Slater orbital

Slater-type orbitals (STOs) are functions used as atomic orbitals in the linear combination of atomic orbitals molecular orbital method. They are named after the physicist John C. Slater, who introduced them in 1930.

STOs have the following radial part:

$R(r)=Nr^{n-1}e^{-\zeta r}\,$

where

n is a natural number that plays the role of principal quantum number, n = 1,2,...,

N is a normalization constant,

r is the distance of the electron from the atomic nucleus, and

$\zeta$ is a constant related to the effective charge of the nucleus, the nuclear charge being partly shielded by electrons.

The normalization constant is computed from the integral

$\int _{0}^{\infty }x^{n}e^{-\alpha x}dx={\frac {n!}{\alpha ^{n+1}}}.$

Hence

$N^{2}\int _{0}^{\infty }\left(r^{n-1}e^{-\zeta r}\right)^{2}r^{2}dr=1\Longrightarrow N=(2\zeta )^{n}{\sqrt {\frac {2\zeta }{(2n)!}}}.$

With this factor the radial part is normalized to unity. That is, explicitly we have,

${\frac {(2\zeta )^{2n+1}}{(2n)!}}\int _{0}^{\infty }r^{2n-2}e^{-2\zeta r}r^{2}dr=1.$

It is common to use the real form of spherical harmonics as the angular part of the Slater orbital. A list of cartesian real spherical harmonics is given in this article. In the article hydrogen-like orbitals is explained that the angular parts can be designated by letters: s, p, d, etc.

The first few Slater type orbitals are given below. We use s for l = 0, p for l = 1 and d for l = 2. Functions between square brackets are normalized real spherical harmonics. These functions are given in cartesian form but, when expressed in spherical polar coordinates, are functions of the colatitude angle θ and longitudinal (azimuthal) angle φ only (they are not functions of r). If we denote such a function by Y(θ, φ), then the following normalization condition is satisfied

$\int _{0}^{\pi }\int _{0}^{2\pi }Y(\theta ,\phi )^{2}\sin \theta \,d\theta \,d\phi =1$

The list up to and including 3d orbitals is,

${\begin{aligned}1s&=2\zeta ^{3/2}e^{-\zeta r}\\2s&={\frac {2}{3}}{\sqrt {3}}\zeta ^{5/2}\,r\,e^{-\zeta r}\\2p_{x}&={\frac {2}{3}}{\sqrt {3}}\zeta ^{5/2}\,r\,e^{-\zeta r}{\Big [}{\sqrt {\frac {3}{4\pi }}}{\frac {x}{r}}{\Big ]}={\sqrt {\frac {\zeta ^{5}}{\pi }}}\,x\,e^{-\zeta r}\\2p_{y}&={\frac {2}{3}}{\sqrt {3}}\zeta ^{5/2}\,r\,e^{-\zeta r}{\Big [}{\sqrt {\frac {3}{4\pi }}}{\frac {y}{r}}{\Big ]}={\sqrt {\frac {\zeta ^{5}}{\pi }}}\,y\,e^{-\zeta r}\\2p_{z}&={\frac {2}{3}}{\sqrt {3}}\zeta ^{5/2}\,r\,e^{-\zeta r}{\Big [}{\sqrt {\frac {3}{4\pi }}}{\frac {z}{r}}{\Big ]}={\sqrt {\frac {\zeta ^{5}}{\pi }}}\,z\,e^{-\zeta r}\\3s&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}\\3p_{x}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {3}{4\pi }}}{\frac {x}{r}}{\Big ]}={\sqrt {\frac {2\zeta ^{7}}{15\pi }}}\,r\,x\,e^{-\zeta r}\\3p_{y}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {3}{4\pi }}}{\frac {y}{r}}{\Big ]}={\sqrt {\frac {2\zeta ^{7}}{15\pi }}}\,r\,y\,e^{-\zeta r}\\3p_{z}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {3}{4\pi }}}{\frac {z}{r}}{\Big ]}={\sqrt {\frac {2\zeta ^{7}}{15\pi }}}\,r\,z\,e^{-\zeta r}\\3d_{z^{2}}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {5}{4\pi }}}{\frac {3z^{2}-r^{2}}{2r^{2}}}{\Big ]}={\frac {1}{3}}{\sqrt {\frac {\zeta ^{7}}{2\pi }}}\,(3z^{2}-r^{2})\,e^{-\zeta r}\\3d_{xz}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {5}{4\pi }}}{\sqrt {3}}{\frac {xz}{r^{2}}}{\Big ]}={\sqrt {\frac {2\zeta ^{7}}{3\pi }}}\,xz\,e^{-\zeta r}\\3d_{yz}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {5}{4\pi }}}{\sqrt {3}}{\frac {yz}{r^{2}}}{\Big ]}={\sqrt {\frac {2\zeta ^{7}}{3\pi }}}\,yz\,e^{-\zeta r}\\3d_{x^{2}-y^{2}}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {5}{4\pi }}}{\frac {1}{2}}{\sqrt {3}}{\frac {x^{2}-y^{2}}{r^{2}}}{\Big ]}={\sqrt {\frac {\zeta ^{7}}{6\pi }}}\,(x^{2}-y^{2})\,e^{-\zeta r}\\3d_{xy}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {5}{4\pi }}}{\sqrt {3}}{\frac {xy}{r^{2}}}{\Big ]}={\sqrt {\frac {2\zeta ^{7}}{3\pi }}}\,xy\,e^{-\zeta r}\\\end{aligned}}$