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## Main Question or Discussion Point

Say we have a Helium atom with the first electron in 1s orbital and second electron in 2s orbital, and if these electrons have parallel spin so that the total spin of the system is S=1, then is this a triplet or a singlet state?

According to LS coupling, multiplicity is defined as:

2S+1 (if L>=S)

and

2L+1 (if L<S)

Since in our case we have L=0 (both electrons in S orbitals) then multiplicity should be 1 (according to the second case) which means we have a singlet. However, I find on many places that states with total spin S=1 are triplets without mentioning the L (total orbital angular moment).

Even for Orthohelium is defined as the Helium state atom in which the spins of the two electrons are parallel. Here http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html and here https://en.wiktionary.org/wiki/orthohelium.

According to LS coupling, multiplicity is defined as:

2S+1 (if L>=S)

and

2L+1 (if L<S)

Since in our case we have L=0 (both electrons in S orbitals) then multiplicity should be 1 (according to the second case) which means we have a singlet. However, I find on many places that states with total spin S=1 are triplets without mentioning the L (total orbital angular moment).

Even for Orthohelium is defined as the Helium state atom in which the spins of the two electrons are parallel. Here http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html and here https://en.wiktionary.org/wiki/orthohelium.