Two electrons in an atomic P state

If we have two atomic electrons in a P state with no external fields applied, states of definite total angular momentum will be the energy eigenstates. We will learn later that closed shells in atoms (or nuclei) have a total angular momentum of zero, allowing us to treat only the valence electrons. Examples of atoms like this would be Carbon, Silicon, and Germanium.

Our two electrons each have \bgroup\color{black}$\ell=1$\egroup (P state) and \bgroup\color{black}$s={1\over 2}$\egroup (electrons). We need to add four angluar momenta together to get the total.

\begin{displaymath}\bgroup\color{black} \vec{J}=\vec{L_1}+\vec{L_2}+\vec{S_1}+\vec{S_2} \egroup\end{displaymath}

We will find it useful to do this addition in two steps. For low Z atoms, it is most useful to add \bgroup\color{black}$\vec{L_1}+\vec{L_2}=\vec{L}$\egroup and \bgroup\color{black}$\vec{S_1}+\vec{S_2}=\vec{S}$\egroup then to add these results \bgroup\color{black}$\vec{L}+\vec{S}=\vec{J}$\egroup.

Since the electrons are identical particles and they are in the same radial state, the angular momentum part of the wavefunction must be antisymmetric under interchange. This will limit the allowed states. So let's do the spinor arithmetic.

\vert\ell_1-\ell_2\vert\leq &\ell& \leq \ell_1+\ell_2 \\
\ell &=& 0,1,2 \\
s &=& 0,1 \\

These states have a definite symmetry under interchange. Before going on to make the total angular momentum states, lets note the symmetry of each of the above states. The maximum allowed state will always need to be symmetric in order to achieve the maximum. The symmetry will alternate as we go down in the quantum number. So, for example, the \bgroup\color{black}$\ell=2$\egroup and \bgroup\color{black}$\ell=0$\egroup states are symmetric, while the \bgroup\color{black}$\ell=1$\egroup state is antisymmetric. The \bgroup\color{black}$s=1$\egroup state is symmetric and the \bgroup\color{black}$s=0$\egroup state is antisymmetric. The overall symmetry of a state will be a product of the these two symmetries (since when we add \bgroup\color{black}$\ell$\egroup and \bgroup\color{black}$s$\egroup to give \bgroup\color{black}$j$\egroup we are not adding identical things anymore). The overall state must be antisymmetic so we can use:

\ell&=&1  s=1  j=0,1,2 \qquad  ^3P_0,  ^3P_1,  ^3P_2 \\...
...uad\qquad  ^1D_2\\
\ell&=&0  s=0  j=0 \qquad\qquad  ^1S_0

Each atomic state will have the angular momentum quantum numbers

\begin{displaymath}\bgroup\color{black} \ell_1, \ell_2, s_1, s_2, \ell , s, j, m. \egroup\end{displaymath}

Normally we will not bother to include that the spins are one half since that's always true for electrons. We will (and must) keep track of the intermediate \bgroup\color{black}$\ell$\egroup and \bgroup\color{black}$s$\egroup quantum numbers. As can be seen above, we need them to identify the states.

In the atomic physics section, we will even deal with more than two electrons outside a closed shell.

Jim Branson 2013-04-22