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

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

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 $e(𝐙-1)$, where $𝐙$ 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
${\widehat{𝐇}}_e=-\frac{{\mathrm{\nabla }}^2}{2}-\frac{𝐙}{r}$, where $𝐙$ 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:
${\mathit{\psi }}_{1s}={\left(\frac{{\zeta }^3}{\pi }\right)}^{0.50}{e}^{-\zeta r}$, where $\mathit{\zeta }$ 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 :
${𝐄}_{1s}=\frac{⟨{\psi }_{1s}|{\widehat{𝐇}}_e|{\psi }_{1s}⟩}{⟨{\psi }_{1s}|{\psi }_{1s}⟩}$, where $\mathbf{⟨}{\mathit{\psi }}_{\mathrm{𝟏}𝐬}\mathbf{|}{\mathit{\psi }}_{\mathrm{𝟏}𝐬}\mathbf{⟩}=1$
${𝐄}_{1s}=⟨{\psi }_{1s}|\mathbf{-}\frac{{\mathrm{\nabla }}^2}{2}-\frac{𝐙}{r}|{\psi }_{1s}⟩$
${𝐄}_{1s}=⟨{\psi }_{1s}|\mathbf{-}\frac{{\mathrm{\nabla }}^2}{2}|{\psi }_{1s}⟩+⟨{\psi }_{1s}|-\frac{𝐙}{r}|{\psi }_{1s}⟩$
${𝐄}_{1s}=⟨{\psi }_{1s}|\mathbf{-}\frac{1}{2r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)|{\psi }_{1s}⟩+⟨{\psi }_{1s}|-\frac{𝐙}{r}|{\psi }_{1s}⟩$. Using the expression for Slater orbital, ${\mathit{\psi }}_{1s}={\left(\frac{{\zeta }^3}{\pi }\right)}^{0.50}{e}^{-\zeta r}$ the integrals can be exactly solved. Thus,
${𝐄}_{1s}=⟨{\left(\frac{{\zeta }^3}{\pi }\right)}^{0.50}{e}^{-\zeta r}|-{\left(\frac{{\zeta }^3}{\pi }\right)}^{0.50}{e}^{-\zeta r}\left[\frac{-2r\zeta +r^2{\zeta }^2}{2r^2}\right]⟩+⟨{\psi }_{1s}|-\frac{𝐙}{r}|{\psi }_{1s}⟩$
${𝐄}_{1s}=\frac{{\zeta }^2}{2}-\zeta 𝐙.$

The optimum value for $\mathit{\zeta }$ is obtained by equating the differential of the energy with respect to $\mathit{\zeta }$ as zero.
$\frac{d{𝐄}_{1s}}{d\zeta }=\zeta -𝐙=0$. Thus $\mathit{\zeta }=𝐙.$

Non-relativistic energy

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

Hydrogen : H
$𝐙=1$ and $\mathit{\zeta }=1$
${𝐄}_{1s}=$−0.5 E_h
${𝐄}_{1s}=$−13.60569850 eV
${𝐄}_{1s}=$−313.75450000 kcal/mol

Gold : Au(78+)
$𝐙=79$ and $\mathit{\zeta }=79$
${𝐄}_{1s}=$−3120.5 E_h
${𝐄}_{1s}=$−84913.16433850 eV
${𝐄}_{1s}=$−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 $\mathit{\zeta }$. The relativistically corrected Slater exponent ${\mathit{\zeta }}_{rel}$ is given as
${\mathit{\zeta }}_{rel}=\frac{𝐙}{\sqrt{1-{𝐙}^2/c^2}}$.
The relativistic energy of an electron in 1s orbital of a hydrogenic atomic systems is obtained by solving the Dirac equation.
${𝐄}_{1s}^{rel}=-(c^2+𝐙\zeta )+\sqrt{c^4+{𝐙}^2{\zeta }^2}$.
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.

Atomic system	$𝐙$	${\mathit{\zeta }}_{nonrel}$	${\mathit{\zeta }}_{rel}$	${𝐄}_{1s}^{nonrel}$	${𝐄}_{1s}^{rel}$using ${\mathit{\zeta }}_{nonrel}$	${𝐄}_{1s}^{rel}$using ${\mathit{\zeta }}_{rel}$
H	1	1.00000000	1.00002663	−0.50000000 Eh	−0.50000666 Eh	−0.50000666 Eh
				−13.60569850 eV	−13.60587963 eV	−13.60587964 eV
				−313.75450000 kcal/mol	−313.75867685 kcal/mol	−313.75867708 kcal/mol
Au(78+)	79	79.00000000	96.68296596	−3120.50000000 Eh	−3343.96438929 Eh	−3434.58676969 Eh
				−84913.16433850 eV	−90993.94255075 eV	−93459.90412098 eV
				−1958141.83450000 kcal/mol	−2098367.74995699 kcal/mol	−2155234.10926142 kcal/mol