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====== Tesseral Harmonics (Z) ======
~~NOTOC~~

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###
The complex spherical Harmonics are, as their name suggests 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 $\cos(m\phi)$ dependence, for $m<0$ they have a $\sin(m\phi)$ dependence.
The tesseral harmonics are defined as:
$$
   Z_l^{(m)}=\left\{\begin{array}{ll}
               Y_l^{(0)} & m=0\\
               \frac{1}{\sqrt{2}}          ( Y_l^{(-m)} + (-1)^m Y_l^{( m)} ) & m>0\\
               \frac{\mathrm{i}}{\sqrt{2}} ( Y_l^{( m)} - (-1)^m Y_l^{(-m)} ) & m<0
            \end{array}\right.
$$
These are the combined eigenstates of the operator $L^2$, $L_z^2$ and the vertical mirror plane $xz$ and $yz$. The tesseral harmonics are also known as the real spherical harmonics. The name \emph{tesseral} derives from \emph{tessera}, the small square tile of a mosaic. The real harmonics $P_l^m(\cos\theta)\cos(m\phi)$ and $P_l^m(\cos\theta)\sin(m\phi)$ have nodes on two families of curves: the $l-m$ circles of latitude where $P_l^m$ vanishes, and the $2m$ meridians where $\cos(m\phi)$ or $\sin(m\phi)$ vanishes. On the sphere, parallels and meridians cross at right angles, so for $0<m<l$ these two families divide the surface into a checkerboard of curvilinear quadrangles with right-angled corners, the \emph{tesserae} that name the functions. For the case where $m=0$ there are no meridian nodes, only latitude circles, and
the surface is split into latitudinal bands. These functions are also known as zonal harmonics. For $|m|=l$ the function $P_l^l\propto\sin^l\theta$ has no interior zeros, so there are no latitude circles, only meridians, and the surface is split into pole-to-pole sectors. These functions are also known as sectorial harmonics. In the strict naming convention the zonal and sectorial harmonics are not part of the tesseral harmonics, since neither produces the quadrangular tiling. We here adopt the more inclusive naming convention and include both the zonal and sectorial harmonics as subclass of the tesseral harmonics. As such the $2l+1$ tesseral harmonics with total angular momentum $l$ from a basis for the complex spherical harmonics with the same $l$. 
###


###
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 $r={Z_l^{(m)}}^* Z_l^{(m)}$. 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.
###

----
----
===== $l=0$ =====
==== $m_l=0$ ====
$$
Z_{0}^{(0)}=\frac{1}{2 \sqrt{\pi }}\\
\phantom{Z_{0}^{(0)}}=\frac{1}{2 \sqrt{\pi }}
$$

{{:physics_chemistry:orbitals:plorbz01.png?nolink |}}

----
----
===== $l=1$ =====
==== $m_l=-1$ ====
$$
Z_{1}^{(-1)}=\frac{1}{2} \sqrt{\frac{3}{\pi }} \sin (\theta ) \sin (\phi )\\
\phantom{Z_{1}^{(-1)}}=\frac{1}{2} \sqrt{\frac{3}{\pi }} y
$$

{{:physics_chemistry:orbitals:plorbz11.png?nolink |}}

==== $m_l=0$ ====
$$
Z_{1}^{(0)}=\frac{1}{2} \sqrt{\frac{3}{\pi }} \cos (\theta )\\
\phantom{Z_{1}^{(0)}}=\frac{1}{2} \sqrt{\frac{3}{\pi }} z
$$

{{:physics_chemistry:orbitals:plorbz12.png?nolink |}}

==== $m_l=1$ ====
$$
Z_{1}^{(1)}=\frac{1}{2} \sqrt{\frac{3}{\pi }} \sin (\theta ) \cos (\phi )\\
\phantom{Z_{1}^{(1)}}=\frac{1}{2} \sqrt{\frac{3}{\pi }} x
$$

{{:physics_chemistry:orbitals:plorbz13.png?nolink |}}

----
----
===== $l=2$ =====
==== $m_l=-2$ ====
$$
Z_{2}^{(-2)}=\frac{1}{4} \sqrt{\frac{15}{\pi }} \sin ^2(\theta ) \sin (2 \phi )\\
\phantom{Z_{2}^{(-2)}}=\frac{1}{2} \sqrt{\frac{15}{\pi }} x y
$$

