ,表示出它們之間的對易關係

角動量算符與自己的對易關係

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思考L^x{\displaystyle {\hat {L}}_{x}\,\!}L^y{\displaystyle {\hat {L}}_{y}\,\!}交換算符

[L^x, L^y]=[y^p^zz^p^y, z^p^xx^p^z]=[y^p^z, z^p^x][z^p^y, z^p^x][y^p^z, x^p^z]+[z^p^y, x^p^z]=i(x^p^yy^p^x)=iL^z{\displaystyle {\begin{aligned}\left.\right.[{\hat {L}}_{x},\ {\hat {L}}_{y}]&=[{\hat {y}}{\hat {p}}_{z}-{\hat {z}}{\hat {p}}_{y},\ {\hat {z}}{\hat {p}}_{x}-{\hat {x}}{\hat {p}}_{z}]\\&=[{\hat {y}}{\hat {p}}_{z},\ {\hat {z}}{\hat {p}}_{x}]-[{\hat {z}}{\hat {p}}_{y},\ {\hat {z}}{\hat {p}}_{x}]-[{\hat {y}}{\hat {p}}_{z},\ {\hat {x}}{\hat {p}}_{z}]+[{\hat {z}}{\hat {p}}_{y},\ {\hat {x}}{\hat {p}}_{z}]\\&=i\hbar ({\hat {x}}{\hat {p}}_{y}-{\hat {y}}{\hat {p}}_{x})\\&=i\hbar {\hat {L}}_{z}\\\end{aligned}}\,\!}

由於兩者的對易關係不等於 0 ,Lx{\displaystyle L_{x}\,\!}Ly{\displaystyle L_{y}\,\!} 彼此是不相容可觀察量L^x{\displaystyle {\hat {L}}_{x}\,\!}L^y{\displaystyle {\hat {L}}_{y}\,\!} 絕對不會有共同的基底量子態。一般而言,L^x{\displaystyle {\hat {L}}_{x}\,\!}本徵態L^y{\displaystyle {\hat {L}}_{y}\,\!} 的本徵態不同。

給予一個量子系統,量子態為|ψ{\displaystyle |\psi \rangle \,\!} 。對於可觀察量算符L^x{\displaystyle {\hat {L}}_{x}\,\!} ,所有本徵值為xi{\displaystyle \ell _{xi}\,\!} 的本徵態|fi,i=1, 2, 3, {\displaystyle |f_{i}\rangle ,\quad i=1,\ 2,\ 3,\ \cdots \,\!} ,形成了一組基底量子態。量子態|ψ{\displaystyle |\psi \rangle \,\!} 可以表達為這基底量子態的線性組合|ψ=i |fifi|ψ{\displaystyle |\psi \rangle =\sum _{i}\ |f_{i}\rangle \langle f_{i}|\psi \rangle \,\!} 。對於可觀察量算符L^y{\displaystyle {\hat {L}}_{y}\,\!} ,所有本徵值為yi{\displaystyle \ell _{yi}\,\!} 的本徵態|gi,i=1, 2, 3, {\displaystyle |g_{i}\rangle ,\quad i=1,\ 2,\ 3,\ \cdots \,\!} ,形成了另外一組基底量子態。量子態|ψ{\displaystyle |\psi \rangle \,\!} 可以表達為這基底量子態的線性組合:|ψ=i |gigi|ψ{\displaystyle |\psi \rangle =\sum _{i}\ |g_{i}\rangle \langle g_{i}|\psi \rangle \,\!}

根據哥本哈根詮釋量子測量可以用量子態塌縮機制來詮釋。假若,我們測量可觀察量Lx{\displaystyle L_{x}\,\!} ,得到的測量值為其本徵值xi{\displaystyle \ell _{xi}\,\!} ,則量子態機率塌縮為本徵態|fi{\displaystyle |f_{i}\rangle \,\!} 。假若,我們立刻再測量可觀察量Lx{\displaystyle L_{x}\,\!} ,得到的答案必定是xi{\displaystyle \ell _{xi}\,\!} ,量子態仍舊處於|fi{\displaystyle |f_{i}\rangle \,\!} 。可是,假若,我們改為測量可觀察量Ly{\displaystyle L_{y}\,\!} ,則量子態不會停留於本徵態|fi{\displaystyle |f_{i}\rangle \,\!} ,而會塌縮為L^y{\displaystyle {\hat {L}}_{y}\,\!} 的本徵態。假若,得到的測量值為其本徵值yj{\displaystyle \ell _{yj}\,\!} ,則量子態機率塌縮為本徵態|gj{\displaystyle |g_{j}\rangle \,\!}

