球諧函數是拉普拉斯方程式的球坐標系形式解的角度部分。在古典場論、量子力學等領域廣泛應用。 本微分方程式的推導 球坐標下的拉普拉斯方程式: ∇ 2 f = 1 r 2 ∂ ∂ r ( r 2 ∂ f ∂ r ) + 1 r 2 sin θ ∂ ∂ θ ( sin θ ∂ f ∂ θ ) + 1 r 2 sin 2 θ ∂ 2 f ∂ φ 2 = 0 {\displaystyle \nabla ^{2}f={1 \over r^{2}}{\partial \over \partial r}\left(r^{2}{\partial f \over \partial r}\right)+{1 \over r^{2}\sin \theta }{\partial \over \partial \theta }\left(\sin \theta {\partial f \over \partial \theta }\right)+{1 \over r^{2}\sin ^{2}\theta }{\partial ^{2}f \over \partial \varphi ^{2}}=0\,\!} 。 實值的球諧函數 Ylm，l = 0 到 4 （由上至下），m=0 到 4（由左至右）。負數階球諧函數 Yl,-m 可由正數階函數對 z-軸轉 90/m 度得到。 利用分離變量法，設定 f ( r , θ , φ ) = R ( r ) Y ( θ , φ ) = R ( r ) Θ ( θ ) Φ ( φ ) {\displaystyle f(r,\ \theta ,\ \varphi )=R(r)Y(\theta ,\ \varphi )=R(r)\Theta (\theta )\Phi (\varphi )} 。其中 Y ( θ , φ ) {\displaystyle Y(\theta ,\ \varphi )} 代表角度部分的解，也就是球諧函數。 代入拉普拉斯方程式，得到： Θ Φ r 2 d d r ( r 2 d R d r ) + R Φ r 2 sin θ d d θ ( sin θ d Θ d θ ) + R Θ r 2 sin 2 θ d 2 Φ d φ 2 = 0 {\displaystyle {\Theta \Phi \over r^{2}}{d \over dr}\left(r^{2}{dR \over dr}\right)+{R\Phi \over r^{2}\sin \theta }{d \over d\theta }\left(\sin \theta {d\Theta \over d\theta }\right)+{R\Theta \over r^{2}\sin ^{2}\theta }{d^{2}\Phi \over d\varphi ^{2}}=0\,\!} 分離變量後得： { 1 R d d r ( r 2 d R d r ) = λ 1 Φ d 2 Φ d φ 2 = − m 2 λ + 1 Θ sin θ d d θ ( sin θ d Θ d θ ) − m 2 sin 2 θ = 0 {\displaystyle {\begin{cases}{\dfrac {1}{R}}{\dfrac {d}{dr}}\left(r^{2}{\dfrac {dR}{dr}}\right)=\lambda \\{\dfrac {1}{\Phi }}{\dfrac {d^{2}\Phi }{d\varphi ^{2}}}=-m^{2}\\\lambda +{\dfrac {1}{\Theta \sin \theta }}{\dfrac {d}{d\theta }}\left(\sin \theta {\dfrac {d\Theta }{d\theta }}\right)-{\dfrac {m^{2}}{\sin ^{2}\theta }}=0\end{cases}}} ，整理得 { r 2 R ″ + 2 r R ′ − λ R = 0 Φ ″ + m 2 Φ = 0 sin θ d d θ ( sin θ Θ ′ ) + ( λ sin 2 θ − m 2 ) Θ = 0 {\displaystyle {\begin{cases}r^{2}R''+2rR'-\lambda R=0\\\Phi ''+m^{2}\Phi =0\\\sin \theta {\dfrac {d}{d\theta }}(\sin \theta \Theta ')+(\lambda \sin ^{2}\theta -m^{2})\Theta =0\end{cases}}} 本徵方程式的求解 這裡， Φ {\displaystyle \Phi } 是一個以 2 π {\displaystyle 2\pi } 為週期的函數，即滿足週期性邊界條件 Φ ( φ ) = Φ ( φ + 2 π ) {\displaystyle \Phi (\varphi )=\Phi (\varphi +2\pi )} ，因此 m {\displaystyle m} 必須為整數。而且可以解出： Φ = e i m ϕ {\displaystyle \Phi =e^{im\phi }} ， m ∈ Z {\displaystyle m\in \mathbb {Z} } 而對於 Θ {\displaystyle \Theta } 的方程式，進行變量替換 t = cos θ {\displaystyle t=\cos \theta } ， d t = − sin θ d θ {\displaystyle dt=-\sin \theta d\theta } ， | t | ⩽ 1 {\displaystyle |t|\leqslant 1} ，得到關於 t {\displaystyle t} 的伴隨勒讓德方程式。方程式的解應滿足在 [ − 1 , 1 ] {\displaystyle [-1,1]} 區間上取有限值，此時必須有 λ = l ( l + 1 ) {\displaystyle \lambda =l(l+1)} ，其中 l {\displaystyle l} 為自然數，且 l ⩾ | m | {\displaystyle l\geqslant |m|} 。對應方程式的解為 P ℓ m ( t ) {\displaystyle P_{\ell }^{m}(t)} 。即可以解出： Θ = P ℓ m ( cos θ ) {\displaystyle \Theta =P_{\ell }^{m}(\cos \theta )} ， l ∈ N , l ⩾ | m | {\displaystyle l\in \mathbb {N} ,l\geqslant |m|} 故球諧函數可以表達為： Y ℓ m ( θ , φ ) = N Φ ( φ ) Θ ( θ ) = N e i m φ P ℓ m ( cos θ ) {\displaystyle Y_{\ell }^{m}(\theta ,\varphi )=N\Phi (\varphi )\Theta (\theta )=N\,e^{im\varphi }\,P_{\ell }^{m}(\cos {\theta })\,\!