# Confusion regarding the possible term symbols/spectroscopic terms for (2p)(3p) configuration

Note: I have read this answer but it still doesn't address my question in terms of the permitted symmetry.

I am learning about atomic states, and possible electron configurations. The first example given in my lecture notes was (2p)^2. Using the Russel-Saunders (LS) coupling scheme, I see that the total L=0,1,2 and S=0,1. The S=0 corresponds to an antisymmetric spin state, and S=1 to symmetric spin state. Looking up the table of Clebsh-Gordon coefficients, it is clear that L=0,2 correspond to antisymmetric spin states, whereas L=1 to a symmetric spin state. Thus you can only have S=0,L=0,2 (J=0,2) and S=1, L=1 (J=0,1,2). These correspond to possible spectroscopic terms $$^1S_0, ^1D_2, ^3P_{0,1,2}$$. This is in line with the answer given in my lecture notes.

I used exactly the same reasoning to arrive at the possible spectroscopic terms for (2p)(3p) as $$^1S_0, ^1D_2, ^3P_{0,1,2}$$. After all, the electrons must still obey the same exchange symmetry so we would want each state to be a product of a symmetric and antisymmetric state (position, spin). However the states listed in my lecture notes are $$^1S_0, ^1P_1, ^1D_2, ^3S_1, ^3P_{0,1,2}, ^3D_{1,2,3}$$ which includes, for example, an S=0 and L=1 state, which is symmetric overall!

I don't understand this! If it has to do with the coupling scheme (which is only introduced later in my lecture notes, so I should think there is a different reason), I still do not understand why the coupling scheme would make a difference for (2p)(3p) and not for (2p)^2. In essence, I do not see the difference between these two cases.

EDIT: After some more digging around, it seems to have to do with whether the eletcrons are 'equivalent' or not, with equivalent electrons being those that have the same principal quantum number n, and angular momentum quantum number l. However I am still unsure as to why not having the same n and l translates to 'distinguishaility' such that the requirmenet for a totally antisymmetric wavefunction is dropped. I might reason that electrons with different n and/or l are on average at different distances from the nucleus, and therefore we don't 'care' so much about the exchange statistics, but that seems very handwavy to me. I shoudl still think the wavefunctions should need to be totally antisymmetric. There is still a great deal of overlap between the wavefucntions, and even if there wasn't these are still two fermions we are describing!

EDIT 2: I think I may have clarified part of the problem. In the lecture note here, it considers the case of 'equivalet' electrons as defined above as special based on the fact that

Equivalent electrons have the same n and l values, so the possiblity exists that they might end up with all four quantum numbers the same, which is forbidden by the Pauli Principle. In this case you have to look at all allowable combinations of ML and MS values, and from those values infer the L and S values for the whole atom

I am not happy with this for the following reason; the Pauli Exclusion principle is a consequence of the antisymmetry of the fermionic wavefunctions. In that sense, it DROPS some of the restriv=ctions on the permitted wavefunctions, by only saying that $$\psi$$ vanishes for electrons in the same quantum state, rather than it being antisymmetric for anychosen states. This directly results in the discrepancy- where I should think that the requirement for an antisymmetric wavefunction would apply to non-equivalent electrons, thus restricting the possible spectroscopic terms, whereas the other sources only seem to consider the Pauli Exclusion rinciple, which would only restrict the spectroscopic terms for equivalent electrons (in fact, I don;t think PEP by itself would, since the electrons can occupy different m_l states)