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Spherical Harmonic (Y)

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The spherical harmonics are defined as Y(m)l=2l+12π(lm)!(l+m)!eimϕP(m)l(cos(θ)), with l the angular momentum and m the z projection of the angular momentum, lml. 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(1x2)m/2dmdxmPl(x)=(1)m2ll!(1x2)m/2dl+mdxl+m(x21)l, using in the second line that: Pl(x)=12ll!dldxl(x21)l. For negative m the associated Legendre polynomials are defined as: P(m)l(x)=(1)m(lm)!(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)lY(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=12πY(0)0=12π



l=1

ml=1

Y(1)1=1232πeiϕsin(θ)Y(1)1=1232π(xiy)

ml=0

Y(0)1=123πcos(θ)Y(0)1=123πz

ml=1

Y(1)1=1232πeiϕsin(θ)Y(1)1=1232π(x+iy)



l=2

ml=2

Y(2)2=14152πe2iϕsin2(θ)Y(2)2=14152π(xiy)2

ml=1

Y(1)2=12152πeiϕsin(θ)cos(θ)Y(1)2=12152πz(xiy)

ml=0

Y(0)2=145π(3cos2(θ)1)Y(0)2=145π(x2+y22z2)

ml=1

Y(1)2=12152πeiϕsin(θ)cos(θ)Y(1)2=12152πz(x+iy)

ml=2

Y(2)2=14152πe2iϕsin2(θ)Y(2)2=14152π(x+iy)2



l=3

ml=3

Y(3)3=1835πe3iϕsin3(θ)Y(3)3=1835π(xiy)3

ml=2

Y(2)3=141052πe2iϕsin2(θ)cos(θ)Y(2)3=141052πz(xiy)2

ml=1

Y(1)3=1821πeiϕsin(θ)(5cos2(θ)1)Y(1)3=1821π(xiy)(x2+y24z2)

ml=0

Y(0)3=147π(5cos3(θ)3cos(θ))Y(0)3=147πz(2z23(x2+y2))

ml=1

Y(1)3=1821πeiϕsin(θ)(5cos2(θ)1)Y(1)3=1821π(x+iy)(x2+y24z2)

ml=2

Y(2)3=141052πe2iϕsin2(θ)cos(θ)Y(2)3=141052πz(x+iy)2

ml=3

Y(3)3=1835πe3iϕsin3(θ)Y(3)3=1835π(x+iy)3



l=4

ml=4

Y(4)4=316352πe4iϕsin4(θ)Y(4)4=316352π(xiy)4

ml=3

Y(3)4=3835πe3iϕsin3(θ)cos(θ)Y(3)4=3835πz(xiy)3

ml=2

Y(2)4=3852πe2iϕsin2(θ)(7cos2(θ)1)Y(2)4=3852π(xiy)2(x2+y26z2)

ml=1

Y(1)4=385πeiϕsin(θ)cos(θ)(7cos2(θ)3)Y(1)4=385πz(xiy)(3(x2+y2)4z2)

ml=0

Y(0)4=3(35cos4(θ)30cos2(θ)+3)16πY(0)4=72z2(x2+y2)+9(x2+y2)2+24z416π

ml=1

Y(1)4=385πeiϕsin(θ)cos(θ)(7cos2(θ)3)Y(1)4=385πz(x+iy)(3(x2+y2)4z2)

ml=2

Y(2)4=3852πe2iϕsin2(θ)(7cos2(θ)1)Y(2)4=3852π(x+iy)2(x2+y26z2)

ml=3

Y(3)4=3835πe3iϕsin3(θ)cos(θ)Y(3)4=3835πz(x+iy)3

ml=4

Y(4)4=316352πe4iϕsin4(θ)Y(4)4=316352π(x+iy)4



l=5

ml=5

Y(5)5=33277πe5iϕsin5(θ)Y(5)5=33277π(xiy)5

ml=4

Y(4)5=3163852πe4iϕsin4(θ)cos(θ)Y(4)5=3163852πz(xiy)4

ml=3

Y(3)5=132385πe3iϕsin3(θ)(9cos2(θ)1)Y(3)5=132385π(xiy)3(x2+y28z2)