} ， l ∈ N , m = 0 , ± 1 , ± 2 , … ± l {\displaystyle l\in \mathbb {N} ,m=0,\pm 1,\pm 2,\ldots \pm l} ； 其中N 是歸一化因子。 經過歸一化後，球諧函數表達為： Y ℓ m ( θ , φ ) = ( − 1 ) m ( 2 ℓ + 1 ) 4 π ( ℓ − | m | ) ! ( ℓ + | m | ) ! P ℓ m ( cos θ ) e i m φ {\displaystyle Y_{\ell }^{m}(\theta ,\ \varphi )=(-1)^{m}{\sqrt {{(2\ell +1) \over 4\pi }{(\ell -|m|)! \over (\ell +|m|)!}}}\,P_{\ell }^{m}(\cos {\theta })\,e^{im\varphi }\,\!} ； 這裡的 Y ℓ m {\displaystyle Y_{\ell }^{m}\,\!} 稱為 ℓ {\displaystyle \ell \,\!} 和 m {\displaystyle m\,\!} 的球諧函數。以上推導過程中， i {\displaystyle i\,\!} 是虛數單位， P ℓ m {\displaystyle P_{\ell }^{m}\,\!} 是伴隨勒讓德多項式 。 其中 P ℓ m ( x ) {\displaystyle P_{\ell }^{m}(x)\,\!} 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_{\ell }^{m}(x)=(1-x^{2})^{|m|/2}\ {\frac {d^{|m|}}{dx^{|m|}}}P_{\ell }(x)\,} ； 而 P ℓ ( x ) {\displaystyle P_{\ell }(x)\,\!} 是 l {\displaystyle l} 階勒讓德多項式，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_{\ell }(x)={1 \over 2^{\ell }\ell !}{d^{\ell } \over dx^{\ell }}(x^{2}-1)^{l}} 。 主條目：球諧函數表 More information , ... l {\displaystyle l} m {\displaystyle m} Φ ( φ ) {\displaystyle \Phi (\varphi )} Θ ( θ ) {\displaystyle \Theta (\theta )} 極坐標中的表達式 直角坐標中的表達式 量子力學中的記號 0 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} 1 2 {\displaystyle {\frac {1}{\sqrt {2}}}} 1 2 π {\displaystyle {\frac {1}{2{\sqrt {\pi }}}}} 1 2 π {\displaystyle {\frac {1}{2{\sqrt {\pi }}}}} s {\displaystyle {\mbox{s}}\,} 1 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} 3 2 cos θ {\displaystyle {\sqrt {\frac {3}{2}}}\cos \theta } 1 2 3 π cos θ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\cos \theta } 1 2 3 π z r {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {z}{r}}} p z {\displaystyle {\mbox{p}}_{z}\,} 1 +1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )} 3 2 sin θ {\displaystyle {\frac {\sqrt {3}}{2}}\sin \theta } { {\displaystyle {\Bigg \{}} 1 2 3 π sin θ cos φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \cos \varphi } 1 2 3 π x r {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {x}{r}}} p x {\displaystyle {\mbox{p}}_{x}\,} 1 -1 1 2 π exp ( − i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )} 3 2 sin θ {\displaystyle {\frac {\sqrt {3}}{2}}\sin \theta } 1 2 3 π sin θ sin φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \sin \varphi } 1 2 3 π y r {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {y}{r}}} p y {\displaystyle {\mbox{p}}_{y}\,} 2 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} 1 2 5 2 ( 3 cos 2 θ − 1 ) {\displaystyle {\frac {1}{2}}{\sqrt {\frac {5}{2}}}(3\cos ^{2}\theta -1)} 1 4 5 π ( 3 cos 2 θ − 1 ) {\displaystyle {\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}(3\cos ^{2}\theta -1)} 1 4 5 π 2 z 2 − x 2 − y 2 r 2 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}{\frac {2z^{2}-x^{2}-y^{2}}{r^{2}}}} d 3 z 2 − r 2 {\displaystyle {\mbox{d}}_{3z^{2}-r^{2}}} 2 +1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )} 15 2 sin θ cos θ {\displaystyle {\frac {\sqrt {15}}{2}}\sin \theta \cos \theta } { {\displaystyle {\Bigg \{}} 1 2 15 π sin θ cos θ cos φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \cos \varphi } 1 2 15 π z x r 2 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {zx}{r^{2}}}} d z x {\displaystyle {\mbox{d}}_{zx}\,} 2 -1 1 2 π exp ( − i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )} 15 2 sin θ cos θ {\displaystyle {\frac {\sqrt {15}}{2}}\sin \theta \cos \theta } 1 2 15 π sin θ cos θ sin φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \sin \varphi } 1 2 15 π y z r 2 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {yz}{r^{2}}}} d y z {\displaystyle {\mbox{d}}_{yz}\,} 2 +2 1 2 π exp ( 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )} 15 4 sin 2 θ {\displaystyle {\frac {\sqrt {15}}{4}}\sin ^{2}\theta } { {\displaystyle {\Bigg \{}} 1 4 15 π sin 2 θ cos 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \cos 2\varphi } 1 4 15 π x 2 − y 2 r 2 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}{\frac {x^{2}-y^{2}}{r^{2}}}} d x 2 − y 2 {\displaystyle {\mbox{d}}_{x^{2}-y^{2}}} 2 -2 1 2 π exp ( − 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )} 15 4 sin 2 θ {\displaystyle {\frac {\sqrt {15}}{4}}\sin ^{2}\theta } 1 4 15 π sin 2 θ sin 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \sin 2\varphi } 1 2 15 π x y r 2 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {xy}{r^{2}}}} d x y {\displaystyle {\mbox{d}}_{xy}\,} 3 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} 1 2 7 2 ( 5 cos 3 θ − 3 cos θ ) {\displaystyle {\frac {1}{2}}{\sqrt {\frac {7}{2}}}(5\cos ^{3}\theta -3\cos \theta )} 1 4 7 π ( 5 cos 3 θ − 3 cos θ ) {\displaystyle {\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}(5\cos ^{3}\theta -3\cos \theta )} 1 4 7 π z ( 2 z 2 − 3 x 2 − 3 y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}{\frac {z(2z^{2}-3x^{2}-3y^{2})}{r^{3}}}} f z ( 5 z 2 − 3 r 2 ) {\displaystyle {\mbox{f}}_{z(5z^{2}-3r^{2})}} 3 +1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )} 1 4 21 2 ( 5 cos 2 θ − 1 ) sin θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta } { {\displaystyle {\Bigg \{}} 1 4 21 2 π ( 5 cos 2 θ − 1 ) sin θ cos φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}(5\cos ^{2}\theta -1)\sin \theta \cos \varphi } 1 4 21 2 π x ( 5 z 2 − r 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {x(5z^{2}-r^{2})}{r^{3}}}} f x ( 5 z 2 − r 2 ) {\displaystyle {\mbox{f}}_{x(5z^{2}-r^{2})}} 3 -1 1 2 π exp ( − i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )} 1 4 21 2 ( 5 cos 2 θ − 1 ) sin θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta } 1 4 21 2 π ( 5 cos 2 θ − 1 ) sin θ sin φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}(5\cos ^{2}\theta -1)\sin \theta \sin \varphi } 1 4 21 2 π y ( 5 z 2 − r 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {y(5z^{2}-r^{2})}{r^{3}}}} f y ( 5 z 2 − r 2 ) {\displaystyle {\mbox{f}}_{y(5z^{2}-r^{2})}} 3 +2 1 2 π exp ( 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )} 105 4 cos θ sin 2 θ {\displaystyle {\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta } { {\displaystyle {\Bigg \{}} 1 4 105 π cos θ sin 2 θ cos 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \cos 