根據不確定性原理

ΔLx ΔLy|[L^x, L^y]2i|=|L^z|2{\displaystyle \Delta L_{x}\ \Delta L_{y}\geq \left|{\frac {\langle [{\hat {L}}_{x},\ {\hat {L}}_{y}]\rangle }{2i}}\right|={\frac {\hbar |\langle {\hat {L}}_{z}\rangle |}{2}}\,\!}

Lx{\displaystyle L_{x}\,\!} 的不確定性與Ly{\displaystyle L_{y}\,\!} 的不確定性的乘積ΔLx ΔLy{\displaystyle \Delta L_{x}\ \Delta L_{y}\,\!} ,必定大於或等於|Lz|2{\displaystyle {\frac {\hbar |\langle L_{z}\rangle |}{2}}\,\!}

Lx{\displaystyle L_{x}\,\!}Lz{\displaystyle L_{z}\,\!} 之間,Ly{\displaystyle L_{y}\,\!}Lz{\displaystyle L_{z}\,\!} 之間,也有類似的特性。

角動量平方算符與角動量算符之間的對易關係

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思考L^2{\displaystyle {\hat {L}}^{2}\,\!}L^z{\displaystyle {\hat {L}}_{z}\,\!} 的交換算符,

[L^2, L^z]=[L^x2+L^y2+L^z2, L^z]=L^xL^xL^zL^zL^xL^x+L^yL^yL^zL^zL^yL^y=L^x(L^zL^xiL^y)(L^xL^z+iL^y)L^x+L^y(L^zL^y+iL^x)(L^yL^ziL^x)L^y=0{\displaystyle {\begin{aligned}\left.\right.[{\hat {L}}^{2},\ {\hat {L}}_{z}]&=[{\hat {L}}_{x}^{2}+{\hat {L}}_{y}^{2}+{\hat {L}}_{z}^{2},\ {\hat {L}}_{z}]\\&={\hat {L}}_{x}{\hat {L}}_{x}{\hat {L}}_{z}-{\hat {L}}_{z}{\hat {L}}_{x}{\hat {L}}_{x}+{\hat {L}}_{y}{\hat {L}}_{y}{\hat {L}}_{z}-{\hat {L}}_{z}{\hat {L}}_{y}{\hat {L}}_{y}\\&={\hat {L}}_{x}({\hat {L}}_{z}{\hat {L}}_{x}-i\hbar {\hat {L}}_{y})-({\hat {L}}_{x}{\hat {L}}_{z}+i\hbar {\hat {L}}_{y}){\hat {L}}_{x}+{\hat {L}}_{y}({\hat {L}}_{z}{\hat {L}}_{y}+i\hbar {\hat {L}}_{x})-({\hat {L}}_{y}{\hat {L}}_{z}-i\hbar {\hat {L}}_{x}){\hat {L}}_{y}\\&=0\\\end{aligned}}\,\!}

L^2{\displaystyle {\hat {L}}^{2}\,\!}L^z{\displaystyle {\hat {L}}_{z}\,\!}對易的L2{\displaystyle L^{2}\,\!}Lz{\displaystyle L_{z}\,\!} 彼此是相容可觀察量,兩個算符有共同的本徵態。根據不確定性原理,我們可以同時地測量到L2{\displaystyle L^{2}\,\!}Lz{\displaystyle L_{z}\,\!} 的本徵值。

類似地,

[L^2, L^x]=0{\displaystyle [{\hat {L}}^{2},\ {\hat {L}}_{x}]=0\,\!}
[L^2, L^y]=0{\displaystyle [{\hat {L}}^{2},\ {\hat {L}}_{y}]=0\,\!}