{{:physics_chemistry:orbitals:plorbz21.png?nolink |}}

==== $m_l=-1$ ====
$$
Z_{2}^{(-1)}=\frac{1}{2} \sqrt{\frac{15}{\pi }} \sin (\theta ) \cos (\theta ) \sin (\phi )\\
\phantom{Z_{2}^{(-1)}}=\frac{1}{2} \sqrt{\frac{15}{\pi }} y z
$$

{{:physics_chemistry:orbitals:plorbz22.png?nolink |}}

==== $m_l=0$ ====
$$
Z_{2}^{(0)}=\frac{1}{8} \sqrt{\frac{5}{\pi }} (3 \cos (2 \theta )+1)\\
\phantom{Z_{2}^{(0)}}=-\frac{1}{4} \sqrt{\frac{5}{\pi }} \left(x^2+y^2-2 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz23.png?nolink |}}

==== $m_l=1$ ====
$$
Z_{2}^{(1)}=\frac{1}{2} \sqrt{\frac{15}{\pi }} \sin (\theta ) \cos (\theta ) \cos (\phi )\\
\phantom{Z_{2}^{(1)}}=\frac{1}{2} \sqrt{\frac{15}{\pi }} x z
$$

{{:physics_chemistry:orbitals:plorbz24.png?nolink |}}

==== $m_l=2$ ====
$$
Z_{2}^{(2)}=\frac{1}{4} \sqrt{\frac{15}{\pi }} \sin ^2(\theta ) \cos (2 \phi )\\
\phantom{Z_{2}^{(2)}}=\frac{1}{4} \sqrt{\frac{15}{\pi }} (x-y) (x+y)
$$

{{:physics_chemistry:orbitals:plorbz25.png?nolink |}}

----
----
===== $l=3$ =====
==== $m_l=-3$ ====
$$
Z_{3}^{(-3)}=\frac{1}{4} \sqrt{\frac{35}{2 \pi }} \sin ^3(\theta ) \sin (3 \phi )\\
\phantom{Z_{3}^{(-3)}}=-\frac{1}{4} \sqrt{\frac{35}{2 \pi }} y \left(y^2-3 x^2\right)
$$

{{:physics_chemistry:orbitals:plorbz31.png?nolink |}}

==== $m_l=-2$ ====
$$
Z_{3}^{(-2)}=\frac{1}{4} \sqrt{\frac{105}{\pi }} \sin ^2(\theta ) \cos (\theta ) \sin (2 \phi )\\
\phantom{Z_{3}^{(-2)}}=\frac{1}{2} \sqrt{\frac{105}{\pi }} x y z
$$

{{:physics_chemistry:orbitals:plorbz32.png?nolink |}}

==== $m_l=-1$ ====
$$
Z_{3}^{(-1)}=\frac{1}{16} \sqrt{\frac{21}{2 \pi }} (\sin (\theta )+5 \sin (3 \theta )) \sin (\phi )\\
\phantom{Z_{3}^{(-1)}}=-\frac{1}{4} \sqrt{\frac{21}{2 \pi }} y \left(x^2+y^2-4 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz33.png?nolink |}}

==== $m_l=0$ ====
$$
Z_{3}^{(0)}=\frac{1}{8} \sqrt{\frac{7}{\pi }} \cos (\theta ) (5 \cos (2 \theta )-1)\\
\phantom{Z_{3}^{(0)}}=\frac{1}{4} \sqrt{\frac{7}{\pi }} z \left(2 z^2-3 \left(x^2+y^2\right)\right)
$$

{{:physics_chemistry:orbitals:plorbz34.png?nolink |}}

==== $m_l=1$ ====
$$
Z_{3}^{(1)}=\frac{1}{16} \sqrt{\frac{21}{2 \pi }} (\sin (\theta )+5 \sin (3 \theta )) \cos (\phi )\\
\phantom{Z_{3}^{(1)}}=-\frac{1}{4} \sqrt{\frac{21}{2 \pi }} x \left(x^2+y^2-4 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz35.png?nolink |}}

==== $m_l=2$ ====
$$
Z_{3}^{(2)}=\frac{1}{4} \sqrt{\frac{105}{\pi }} \sin ^2(\theta ) \cos (\theta ) \cos (2 \phi )\\
\phantom{Z_{3}^{(2)}}=\frac{1}{4} \sqrt{\frac{105}{\pi }} z (x-y) (x+y)
$$

