Tesseral Harmonics (Z)
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The spherical harmonics are complex functions. For many cases one does not need to work with complex numbers and by making a suitable linear combination of the complex orbitals one can get a real basis. The tesseral harmonics are linear combinations of the spherical harmonics with +m and −m such that the result is a pure real function. For m>0 the tesseral harmonics have a cos(mϕ) dependence, for m<0 they have a sin(mϕ) dependence. The tesseral harmonics are defined as: Z(m)l={Y(0)lm=01√2(Y(−m)l+(−1)mY(m)l)m>0i√2(Y(m)l−(−1)mY(−m)l)m<0
The following table shows the tesseral harmonics up to l=6. We list the explicit function in terms of the directional cosines x, y and z. The plots show the surface defined by the equation r=Z(m)l∗Z(m)l. The color of the surface is according to the phase with red for positive and cyan for negative. We show a 3D image as well as a projection along the x, y and z direction.
l=0
ml=0
Z(0)0=12√πZ(0)0=12√π
l=1
ml=−1
Z(−1)1=12√3πsin(θ)sin(ϕ)Z(−1)1=12√3πy
ml=0
Z(0)1=12√3πcos(θ)Z(0)1=12√3πz
ml=1
Z(1)1=12√3πsin(θ)cos(ϕ)Z(1)1=12√3πx
l=2
ml=−2
Z(−2)2=14√15πsin2(θ)sin(2ϕ)Z(−2)2=12√15πxy
ml=−1
Z(−1)2=12√15πsin(θ)cos(θ)sin(ϕ)Z(−1)2=12√15πyz
ml=0
Z(0)2=18√5π(3cos(2θ)+1)Z(0)2=−14√5π(x2+y2−2z2)
ml=1
Z(1)2=12√15πsin(θ)cos(θ)cos(ϕ)Z(1)2=12√15πxz
ml=2
Z(2)2=14√15πsin2(θ)cos(2ϕ)Z(2)2=14√15π(x−y)(x+y)
l=3
ml=−3
Z(−3)3=14√352πsin3(θ)sin(3ϕ)Z(−3)3=−14√352πy(y2−3x2)
ml=−2
Z(−2)3=14√105πsin2(θ)cos(θ)sin(2ϕ)Z(−2)3=12√105πxyz
ml=−1
Z(−1)3=116√212π(sin(θ)+5sin(3θ))sin(ϕ)Z(−1)3=−14√212πy(x2+y2−4z2)
ml=0
Z(0)3=18√7πcos(θ)(5cos(2θ)−1)Z(0)3=14√7πz(2z2−3(x2+y2))
ml=1
Z(1)3=116√212π(sin(θ)+5sin(3θ))cos(ϕ)Z(1)3=−14√212πx(x2+y2−4z2)
ml=2
Z(2)3=14√105πsin2(θ)cos(θ)cos(2ϕ)Z(2)3=14√105πz(x−y)(x+y)
ml=3
Z(3)3=14√352πsin3(θ)cos(3ϕ)Z(3)3=14√352πx(x2−3y2)
l=4
ml=−4
Z(−4)4=316√35πsin4(θ)sin(4ϕ)Z(−4)4=34√35πxy(x−y)(x+y)
ml=−3
Z(−3)4=34√352πsin3(θ)cos(θ)sin(3ϕ)Z(−3)4=−34√352πyz(y2−3x2)
ml=−2
Z(−2)4=316√5πsin2(θ)(7cos(2θ)+5)sin(2ϕ)Z(−2)4=−34√5πxy(x2+y2−6z2)
ml=−1
Z(−1)4=332√52π(2sin(2θ)+7sin(4θ))sin(ϕ)Z(−1)4=−34√52πyz(3(x2+y2)−4z2)
ml=0
Z(0)4=3(20cos(2θ)+35cos(4θ)+9)128√πZ(0)4=−72z2(x2+y2)+9(x2+y2)2+24z416√π
ml=1
Z(1)4=332√52π(2sin(2θ)+7sin(4θ))cos(ϕ)Z(1)4=−34√52πxz(3(x2+y2)−4z2)
ml=2