ml=2

Y(2)5=1811552πe2iϕsin2(θ)cos(θ)(3cos2(θ)1)Y(2)5=1811552πz(xiy)2(x2+y22z2)

ml=1

Y(1)5=1161652πeiϕsin(θ)(21cos4(θ)14cos2(θ)+1)Y(1)5=1161652π(xiy)(12z2(x2+y2)+(x2+y2)2+8z4)

ml=0

Y(0)5=11611π(63cos5(θ)70cos3(θ)+15cos(θ))Y(0)5=11611πz(40z2(x2+y2)+15(x2+y2)2+8z4)

ml=1

Y(1)5=1161652πeiϕsin(θ)(21cos4(θ)14cos2(θ)+1)Y(1)5=1161652π(x+iy)(12z2(x2+y2)+(x2+y2)2+8z4)

ml=2

Y(2)5=1811552πe2iϕsin2(θ)cos(θ)(3cos2(θ)1)Y(2)5=1811552πz(x+iy)2(x2+y22z2)

ml=3

Y(3)5=132385πe3iϕsin3(θ)(9cos2(θ)1)Y(3)5=132385π(x+iy)3(x2+y28z2)

ml=4

Y(4)5=3163852πe4iϕsin4(θ)cos(θ)Y(4)5=3163852πz(x+iy)4

ml=5

Y(5)5=33277πe5iϕsin5(θ)Y(5)5=33277π(x+iy)5



l=6

ml=6

Y(6)6=1643003πe6iϕsin6(θ)Y(6)6=1643003π(xiy)6

ml=5

Y(5)6=3321001πe5iϕsin5(θ)cos(θ)Y(5)6=3321001πz(xiy)5

ml=4

Y(4)6=332912πe4iϕsin4(θ)(11cos2(θ)1)Y(4)6=332912π(xiy)4(x2+y210z2)

ml=3

Y(3)6=1321365πe3iϕsin3(θ)cos(θ)(11cos2(θ)3)Y(3)6=1321365πz(xiy)3(3(x2+y2)8z2)

ml=2

Y(2)6=1641365πe2iϕsin2(θ)(33cos4(θ)18cos2(θ)+1)Y(2)6=1641365π(xiy)2(16z2(x2+y2)+(x2+y2)2+16z4)

ml=1

Y(1)6=1162732πeiϕsin(θ)cos(θ)(33cos4(θ)30cos2(θ)+5)Y(1)6=1162732πz(xiy)(20z2(x2+y2)+5(x2+y2)2+8z4)

ml=0

Y(0)6=13213π(231cos6(θ)315cos4(θ)+105cos2(θ)5)Y(0)6=13213π(120z4(x2+y2)+90z2(x2+y2)25(x2+y2)3+16z6)

ml=1

Y(1)6=1162732πeiϕsin(θ)cos(θ)(33cos4(θ)30cos2(θ)+5)Y(1)6=1162732πz(x+iy)(20z2(x2+y2)+5(x2+y2)2+8z4)

ml=2

Y(2)6=1641365πe2iϕsin2(θ)(33cos4(θ)18cos2(θ)+1)Y(2)6=1641365π(x+iy)2(16z2(x2+y2)+(x2+y2)2+16z4)

ml=3

Y(3)6=1321365πe3iϕsin3(θ)cos(θ)(11cos2(θ)3)Y(3)6=1321365πz(x+iy)3(3(x2+y2)8z2)

ml=4

Y(4)6=332912πe4iϕsin4(θ)(11cos2(θ)1)Y(4)6=332912π(x+iy)4(x2+y210z2)

ml=5

Y(5)6=3321001πe5iϕsin5(θ)cos(θ)Y(5)6=3321001πz(x+iy)5

ml=6

Y(6)6=1643003πe6iϕsin6(θ)Y(6)6=1643003π(x+iy)6

Different orbital basis sets used

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