{{:physics_chemistry:orbitals:plorbz36.png?nolink |}}

==== $m_l=3$ ====
$$
Z_{3}^{(3)}=\frac{1}{4} \sqrt{\frac{35}{2 \pi }} \sin ^3(\theta ) \cos (3 \phi )\\
\phantom{Z_{3}^{(3)}}=\frac{1}{4} \sqrt{\frac{35}{2 \pi }} x \left(x^2-3 y^2\right)
$$

{{:physics_chemistry:orbitals:plorbz37.png?nolink |}}

----
----
===== $l=4$ =====
==== $m_l=-4$ ====
$$
Z_{4}^{(-4)}=\frac{3}{16} \sqrt{\frac{35}{\pi }} \sin ^4(\theta ) \sin (4 \phi )\\
\phantom{Z_{4}^{(-4)}}=\frac{3}{4} \sqrt{\frac{35}{\pi }} x y (x-y) (x+y)
$$

{{:physics_chemistry:orbitals:plorbz41.png?nolink |}}

==== $m_l=-3$ ====
$$
Z_{4}^{(-3)}=\frac{3}{4} \sqrt{\frac{35}{2 \pi }} \sin ^3(\theta ) \cos (\theta ) \sin (3 \phi )\\
\phantom{Z_{4}^{(-3)}}=-\frac{3}{4} \sqrt{\frac{35}{2 \pi }} y z \left(y^2-3 x^2\right)
$$

{{:physics_chemistry:orbitals:plorbz42.png?nolink |}}

==== $m_l=-2$ ====
$$
Z_{4}^{(-2)}=\frac{3}{16} \sqrt{\frac{5}{\pi }} \sin ^2(\theta ) (7 \cos (2 \theta )+5) \sin (2 \phi )\\
\phantom{Z_{4}^{(-2)}}=-\frac{3}{4} \sqrt{\frac{5}{\pi }} x y \left(x^2+y^2-6 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz43.png?nolink |}}

==== $m_l=-1$ ====
$$
Z_{4}^{(-1)}=\frac{3}{32} \sqrt{\frac{5}{2 \pi }} (2 \sin (2 \theta )+7 \sin (4 \theta )) \sin (\phi )\\
\phantom{Z_{4}^{(-1)}}=-\frac{3}{4} \sqrt{\frac{5}{2 \pi }} y z \left(3 \left(x^2+y^2\right)-4 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz44.png?nolink |}}

==== $m_l=0$ ====
$$
Z_{4}^{(0)}=\frac{3 (20 \cos (2 \theta )+35 \cos (4 \theta )+9)}{128 \sqrt{\pi }}\\
\phantom{Z_{4}^{(0)}}=\frac{-72 z^2 \left(x^2+y^2\right)+9 \left(x^2+y^2\right)^2+24 z^4}{16 \sqrt{\pi }}
$$

{{:physics_chemistry:orbitals:plorbz45.png?nolink |}}

==== $m_l=1$ ====
$$
Z_{4}^{(1)}=\frac{3}{32} \sqrt{\frac{5}{2 \pi }} (2 \sin (2 \theta )+7 \sin (4 \theta )) \cos (\phi )\\
\phantom{Z_{4}^{(1)}}=-\frac{3}{4} \sqrt{\frac{5}{2 \pi }} x z \left(3 \left(x^2+y^2\right)-4 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz46.png?nolink |}}

==== $m_l=2$ ====
$$
Z_{4}^{(2)}=\frac{3}{16} \sqrt{\frac{5}{\pi }} \sin ^2(\theta ) (7 \cos (2 \theta )+5) \cos (2 \phi )\\
\phantom{Z_{4}^{(2)}}=-\frac{3}{8} \sqrt{\frac{5}{\pi }} (x-y) (x+y) \left(x^2+y^2-6 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz47.png?nolink |}}

==== $m_l=3$ ====
$$
Z_{4}^{(3)}=\frac{3}{4} \sqrt{\frac{35}{2 \pi }} \sin ^3(\theta ) \cos (\theta ) \cos (3 \phi )\\
\phantom{Z_{4}^{(3)}}=\frac{3}{4} \sqrt{\frac{35}{2 \pi }} x z \left(x^2-3 y^2\right)
$$

