1s Slater-type function

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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

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

It is a particular case of a Slater-type orbital (STO) in which the principal quantum number n is 1. The parameter ${\displaystyle \zeta }$ 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.

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 ${\displaystyle e(\mathbf {Z} -1)}$, where ${\displaystyle \mathbf {Z} }$ 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.^[2] The electronic Hamiltonian (in atomic units) of a Hydrogenic system is given by
${\displaystyle \mathbf {\hat {H}} _{e}=-{\frac {\nabla ^{2}}{2}}-{\frac {\mathbf {Z} }{r}}}$, where ${\displaystyle \mathbf {Z} }$ 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:
${\displaystyle \mathbf {\psi } _{1s}=\left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}}$, where ${\displaystyle \mathbf {\zeta } }$ 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.

The energy of a hydrogenic system can be exactly calculated analytically as follows :
${\displaystyle \mathbf {E} _{1s}={\frac {\langle \psi _{1s}|\mathbf {\hat {H}} _{e}|\psi _{1s}\rangle }{\langle \psi _{1s}|\psi _{1s}\rangle }}}$, where ${\displaystyle \mathbf {\langle \psi _{1s}|\psi _{1s}\rangle } =1}$
${\displaystyle \mathbf {E} _{1s}=\langle \psi _{1s}|\mathbf {-} {\frac {\nabla ^{2}}{2}}-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle }$
${\displaystyle \mathbf {E} _{1s}=\langle \psi _{1s}|\mathbf {-} {\frac {\nabla ^{2}}{2}}|\psi _{1s}\rangle +\langle \psi _{1s}|-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle }$
${\displaystyle \mathbf {E} _{1s}=\langle \psi _{1s}|\mathbf {-} {\frac {1}{2r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial }{\partial r}}\right)|\psi _{1s}\rangle +\langle \psi _{1s}|-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle }$. Using the expression for Slater orbital, ${\displaystyle \mathbf {\psi } _{1s}=\left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}}$ the integrals can be exactly solved. Thus,
${\displaystyle \mathbf {E} _{1s}=\left\langle \left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}\right|\left.-\left({\frac {\zeta ^{3}}{\pi }}\right)^{0.50}e^{-\zeta r}\left[{\frac {-2r\zeta +r^{2}\zeta ^{2}}{2r^{2}}}\right]\right\rangle +\langle \psi _{1s}|-{\frac {\mathbf {Z} }{r}}|\psi _{1s}\rangle }$
${\displaystyle \mathbf {E} _{1s}={\frac {\zeta ^{2}}{2}}-\zeta \mathbf {Z} .}$

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

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

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

Gold : Au(78+)
${\displaystyle \mathbf {Z} =79}$ and ${\displaystyle \mathbf {\zeta } =79}$
${\displaystyle \mathbf {E} _{1s}=}$−3120.5 E_h
${\displaystyle \mathbf {E} _{1s}=}$−84913.16433850 eV
${\displaystyle \mathbf {E} _{1s}=}$−1958141.8345 kcal/mol.

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 ${\displaystyle \mathbf {\zeta } }$. The relativistically corrected Slater exponent ${\displaystyle \mathbf {\zeta } _{rel}}$ is given as
${\displaystyle \mathbf {\zeta } _{rel}={\frac {\mathbf {Z} }{\sqrt {1-\mathbf {Z} ^{2}/c^{2}}}}}$.
The relativistic energy of an electron in 1s orbital of a hydrogenic atomic systems is obtained by solving the Dirac equation.
${\displaystyle \mathbf {E} _{1s}^{rel}=-(c^{2}+\mathbf {Z} \zeta )+{\sqrt {c^{4}+\mathbf {Z} ^{2}\zeta ^{2}}}}$.
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.

Atomic system	${\displaystyle \mathbf {Z} }$	${\displaystyle \mathbf {\zeta } _{nonrel}}$	${\displaystyle \mathbf {\zeta } _{rel}}$	${\displaystyle \mathbf {E} _{1s}^{nonrel}}$	${\displaystyle \mathbf {E} _{1s}^{rel}}$using ${\displaystyle \mathbf {\zeta } _{nonrel}}$	${\displaystyle \mathbf {E} _{1s}^{rel}}$using ${\displaystyle \mathbf {\zeta } _{rel}}$
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