# How to calculate the potential energy of an electron in the "nth" shell (also take example "n=3" shell) of a copper atom using the Hartree Fock Theory? Also calculate its kinetic energy

Jun 29, 2017

You can't calculate the kinetic energy of an electron that hasn't left the atom. So we could only calculate the orbital potential energy.

Since copper is not that heavy, we don't have to worry too much about scalar relativistic effects or spin-orbit coupling. In fact, its ground electronic state is a $\text{^2 S_"1/2}$, with no low-lying excited states to include in a more multi-referenced level of theory like MCSCF or CCSD(T).

The ground-state electron configuration would be:

$\left[A r\right] 3 {d}^{10} 4 {s}^{1}$

INPUT FILE SETUP

At the Hartree-Fock level of theory (a single-reference method), we assume a single configuration with doubly-occupied core and $3 d$ valence orbitals, and a singly-occupied $4 s$ valence orbital. We don't expect good agreement with experiment with this level of theory...

MolPro, Psi4, and other computational software could do this, but I have MolPro with me right now. A basic input for copper atom would look like this:

[

Here,

$\text{wf, 29, 1, 1}$

VERIFYING THE ORBITAL OCCUPATION

The orbitals in the basis set are being printed in order by energy within each irreducible representation of the ${D}_{2 h}$ point group, whose irreducible representations are listed as ${A}_{g}$, ${B}_{3 u}$, ${B}_{2 u}$, ${B}_{1 g}$, ${B}_{1 u}$, ${B}_{2 g}$, ${B}_{3 g}$, ${A}_{u}$.

So, orbital $6.1$ is the sixth orbital (starting from the lowest energy) of ${A}_{g}$ symmetry, orbital $2.5$ is the second orbital of ${B}_{1 u}$ symmetry, etc.

We expect the orbital occupation is ordered as follows:

$\underline{{A}_{g} \text{ "B_(3u)" "B_(2u)" "B_(1g)" "B_(1u)" "B_(2g)" "B_(3g)" } {A}_{u}}$
$1 s \text{ "2p_x" "2p_y" "3d_(xy)" "2p_z" "3d_(xz)" "3d_(yz)" } -$
$2 s \text{ "3p_x" "3p_y" "" "" } \textcolor{w h i t e}{.} 3 {p}_{z}$
$3 s \text{ }$
$3 {d}_{z}^{2} \text{ }$
$3 {d}_{{x}^{2} - {y}^{2}}$
$\textcolor{red}{4 s} \text{ }$

where red indicates the singly-occupied orbital with one spin-up electron.

MolPro believes it to be:

[

The orbital occupation looks correct on the first go, but in practice, you should specify the occ and closed cards anyway. It should be an occupied configuration of:

$\left({A}_{g} , {B}_{3 u} , {B}_{2 u} , {B}_{1 g} , {B}_{1 u} , {B}_{2 g} , {B}_{3 g} , {A}_{u}\right) = \left(6 , 2 , 2 , 1 , 2 , 1 , 1 , 0\right)$, specified by

$\text{occ, 6, 2, 2, 1, 2, 1, 1, 0}$ within the brackets for the HF command.

and a closed-shell (doubly-occupied) configuration of:

$\left({A}_{g} , {B}_{3 u} , {B}_{2 u} , {B}_{1 g} , {B}_{1 u} , {B}_{2 g} , {B}_{3 g} , {A}_{u}\right) = \left(5 , 2 , 2 , 1 , 2 , 1 , 1 , 0\right)$, specified by

$\text{closed, 5, 2, 2, 1, 2, 1, 1, 0}$ within the brackets for the HF command

indicating one spin-up electron in the $4 s$ orbital (since it is the sixth orbital in symmetry 1).

ORBITAL ENERGIES (BE CRITICAL!)

The orbital output was:

[

The default energies listed are in Hartrees, so you'll have to convert to $\text{eV}$s if you want them in $\text{eV}$s. There are $\text{27.2114 eV}$s per Hartree.

Apparently, the HOMO, the $4 s$ (orbital $6.1$), was (poorly) calculated to have an energy of $- \text{6.49 eV}$, which is too high in energy by about $\text{2 eV}$...

The $3 d$, orbital $5.1$ (orbital five of symmetry ${A}_{g}$), lists an energy of $\boldsymbol{- \text{13.37 eV}}$, which is quite close to the actual $- \text{13.47 eV}$ here.

The $3 p$ orbitals are the last orbitals in symmetry 2, 3, and 5, so they are orbitals $2.2 , 2.3 , 2.5$. These have calculated energies of $- \text{90.48 eV}$, which is quite low and certainly core-like. I am not at all sure if these $3 p$ energies are correct, but who really cares about the core orbitals in a light atom like this (especially since the $3 d$ subshell is full)?

For a basic calculation at a low level of theory, this isn't that bad...