Tabla matemática
Esta es una tabla de armónicos esféricos ortonormalizados que emplean la fase de Condon-Shortley hasta el grado . Algunas de estas fórmulas se expresan en términos de la expansión cartesiana de los armónicos esféricos en polinomios en x , y , z y r . Para los fines de esta tabla, es útil expresar las transformaciones esféricas a cartesianas habituales que relacionan estos componentes cartesianos con y como ℓ = 10 {\displaystyle \ell = 10} θ {\estilo de visualización \theta} φ {\estilo de visualización \varphi}
{ porque ( θ ) = el / a mi ± i φ ⋅ pecado ( θ ) = ( incógnita ± i y ) / a {\displaystyle {\begin{cases}\cos(\theta )&=z/r\\e^{\pm i\varphi }\cdot \sin(\theta )&=(x\pm iy)/r\end{cases}}}
Armónicos esféricos complejos Para ℓ = 0, …, 5, véase. [1]
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 Y 1 − 1 ( θ , φ ) = 1 2 3 2 π ⋅ mi − i φ ⋅ pecado θ = 1 2 3 2 π ⋅ ( incógnita − i y ) a Y 1 0 ( θ , φ ) = 1 2 3 π ⋅ porque θ = 1 2 3 π ⋅ el a Y 1 1 ( θ , φ ) = − 1 2 3 2 π ⋅ mi i φ ⋅ pecado θ = − 1 2 3 2 π ⋅ ( incógnita + i y ) a {\displaystyle {\begin{aligned}Y_{1}^{-1}(\theta ,\varphi )&=&&{1 \sobre 2}{\sqrt {3 \sobre 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta &&=&&{1 \sobre 2}{\sqrt {3 \sobre 2\pi }}\cdot {(x-iy) \sobre r}\\Y_{1}^{0}(\theta ,\varphi )&=&&{1 \sobre 2}{\sqrt {3 \sobre \pi }}\cdot \cos \theta &&=&&{1 \sobre 2}{\sqrt {3 \sobre \pi }}\cdot {z \sobre r}\\Y_{1}^{1}(\theta ,\varphi )&=&-&{1 \sobre 2}{\sqrt {3 \sobre 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta &&=&-&{1 \sobre 2}{\sqrt {3 \sobre 2\pi }}\cdot {(x+iy) \sobre r}\end{alineado}}}
ℓ= 2 Y 2 − 2 ( θ , φ ) = 1 4 15 2 π ⋅ mi − 2 i φ ⋅ pecado 2 θ = 1 4 15 2 π ⋅ ( incógnita − i y ) 2 a 2 Y 2 − 1 ( θ , φ ) = 1 2 15 2 π ⋅ mi − i φ ⋅ pecado θ ⋅ porque θ = 1 2 15 2 π ⋅ ( incógnita − i y ) ⋅ el a 2 Y 2 0 ( θ , φ ) = 1 4 5 π ⋅ ( 3 porque 2 θ − 1 ) = 1 4 5 π ⋅ ( 3 el 2 − a 2 ) a 2 Y 2 1 ( θ , φ ) = − 1 2 15 2 π ⋅ mi i φ ⋅ pecado θ ⋅ porque θ = − 1 2 15 2 π ⋅ ( incógnita + i y ) ⋅ el a 2 Y 2 2 ( θ , φ ) = 1 4 15 2 π ⋅ mi 2 i φ ⋅ pecado 2 θ = 1 4 15 2 π ⋅ ( incógnita + i y ) 2 a 2 {\displaystyle {\begin{aligned}Y_{2}^{-2}(\theta ,\varphi )&=&&{1 \over 4}{\sqrt {15 \over 2\pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \quad &&=&&{1 \over 4}{\sqrt {15 \over 2\pi }}\cdot {(x-iy)^{2} \over r^{2}}&\\Y_{2}^{-1}(\theta ,\varphi )&=&&{1 \over 2}{\sqrt {15 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot \cos \theta \quad &&=&&{1 \over 2}{\sqrt {15 \over 2\pi }}\cdot {(x-iy)\cdot z \over r^{2}}&\\Y_{2}^{0}(\theta ,\varphi )&=&&{1 \over 4}{\sqrt {5 \over \pi }}\cdot (3\cos ^{2}\theta -1)\quad &&=&&{1 \over 4}{\sqrt {5 \over \pi }}\cdot {(3z^{2}-r^{2}) \over r^{2}}&\\Y_{2}^{1}(\theta ,\varphi )&=&-&{1 \over 2}{\sqrt {15 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot \cos \theta \quad &&=&-&{1 \over 2}{\sqrt {15 \over 2\pi }}\cdot {(x+iy)\cdot z \over r^{2}}&\\Y_{2}^{2}(\theta ,\varphi )&=&&{1 \over 4}{\sqrt {15 \over 2\pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \quad &&=&&{1 \over 4}{\sqrt {15 \over 2\pi }}\cdot {(x+iy)^{2} \over r^{2}}&\end{aligned}}}
ℓ= 3 Y 3 − 3 ( θ , φ ) = 1 8 35 π ⋅ e − 3 i φ ⋅ sin 3 θ = 1 8 35 π ⋅ ( x − i y ) 3 r 3 Y 3 − 2 ( θ , φ ) = 1 4 105 2 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ cos θ = 1 4 105 2 π ⋅ ( x − i y ) 2 ⋅ z r 3 Y 3 − 1 ( θ , φ ) = 1 8 21 π ⋅ e − i φ ⋅ sin θ ⋅ ( 5 cos 2 θ − 1 ) = 1 8 21 π ⋅ ( x − i y ) ⋅ ( 5 z 2 − r 2 ) r 3 Y 3 0 ( θ , φ ) = 1 4 7 π ⋅ ( 5 cos 3 θ − 3 cos θ ) = 1 4 7 π ⋅ ( 5 z 3 − 3 z r 2 ) r 3 Y 3 1 ( θ , φ ) = − 1 8 21 π ⋅ e i φ ⋅ sin θ ⋅ ( 5 cos 2 θ − 1 ) = − 1 8 21 π ⋅ ( x + i y ) ⋅ ( 5 z 2 − r 2 ) r 3 Y 3 2 ( θ , φ ) = 1 4 105 2 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ cos θ = 1 4 105 2 π ⋅ ( x + i y ) 2 ⋅ z r 3 Y 3 3 ( θ , φ ) = − 1 8 35 π ⋅ e 3 i φ ⋅ sin 3 θ = − 1 8 35 π ⋅ ( x + i y ) 3 r 3 {\displaystyle {\begin{aligned}Y_{3}^{-3}(\theta ,\varphi )&=&&{1 \over 8}{\sqrt {35 \over \pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \quad &&=&&{1 \over 8}{\sqrt {35 \over \pi }}\cdot {(x-iy)^{3} \over r^{3}}&\\Y_{3}^{-2}(\theta ,\varphi )&=&&{1 \over 4}{\sqrt {105 \over 2\pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot \cos \theta \quad &&=&&{1 \over 4}{\sqrt {105 \over 2\pi }}\cdot {(x-iy)^{2}\cdot z \over r^{3}}&\\Y_{3}^{-1}(\theta ,\varphi )&=&&{1 \over 8}{\sqrt {21 \over \pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (5\cos ^{2}\theta -1)\quad &&=&&{1 \over 8}{\sqrt {21 \over \pi }}\cdot {(x-iy)\cdot (5z^{2}-r^{2}) \over r^{3}}&\\Y_{3}^{0}(\theta ,\varphi )&=&&{1 \over 4}{\sqrt {7 \over \pi }}\cdot (5\cos ^{3}\theta -3\cos \theta )\quad &&=&&{1 \over 4}{\sqrt {7 \over \pi }}\cdot {(5z^{3}-3zr^{2}) \over r^{3}}&\\Y_{3}^{1}(\theta ,\varphi )&=&-&{1 \over 8}{\sqrt {21 \over \pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (5\cos ^{2}\theta -1)\quad &&=&&{-1 \over 8}{\sqrt {21 \over \pi }}\cdot {(x+iy)\cdot (5z^{2}-r^{2}) \over r^{3}}&\\Y_{3}^{2}(\theta ,\varphi )&=&&{1 \over 4}{\sqrt {105 \over 2\pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot \cos \theta \quad &&=&&{1 \over 4}{\sqrt {105 \over 2\pi }}\cdot {(x+iy)^{2}\cdot z \over r^{3}}&\\Y_{3}^{3}(\theta ,\varphi )&=&-&{1 \over 8}{\sqrt {35 \over \pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \quad &&=&&{-1 \over 8}{\sqrt {35 \over \pi }}\cdot {(x+iy)^{3} \over r^{3}}&\end{aligned}}}
