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

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

It is a particular case of a Slater-type orbital (STO) in which the principal quantum number n is 1. The parameter [math]\displaystyle{ \zeta }[/math] 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 [math]\displaystyle{ e(\mathbf Z-1) }[/math], where [math]\displaystyle{ \mathbf Z }[/math] 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
[math]\displaystyle{ \mathbf{\hat{H}}_e = - \frac{\nabla^2}{2} - \frac{\mathbf Z}{r} }[/math], where [math]\displaystyle{ \mathbf Z }[/math] 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:
[math]\displaystyle{ \mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r} }[/math], where [math]\displaystyle{ \mathbf \zeta }[/math] 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 :
[math]\displaystyle{ \mathbf E_{1s} = \frac{\langle\psi_{1s}|\mathbf{\hat{H}}_e|\psi_{1s}\rangle}{\langle\psi_{1s}|\psi_{1s}\rangle} }[/math], where [math]\displaystyle{ \mathbf{\langle\psi_{1s}|\psi_{1s}\rangle} = 1 }[/math]
[math]\displaystyle{ \mathbf E_{1s} = \langle\psi_{1s}|\mathbf - \frac{\nabla^2}{2} - \frac{\mathbf Z}{r}|\psi_{1s}\rangle }[/math]
[math]\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 }[/math]
[math]\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 }[/math]. Using the expression for Slater orbital, [math]\displaystyle{ \mathbf \psi_{1s} = \left (\frac{\zeta^3}{\pi} \right ) ^{0.50}e^{-\zeta r} }[/math] the integrals can be exactly solved. Thus,
[math]\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 }[/math]
[math]\displaystyle{ \mathbf E_{1s} = \frac{\zeta^2}{2}-\zeta \mathbf Z. }[/math]

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

Non-relativistic energy

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

Hydrogen : H
[math]\displaystyle{ \mathbf Z=1 }[/math] and [math]\displaystyle{ \mathbf \zeta=1 }[/math]
[math]\displaystyle{ \mathbf E_{1s}= }[/math]−0.5 E_h
[math]\displaystyle{ \mathbf E_{1s}= }[/math]−13.60569850 eV
[math]\displaystyle{ \mathbf E_{1s}= }[/math]−313.75450000 kcal/mol

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

References

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