L^2{\displaystyle {\hat {L}}^{2}\,\!}L^x{\displaystyle {\hat {L}}_{x}\,\!} 之間、L^2{\displaystyle {\hat {L}}^{2}\,\!}L^y{\displaystyle {\hat {L}}_{y}\,\!} 之間,都分別擁有類似的物理特性。

在經典力學裏的對易關係

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在經典力學裏,角動量算符也遵守類似的對易關係:

{Li, Lj}=ϵijkLk{\displaystyle \{L_{i},\ L_{j}\}=\epsilon _{ijk}L_{k}\,\!}

其中,{ , }{\displaystyle \{\ ,\ \}\,\!}帕松括號ϵijk{\displaystyle \epsilon _{ijk}\,\!}列維-奇維塔符號i{\displaystyle i\,\!}j{\displaystyle j\,\!}k{\displaystyle k\,\!} ,代表直角坐標(x, y, z){\displaystyle (x,\ y,\ z)\,\!}

本徵值與本徵函數

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採用球坐標。展開角動量算符的方程式:

L^=ir^×=irer×(err+eθ1rθ+eϕ1rsinθϕ)=i(eθ1sinθϕ+eϕθ){\displaystyle {\begin{aligned}{\hat {\mathbf {L} }}&={\frac {\hbar }{i}}{\hat {\mathbf {r} }}\times \nabla \\&={\frac {\hbar }{i}}r\mathbf {e} _{r}\times \left(\mathbf {e} _{r}{\frac {\partial }{\partial r}}+\mathbf {e} _{\theta }{\frac {1}{r}}{\frac {\partial }{\partial \theta }}+\mathbf {e} _{\phi }{\frac {1}{r\sin \theta }}{\frac {\partial }{\partial \phi }}\right)\\&={\frac {\hbar }{i}}\left(-\mathbf {e} _{\theta }{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \phi }}+\mathbf {e} _{\phi }{\frac {\partial }{\partial \theta }}\right)\\\end{aligned}}\,\!}

其中,er{\displaystyle \mathbf {e} _{r}\,\!}eθ{\displaystyle \mathbf {e} _{\theta }\,\!}eϕ{\displaystyle \mathbf {e} _{\phi }\,\!} ,分別為徑向單位向量、天頂角單位向量、與方位角單位向量。

轉換回直角坐標

L^=i[ex(sinϕθcotθcosϕϕ)+ey(cosϕθcotθsinϕϕ)+ezϕ]{\displaystyle {\hat {\mathbf {L} }}={\frac {\hbar }{i}}\left[\mathbf {e} _{x}\left(-\sin \phi {\frac {\partial }{\partial \theta }}-\cot \theta \cos \phi {\frac {\partial }{\partial \phi }}\right)+\mathbf {e} _{y}\left(\cos \phi {\frac {\partial }{\partial \theta }}-\cot \theta \sin \phi {\frac {\partial }{\partial \phi }}\right)+\mathbf {e} _{z}{\frac {\partial }{\partial \phi }}\right]\,\!}

其中,ex{\displaystyle \mathbf {e} _{x}\,\!}ey{\displaystyle \mathbf {e} _{y}\,\!}ez{\displaystyle \mathbf {e} _{z}\,\!} ,分別為 x-單位向量、y-單位向量、與 z-單位向量。

所以,L^x{\displaystyle {\hat {L}}_{x}\,\!}L^y{\displaystyle {\hat {L}}_{y}\,\!}L^z{\displaystyle {\hat {L}}_{z}\,\!} 分別是

L^x=i(sinϕθcotθcosϕϕ){\displaystyle {\hat {L}}_{x}={\frac {\hbar }{i}}\left(-\sin \phi {\frac {\partial }{\partial \theta }}-\cot \theta \cos \phi {\frac {\partial }{\partial \phi }}\right)\,\!}
L^y=i(cosϕθcotθsinϕϕ){\displaystyle {\hat {L}}_{y}={\frac {\hbar }{i}}\left(\cos \phi {\frac {\partial }{\partial \theta }}-\cot \theta \sin \phi {\frac {\partial }{\partial \phi }}\right)\,\!}
L^z=iϕ{\displaystyle {\hat {L}}_{z}={\frac {\hbar }{i}}{\frac {\partial }{\partial \phi }}\,\!}

