Ordering energy levels of (excited) electron configurations

In the Russel-Sanders LS-coupling scheme, Hund's rules provide a mechanism for finding the ground-state term $^{2S+1}L_J$ of a certain element. I'm asked to consider optical transitions between specific (excited) terms of a neutral element. Hund's rules seem to only hold for determining the ground state.

How can I order the energy levels of the excited terms?

Say, we have $D_J$ and $P_J$. Since $D_J$ has a higher angular momentum $L$, Hund's rules dictate that $D_J$ should lie lower in energy.

• In the LS coupling, the configuration with the highest orbital angular momentum(L) and highest spin angular momentum(S) but with the lowest TOTAL angular momentum(J) has the lowest energy What exactly are you asking? – Prasad Mani Oct 8 '16 at 9:58
• So, you're saying that the configuration with the highest L and S but lowest J is always the lowest in energy, in LS-coupling? So it doesn't only hold for determining the ground state? My confusion arises from the restriction to determination of ground states. – Mussé Redi Oct 8 '16 at 10:02
• Or in other words, Hund's rules --- (1) highest L (2) highest S (3) highest J if shell is less than half-full, otherwise lowest J --> ground state --- only hold for ground states. Can you also use it for the ordering of excited states? – Mussé Redi Oct 8 '16 at 10:06
• No it does not hold for determining the ground state configuration. In fact determining the ground state configuration gets exponentially difficult as the atomic number increases. Even for nitrogen, you have to check for all sorts of anti-symmetrization requirements (symmetric spin and antisymmetric position state(which again has to be checked in spherical harmonics for every valid $l$) and vice versa) so the simple formula of highest L and S (and lowest J) doesnt apply here – Prasad Mani Oct 8 '16 at 10:06
• In the case of Calcium (Z=20), how would you then order the two configurations (4s4d) and (4s4p), when considering optical transitions between them? – Mussé Redi Oct 8 '16 at 10:15

You probably got confused while reading the LS coupling for carbon. When two electrons have two different radial wavefunctions eg, $2p^13s^1$, you can combine any of the $S(=0,1)$ with any of the $L=(0,1,2)$. For this configuration, $3D_1$ has the lowest energy, hope you can work that out. But if the two electrons have same radial wavefunction $2p^12p^1$ (in other words, $2p^2$), you cannot combine any $S(=0,1)$ with any $L(=0,1,2)$. Usually $S$ dominance is more hence you try to ensure $S$ takes the highest value. Since the spins are aligned,(in other words $S=1$), the position states have to be anti-symmetric (in other words,$L$ can only be 1,....check it with spherical harmonic function, you will see that $l=1$ is odd(antisymmetric)) so as to make the overall wavefunction antisymmetric
EDIT- $3D_0$ is not possible because you first get $3$ by taking $S=1$ and '$D$' by taking '$l'=2$' and then $j'=$ all the values in integer steps from $l'+s'$ to |$l'-s'$| ie $j'=3,2,1$