ℓ= 4 Y 4 − 4 ( θ , φ ) = 3 16 35 2 π ⋅ e − 4 i φ ⋅ sin 4 θ = 3 16 35 2 π ⋅ ( x − i y ) 4 r 4 Y 4 − 3 ( θ , φ ) = 3 8 35 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ cos θ = 3 8 35 π ⋅ ( x − i y ) 3 z r 4 Y 4 − 2 ( θ , φ ) = 3 8 5 2 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 7 cos 2 θ − 1 ) = 3 8 5 2 π ⋅ ( x − i y ) 2 ⋅ ( 7 z 2 − r 2 ) r 4 Y 4 − 1 ( θ , φ ) = 3 8 5 π ⋅ e − i φ ⋅ sin θ ⋅ ( 7 cos 3 θ − 3 cos θ ) = 3 8 5 π ⋅ ( x − i y ) ⋅ ( 7 z 3 − 3 z r 2 ) r 4 Y 4 0 ( θ , φ ) = 3 16 1 π ⋅ ( 35 cos 4 θ − 30 cos 2 θ + 3 ) = 3 16 1 π ⋅ ( 35 z 4 − 30 z 2 r 2 + 3 r 4 ) r 4 Y 4 1 ( θ , φ ) = − 3 8 5 π ⋅ e i φ ⋅ sin θ ⋅ ( 7 cos 3 θ − 3 cos θ ) = − 3 8 5 π ⋅ ( x + i y ) ⋅ ( 7 z 3 − 3 z r 2 ) r 4 Y 4 2 ( θ , φ ) = 3 8 5 2 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 7 cos 2 θ − 1 ) = 3 8 5 2 π ⋅ ( x + i y ) 2 ⋅ ( 7 z 2 − r 2 ) r 4 Y 4 3 ( θ , φ ) = − 3 8 35 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ cos θ = − 3 8 35 π ⋅ ( x + i y ) 3 z r 4 Y 4 4 ( θ , φ ) = 3 16 35 2 π ⋅ e 4 i φ ⋅ sin 4 θ = 3 16 35 2 π ⋅ ( x + i y ) 4 r 4 {\displaystyle {\begin{aligned}Y_{4}^{-4}(\theta ,\varphi )&={3 \over 16}{\sqrt {35 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta ={\frac {3}{16}}{\sqrt {\frac {35}{2\pi }}}\cdot {\frac {(x-iy)^{4}}{r^{4}}}\\Y_{4}^{-3}(\theta ,\varphi )&={3 \over 8}{\sqrt {35 \over \pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot \cos \theta ={\frac {3}{8}}{\sqrt {\frac {35}{\pi }}}\cdot {\frac {(x-iy)^{3}z}{r^{4}}}\\Y_{4}^{-2}(\theta ,\varphi )&={3 \over 8}{\sqrt {5 \over 2\pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (7\cos ^{2}\theta -1)={\frac {3}{8}}{\sqrt {\frac {5}{2\pi }}}\cdot {\frac {(x-iy)^{2}\cdot (7z^{2}-r^{2})}{r^{4}}}\\Y_{4}^{-1}(\theta ,\varphi )&={3 \over 8}{\sqrt {5 \over \pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (7\cos ^{3}\theta -3\cos \theta )={\frac {3}{8}}{\sqrt {\frac {5}{\pi }}}\cdot {\frac {(x-iy)\cdot (7z^{3}-3zr^{2})}{r^{4}}}\\Y_{4}^{0}(\theta ,\varphi )&={3 \over 16}{\sqrt {1 \over \pi }}\cdot (35\cos ^{4}\theta -30\cos ^{2}\theta +3)={\frac {3}{16}}{\sqrt {\frac {1}{\pi }}}\cdot {\frac {(35z^{4}-30z^{2}r^{2}+3r^{4})}{r^{4}}}\\Y_{4}^{1}(\theta ,\varphi )&={-3 \over 8}{\sqrt {5 \over \pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (7\cos ^{3}\theta -3\cos \theta )={\frac {-3}{8}}{\sqrt {\frac {5}{\pi }}}\cdot {\frac {(x+iy)\cdot (7z^{3}-3zr^{2})}{r^{4}}}\\Y_{4}^{2}(\theta ,\varphi )&={3 \over 8}{\sqrt {5 \over 2\pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (7\cos ^{2}\theta -1)={\frac {3}{8}}{\sqrt {\frac {5}{2\pi }}}\cdot {\frac {(x+iy)^{2}\cdot (7z^{2}-r^{2})}{r^{4}}}\\Y_{4}^{3}(\theta ,\varphi )&={-3 \over 8}{\sqrt {35 \over \pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot \cos \theta ={\frac {-3}{8}}{\sqrt {\frac {35}{\pi }}}\cdot {\frac {(x+iy)^{3}z}{r^{4}}}\\Y_{4}^{4}(\theta ,\varphi )&={3 \over 16}{\sqrt {35 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta ={\frac {3}{16}}{\sqrt {\frac {35}{2\pi }}}\cdot {\frac {(x+iy)^{4}}{r^{4}}}\end{aligned}}}
ℓ= 5 Y 5 − 5 ( θ , φ ) = 3 32 77 π ⋅ e − 5 i φ ⋅ sin 5 θ Y 5 − 4 ( θ , φ ) = 3 16 385 2 π ⋅ e − 4 i φ ⋅ sin 4 θ ⋅ cos θ Y 5 − 3 ( θ , φ ) = 1 32 385 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ ( 9 cos 2 θ − 1 ) Y 5 − 2 ( θ , φ ) = 1 8 1155 2 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 3 cos 3 θ − cos θ ) Y 5 − 1 ( θ , φ ) = 1 16 165 2 π ⋅ e − i φ ⋅ sin θ ⋅ ( 21 cos 4 θ − 14 cos 2 θ + 1 ) Y 5 0 ( θ , φ ) = 1 16 11 π ⋅ ( 63 cos 5 θ − 70 cos 3 θ + 15 cos θ ) Y 5 1 ( θ , φ ) = − 1 16 165 2 π ⋅ e i φ ⋅ sin θ ⋅ ( 21 cos 4 θ − 14 cos 2 θ + 1 ) Y 5 2 ( θ , φ ) = 1 8 1155 2 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 3 cos 3 θ − cos θ ) Y 5 3 ( θ , φ ) = − 1 32 385 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ ( 9 cos 2 θ − 1 ) Y 5 4 ( θ , φ ) = 3 16 385 2 π ⋅ e 4 i φ ⋅ sin 4 θ ⋅ cos θ Y 5 5 ( θ , φ ) = − 3 32 77 π ⋅ e 5 i φ ⋅ sin 5 θ {\displaystyle {\begin{aligned}Y_{5}^{-5}(\theta ,\varphi )&={3 \over 32}{\sqrt {77 \over \pi }}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \\Y_{5}^{-4}(\theta ,\varphi )&={3 \over 16}{\sqrt {385 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot \cos \theta \\Y_{5}^{-3}(\theta ,\varphi )&={1 \over 32}{\sqrt {385 \over \pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (9\cos ^{2}\theta -1)\\Y_{5}^{-2}(\theta ,\varphi )&={1 \over 8}{\sqrt {1155 \over 2\pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (3\cos ^{3}\theta -\cos \theta )\\Y_{5}^{-1}(\theta ,\varphi )&={1 \over 16}{\sqrt {165 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (21\cos ^{4}\theta -14\cos ^{2}\theta +1)\\Y_{5}^{0}(\theta ,\varphi )&={1 \over 16}{\sqrt {11 \over \pi }}\cdot (63\cos ^{5}\theta -70\cos ^{3}\theta +15\cos \theta )\\Y_{5}^{1}(\theta ,\varphi )&={-1 \over 16}{\sqrt {165 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (21\cos ^{4}\theta -14\cos ^{2}\theta +1)\\Y_{5}^{2}(\theta ,\varphi )&={1 \over 8}{\sqrt {1155 \over 2\pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (3\cos ^{3}\theta -\cos \theta )\\Y_{5}^{3}(\theta ,\varphi )&={-1 \over 32}{\sqrt {385 \over \pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot (9\cos ^{2}\theta -1)\\Y_{5}^{4}(\theta ,\varphi )&={3 \over 16}{\sqrt {385 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta \cdot \cos \theta \\Y_{5}^{5}(\theta ,\varphi )&={-3 \over 32}{\sqrt {77 \over \pi }}\cdot e^{5i\varphi }\cdot \sin ^{5}\theta \end{aligned}}}