角動量平方算符是

L^2=L^x2+L^y2+L^z2{\displaystyle {\hat {L}}^{2}={\hat {L}}_{x}^{2}+{\hat {L}}_{y}^{2}+{\hat {L}}_{z}^{2}\,\!}

其中,

L^x2=2(sinϕθcotθcosϕϕ)(sinϕθcotθcosϕϕ)=2(sin2ϕ2θ2+cotθcos2ϕθ+cotθsinϕcosϕ2θϕcsc2θsinϕcosϕϕ{\displaystyle {\begin{aligned}{\hat {L}}_{x}^{2}&=-\hbar ^{2}\left(-\sin \phi {\frac {\partial }{\partial \theta }}-\cot \theta \cos \phi {\frac {\partial }{\partial \phi }}\right)\left(-\sin \phi {\frac {\partial }{\partial \theta }}-\cot \theta \cos \phi {\frac {\partial }{\partial \phi }}\right)\\&=-\hbar ^{2}\left(\sin ^{2}\phi {\frac {\partial ^{2}}{\partial \theta ^{2}}}+\cot \theta \cos ^{2}\phi {\frac {\partial }{\partial \theta }}+\cot \theta \sin \phi \cos \phi {\frac {\partial ^{2}}{\partial \theta \partial \phi }}-\csc ^{2}\theta \sin \phi \cos \phi {\frac {\partial }{\partial \phi }}\right.\\\end{aligned}}\,\!}
+cotθsinϕcosϕ2θϕcot2θsinϕcosϕϕ+cot2θcos2ϕ2ϕ2){\displaystyle \left.+\cot \theta \sin \phi \cos \phi {\frac {\partial ^{2}}{\partial \theta \partial \phi }}-\cot ^{2}\theta \sin \phi \cos \phi {\frac {\partial }{\partial \phi }}+\cot ^{2}\theta \cos ^{2}\phi {\frac {\partial ^{2}}{\partial \phi ^{2}}}\right)\,\!}
L^y2=2(cosϕθcotθsinϕϕ)(cosϕθcotθsinϕϕ)=2(cos2ϕ2θ2+cotθsin2ϕθcotθsinϕcosϕ2θϕ+csc2θsinϕcosϕϕ{\displaystyle {\begin{aligned}{\hat {L}}_{y}^{2}&=-\hbar ^{2}\left(\cos \phi {\frac {\partial }{\partial \theta }}-\cot \theta \sin \phi {\frac {\partial }{\partial \phi }}\right)\left(\cos \phi {\frac {\partial }{\partial \theta }}-\cot \theta \sin \phi {\frac {\partial }{\partial \phi }}\right)\\&=-\hbar ^{2}\left(\cos ^{2}\phi {\frac {\partial ^{2}}{\partial \theta ^{2}}}+\cot \theta \sin ^{2}\phi {\frac {\partial }{\partial \theta }}-\cot \theta \sin \phi \cos \phi {\frac {\partial ^{2}}{\partial \theta \partial \phi }}+\csc ^{2}\theta \sin \phi \cos \phi {\frac {\partial }{\partial \phi }}\right.\\\end{aligned}}\,\!}
cotθsinϕcosϕ2θϕ+cot2θsinϕcosϕϕ+cot2θsin2ϕ2ϕ2){\displaystyle \left.-\cot \theta \sin \phi \cos \phi {\frac {\partial ^{2}}{\partial \theta \partial \phi }}+\cot ^{2}\theta \sin \phi \cos \phi {\frac {\partial }{\partial \phi }}+\cot ^{2}\theta \sin ^{2}\phi {\frac {\partial ^{2}}{\partial \phi ^{2}}}\right)\,\!}
L^z2=22ϕ2{\displaystyle {\hat {L}}_{z}^{2}=-\hbar ^{2}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\,\!}

經過一番繁雜的運算,終於得到想要的方程式[2]:169

L^2=2(2θ2+cotθθ+(1+cot2θ)2ϕ2)=2(1sinθθ(sinθθ)+1sin2θ2ϕ2){\displaystyle {\hat {L}}^{2}=-\hbar ^{2}\left({\frac {\partial ^{2}}{\partial \theta ^{2}}}+\cot \theta {\frac {\partial }{\partial \theta }}+(1+\cot ^{2}\theta ){\frac {\partial ^{2}}{\partial \phi ^{2}}}\right)=-\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 \phi ^{2}}}\right)\,\!}