{{:physics_chemistry:orbitals:plorbz48.png?nolink |}}

==== $m_l=4$ ====
$$
Z_{4}^{(4)}=\frac{3}{16} \sqrt{\frac{35}{\pi }} \sin ^4(\theta ) \cos (4 \phi )\\
\phantom{Z_{4}^{(4)}}=\frac{3}{16} \sqrt{\frac{35}{\pi }} \left(x^4-6 x^2 y^2+y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz49.png?nolink |}}

----
----
===== $l=5$ =====
==== $m_l=-5$ ====
$$
Z_{5}^{(-5)}=\frac{3}{16} \sqrt{\frac{77}{2 \pi }} \sin ^5(\theta ) \sin (5 \phi )\\
\phantom{Z_{5}^{(-5)}}=\frac{3}{16} \sqrt{\frac{77}{2 \pi }} y \left(5 x^4-10 x^2 y^2+y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz51.png?nolink |}}

==== $m_l=-4$ ====
$$
Z_{5}^{(-4)}=\frac{3}{16} \sqrt{\frac{385}{\pi }} \sin ^4(\theta ) \cos (\theta ) \sin (4 \phi )\\
\phantom{Z_{5}^{(-4)}}=\frac{3}{4} \sqrt{\frac{385}{\pi }} x y z (x-y) (x+y)
$$

{{:physics_chemistry:orbitals:plorbz52.png?nolink |}}

==== $m_l=-3$ ====
$$
Z_{5}^{(-3)}=\frac{1}{32} \sqrt{\frac{385}{2 \pi }} \sin ^3(\theta ) (9 \cos (2 \theta )+7) \sin (3 \phi )\\
\phantom{Z_{5}^{(-3)}}=\frac{1}{16} \sqrt{\frac{385}{2 \pi }} y \left(y^2-3 x^2\right) \left(x^2+y^2-8 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz53.png?nolink |}}

==== $m_l=-2$ ====
$$
Z_{5}^{(-2)}=\frac{1}{16} \sqrt{\frac{1155}{\pi }} \sin ^2(\theta ) \cos (\theta ) (3 \cos (2 \theta )+1) \sin (2 \phi )\\
\phantom{Z_{5}^{(-2)}}=-\frac{1}{4} \sqrt{\frac{1155}{\pi }} x y z \left(x^2+y^2-2 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz54.png?nolink |}}

==== $m_l=-1$ ====
$$
Z_{5}^{(-1)}=\frac{1}{256} \sqrt{\frac{165}{\pi }} (2 \sin (\theta )+7 (\sin (3 \theta )+3 \sin (5 \theta ))) \sin (\phi )\\
\phantom{Z_{5}^{(-1)}}=\frac{1}{16} \sqrt{\frac{165}{\pi }} y \left(-12 z^2 \left(x^2+y^2\right)+\left(x^2+y^2\right)^2+8 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz55.png?nolink |}}

==== $m_l=0$ ====
$$
Z_{5}^{(0)}=\frac{1}{256} \sqrt{\frac{11}{\pi }} (30 \cos (\theta )+35 \cos (3 \theta )+63 \cos (5 \theta ))\\
\phantom{Z_{5}^{(0)}}=\frac{1}{16} \sqrt{\frac{11}{\pi }} z \left(-40 z^2 \left(x^2+y^2\right)+15 \left(x^2+y^2\right)^2+8 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz56.png?nolink |}}

==== $m_l=1$ ====
$$
Z_{5}^{(1)}=\frac{1}{256} \sqrt{\frac{165}{\pi }} (2 \sin (\theta )+7 (\sin (3 \theta )+3 \sin (5 \theta ))) \cos (\phi )\\
\phantom{Z_{5}^{(1)}}=\frac{1}{16} \sqrt{\frac{165}{\pi }} x \left(-12 z^2 \left(x^2+y^2\right)+\left(x^2+y^2\right)^2+8 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz57.png?nolink |}}

==== $m_l=2$ ====
$$
Z_{5}^{(2)}=\frac{1}{16} \sqrt{\frac{1155}{\pi }} \sin ^2(\theta ) \cos (\theta ) (3 \cos (2 \theta )+1) \cos (2 \phi )\\
\phantom{Z_{5}^{(2)}}=-\frac{1}{8} \sqrt{\frac{1155}{\pi }} z (x-y) (x+y) \left(x^2+y^2-2 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz58.png?nolink |}}

