title: 1s Slater-type function
text: # 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 $E_h$ $\mathbf {E} _{1s}=$−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 $E_h$ $\mathbf {E} _{1s}=$−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.
  
  | Atomic system | $\mathbf {Z}$ | $\mathbf {\zeta } _{nonrel}$ | $\mathbf {\zeta } _{rel}$ | $\mathbf {E} _{1s}^{nonrel}$ | $\mathbf {E} _{1s}^{rel}$using $\mathbf {\zeta } _{nonrel}$ | $\mathbf {E} _{1s}^{rel}$using $\mathbf {\zeta } _{rel}$ |
  | - | - | - | - | - | - | - |
  | H | 1 | 1.00000000 | 1.00002663 | −0.50000000 $E_h$ | −0.50000666 $E_h$ | −0.50000666 $E_h$ |
  | H | 1 | 1.00000000 | 1.00002663 | −13.60569850 eV | −13.60587963 eV | −13.60587964 eV |
  | H | 1 | 1.00000000 | 1.00002663 | −313.75450000 kcal/mol | −313.75867685 kcal/mol | −313.75867708 kcal/mol |
  | Au(78+) | 79 | 79.00000000 | 96.68296596 | −3120.50000000 $E_h$ | −3343.96438929 $E_h$ | −3434.58676969 $E_h$ |
  | Au(78+) | 79 | 79.00000000 | 96.68296596 | −84913.16433850 eV | −90993.94255075 eV | −93459.90412098 eV |
  | Au(78+) | 79 | 79.00000000 | 96.68296596 | −1958141.83450000 kcal/mol | −2098367.74995699 kcal/mol | −2155234.10926142 kcal/mol |
  