I'm using these rules, $^{2S+1}L_J$ with $J=L+S$ and it seems to imply that $^3P_1$ means $S=1$, $L=1$, and $J=1$ but clearly $1+1 \ne 1$ so according to me this shouldn't be an atomic term symbol but it is. Why?
2 Answers
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with $J=L+S$
Nope. In general, it can be shown that the permitted values of the $J$ quantum number for the total angular momentum that arise from two sources characterized by quantum numbers $J_1$ and $J_2$ are given by $$ J = J_1 + J_2, J_1 + J_2 - 1, \dotsc, |J_1 - J_2| \, . $$ In this particular case, $J_1 = L$ and $J_2 = S$, so $J$ ranges from $L + S$ to $|L - S|$ in integral steps and not necessarily equal to $L + S$.
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You are correct in saying that the total spin for two electrons is S = 1 as its spin multiplicity is 3 and that L =1. The J value is found in the same way as S and L values were using the Clebsch-Gordon series, $$J= |L+S|...|L-S| $$ This means find the maximum value, then the minimum make it positive, and separate any other values by 1. Sometimes there is only 1 value so be careful. We use this type of series as a short-cut for vector addition of angular momentum which is why you question '1+1 = 1'.
In your case J = 1+1 to 0 , which gives three values J = 2, 1 and 0 so three possible term symbols. There will also be singlet terms where the total spin S = 0.
The triplet term means that electrons are in different orbitals, generally this lowers the energy, and also means that there is no Pauli principle to consider. Hund's rules are used to determine which levels and states are lowest in energy. (See for example Steinfeld, 'Molecules & radiation')
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$\begingroup$ Ah ok, so it seems I was just told some quick short cut for some actual process. What's the actual process called if I want to do it "the real way"; Clebsch-Gordon addition of angular momentum or something I guess? $\endgroup$– user19026Commented Jul 9, 2016 at 23:47
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$\begingroup$ I'm not sure it has a name as such, but its based on the Russell Saunders coupling scheme and vector addition of angular momentum. This is described in most undergraduate phys chem text books. $\endgroup$ Commented Jul 10, 2016 at 8:54
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$\begingroup$ Yeah, for reference I was reading McQuarrie Physical Chemistry and this seemed to be just thrown at me without justification so I guess I need to refer to a different book for the explanation on that. $\endgroup$– user19026Commented Jul 14, 2016 at 5:59