体球調和関数
物理学と数学において、体球調和関数(たいきゅうちょうわかんすう、英: solid harmonics)は球面座標系でのラプラス方程式の解を指す。原点で0になる正則な(regular)体球調和関数 R l m ( r ) {\displaystyle R_{\ell }^{m}({\boldsymbol {r}})} と、原点が特異点となる非正則な(irregular)体球調和関数 I l m ( r ) {\displaystyle I_{\ell }^{m}({\boldsymbol {r}})} の2種がある。いずれの関数集合もポテンシャル論で重要な役割を演じ、また適当にスケーリングすることで球面調和関数が得られる。
R l m ( r ) ≡ 4 π 2 l + 1 r l Y l m ( θ , φ ) {\displaystyle R_{\ell }^{m}({\boldsymbol {r}})\equiv {\sqrt {\frac {4\pi }{2\ell +1}}}\;r^{\ell }Y_{\ell }^{m}(\theta ,\varphi )}
I l m ( r ) ≡ 4 π 2 l + 1 Y l m ( θ , φ ) r l + 1 {\displaystyle I_{\ell }^{m}({\boldsymbol {r}})\equiv {\sqrt {\frac {4\pi }{2\ell +1}}}\;{\frac {Y_{\ell }^{m}(\theta ,\varphi )}{r^{\ell +1}}}}
C l m ( x , y , z ) = [ ( 2 − δ m 0 ) ( l − m ) ! ( l + m ) ! ] 1 / 2 Π l m ( z ) A m ( x , y ) , m = 0 , 1 , ... , l {\displaystyle C_{\ell }^{m}(x,y,z)=\left[{\frac {(2-\delta _{m0})(\ell -m)!}{(\ell +m)!}}\right]^{1/2}\Pi _{\ell }^{m}(z)\;A_{m}(x,y),\qquad m=0,1,\ldots ,\ell }
S l m ( x , y , z ) = [ 2 ( l − m ) ! ( l + m ) ! ] 1 / 2 Π l m ( z ) B m ( x , y ) , m = 1 , 2 , ... , l . {\displaystyle S_{\ell }^{m}(x,y,z)=\left[{\frac {2(\ell -m)!}{(\ell +m)!}}\right]^{1/2}\Pi _{\ell }^{m}(z)\;B_{m}(x,y),\qquad m=1,2,\ldots ,\ell .}
ここで | r | ≤ | a | {\displaystyle |r|\leq |a|} とする。ブラケットで囲まれた因子は再びクレブシュ–ゴルダン係数である。
⟨ λ , μ ; l + λ , m − μ | l m ⟩ = ( − 1 ) λ + μ ( l + λ − m + μ λ + μ ) 1 / 2 ( l + λ + m − μ λ − μ ) 1 / 2 ( 2 l + 2 λ + 1 2 λ ) − 1 / 2 . {\displaystyle \langle \lambda ,\mu ;\ell +\lambda ,m-\mu |\ell m\rangle =(-1)^{\lambda +\mu }{\binom {\ell +\lambda -m+\mu }{\lambda +\mu }}^{1/2}{\binom {\ell +\lambda +m-\mu }{\lambda -\mu }}^{1/2}{\binom {2\ell +2\lambda +1}{2\lambda }}^{-1/2}.}
Π ̄ 0 0 = 1 Π ̄ 3 1 = 1 4 6 ( 5 z 2 − r 2 ) Π ̄ 4 4 = 1 8 35 Π ̄ 1 0 = z Π ̄ 3 2 = 1 2 15 z Π ̄ 5 0 = 1 8 z ( 63 z 4 − 70 z 2 r 2 + 15 r 4 ) Π ̄ 1 1 = 1 Π ̄ 3 3 = 1 4 10 Π ̄ 5 1 = 1 8 15 ( 21 z 4 − 14 z 2 r 2 + r 4 ) Π ̄ 2 0 = 1 2 ( 3 z 2 − r 2 ) Π ̄ 4 0 = 1 8 ( 35 z 4 − 30 r 2 z 2 + 3 r 4 ) Π ̄ 5 2 = 1 4 105 ( 3 z 2 − r 2 ) z Π ̄ 2 1 = 3 z Π ̄ 4 1 = 10 4 z ( 7 z 2 − 3 r 2 ) Π ̄ 5 3 = 1 16 70 ( 9 z 2 − r 2 ) Π ̄ 2 2 = 1 2 3 Π ̄ 4 2 = 1 4 5 ( 7 z 2 − r 2 ) Π ̄ 5 4 = 3 8 35 z Π ̄ 3 0 = 1 2 z ( 5 z 2 − 3 r 2 ) Π ̄ 4 3 = 1 4 70 z Π ̄ 5 5 = 3 16 14 {\displaystyle {\begin{aligned}{\bar {\Pi }}_{0}^{0}&=1&{\bar {\Pi }}_{3}^{1}&={\frac {1}{4}}{\sqrt {6}}(5z^{2}-r^{2})&{\bar {\Pi }}_{4}^{4}&={\frac {1}{8}}{\sqrt {35}}\\{\bar {\Pi }}_{1}^{0}&=z&{\bar {\Pi }}_{3}^{2}&={\frac {1}{2}}{\sqrt {15}}\;z&{\bar {\Pi }}_{5}^{0}&={\frac {1}{8}}z(63z^{4}-70z^{2}r^{2}+15r^{4})\\{\bar {\Pi }}_{1}^{1}&=1&{\bar {\Pi }}_{3}^{3}&={\frac {1}{4}}{\sqrt {10}}&{\bar {\Pi }}_{5}^{1}&={\frac {1}{8}}{\sqrt {15}}(21z^{4}-14z^{2}r^{2}+r^{4})\\{\bar {\Pi }}_{2}^{0}&={\frac {1}{2}}(3z^{2}-r^{2})&{\bar {\Pi }}_{4}^{0}&={\frac {1}{8}}(35z^{4}-30r^{2}z^{2}+3r^{4})&{\bar {\Pi }}_{5}^{2}&={\frac {1}{4}}{\sqrt {105}}(3z^{2}-r^{2})z\\{\bar {\Pi }}_{2}^{1}&={\sqrt {3}}z&{\bar {\Pi }}_{4}^{1}&={\frac {\sqrt {10}}{4}}z(7z^{2}-3r^{2})&{\bar {\Pi }}_{5}^{3}&={\frac {1}{16}}{\sqrt {70}}(9z^{2}-r^{2})\\{\bar {\Pi }}_{2}^{2}&={\frac {1}{2}}{\sqrt {3}}&{\bar {\Pi }}_{4}^{2}&={\frac {1}{4}}{\sqrt {5}}(7z^{2}-r^{2})&{\bar {\Pi }}_{5}^{4}&={\frac {3}{8}}{\sqrt {35}}z\\{\bar {\Pi }}_{3}^{0}&={\frac {1}{2}}z(5z^{2}-3r^{2})&{\bar {\Pi }}_{4}^{3}&={\frac {1}{4}}{\sqrt {70}}\;z&{\bar {\Pi }}_{5}^{5}&={\frac {3}{16}}{\sqrt {14}}\\\end{aligned}}}