ℓ= 6 Y 6 − 6 ( θ , φ ) = 1 64 3003 π ⋅ e − 6 i φ ⋅ sin 6 θ Y 6 − 5 ( θ , φ ) = 3 32 1001 π ⋅ e − 5 i φ ⋅ sin 5 θ ⋅ cos θ Y 6 − 4 ( θ , φ ) = 3 32 91 2 π ⋅ e − 4 i φ ⋅ sin 4 θ ⋅ ( 11 cos 2 θ − 1 ) Y 6 − 3 ( θ , φ ) = 1 32 1365 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ ( 11 cos 3 θ − 3 cos θ ) Y 6 − 2 ( θ , φ ) = 1 64 1365 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 33 cos 4 θ − 18 cos 2 θ + 1 ) Y 6 − 1 ( θ , φ ) = 1 16 273 2 π ⋅ e − i φ ⋅ sin θ ⋅ ( 33 cos 5 θ − 30 cos 3 θ + 5 cos θ ) Y 6 0 ( θ , φ ) = 1 32 13 π ⋅ ( 231 cos 6 θ − 315 cos 4 θ + 105 cos 2 θ − 5 ) Y 6 1 ( θ , φ ) = − 1 16 273 2 π ⋅ e i φ ⋅ sin θ ⋅ ( 33 cos 5 θ − 30 cos 3 θ + 5 cos θ ) Y 6 2 ( θ , φ ) = 1 64 1365 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 33 cos 4 θ − 18 cos 2 θ + 1 ) Y 6 3 ( θ , φ ) = − 1 32 1365 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ ( 11 cos 3 θ − 3 cos θ ) Y 6 4 ( θ , φ ) = 3 32 91 2 π ⋅ e 4 i φ ⋅ sin 4 θ ⋅ ( 11 cos 2 θ − 1 ) Y 6 5 ( θ , φ ) = − 3 32 1001 π ⋅ e 5 i φ ⋅ sin 5 θ ⋅ cos θ Y 6 6 ( θ , φ ) = 1 64 3003 π ⋅ e 6 i φ ⋅ sin 6 θ {\displaystyle {\begin{aligned}Y_{6}^{-6}(\theta ,\varphi )&={1 \over 64}{\sqrt {3003 \over \pi }}\cdot e^{-6i\varphi }\cdot \sin ^{6}\theta \\Y_{6}^{-5}(\theta ,\varphi )&={3 \over 32}{\sqrt {1001 \over \pi }}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \cdot \cos \theta \\Y_{6}^{-4}(\theta ,\varphi )&={3 \over 32}{\sqrt {91 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot (11\cos ^{2}\theta -1)\\Y_{6}^{-3}(\theta ,\varphi )&={1 \over 32}{\sqrt {1365 \over \pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (11\cos ^{3}\theta -3\cos \theta )\\Y_{6}^{-2}(\theta ,\varphi )&={1 \over 64}{\sqrt {1365 \over \pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (33\cos ^{4}\theta -18\cos ^{2}\theta +1)\\Y_{6}^{-1}(\theta ,\varphi )&={1 \over 16}{\sqrt {273 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (33\cos ^{5}\theta -30\cos ^{3}\theta +5\cos \theta )\\Y_{6}^{0}(\theta ,\varphi )&={1 \over 32}{\sqrt {13 \over \pi }}\cdot (231\cos ^{6}\theta -315\cos ^{4}\theta +105\cos ^{2}\theta -5)\\Y_{6}^{1}(\theta ,\varphi )&=-{1 \over 16}{\sqrt {273 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (33\cos ^{5}\theta -30\cos ^{3}\theta +5\cos \theta )\\Y_{6}^{2}(\theta ,\varphi )&={1 \over 64}{\sqrt {1365 \over \pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (33\cos ^{4}\theta -18\cos ^{2}\theta +1)\\Y_{6}^{3}(\theta ,\varphi )&=-{1 \over 32}{\sqrt {1365 \over \pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot (11\cos ^{3}\theta -3\cos \theta )\\Y_{6}^{4}(\theta ,\varphi )&={3 \over 32}{\sqrt {91 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta \cdot (11\cos ^{2}\theta -1)\\Y_{6}^{5}(\theta ,\varphi )&=-{3 \over 32}{\sqrt {1001 \over \pi }}\cdot e^{5i\varphi }\cdot \sin ^{5}\theta \cdot \cos \theta \\Y_{6}^{6}(\theta ,\varphi )&={1 \over 64}{\sqrt {3003 \over \pi }}\cdot e^{6i\varphi }\cdot \sin ^{6}\theta \end{aligned}}}
ℓ= 7 Y 7 − 7 ( θ , φ ) = 3 64 715 2 π ⋅ e − 7 i φ ⋅ sin 7 θ Y 7 − 6 ( θ , φ ) = 3 64 5005 π ⋅ e − 6 i φ ⋅ sin 6 θ ⋅ cos θ Y 7 − 5 ( θ , φ ) = 3 64 385 2 π ⋅ e − 5 i φ ⋅ sin 5 θ ⋅ ( 13 cos 2 θ − 1 ) Y 7 − 4 ( θ , φ ) = 3 32 385 2 π ⋅ e − 4 i φ ⋅ sin 4 θ ⋅ ( 13 cos 3 θ − 3 cos θ ) Y 7 − 3 ( θ , φ ) = 3 64 35 2 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ ( 143 cos 4 θ − 66 cos 2 θ + 3 ) Y 7 − 2 ( θ , φ ) = 3 64 35 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 143 cos 5 θ − 110 cos 3 θ + 15 cos θ ) Y 7 − 1 ( θ , φ ) = 1 64 105 2 π ⋅ e − i φ ⋅ sin θ ⋅ ( 429 cos 6 θ − 495 cos 4 θ + 135 cos 2 θ − 5 ) Y 7 0 ( θ , φ ) = 1 32 15 π ⋅ ( 429 cos 7 θ − 693 cos 5 θ + 315 cos 3 θ − 35 cos θ ) Y 7 1 ( θ , φ ) = − 1 64 105 2 π ⋅ e i φ ⋅ sin θ ⋅ ( 429 cos 6 θ − 495 cos 4 θ + 135 cos 2 θ − 5 ) Y 7 2 ( θ , φ ) = 3 64 35 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 143 cos 5 θ − 110 cos 3 θ + 15 cos θ ) Y 7 3 ( θ , φ ) = − 3 64 35 2 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ ( 143 cos 4 θ − 66 cos 2 θ + 3 ) Y 7 4 ( θ , φ ) = 3 32 385 2 π ⋅ e 4 i φ ⋅ sin 4 θ ⋅ ( 13 cos 3 θ − 3 cos θ ) Y 7 5 ( θ , φ ) = − 3 64 385 2 π ⋅ e 5 i φ ⋅ sin 5 θ ⋅ ( 13 cos 2 θ − 1 ) Y 7 6 ( θ , φ ) = 3 64 5005 π ⋅ e 6 i φ ⋅ sin 6 θ ⋅ cos θ Y 7 7 ( θ , φ ) = − 3 64 715 2 π ⋅ e 7 i φ ⋅ sin 7 θ {\displaystyle {\begin{aligned}Y_{7}^{-7}(\theta ,\varphi )&={3 \over 64}{\sqrt {715 \over 2\pi }}\cdot e^{-7i\varphi }\cdot \sin ^{7}\theta \\Y_{7}^{-6}(\theta ,\varphi )&={3 \over 64}{\sqrt {5005 \over \pi }}\cdot e^{-6i\varphi }\cdot \sin ^{6}\theta \cdot \cos \theta \\Y_{7}^{-5}(\theta ,\varphi )&={3 \over 64}{\sqrt {385 \over 2\pi }}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \cdot (13\cos ^{2}\theta -1)\\Y_{7}^{-4}(\theta ,\varphi )&={3 \over 32}{\sqrt {385 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot (13\cos ^{3}\theta -3\cos \theta )\\Y_{7}^{-3}(\theta ,\varphi )&={3 \over 64}{\sqrt {35 \over 2\pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (143\cos ^{4}\theta -66\cos ^{2}\theta +3)\\Y_{7}^{-2}(\theta ,\varphi )&={3 \over 64}{\sqrt {35 \over \pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (143\cos ^{5}\theta -110\cos ^{3}\theta +15\cos \theta )\\Y_{7}^{-1}(\theta ,\varphi )&={1 \over 64}{\sqrt {105 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (429\cos ^{6}\theta -495\cos ^{4}\theta +135\cos ^{2}\theta -5)\\Y_{7}^{0}(\theta ,\varphi )&={1 \over 32}{\sqrt {15 \over \pi }}\cdot (429\cos ^{7}\theta -693\cos ^{5}\theta +315\cos ^{3}\theta -35\cos \theta )\\Y_{7}^{1}(\theta ,\varphi )&=-{1 \over 64}{\sqrt {105 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (429\cos ^{6}\theta -495\cos ^{4}\theta +135\cos ^{2}\theta -5)\\Y_{7}^{2}(\theta ,\varphi )&={3 \over 64}{\sqrt {35 \over \pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (143\cos ^{5}\theta -110\cos ^{3}\theta +15\cos \theta )\\Y_{7}^{3}(\theta ,\varphi )&=-{3 \over 64}{\sqrt {35 \over 2\pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot (143\cos ^{4}\theta -66\cos ^{2}\theta +3)\\Y_{7}^{4}(\theta ,\varphi )&={3 \over 32}{\sqrt {385 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta \cdot (13\cos ^{3}\theta -3\cos \theta )\\Y_{7}^{5}(\theta ,\varphi )&=-{3 \over 64}{\sqrt {385 \over 2\pi }}\cdot e^{5i\varphi }\cdot \sin ^{5}\theta \cdot (13\cos ^{2}\theta -1)\\Y_{7}^{6}(\theta ,\varphi )&={3 \over 64}{\sqrt {5005 \over \pi }}\cdot e^{6i\varphi }\cdot \sin ^{6}\theta \cdot \cos \theta \\Y_{7}^{7}(\theta ,\varphi )&=-{3 \over 64}{\sqrt {715 \over 2\pi }}\cdot e^{7i\varphi }\cdot \sin ^{7}\theta \end{aligned}}}
ℓ= 8 Y 8 − 8 ( θ , φ ) = 3 256 12155 2 π ⋅ e − 8 i φ ⋅ sin 8 θ Y 8 − 7 ( θ , φ ) = 3 64 12155 2 π ⋅ e − 7 i φ ⋅ sin 7 θ ⋅ cos θ Y 8 − 6 ( θ , φ ) = 1 128 7293 π ⋅ e − 6 i φ ⋅ sin 6 θ ⋅ ( 15 cos 2 θ − 1 ) Y 8 − 5 ( θ , φ ) = 3 64 17017 2 π ⋅ e − 5 i φ ⋅ sin 5 θ ⋅ ( 5 cos 3 θ − cos θ ) Y 8 − 4 ( θ , φ ) = 3 128 1309 2 π ⋅ e − 4 i φ ⋅ sin 4 θ ⋅ ( 65 cos 4 θ − 26 cos 2 θ + 1 ) Y 8 − 3 ( θ , φ ) = 1 64 19635 2 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ ( 39 cos 5 θ − 26 cos 3 θ + 3 cos θ ) Y 8 − 2 ( θ , φ ) = 3 128 595 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 143 cos 6 θ − 143 cos 4 θ + 33 cos 2 θ − 1 ) Y 8 − 1 ( θ , φ ) = 3 64 17 2 π ⋅ e − i φ ⋅ sin θ ⋅ ( 715 cos 7 θ − 1001 cos 5 θ + 385 cos 3 θ − 35 cos θ ) Y 8 0 ( θ , φ ) = 1 256 17 π ⋅ ( 6435 cos 8 θ − 12012 cos 6 θ + 6930 cos 4 θ − 1260 cos 2 θ + 35 ) Y 8 1 ( θ , φ ) = − 3 64 17 2 π ⋅ e i φ ⋅ sin θ ⋅ ( 715 cos 7 θ − 1001 cos 5 θ + 385 cos 3 θ − 35 cos θ ) Y 8 2 ( θ , φ ) = 3 128 595 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 143 cos 6 θ − 143 cos 4 θ + 33 cos 2 θ − 1 ) Y 8 3 ( θ , φ ) = − 1 64 19635 2 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ ( 39 cos 5 θ − 26 cos 3 θ + 3 cos θ ) Y 8 4 ( θ , φ ) = 3 128 1309 2 π ⋅ e 4 i φ ⋅ sin 4 θ ⋅ ( 65 cos 4 θ − 26 cos 2 θ + 1 ) Y 8 5 ( θ , φ ) = − 3 64 17017 2 π ⋅ e 5 i φ ⋅ sin 5 θ ⋅ ( 5 cos 3 θ − cos θ ) Y 8 6 ( θ , φ ) = 1 128 7293 π ⋅ e 6 i φ ⋅ sin 6 θ ⋅ ( 15 cos 2 θ − 1 ) Y 8 7 ( θ , φ ) = − 3 64 12155 2 π ⋅ e 7 i φ ⋅ sin 7 θ ⋅ cos θ Y 8 8 ( θ , φ ) = 3 256 12155 2 π ⋅ e 8 i φ ⋅ sin 8 θ {\displaystyle {\begin{aligned}Y_{8}^{-8}(\theta ,\varphi )&={3 \over 256}{\sqrt {12155 \over 2\pi }}\cdot e^{-8i\varphi }\cdot \sin ^{8}\theta \\Y_{8}^{-7}(\theta ,\varphi )&={3 \over 64}{\sqrt {12155 \over 2\pi }}\cdot e^{-7i\varphi }\cdot \sin ^{7}\theta \cdot \cos \theta \\Y_{8}^{-6}(\theta ,\varphi )&={1 \over 128}{\sqrt {7293 \over \pi }}\cdot e^{-6i\varphi }\cdot \sin ^{6}\theta \cdot (15\cos ^{2}\theta -1)\\Y_{8}^{-5}(\theta ,\varphi )&={3 \over 64}{\sqrt {17017 \over 2\pi }}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \cdot (5\cos ^{3}\theta -\cos \theta )\\Y_{8}^{-4}(\theta ,\varphi )&={3 \over 128}{\sqrt {1309 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot (65\cos ^{4}\theta -26\cos ^{2}\theta +1)\\Y_{8}^{-3}(\theta ,\varphi )&={1 \over 64}{\sqrt {19635 \over 2\pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (39\cos ^{5}\theta -26\cos ^{3}\theta +3\cos \theta )\\Y_{8}^{-2}(\theta ,\varphi )&={3 \over 128}{\sqrt {595 \over \pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (143\cos ^{6}\theta -143\cos ^{4}\theta +33\cos ^{2}\theta -1)\\Y_{8}^{-1}(\theta ,\varphi )&={3 \over 64}{\sqrt {17 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (715\cos ^{7}\theta -1001\cos ^{5}\theta +385\cos ^{3}\theta -35\cos \theta )\\Y_{8}^{0}(\theta ,\varphi )&={1 \over 256}{\sqrt {17 \over \pi }}\cdot (6435\cos ^{8}\theta -12012\cos ^{6}\theta +6930\cos ^{4}\theta -1260\cos ^{2}\theta +35)\\Y_{8}^{1}(\theta ,\varphi )&={-3 \over 64}{\sqrt {17 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (715\cos ^{7}\theta -1001\cos ^{5}\theta +385\cos ^{3}\theta -35\cos \theta )\\Y_{8}^{2}(\theta ,\varphi )&={3 \over 128}{\sqrt {595 \over \pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (143\cos ^{6}\theta -143\cos ^{4}\theta +33\cos ^{2}\theta -1)\\Y_{8}^{3}(\theta ,\varphi )&={-1 \over 64}{\sqrt {19635 \over 2\pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot (39\cos ^{5}\theta -26\cos ^{3}\theta +3\cos \theta )\\Y_{8}^{4}(\theta ,\varphi )&={3 \over 128}{\sqrt {1309 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta \cdot (65\cos ^{4}\theta -26\cos ^{2}\theta +1)\\Y_{8}^{5}(\theta ,\varphi )&={-3 \over 64}{\sqrt {17017 \over 2\pi }}\cdot e^{5i\varphi }\cdot \sin ^{5}\theta \cdot (5\cos ^{3}\theta -\cos \theta )\\Y_{8}^{6}(\theta ,\varphi )&={1 \over 128}{\sqrt {7293 \over \pi }}\cdot e^{6i\varphi }\cdot \sin ^{6}\theta \cdot (15\cos ^{2}\theta -1)\\Y_{8}^{7}(\theta ,\varphi )&={-3 \over 64}{\sqrt {12155 \over 2\pi }}\cdot e^{7i\varphi }\cdot \sin ^{7}\theta \cdot \cos \theta \\Y_{8}^{8}(\theta ,\varphi )&={3 \over 256}{\sqrt {12155 \over 2\pi }}\cdot e^{8i\varphi }\cdot \sin ^{8}\theta \end{aligned}}}
