Spherical Harmonic (Y)
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The spherical harmonics are defined as Y(m)l=√2l+12√π√(l−m)!(l+m)!eimϕP(m)l(cos(θ)), with l the angular momentum and m the z projection of the angular momentum, −l≤m≤l. P(m)l are the associated Legendre polynomials. For positive m these are defined in terms of the unassociated Legendre polynomials as: P(m)l(x)=(−1)m(1−x2)m/2dmdxmPl(x)=(−1)m2ll!(1−x2)m/2dl+mdxl+m(x2−1)l, using in the second line that: Pl(x)=12ll!dldxl(x2−1)l. For negative m the associated Legendre polynomials are defined as: P(−m)l(x)=(−1)m(l−m)!(l+m)!P(m)l(x). We thus include the Condon-Shortley phase (−1)m in our definitions of the associated Legendre polynomials.
The following table shows the spherical harmonics up to l=6. We list both the explicit function in terms of the angular coordinates θ and ϕ as well as the function in terms of the directional cosines x, y and z. Using a right hand coordinate system the relation is: x=cos(ϕ)sin(θ), y=sin(ϕ)sin(θ) and z=cos(θ). The plots show the surface defined by the equation r=Y(m)l∗Y(m)l. The color of the surface is according to the phase with red for positive, cyan for negative, a mixture of yellow and green for pure positive imaginary values and a mixture of blue and magenta for pure negative imaginary values. We show a 3D image as well as a projection along the x, y and z direction.
l=0
ml=0
Y(0)0=12√πY(0)0=12√π
l=1
ml=−1
Y(−1)1=12√32πe−iϕsin(θ)Y(−1)1=12√32π(x−iy)
ml=0
Y(0)1=12√3πcos(θ)Y(0)1=12√3πz
ml=1
Y(1)1=−12√32πeiϕsin(θ)Y(1)1=−12√32π(x+iy)
l=2
ml=−2
Y(−2)2=14√152πe−2iϕsin2(θ)Y(−2)2=14√152π(x−iy)2
ml=−1
Y(−1)2=12√152πe−iϕsin(θ)cos(θ)Y(−1)2=12√152πz(x−iy)
ml=0
Y(0)2=14√5π(3cos2(θ)−1)Y(0)2=−14√5π(x2+y2−2z2)
ml=1
Y(1)2=−12√152πeiϕsin(θ)cos(θ)Y(1)2=−12√152πz(x+iy)
ml=2
Y(2)2=14√152πe2iϕsin2(θ)Y(2)2=14√152π(x+iy)2
l=3
ml=−3
Y(−3)3=18√35πe−3iϕsin3(θ)Y(−3)3=18√35π(x−iy)3
ml=−2
Y(−2)3=14√1052πe−2iϕsin2(θ)cos(θ)Y(−2)3=14√1052πz(x−iy)2
ml=−1
Y(−1)3=18√21πe−iϕsin(θ)(5cos2(θ)−1)Y(−1)3=−18√21π(x−iy)(x2+y2−4z2)
ml=0
Y(0)3=14√7π(5cos3(θ)−3cos(θ))Y(0)3=14√7πz(2z2−3(x2+y2))
ml=1
Y(1)3=−18√21πeiϕsin(θ)(5cos2(θ)−1)Y(1)3=18√21π(x+iy)(x2+y2−4z2)
ml=2
Y(2)3=14√1052πe2iϕsin2(θ)cos(θ)Y(2)3=14√1052πz(x+iy)2
ml=3
Y(3)3=−18√35πe3iϕsin3(θ)Y(3)3=−18√35π(x+iy)3
l=4
ml=−4
Y(−4)4=316√352πe−4iϕsin4(θ)Y(−4)4=316√352π(x−iy)4
ml=−3
Y(−3)4=38√35πe−3iϕsin3(θ)cos(θ)Y(−3)4=38√35πz(x−iy)3
ml=−2
Y(−2)4=38√52πe−2iϕsin2(θ)(7cos2(θ)−1)Y(−2)4=−38√52π(x−iy)2(x2+y2−6z2)
