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

1. Nov 9, 2017

### Twigg

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?

2. Nov 9, 2017

### mikeyork

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.

3. Nov 9, 2017

### Twigg

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?

4. Nov 9, 2017

### mikeyork

Yes.

5. Nov 10, 2017

### DrDu

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.