ℓ= 9 Y 9 − 9 ( θ , φ ) = 1 512 230945 π ⋅ e − 9 i φ ⋅ sin 9 θ Y 9 − 8 ( θ , φ ) = 3 256 230945 2 π ⋅ e − 8 i φ ⋅ sin 8 θ ⋅ cos θ Y 9 − 7 ( θ , φ ) = 3 512 13585 π ⋅ e − 7 i φ ⋅ sin 7 θ ⋅ ( 17 cos 2 θ − 1 ) Y 9 − 6 ( θ , φ ) = 1 128 40755 π ⋅ e − 6 i φ ⋅ sin 6 θ ⋅ ( 17 cos 3 θ − 3 cos θ ) Y 9 − 5 ( θ , φ ) = 3 256 2717 π ⋅ e − 5 i φ ⋅ sin 5 θ ⋅ ( 85 cos 4 θ − 30 cos 2 θ + 1 ) Y 9 − 4 ( θ , φ ) = 3 128 95095 2 π ⋅ e − 4 i φ ⋅ sin 4 θ ⋅ ( 17 cos 5 θ − 10 cos 3 θ + cos θ ) Y 9 − 3 ( θ , φ ) = 1 256 21945 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ ( 221 cos 6 θ − 195 cos 4 θ + 39 cos 2 θ − 1 ) Y 9 − 2 ( θ , φ ) = 3 128 1045 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 221 cos 7 θ − 273 cos 5 θ + 91 cos 3 θ − 7 cos θ ) Y 9 − 1 ( θ , φ ) = 3 256 95 2 π ⋅ e − i φ ⋅ sin θ ⋅ ( 2431 cos 8 θ − 4004 cos 6 θ + 2002 cos 4 θ − 308 cos 2 θ + 7 ) Y 9 0 ( θ , φ ) = 1 256 19 π ⋅ ( 12155 cos 9 θ − 25740 cos 7 θ + 18018 cos 5 θ − 4620 cos 3 θ + 315 cos θ ) Y 9 1 ( θ , φ ) = − 3 256 95 2 π ⋅ e i φ ⋅ sin θ ⋅ ( 2431 cos 8 θ − 4004 cos 6 θ + 2002 cos 4 θ − 308 cos 2 θ + 7 ) Y 9 2 ( θ , φ ) = 3 128 1045 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 221 cos 7 θ − 273 cos 5 θ + 91 cos 3 θ − 7 cos θ ) Y 9 3 ( θ , φ ) = − 1 256 21945 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ ( 221 cos 6 θ − 195 cos 4 θ + 39 cos 2 θ − 1 ) Y 9 4 ( θ , φ ) = 3 128 95095 2 π ⋅ e 4 i φ ⋅ sin 4 θ ⋅ ( 17 cos 5 θ − 10 cos 3 θ + cos θ ) Y 9 5 ( θ , φ ) = − 3 256 2717 π ⋅ e 5 i φ ⋅ sin 5 θ ⋅ ( 85 cos 4 θ − 30 cos 2 θ + 1 ) Y 9 6 ( θ , φ ) = 1 128 40755 π ⋅ e 6 i φ ⋅ sin 6 θ ⋅ ( 17 cos 3 θ − 3 cos θ ) Y 9 7 ( θ , φ ) = − 3 512 13585 π ⋅ e 7 i φ ⋅ sin 7 θ ⋅ ( 17 cos 2 θ − 1 ) Y 9 8 ( θ , φ ) = 3 256 230945 2 π ⋅ e 8 i φ ⋅ sin 8 θ ⋅ cos θ Y 9 9 ( θ , φ ) = − 1 512 230945 π ⋅ e 9 i φ ⋅ sin 9 θ {\displaystyle {\begin{aligned}Y_{9}^{-9}(\theta ,\varphi )&={1 \over 512}{\sqrt {230945 \over \pi }}\cdot e^{-9i\varphi }\cdot \sin ^{9}\theta \\Y_{9}^{-8}(\theta ,\varphi )&={3 \over 256}{\sqrt {230945 \over 2\pi }}\cdot e^{-8i\varphi }\cdot \sin ^{8}\theta \cdot \cos \theta \\Y_{9}^{-7}(\theta ,\varphi )&={3 \over 512}{\sqrt {13585 \over \pi }}\cdot e^{-7i\varphi }\cdot \sin ^{7}\theta \cdot (17\cos ^{2}\theta -1)\\Y_{9}^{-6}(\theta ,\varphi )&={1 \over 128}{\sqrt {40755 \over \pi }}\cdot e^{-6i\varphi }\cdot \sin ^{6}\theta \cdot (17\cos ^{3}\theta -3\cos \theta )\\Y_{9}^{-5}(\theta ,\varphi )&={3 \over 256}{\sqrt {2717 \over \pi }}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \cdot (85\cos ^{4}\theta -30\cos ^{2}\theta +1)\\Y_{9}^{-4}(\theta ,\varphi )&={3 \over 128}{\sqrt {95095 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot (17\cos ^{5}\theta -10\cos ^{3}\theta +\cos \theta )\\Y_{9}^{-3}(\theta ,\varphi )&={1 \over 256}{\sqrt {21945 \over \pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (221\cos ^{6}\theta -195\cos ^{4}\theta +39\cos ^{2}\theta -1)\\Y_{9}^{-2}(\theta ,\varphi )&={3 \over 128}{\sqrt {1045 \over \pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (221\cos ^{7}\theta -273\cos ^{5}\theta +91\cos ^{3}\theta -7\cos \theta )\\Y_{9}^{-1}(\theta ,\varphi )&={3 \over 256}{\sqrt {95 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (2431\cos ^{8}\theta -4004\cos ^{6}\theta +2002\cos ^{4}\theta -308\cos ^{2}\theta +7)\\Y_{9}^{0}(\theta ,\varphi )&={1 \over 256}{\sqrt {19 \over \pi }}\cdot (12155\cos ^{9}\theta -25740\cos ^{7}\theta +18018\cos ^{5}\theta -4620\cos ^{3}\theta +315\cos \theta )\\Y_{9}^{1}(\theta ,\varphi )&={-3 \over 256}{\sqrt {95 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (2431\cos ^{8}\theta -4004\cos ^{6}\theta +2002\cos ^{4}\theta -308\cos ^{2}\theta +7)\\Y_{9}^{2}(\theta ,\varphi )&={3 \over 128}{\sqrt {1045 \over \pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (221\cos ^{7}\theta -273\cos ^{5}\theta +91\cos ^{3}\theta -7\cos \theta )\\Y_{9}^{3}(\theta ,\varphi )&={-1 \over 256}{\sqrt {21945 \over \pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot (221\cos ^{6}\theta -195\cos ^{4}\theta +39\cos ^{2}\theta -1)\\Y_{9}^{4}(\theta ,\varphi )&={3 \over 128}{\sqrt {95095 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta \cdot (17\cos ^{5}\theta -10\cos ^{3}\theta +\cos \theta )\\Y_{9}^{5}(\theta ,\varphi )&={-3 \over 256}{\sqrt {2717 \over \pi }}\cdot e^{5i\varphi }\cdot \sin ^{5}\theta \cdot (85\cos ^{4}\theta -30\cos ^{2}\theta +1)\\Y_{9}^{6}(\theta ,\varphi )&={1 \over 128}{\sqrt {40755 \over \pi }}\cdot e^{6i\varphi }\cdot \sin ^{6}\theta \cdot (17\cos ^{3}\theta -3\cos \theta )\\Y_{9}^{7}(\theta ,\varphi )&={-3 \over 512}{\sqrt {13585 \over \pi }}\cdot e^{7i\varphi }\cdot \sin ^{7}\theta \cdot (17\cos ^{2}\theta -1)\\Y_{9}^{8}(\theta ,\varphi )&={3 \over 256}{\sqrt {230945 \over 2\pi }}\cdot e^{8i\varphi }\cdot \sin ^{8}\theta \cdot \cos \theta \\Y_{9}^{9}(\theta ,\varphi )&={-1 \over 512}{\sqrt {230945 \over \pi }}\cdot e^{9i\varphi }\cdot \sin ^{9}\theta \end{aligned}}}