滿足算符L^2{\displaystyle {\hat {L}}^{2}\,\!}本徵函數球諧函數Ym{\displaystyle Y_{\ell m}\,\!}

L^2Ym=(+1)2Ym{\displaystyle {\hat {L}}^{2}Y_{\ell m}=\ell (\ell +1)\hbar ^{2}Y_{\ell m}\,\!}

其中,本徵值{\displaystyle \ell \,\!} 是正整數。

球諧函數也是滿足算符L^z{\displaystyle {\hat {L}}_{z}\,\!} 微分方程式的本徵函數:

L^zYm=mYm{\displaystyle {\hat {L}}_{z}Y_{\ell m}=m\hbar Y_{\ell m}\,\!}

其中,本徵值m{\displaystyle m\,\!} 是整數,m0{\displaystyle -\ell \leq m\leq 0\,\!}

因為這兩個算符的正則對易關係是 0 ,它們可以有共同的本徵函數。

球諧函數Ym{\displaystyle Y_{\ell m}\,\!} 表達為

Ym(θ, ϕ)=(i)m+|m|(2+1)4π(m)!(+m)!Pm(cosθ)eimϕ{\displaystyle Y_{\ell m}(\theta ,\ \phi )=(i)^{m+|m|}{\sqrt {{(2\ell +1) \over 4\pi }{(\ell -m)! \over (\ell +m)!}}}\,P_{\ell m}(\cos {\theta })\,e^{im\phi }\,\!}

其中,i{\displaystyle i\,\!}虛數單位Pm(cosθ){\displaystyle P_{\ell m}(\cos {\theta })\,\!}伴隨勒讓德多項式,用方程式定義為

Pm(x)=(1x2)|m|/2 d|m|dx|m|P(x){\displaystyle P_{\ell m}(x)=(1-x^{2})^{|m|/2}\ {\frac {d^{|m|}}{dx^{|m|}}}P_{\ell }(x)\,}

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

P(x)=12!ddx(x21){\displaystyle P_{\ell }(x)={1 \over 2^{\ell }\ell !}{d^{\ell } \over dx^{\ell }}(x^{2}-1)^{\ell }}

球諧函數滿足正交歸一性

02π0π Y1m1Y2m2sin(θ)dθdϕ=δ12δm1m2{\displaystyle \int _{0}^{2\pi }\int _{0}^{\pi }\ Y_{\ell _{1}m_{1}}Y_{\ell _{2}m_{2}}\sin(\theta )d\theta d\phi =\delta _{\ell _{1}\ell _{2}}\delta _{m_{1}m_{2}}\,\!}

這樣,角動量算符的本徵函數,形成一組單範正交基。任意波函數ψ(θ,ϕ){\displaystyle \psi (\theta ,\,\phi )\,\!} 都可以表達為這單範正交基的線性組合

ψ(θ,ϕ)=,m AmYm(θ,ϕ){\displaystyle \psi (\theta ,\,\phi )=\sum _{\ell ,m}\ A_{\ell m}Y_{\ell m}(\theta ,\,\phi )\,\!}

其中,Am=02π0π Ym(θ,ϕ)ψ(θ,ϕ)sin(θ)dθdϕ{\displaystyle A_{\ell m}=\int _{0}^{2\pi }\int _{0}^{\pi }\ Y_{\ell m}^{*}(\theta ,\,\phi )\psi (\theta ,\,\phi )\sin(\theta )d\theta d\phi \,\!}

參閱

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參考文獻

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  1. ^Introductory Quantum Mechanics, Richard L. Liboff, 2nd Edition,ISBN 0201547155
  2. ^Griffiths, David J., Introduction to Quantum Mechanics (2nd ed.), Prentice Hall, 2004,ISBN 0-13-111892-7 

外部連結

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  • 圣地牙哥加州大学物理系量子力学視聽教學:角動量加法
检索自“https://zh.wikipedia.org/w/index.php?title=角動量算符&oldid=88308498
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