Atomic Physics

The Hamiltonian for an atom with Z electrons and protons has many terms representing the repulsion between each pair of electrons.

\begin{displaymath}\bgroup\color{black}\left[ \sum\limits^Z_{i=1} \left({p^2_i\o...
...\vert\vec{r_i}-\vec{r_j}\right\vert}} \right]\psi=E\psi.\egroup\end{displaymath}

We have seen that the coulomb repulsion between electrons is a very large correction in Helium and that the three body problem in quantum mechanics is only solved by approximation.

The physics of closed shells and angular momentum enable us to make sense of even the most complex atoms. When we have enough electrons to fill a shell, say the 1s or 2p, The resulting electron distribution is spherically symmetric because

\begin{displaymath}\bgroup\color{black}\sum^\ell_{m=-\ell}\left\vert Y_{\ell m}\...
...a , \phi\right)\right\vert^2
= {2\ell + 1\over{4\pi}}. \egroup\end{displaymath}

With all the states filled and the relative phases determined by the antisymmetry required by Pauli, the quantum numbers of the closed shell are determined. There is only one possible state representing a closed shell and the quantum numbers are

& s=0 \\
& \ell=0 \\
& j=0

The closed shell screens the nuclear charge. Because of the screening, the potential no longer has a pure \bgroup\color{black}${1\over r}$\egroup behavior. Electrons which are far away from the nucleus see less of the nuclear charge and shift up in energy. We see that the atomic shells fill up in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p. The effect of screening increasing the energy of higher \bgroup\color{black}$\ell$\egroup states is clear. Its no wonder that the periodic table is not completely periodic.

A set of guidelines, known as Hund's rules, help us determine the quantum numbers for the ground states of atoms. The hydrogenic shells fill up giving well defined \bgroup\color{black}$j=0$\egroup states for the closed shells. As we add valence electrons we follow Hund's rules to determine the ground state. We get a great simplification by treating nearly closed shells as a closed shell plus positively charged, spin \bgroup\color{black}${1\over 2}$\egroup holes. For example, if an atom is two electrons short of a closed shell, we treat it as a closed shell plus two positive holes.)

  1. Couple the valence electrons (or holes) to give maximum total spin.
  2. Now choose the state of maximum $\ell $ (subject to the Pauli principle. The Pauli principle rather than the rule, often determines everything here.)
  3. If the shell is more than half full, pick the highest total angular momentum state $j=\ell+s$ otherwise pick the lowest $j=\vert\ell-s\vert$.

Jim Branson 2013-04-22