The following functions are normalized to unity and have the Condon & Shortley phase.
Y 0 0 ( θ , φ ) = 1 4 π Y 1 0 ( θ , φ ) = 3 4 π cos θ = 3 4 π z r Y 1 ± 1 ( θ , φ ) = ∓ 3 4 π 1 2 sin θ e ± i φ = ∓ 3 4 π 1 2 x ± i y r Y 2 0 ( θ , φ ) = 5 4 π 1 2 ( 3 cos 2 θ − 1 ) = 5 4 π 1 2 3 z 2 − r 2 r 2 Y 2 ± 1 ( θ , φ ) = ∓ 5 4 π 3 2 sin θ cos θ e ± i φ = ∓ 5 4 π 3 2 z ( x ± i y ) r 2 Y 2 ± 2 ( θ , φ ) = 5 4 π 3 8 sin 2 θ e ± 2 i φ = 5 4 π 3 8 ( x ± i y ) 2 r 2 Y 3 0 ( θ , φ ) = 7 4 π 1 2 ( 5 cos 3 θ − 3 cos θ ) = 7 4 π 1 2 5 z 3 − 3 z r 2 r 3 Y 3 ± 1 ( θ , φ ) = ∓ 7 4 π 3 16 sin θ ( 5 cos 2 θ − 1 ) e ± i φ = ∓ 7 4 π 3 16 ( x ± i y ) ( 5 z 2 − r 2 ) r 3 Y 3 ± 2 ( θ , φ ) = 7 4 π 15 8 sin 2 θ cos θ e ± 2 i φ = 7 4 π 15 8 z ( x ± i y ) 2 r 3 Y 3 ± 3 ( θ , φ ) = ∓ 7 4 π 5 16 sin 3 θ e ± 3 i φ = ∓ 7 4 π 5 16 ( x ± i y ) 3 r 3 Y 4 0 ( θ , φ ) = 9 4 π 1 8 ( 35 cos 4 θ − 30 cos 2 θ + 3 ) = 9 4 π 1 8 35 z 4 − 30 z 2 r 2 + 3 r 4 r 4 Y 4 ± 1 ( θ , φ ) = ∓ 9 4 π 5 16 sin θ ( 7 cos 3 θ − 3 cos θ ) e ± i φ = ∓ 9 4 π 5 16 ( x ± i y ) ( 7 z 3 − 3 z r 2 ) r 4 Y 4 ± 2 ( θ , φ ) = 9 4 π 5 32 sin 2 θ ( 7 cos 2 θ − 1 ) e ± 2 i φ = 9 4 π 5 32 ( x ± i y ) 2 ( 7 z 2 − r 2 ) r 4 Y 4 ± 3 ( θ , φ ) = ∓ 9 4 π 35 16 sin 3 θ cos θ e ± 3 i φ = ∓ 9 4 π 35 16 z ( x ± i y ) 3 r 4 Y 4 ± 4 ( θ , φ ) = 9 4 π 35 128 sin 4 e ± 4 i φ = 9 4 π 35 128 ( x ± i y ) 4 r 4 {\displaystyle {\begin{aligned}Y_{0}^{0}(\theta ,\varphi )&={\sqrt {1 \over 4\pi }}\\&\\Y_{1}^{0}(\theta ,\varphi )&={\sqrt {3 \over 4\pi }}\,\cos \theta ={\sqrt {3 \over 4\pi }}\,{\frac {z}{r}}\\Y_{1}^{\pm 1}(\theta ,\varphi )&=\mp {\sqrt {3 \over 4\pi }}{\sqrt {\frac {1}{2}}}\,\sin \theta \,e^{\pm i\varphi }=\mp {\sqrt {3 \over 4\pi }}{\sqrt {\frac {1}{2}}}\,{\frac {x\pm iy}{r}}\\&\\Y_{2}^{0}(\theta ,\varphi )&={\sqrt {5 \over 4\pi }}\,{\frac {1}{2}}(3\cos ^{2}\theta -1)={\sqrt {5 \over 4\pi }}\,{\frac {1}{2}}{\frac {3z^{2}-r^{2}}{r^{2}}}\\Y_{2}^{\pm 1}(\theta ,\varphi )&=\mp {\sqrt {5 \over 4\pi }}{\sqrt {\frac {3}{2}}}\,\sin \theta \,\cos \theta \,e^{\pm i\varphi }=\mp {\sqrt {5 \over 4\pi }}{\sqrt {\frac {3}{2}}}\,{\frac {z(x\pm iy)}{r^{2}}}\\Y_{2}^{\pm 2}(\theta ,\varphi )&={\sqrt {5 \over 4\pi }}\,{\sqrt {\frac {3}{8}}}\sin ^{2}\theta \,e^{\pm 2i\varphi }={\sqrt {5 \over 4\pi }}\,{\sqrt {\frac {3}{8}}}{\frac {(x\pm iy)^{2}}{r^{2}}}\\&\\Y_{3}^{0}(\theta ,\varphi )&={\sqrt {\frac {7}{4\pi }}}\,{\frac {1}{2}}\,(5\cos ^{3}\theta -3\cos \theta )={\sqrt {\frac {7}{4\pi }}}\,{\frac {1}{2}}\,{\frac {5z^{3}-3zr^{2}}{r^{3}}}\\Y_{3}^{\pm 1}(\theta ,\varphi )&=\mp {\sqrt {\frac {7}{4\pi }}}\,{\sqrt {\frac {3}{16}}}\,\sin \theta (5\cos ^{2}\theta -1)e^{\pm i\varphi }=\mp {\sqrt {\frac {7}{4\pi }}}\,{\sqrt {\frac {3}{16}}}\,{\frac {(x\pm iy)(5z^{2}-r^{2})}{r^{3}}}\\Y_{3}^{\pm 2}(\theta ,\varphi )&={\sqrt {\frac {7}{4\pi }}}\,{\sqrt {\frac {15}{8}}}\,\sin ^{2}\theta \cos \theta e^{\pm 2i\varphi }={\sqrt {\frac {7}{4\pi }}}\,{\sqrt {\frac {15}{8}}}\,{\frac {z(x\pm iy)^{2}}{r^{3}}}\\Y_{3}^{\pm 3}(\theta ,\varphi )&=\mp {\sqrt {\frac {7}{4\pi }}}\,{\sqrt {\frac {5}{16}}}\,\sin ^{3}\theta e^{\pm 3i\varphi }=\mp {\sqrt {\frac {7}{4\pi }}}\,{\sqrt {\frac {5}{16}}}\,{\frac {(x\pm iy)^{3}}{r^{3}}}\\&\\Y_{4}^{0}(\theta ,\varphi )&={\sqrt {\frac {9}{4\pi }}}\,{\frac {1}{8}}\,(35\cos ^{4}\theta -30\cos ^{2}\theta +3)={\sqrt {\frac {9}{4\pi }}}\,{\frac {1}{8}}\,{\frac {35z^{4}-30z^{2}r^{2}+3r^{4}}{r^{4}}}\\Y_{4}^{\pm 1}(\theta ,\varphi )&=\mp {\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {5}{16}}}\,\sin \theta (7\cos ^{3}\theta -3\cos \theta )e^{\pm i\varphi }=\mp {\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {5}{16}}}\,{\frac {(x\pm iy)(7z^{3}-3zr^{2})}{r^{4}}}\\Y_{4}^{\pm 2}(\theta ,\varphi )&={\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {5}{32}}}\,\sin ^{2}\theta (7\cos ^{2}\theta -1)e^{\pm 2i\varphi }={\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {5}{32}}}\,{\frac {(x\pm iy)^{2}(7z^{2}-r^{2})}{r^{4}}}\\Y_{4}^{\pm 3}(\theta ,\varphi )&=\mp {\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {35}{16}}}\,\sin ^{3}\theta \cos \theta e^{\pm 3i\varphi }=\mp {\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {35}{16}}}\,{\frac {z(x\pm iy)^{3}}{r^{4}}}\\Y_{4}^{\pm 4}(\theta ,\varphi )&={\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {35}{128}}}\,\sin ^{4}e^{\pm 4i\varphi }={\sqrt {\frac {9}{4\pi }}}\,{\sqrt {\frac {35}{128}}}\,{\frac {(x\pm iy)^{4}}{r^{4}}}\\\end{aligned}}}
