How to determine which levels correspond to each other in LS coupling and jj-coupling?

Question

In the LS coupling scheme, L, S and J are good quantum numbers; whereas in jj-coupling scheme, j₁, j₂ and J are good quantum numbers. I have learnt how to get the term symbols for LS coupling, and for jj-coupling for a particular electronic configuration. However, what I can't figure out is how the term symbols from each of these schemes are related to each other. None of the textbooks I have cover this topic in any way.

For example, in case of a $\mathrm{p^2}$ configuration, the LS terms are $\mathrm{^1S_0, ^1D_2, ^3P_2, ^3P_1, ^3P_0}$, and the corresponding jj-terms are $(3/2,3/2)_0, (3/2,3/2)_2, (3/2,1/2)_2, (3/2,1/2)_1, (1/2,1/2)_0~$ respectively. The LS terms are more useful when the atomic number is low e.g. C, whereas the jj terms are more useful when the atomic number is high e.g. Pb.

How do I know which LS terms are related to which jj-terms? (Obviously, we can look at the J and identify some terms but that does not work for every term)

For example, from the diagram we can tell that $\mathrm{^1S_0}$ is the LS-term that corresponds to $(3/2,3/2)_0$ in jj-term.

[P.S. I would also be happy if you could recommend a book where I can read around this topic]

Answer 1

In the jj-representation, each electron from $i$ to $N$ will have:

• $\vec{l}_i$ (orbital angular momentum),
• $\vec{s}_i$ (spin angular momentum), and
• $\vec{j}_i=\vec{l}_i + \vec{s}_i$ (total angular momentum).

In your example we have $N=2$ and both electrons are $p$-type so we have:

$$\tag{1} {l}_1=1,~~~~~~~{l}_2=1. $$

Let's also assign the spins for each electron as:

$$\tag{2} {s}_1=1/2,~~~~~~~{s}_2=1/2. $$

We now have the following possibilities for ${j}_i$:

\begin{eqnarray} ~&|l_i - s_i| \tag{3}\\ ~&~~~~~~~ |l_i - s_i| +1 \tag{4}\\ ~& ~\vdots \tag{5}\\ ~& ~l_i + s_i. \tag{6}\\ \end{eqnarray}

Plugging in $l_i$ and $s_i$ we get the following possibilities for $j_i$:

$$\tag{7} j_i = 1/2 , ~~~~~~~j_i=3/2. $$

Therefore we have 4 possibilities for the $(j_1,j_2)$ pair:

$$\tag{8} (1/2,1/2),(1/2,3/2),(3/2,1/2),(3/2,3/2). $$

These are all included in your example. Let's now look at the total $J$ values, which come from adding up the total angular momenta $j_i$ from each of the electrons. Since we have only 2 electrons, $J$ can be:

\begin{eqnarray} ~&|j_1 - j_2| \tag{9}\\ ~&~~~~~~~ |j_1 - j_2| +1 \tag{10}\\ ~& ~\vdots \tag{11}\\ ~& ~j_1 + j_2. \tag{12}\\ \end{eqnarray}

So the possible $J$ values in our case are (even including some which you can eliminate due to the Pauli exclusion principle):

• for $(1/2,1/2)$: $J=0$ or $1$, which means we can have $(1/2,1/2)_0$ and $(1/2,1/2)_1$.
• for $(1/2,3/2)$: $J=1$ or $2$, which means we can have $(1/2,3/2)_1$ and $(1/2,3/2)_2$.
• for $(3/2,1/2)$: $J=1$ or $2$, which means we can have $(3/2,1/2)_1$ and $(3/2,1/2)_2$.
• for $(3/2,3/2)$: $J=0,1,2,3$, so: $(3/2,3/2)_0$, $(3/2,3/2)_1$, $(3/2,3/2)_2$, $(3/2,3/2)_3$.

For the LS-representation, we have the total orbital angular momentum:

$$\tag{13} \vec{L}=\vec{l}_1 + \vec{l}_2, $$

and total spin angular momentum:

$$\tag{14} \vec{S}=\vec{s}_1 + \vec{s}_2. $$

The possible values of $L$ are:

\begin{eqnarray} ~&|l_1 - l_2| \tag{15}\\ ~&~~~~~~~ |l_1 - l_2| +1 \tag{16}\\ ~& ~\vdots \tag{17}\\ ~& ~l_1 + l_2, \tag{18}\\ \end{eqnarray}

and we have the same structure for $\vec{S}$.

To correspond term symbols from the LS-representation with term symbols from the jj-representation, you can assign each electron an $\vec{l}_i$ and $\vec{s}_i$ and work out the corresponding $\vec{j}_i,J,L$ and $S$ values.

For example, in $(3/2,3/2)_0$ we have an overall total angular momentum of $J=0$ despite each individual electron having a total angular momentum of $j_i=3/2$. The only way that can happen when each electron has $l_i=1$ (since you wanted them to both be $p$) and the same principal quantum number, is if the spins are in opposite directions: i.e. the singlet state $^1 S_0$ rather than the other $J=$ option which was $^3P_0$. In the other $J=0$ case, which is $(1/2,1/2)_0$, when $l_i=1$ the only way to get $j_i=1/2$ is for $\vec{s}_i=-1/2$ meaning that both spins are the same, and hence the triplet $^3 P_0$ rather than the $^1 S_0$.