Z(2)4=316√5πsin2(θ)(7cos(2θ)+5)cos(2ϕ)Z(2)4=−38√5π(x−y)(x+y)(x2+y2−6z2)
ml=3
Z(3)4=34√352πsin3(θ)cos(θ)cos(3ϕ)Z(3)4=34√352πxz(x2−3y2)
ml=4
Z(4)4=316√35πsin4(θ)cos(4ϕ)Z(4)4=316√35π(x4−6x2y2+y4)
l=5
ml=−5
Z(−5)5=316√772πsin5(θ)sin(5ϕ)Z(−5)5=316√772πy(5x4−10x2y2+y4)
ml=−4
Z(−4)5=316√385πsin4(θ)cos(θ)sin(4ϕ)Z(−4)5=34√385πxyz(x−y)(x+y)
ml=−3
Z(−3)5=132√3852πsin3(θ)(9cos(2θ)+7)sin(3ϕ)Z(−3)5=116√3852πy(y2−3x2)(x2+y2−8z2)
ml=−2
Z(−2)5=116√1155πsin2(θ)cos(θ)(3cos(2θ)+1)sin(2ϕ)Z(−2)5=−14√1155πxyz(x2+y2−2z2)
ml=−1
Z(−1)5=1256√165π(2sin(θ)+7(sin(3θ)+3sin(5θ)))sin(ϕ)Z(−1)5=116√165πy(−12z2(x2+y2)+(x2+y2)2+8z4)
ml=0
Z(0)5=1256√11π(30cos(θ)+35cos(3θ)+63cos(5θ))Z(0)5=116√11πz(−40z2(x2+y2)+15(x2+y2)2+8z4)
ml=1
Z(1)5=1256√165π(2sin(θ)+7(sin(3θ)+3sin(5θ)))cos(ϕ)Z(1)5=116√165πx(−12z2(x2+y2)+(x2+y2)2+8z4)
ml=2
Z(2)5=116√1155πsin2(θ)cos(θ)(3cos(2θ)+1)cos(2ϕ)Z(2)5=−18√1155πz(x−y)(x+y)(x2+y2−2z2)
ml=3
Z(3)5=132√3852πsin3(θ)(9cos(2θ)+7)cos(3ϕ)Z(3)5=−116√3852πx(x2−3y2)(x2+y2−8z2)
ml=4
Z(4)5=316√385πsin4(θ)cos(θ)cos(4ϕ)Z(4)5=316√385πz(x4−6x2y2+y4)
ml=5
Z(5)5=316√772πsin5(θ)cos(5ϕ)Z(5)5=316√772πx(x4−10x2y2+5y4)
l=6
ml=−6
Z(−6)6=132√30032πsin6(θ)sin(6ϕ)Z(−6)6=116√30032πxy(3x4−10x2y2+3y4)
ml=−5
Z(−5)6=316√10012πsin5(θ)cos(θ)sin(5ϕ)Z(−5)6=316√10012πyz(5x4−10x2y2+y4)
ml=−4
Z(−4)6=364√91πsin4(θ)(11cos(2θ)+9)sin(4ϕ)Z(−4)6=−38√91πxy(x−y)(x+y)(x2+y2−10z2)
ml=−3
Z(−3)6=164√13652πsin3(θ)(21cos(θ)+11cos(3θ))sin(3ϕ)Z(−3)6=116√13652πyz(y2−3x2)(3(x2+y2)−8z2)
ml=−2
Z(−2)6=1256√13652πsin2(θ)(60cos(2θ)+33cos(4θ)+35)sin(2ϕ)Z(−2)6=116√13652πxy(−16z2(x2+y2)+(x2+y2)2+16z4)
ml=−1
Z(−1)6=1512√273π(5sin(2θ)+12sin(4θ)+33sin(6θ))sin(ϕ)Z(−1)6=116√273πyz(−20z2(x2+y2)+5(x2+y2)2+8z4)
ml=0
Z(0)6=132√13π(21cos2(θ)(11cos4(θ)−15cos2(θ)+5)−5)Z(0)6=132√13π(−120z4(x2+y2)+90z2(x2+y2)2−5(x2+y2)3+16z6)
ml=1
Z(1)6=1512√273π(5sin(2θ)+12sin(4θ)+33sin(6θ))cos(ϕ)Z(1)6=116√273πxz(−20z2(x2+y2)+5(x2+y2)2+8z4)
ml=2
Z(2)6=1256√13652πsin2(θ)(60cos(2θ)+33cos(4θ)+35)cos(2ϕ)Z(2)6=132√13652π(x−y)(x+y)(−16z2(x2+y2)+(x2+y2)2+16z4)
ml=3
Z(3)6=164√13652πsin3(θ)(21cos(θ)+11cos(3θ))cos(3ϕ)Z(3)6=−116√13652πxz(x2−3y2)(3(x2+y2)−8z2)
ml=4
Z(4)6=364√91πsin4(θ)(11cos(2θ)+9)cos(4ϕ)Z(4)6=−332√91π(x4−6x2y2+y4)(x2+y2−10z2)
ml=5
Z(5)6=316√10012πsin5(θ)cos(θ)cos(5ϕ)Z(5)6=316√10012πxz(x4−10x2y2+5y4)
ml=6
Z(6)6=132√30032πsin6(θ)cos(6ϕ)Z(6)6=132√30032π(x6−15x4y2+15x2y4−y6)