2\varphi } 1 4 105 π z ( x 2 − y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}{\frac {z(x^{2}-y^{2})}{r^{3}}}} f z ( x 2 − y 2 ) {\displaystyle {\mbox{f}}_{z(x^{2}-y^{2})}} 3 -2 1 2 π exp ( − 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )} 105 4 cos θ sin 2 θ {\displaystyle {\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta } 1 4 105 π cos θ sin 2 θ sin 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \sin 2\varphi } 1 2 105 π x y z r 3 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {105}{\pi }}}{\frac {xyz}{r^{3}}}} f x y z {\displaystyle {\mbox{f}}_{xyz}\,} 3 +3 1 2 π exp ( 3 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(3i\varphi )} 1 4 35 2 sin 3 θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta } { {\displaystyle {\Bigg \{}} 1 4 35 2 π sin 3 θ cos 3 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \cos 3\varphi } 1 4 35 2 π x ( x 2 − 3 y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {x(x^{2}-3y^{2})}{r^{3}}}} f x ( x 2 − 3 y 2 ) {\displaystyle {\mbox{f}}_{x(x^{2}-3y^{2})}} 3 -3 1 2 π exp ( − 3 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-3i\varphi )} 1 4 35 2 sin 3 θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta } 1 4 35 2 π sin 3 θ sin 3 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \sin 3\varphi } 1 4 35 2 π y ( 3 x 2 − y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {y(3x^{2}-y^{2})}{r^{3}}}} f y ( 3 x 2 − y 2 ) {\displaystyle {\mbox{f}}_{y(3x^{2}-y^{2})}} Close l = 0 {\displaystyle l=0} Y 0 0 ( θ , φ ) = 1 2 1 π {\displaystyle Y_{0}^{0}(\theta ,\varphi )={1 \over 2}{\sqrt {1 \over \pi }}} l = 1 {\displaystyle l=1} Y 1 − 1 ( θ , φ ) = 1 2 3 2 π sin θ e − i φ {\displaystyle Y_{1}^{-1}(\theta ,\varphi )={1 \over 2}{\sqrt {3 \over 2\pi }}\,\sin \theta \,e^{-i\varphi }} Y 1 0 ( θ , φ ) = 1 2 3 π cos θ {\displaystyle Y_{1}^{0}(\theta ,\varphi )={1 \over 2}{\sqrt {3 \over \pi }}\,\cos \theta } Y 1 1 ( θ , φ ) = − 1 2 3 2 π sin θ e i φ {\displaystyle Y_{1}^{1}(\theta ,\varphi )={-1 \over 2}{\sqrt {3 \over 2\pi }}\,\sin \theta \,e^{i\varphi }} l = 2 {\displaystyle l=2} Y 2 − 2 ( θ , φ ) = 1 4 15 2 π sin 2 θ e − 2 i φ {\displaystyle Y_{2}^{-2}(\theta ,\varphi )={1 \over 4}{\sqrt {15 \over 2\pi }}\,\sin ^{2}\theta \,e^{-2i\varphi }} Y 2 − 1 ( θ , φ ) = 1 2 15 2 π sin θ cos θ e − i φ {\displaystyle Y_{2}^{-1}(\theta ,\varphi )={1 \over 2}{\sqrt {15 \over 2\pi }}\,\sin \theta \,\cos \theta \,e^{-i\varphi }} Y 2 0 ( θ , φ ) = 1 4 5 π ( 3 cos 2 θ − 1 ) {\displaystyle Y_{2}^{0}(\theta ,\varphi )={1 \over 4}{\sqrt {5 \over \pi }}\,(3\cos ^{2}\theta -1)} Y 2 1 ( θ , φ ) = − 1 2 15 2 π sin θ cos θ e i φ {\displaystyle Y_{2}^{1}(\theta ,\varphi )={-1 \over 2}{\sqrt {15 \over 2\pi }}\,\sin \theta \,\cos \theta \,e^{i\varphi }} Y 2 2 ( θ , φ ) = 1 4 15 2 π sin 2 θ e 2 i φ {\displaystyle Y_{2}^{2}(\theta ,\varphi )={1 \over 4}{\sqrt {15 \over 2\pi }}\,\sin ^{2}\theta \,e^{2i\varphi }} 勒讓德多項式 伴隨勒讓德多項式 施圖姆-劉維爾理論 柱諧函數 向量球諧函數 Wikiwand in your browser!Seamless Wikipedia browsing. 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