==== $m_l=3$ ====
$$
Z_{5}^{(3)}=\frac{1}{32} \sqrt{\frac{385}{2 \pi }} \sin ^3(\theta ) (9 \cos (2 \theta )+7) \cos (3 \phi )\\
\phantom{Z_{5}^{(3)}}=-\frac{1}{16} \sqrt{\frac{385}{2 \pi }} x \left(x^2-3 y^2\right) \left(x^2+y^2-8 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz59.png?nolink |}}

==== $m_l=4$ ====
$$
Z_{5}^{(4)}=\frac{3}{16} \sqrt{\frac{385}{\pi }} \sin ^4(\theta ) \cos (\theta ) \cos (4 \phi )\\
\phantom{Z_{5}^{(4)}}=\frac{3}{16} \sqrt{\frac{385}{\pi }} z \left(x^4-6 x^2 y^2+y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz510.png?nolink |}}

==== $m_l=5$ ====
$$
Z_{5}^{(5)}=\frac{3}{16} \sqrt{\frac{77}{2 \pi }} \sin ^5(\theta ) \cos (5 \phi )\\
\phantom{Z_{5}^{(5)}}=\frac{3}{16} \sqrt{\frac{77}{2 \pi }} x \left(x^4-10 x^2 y^2+5 y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz511.png?nolink |}}

----
----
===== $l=6$ =====
==== $m_l=-6$ ====
$$
Z_{6}^{(-6)}=\frac{1}{32} \sqrt{\frac{3003}{2 \pi }} \sin ^6(\theta ) \sin (6 \phi )\\
\phantom{Z_{6}^{(-6)}}=\frac{1}{16} \sqrt{\frac{3003}{2 \pi }} x y \left(3 x^4-10 x^2 y^2+3 y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz61.png?nolink |}}

==== $m_l=-5$ ====
$$
Z_{6}^{(-5)}=\frac{3}{16} \sqrt{\frac{1001}{2 \pi }} \sin ^5(\theta ) \cos (\theta ) \sin (5 \phi )\\
\phantom{Z_{6}^{(-5)}}=\frac{3}{16} \sqrt{\frac{1001}{2 \pi }} y z \left(5 x^4-10 x^2 y^2+y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz62.png?nolink |}}

==== $m_l=-4$ ====
$$
Z_{6}^{(-4)}=\frac{3}{64} \sqrt{\frac{91}{\pi }} \sin ^4(\theta ) (11 \cos (2 \theta )+9) \sin (4 \phi )\\
\phantom{Z_{6}^{(-4)}}=-\frac{3}{8} \sqrt{\frac{91}{\pi }} x y (x-y) (x+y) \left(x^2+y^2-10 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz63.png?nolink |}}

==== $m_l=-3$ ====
$$
Z_{6}^{(-3)}=\frac{1}{64} \sqrt{\frac{1365}{2 \pi }} \sin ^3(\theta ) (21 \cos (\theta )+11 \cos (3 \theta )) \sin (3 \phi )\\
\phantom{Z_{6}^{(-3)}}=\frac{1}{16} \sqrt{\frac{1365}{2 \pi }} y z \left(y^2-3 x^2\right) \left(3 \left(x^2+y^2\right)-8 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz64.png?nolink |}}

==== $m_l=-2$ ====
$$
Z_{6}^{(-2)}=\frac{1}{256} \sqrt{\frac{1365}{2 \pi }} \sin ^2(\theta ) (60 \cos (2 \theta )+33 \cos (4 \theta )+35) \sin (2 \phi )\\
\phantom{Z_{6}^{(-2)}}=\frac{1}{16} \sqrt{\frac{1365}{2 \pi }} x y \left(-16 z^2 \left(x^2+y^2\right)+\left(x^2+y^2\right)^2+16 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz65.png?nolink |}}

==== $m_l=-1$ ====
$$
Z_{6}^{(-1)}=\frac{1}{512} \sqrt{\frac{273}{\pi }} (5 \sin (2 \theta )+12 \sin (4 \theta )+33 \sin (6 \theta )) \sin (\phi )\\
\phantom{Z_{6}^{(-1)}}=\frac{1}{16} \sqrt{\frac{273}{\pi }} y z \left(-20 z^2 \left(x^2+y^2\right)+5 \left(x^2+y^2\right)^2+8 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz66.png?nolink |}}