ℓ= 10 Y 10 − 10 ( θ , φ ) = 1 1024 969969 π ⋅ e − 10 i φ ⋅ sin 10 θ Y 10 − 9 ( θ , φ ) = 1 512 4849845 π ⋅ e − 9 i φ ⋅ sin 9 θ ⋅ cos θ Y 10 − 8 ( θ , φ ) = 1 512 255255 2 π ⋅ e − 8 i φ ⋅ sin 8 θ ⋅ ( 19 cos 2 θ − 1 ) Y 10 − 7 ( θ , φ ) = 3 512 85085 π ⋅ e − 7 i φ ⋅ sin 7 θ ⋅ ( 19 cos 3 θ − 3 cos θ ) Y 10 − 6 ( θ , φ ) = 3 1024 5005 π ⋅ e − 6 i φ ⋅ sin 6 θ ⋅ ( 323 cos 4 θ − 102 cos 2 θ + 3 ) Y 10 − 5 ( θ , φ ) = 3 256 1001 π ⋅ e − 5 i φ ⋅ sin 5 θ ⋅ ( 323 cos 5 θ − 170 cos 3 θ + 15 cos θ ) Y 10 − 4 ( θ , φ ) = 3 256 5005 2 π ⋅ e − 4 i φ ⋅ sin 4 θ ⋅ ( 323 cos 6 θ − 255 cos 4 θ + 45 cos 2 θ − 1 ) Y 10 − 3 ( θ , φ ) = 3 256 5005 π ⋅ e − 3 i φ ⋅ sin 3 θ ⋅ ( 323 cos 7 θ − 357 cos 5 θ + 105 cos 3 θ − 7 cos θ ) Y 10 − 2 ( θ , φ ) = 3 512 385 2 π ⋅ e − 2 i φ ⋅ sin 2 θ ⋅ ( 4199 cos 8 θ − 6188 cos 6 θ + 2730 cos 4 θ − 364 cos 2 θ + 7 ) Y 10 − 1 ( θ , φ ) = 1 256 1155 2 π ⋅ e − i φ ⋅ sin θ ⋅ ( 4199 cos 9 θ − 7956 cos 7 θ + 4914 cos 5 θ − 1092 cos 3 θ + 63 cos θ ) Y 10 0 ( θ , φ ) = 1 512 21 π ⋅ ( 46189 cos 10 θ − 109395 cos 8 θ + 90090 cos 6 θ − 30030 cos 4 θ + 3465 cos 2 θ − 63 ) Y 10 1 ( θ , φ ) = − 1 256 1155 2 π ⋅ e i φ ⋅ sin θ ⋅ ( 4199 cos 9 θ − 7956 cos 7 θ + 4914 cos 5 θ − 1092 cos 3 θ + 63 cos θ ) Y 10 2 ( θ , φ ) = 3 512 385 2 π ⋅ e 2 i φ ⋅ sin 2 θ ⋅ ( 4199 cos 8 θ − 6188 cos 6 θ + 2730 cos 4 θ − 364 cos 2 θ + 7 ) Y 10 3 ( θ , φ ) = − 3 256 5005 π ⋅ e 3 i φ ⋅ sin 3 θ ⋅ ( 323 cos 7 θ − 357 cos 5 θ + 105 cos 3 θ − 7 cos θ ) Y 10 4 ( θ , φ ) = 3 256 5005 2 π ⋅ e 4 i φ ⋅ sin 4 θ ⋅ ( 323 cos 6 θ − 255 cos 4 θ + 45 cos 2 θ − 1 ) Y 10 5 ( θ , φ ) = − 3 256 1001 π ⋅ e 5 i φ ⋅ sin 5 θ ⋅ ( 323 cos 5 θ − 170 cos 3 θ + 15 cos θ ) Y 10 6 ( θ , φ ) = 3 1024 5005 π ⋅ e 6 i φ ⋅ sin 6 θ ⋅ ( 323 cos 4 θ − 102 cos 2 θ + 3 ) Y 10 7 ( θ , φ ) = − 3 512 85085 π ⋅ e 7 i φ ⋅ sin 7 θ ⋅ ( 19 cos 3 θ − 3 cos θ ) Y 10 8 ( θ , φ ) = 1 512 255255 2 π ⋅ e 8 i φ ⋅ sin 8 θ ⋅ ( 19 cos 2 θ − 1 ) Y 10 9 ( θ , φ ) = − 1 512 4849845 π ⋅ e 9 i φ ⋅ sin 9 θ ⋅ cos θ Y 10 10 ( θ , φ ) = 1 1024 969969 π ⋅ e 10 i φ ⋅ sin 10 θ {\displaystyle {\begin{aligned}Y_{10}^{-10}(\theta ,\varphi )&={1 \over 1024}{\sqrt {969969 \over \pi }}\cdot e^{-10i\varphi }\cdot \sin ^{10}\theta \\Y_{10}^{-9}(\theta ,\varphi )&={1 \over 512}{\sqrt {4849845 \over \pi }}\cdot e^{-9i\varphi }\cdot \sin ^{9}\theta \cdot \cos \theta \\Y_{10}^{-8}(\theta ,\varphi )&={1 \over 512}{\sqrt {255255 \over 2\pi }}\cdot e^{-8i\varphi }\cdot \sin ^{8}\theta \cdot (19\cos ^{2}\theta -1)\\Y_{10}^{-7}(\theta ,\varphi )&={3 \over 512}{\sqrt {85085 \over \pi }}\cdot e^{-7i\varphi }\cdot \sin ^{7}\theta \cdot (19\cos ^{3}\theta -3\cos \theta )\\Y_{10}^{-6}(\theta ,\varphi )&={3 \over 1024}{\sqrt {5005 \over \pi }}\cdot e^{-6i\varphi }\cdot \sin ^{6}\theta \cdot (323\cos ^{4}\theta -102\cos ^{2}\theta +3)\\Y_{10}^{-5}(\theta ,\varphi )&={3 \over 256}{\sqrt {1001 \over \pi }}\cdot e^{-5i\varphi }\cdot \sin ^{5}\theta \cdot (323\cos ^{5}\theta -170\cos ^{3}\theta +15\cos \theta )\\Y_{10}^{-4}(\theta ,\varphi )&={3 \over 256}{\sqrt {5005 \over 2\pi }}\cdot e^{-4i\varphi }\cdot \sin ^{4}\theta \cdot (323\cos ^{6}\theta -255\cos ^{4}\theta +45\cos ^{2}\theta -1)\\Y_{10}^{-3}(\theta ,\varphi )&={3 \over 256}{\sqrt {5005 \over \pi }}\cdot e^{-3i\varphi }\cdot \sin ^{3}\theta \cdot (323\cos ^{7}\theta -357\cos ^{5}\theta +105\cos ^{3}\theta -7\cos \theta )\\Y_{10}^{-2}(\theta ,\varphi )&={3 \over 512}{\sqrt {385 \over 2\pi }}\cdot e^{-2i\varphi }\cdot \sin ^{2}\theta \cdot (4199\cos ^{8}\theta -6188\cos ^{6}\theta +2730\cos ^{4}\theta -364\cos ^{2}\theta +7)\\Y_{10}^{-1}(\theta ,\varphi )&={1 \over 256}{\sqrt {1155 \over 2\pi }}\cdot e^{-i\varphi }\cdot \sin \theta \cdot (4199\cos ^{9}\theta -7956\cos ^{7}\theta +4914\cos ^{5}\theta -1092\cos ^{3}\theta +63\cos \theta )\\Y_{10}^{0}(\theta ,\varphi )&={1 \over 512}{\sqrt {21 \over \pi }}\cdot (46189\cos ^{10}\theta -109395\cos ^{8}\theta +90090\cos ^{6}\theta -30030\cos ^{4}\theta +3465\cos ^{2}\theta -63)\\Y_{10}^{1}(\theta ,\varphi )&={-1 \over 256}{\sqrt {1155 \over 2\pi }}\cdot e^{i\varphi }\cdot \sin \theta \cdot (4199\cos ^{9}\theta -7956\cos ^{7}\theta +4914\cos ^{5}\theta -1092\cos ^{3}\theta +63\cos \theta )\\Y_{10}^{2}(\theta ,\varphi )&={3 \over 512}{\sqrt {385 \over 2\pi }}\cdot e^{2i\varphi }\cdot \sin ^{2}\theta \cdot (4199\cos ^{8}\theta -6188\cos ^{6}\theta +2730\cos ^{4}\theta -364\cos ^{2}\theta +7)\\Y_{10}^{3}(\theta ,\varphi )&={-3 \over 256}{\sqrt {5005 \over \pi }}\cdot e^{3i\varphi }\cdot \sin ^{3}\theta \cdot (323\cos ^{7}\theta -357\cos ^{5}\theta +105\cos ^{3}\theta -7\cos \theta )\\Y_{10}^{4}(\theta ,\varphi )&={3 \over 256}{\sqrt {5005 \over 2\pi }}\cdot e^{4i\varphi }\cdot \sin ^{4}\theta \cdot (323\cos ^{6}\theta -255\cos ^{4}\theta +45\cos ^{2}\theta -1)\\Y_{10}^{5}(\theta ,\varphi )&={-3 \over 256}{\sqrt {1001 \over \pi }}\cdot e^{5i\varphi }\cdot \sin ^{5}\theta \cdot (323\cos ^{5}\theta -170\cos ^{3}\theta +15\cos \theta )\\Y_{10}^{6}(\theta ,\varphi )&={3 \over 1024}{\sqrt {5005 \over \pi }}\cdot e^{6i\varphi }\cdot \sin ^{6}\theta \cdot (323\cos ^{4}\theta -102\cos ^{2}\theta +3)\\Y_{10}^{7}(\theta ,\varphi )&={-3 \over 512}{\sqrt {85085 \over \pi }}\cdot e^{7i\varphi }\cdot \sin ^{7}\theta \cdot (19\cos ^{3}\theta -3\cos \theta )\\Y_{10}^{8}(\theta ,\varphi )&={1 \over 512}{\sqrt {255255 \over 2\pi }}\cdot e^{8i\varphi }\cdot \sin ^{8}\theta \cdot (19\cos ^{2}\theta -1)\\Y_{10}^{9}(\theta ,\varphi )&={-1 \over 512}{\sqrt {4849845 \over \pi }}\cdot e^{9i\varphi }\cdot \sin ^{9}\theta \cdot \cos \theta \\Y_{10}^{10}(\theta ,\varphi )&={1 \over 1024}{\sqrt {969969 \over \pi }}\cdot e^{10i\varphi }\cdot \sin ^{10}\theta \end{aligned}}}
Visualización de armónicos esféricos complejos
Mapas de ángulos polares/azimutales 2D A continuación, se representan los armónicos esféricos complejos en gráficos 2D con el ángulo azimutal, , en el eje horizontal y el ángulo polar, , en el eje vertical. La saturación del color en cualquier punto representa la magnitud del armónico esférico y el tono representa la fase. ϕ {\displaystyle \phi } θ {\displaystyle \theta }
Las líneas nodales de latitud se ven como líneas blancas horizontales. Las líneas nodales de longitud se ven como líneas blancas verticales.
