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

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

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.

The first few Slater type orbitals are (where we use s for l = 0, p for l = 1 and d for l = 2. Functions between square brackets are normalized real spherical harmonics):

${\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 }}}e^{-\zeta r}(3z^{2}-r^{2})\\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_{xy}&={\frac {2}{15}}{\sqrt {10}}\zeta ^{7/2}\,r^{2}\,e^{-\zeta r}{\Big [}{\sqrt {\frac {5}{4\pi }}}{\frac {2}{3}}{\sqrt {3}}{\frac {xy}{r^{2}}}{\Big ]}={\frac {4}{3}}{\sqrt {\frac {2\zeta ^{7}}{15\pi }}}\,xy\,e^{-\zeta r}\\\end{aligned}}$