# LS-Coupling: intuition why 2p^2 has no singlet P?

• I
Science Advisor
Gold Member
I get the microstate-counting approach to finding the term symbols for a given configuration. But based on what I know about addition of angular momentum in quantum mechanics, I feel like there's a conceptual gap. When I do the microstate counting on the 2p##^{2}## configuration, I get singlet S, triplet P, and singlet D states. But when I add two electrons with l = 1 and s = 1/2, I get singlet S, singlet P, triplet P, and singlet D.
To clarify what I mean by adding these two electrons, I mean that I add the two orbital angular momenta l =1 to get values of L = 0,1,2, and I add the two spin angular momenta s = 1/2 to get values of S = 0,1. I then look at all six of these hypothetical states and eliminate the ones that violate the Pauli principle, namely the triplet D and triplet S. (These would violate the Pauli principle because they would require electrons with parallel spin AND parallel orbital angular momenta.) However, this leaves me with the extra singlet P state that does not appear in the microstate counting treatment. Where is my misconception here? What is forbidden about the singlet P state?

## Answers and Replies

Are you referring to atomic electron orbitals or the CM system of two electrons? In general, two electrons in their CM frame obey the rule L+S is even. (It can be derived from the Pauli rule.) I think its also true in any other frame but my mind is playing tricks on me (it happens when you get to 70!) and I can't be sure at this moment.

Science Advisor
Gold Member
That actually makes sense, thanks! Even L + S would give you an antisymmetric spin state, and that would guarantee Pauli exclusion. Does that sound about right?

Yes.

DrDu
Science Advisor
Yes, this is due to the Pauli principle. Namely the orbital part of the wavefunction of a P state is antisymmetric with respect to exchange of the electrons, so the spin function has to be symmetric, i.e. a triplett. For example the P state with M= 1 is ##p_{+1}(1)p_{0}(2)-p_{0}(1)p_{+1}(2)##, while the state ##p_{+1}(1)p_{0}(2)+p_{0}(1)p_{+1}(2)## is the D-state with M=1.

• Twigg, DrClaude and vanhees71