Matriz visual de armónicos esféricos complejos representados como mapas theta/phi 2D
Diagramas polares A continuación, se representan los armónicos esféricos complejos en gráficos polares. La magnitud del armónico esférico en ángulos polares y azimutales particulares se representa mediante la saturación del color en ese punto y la fase se representa mediante el tono en ese punto.
Matriz visual de armónicos esféricos complejos representados con un diagrama polar
Diagramas polares con magnitud como radio A continuación, se representan los armónicos esféricos complejos en gráficos polares. La magnitud del armónico esférico en ángulos polares y azimutales particulares se representa mediante el radio del gráfico en ese punto y la fase se representa mediante el tono en ese punto.
Matriz visual de armónicos esféricos complejos representados con un diagrama polar con magnitud asignada al radio
Armónicos esféricos reales Para cada armónico esférico real, también se informa el símbolo orbital atómico correspondiente ( s , p , d , f ). [2] [3]
Para ℓ = 0, …, 3, véase. [4] [5]
ℓ= 0 Y 00 = s = Y 0 0 = 1 2 1 π {\displaystyle Y_{00}=s=Y_{0}^{0}={\frac {1}{2}}{\sqrt {\frac {1}{\pi }}}}
ℓ= 1 Y 1 , − 1 = p y = i 1 2 ( Y 1 − 1 + Y 1 1 ) = 3 4 π ⋅ y r = 3 4 π sin ( θ ) sin φ Y 1 , 0 = p z = Y 1 0 = 3 4 π ⋅ z r = 3 4 π cos ( θ ) Y 1 , 1 = p x = 1 2 ( Y 1 − 1 − Y 1 1 ) = 3 4 π ⋅ x r = 3 4 π sin ( θ ) cos φ {\displaystyle {\begin{aligned}Y_{1,-1}&=p_{y}=i{\sqrt {\frac {1}{2}}}\left(Y_{1}^{-1}+Y_{1}^{1}\right)={\sqrt {\frac {3}{4\pi }}}\cdot {\frac {y}{r}}={\sqrt {\frac {3}{4\pi }}}\sin(\theta )\sin \varphi \\Y_{1,0}&=p_{z}=Y_{1}^{0}={\sqrt {\frac {3}{4\pi }}}\cdot {\frac {z}{r}}={\sqrt {\frac {3}{4\pi }}}\cos(\theta )\\Y_{1,1}&=p_{x}={\sqrt {\frac {1}{2}}}\left(Y_{1}^{-1}-Y_{1}^{1}\right)={\sqrt {\frac {3}{4\pi }}}\cdot {\frac {x}{r}}={\sqrt {\frac {3}{4\pi }}}\sin(\theta )\cos \varphi \end{aligned}}}
ℓ= 2 Y 2 , − 2 = d x y = i 1 2 ( Y 2 − 2 − Y 2 2 ) = 1 2 15 π ⋅ x y r 2 = 1 4 15 π sin 2 θ sin ( 2 φ ) Y 2 , − 1 = d y z = i 1 2 ( Y 2 − 1 + Y 2 1 ) = 1 2 15 π ⋅ y ⋅ z r 2 = 1 4 15 π sin ( 2 θ ) sin φ Y 2 , 0 = d z 2 = Y 2 0 = 1 4 5 π ⋅ 3 z 2 − r 2 r 2 = 1 4 5 π ( 3 cos 2 θ − 1 ) Y 2 , 1 = d x z = 1 2 ( Y 2 − 1 − Y 2 1 ) = 1 2 15 π ⋅ x ⋅ z r 2 = 1 4 15 π sin ( 2 θ ) cos φ Y 2 , 2 = d x 2 − y 2 = 1 2 ( Y 2 − 2 + Y 2 2 ) = 1 4 15 π ⋅ x 2 − y 2 r 2 = 1 4 15 π sin 2 θ cos ( 2 φ ) {\displaystyle {\begin{aligned}Y_{2,-2}&=d_{xy}=i{\sqrt {\frac {1}{2}}}\left(Y_{2}^{-2}-Y_{2}^{2}\right)={\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\cdot {\frac {xy}{r^{2}}}={\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \sin(2\varphi )\\Y_{2,-1}&=d_{yz}=i{\sqrt {\frac {1}{2}}}\left(Y_{2}^{-1}+Y_{2}^{1}\right)={\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\cdot {\frac {y\cdot z}{r^{2}}}={\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin(2\theta )\sin \varphi \\Y_{2,0}&=d_{z^{2}}=Y_{2}^{0}={\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}\cdot {\frac {3z^{2}-r^{2}}{r^{2}}}={\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}(3\cos ^{2}\theta -1)\\Y_{2,1}&=d_{xz}={\sqrt {\frac {1}{2}}}\left(Y_{2}^{-1}-Y_{2}^{1}\right)={\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\cdot {\frac {x\cdot z}{r^{2}}}={\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin(2\theta )\cos \varphi \\Y_{2,2}&=d_{x^{2}-y^{2}}={\sqrt {\frac {1}{2}}}\left(Y_{2}^{-2}+Y_{2}^{2}\right)={\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\cdot {\frac {x^{2}-y^{2}}{r^{2}}}={\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \cos(2\varphi )\end{aligned}}}
ℓ= 3 Y 3 , − 3 = f y ( 3 x 2 − y 2 ) = i 1 2 ( Y 3 − 3 + Y 3 3 ) = 1 4 35 2 π ⋅ y ( 3 x 2 − y 2 ) r 3 Y 3 , − 2 = f x y z = i 1 2 ( Y 3 − 2 − Y 3 2 ) = 1 2 105 π ⋅ x y ⋅ z r 3 Y 3 , − 1 = f y z 2 = i 1 2 ( Y 3 − 1 + Y 3 1 ) = 1 4 21 2 π ⋅ y ⋅ ( 5 z 2 − r 2 ) r 3 Y 3 , 0 = f z 3 = Y 3 0 = 1 4 7 π ⋅ 5 z 3 − 3 z r 2 r 3 Y 3 , 1 = f x z 2 = 1 2 ( Y 3 − 1 − Y 3 1 ) = 1 4 21 2 π ⋅ x ⋅ ( 5 z 2 − r 2 ) r 3 Y 3 , 2 = f z ( x 2 − y 2 ) = 1 2 ( Y 3 − 2 + Y 3 2 ) = 1 4 105 π ⋅ ( x 2 − y 2 ) ⋅ z r 3 Y 3 , 3 = f x ( x 2 − 3 y 2 ) = 1 2 ( Y 3 − 3 − Y 3 3 ) = 1 4 35 2 π ⋅ x ( x 2 − 3 y 2 ) r 3 {\displaystyle {\begin{aligned}Y_{3,-3}&=f_{y(3x^{2}-y^{2})}=i{\sqrt {\frac {1}{2}}}\left(Y_{3}^{-3}+Y_{3}^{3}\right)={\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\cdot {\frac {y\left(3x^{2}-y^{2}\right)}{r^{3}}}\\Y_{3,-2}&=f_{xyz}=i{\sqrt {\frac {1}{2}}}\left(Y_{3}^{-2}-Y_{3}^{2}\right)={\frac {1}{2}}{\sqrt {\frac {105}{\pi }}}\cdot {\frac {xy\cdot z}{r^{3}}}\\Y_{3,-1}&=f_{yz^{2}}=i{\sqrt {\frac {1}{2}}}\left(Y_{3}^{-1}+Y_{3}^{1}\right)={\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}\cdot {\frac {y\cdot (5z^{2}-r^{2})}{r^{3}}}\\Y_{3,0}&=f_{z^{3}}=Y_{3}^{0}={\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}\cdot {\frac {5z^{3}-3zr^{2}}{r^{3}}}\\Y_{3,1}&=f_{xz^{2}}={\sqrt {\frac {1}{2}}}\left(Y_{3}^{-1}-Y_{3}^{1}\right)={\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}\cdot {\frac {x\cdot (5z^{2}-r^{2})}{r^{3}}}\\Y_{3,2}&=f_{z(x^{2}-y^{2})}={\sqrt {\frac {1}{2}}}\left(Y_{3}^{-2}+Y_{3}^{2}\right)={\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cdot {\frac {\left(x^{2}-y^{2}\right)\cdot z}{r^{3}}}\\Y_{3,3}&=f_{x(x^{2}-3y^{2})}={\sqrt {\frac {1}{2}}}\left(Y_{3}^{-3}-Y_{3}^{3}\right)={\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\cdot {\frac {x\left(x^{2}-3y^{2}\right)}{r^{3}}}\end{aligned}}}
