1s Slater-type function

A normalized 1s Slater-type function is a function which is used in the descriptions of atoms and in a broader way in the description of atoms in molecules. It is particularly important as the accurate quantum theory description of the smallest free atom, hydrogen. It has the form


 * $$\psi_{1s}(\zeta, \mathbf{r - R}) = \left(\frac{\zeta^3}{\pi}\right)^{1 \over 2} \, e^{-\zeta |\mathbf{r - R}|}.$$

It is a particular case of a Slater-type orbital (STO) in which the principal quantum number n is 1. The parameter $$\zeta$$ is called the Slater orbital exponent. Related sets of functions can be used to construct STO-nG basis sets which are used in quantum chemistry.

Applications for hydrogen-like atomic systems
A hydrogen-like atom or a hydrogenic atom is an atom with one  electron. Except for the hydrogen atom itself (which is neutral) these atoms carry positive charge $$e(\mathbf Z-1)$$, where $$\mathbf Z$$ is the atomic number of the atom. Because hydrogen-like atoms are two-particle systems with an interaction depending only on the distance between the two particles, their (non-relativistic) Schrödinger equation can be exactly solved in analytic form. The solutions are one-electron functions and are referred to as hydrogen-like atomic orbitals. The electronic Hamiltonian (in atomic units) of a Hydrogenic system is given by

$$\mathbf{\hat{H}}_e = - \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}$$, where $$\mathbf Z$$ is the nuclear charge of the hydrogenic atomic system. The 1s electron of a hydrogenic systems can be accurately described by the corresponding Slater orbital:

$$\mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r}$$, where $$\mathbf \zeta$$ is the Slater exponent. This state, the ground state, is the only state that can be described by a Slater orbital. Slater orbitals have no radial nodes, while the excited states of the hydrogen atom have radial nodes.

Exact energy of a hydrogen-like atom
The energy of a hydrogenic system can be exactly calculated analytically as follows :

$$\mathbf E_{1s} = \frac{\langle\psi_{1s}|\mathbf{\hat{H}}_e|\psi_{1s}\rangle}{\langle\psi_{1s}|\psi_{1s}\rangle}$$, where $$\mathbf{\langle\psi_{1s}|\psi_{1s}\rangle} = 1$$

$$\mathbf E_{1s} = \langle\psi_{1s}|\mathbf - \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}|\psi_{1s}\rangle$$

$$\mathbf E_{1s} = \langle\psi_{1s}|\mathbf - \frac{\nabla^2}{2}|\psi_{1s}\rangle+\langle\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}\rangle$$

$$\mathbf E_{1s} = \langle\psi_{1s}|\mathbf - \frac{1}{2r^2}\frac{\partial}{\partial r}\left (r^2 \frac{\partial}{\partial r}\right )|\psi_{1s}\rangle+\langle\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}\rangle$$. Using the expression for Slater orbital, $$\mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r}$$ the integrals can be exactly solved. Thus,

$$\mathbf E_{1s} = \left\langle \left(\frac{\zeta^3}{\pi} \right)^{0.50} e^{-\zeta r} \right|\left. -\left(\frac{\zeta^3}{\pi} \right)^{0.50}e^{-\zeta r}\left[\frac{-2r\zeta+r^2\zeta^2}{2r^2}\right]\right\rangle+\langle\psi_{1s}| - \frac{\mathbf Z}{r}|\psi_{1s}\rangle$$

$$\mathbf E_{1s} = \frac{\zeta^2}{2}-\zeta \mathbf Z.$$

The optimum value for $$\mathbf \zeta$$ is obtained by equating the differential of the energy with respect to $$\mathbf \zeta$$ as zero.

$$ \frac{d\mathbf E_{1s}}{d\zeta}=\zeta-\mathbf Z=0$$. Thus $$ \mathbf \zeta=\mathbf Z.$$

Non-relativistic energy
The following energy values are thus calculated by using the expressions for energy and for the Slater exponent.

Hydrogen : H

$$ \mathbf Z=1$$ and $$ \mathbf \zeta=1$$

$$ \mathbf E_{1s}=$$−0.5 Eh

$$ \mathbf E_{1s}=$$−13.60569850 eV

$$ \mathbf E_{1s}=$$−313.75450000 kcal/mol

Gold : Au(78+)

$$ \mathbf Z=79$$ and $$ \mathbf \zeta=79$$

$$ \mathbf E_{1s}=$$−3120.5 Eh

$$ \mathbf E_{1s}=$$−84913.16433850 eV

$$ \mathbf E_{1s}=$$−1958141.8345 kcal/mol.

Relativistic energy of Hydrogenic atomic systems
Hydrogenic atomic systems are suitable models to demonstrate the relativistic effects in atomic systems in a simple way. The energy expectation value can calculated by using the Slater orbitals with or without considering the relativistic correction for the Slater exponent $$ \mathbf \zeta $$. The relativistically corrected Slater exponent $$ \mathbf \zeta_{rel} $$ is given as

$$ \mathbf \zeta_{rel}= \frac{\mathbf Z}{\sqrt {1-\mathbf Z^2/c^2}}$$.

The relativistic energy of an electron in 1s orbital of a hydrogenic atomic systems is obtained by solving the Dirac equation.

$$\mathbf E_{1s}^{rel} = -(c^2+\mathbf Z\zeta)+\sqrt{c^4+\mathbf Z^2\zeta^2}$$.

Following table illustrates the relativistic corrections in energy and it can be seen how the relativistic correction scales with the atomic number of the system.