ml=−1
Y(−1)4=38√5πe−iϕsin(θ)cos(θ)(7cos2(θ)−3)Y(−1)4=−38√5πz(x−iy)(3(x2+y2)−4z2)
ml=0
Y(0)4=3(35cos4(θ)−30cos2(θ)+3)16√πY(0)4=−72z2(x2+y2)+9(x2+y2)2+24z416√π
ml=1
Y(1)4=−38√5πeiϕsin(θ)cos(θ)(7cos2(θ)−3)Y(1)4=38√5πz(x+iy)(3(x2+y2)−4z2)
ml=2
Y(2)4=38√52πe2iϕsin2(θ)(7cos2(θ)−1)Y(2)4=−38√52π(x+iy)2(x2+y2−6z2)
ml=3
Y(3)4=−38√35πe3iϕsin3(θ)cos(θ)Y(3)4=−38√35πz(x+iy)3
ml=4
Y(4)4=316√352πe4iϕsin4(θ)Y(4)4=316√352π(x+iy)4
l=5
ml=−5
Y(−5)5=332√77πe−5iϕsin5(θ)Y(−5)5=332√77π(x−iy)5
ml=−4
Y(−4)5=316√3852πe−4iϕsin4(θ)cos(θ)Y(−4)5=316√3852πz(x−iy)4
ml=−3
Y(−3)5=132√385πe−3iϕsin3(θ)(9cos2(θ)−1)Y(−3)5=−132√385π(x−iy)3(x2+y2−8z2)
ml=−2
Y(−2)5=18√11552πe−2iϕsin2(θ)cos(θ)(3cos2(θ)−1)Y(−2)5=−18√11552πz(x−iy)2(x2+y2−2z2)
ml=−1
Y(−1)5=116√1652πe−iϕsin(θ)(21cos4(θ)−14cos2(θ)+1)Y(−1)5=116√1652π(x−iy)(−12z2(x2+y2)+(x2+y2)2+8z4)
ml=0
Y(0)5=116√11π(63cos5(θ)−70cos3(θ)+15cos(θ))Y(0)5=116√11πz(−40z2(x2+y2)+15(x2+y2)2+8z4)
ml=1
Y(1)5=−116√1652πeiϕsin(θ)(21cos4(θ)−14cos2(θ)+1)Y(1)5=−116√1652π(x+iy)(−12z2(x2+y2)+(x2+y2)2+8z4)
ml=2
Y(2)5=18√11552πe2iϕsin2(θ)cos(θ)(3cos2(θ)−1)Y(2)5=−18√11552πz(x+iy)2(x2+y2−2z2)
ml=3
Y(3)5=−132√385πe3iϕsin3(θ)(9cos2(θ)−1)Y(3)5=132√385π(x+iy)3(x2+y2−8z2)
ml=4
Y(4)5=316√3852πe4iϕsin4(θ)cos(θ)Y(4)5=316√3852πz(x+iy)4
ml=5
Y(5)5=−332√77πe5iϕsin5(θ)Y(5)5=−332√77π(x+iy)5
l=6
ml=−6
Y(−6)6=164√3003πe−6iϕsin6(θ)Y(−6)6=164√3003π(x−iy)6
ml=−5
Y(−5)6=332√1001πe−5iϕsin5(θ)cos(θ)Y(−5)6=332√1001πz(x−iy)5
ml=−4
Y(−4)6=332√912πe−4iϕsin4(θ)(11cos2(θ)−1)Y(−4)6=−332√912π(x−iy)4(x2+y2−10z2)
ml=−3
Y(−3)6=132√1365πe−3iϕsin3(θ)cos(θ)(11cos2(θ)−3)Y(−3)6=−132√1365πz(x−iy)3(3(x2+y2)−8z2)
ml=−2
Y(−2)6=164√1365πe−2iϕsin2(θ)(33cos4(θ)−18cos2(θ)+1)Y(−2)6=164√1365π(x−iy)2(−16z2(x2+y2)+(x2+y2)2+16z4)
ml=−1
Y(−1)6=116√2732πe−iϕsin(θ)cos(θ)(33cos4(θ)−30cos2(θ)+5)Y(−1)6=116√2732πz(x−iy)(−20z2(x2+y2)+5(x2+y2)2+8z4)
ml=0
Y(0)6=132√13π(231cos6(θ)−315cos4(θ)+105cos2(θ)−5)Y(0)6=132√13π(−120z4(x2+y2)+90z2(x2+y2)2−5(x2+y2)3+16z6)
ml=1
Y(1)6=−116√2732πeiϕsin(θ)cos(θ)(33cos4(θ)−30cos2(θ)+5)Y(1)6=−116√2732πz(x+iy)(−20z2(x2+y2)+5(x2+y2)2+8z4)
ml=2
Y(2)6=164√1365πe2iϕsin2(θ)(33cos4(θ)−18cos2(θ)+1)Y(2)6=164√1365π(x+iy)2(−16z2(x2+y2)+(x2+y2)2+16z4)
ml=3
Y(3)6=−132√1365πe3iϕsin3(θ)cos(θ)(11cos2(θ)−3)Y(3)6=132√1365πz(x+iy)3(3(x2+y2)−8z2)
ml=4
Y(4)6=332√912πe4iϕsin4(θ)(11cos2(θ)−1)Y(4)6=−332√912π(x+iy)4(x2+y2−10z2)
ml=5
Y(5)6=−332√1001πe5iϕsin5(θ)cos(θ)Y(5)6=−332√1001πz(x+iy)5
ml=6
Y(6)6=164√3003πe6iϕsin6(θ)Y(6)6=164√3003π(x+iy)6