ℓ= 4 Y 4 , − 4 = i 1 2 ( Y 4 − 4 − Y 4 4 ) = 3 4 35 π ⋅ x y ( x 2 − y 2 ) r 4 Y 4 , − 3 = i 1 2 ( Y 4 − 3 + Y 4 3 ) = 3 4 35 2 π ⋅ y ( 3 x 2 − y 2 ) ⋅ z r 4 Y 4 , − 2 = i 1 2 ( Y 4 − 2 − Y 4 2 ) = 3 4 5 π ⋅ x y ⋅ ( 7 z 2 − r 2 ) r 4 Y 4 , − 1 = i 1 2 ( Y 4 − 1 + Y 4 1 ) = 3 4 5 2 π ⋅ y ⋅ ( 7 z 3 − 3 z r 2 ) r 4 Y 4 , 0 = Y 4 0 = 3 16 1 π ⋅ 35 z 4 − 30 z 2 r 2 + 3 r 4 r 4 Y 4 , 1 = 1 2 ( Y 4 − 1 − Y 4 1 ) = 3 4 5 2 π ⋅ x ⋅ ( 7 z 3 − 3 z r 2 ) r 4 Y 4 , 2 = 1 2 ( Y 4 − 2 + Y 4 2 ) = 3 8 5 π ⋅ ( x 2 − y 2 ) ⋅ ( 7 z 2 − r 2 ) r 4 Y 4 , 3 = 1 2 ( Y 4 − 3 − Y 4 3 ) = 3 4 35 2 π ⋅ x ( x 2 − 3 y 2 ) ⋅ z r 4 Y 4 , 4 = 1 2 ( Y 4 − 4 + Y 4 4 ) = 3 16 35 π ⋅ x 2 ( x 2 − 3 y 2 ) − y 2 ( 3 x 2 − y 2 ) r 4 {\displaystyle {\begin{aligned}Y_{4,-4}&=i{\sqrt {\frac {1}{2}}}\left(Y_{4}^{-4}-Y_{4}^{4}\right)={\frac {3}{4}}{\sqrt {\frac {35}{\pi }}}\cdot {\frac {xy\left(x^{2}-y^{2}\right)}{r^{4}}}\\Y_{4,-3}&=i{\sqrt {\frac {1}{2}}}\left(Y_{4}^{-3}+Y_{4}^{3}\right)={\frac {3}{4}}{\sqrt {\frac {35}{2\pi }}}\cdot {\frac {y(3x^{2}-y^{2})\cdot z}{r^{4}}}\\Y_{4,-2}&=i{\sqrt {\frac {1}{2}}}\left(Y_{4}^{-2}-Y_{4}^{2}\right)={\frac {3}{4}}{\sqrt {\frac {5}{\pi }}}\cdot {\frac {xy\cdot (7z^{2}-r^{2})}{r^{4}}}\\Y_{4,-1}&=i{\sqrt {\frac {1}{2}}}\left(Y_{4}^{-1}+Y_{4}^{1}\right)={\frac {3}{4}}{\sqrt {\frac {5}{2\pi }}}\cdot {\frac {y\cdot (7z^{3}-3zr^{2})}{r^{4}}}\\Y_{4,0}&=Y_{4}^{0}={\frac {3}{16}}{\sqrt {\frac {1}{\pi }}}\cdot {\frac {35z^{4}-30z^{2}r^{2}+3r^{4}}{r^{4}}}\\Y_{4,1}&={\sqrt {\frac {1}{2}}}\left(Y_{4}^{-1}-Y_{4}^{1}\right)={\frac {3}{4}}{\sqrt {\frac {5}{2\pi }}}\cdot {\frac {x\cdot (7z^{3}-3zr^{2})}{r^{4}}}\\Y_{4,2}&={\sqrt {\frac {1}{2}}}\left(Y_{4}^{-2}+Y_{4}^{2}\right)={\frac {3}{8}}{\sqrt {\frac {5}{\pi }}}\cdot {\frac {(x^{2}-y^{2})\cdot (7z^{2}-r^{2})}{r^{4}}}\\Y_{4,3}&={\sqrt {\frac {1}{2}}}\left(Y_{4}^{-3}-Y_{4}^{3}\right)={\frac {3}{4}}{\sqrt {\frac {35}{2\pi }}}\cdot {\frac {x(x^{2}-3y^{2})\cdot z}{r^{4}}}\\Y_{4,4}&={\sqrt {\frac {1}{2}}}\left(Y_{4}^{-4}+Y_{4}^{4}\right)={\frac {3}{16}}{\sqrt {\frac {35}{\pi }}}\cdot {\frac {x^{2}\left(x^{2}-3y^{2}\right)-y^{2}\left(3x^{2}-y^{2}\right)}{r^{4}}}\end{aligned}}}
Visualización de armónicos esféricos reales
Mapas de ángulos polares/azimutales 2D A continuación, se representan los armónicos esféricos reales en gráficos 2D con el ángulo azimutal, , en el eje horizontal y el ángulo polar, , en el eje vertical. La saturación del color en cualquier punto representa la magnitud del armónico esférico. Los valores positivos son rojos y los negativos, verde azulado. ϕ {\displaystyle \phi } θ {\displaystyle \theta }
Las líneas nodales de latitud se ven como líneas blancas horizontales. Las líneas nodales de longitud se ven como líneas blancas verticales.
Matriz visual de armónicos esféricos reales representados como mapas theta/phi 2D
Diagramas polares A continuación se representan los armónicos esféricos reales en gráficos polares. La magnitud del armónico esférico en ángulos polares y azimutales particulares se representa mediante la saturación del color en ese punto y la fase se representa mediante el tono en ese punto.
Matriz visual de armónicos esféricos reales representados con un diagrama polar
Diagramas polares con magnitud como radio A continuación se muestran los armónicos esféricos reales en gráficos polares. La magnitud del armónico esférico en ángulos polares y azimutales particulares se representa mediante el radio del gráfico en ese punto y la fase se representa mediante el tono en ese punto.
Matriz visual de armónicos esféricos reales representados con un diagrama polar con magnitud asignada al radio
Diagramas polares con amplitud como elevación A continuación, se representan los armónicos esféricos reales en diagramas polares. La amplitud del armónico esférico (magnitud y signo) en un ángulo polar y azimutal particular se representa mediante la elevación del diagrama en ese punto por encima o por debajo de la superficie de una esfera uniforme. La magnitud también se representa mediante la saturación del color en un punto determinado. La fase se representa mediante el tono en un punto determinado.
Matriz visual de armónicos esféricos reales representados con un diagrama polar con amplitud asignada a la elevación y la saturación
Véase también
Enlaces externos Armónicos esféricos en MathWorld Representación 3D de armónicos esféricos
Referencias
Referencias citadas ^ DA Varshalovich; AN Moskalev; VK Khersonskii (1988). Teoría cuántica del momento angular: tensores irreducibles, armónicos esféricos, coeficientes de acoplamiento vectorial, símbolos 3nj (1.ª ed. repr.). Singapur: World Scientific Pub. págs. 155-156. ISBN 9971-50-107-4 ^ Petrucci (2016). Química general: principios y aplicaciones modernas . Prentice Hall. ISBN 0133897311 ^ Friedman (1964). "Las formas de los orbitales f". J. Chem. Educ . 41 (7): 354. ^ CDH Chisholm (1976). Técnicas teóricas de grupo en química cuántica . Nueva York: Academic Press. ISBN 0-12-172950-8 ^ Blanco, Miguel A.; Flórez, M.; Bermejo, M. (1 de diciembre de 1997). "Evaluación de las matrices de rotación en base a armónicos esféricos reales". Journal of Molecular Structure: THEOCHEM . 419 (1–3): 19–27. doi :10.1016/S0166-1280(97)00185-1.
Referencias generales Véase la sección 3 en Mathar, RJ (2009). "Base de Zernike para transformaciones cartesianas". Revista astronómica serbia . 179 (179): 107–120. arXiv : 0809.2368 . Código Bibliográfico :2009SerAJ.179..107M. doi :10.2298/SAJ0979107M. Para armónicos esféricos complejos, consulte también SphericalHarmonicY[l,m,theta,phi] en Wolfram Alpha, especialmente para valores específicos de l y m. 
