球面調和関数表
Y 2 − 2 ( x ) = 1 4 15 2 π ⋅ e − 2 i φ ⋅ sin 2  θ = 1 4 15 2 π ⋅ x 2 − 2 i x y − y 2 r 2 {\displaystyle Y_{2}^{-2}(x)={\frac {1}{4}}{\sqrt {\frac {15}{2\pi }}}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta ={\frac {1}{4}}{\sqrt {\frac {15}{2\pi }}}\cdot {\frac {x^{2}-2ixy-y^{2}}{r^{2}}}}
Y 2 − 1 ( x ) = 1 2 15 2 π ⋅ e − i φ ⋅ sin  θ ⋅ cos  θ = 1 2 15 2 π ⋅ x z − i y z r 2 {\displaystyle Y_{2}^{-1}(x)={\frac {1}{2}}{\sqrt {\frac {15}{2\pi }}}\cdot e^{-i\varphi }\cdot \sin \theta \cdot \cos \theta ={\frac {1}{2}}{\sqrt {\frac {15}{2\pi }}}\cdot {\frac {xz-iyz}{r^{2}}}}
Y 6 − 6 ( x ) = 1 64 3003 π ⋅ e − 6 i φ ⋅ sin 6  θ {\displaystyle Y_{6}^{-6}(x)={\frac {1}{64}}{\sqrt {\frac {3003}{\pi }}}\cdot e^{-6i\varphi }\cdot \sin ^{6}\theta }
Y 6 − 5 ( x ) = 3 32 1001 π ⋅ e − 5 i φ ⋅ sin 5  θ ⋅ cos  θ {\displaystyle Y_{6}^{-5}(x)={\frac {3}{32}}{\sqrt {\frac {1001}{\pi }}}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \cdot \cos \theta }
Y 6 − 4 ( x ) = 3 32 91 2 π ⋅ e − 4 i φ ⋅ sin 4  θ ⋅ ( 11 cos 2  θ − 1 ) {\displaystyle Y_{6}^{-4}(x)={\frac {3}{32}}{\sqrt {\frac {91}{2\pi }}}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot (11\cos ^{2}\theta -1)}
Y 6 − 3 ( x ) = 1 32 1365 π ⋅ e − 3 i φ ⋅ sin 3  θ ⋅ ( 11 cos 3  θ − 3 cos  θ ) {\displaystyle Y_{6}^{-3}(x)={\frac {1}{32}}{\sqrt {\frac {1365}{\pi }}}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (11\cos ^{3}\theta -3\cos \theta )}
Y 3 2 ( x ) = 1 4 105 2 π ⋅ e 2 i φ ⋅ sin 2  θ ⋅ cos  θ = 1 4 105 2 π ⋅ x 2 z + 2 i x y z − y 2 z r 3 {\displaystyle Y_{3}^{2}(x)={\frac {1}{4}}{\sqrt {\frac {105}{2\pi }}}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot \cos \theta ={\frac {1}{4}}{\sqrt {\frac {105}{2\pi }}}\cdot {\frac {x^{2}z+2ixyz-y^{2}z}{r^{3}}}}
Y 3 3 ( x ) = − 1 8 35 π ⋅ e 3 i φ ⋅ sin 3  θ = − 1 8 35 π ⋅ x 3 + 3 i x 2 y − 3 x y 2 − i y 3 r 3 {\displaystyle Y_{3}^{3}(x)=-{\frac {1}{8}}{\sqrt {\frac {35}{\pi }}}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta =-{\frac {1}{8}}{\sqrt {\frac {35}{\pi }}}\cdot {\frac {x^{3}+3ix^{2}y-3xy^{2}-iy^{3}}{r^{3}}}}