Yesterday I asked this question concerning the existence of terms. This question is on the same topic. I have been looking at some tables (in Yang and Hamilton (2010), pg 199) at the terms that exist for two electrons in the same configuration. I noticed that for example if the state $^1S$ exists the state $^3S$ would not and vice versa. Is this a general rule? I.e. can we say:

For two electrons in the configuration $nl^2$ if the term $^1L$ exists then the term $^3L$ will not and vice versa.

If so can it be proved easily and if not why not?


The general rule is correct as you have stated it. This is because there is only one way to add two $l$ angular momenta to make a total angular momentum of $L$; this combination will be either even or odd under exchange of the two sectors. Since the total electron state needs to be exchange-antisymmetric, and the singlet and triplet states are respectively antisymmetric and symmetric, only one of the combinations will be possible.

The key part of the argument is that the combination of $l\oplus l=L$ will always give some definite exchange symmetry. The simplest way I can see of proving this is reducing the couplings to the Wigner $3j$ symbols, which have definite symmetries under exchange (cf. the DLMF here), which specify to $$ \begin{pmatrix} l&l&L \\ m_2 & m_1 & -M \end{pmatrix} = (-1)^{l+l+L} \begin{pmatrix} l&l&L \\ m_1 & m_2 & -M \end{pmatrix} $$ i.e. the coupling will have $(-1)^L$ parity under exchange, independently of the value $l$.

This means that you can strengthen your statement to read

the coupling of two electrons from the same shell, $nl^2$, can only form the term $^3L$ if $L$ is odd and the term $^1L$ if $L$ is even.

On the other hand, it is perfectly possible for an atom to exhibit both $^1L$ and $^3L$ terms for any $L$ - you just need to combine electrons from different shells. Thus, you can have both $^1P$ and $^3P$ terms in the $1s2p$ shell of neutral helium, along with a slew of other similar states which you can explore via the NIST energy levels database.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.