# How do we know that $^1S$ is the ground state of the helium atom?

Let $$\psi=a_1\phi(1s(2) \ ^1S)+a_2\phi(1s(1)2s(1) \ ^1S)+a_3\phi(2s(2) \ ^1S) +...$$ be a state of the helium atom. Applying variationally calculus we can found the energy expectation value of this state is almost exact to the experimental value. Is by this comparation that we know that the total angular momentum $$J$$ of the helium atom is zero or can we proof it theoretically alone, or experimentally alone ?

## 1 Answer

The answer to this question can be given with a much simpler explanation. Hund's rules state that:

1)For a given electron configuration, the term with maximum multiplicity has the lowest energy. The multiplicity is equal to $$2S+1$$ , where $$S$$ is the total spin angular momentum for all electrons. The multiplicity is also equal to the number of unpaired electrons plus one. Therefore, the term with lowest energy is also the term with maximum $$S$$, and maximum number of unpaired electrons.

2)For a given multiplicity, the term with the largest value of the total orbital angular momentum quantum number $$L$$ , has the lowest energy.

3)For a given term, in an atom with outermost subshell half-filled or less, the level with the lowest value of the total angular momentum quantum number $$J$$ (for the operator $$J=L+S$$ lies lowest in energy. If the outermost shell is more than half-filled, the level with the highest value of $$J$$ is lowest in energy.

Since the total spin is 0 and the angular momentum is 0, you have the ground state that you have written.