==== $m_l=0$ ====
$$
Z_{6}^{(0)}=\frac{1}{32} \sqrt{\frac{13}{\pi }} \left(21 \cos ^2(\theta ) \left(11 \cos ^4(\theta )-15 \cos ^2(\theta )+5\right)-5\right)\\
\phantom{Z_{6}^{(0)}}=\frac{1}{32} \sqrt{\frac{13}{\pi }} \left(-120 z^4 \left(x^2+y^2\right)+90 z^2 \left(x^2+y^2\right)^2-5 \left(x^2+y^2\right)^3+16 z^6\right)
$$

{{:physics_chemistry:orbitals:plorbz67.png?nolink |}}

==== $m_l=1$ ====
$$
Z_{6}^{(1)}=\frac{1}{512} \sqrt{\frac{273}{\pi }} (5 \sin (2 \theta )+12 \sin (4 \theta )+33 \sin (6 \theta )) \cos (\phi )\\
\phantom{Z_{6}^{(1)}}=\frac{1}{16} \sqrt{\frac{273}{\pi }} x z \left(-20 z^2 \left(x^2+y^2\right)+5 \left(x^2+y^2\right)^2+8 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz68.png?nolink |}}

==== $m_l=2$ ====
$$
Z_{6}^{(2)}=\frac{1}{256} \sqrt{\frac{1365}{2 \pi }} \sin ^2(\theta ) (60 \cos (2 \theta )+33 \cos (4 \theta )+35) \cos (2 \phi )\\
\phantom{Z_{6}^{(2)}}=\frac{1}{32} \sqrt{\frac{1365}{2 \pi }} (x-y) (x+y) \left(-16 z^2 \left(x^2+y^2\right)+\left(x^2+y^2\right)^2+16 z^4\right)
$$

{{:physics_chemistry:orbitals:plorbz69.png?nolink |}}

==== $m_l=3$ ====
$$
Z_{6}^{(3)}=\frac{1}{64} \sqrt{\frac{1365}{2 \pi }} \sin ^3(\theta ) (21 \cos (\theta )+11 \cos (3 \theta )) \cos (3 \phi )\\
\phantom{Z_{6}^{(3)}}=-\frac{1}{16} \sqrt{\frac{1365}{2 \pi }} x z \left(x^2-3 y^2\right) \left(3 \left(x^2+y^2\right)-8 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz610.png?nolink |}}

==== $m_l=4$ ====
$$
Z_{6}^{(4)}=\frac{3}{64} \sqrt{\frac{91}{\pi }} \sin ^4(\theta ) (11 \cos (2 \theta )+9) \cos (4 \phi )\\
\phantom{Z_{6}^{(4)}}=-\frac{3}{32} \sqrt{\frac{91}{\pi }} \left(x^4-6 x^2 y^2+y^4\right) \left(x^2+y^2-10 z^2\right)
$$

{{:physics_chemistry:orbitals:plorbz611.png?nolink |}}

==== $m_l=5$ ====
$$
Z_{6}^{(5)}=\frac{3}{16} \sqrt{\frac{1001}{2 \pi }} \sin ^5(\theta ) \cos (\theta ) \cos (5 \phi )\\
\phantom{Z_{6}^{(5)}}=\frac{3}{16} \sqrt{\frac{1001}{2 \pi }} x z \left(x^4-10 x^2 y^2+5 y^4\right)
$$

{{:physics_chemistry:orbitals:plorbz612.png?nolink |}}

==== $m_l=6$ ====
$$
Z_{6}^{(6)}=\frac{1}{32} \sqrt{\frac{3003}{2 \pi }} \sin ^6(\theta ) \cos (6 \phi )\\
\phantom{Z_{6}^{(6)}}=\frac{1}{32} \sqrt{\frac{3003}{2 \pi }} \left(x^6-15 x^4 y^2+15 x^2 y^4-y^6\right)
$$

{{:physics_chemistry:orbitals:plorbz613.png?nolink |}}



====== Different orbital basis sets used ======
{{indexmenu>.#1|msort}}